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Pigments in Our Lives Whenever you paint a picture or a room, or use inks or cosmetics, you use pigments. A pigment is a substance that gives a color to another material. Most pigments are a mixture of a powder and a colorless solvent.

On the front cover of this book, you can see examples of inorganic pigments that are obtained from powdered minerals of different chemical elements. Cadmium in cadmium orange, chromium in chrome yellow, cobalt in cobalt violet, iron in red ocher, manganese in manganese purple. However, the elements chromium and cadmium are toxic and have been replaced by nontoxic pigments.

Organic pigments containing chains of the element carbon are obtained from plants and animals. Indigo dye from the plant Indigofera tinctoria is an organic compound with a deep, blue color used to color blue jeans and fabrics. Carbon black is a powdered form of the element carbon used to color plastics, tires, inks, and paints.

Color is seen when a pigment absorbs certain wavelengths of visible light and reflects the remaining wavelengths as a color. For example, a pigment that absorbs red and green light, and reflects the blue wavelengths, has a blue color. If a substance absorbs green light but relects red and blue wavelengths, it appears to have a violet color. A leaf with chlorophyll pigment is green because all wavelengths in sunlight are absorbed except green.

Mastering™ Chemistry is a learning platform designed with you in mind, offering:

• A complete eText! More than a PDF, the Pearson eText includes embedded videos, interactive self-assessments, and more—all accessible on any device via the Pearson eText app and available even when offline

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Sixth Edition

Timberlake &

Timberlake

Sixth Edition

ISBN-13: 978-0-13-487811-9 ISBN-10: 0-13-487811-6

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Atomic Masses of the Elements

aValues for atomic masses are given to four significant figures. bValues in parentheses are the mass number of an important radioactive isotope.

Name Symbol Atomic Number Atomic Massa

Actinium Ac 89 (227)b

Aluminum Al 13 26.98 Americium Am 95 (243) Antimony Sb 51 121.8 Argon Ar 18 39.95 Arsenic As 33 74.92 Astatine At 85 (210) Barium Ba 56 137.3 Berkelium Bk 97 (247) Beryllium Be 4 9.012 Bismuth Bi 83 209.0 Bohrium Bh 107 (264) Boron B 5 10.81 Bromine Br 35 79.90 Cadmium Cd 48 112.4 Calcium Ca 20 40.08 Californium Cf 98 (251) Carbon C 6 12.01 Cerium Ce 58 140.1 Cesium Cs 55 132.9 Chlorine Cl 17 35.45 Chromium Cr 24 52.00 Cobalt Co 27 58.93 Copernicium Cn 112 (285) Copper Cu 29 63.55 Curium Cm 96 (247) Darmstadtium Ds 110 (271) Dubnium Db 105 (262) Dysprosium Dy 66 162.5 Einsteinium Es 99 (252) Erbium Er 68 167.3 Europium Eu 63 152.0 Fermium Fm 100 (257) Flerovium Fl 114 (289) Fluorine F 9 19.00 Francium Fr 87 (223) Gadolinium Gd 64 157.3 Gallium Ga 31 69.72 Germanium Ge 32 72.64 Gold Au 79 197.0 Hafnium Hf 72 178.5 Hassium Hs 108 (265) Helium He 2 4.003 Holmium Ho 67 164.9 Hydrogen H 1 1.008 Indium In 49 114.8 Iodine I 53 126.9 Iridium Ir 77 192.2 Iron Fe 26 55.85 Krypton Kr 36 83.80 Lanthanum La 57 138.9 Lawrencium Lr 103 (262) Lead Pb 82 207.2 Lithium Li 3 6.941 Livermorium Lv 116 (293) Lutetium Lu 71 175.0 Magnesium Mg 12 24.31 Manganese Mn 25 54.94 Meitnerium Mt 109 (268)

Name Symbol Atomic Number Atomic Massa

Mendelevium Md 101 (258) Mercury Hg 80 200.6 Molybdenum Mo 42 95.94 Moscovium Mc 115 (289) Neodymium Nd 60 144.2 Neon Ne 10 20.18 Neptunium Np 93 (237) Nickel Ni 28 58.69 Nihonium Nh 113 (286) Niobium Nb 41 92.91 Nitrogen N 7 14.01 Nobelium No 102 (259) Oganesson Og 118 (294) Osmium Os 76 190.2 Oxygen O 8 16.00 Palladium Pd 46 106.4 Phosphorus P 15 30.97 Platinum Pt 78 195.1 Plutonium Pu 94 (244) Polonium Po 84 (209) Potassium K 19 39.10 Praseodymium Pr 59 140.9 Promethium Pm 61 (145) Protactinium Pa 91 231.0 Radium Ra 88 (226) Radon Rn 86 (222) Rhenium Re 75 186.2 Rhodium Rh 45 102.9 Roentgenium Rg 111 (272) Rubidium Rb 37 85.47 Ruthenium Ru 44 101.1 Rutherfordium Rf 104 (261) Samarium Sm 62 150.4 Scandium Sc 21 44.96 Seaborgium Sg 106 (266) Selenium Se 34 78.96 Silicon Si 14 28.09 Silver Ag 47 107.9 Sodium Na 11 22.99 Strontium Sr 38 87.62 Sulfur S 16 32.07 Tantalum Ta 73 180.9 Technetium Tc 43 (99) Tellurium Te 52 127.6 Tennessine Ts 117 (294) Terbium Tb 65 158.9 Thallium Tl 81 204.4 Thorium Th 90 232.0 Thulium Tm 69 168.9 Tin Sn 50 118.7 Titanium Ti 22 47.87 Tungsten W 74 183.8 Uranium U 92 238.0 Vanadium V 23 50.94 Xenon Xe 54 131.3 Ytterbium Yb 70 173.0 Yttrium Y 39 88.91 Zinc Zn 30 65.41 Zirconium Zr 40 91.22

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BASIC CHEMISTRY

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BASIC CHEMISTRY

Sixth Edition

Karen Timberlake William Timberlake

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Library of Congress Cataloging-in-Publication Data

Names: Timberlake, Karen, author. | Timberlake, William, author. Title: Basic chemistry / Karen Timberlake, William Timberlake. Description: Sixth edition. | New York, NY : Pearson, [2019] | Includes index. Identifiers: LCCN 2018048212 | ISBN 9780134878119 Subjects: LCSH: Chemistry--Textbooks. Classification: LCC QD31.3 .T54 2019 | DDC 540--dc23 LC record available at https://lccn.loc.gov/2018048212

ISBN 10: 0-134-87811-6; ISBN 13: 978-0-134-87811-9 (Student edition)

ISBN 10: 0-134-98699-7; ISBN 13: 978-0-13498699-9 (Looseleaf Edition)

ISBN 10: 0-135-24461-7; ISBN 13: 978-0-13524461-6 (NASTA)

www.pearson.com

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v

Brief Contents

1 Chemistry in Our Lives 1 2 Chemistry and Measurements 27 3 Matter and Energy 68 4 Atoms and Elements 100 5 Electronic Structure of Atoms and Periodic Trends 125 6 Ionic and Molecular Compounds 156 7 Chemical Quantities 183 8 Chemical Reactions 213 9 Chemical Quantities in Reactions 239 10 Bonding and Properties of Solids and Liquids 269 11 Gases 311 12 Solutions 350 13 Reaction Rates and Chemical Equilibrium 398 14 Acids and Bases 431 15 Oxidation and Reduction 476 16 Nuclear Chemistry 508 17 Organic Chemistry 540 18 Biochemistry 592

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vii

1 Chemistry in Our Lives 1

CAREER Forensic Scientist 1 1.1 Chemistry and Chemicals 2 1.2 Scientific Method: Thinking Like a Scientist 3

CHEMISTRY LINK TO HEALTH Early Chemist: Paracelsus 4

1.3 Studying and Learning Chemistry 6 1.4 Key Math Skills for Chemistry 9 1.5 Writing Numbers in Scientific Notation 17

UPDATE Forensic Evidence Helps Solve the Crime 21

Concept Map 21 Chapter Review 22 Key Terms 22 Key Math Skills 22 Understanding the Concepts 24 Additional Practice Problems 24 Challenge Problems 25 Answers to Engage Questions 25 Answers to Selected Problems 26

2 Chemistry and Measurements 27

CAREER Registered Nurse 27 2.1 Units of Measurement 28 2.2 Measured Numbers and Significant Figures 31 2.3 Significant Figures in Calculations 34 2.4 Prefixes and Equalities 39 2.5 Writing Conversion Factors 42 2.6 Problem Solving Using Unit Conversion 47

CHEMISTRY LINK TO HEALTH Toxicology and Risk–Benefit Assessment 52

Contents

2.7 Density 53 CHEMISTRY LINK TO HEALTH Bone Density 56

UPDATE Greg’s Visit with His Doctor 59

Concept Map 59 Chapter Review 60 Key Terms 61 Key Math Skills 61 Core Chemistry Skills 61 Understanding the Concepts 62 Additional Practice Problems 64 Challenge Problems 65 Answers to Engage Questions 65 Answers to Selected Problems 65

3 Matter and Energy 68

CAREER Dietitian 68 3.1 Classification of Matter 69

CHEMISTRY LINK TO HEALTH Breathing Mixtures 70

3.2 States and Properties of Matter 72 3.3 Temperature 75

CHEMISTRY LINK TO HEALTH Variation in Body Temperature 79

3.4 Energy 79 3.5 Specific Heat 82 3.6 Energy and Nutrition 87

CHEMISTRY LINK TO HEALTH Losing and Gaining Weight 89

UPDATE A Diet and Exercise Program 90

Concept Map 91 Chapter Review 91 Key Terms 92 Core Chemistry Skills 92 Understanding the Concepts 93 Additional Practice Problems 94 Challenge Problems 95 Answers to Engage Questions 96 Answers to Selected Problems 96 Combining Ideas from Chapters 1 to 3 98

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viii Contents

4 Atoms and Elements 100

CAREER Farmer 100 4.1 Elements and Symbols 101 4.2 The Periodic Table 103

CHEMISTRY LINK TO HEALTH Elements Essential to Health 106

4.3 The Atom 108 4.4 Atomic Number and Mass Number 111 4.5 Isotopes and Atomic Mass 113

UPDATE Improving Crop Production 117

Concept Map 118 Chapter Review 118 Key Terms 119 Core Chemistry Skills 119 Understanding the Concepts 120 Additional Practice Problems 121 Challenge Problems 122 Answers to Engage Questions 123 Answers to Selected Problems 123

5 Electronic Structure of Atoms and Periodic Trends 125

CAREER Materials Engineer 125 5.1 Electromagnetic Radiation 126

CHEMISTRY LINK TO HEALTH Biological Reactions to UV Light 128

5.2 Atomic Spectra and Energy Levels 129 5.3 Sublevels and Orbitals 131 5.4 Orbital Diagrams and Electron

Configurations 135 5.5 Electron Configurations and the Periodic

Table 139 5.6 Trends in Periodic Properties 143

UPDATE Developing New Materials for Computer Chips 148

Concept Map 149 Chapter Review 149 Key Terms 150 Core Chemistry Skills 151 Understanding The Concepts 151 Additional Practice Problems 152 Challenge Problems 153 Answers to Engage Questions 153 Answers to Selected Problems 154

6 Ionic and Molecular Compounds 156

CAREER Pharmacist 156 6.1 Ions: Transfer of Electrons 157

CHEMISTRY LINK TO HEALTH Some Important Ions in the Body 161

6.2 Ionic Compounds 161 6.3 Naming and Writing Ionic Formulas 164 6.4 Polyatomic Ions 168 6.5 Molecular Compounds: Sharing Electrons 172

UPDATE Compounds at the Pharmacy 176

Concept Map 177 Chapter Review 177 Key Terms 178 Core Chemistry Skills 178 Understanding the Concepts 179 Additional Practice Problems 179 Challenge Problems 180 Answers to Engage Questions 181 Answers to Selected Problems 181

7 Chemical Quantities 183

CAREER Veterinarian 183 7.1 The Mole 184 7.2 Molar Mass 188 7.3 Calculations Using Molar Mass 190 7.4 Mass Percent Composition 194

CHEMISTRY LINK TO THE ENVIRONMENT Fertilizers 196

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Contents ix

7.5 Empirical Formulas 197 7.6 Molecular Formulas 201

UPDATE Prescriptions for Max 204

Concept Map 204 Chapter Review 205 Key Terms 205 Core Chemistry Skills 205 Understanding the Concepts 207 Additional Practice Problems 207 Challenge Problems 208 Answers to Engage Questions 209 Answers to Selected Problems 209 Combining Ideas from Chapters 4 to 7 211

8 Chemical Reactions 213

CAREER Exercise Physiologist 213 8.1 Equations for Chemical Reactions 214 8.2 Balancing a Chemical Equation 217 8.3 Types of Chemical Reactions 223

CHEMISTRY LINK TO HEALTH Incomplete Combustion: Toxicity of Carbon Monoxide 227

8.4 Oxidation–Reduction Reactions 228 UPDATE Improving Natalie’s Overall Fitness 232

Concept Map 232 Chapter Review 233 Key Terms 233 Core Chemistry Skills 233 Understanding the Concepts 234 Additional Practice Problems 235 Challenge Problems 236 Answers to Engage Questions 237 Answers to Selected Problems 237

9 Chemical Quantities in Reactions 239

CAREER Environmental Scientist 239 9.1 Conservation of Mass 240 9.2 Mole Relationships in Chemical Equations 241 9.3 Mass Calculations for Chemical Reactions 245 9.4 Limiting Reactants 247

9.5 Percent Yield 252 9.6 Energy in Chemical Reactions 254

CHEMISTRY LINK TO HEALTH Cold Packs and Hot Packs 257

UPDATE Testing Water Samples for Insecticides 260

Concept Map 261 Chapter Review 261 Key Terms 262 Core Chemistry Skills 262 Understanding the Concepts 264 Additional Practice Problems 265 Challenge Problems 266 Answers to Engage Questions 267 Answers to Selected Problems 267

10 Bonding and Properties of Solids and Liquids 269

CAREER Histologist 269

10.1 Lewis Structures for Molecules and Polyatomic Ions 270

10.2 Resonance Structures 276 10.3 Shapes of Molecules and Polyatomic Ions

(VSEPR Theory) 279

10.4 Electronegativity and Bond Polarity 283 10.5 Polarity of Molecules 287 10.6 Intermolecular Forces Between Atoms

or Molecules 288

10.7 Changes of State 291 CHEMISTRY LINK TO HEALTH Steam Burns 297

UPDATE Histologist Stains Tissue with Dye 298

Concept Map 299 Chapter Review 299 Key Terms 300 Core Chemistry Skills 301 Understanding the Concepts 303 Additional Practice Problems 304 Challenge Problems 305 Answers to Engage Questions 306 Answers to Selected Problems 306 Combining Ideas from Chapters 8 to 10 309

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11 Gases 311

CAREER Respiratory Therapist 311 11.1 Properties of Gases 312

CHEMISTRY LINK TO HEALTH Measuring Blood Pressure 316

11.2 Pressure and Volume (Boyle’s Law) 317 CHEMISTRY LINK TO HEALTH Pressure–Volume Relationship in Breathing 318

11.3 Temperature and Volume (Charles’s Law) 320 11.4 Temperature and Pressure (Gay-Lussac’s

Law) 322 11.5 The Combined Gas Law 325 11.6 Volume and Moles (Avogadro’s Law) 327 11.7 The Ideal Gas Law 329 11.8 Gas Laws and Chemical Reactions 334 11.9 Partial Pressures (Dalton’s Law) 335

CHEMISTRY LINK TO HEALTH Blood Gases 337

CHEMISTRY LINK TO HEALTH Hyperbaric Chambers 340

UPDATE Exercise-Induced Asthma 341

Concept Map 342 Chapter Review 342 Key Terms 343 Core Chemistry Skills 344 Understanding the Concepts 345 Additional Practice Problems 346 Challenge Problems 347 Answers to Engage Questions 347 Answers to Selected Problems 348

12 Solutions 350

CAREER Dialysis Nurse 350 12.1 Solutions 351

CHEMISTRY LINK TO HEALTH Water in the Body 352

12.2 Electrolytes and Nonelectrolytes 355 CHEMISTRY LINK TO HEALTH Electrolytes in Body Fluids 356

12.3 Solubility 357 CHEMISTRY LINK TO HEALTH Gout and Kidney Stones: Saturation in Body Fluids 358

12.4 Solution Concentrations 363 12.5 Dilution of Solutions 371 12.6 Chemical Reactions in Solution 374 12.7 Molality and Freezing Point Lowering/Boiling

Point Elevation 378 12.8 Properties of Solutions: Osmosis 385

CHEMISTRY LINK TO HEALTH Hemodialysis and the Artificial Kidney 387

UPDATE Using Dialysis for Renal Failure 388

Concept Map 389 Chapter Review 389 Key Terms 390 Core Chemistry Skills 391 Understanding the Concepts 392 Additional Practice Problems 393 Challenge Problems 395 Answers to Engage Questions 396 Answers to Selected Problems 396

13 Reaction Rates and Chemical Equilibrium 398

CAREER Chemical Oceanographer 398 13.1 Rates of Reactions 399 13.2 Chemical Equilibrium 403 13.3 Equilibrium Constants 406 13.4 Using Equilibrium Constants 410 13.5 Changing Equilibrium Conditions:

Le Châtelier’s Principle 414 CHEMISTRY LINK TO HEALTH Oxygen– Hemoglobin Equilibrium and Hypoxia 417

CHEMISTRY LINK TO HEALTH Homeostasis: Regulation of Body Temperature 419

13.6 Equilibrium in Saturated Solutions 420 UPDATE Equilibrium of CO2 in the Ocean 424

Concept Map 424 Chapter Review 425 Key Terms 425 Core Chemistry Skills 426 Understanding the Concepts 427 Additional Practice Problems 428 Challenge Problems 429 Answers to Engage Questions 429 Answers to Selected Problems 430

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14 Acids and Bases 431

CAREER Medical Laboratory Technologist 431 14.1 Acids and Bases 432 14.2 Brønsted–Lowry Acids and Bases 434 14.3 Strengths of Acids and Bases 437 14.4 Dissociation of Weak Acids and Bases 442 14.5 Dissociation of Water 444 14.6 The pH Scale 447

CHEMISTRY LINK TO HEALTH Stomach Acid, HCl 453

14.7 Reactions of Acids and Bases 454 CHEMISTRY LINK TO HEALTH Antacids 456

14.8 Acid–Base Titration 457 14.9 Buffers 459

CHEMISTRY LINK TO HEALTH Buffers in the Blood Plasma 462

UPDATE Acid Reflux Disease 463

Concept Map 464 Chapter Review 465 Key Terms 466 Key Math Skills 466 Core Chemistry Skills 467 Understanding the Concepts 468 Additional Practice Problems 468 Challenge Problems 469 Answers to Engage Questions 470 Answers to Selected Problems 471 Combining Ideas from Chapters 11 to 14 473

15 Oxidation and Reduction 476

CAREER Dentist 476 15.1 Oxidation and Reduction 477 15.2 Balancing Oxidation–Reduction Equations

Using Half-Reactions 483 15.3 Electrical Energy from Oxidation– Reduction

Reactions 488 CHEMISTRY LINK TO THE ENVIRONMENT Corrosion: Oxidation of Metals 494

CHEMISTRY LINK TO THE ENVIRONMENT Fuel Cells: Clean Energy for the Future 496

15.4 Oxidation–Reduction Reactions That Require Electrical Energy 497 UPDATE Whitening Kimberly’s Teeth 499

Concept Map 500 Chapter Review 500 Key Terms 501 Core Chemistry Skills 501 Understanding the Concepts 502 Additional Practice Problems 503 Challenge Problems 504 Answers to Engage Questions 505 Answers to Selected Problems 505

16 Nuclear Chemistry 508

CAREER Radiation Technologist 508 16.1 Natural Radioactivity 509 16.2 Nuclear Reactions 512

CHEMISTRY LINK TO HEALTH Radon in Our Homes 514

16.3 Radiation Measurement 519 CHEMISTRY LINK TO HEALTH Radiation and Food 520

16.4 Half-Life of a Radioisotope 522 CHEMISTRY LINK TO THE ENVIRONMENT Dating Ancient Objects 524

16.5 Medical Applications Using Radioactivity 526 CHEMISTRY LINK TO HEALTH Brachytherapy 528

16.6 Nuclear Fission and Fusion 529 UPDATE Cardiac Imaging Using a Radioisotope 532

Concept Map 532 Chapter Review 533 Key Terms 533 Core Chemistry Skills 534 Understanding the Concepts 534 Additional Practice Problems 535 Challenge Problems 536 Answers to Engage Questions 536 Answers to Selected Problems 537 Combining Ideas from Chapters 15 and 16 538

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17 Organic Chemistry 540

CAREER Firefighter/Emergency Medical Technician 540

17.1 Alkanes 541 17.2 Alkenes, Alkynes, and Polymers 551

CHEMISTRY LINK TO HEALTH Hydrogenation of Unsaturated Fats 554

17.3 Aromatic Compounds 557 CHEMISTRY LINK TO HEALTH Some Common Aromatic Compounds 559

CHEMISTRY LINK TO HEALTH Polycyclic Aromatic Hydrocarbons (PAHs) 560

17.4 Alcohols and Ethers 560 CHEMISTRY LINK TO HEALTH Some Important Alcohols, Phenols, and Ethers 562

17.5 Aldehydes and Ketones 564 CHEMISTRY LINK TO HEALTH Some Important Aldehydes and Ketones 567

17.6 Carboxylic Acids and Esters 568 CHEMISTRY LINK TO HEALTH Carboxylic Acids in Metabolism 570

17.7 Amines and Amides 575 CHEMISTRY LINK TO THE ENVIRONMENT Alkaloids: Amines in Plants 576

UPDATE Diane’s Treatment in the Burn Unit 580

Concept Map 581 Chapter Review 581 Summary of Naming 582 Summary of Reactions 583 Key Terms 583 Core Chemistry Skills 584 Understanding the Concepts 585 Additional Practice Problems 585 Challenge Problems 587 Answers to Engage Questions 588 Answers to Selected Problems 588

18 Biochemistry 592

CAREER Diabetes Nurse 592 18.1 Carbohydrates 593

CHEMISTRY LINK TO HEALTH Hyperglycemia and Hypoglycemia 595

18.2 Disaccharides and Polysaccharides 598 CHEMISTRY LINK TO HEALTH How Sweet is My Sweetener? 600

18.3 Lipids 605 18.4 Amino Acids and Proteins 612

CHEMISTRY LINK TO HEALTH Essential Amino Acids and Complete Proteins 615

18.5 Protein Structure 617 18.6 Proteins as Enzymes 622 18.7 Nucleic Acids 624 18.8 Protein Synthesis 629

UPDATE Kate’s Program for Type 2 Diabetes 633

Concept Map 634 Chapter Review 635 Key Terms 636 Core Chemistry Skills 637 Understanding the Concepts 638 Additional Practice Problems 639 Challenge Problems 640 Answers to Engage Questions 641 Answers to Selected Problems 642 Combining Ideas from Chapters 17 and 18 645

Credits C-1

Glossary/Index I-1

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KEY MATH SKILLS Identifying Place Values 10 Using Positive and Negative Numbers in Calculations 11 Calculating Percentages 12 Solving Equations 13 Interpreting Graphs 14 Writing Numbers in Scientific Notation 17 Rounding Off 35 Calculating pH from [H3O

+] 449 Calculating [H3O

+] from pH 452

CORE CHEMISTRY SKILLS Counting Significant Figures 31 Using Significant Figures in Calculations 36 Using Prefixes 39 Writing Conversion Factors from Equalities 42 Using Conversion Factors 49 Using Density as a Conversion Factor 55 Identifying Physical and Chemical Changes 74 Converting Between Temperature Scales 76 Using Energy Units 80 Calculating Specific Heat 83 Using the Heat Equation 83 Counting Protons and Neutrons 111 Writing Atomic Symbols for Isotopes 113 Calculating Atomic Mass 115 Writing Electron Configurations 136 Using the Periodic Table to Write Electron

Configurations 140 Identifying Trends in Periodic Properties 144 Writing Positive and Negative Ions 158 Writing Ionic Formulas 163 Naming Ionic Compounds 164 Writing the Names and Formulas for Molecular

Compounds 173 Converting Particles to Moles 184 Calculating Molar Mass 188 Using Molar Mass as a Conversion Factor 190 Calculating Mass Percent Composition 195 Calculating an Empirical Formula 197 Calculating a Molecular Formula 202 Balancing a Chemical Equation 217 Classifying Types of Chemical Reactions 223 Identifying Oxidized and Reduced Substances 229 Using Mole–Mole Factors 242

Applications and Activities

Converting Grams to Grams 245 Calculating Quantity of Product from a Limiting

Reactant 248 Calculating Percent Yield 252 Using the Heat of Reaction 256 Drawing Lewis Symbols 270 Drawing Lewis Structures 271 Drawing Resonance Structures 276 Predicting Shape 279 Using Electronegativity 284 Identifying Polarity of Molecules 287 Identifying Intermolecular Forces 288 Calculating Heat for Change of State 292 Using the Gas Laws 317 Using the Ideal Gas Law 330 Calculating Mass or Volume of a Gas in a Chemical

Reaction 334 Calculating Partial Pressure 336 Using Solubility Rules 360 Calculating Concentration 363 Using Concentration as a Conversion Factor 365 Calculating the Quantity of a Reactant or Product

for a Chemical Reaction in Solution 375 Calculating the Freezing Point/Boiling Point of a

Solution 381 Writing the Equilibrium Expression 406 Calculating an Equilibrium Constant 407 Calculating Equilibrium Concentrations 412 Using Le Châtelier’s Principle 415 Writing the Solubility Product Expression 420 Calculating a Solubility Product Constant 421 Calculating the Molar Solubility 423 Identifying Conjugate Acid–Base Pairs 436 Calculating [H3O

+] and [OH-] in Solutions 446 Writing Equations for Reactions of Acids

and Bases 455 Calculating Molarity or Volume of an Acid or Base

in a Titration 457 Calculating the pH of a Buffer 460 Assigning Oxidation Numbers 478 Using Oxidation Numbers 480 Identifying Oxidizing and Reducing Agents 481 Using Half-Reactions to Balance Redox Equations 483 Identifying Spontaneous Reactions 488 Writing Nuclear Equations 512 Using Half-Lives 523 Naming and Drawing Alkanes 544

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xiv Applications and Activities

Writing Equations for Hydrogenation and Polymerization 553

Naming Aldehydes and Ketones 565 Naming Carboxylic Acids 569 Forming Esters 571 Forming Amides 577 Drawing Haworth Structures 595 Identifying Fatty Acids 605 Drawing Structures for Triacylglycerols 607 Drawing the Structure for an Amino Acid at

Physiological pH 613 Identifying the Primary, Secondary, Tertiary, and

Quaternary Structures of Proteins 620 Writing the Complementary DNA Strand 626 Writing the mRNA Segment for a DNA Template 630 Writing the Amino Acid for an mRNA Codon 631

Interactive Videos Solving Equations 14 Conversion Factors 48 Chemical vs. Physical Changes 75 Rutherford’s Gold-Foil Experiment 109 Isotopes and Atomic Mass 116 Naming and Writing Ionic Formulas 167 Drawing Lewis Structures with Multiple Bonds 275 Kinetic Molecular Theory 312 Solutions 374 Calculations Involving Solutions in Reactions 376 Acid–Base Titration 458 Calculation the pH of a Buffer 460 Writing Equations for an Isotope Produced by

Bombardment 519 Half-Lives 525 Naming Alkanes 547 Haworth Structures of Monosaccharides 596 Amino Acids at Physiological pH 613 Different Levels of Protein Structure 621 Protein Synthesis 630

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KAREN TIMBERLAKE is Professor Emerita of Chemistry at Los Angeles Valley College, where she taught chemistry for allied health and preparatory chemistry for 36 years. She received her bachelor’s degree in chemistry from the University of Washington and her master’s degree in biochemistry from the University of California at Los Angeles.

Professor Timberlake has been writing chemistry textbooks for more than 40 years. During that time, her name has become associated with the strategic use of pedagogical tools that promote student success in chemistry and the application of chemistry to real-life situations. More than one million students have learned chemistry using texts, laboratory manuals, and study guides written by Karen Timberlake. In addition to Basic Chemistry, sixth edition, she is also the author of General, Organic, and Biological Chemistry: Structures of Life, sixth edition, with the accompanying Study Guide and Selected Solutions Manual, and Chemistry: An Introduction to General, Organic, and Biological Chemistry, thirteenth edition, with the accompanying Study Guide and Selected Solutions Manual, Laboratory Manual, and Essential Laboratory Manual.

Professor Timberlake belongs to numerous scientific and educational organizations including the American Chemical

About the Authors

Society (ACS) and the National Science Te a c h e r s A s s o c i a t i o n ( N S TA ) . S h e has been the Western Regional Winner of Excellence in College Chemistry Teaching Award given by the Chemical Manufacturers Association. She received the McGuffey Award in Physical Sciences from the Textbook Authors Association for her textbook Chemistry: An Introduction to General, Organic, and Biological Chemistry, eighth edition. She received the “Texty” Textbook Excellence Award from the Textbook Authors Association for the first edition of Basic Chemistry. She has participated in education grants for science teaching including the Los Angeles Collaborative for Teaching Excellence (LACTE) and a Title III grant

at her college. She speaks at conferences and educational meetings on the use of student-centered teaching methods in chemistry to promote the learning success of students.

Her husband, William Timberlake, who is the coauthor of this text, is Professor Emeritus of Chemistry at Los Angeles Harbor College, where he taught preparatory and organic chemistry for 36 years. He received his bachelor’s degree in chemistry from Carnegie Mellon University and his master’s degree in organic chemistry from the University of California at Los Angeles.

When the Professors Timberlake are not writing textbooks, they relax by playing tennis, ballroom dancing, hiking, traveling, trying new restaurants, cooking, and enjoying care of their grandchildren, Daniel and Emily.

DEDICATION • Our son, John, daughter-in-law, Cindy, grandson, Daniel,

and granddaughter, Emily, for the precious things in life

• The wonderful students over many years whose hard work and commitment always motivated us and put purpose in our writing

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W elcome to the sixth edition of Basic Chemistry. This chemistry text was written and designed to prepare you for science-related professions, such as

engineering, nursing, medicine, environmental or agricultural science, or for careers such as laboratory technology. This text assumes no prior knowledge of chemistry. Our main objective in writing this text is to make the study of chemistry an engaging and positive experience for you by relating the structure and behavior of matter to real life. This new edition introduces more problem-solving strategies, Analyze the Problem with Connect features, Try It First and Engage features, and conceptual and challenge problems.

It is our goal to help you become a critical thinker by understanding scientific concepts that will form a basis for making important decisions about issues concerning health and the environment. Thus, we have utilized materials that

• help you to learn and enjoy chemistry • relate chemistry to careers that interest you • develop problem-solving skills • promote learning and success in chemistry

Active Reading Features for Successful Learning In this sixth edition of Basic Chemistry, we have increased our emphasis on providing features that promote student interaction with the textual content. We continue to develop features based on new research on learning theory and extend them throughout the text as a part of our study plan Strategies and Practices for Active Reading in Chemistry (SPARC).

With the success of students involved in active learning in the classroom, we see the opportunity to develop a parallel plan of reading and learning experiences using our textbook. As chemistry textbook authors we are interested in connecting cognitive science and learning research to improve student reading and learning. SPARC is the combination of utilizing reading strategies that increase learning and success in chemistry.

Strategies for Learning New Information

1. Combine graphics with words Students improve learning by receiving information in different ways. In this text, we combine text and ques- tions with illustrations using macro-to-micro art, tables, graphs, diagrams, videos, photos, and concept maps.

2. Connect abstract concepts with concrete representations New concepts are illustrated and explained with real-life examples, career stories and updates, Chemical Links to Health and the Environment, and Applications. Prompts of Key Math Skills and Core Chemistry Skills alert stu- dents to the fundamental ideas in each chapter.

Strategies for Connecting New Information

3. Ask inquiring questions Engage questions throughout each chapter ask students “why,” “how,” and “what if,” requiring them to link new information with prior knowledge.

4. Alternate problems containing solutions with similar problems that students must solve Many Sample Problems with Try It First reminders throughout each chapter contain step-by-step solutions that guide students through the process of problem-solving. An abundance of Practice Problems, Understanding the Concepts, Additional Problems, and Challenge Problems provide students with similar problem-solving experience. Answers are provided for immediate feedback.

Strategies for Recalling and Retrieving Information

5. Provide opportunities to practice recall and retrieval Throughout each chapter, students are encouraged to prac- tice the recall and retrieval of material by repeating practice every few days, weeks, and even months using Self Tests, Practice Problems, Understanding the Concepts, Additional Problems, and Challenge Problems. Review prompts remind students of key ideas in previous chapters.

6. Combine different but related topics and skills Better learning can be achieved by alternating different topics and types of content. In this text, problem sets including Understanding the Concepts, Additional Practice Problems, Challenge Problems, and Combining Ideas provide students practice of different topics or skills, rather than focusing on one topic or skill.

7. Assess to maintain and improve retention Every chapter provides many types of assessment such as Self Tests, Practice Problems, Understanding the Concepts, Additional Practice Problems, Challenge Problems, and Combining Ideas. By practicing informa- tion retrieval, checking progress, and reviewing, students improve their success on exams.

Preface

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xviii Preface

New and Updated for the Sixth Edition New and updated features have been added throughout this sixth edition, including the following:

• NEW! Chapter Openers provide timely examples and engaging, topical examples of the chemistry that is part of contemporary professions.

• NEW! Chapter Openers include references to new Update features at the end of the chapter that continue the story.

• NEW! Review heads are now listed at the beginning of each section to emphasize the Key Math Skills and Core Chemistry Skills from previous chapters required for learning new chemistry principles.

• NEW! Pictorial Representations using photos and graphs are added to increase the understanding of new topics.

• NEW! Sample Problems show Steps to guide the stu- dent through problem solving.

• NEW! Self Test icons in Sample Problems encourage students to use problem-solving strategies immediately as they review the content in a section.

• NEW! Expanded Self Test questions in Sample Problems provide students with additional self-testing practice.

• NEW! Margin icons Practice Problems encourage stu- dents to work related Practice Problems and self-assess as they study.

• NEW! Expanded Engage Questions and Answers are included in the chapter.

• NEW! Additional Practice Problems and Challenge Problems are added to help student practice testing and increase understanding of the concepts in the chapter.

• NEW! Three-Dimensional Representations, including ball-and-stick and space-filling models, are included to illustrate the shapes of molecules and polyatomic ions.

• NEW! Foreground colors in color palette are now ADA accessible.

• NEW! Multiple art pieces contain separate captions in boxes for each art.

• NEW! Concept Maps are tinted with color screens for emphasis of topics.

• NEW! Interactive Videos are added to illustrate more step-by-step problem-solving strategies.

Chapter Organization of the Sixth Edition In each textbook we write, we consider it essential to relate every chemical concept to real-life issues. Because a chemistry course may be taught in different time frames, it may be difficult to cover all the chapters in this text. However, each chapter is a complete package, which allows some chapters to be skipped or the order of presentation to be changed.

Chapter 1, Chemistry in Our Lives, discusses the Scientific Method in everyday terms and guides students in developing a study plan for learning chemistry, with a section of Key Math Skills that reviews the basic math, including scientific notation, needed in chemistry calculations.

• The Chapter Opener tells the story of a murder and features the work and career of forensic scientists.

• The Update feature describes the forensic evidence that helps to solve the murder and includes Applications.

• An updated Section 1.3, Studying and Learning Chemistry, expands the discussion of study strategies that improve learning and understanding of content.

• A new Decimal Place Value Chart is added in Section 1.4, Key Math Skills for Chemistry, to clarify decimal place values.

• In Section 1.4, Interpreting Graphs, the format for the x and y axes is standardized.

• Key Math Skills are: Identifying Place Values, Using Positive and Negative Numbers in Calculations, Calculating Percentages, Solving Equations, Interpreting Graphs, and Writing Numbers in Scientific Notation.

Chapter 2, Chemistry and Measurements, looks at measurement and emphasizes the need to understand numeri- cal relationships of the metric system. Significant figures are discussed in the determination of final answers. Prefixes from the metric system are used to write equalities and conversion factors for problem-solving strategies. Density is discussed and used as a conversion factor.

• The Chapter Opener tells the story of a patient with high blood pressure and features the work and career of a reg- istered nurse.

• The Update describes the patient’s status and follow-up visit with his doctor.

• In Section 2.5, conversion factors with multiple units are added in the Practice Problems.

• In Section 2.6, Steps as guides to problem solving are added to Sample Problems 2.3, 2.4, and 2.5.

• In Section 2.6, a new type of Sample Problem and new Practice Problems for the conversion of units in a frac- tion are added.

• Sample Problems relate problem solving to health- related topics such as the measurements of blood volume, omega-3 fatty acids, radiological imaging, body fat, cholesterol, and medication orders.

• Applications feature questions about measurements, daily values for minerals and vitamins, and equalities and conversion factors for medications.

• Key Math Skill is: Rounding Off. • Core Chemistry Skills are: Counting Significant Figures,

Using Significant Figures in Calculations, Using Prefixes, Writing Conversion Factors from Equalities, Using Con- version Factors, and Using Density as a Conversion Factor.

Chapter 3, Matter and Energy, classifies matter and states of matter, describes temperature measurement, and discusses energy, specific heat, and energy in nutrition. Physical and

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Preface xix

chemical properties and physical and chemical changes are discussed.

• The Chapter Opener describes diet and exercise for an overweight adolescent at risk for type 2 diabetes and fea- tures the work and career of a dietitian.

• The Update describes the diet prepared with a dietitian for weight loss.

• Figures of Decomposition of Salt, and Separation of Mixtures by Filtration are moved to Section 3.2 for clarity of content.

• In Section 3.5, Specific Heat, a new Sample Problem using heat exchange data is added along with new Practice Problems.

• Practice Problems and Sample Problems include high temperatures used in cancer treatment, the energy pro- duced by a high-energy shock output of a defibrillator, body temperature lowering using a cooling cap, ice bag therapy for muscle injury, dental implants, and energy values for food.

• Core Chemistry Skills are: Identifying Physical and Chemical Changes, Converting Between Temperature Scales, Using Energy Units, Calculating Specific Heat, and Using the Heat Equation.

• The interchapter problem set, Combining Ideas from Chapters 1 to 3, completes the chapter.

Chapter 4, Atoms and Elements, introduces elements and atoms and the periodic table. The names and symbols for the newest elements 113, Nihonium, Nh, 115, Moscovium, Mc, 117, Tennessine, Ts, and 118, Oganesson, Og, are included on the periodic table. Atomic numbers and mass numbers are deter- mined for isotopes. Atomic mass is calculated using the masses of the naturally occurring isotopes and their abundances.

• The Chapter Opener and Update discuss the improve- ment in crop production and feature the work and career of a farmer.

• Atomic number and mass number are used to calculate the number of protons and neutrons in an atom.

• The number of protons and neutrons are used to calculate the mass number and to write the atomic symbol for an isotope.

• Figure 4.3 now includes Francium (Fr) in Group 1A (1), and Figure 4.4 now includes Tennessine (Ts).

• Core Chemistry Skills are: Counting Protons and Neutrons, Writing Atomic Symbols for Isotopes, and Calculating Atomic Mass.

Chapter 5, Electronic Structure of Atoms and Periodic Trends, uses the electromagnetic spectrum to explain atomic spectra and develop the concept of energy levels and sublevels. Electrons in sublevels and orbitals are represented using orbital diagrams and electron configurations. Periodic properties of elements, including atomic size, ionization energy and metallic character, are related to their valence electrons. Small periodic tables illustrate the trends of periodic properties.

• The Chapter Opener and Update discuss the development of new products of metals, plastics, and semiconductors, and career of a materials engineer.

• The electromagnetic spectrum is described with everyday examples and a diagram.

• The three-dimensional representations of the s, p, and d orbitals are drawn.

• The trends in periodic properties are described for valence electrons, atomic size, ionization energy, and metallic character.

• A photo of infrared radiation used to keep food warm, and a photo of gamma knife radiation used to kill cancer cells are added.

• Table 5.2 for electron capacity in sublevels is reordered with Energy Level n = 1 at the top and Energy Level n = 4 at the bottom.

• Core Chemistry Skills are: Writing Electron Configurations, Using the Periodic Table to Write Electron Configurations, and Identifying Trends in Periodic Properties.

Chapter 6, Ionic and Molecular Compounds, describes the formation of ionic and covalent bonds. Chemical formulas are written, and ionic compounds—including those with poly- atomic ions—and molecular compounds are named.

• The Chapter Opener describes the chemistry of aspirin and features the work and career of a pharmacist.

• The Update describes several types of compounds at a pharmacy and includes Applications.

• New art is added or updated to provide everyday exam- ples of the content.

• New material on polyatomic ions compares the names of ate ions and ite ions, the charge of sulfate and sulfite, phosphate and phosphite, carbonate and hydrogen carbonate, and the formulas and charges of halogen polyatomic ions with oxygen.

• Core Chemistry Skills are: Writing Positive and Negative Ions, Writing Ionic Formulas, Naming Ionic Compounds, and Writing the Names and Formulas for Molecular Compounds.

Chapter 7, Chemical Quantities, discusses Avogadro’s number, the mole, and molar masses of compounds, which are used in calculations to determine the mass or number of parti- cles in a quantity of a substance. The mass percent composition of a compound is calculated and used to determine its empiri- cal and molecular formula.

• The Chapter Opener and Update describe the diagno- sis and treatment of a pet and the work and career of a veterinarian.

• Core Chemistry Skills are: Converting Particles to Moles, Calculating Molar Mass, Using Molar Mass as a Conversion Factor, Calculating Mass Percent Composi- tion, Calculating an Empirical Formula, and Calculating a Molecular Formula.

• The interchapter problem set, Combining Ideas from Chapters 4 to 7, completes the chapter.

Chapter 8, Chemical Reactions, shows students how to balance chemical equations, and discusses how to classify chemical reactions into types: combination, decomposition, single replacement, double replacement, combustion, and oxidation–reduction.

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xx Preface

• The Chapter Opener and Update discuss tests and treatment for emphysema and the work and career of an exercise physiologist.

• Core Chemistry Skills are: Balancing a Chemical Equation, Classifying Types of Chemical Reactions, and Identifying Oxidized and Reduced Substances.

Chapter 9, Chemical Quantities in Reactions, describes the mole and mass relationships among the reactants and prod- ucts and provides calculations of limiting reactants and percent yields. The chapter concludes with a discussion of energy in reactions.

• The Chapter Opener describes insecticides and pharma- ceuticals used on a ranch and discusses the career of an environmental scientist.

• The Update describes the collection of soil and water samples for testing for insecticides.

• Mole and mass relationships among the reactants and products are examined along with calculations of percent yield and limiting reactants.

• Material including new problems with three equations and calculations using Hess’s Law is rewritten for clarity.

• Core Chemistry Skills are: Using Mole–Mole Factors, Converting Grams to Grams, Calculating Quantity of Product from a Limiting Reactant, Calculating Percent Yield, and Using the Heat of Reaction.

Chapter 10, Bonding and Properties of Solids and Liquids, introduces Lewis structures for molecules and ions with single and multiple bonds as well as resonance structures. Electronegativity leads to a discussion of the polarity of bonds and molecules. Lewis structures and VSEPR theory illustrate covalent bonding and the three-dimensional shapes of molecules and ions. The intermolecular forces between particles and their impact on states of matter and changes of state are described. The energy involved with changes of state is calculated.

• The Chapter Opener and Update describe the process- ing of a tissue sample and the work and career of a histologist.

• New three-dimensional representations of ball-and-stick models and space-filling models are added to illustrate shapes of molecules and polyatomic ions.

• Lewis structures are drawn for molecules and ions with single, double, and triple bonds. Resonance structures are drawn if two or more Lewis structures are possible.

• Shapes and polarity of bonds and molecules are predicted using VSEPR theory.

• Intermolecular forces in compounds are discussed including ionic bonds, hydrogen bonds, dipole–dipole attractions, and dispersion forces.

• Core Chemistry Skills are: Drawing Lewis Symbols, Drawing Lewis Structures, Drawing Resonance Structures, Predicting Shape, Using Electronegativity, Identifying Polarity of Molecules, Identifying Intermolecular Forces, and Calculating Heat for Change of State.

• The interchapter problem set, Combining Ideas from Chapters 8 to 10, completes the chapter.

Chapter 11, Gases, discusses the properties of gases and calculates changes in gases using the gas laws: Boyle’s, Charles’s, Gay-Lussac’s, Avogadro’s, Dalton’s, and the Ideal Gas Law. Problem-solving strategies enhance the discussion and calculations with gas laws including chemical reactions using the ideal gas law.

• The Chapter Opener and Update feature the work and career of a respiratory therapist, who uses oxygen to treat a child with asthma.

• Applications include calculations of mass or pressure of oxygen in uses of hyperbaric chambers.

• Core Chemistry Skills are: Using the Gas Laws, Using the Ideal Gas Law, Calculating Mass or Volume of a Gas in a Chemical Reaction, and Calculating Partial Pressure.

Chapter 12, Solutions, describes solutions, electrolytes, saturation and solubility, insoluble salts, concentrations, and osmosis. The concentrations of solutions are used to determine volume or mass of solute. The volumes and molarities of solutions are used in calculations for dilutions and titration. Properties of solutions, freezing and boiling points, osmosis, and dialysis are discussed.

• The Chapter Opener describes a patient with kidney failure and dialysis treatment and features the work and career of a dialysis nurse.

• The Update discusses dialysis treatment and electrolyte levels in the dialysate fluid.

• A new example of suspensions used to purify water in treatment plants is added.

• New art illustrates the freezing point decrease and boil- ing point increase for aqueous solutions with increasing number of moles of solute in one kilogram of water.

• Core Chemistry Skills are: Using Solubility Rules, Calcu- lating Concentration, Using Concentration as a Conversion Factor, Calculating the Quantity of a Reactant or Product for a Chemical Reaction in Solution, and Calculating the Freezing Point/Boiling Point of a Solution.

Chapter 13, Reaction Rates and Chemical Equilibrium, looks at the rates of reactions and the equilibrium condition when forward and reverse rates for a reaction become equal. Equilibrium expressions for reactions are written and equilib- rium constants are calculated. The equilibrium constant is used to calculate the concentration of a reactant or product at equi- librium. Le Châtelier’s principle is used to evaluate the impact on concentrations when stress is placed on a system at equi- librium. The concentrations of solutes in a solution is used to calculate the solubility product constant (Ksp).

• The Chapter Opener and Update discuss the equilibrium of CO2 in the ocean and feature the work and career of a chemical oceanographer.

• Core Chemistry Skills are: Writing the Equilibrium Expression, Calculating an Equilibrium Constant, Calculating Equilibrium Concentrations, Using Le Châtelier’s Principle, Writing the Solubility Product Expression, Calculating a Solubility Product Constant, and Calculating the Molar Solubility.

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Chapter 14, Acids and Bases, discusses acids and bases and their strengths, and conjugate acid–base pairs. The dissociation of strong and weak acids and bases is related to their strengths as acids or bases. The dissociation of water leads to the water dissociation expression, Kw, the pH scale, and the calculation of pH. Chemical equations for acids in reactions are balanced and titration of an acid is illustrated. Buffers are discussed along with their role in the blood. The pH of a buffer is calculated.

• The Chapter Opener describes a blood sample for an emergency room patient sent to the clinical laboratory for analysis of blood pH and CO2 gas and describes the work and career of a medical laboratory technologist.

• The Update describes the symptoms and treatment for acid reflux disease (GERD).

• Key Math Skills are: Calculating pH from [H3O +], and

Calculating [H3O +] from pH.

• Core Chemistry Skills are: Identifying Conjugate Acid– Base Pairs, Calculating [H3O

+] and [OH-] in Solutions, Writing Equations for Reactions of Acids and Bases, Calculating Molarity or Volume of an Acid or Base in a Titration, and Calculating the pH of a Buffer.

• The interchapter problem set, Combining Ideas from Chapters 11 to 14, completes the chapter.

Chapter 15, Oxidation and Reduction, looks at the characteristics of oxidation and reduction reactions. Oxidation numbers are assigned to the atoms in elements, molecules, and ions to determine the components that lose electrons during oxidation and gain electrons during reduction. The half-reaction method is utilized to balance oxidation–reduction reactions. The production of electrical energy in voltaic cells and the requirement of electrical energy in electrolytic cells are diagrammed using half-cells. The activity series is used to determine the spotaneous direction of an oxidation–reduction reaction.

• The Chapter Opener and Update discuss the reactions involved in teeth whitening and the work and career of a dentist.

• New material and art on lithium-ion batteries is added. • Core Chemistry Skills are: Assigning Oxidation Numbers,

Using Oxidation Numbers, Identifying Oxidizing and Reducing Agents, Using Half-Reactions to Balance Redox Equations, and Identifying Spontaneous Reactions.

Chapter 16, Nuclear Chemistry, looks at the types of radiation emitted from the nuclei of radioactive atoms. Nuclear equations are written and balanced for both naturally occurring radioactivity and artificially produced radioactivity. The half- lives of radioisotopes are discussed, and the amount of time for a sample to decay is calculated. Radioisotopes important in the field of nuclear medicine are described. Fission and fusion and their role in energy production are discussed.

• The Chapter Opener and Update describe a stress test using a radioactive isotope and feature the work and career of a radiation technologist.

• Core Chemistry Skills are: Writing Nuclear Equations, and Using Half-Lives.

• The interchapter problem set, Combining Ideas from Chapters 15 and 16, completes the chapter.

Chapter 17, Organic Chemistry, compares inorganic and organic compounds, and describes the condensed structural and line-angle formulas of alkanes, alkenes, alcohols, ethers, aldehydes, ketones, carboxylic acids, esters, amines, and amides.

• The Chapter Opener and Update describe emergency treatment for burns and feature the work and career of a firefighter/emergency medical technician.

• The properties of organic and inorganic compounds are compared in Table 17.1.

• Line-angle formulas are added to Table 17.2 IUPAC Names and Formulas of the First 10 Alkanes.

• More line-angle structures are included in text examples, sample problems, questions, and problems.

• The two-dimensional and three-dimensional repre- sentations of methane and ethane are illustrated using condensed structural formulas, expanded structural for- mulas, ball-and-stick models, space-filling models, and wedge–dash models.

• Core Chemistry Skills are: Naming and Drawing Alkanes, Writing Equations for Hydrogenation and Polymeriza- tion, Naming Aldehydes and Ketones, Naming Carboxylic Acids, Forming Esters, and Forming Amides.

Chapter 18, Biochemistry, looks at the chemical structures and reactions of chemicals that occur in living systems. We focus on four types of biomolecules—carbohydrates, lipids, proteins, and nucleic acids—as well as their biochemical reactions.

• The Chapter Opener and Update describe diagnosis and treatment of diabetes and feature the work and career of a diabetes nurse.

• Monosaccharides are classified as aldo or keto pentoses or hexoses.

• Haworth structures are drawn for monosaccharides, disaccharides, and polysaccharides.

• The shapes of proteins are related to the activity and regulation of enzyme activity.

• The genetic code is described and utilized in the process of protein synthesis.

• Core Chemistry Skills are: Drawing Haworth Structures, Identifying Fatty Acids, Drawing Structures for Triacyl- glycerols, Drawing the Structure for an Amino Acid at Physiological pH, Identifying the Primary, Secondary, Tertiary, and Quaternary Structures of Proteins, Writing the Complementary DNA Strand, Writing the mRNA Segment for a DNA Template, and Writing the Amino Acid for an mRNA Codon.

• The interchapter problem set, Combining Ideas from Chapters 17 and 18, completes the chapter.

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xxii Preface

Acknowledgments The preparation of a new text is a continuous effort of many people. We are thankful for the support, encouragement, and dedication of many people who put in hours of tireless effort to produce a high-quality book that provides an outstanding learning package. The editorial team at Pearson has done an exceptional job. We want to thank Jeanne Zalesky, Director, Courseware Portfolio Management, and Editor Jessica Moro, who supported our vision of this sixth edition and the development of strategies based on learning theory research.

We appreciate all the wonderful work of Melanie Field, Content Producer, who skillfully brought together files, art, web site materials, and all the things it takes to prepare a book for production. We appreciate the work of Rose Kernan at SPi Global, who brilliantly coordinated all phases of the manuscript to the final pages of a beautiful book. Thanks to Mark Quirie, manuscript and accuracy reviewer, and Karen Slaght, who analyzed and edited the manuscripts and pages to make sure the words and problems were correct to help students learn chemistry. Their keen eyes and thoughtful comments were extremely helpful in the development of this text.

We appreciate the contributions from Dr. John Timberlake that connected recent learning theory research with our effort to encourage students to incorporate active reading in their study plan.

Thanks to Kristen Flathman, Managing Producer, Coleen Morrison, Courseware Analyst, and Barbara Yien, Course- ware Director, for their excellent review of pages and helpful suggestions.

We am especially proud of the art program in this text, which lends beauty and understanding to chemistry. We would like to thank Jay McElroy, Art Courseware Analyst, and Stephanie Marquez and Alicia Elliott, Photo and Illustration Project Managers, Mark Ong, Design Manager, and Tamara

Newnam, Cover and Interior Designer, whose creative ideas provided the outstanding design for the cover and pages of the book. We appreciate the tireless efforts of Namrata Aggarwal, Photo Researcher, and Matt Perry, Rights and Permissions Project Manager, in researching and selecting vivid photos for the text so that students can see the beauty of chemistry. Thanks also to Bio-Rad Laboratories for their courtesy and use of KnowItAll ChemWindows, drawing software that helped us produce chemical structures for the manuscript. The macro-to- micro illustrations designed by Jay McElroy and Imagineering Art give students visual impressions of the atomic and molecular organization of everyday things and are a fantastic learning tool. We also appreciate all the hard work in the field put in by the marketing team and Allison Rona, Marketing Manager.

We am extremely grateful to an incredible group of peers for their careful assessment of all the new ideas for the text; for their suggested additions, corrections, changes, and deletions; and for providing an incredible amount of feedback about improvements for the book. We admire and appreciate every one of you.

If you would like to share your experience with chemistry, or have questions and comments about this text, We would appreciate hearing from you.

Karen Timberlake William Timberlake

Email: [email protected]

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xxiii

Sixth Edition Reviewers Jennifer Bell Daytona State College, Main

Karen Endebrock Daytona State College, Deland

Paul Forster University of Nevada, Las Vegas

Ann Mills Minnesota West Community and Technical College, Worthington

Phil Nubel Waubonsee Community College

Amy Whiting North Central Texas College

Accuracy Reviewer Mark Quirie Algonquin College

Reviewers

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BASIC CHEMISTRY

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Help students master the math and problem solving they will use in their future careers

Basic Chemistry introduces students to the essential scientific and mathematical concepts of chemistry while providing the scaffolded support they need. With accessible language and a moderate pace, the text is easy- to-follow for first-time chemistry students. The 6th Edition incorporates sound pedagogy and the best principles from learning design theory to create an updated learning program designed for today’s students. The applied focus helps students connect chemistry with their interests and potential careers. Enhanced digital tools and additional practice problems in Mastering Chemistry ensure students master the basic quantitative and science skills needed to succeed in this course and beyond.

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Art and Videos that are more understandable than ever before

P. 74

Updated! Art program incorporates sound pedagogy and the best learning design principles based on the way today’s students learn.

74 CHAPTER 3 Matter and Energy

TABLE 3.3 Summary of Physical and Chemical Properties and Changes Physical Chemical

Property A characteristic of a substance: color, shape, odor, luster, size, melting point, or density.

A characteristic that indicates the ability of a substance to form another substance: paper can burn, iron can rust, silver can tarnish.

Change A change in a physical property that retains the identity of the substance: a change of state, a change in size, or a change in shape.

A change in which the original substance is converted to one or more new substances: paper burns, iron rusts, silver tarnishes.

TABLE 3.4 Examples of Some Physical and Chemical Changes Physical Changes Chemical Changes

Water boils to form water vapor. Water and cesium combine explosively.

Paper is cut into tiny pieces of confetti. Paper burns with a bright flame and produces heat, ashes, carbon dioxide, and water vapor.

Sugar dissolves in water to form a sugar solution.

Heating sugar forms a smooth, caramel- colored substance.

Iron has a melting point of 1538 °C. Iron, which is gray and shiny, combines with oxygen to form orange-red iron oxide (rust).

A chemical change occurs when sugar is heated, forming a caramelized topping for flan.

CORE CHEMISTRY SKILL Identifying Physical and Chemical

Changes

ENGAGE 3.5 Why is the melting point of iron, 1538 °C, a physical property, whereas the heating of iron with oxygen to form rust, Fe2O3, is a chemical property?

FIGURE 3.3 The decomposition of salt, NaCl, produces the elements sodium and chlorine.

Sodium metal

Sodium chloride

Chemical change

Chlorine gasand

+

ENGAGE 3.4 Why is the decomposition of salt a chemical change?

SAMPLE PROBLEM 3.2 Physical and Chemical Changes

TRY IT FIRST

Classify each of the following as a physical or chemical change:

a. A gold ingot is hammered to form gold leaf. b. Gasoline burns in air. c. Garlic is chopped into small pieces. d. Milk left in a warm room turns sour. e. A mixture of oil and water is separated.

A gold ingot is hammered to form gold leaf.

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In-art captions replace long legends, and the flow and size of the art is updated to help increase student understanding.

3.1 Classification of Matter 69

3.1 Classification of Matter LEARNING GOAL Classify examples of matter as pure substances or mixtures.

Matter is anything that has mass and occupies space. Matter is everywhere around us: the orange juice we had for breakfast, the water we put in the coffee maker, the plastic bag we put our sandwich in, our toothbrush and toothpaste, the oxygen we inhale, and the carbon dioxide we exhale. All of this material is matter. The different types of matter are classified by their composition.

Pure Substances: Elements and Compounds All matter is made of extremely small particles called atoms. Much of matter is made of atoms bonded together in definite arrangements called molecules. A pure substance is matter that consists of just one type of atom or one type of molecule. An element, the simplest type of a pure substance, is composed of only one type of atom such as silver, iron, or aluminum. Silver is composed of silver atoms, iron of iron atoms, and aluminum of aluminum atoms. A full list of the elements is found on the inside front cover of this text.

A compound is also a pure substance, but it consists of atoms of two or more elements always chemically combined in the same proportion. For example, in the compound water, there are two hydrogen atoms for every one oxygen atom, which is represented by the formula H2O. This means that water always has the same composi- tion of H2O. Another compound that consists of a chemical combination of hydrogen and oxygen is hydrogen peroxide. It has two hydrogen atoms for every two oxygen atoms and is represented by the formula H2O2. Thus, water (H2O) and hydrogen peroxide (H2O2) are different compounds even though they contain the same elements, hydrogen and oxygen.

ENGAGE 3.1 Why are elements and compounds both pure substances?

PRACTICE PROBLEMS Try Practice Problems 3.1 and 3.2

LOOKING AHEAD

3.1 Classification of Matter 69

3.2 States and Properties of Matter 72

3.3 Temperature 75 3.4 Energy 79 3.5 Specific Heat 82 3.6 Energy and Nutrition 87

Aluminum atom

An aluminum can consists of many atoms of aluminum.

H H O

Water molecule

A water molecule, H2O, consists of two atoms of hydrogen (white) for one atom of oxygen (red).

H

H O

O Hydroxide peroxide molecule

A hydroxide peroxide molecule, H2O2, consists of two atoms of hydrogen (white) for every two atoms of oxygen (red).

Mixtures In a mixture, two or more different substances are physically mixed. Much of the matter in our everyday lives consists of mixtures. The air we breathe is a mixture of mostly oxygen and nitrogen gases. The steel in buildings and railroad tracks is a mixture of iron, nickel, carbon, and chromium. The brass in doorknobs and musical instruments is a mixture of

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P. 69

3.1 Classification of Matter 69

3.1 Classification of Matter LEARNING GOAL Classify examples of matter as pure substances or mixtures.

Matter is anything that has mass and occupies space. Matter is everywhere around us: the orange juice we had for breakfast, the water we put in the coffee maker, the plastic bag we put our sandwich in, our toothbrush and toothpaste, the oxygen we inhale, and the carbon dioxide we exhale. All of this material is matter. The different types of matter are classified by their composition.

Pure Substances: Elements and Compounds All matter is made of extremely small particles called atoms. Much of matter is made of atoms bonded together in definite arrangements called molecules. A pure substance is matter that consists of just one type of atom or one type of molecule. An element, the simplest type of a pure substance, is composed of only one type of atom such as silver, iron, or aluminum. Silver is composed of silver atoms, iron of iron atoms, and aluminum of aluminum atoms. A full list of the elements is found on the inside front cover of this text.

A compound is also a pure substance, but it consists of atoms of two or more elements always chemically combined in the same proportion. For example, in the compound water, there are two hydrogen atoms for every one oxygen atom, which is represented by the formula H2O. This means that water always has the same composi- tion of H2O. Another compound that consists of a chemical combination of hydrogen and oxygen is hydrogen peroxide. It has two hydrogen atoms for every two oxygen atoms and is represented by the formula H2O2. Thus, water (H2O) and hydrogen peroxide (H2O2) are different compounds even though they contain the same elements, hydrogen and oxygen.

ENGAGE 3.1 Why are elements and compounds both pure substances?

PRACTICE PROBLEMS Try Practice Problems 3.1 and 3.2

LOOKING AHEAD

3.1 Classification of Matter 69

3.2 States and Properties of Matter 72

3.3 Temperature 75 3.4 Energy 79 3.5 Specific Heat 82 3.6 Energy and Nutrition 87

Aluminum atom

An aluminum can consists of many atoms of aluminum.

H H O

Water molecule

A water molecule, H2O, consists of two atoms of hydrogen (white) for one atom of oxygen (red).

H

H O

O Hydroxide peroxide molecule

A hydroxide peroxide molecule, H2O2, consists of two atoms of hydrogen (white) for every two atoms of oxygen (red).

Mixtures In a mixture, two or more different substances are physically mixed. Much of the matter in our everyday lives consists of mixtures. The air we breathe is a mixture of mostly oxygen and nitrogen gases. The steel in buildings and railroad tracks is a mixture of iron, nickel, carbon, and chromium. The brass in doorknobs and musical instruments is a mixture of

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Engage students in learning chemistry

Updated! Chemistry Links to Health and Chemistry Links to the Environment appear throughout the text and relate chemistry concepts to real-life topics in health, the environment, and medicine. High-interest topics include weight loss and weight gain, hyperglycemia and hypoglycemia, antacids, gout and kidney stones, sweeteners, and essential amino acids. Follow-up questions also appear throughout the text.

Updated! Interactive Videos give students an opportunity to reinforce what they just learned by showing how chemistry works in real life and introducing a bit of humor into chemical problem solving and demonstrations. Topics include Using Conversion Factors, Mass Calculations for Reactions, Concentration of Solutions, Balancing Nuclear Equations, and Chemical v. Physical Change.

52 CHAPTER 2 Chemistry and Measurements

Chemistry Link to Health Toxicology and Risk–Benefit Assessment

Each day, we make choices about what we do or what we eat, often without thinking about the risks associated with these choices. We are aware of the risks of cancer from smoking or the risks of lead poisoning, and we know there is a greater risk of having an accident if we cross a street where there is no light or crosswalk.

A basic concept of toxicology is the statement of Paracelsus that the dose is the difference between a poison and a cure. To evaluate the level of danger from various substances, natural or synthetic, a risk assessment is made by exposing laboratory animals to the sub- stances and monitoring the health effects. Often, doses very much greater than humans might ordinarily encounter are given to the test animals.

Many hazardous chemicals or substances have been identified by these tests. One measure of toxicity is the LD50, or lethal dose, which is the con- centration of the substance that causes death in 50% of the test animals. A dosage is typically measured in milligrams per kilogram (mg/kg) of body mass or micrograms per kilogram (mcg/kg) of body mass.

Other evaluations need to be made, but it is easy to compare LD50 values. Parathion, a pesticide, with an LD50 of 3 mg/kg, would be highly toxic. This means that 3 mg of parathion per kg of body mass would be fatal to half the test animals. Table salt (sodium chloride) with an LD50 of 3300 mg/kg would have a much lower toxicity. You would need to ingest a huge amount of salt before any toxic effect

would be observed. Although the risk to animals can be evaluated in the laboratory, it is more difficult to determine the impact in the envi- ronment since there is also a difference between continued exposure and a single, large dose of the substance.

TABLE 2.9 lists some LD50 values and compares substances in order of increasing toxicity.

Substance LD50 (mg/kg)

Table sugar 29 700

Boric acid 5140

Baking soda 4220

Table salt 3300

Ethanol 2080

Aspirin 1100

Codeine 800

Oxycodone 480

Caffeine 192

DDT 113

Cocaine (injected) 95

Dichlorvos (pesticide strips) 56

Ricin 30

Sodium cyanide 6

Parathion 3

TABLE 2.9 Some LD50 Values for Substances Tested in Rats

The LD50 of caffeine is 192 mg/kg.

PRACTICE PROBLEMS

2.6 Problem Solving Using Unit Conversion

2.57 Perform each of the following conversions using metric conversion factors:

a. 44.2 mL to liters b. 8.65 m to nanometers c. 5.2 * 108 g to megagrams d. 0.72 ks to milliseconds

2.58 Perform each of the following conversions using metric conversion factors:

a. 4.82 * 10-5 L to picoliters b. 575.2 dm to kilometers c. 5 * 10-4 kg to micrograms d. 6.4 * 1010 ps to seconds

2.59 Perform each of the following conversions using metric and U.S. conversion factors:

a. 3.428 lb to kilograms b. 1.6 m to inches c. 4.2 L to quarts d. 0.672 ft to millimeters

2.60 Perform each of the following conversions using metric and U.S. conversion factors:

a. 0.21 lb to grams b. 11.6 in. to centimeters c. 0.15 qt to milliliters d. 35.41 kg to pounds

2.61 Use metric conversion factors to solve each of the following problems: a. If a student is 175 cm tall, how tall is the student in

meters? b. A cooler has a volume of 5000 mL. What is the capacity of

the cooler in liters? c. A hummingbird has a mass of 0.0055 kg. What is the mass,

in grams, of the hummingbird? d. A balloon has a volume of 3500 cm3. What is the volume in

liters?

2.62 Use metric conversion factors to solve each of the following problems: a. The Daily Value (DV) for phosphorus is 800 mg. How many

grams of phosphorus are recommended? b. A glass of orange juice contains 3.2 dL of juice. How many

milliliters of orange juice are in the glass? c. A package of chocolate instant pudding contains 2840 mg of

sodium. How many grams of sodium are in the pudding? d. A jar contains 0.29 kg of olives. How many grams of olives

are in the jar?

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Tools to help students succeed

Marginal notes and end-of-chapter problems deepen the connection between key math skills, core chemistry skills, textual content, practice problems, and why they are so important to success in the course.

The Chemistry Primer in Mastering Chemistry helps students remediate their chemistry math skills and prepare for their first college chemistry course. Scaled to students’ needs, remediation is only suggested to students that perform poorly on an initial assessment. Remediation includes tutorials, wrong-answer specific feedback, video instruction, and stepwise scaffolding to build students’ abilities.

1.3 Studying and Learning Chemistry 7

5. Study different topics in a chapter, and relate the new concepts to concepts you know. We learn material more efficiently by relating it to information we already know. By increasing connections between concepts, we can retrieve information when we need it.

Helpful Not helpful

Testing practice Highlighting

Studying different ideas at the same time

Underlining Reading the chapter many times

Retesting a few days later Memorizing the key words

Cramming

ENGAGE 1.3 Why is self-testing helpful for learning new concepts?

SAMPLE PROBLEM 1.2 Strategies for Learning Chemistry

TRY IT FIRST

Predict which student, a, b, or c, will be most successful on the exam.

a. Bill, who reads the chapter four times b. Jennifer, who reads the chapter two times and works all the problems at the end of each

section c. Mark, who reads the chapter the night before the exam

SOLUTION

b. Jennifer, who reads the chapter two times and works all the problems at the end of each section, interacts with the content in the chapter using self-testing to make connections between concepts and practicing retrieving information learned previously.

SELF TEST 1.2

What are two more ways that Jennifer could improve her retrieval of information?

ANSWER

1. Jennifer could wait two or three days and practice working the problems in each section again to determine how much she has learned. Retesting strengthens connections between new and previously learned information for longer lasting memory and more efficient retrieval.

2. Jennifer could also ask questions as she reads and try to study at a regular pace to avoid cramming.

Features in This Text That Help You Study and Learn Chemistry This text has been designed with study features to support your learning. On the inside of the front cover is a periodic table of the elements. On the inside of the back cover are tables that summarize useful information needed throughout your study of chemistry. Each chapter begins with Looking Ahead, which outlines the topics in the chapter. At the beginning of each section, a Learning Goal describes the topics to learn. Review icons in the margins refer to Key Math Skills or Core Chemistry Skills from previous chapters that relate to new material in the chapter. Key Terms are bolded when they first appear in the text and are summarized at the end of each chapter. They are also listed and defined in the comprehensive Glossary and Index, which appears at the end of the text. Key Math Skills and Core Chemistry Skills that are critical to learning chemistry are indicated by icons in the margin and summarized at the end of each chapter.

Before you begin reading, obtain an overview of a chapter by reviewing the topics in Looking Ahead. As you prepare to read a section of the chapter, look at the section title, and turn it into a question. Asking yourself questions about new topics builds new connections to material you have already learned. For example, for Section 1.1, “Chemistry and Chemi- cals,” you could ask, “What is chemistry?” or “What are chemicals?” At the beginning of

REVIEW

KEY MATH SKILL

CORE CHEMISTRY SKILL

ENGAGE

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Build students’ problem-solving skills

Core Chemistry Skills and Key Math Skills Tutorials in Mastering Chemistry provide assignable practice problems related to the in-text feature boxes, ensuring that students master the basic quantitative and science skills they need to succeed in the course.

NEW! Pedagogical features in worked Sample Problems throughout the text help students build stronger problem-solving skills, setting them up for success in this and future courses. TRY IT FIRST! Feature encourages students to solve the problem before looking at the solution.

UPDATED! Connect feature added to Analyze the Problem boxes specify information that relates the Given and Need boxes to help students identify and connect the components within a word problem and show STEPs as guides that set up a solution strategy.

NEW! Engage questions remind students to pause and answer a question related to the material.

UPDATED! Self Tests provide students with immediate problem solving and feedback with answers.

NEW! Practice Problems suggest problems to work as students study the section.

446 CHAPTER 14 Acids and Bases

in [H3O +] and a decrease in [OH-], which makes an acidic solution. If base is added, [OH-]

increases and [H3O +] decreases, which gives a basic solution. However, for any aqueous

solution, whether it is neutral, acidic, or basic, the product [H3O +][OH-] is equal to Kw

(1.0 * 10-14 at 25 °C) (see TABLE 14.6).

ENGAGE 14.7 Is a solution that has a [H3O

+] of 1.0 * 10-3 M acidic, basic, or neutral?

PRACTICE PROBLEMS Try Practice Problems 14.35 and 14.36

Type of Solution [H3O +] [OH−] Kw (25 °C)

Neutral 1.0 * 10-7 M 1.0 * 10-7 M 1.0 * 10-14

Acidic 1.0 * 10-2 M 1.0 * 10-12 M 1.0 * 10-14

Acidic 2.5 * 10-5 M 4.0 * 10-10 M 1.0 * 10-14

Basic 1.0 * 10-8 M 1.0 * 10-6 M 1.0 * 10-14

Basic 5.0 * 10-11 M 2.0 * 10-4 M 1.0 * 10-14

TABLE 14.6 Examples of [H3O +] and [OH−] in Neutral, Acidic, and Basic

Solutions

Using the Kw to Calculate [H3O +] and [OH−] in a Solution

If we know the [H3O +] of a solution, we can use the Kw to calculate [OH

-]. If we know the [OH-] of a solution, we can calculate [H3O

+] from their relationship in the Kw, as shown in Sample Problem 14.6.

Kw = [H3O +][OH-]

[OH-] = Kw

[H3O +] [H3O

+] = Kw

[OH-]

ENGAGE 14.9 Why does the [H3O

+] of an aqueous solution increase if the [OH-] decreases?

SAMPLE PROBLEM 14.6 Calculating the [H3O +] of a Solution

TRY IT FIRST

A vinegar solution has a [OH-] = 5.0 * 10-12 M at 25 °C. What is the [H3O+] of the vinegar solution? Is the solution acidic, basic, or neutral?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

[OH-] = 5.0 * 10-12 M [H3O+] Kw = [H3O+][OH-]

STEP 2 Write the Kw for water and solve for the unknown [H3O +].

Kw = [H3O +][OH-] = 1.0 * 10-14

Solve for [H3O +] by dividing both sides by [OH-].

Kw

[OH-] =

[H3O +][OH-]

[OH-]

[H3O +] =

1.0 * 10-14

[OH-]

STEP 3 Substitute the known [OH−] into the equation and calculate.

[H3O +] =

1.0 * 10-14

[5.0 * 10-12] = 2.0 * 10-3 M

Because the [H3O +] of 2.0 * 10-3 M is larger than the [OH-] of 5.0 * 10-12 M,

the solution is acidic.

ENGAGE 14.8 If you know the [H3O

+] of a solution, how do you use the Kw to calculate the [OH-]?

CORE CHEMISTRY SKILL Calculating [H3O

+] and [OH-] in Solutions

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14.6 The pH Scale 447

SELF TEST 14.6

a. What is the [H3O +] of an ammonia cleaning solution with [OH-] = 4.0 * 10-4 M? Is

the solution acidic, basic, or neutral? b. The [H3O

+] of tomato juice is 6.3 * 10-5 M. What is the [OH-] of the juice? Is the tomato juice acidic, basic, or neutral?

ANSWER

a. [H3O +] = 2.5 * 10-11 M, basic b. [OH-] = 1.6 * 10-10 M, acidic

PRACTICE PROBLEMS

14.5 Dissociation of Water

14.33 Why are the concentrations of H3O + and OH- equal in pure water?

14.34 What is the meaning and value of Kw at 25 °C?

14.35 In an acidic solution, how does the concentration of H3O +

compare to the concentration of OH-?

14.36 If a base is added to pure water, why does the [H3O +] decrease?

14.37 Indicate whether each of the following solutions is acidic, basic, or neutral:

a. [H3O +] = 2.0 * 10-5 M

b. [H3O +] = 1.4 * 10-9 M

c. [OH-] = 8.0 * 10-3 M d. [OH-] = 3.5 * 10-10 M 14.38 Indicate whether each of the following solutions is acidic, basic,

or neutral: a. [H3O

+] = 6.0 * 10-12 M b. [H3O

+] = 1.4 * 10-4 M c. [OH-] = 5.0 * 10-12 M d. [OH-] = 4.5 * 10-2 M 14.39 Calculate the [H3O

+] of each aqueous solution with the following [OH-]:

a. coffee, 1.0 * 10-9 M b. soap, 1.0 * 10-6 M c. cleanser, 2.0 * 10-5 M d. lemon juice, 4.0 * 10-13 M

14.40 Calculate the [H3O +] of each aqueous solution with the

following [OH-]: a. NaOH solution, 1.0 * 10-2 M b. milk of magnesia, 1.0 * 10-5 M c. aspirin, 1.8 * 10-11 M d. seawater, 2.5 * 10-6 M

Applications

14.41 Calculate the [OH-] of each aqueous solution with the following [H3O

+]: a. stomach acid, 4.0 * 10-2 M b. urine, 5.0 * 10-6 M c. orange juice, 2.0 * 10-4 M d. bile, 7.9 * 10-9 M 14.42 Calculate the [OH-] of each aqueous solution with the

following [H3O +]:

a. baking soda, 1.0 * 10-8 M b. blood, 4.2 * 10-8 M c. milk, 5.0 * 10-7 M d. pancreatic juice, 4.0 * 10-9 M

14.6 The pH Scale LEARNING GOAL Calculate pH from [H3O

+]; given the pH, calculate the [H3O +] and

[OH-] of a solution.

In the environment, the acidity, or pH, of rain can have significant effects. When rain becomes too acidic, it can dissolve marble statues and accelerate the corrosion of metals. In lakes and ponds, the acidity of water can affect the ability of plants and fish to survive. The acidity of soil around plants affects their growth. If the soil pH is too acidic or too basic, the roots of the plant cannot take up some nutrients. Most plants thrive in soil with a nearly neutral pH, although certain plants, such as orchids, camellias, and blueberries, require a more acidic soil.

Although we have expressed H3O + and OH- as molar concentrations, it is more conve-

nient to describe the acidity of solutions using the pH scale. On this scale, a number between 0 and 14 represents the H3O

+ concentration for common solutions. A neutral solution has a pH of 7.0 at 25 °C. An acidic solution has a pH less than 7.0; a basic solution has a pH greater than 7.0 (see FIGURE 14.4).

When we relate acidity and pH, we are using an inverse relationship, which is when one component increases while the other component decreases. When an acid is added to pure water, the [H3O

+] (acidity) of the solution increases but its pH decreases. When a base is added to pure water, it becomes more basic, which means its acidity decreases and the pH increases.

PRACTICE PROBLEMS Try Practice Problems 14.37 to 14.42

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Give students anytime, anywhere access with Pearson eText

Pearson eText is a simple-to-use, mobile-optimized, personalized reading experience available within Mastering. It allows students to easily highlight, take notes, and review key vocabulary all in one place—even when offline. Seamlessly integrated videos, rich media, and interactive self-assessment questions engage students and give them access to the help they need, when they need it. Pearson eText is available within Mastering when packaged with a new book; students can also purchase Mastering with Pearson eText online.

For instructors not using Mastering, Pearson eText can also be adopted on its own as the main course material.

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Improve learning with Dynamic Study Modules

Dynamic Study Modules in Mastering Chemistry help students study effectively— and at their own pace—by keeping them motivated and engaged. The assignable modules rely on the latest research in cognitive science, using methods—such as adaptivity, gamification, and intermittent rewards—to stimulate learning and improve retention.

Each module poses a series of questions about a course topic. These question sets adapt to each student’s performance and offer personalized, targeted feedback to help them master key concepts. With Dynamic Study Modules, students build the confidence they need to deepen their understanding, participate meaningfully, and perform better—in and out of class.

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Instructor support you can rely on

Basic Chemistry includes a full suite of instructor support materials in the Instructor Resources area in Mastering Chemistry.

Resources include customizable PowerPoint lecture and image presentations; all images and worked examples from the text; and a test bank.

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1

In the forensic laboratory, Sarah analyzes Gloria’s stomach contents and blood for toxic compounds. You can view the results of the tests on the forensic evidence in the UPDATE Forensic Evidence Helps Solve the Crime, page 21, and determine if Gloria ingested a toxic level of ethylene glycol (antifreeze).

UPDATE Forensic Evidence Helps Solve the Crime

A call came in to 911 from a man who arrived home from work to find his wife, Gloria, lying on their living room floor. When the police arrive, they pronounce the woman dead. There is no blood at the scene, but the police do find a glass on the side table that contains a small amount of liquid. In an adjacent laundry room, the police find a half-empty bottle of antifreeze, which contains the toxic compound ethylene glycol. The bottle, glass, and liquid are bagged and sent to the forensic laboratory. At the morgue, Gloria’s height is measured as 1.673 m, and her mass is 60.5 kg.

Sarah, a forensic scientist, uses scientific procedures and chemical tests to examine the evidence from law enforcement agencies. She analyzes blood, stomach contents, and the unknown liquid from Gloria’s home, as well as the fingerprints on the glass. She also looks for the presence of drugs, poisons, and alcohol.

CAREER

Forensic Scientist Most forensic scientists work in crime laboratories that are part of city or county legal systems. They analyze bodily fluids and tissue samples collected by crime scene investigators. In analyzing these samples, forensic scientists identify the presence or absence of specific chemicals within the body to help solve criminal cases. Some of the chemicals they look for include alcohol, illegal or prescription drugs, poisons, arson debris, metals, and various gases such as carbon monoxide. To identify these substances, they use a variety of instruments and highly specific methodologies. Forensic scientists analyze samples from criminal suspects, athletes, and potential employees. They also work on cases involving environmental contamination and animal samples for wildlife crimes. Forensic scientists usually have a bachelor’s degree that includes courses in math, chemistry, and biology.

Chemistry in Our Lives 1

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2 CHAPTER 1 Chemistry in Our Lives

1.1 Chemistry and Chemicals LEARNING GOAL Define the term chemistry, and identify chemicals.

Now that you are in a chemistry class, you may be wondering what you will be learning. What questions in science have you been curious about? Perhaps you are interested in what smog is or how aspirin relieves a headache. Just like you, chemists are curious about the world we live in.

How does car exhaust produce the smog that hangs over our cities? One component of car exhaust is nitrogen oxide (NO), which forms in car engines where high temperatures convert nitrogen gas (N2) and oxygen gas (O2) to NO. In the atmosphere, the NO(g) reacts with O2(g) to form NO2(g), which has a reddish brown color of smog. In chemistry, reactions are written in the form of equations:

N2(g) + O2(g) h 2NO(g) 2NO(g) + O2(g) h 2NO2(g)

Smog

Why does aspirin relieve a headache? When a part of the body is injured, substances called prostaglandins are produced, which cause inflammation and pain. Aspirin acts to block the production of prostaglandins, reducing inflammation and pain. Chemists in the medical field develop new treatments for diabetes, genetic defects, cancer, AIDS, and other diseases. For the forensic scientist, the nurse, the dietitian, the chemical engineer, or the agricultural scientist, chemistry plays a central role in understanding problems and assessing possible solutions.

Chemistry Chemistry is the study of the composition, structure, properties, and reactions of matter. Matter is another word for all the substances that make up our world. Perhaps you imagine that chemistry takes place only in a laboratory where a chemist is working in a white coat and goggles. Actually, chemistry happens all around you every day and has an impact on everything you use and do. You are doing chemistry when you cook food, add bleach to your laundry, or start your car. A chemical reaction has taken place when silver tarnishes or an antacid tablet fizzes when dropped into water. Plants grow because chemical reactions convert carbon dioxide, water, and energy to carbohydrates. Chemical reactions take place when you digest food and break it down into substances that you need for energy and health.

Chemicals A chemical is a substance that always has the same composition and properties wherever it is found. All the things you see around you are composed of one or more chemicals. Often the terms chemical and substance are used interchangeably to describe a specific type of matter.

Every day, you use products containing substances that were developed and prepared by chemists. Soaps and shampoos contain chemicals that remove oils on your skin and scalp. In cosmetics and lotions, chemicals are used to moisturize, prevent deterioration of the product, fight bacteria, and thicken the product. Perhaps you wear a ring or watch made of gold, silver, or platinum. Your breakfast cereal is probably fortified with iron, calcium, and phosphorus, whereas the milk you drink is enriched with vitamins A and D. When you brush your teeth, the substances in toothpaste clean your teeth, prevent plaque formation, and stop tooth decay. Some of the chemicals used to make toothpaste are listed in TABLE 1.1.

ENGAGE 1.1 Why is water a chemical?

PRACTICE PROBLEMS Try Practice Problems 1.1 to 1.6

Molecules of NO2

The chemical reaction of NO with oxygen in the air forms NO2, which produces the reddish brown color of smog.

LOOKING AHEAD

1.1 Chemistry and Chemicals 2

1.2 Scientific Method: Thinking Like a Scientist 3

1.3 Studying and Learning Chemistry 6

1.4 Key Math Skills for Chemistry 9

1.5 Writing Numbers in Scientific Notation 17

Antacid tablets undergo a chemical reaction when dropped into water.

Toothpaste is a combination of many chemicals.

Chemical Function

Calcium carbonate Used as an abrasive to remove plaque

Sorbitol Prevents loss of water and hardening of toothpaste

Sodium lauryl sulfate Used to loosen plaque

Titanium dioxide Makes toothpaste white and opaque

Sodium fluorophosphate Prevents formation of cavities by strengthening tooth enamel

Methyl salicylate Gives toothpaste a pleasant wintergreen flavor

TABLE 1.1 Chemicals Commonly Used in Toothpaste

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1.2 Scientific Method: Thinking Like a Scientist 3

Branches of Chemistry The field of chemistry is divided into several branches. General chemistry is the study of the composition, properties, and reactions of matter. Organic chemistry is the study of substances that contain the element carbon. Biological chemistry is the study of the chemical reactions that take place in biological systems. Today chemistry is often combined with other sciences, such as geology and physics, to form cross-disciplines such as geochemistry and physical chemistry. Geochemistry is the study of the chemical composition of ores, soils, and minerals of the surface of the Earth and other planets. Physical chemistry is the study of the physical nature of chemical systems, including energy changes. A geochemist collects newly

erupted lava samples from Kilauea Volcano, Hawaii.

PRACTICE PROBLEMS

1.1 Chemistry and Chemicals

In every chapter, odd-numbered exercises in the Practice Problems are paired with even-numbered exercises. The answers for the orange- shaded, odd-numbered Practice Problems are given at the end of each chapter. The complete solutions to the odd-numbered Practice Problems are in the Study Guide and Student Solutions Manual.

1.1 Write a one-sentence definition for each of the following: a. chemistry b. chemical

1.2 Ask two of your friends (not in this class) to define the terms in problem 1.1. Do their answers agree with the definitions you provided?

Applications

1.3 Obtain a bottle of multivitamins, and read the list of ingredients. What are four chemicals from the list?

1.4 Obtain a box of breakfast cereal, and read the list of ingredients. What are four chemicals from the list?

1.5 Read the labels on some items found in a drugstore. What are the names of some chemicals contained in those items?

1.6 Read the labels on products used to wash your dishes. What are the names of some chemicals contained in those products?

1.2 Scientific Method: Thinking Like a Scientist LEARNING GOAL Describe the scientific method.

When you were very young, you explored the things around you by touching and tasting. As you grew, you asked questions about the world in which you live. What is lightning? Where does a rainbow come from? Why is the sky blue? As an adult, you may have wondered how antibiotics work or why vitamins are important to your health. Every day, you ask questions and seek answers to organize and make sense of the world around you.

When the late Nobel Laureate Linus Pauling (1901–1994) described his student life in Oregon, he recalled that he read many books on chemistry, mineralogy, and physics. “I mulled over the properties of materials: why are some substances colored and others not, why are some minerals or inorganic compounds hard and others soft?” He said, “I was building up this tremendous background of empirical knowledge and at the same time asking a great number of questions.” Linus Pauling won two Nobel Prizes: the first, in 1954, was in chemistry for his work on the nature of chemical bonds and the determination of the structures of complex substances; the second, in 1962, was the Peace Prize, for his opposition to the spread of nuclear weapons.

The Scientific Method The process of trying to understand nature is unique to each scientist. However, the scientific method is a process that scientists use to make observations in nature, gather data, and explain natural phenomena (see the figure on the next page).

1. Make Observations The first step in the scientific method is to make observations about nature and ask questions about what you observe. When an observation always seems to be true, it may be stated as a law that predicts that behavior and is often

Linus Pauling won the Nobel Prize in Chemistry in 1954.

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4 CHAPTER 1 Chemistry in Our Lives

measurable. However, a law does not explain that observation. For example, we can use the Law of Gravity to predict that if we drop our chemistry book, it would fall on the floor, but this law does not explain why our book falls.

2. Form a Hypothesis A scientist forms a hypothesis, which gives a possible explanation of an observation or a law. The hypothesis must be stated in such a way that it can be tested by experiments.

3. Design Experiments To determine if a hypothesis is valid, experiments are done to find a relationship between the hypothesis and the observations. The results of the experiments may confirm the validity of the hypothesis. However, experiments may also show that the hypothesis is invalid, which means that it is modified or discarded. Then new experiments will be designed to test the new hypothesis.

4. Draw a Conclusion When many experiments give consistent results, we may draw the conclusion that the hypothesis is valid. Even then, more experiments are done to test the hypothesis. If new experimental results indicate the hypothesis is not valid, it is modified or replaced.

Law

Form a Hypothesis

The hypothesis is modified if the results of the experiments do not support it.

Design Experiments

Draw a Conclusion

Make Observations

Scientific Method

The scientific method develops a conclusion about nature using observations, hypotheses, and experiments.

Chemistry Link to Health Early Chemist: Paracelsus

For many centuries, chemistry has been the study of changes in matter. From the time of the ancient Greeks to the sixteenth century, alchemists described matter in terms of four components of nature: earth, air, fire, and water. By the eighth century, alchemists believed that they could change metals such as copper and lead into gold and silver. Although these efforts failed, the alchemists provided informa- tion on the chemical reactions involved in the extraction of metals from ores. The alchemists also designed some of the first laboratory equipments and developed early laboratory procedures. These early efforts were some of the first observations and experiments using the scientific method.

Paracelsus (1493–1541) was a physician and an alchemist who thought that alchemy should be about preparing new medicines. Using observation and experimentation, he proposed that a healthy body was regulated by a series of chemical processes that could be unbalanced by certain chemical compounds and rebalanced by using minerals and medicines. For example, he determined that inhaled

dust caused lung disease in miners. He also thought that goiter was a problem caused by contaminated water, and he treated syphilis

with compounds of mercury. His opinion of medicines was that the right dose makes the difference between a poison and a cure. Paracelsus changed alchemy in ways that helped establish mod- ern medicine and chemistry.

Swiss physician and alchemist Paracelsus (1493–1541) believed that chemicals and minerals could be used as medicines.

Using the Scientific Method in Everyday Life You may be surprised to realize that you use the scientific method in your everyday life. Suppose you visit a friend in her home. Soon after you arrive, your eyes start to itch and you begin to sneeze. Then you observe that your friend has a new cat. Perhaps you form the hypothesis that you are allergic to cats. To test your hypothesis, you leave your friend’s home. If the sneezing stops, perhaps your hypothesis is correct. You test your hypothesis further by visiting another friend who also has a cat. If you start to sneeze again, your experimental results support your hypothesis, and you come to the conclusion that you are allergic to cats. However, if you continue sneezing after you leave your friend’s home, your hypothesis is not supported. Now you need to form a new hypothesis, which could be that you have a cold.

ENGAGE 1.2 Why would the following statement, “Today I placed two tomato seedlings in the garden, and two more in a closet. I will give all the plants the same amount of water and fertilizer,” be considered an experiment?

SAMPLE PROBLEM 1.1 Scientific Method

TRY IT FIRST

Identify each of the following as an observation, a hypothesis, an experiment, or a conclusion:

a. During an assessment in the emergency room, a nurse writes that the patient has a resting pulse of 30 beats/min.

Through observation you may think that you are allergic to cats.

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1.2 Scientific Method: Thinking Like a Scientist 5

b. Repeated studies show that lowering sodium in the diet leads to a decrease in blood pressure.

c. A nurse thinks that an incision from a recent surgery that is red and swollen is infected.

SOLUTION

a. observation b. conclusion c. hypothesis

SELF TEST 1.1

Identify each of the following as an observation, a hypothesis, an experiment, or a conclusion:

a. Drinking coffee at night keeps me awake. b. I will try drinking coffee only in the morning. c. If I stop drinking coffee in the afternoon, I will be able to sleep at night. d. When I drink decaffeinated coffee, I sleep better at night. e. I am going to drink only decaffeinated coffee. f. I sleep better at night because I stopped drinking caffeinated drinks.

PRACTICE PROBLEMS Try Practice Problems 1.7 to 1.10

ANSWER

a. observation b. experiment c. hypothesis d. observation e. experiment f. conclusion

Nurses make observations in the hospital.

PRACTICE PROBLEMS

1.2 Scientific Method: Thinking Like a Scientist

1.7 Identify each activity, a to f, as an observation, a hypothesis, an experiment, or a conclusion. At a popular restaurant, where Chang is the head chef, the fol- lowing occurred: a. Chang determined that sales of the house salad had dropped. b. Chang decided that the house salad needed a new dressing. c. In a taste test, Chang prepared four bowls of sliced

cucumber, each with a new dressing: sesame seed, olive oil and balsamic vinegar, creamy Italian, and blue cheese.

d. Tasters rated the sesame seed salad dressing as the favorite. e. After two weeks, Chang noted that the orders for the house

salad with the new sesame seed dressing had doubled. f. Chang decided that the sesame seed dressing improved the

sales of the house salad because the sesame seed dressing enhanced the taste.

1.8 Identify each activity, a to f, as an observation, a hypothesis, an experiment, or a conclusion. Lucia wants to develop a process for dyeing shirts so that the color will not fade when the shirt is washed. She proceeds with the following activities: a. Lucia notices that the dye in a design fades when the shirt is

washed. b. Lucia decides that the dye needs something to help it

combine with the fabric.

c. She places a spot of dye on each of four shirts and then places each one separately in water, salt water, vinegar, and baking soda and water.

d. After one hour, all the shirts are removed and washed with a detergent.

e. Lucia notices that the dye has faded on the shirts in water, salt water, and baking soda, whereas the dye did not fade on the shirt soaked in vinegar.

f. Lucia thinks that the vinegar binds with the dye so it does not fade when the shirt is washed.

Applications

1.9 Identify each of the following as an observation, a hypothesis, an experiment, or a conclusion: a. One hour after drinking a glass of regular milk, Jim

experienced stomach cramps. b. Jim thinks he may be lactose intolerant. c. Jim drinks a glass of lactose-free milk and does not have any

stomach cramps. d. Jim drinks a glass of regular milk to which he has added

lactase, an enzyme that breaks down lactose, and has no stomach cramps.

1.10 Identify each of the following as an observation, a hypothesis, an experiment, or a conclusion: a. Sally thinks she may be allergic to shrimp. b. Yesterday, one hour after Sally ate a shrimp salad, she broke

out in hives. c. Today, Sally had some soup that contained shrimp, but she

did not break out in hives. d. Sally realizes that she does not have an allergy to

shrimp.

Customers rated the sesame seed dressing as the best.

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6 CHAPTER 1 Chemistry in Our Lives

1.3 Studying and Learning Chemistry LEARNING GOAL Identify strategies that are effective for learning. Develop a study plan for learning chemistry.

Here you are taking chemistry, perhaps for the first time. Whatever your reasons for choosing to study chemistry, you can look forward to learning many new and exciting ideas.

Strategies to Improve Learning and Understanding Success in chemistry utilizes good study habits, connecting new information with your knowledge base, rechecking what you have learned and what you have forgotten, and retrieving what you have learned for an exam. Let’s take a look at ways that can help you study and learn chemistry. Suppose you were asked to indicate if you think each of the following common study habits is helpful or not helpful:

Helpful Not helpful

Highlighting

Underlining

Reading the chapter many times

Memorizing the key words

Testing practice

Cramming

Studying different ideas at the same time

Retesting a few days later

Learning chemistry requires us to place new information in our long-term memory, which allows us to remember those ideas for an exam, a process called retrieval. Thus, our study habits need to help us to recall knowledge. The study habits that are not very helpful in retrieval include highlighting, underlining, reading the chapter many times, memorizing key words, and cramming. If we want to recall new information, we need to connect it with prior knowledge. By doing practice tests, you develop the skills needed to retrieve new information. We can determine how much we have learned by going back a few days later and retesting. Another useful learning strategy is to study different ideas at the same time, which allows us to connect those ideas and to differentiate between them. Although these study habits may take more time and seem more difficult, they help us find the gaps in our knowledge and connect new information with what we already know.

Tips for Using New Study Habits for Successful Learning

1. Do not keep rereading text or notes. Reading the same material over and over will make that material seem familiar but does not mean that you have learned it. You need to test yourself to find out what you do and do not know.

2. Ask yourself questions as you read. Asking yourself questions as you read requires you to interact continually with new material. The Engage features in the margin of this text are helpful in asking you to think about new material. By linking new material with long-term knowledge, you make pathways for retrieving new material.

3. Self-test by giving yourself quizzes. Using problems in the text or sample exams, practice taking tests frequently.

4. Study at a regular pace rather than cramming. Once you have tested your- self, go back in a few days and practice testing and retrieving information again. We do not recall all the information when we first read it. By frequent quizzing and retesting, we identify what we still need to learn. Sleep is also important for strengthening the associations between newly learned information. Lack of sleep may interfere with retrieval of information as well. So staying up all night to cram for your chemistry exam is not a good idea. Success in chemistry is a combined effort to learn new information and then to retrieve that information when you need it for an exam.

Students learn by continuously asking and answering questions as they study new material.

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1.3 Studying and Learning Chemistry 7

5. Study different topics in a chapter, and relate the new concepts to concepts you know. We learn material more efficiently by relating it to information we already know. By increasing connections between concepts, we can retrieve information when we need it.

Helpful Not helpful

Testing practice Highlighting

Studying different ideas at the same time

Underlining Reading the chapter many times

Retesting a few days later Memorizing the key words

Cramming

ENGAGE 1.3 Why is self-testing helpful for learning new concepts?

SAMPLE PROBLEM 1.2 Strategies for Learning Chemistry

TRY IT FIRST

Predict which student, a, b, or c, will be most successful on the exam.

a. Bill, who reads the chapter four times b. Jennifer, who reads the chapter two times and works all the problems at the end of each

section c. Mark, who reads the chapter the night before the exam

SOLUTION

b. Jennifer, who reads the chapter two times and works all the problems at the end of each section, interacts with the content in the chapter using self-testing to make connections between concepts and practicing retrieving information learned previously.

SELF TEST 1.2

What are two more ways that Jennifer could improve her retrieval of information?

ANSWER

1. Jennifer could wait two or three days and practice working the problems in each section again to determine how much she has learned. Retesting strengthens connections between new and previously learned information for longer lasting memory and more efficient retrieval.

2. Jennifer could also ask questions as she reads and try to study at a regular pace to avoid cramming.

Features in This Text That Help You Study and Learn Chemistry This text has been designed with study features to support your learning. On the inside of the front cover is a periodic table of the elements. On the inside of the back cover are tables that summarize useful information needed throughout your study of chemistry. Each chapter begins with Looking Ahead, which outlines the topics in the chapter. At the beginning of each section, a Learning Goal describes the topics to learn. Review icons in the margins refer to Key Math Skills or Core Chemistry Skills from previous chapters that relate to new material in the chapter. Key Terms are bolded when they first appear in the text and are summarized at the end of each chapter. They are also listed and defined in the comprehensive Glossary and Index, which appears at the end of the text. Key Math Skills and Core Chemistry Skills that are critical to learning chemistry are indicated by icons in the margin and summarized at the end of each chapter.

Before you begin reading, obtain an overview of a chapter by reviewing the topics in Looking Ahead. As you prepare to read a section of the chapter, look at the section title, and turn it into a question. Asking yourself questions about new topics builds new connections to material you have already learned. For example, for Section 1.1, “Chemistry and Chemi- cals,” you could ask, “What is chemistry?” or “What are chemicals?” At the beginning of

REVIEW

KEY MATH SKILL

CORE CHEMISTRY SKILL

ENGAGE

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8 CHAPTER 1 Chemistry in Our Lives

each section, a Learning Goal states what you need to understand. As you read the text, you will see Engage questions in the margin, which remind you to pause your reading and test yourself with a question related to the material. The Answers to Engage Questions are at the end of each chapter.

Several Sample Problems are included in each chapter. The Try It First feature reminds you to work the problem before you look at the Solution. It is helpful to try to work a problem first because it helps you link what you know to what you need to learn. The Analyze the Problem feature includes Given, the information you have; Need, what you have to accomplish; and Connect, how you proceed. Sample Problems include a Solution that shows the steps you can use for problem solving. Work the associated Self Test problems, and compare your answer to the one provided.

TRY IT FIRST

ANALYZE THE PROBLEM

Given Need Connect

At the end of each chapter section, you will find a set of Practice Problems that allows you to apply problem solving immediately to the new concepts. Throughout each section, Practice Problem icons remind you to try the indicated Practice Problems as you study. The Applications in the Practice Problems relate the content to health, careers, and real-life examples. The problems are paired, which means that each of the odd-numbered problems is matched to the following even-numbered problem. At the end of each chapter, the Answers to all the odd-numbered problems are provided. If the answers match yours, you most likely understand the topic; if not, you need to study the section again.

Throughout each chapter, boxes titled Chemistry Link to Health and Chemistry Link to the Environment help you relate the chemical concepts you are learning to real-life situ- ations. Many of the figures and diagrams use macro-to-micro illustrations to depict the atomic level of organization of ordinary objects, such as the atoms in aluminum foil. These visual models illustrate the concepts described in the text and allow you to “see” the world in a microscopic way. Interactive Video suggestions illustrate content as well as problem solving.

At the end of each chapter, you will find several study aids that complete the chapter. Chapter Reviews provide a summary in easy-to-read bullet points, and Concept Maps visually show the connections between important topics. Understanding the Concepts are problems that use art and models to help you visualize concepts and connect them to your background knowledge. Additional Practice Problems and Challenge Problems provide additional exercises to test your understanding of the topics in the chapter.

After some chapters, problem sets called Combining Ideas test your ability to solve problems containing material from more than one chapter.

Many students find that studying with a group can be beneficial to learning. In a group, students motivate each other to study, fill in gaps, and correct misunderstandings by teaching and learning together. Studying alone does not allow the process of peer correction. In a group, you can cover the ideas more thoroughly as you discuss the reading and problem solve with other students.

Making a Study Plan As you embark on your journey into the world of chemistry, think about your approach to studying and learning chemistry. You might consider some of the ideas in the following list. Check those ideas that will help you successfully learn chemistry. Commit to them now. Your success depends on you.

My study plan for learning chemistry will include the following:

reading the chapter before class going to class reviewing the Learning Goals keeping a problem notebook reading the text working the Practice Problems as I read each section answering the Engage questions trying to work the Sample Problem before looking at the Solution studying different topics at the same time organizing a study group

PRACTICE PROBLEMS

INTERACTIVE VIDEO

Illustrating the atoms of aluminum in aluminum foil is an example of macro-to-micro art.

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1.4 Key Math Skills for Chemistry 9

seeing the professor during office hours reviewing Key Math Skills and Core Chemistry Skills attending review sessions studying as often as I can

PRACTICE PROBLEMS Try Practice Problems 1.11 to 1.14

SAMPLE PROBLEM 1.3 A Study Plan for Learning Chemistry

TRY IT FIRST

Which of the following activities should you include in your study plan for learning chemistry successfully?

a. reading the chapter over and over until you think you understand it b. going to the professor’s office hours c. doing the Self Test associated with each Sample Problem d. waiting to study until the night before the exam e. trying to work the Sample Problem before looking at the Solution f. retesting on new information a few days later

SOLUTION

b, c, e, and f

SELF TEST 1.3

Which of the following will help you learn chemistry?

a. skipping review sessions b. working Practice Problems as you read each section c. staying up all night before an exam d. reading the assignment before class e. highlighting the key ideas in the text

ANSWER

b and d

PRACTICE PROBLEMS

1.3 Studying and Learning Chemistry

1.11 What are four things you can do to help yourself to succeed in chemistry?

1.12 What are four things that would make it difficult for you to learn chemistry?

1.13 A student in your class asks you for advice on learning chemistry. Which of the following might you suggest? a. forming a study group b. skipping class c. asking yourself questions while reading the text

d. waiting until the night before an exam to study e. answering the Engage questions

1.14 A student in your class asks you for advice on learning chemistry. Which of the following might you suggest?

a. studying different topics at the same time b. not reading the text; it’s never on the test c. attending review sessions d. working the problems again after a few days e. keeping a problem notebook

1.4 Key Math Skills for Chemistry LEARNING GOAL Review math concepts used in chemistry: place values, positive and negative numbers, percentages, solving equations, and interpreting graphs.

During your study of chemistry, you will work many problems that involve numbers. You will need various math skills and operations. We will review some of the key math skills that are particularly important for chemistry. As we move through the chapters, we will also reference the key math skills as they apply.

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10 CHAPTER 1 Chemistry in Our Lives

Identifying Place Values The name of each place value depends on its position relative to the decimal point. For the number 482.57 in the Decimal Place Value Chart, the place values increase 10 times moving to the left and decrease 10 times by moving to the right. The digits on the left of the decimal point on the place value chart are 4 hundreds (400), 8 tens (80), and 2 ones (2), or four hundred eighty-two (482). The digits on the right of the decimal point are 5 tenths (.5), and 7 hundredths (.07), or fifty-seven hundredths (.57).

A premature baby has a mass of 2518 g. We can assign the place values for the number 2518 as follows:

Digit Place Value

2 thousands

5 hundreds

1 tens

8 ones

A silver coin has a mass of 6.407 g. We can assign the place values for the number 6.407 as follows:

Digit Place Value

6 ones

4 tenths

0 hundredths

7 thousandths

Note that place values ending with the suffix ths refer to the decimal places to the right of the decimal point.

ENGAGE 1.4 In the number 8.034, how do you know the 0 is in the tenths place?

KEY MATH SKILL Identifying Place Values

PRACTICE PROBLEMS Try Practice Problems 1.15 and 1.16

4 8

Decimal Place Value Chart

2 . 5 7

T ho

us an

ds

H un

dr ed

s

T en

s

O ne

s

D ec

im al

p oi

nt

T en

th s

H un

dr ed

th s

T ho

us an

dt hs

SAMPLE PROBLEM 1.4 Identifying Place Values

TRY IT FIRST

A bullet found at a crime scene has a mass of 15.24 g. What are the place values for each of the digits in the mass of the bullet?

SOLUTION

Digit Place Value

1 tens

5 ones

2 tenths

4 hundredths

SELF TEST 1.4

Identify the place values for the digits in each of the following:

a. the victim’s height of 1.673 m b. the victim’s mass of 60.5 kg

ANSWER

a. Digit Place Value

1 ones

6 tenths

7 hundredths

3 thousandths

Digit Place Value

6 tens

0 ones

5 tenths

b.

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1.4 Key Math Skills for Chemistry 11

Using Positive and Negative Numbers in Calculations A positive number is any number that is greater than zero and has a positive sign ( + ). Often the positive sign is understood and not written in front of the number. For example, the number + 8 is usually written as 8. A negative number is any number that is less than zero and is written with a negative sign ( - ). For example, a negative eight is written as - 8.

Multiplication and Division of Positive and Negative Numbers When two positive numbers or two negative numbers are multiplied, the answer is positive ( + ).

2 * 3 = 6 The + sign ( + 6) is understood. ( - 2) * ( - 3) = 6

When a positive number and a negative number are multiplied, the answer is negative ( - ).

2 * ( - 3) = - 6 ( - 2) * 3 = - 6

The rules for the division of positive and negative numbers are the same as the rules for multiplication. When two positive numbers or two negative numbers are divided, the answer is positive ( + ).

6 3

= 2 - 6 - 3

= 2

When a positive number and a negative number are divided, the answer is negative ( - ).

- 6 3

= - 2 6

- 3 = - 2

Addition of Positive and Negative Numbers When positive numbers are added, the sign of the answer is positive.

3 + 4 = 7 The + sign ( + 7) is understood.

When negative numbers are added, the sign of the answer is negative.

- 3 + ( - 4) = - 7

When a positive number and a negative number are added, the smaller number is subtracted from the larger number, and the result has the same sign as the larger number.

12 + ( - 15) = - 3

Subtraction of Positive and Negative Numbers When two numbers are subtracted, change the sign of the number to be subtracted and follow the rules for addition shown above.

12 - (+5) = 12 + ( - 5) = 7 - 12 - (−5) = - 12 + ( + 5) = - 7

Calculator Operations On your calculator, there are four keys that are used for basic mathematical operations. The change sign + / - key may be used to change the sign of a number.

To practice these basic calculations on the calculator, work through the problem going from the left to the right, doing the operations in the order they occur. On some calculators, a negative number is entered by pressing the number and then pressing the change sign + / - key. However, this may not work on all calculators. Because each type of calculator is slightly different, make sure that you know how your particular calculator works. At the end, press the equals = key or ANS or ENTER.

KEY MATH SKILL Using Positive and Negative

Numbers in Calculations

ENGAGE 1.5 Why does - 5 + 4 = - 1, whereas - 5 + ( - 4) = - 9?

PRACTICE PROBLEMS Try Practice Problems 1.17 and 1.18

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12 CHAPTER 1 Chemistry in Our Lives

Addition and Subtraction Multiplication and Division

Example 1:

Solution:

15 - 8 + 2 = 15 - 8 + 2 = 9

Example 3:

Solution:

2 * ( - 3) = 2 * 3 + / - = - 6

Example 2:

Solution:

4 + ( - 10) - 5 =

4 + 10 + / - - 5 = - 11

Example 4:

Solution:

8 * 3 4

=

8 * 3 , 4 = 6

Calculating Percentages To determine a percentage, divide the parts by the total (whole) and multiply by 100%. For example, if an aspirin tablet contains 325 mg of aspirin (active ingredient) and the tablet has a mass of 545 mg, what is the percentage of aspirin in the tablet?

325 mg aspirin

545 mg tablet * 100% = 59.6% aspirin

When a value is described as a percent (%), it represents the number of parts of an item in 100 of those items. If the percent of red balls is 5, it means there are 5 red balls in every 100 balls. If the percent of green balls is 50, there are 50 green balls in every 100 balls.

5% red balls = 5 red balls 100 balls

50% green balls = 50 green balls

100 balls

KEY MATH SKILL Calculating Percentages

ENGAGE 1.6 Why is the value of 100% used in the calculation of a percentage?

Multiplication

Division

Subtraction

Equals

Change sign Addition

SAMPLE PROBLEM 1.5 Calculating a Percentage

TRY IT FIRST

A bullet found at a crime scene may be used as evidence in a trial if the percentage of metals is a match to the composition of metals in a bullet from the suspect’s ammunition. A forensic scientist’s analysis of the bullet shows that it contains 13.9 g of lead, 0.3 g of tin, and 0.9 g of antimony. What is the percentage of each metal in the bullet? Express your answers to the ones place.

SOLUTION

A percentage is calculated by dividing the mass of each element by the total mass of the metals in the bullet and multiplying by 100%.

Total mass = 13.9 g + 0.3 g + 0.9 g = 15.1 g

Percentage of lead

13.9 g

15.1 g * 100% = 92% lead

Percentage of tin

0.3 g

15.1 g * 100% = 2% tin

Percentage of antimony

0.9 g

15.1 g * 100% = 6% antimony

A bullet casing at a crime scene is marked as evidence.

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1.4 Key Math Skills for Chemistry 13

Solving Equations In chemistry, we use equations that express the relationship between certain variables. Let’s look at how we would solve for x in the following equation:

2x + 8 = 14

Our overall goal is to rearrange the items in the equation to obtain x on one side.

1. Place all like terms on one side. The numbers 8 and 14 are like terms. To remove the 8 from the left side of the equation, we subtract 8. To keep a balance, we need to subtract 8 from the 14 on the other side.

2x + 8 - 8 = 14 - 8 2x = 6

2. Isolate the variable you need to solve for. In this problem, we obtain x by dividing both sides of the equation by 2. The value of x is the result when 6 is divided by 2.

2x 2

= 6 2

x = 3

3. Check your answer. Check your answer by substituting your value for x back into the original equation.

2(3) + 8 = 14 6 + 8 = 14

14 = 14 Your answer x = 3 is correct.

Summary: To solve an equation for a particular variable, be sure you perform the same mathematical operations on both sides of the equation.

If you eliminate a symbol or number by subtracting, you need to subtract that same symbol or number from both sides.

If you eliminate a symbol or number by adding, you need to add that same symbol or number to both sides.

If you cancel a symbol or number by dividing, you need to divide both sides by that same symbol or number.

If you cancel a symbol or number by multiplying, you need to multiply both sides by that same symbol or number.

When we work with temperature, we may need to convert between degrees Celsius and degrees Fahrenheit using the following equation:

TF = 1.8(TC) + 32

KEY MATH SKILL Solving Equations

ENGAGE 1.7 Why is the number 8 subtracted from both sides of this equation?

SELF TEST 1.5

A bullet seized from the suspect’s ammunition has a composition of lead 11.6 g, tin 0.5 g, and antimony 0.4 g.

a. What is the percentage of each metal in the bullet? Express your answers to the ones place.

b. Could the bullet removed from the suspect’s ammunition be considered as evidence that the suspect was at the crime scene mentioned in Sample Problem 1.5?

ANSWER

a. The bullet from the suspect’s ammunition is lead 93%, tin 4%, and antimony 3%. b. The composition of this bullet does not match the bullet from the crime scene and

cannot be used as supporting evidence.

35 36 37 38 39 40

95 96.8 98.6 100.4 102.2 104

°C °C

°F °F

35 36 37 38 39 40

95 96.8 98.6 100.4 102.2 104

°C °C

°F °F

A plastic strip thermometer changes color to indicate body temperature.

PRACTICE PROBLEMS Try Practice Problems 1.23 and 1.24

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14 CHAPTER 1 Chemistry in Our Lives

To obtain the equation for converting degrees Fahrenheit to degrees Celsius, we subtract 32 from both sides.

TF = 1.8(TC) + 32 TF - 32 = 1.8(TC) + 32 - 32 TF - 32 = 1.8(TC)

To obtain TC by itself, we divide both sides by 1.8.

TF - 32 1.8

= 1.8(TC)

1.8 = TC

PRACTICE PROBLEMS Try Practice Problems 1.19 and 1.20

ANSWER

a. m = q

∆T * SH b. m = D * V

SAMPLE PROBLEM 1.6 Solving Equations

TRY IT FIRST

Solve the following equation for V2:

P1V1 = P2V2

SOLUTION

P1V1 = P2V2

To solve for V2, divide both sides by the symbol P2.

P1V1 P2

= P2V2 P2

V2 = P1V1 P2

SELF TEST 1.6

Solve each of the following equations for m:

a. q = m * ∆T * SH b. D = m V

INTERACTIVE VIDEO

Solving Equations

ENGAGE 1.8 Why is the numerator divided by P2 on both sides of the equation?

Interpreting Graphs A graph is a diagram that represents the relationship between two variables. These quantities are plotted along two perpendicular axes, which are the x axis (horizontal) and y axis (vertical).

Example In the graph Volume of a Balloon versus Temperature, the volume of a gas in a balloon is plotted against its temperature.

Title Look at the title. What does it tell us about the graph? The title indicates that the volume of a balloon was measured at different temperatures.

Vertical Axis Look at the label and the numbers on the vertical (y) axis. The label indicates that the volume of the balloon was measured in liters (L). The numbers, which are chosen to include the low and high measurements of the volume of the gas, are evenly spaced from 22.0 L to 32.0 L.

Horizontal Axis The label on the horizontal (x) axis indicates that the temperatures of the balloon, measured in kelvins (K), are evenly spaced from 270 K to 390 K.

KEY MATH SKILL Interpreting Graphs

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1.4 Key Math Skills for Chemistry 15

Points on the Graph Each point on the graph represents a volume in liters that was measured at a specific temperature. When these points are connected, a line is obtained.

Interpreting the Graph From the graph, we see that the volume of the gas increases as the temperature of the gas increases. This is called a direct relationship. Now we use the graph to determine the volume at various temperatures. For example, suppose we want to know the volume of the gas at 320 K. We would start by finding 320 K on the x axis and then drawing a line up to the plotted line. From there, we would draw a horizontal line that intersects the y axis and read the volume value where the line crosses the y axis as shown on the graph above.

ENGAGE 1.9 Why are the numbers on the vertical and horizontal axes placed at regular intervals?

PRACTICE PROBLEMS Try Practice Problems 1.21 and 1.22, 1.25 and 1.26

y axis (vertical axis)

x axis (horizontal axis)

22.0 270 290 310 330 350 370 390

24.0

26.0V ol

um e

(L )

28.0

30.0

32.0

Temperature (K)

Volume of a Balloon versus Temperature

ANSWER

a. 37.6 °C b. 8 min c. 0.9 °C

SAMPLE PROBLEM 1.7 Interpreting a Graph

TRY IT FIRST

A nurse administers Tylenol to lower a child’s fever. The graph shows the body temperature of the child plotted against time.

a. What is measured on the vertical axis? b. What is the range of values on the vertical axis? c. What is measured on the horizontal axis? d. What is the range of values on the horizontal axis?

SOLUTION

a. body temperature, in degrees Celsius b. 37.0 °C to 39.8 °C c. time, in minutes, after Tylenol was given d. 0 min to 30 min

SELF TEST 1.7

a. Using the graph in Sample Problem 1.7, what was the child’s temperature 15 min after Tylenol was given?

b. How many minutes elapsed for the temperature to decrease from 39.4 °C to 38.0 °C? c. What was the decrease, in degrees Celsius, between 5 min and 20 min?

0 5 10 15 20 25 30

37.4

37.0

37.8

38.2

38.6

39.0

39.4

39.8

Time (min) After Tylenol was Given

Child’s Body Temperature versus Time

B od

y T

em pe

ra tu

re (

°C )

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16 CHAPTER 1 Chemistry in Our Lives

PRACTICE PROBLEMS

1.4 Key Math Skills for Chemistry

1.15 What is the place value for the bold digit? a. 7.3288 b. 16.1234 c. 0.280 d. 4675.99 e. 0.0901

1.16 What is the place value for the bold digit? a. 975.689 b. 375.88 c. 46.1000 d. 0.8128 e. 2.710

1.17 Evaluate each of the following: a. 15 - ( - 8) = _______ b. - 8 + ( - 22) = _______ c. 4 * ( - 2) + 6 = ______ d. 75 - ( - 44) - 65 = _____

e. - 25 + ( - 11)

- 3 + 6 = _______

1.18 Evaluate each of the following: a. - 11 - ( - 9) = _______ b. 34 + ( - 55) = _______

c. - 56

8 = _______ d. 0.04 + ( - 0.08) = _______

e. - 2 * ( - 4)

2 + 12 = _______

1.19 Solve each of the following for a:

a. 4a + 4 = 40 b. a 6

= 7

c. 55 - a + 15

4 = a d.

6a - 27 3

= - a 1.20 Solve each of the following for b: a. 2b + 7 = b + 10 b. 3b - 4 = 24 - b

c. - 16 - 2b + 8b

2 = b d.

3 * ( - 4b) + 32 2

= - 4b

Use the following graph for problems 1.21 and 1.22:

Applications 1.23 a. A clinic had 25 patients on Friday morning. If 21 patients

were given flu shots, what percentage of the patients received flu shots? Express your answer to the ones place.

b. An alloy contains 56 g of pure silver and 22 g of pure copper. What is the percentage of silver in the alloy? Express your answer to the ones place.

c. A collection of coins contains 11 nickels, 5 quarters, and 7 dimes. What is the percentage of dimes in the collection? Express your answer to the ones place.

d. A saline solution has a mass of 44 g, of which 2.2 g is sodium chloride. What percent of the solution is sodium chloride? Express your answer to the ones place.

1.24 a. At a local hospital, 35 babies were born in May. If 22 were boys, what percentage of the newborns were boys? Express your answer to the ones place.

b. An alloy contains 67 g of pure gold and 35 g of pure zinc. What is the percentage of zinc in the alloy? Express your answer to the ones place.

c. A collection of coins contains 15 pennies, 14 dimes, and 6 quarters. What is the percentage of pennies in the collection? Express your answer to the ones place.

d. In the hospital storeroom, 18 of 56 intravenous (IV) fluid bags contain dextrose solutions. The bags with dextrose are what percent of the total? Express your answer to the ones place.

Use the following graph for problems 1.25 and 1.26:

T em

pe ra

tu re

( °C

)

20 0 20 40

Time (min)

Temperature of Tea versus Time for Cooling

60 80 100

30

40

50

60

70

80

90

20 0 5 10

Time (h) Since Death

Post-Mortem Body Temperature versus Time

15 20 25

24

28

32

36

40

44

T em

pe ra

tu re

( °C

)

1.25 a. What does the title indicate about the graph? b. What is measured on the vertical axis? c. What is the range of values on the vertical axis? d. Does the temperature increase or decrease with an increase

in time?

1.26 a. What is measured on the horizontal axis? b. What is the range of values on the horizontal axis? c. How many hours were needed to reach a temperature of 28 °C? d. The coroner measured Gloria’s body temperature at 9 p.m. as

34 °C. What was the time of her death?

1.21 a. What does the title indicate about the graph? b. What is measured on the vertical axis? c. What is the range of values on the vertical axis? d. Does the temperature increase or decrease with an increase

in time?

1.22 a. What is measured on the horizontal axis? b. What is the range of values on the horizontal axis? c. What is the temperature of the tea after 20 min? d. How many minutes were needed to reach a temperature of

45 °C?

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1.5 Writing Numbers in Scientific Notation 17

1.5 Writing Numbers in Scientific Notation LEARNING GOAL Write a number in scientific notation.

In chemistry, we often work with numbers that are very large and very small. We might measure something as tiny as the width of a human hair, which is about 0.000 008 m. Or perhaps we want to count the number of hairs on the average human scalp, which is about 100 000 hairs. In this text, we add spaces between sets of three digits when it helps make the places easier to count. However, we will see that it is more convenient to write large and small numbers in scientific notation.

8 * 10-6 m1 * 105 hairs

Humans have an average of 1 * 10 5 hairs on their scalps. Each hair is about 8 * 10 -6 m wide.

A number written in scientific notation has two parts: a coefficient and a power of 10. For example, the number 2400 is written in scientific notation as 2.4 * 103. The coefficient, 2.4, is obtained by moving the decimal point to the left to give a number that is at least 1 but less than 10. Because we moved the decimal point three places to the left, the power of 10 is a positive 3, which is written as 103. When a number greater than 1 is converted to scientific notation, the power of 10 is positive.

KEY MATH SKILL Writing Numbers in Scientific

Notation

ENGAGE 1.10 Why is 530 000 written as 5.3 * 10 5 in scientific notation?

Standard Number Scientific Notation

0.000 008 m 8 * 10-6 m

100 000 hairs 1 * 105 hairs

= 3 places Coefficient Power

of 10

Standard Number Scientific Notation

2 4 0 0. 2.4 * 103

Standard Number Scientific Notation

0.0 0 0 8 6 = 8.6 10-4

4 places Coefficient Power of 10

*

In another example, 0.000 86 is written in scientific notation as 8.6 * 10-4. The coefficient, 8.6, is obtained by moving the decimal point to the right. Because the decimal point is moved four places to the right, the power of 10 is a negative 4, written as 10-4. When a number less than 1 is written in scientific notation, the power of 10 is negative.

ENGAGE 1.11 Why is 0.000 053 written as 5.3 * 10 -5 in scientific notation?

TABLE 1.2 gives some examples of numbers written as positive and negative powers of 10. The powers of 10 are a way of keeping track of the decimal point in the number. TABLE 1.3 gives several examples of writing measurements in scientific notation.

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18 CHAPTER 1 Chemistry in Our Lives

Standard Number Multiples of 10 Scientific Notation

10 000 10 * 10 * 10 * 10 1 * 104 Some positive powers of 10

1000 10 * 10 * 10 1 * 103

100 10 * 10 1 * 102

10 10 1 * 101

1 0 1 * 100

0.1 1 10

1 * 10-1 Some negative powers of 10

0.01 1 10

* 1 10

= 1

100 1 * 10-2

0.001 1 10

* 1 10

* 1 10

= 1

1000 1 * 10-3

0.0001 1 10

* 1 10

* 1 10

* 1 10

= 1

10 000 1 * 10-4

TABLE 1.2 Some Powers of 10

Measured Quantity Standard Number Scientific Notation

Volume of gasoline used in the United States each year 550 000 000 000 L 5.5 * 1011 L

Diameter of Earth 12 800 000 m 1.28 * 107 m

Average volume of blood pumped in 1 day 8500 L 8.5 * 103 L

Time for light to travel from the Sun to Earth 500 s 5 * 102 s

Mass of a typical human 68 kg 6.8 * 101 kg

Mass of stirrup bone in ear 0.003 g 3 * 10-3 g

Diameter of a chickenpox (Varicella zoster) virus 0.000 000 3 m 3 * 10-7 m

Mass of bacterium (mycoplasma) 0.000 000 000 000 000 000 1 kg 1 * 10-19 kg

TABLE 1.3 Some Measurements Written as Standard Numbers and in Scientific Notation

A chickenpox virus has a diameter of 3 * 10 -7 m .

SAMPLE PROBLEM 1.8 Writing a Number in Scientific Notation

TRY IT FIRST

Write each of the following in scientific notation:

a. 3500 b. 0.000 016

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

standard number scientific notation coefficient is at least 1 but less than 10

a. 3500

STEP 1 Move the decimal point to obtain a coefficient that is at least 1 but less than 10. For a number greater than 1, the decimal point is moved to the left three places to give a coefficient of 3.5.

STEP 2 Express the number of places moved as a power of 10. Moving the decimal point three places to the left gives a power of 3, written as 103.

STEP 3 Write the product of the coefficient multiplied by the power of 10. 3.5 * 103

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1.5 Writing Numbers in Scientific Notation 19

Scientific Notation and Calculators You can enter a number in scientific notation on many calculators using the EE or EXP key. After you enter the coefficient, press the EE or EXP key and enter the power 10. To enter a negative power of 10, press the + / - key or the - key, depending on your calculator.

Number to Enter Procedure Calculator Display

4 * 106 4 EE or EXP 6

2.5 * 10-4 2.5 EE or EXP + / - 4

When a calculator answer appears in scientific notation, the coefficient is shown as a number that is at least 1 but less than 10, followed by a space or E and the power of 10. To express this display in scientific notation, write the coefficient value, write * 10, and use the power of 10 as an exponent.

Calculator Display Expressed in Scientific Notation

7.52 * 104

5.8 * 10-2

On many calculators, a number is converted into scientific notation using the appropriate keys. For example, the number 0.000 52 is entered, followed by pressing the 2nd or 3rd function key (2nd F) and the SCI key. The scientific notation appears in the calculator display as a coefficient and the power of 10.

or or

or or

4 06

2.5 — 04 2.5 E — 042.5 —04

4 E064 06

ENGAGE 1.12 Describe how you enter a number in scientific notation on your calculator.

or or

or or

7.52 04

5.8 — 02 5.8 E — 025.8 —02

7.52 E047.52 04

b. 0.000 016

STEP 1 Move the decimal point to obtain a coefficient that is at least 1 but less than 10. For a number less than 1, the decimal point is moved to the right five places to give a coefficient of 1.6.

STEP 2 Express the number of places moved as a power of 10. Moving the decimal point five places to the right gives a power of negative 5, written as 10-5.

STEP 3 Write the product of the coefficient multiplied by the power of 10. 1.6 * 10-5

SELF TEST 1.8

Write each of the following in scientific notation:

a. 425 000 b. 0.000 000 86 c. 0.007 30 d. 978 * 105

ANSWER

a. 4.25 * 105 b. 8.6 * 10-7 c. 7.30 * 10-3 d. 9.78 * 107 PRACTICE PROBLEMS

Try Practice Problems 1.27 and 1.28

0.000 52 5.2 10-4* Calculator display

2ndF SCI oror == 5.2— 04 5.2 E— 045.2 —04

Converting Scientific Notation to a Standard Number When a number written in scientific notation has a positive power of 10, the standard number is obtained by moving the decimal point to the right for the same number of places as the power of 10. Placeholder zeros are used, as needed, to give additional places.

Scientific Notation Standard Number

4.3 0 430= =4.3 102*

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20 CHAPTER 1 Chemistry in Our Lives

For a number with a negative power of 10, the standard number is obtained by moving the decimal point to the left for the same number of places as the power of 10. Placeholder zeros are added in front of the coefficient as needed.

PRACTICE PROBLEMS Try Practice Problems 1.29 and 1.30

Scientific Notation Standard Number

0 0 0 0 0 2.5= 0.000 025=2.5 10-5*

SAMPLE PROBLEM 1.9 Writing Scientific Notation as a Standard Number

TRY IT FIRST

Write each of the following as a standard number:

a. 7.2 * 10-3 b. 2.4 * 105

SOLUTION

a. To write the standard number for an exponential number with a negative power of 10, move the decimal point to the left the same number of places (three) as the power of 10. Add placeholder zeros before the coefficient as needed.

ANSWER

a. 0.000 725 b. 3 050 000

2.4 0 0 0 0= 240 000=2.4 105*

0 0 0 7.2 m= 0.0072=7.2 10-3*

b. To write the standard number for an exponential number with a positive power of 10, move the decimal point to the right the same number of places (five) as the power of 10. Add placeholder zeros following the coefficient as needed.

SELF TEST 1.9

Write each of the following as a standard number:

a. 7.25 * 10-4 b. 3.05 * 106

PRACTICE PROBLEMS

1.5 Writing Numbers in Scientific Notation

1.27 Write each of the following in scientific notation: a. 55 000 b. 480 c. 0.000 005 d. 0.000 14 e. 0.0072 f. 670 000

1.28 Write each of the following in scientific notation: a. 180 000 000 b. 0.000 06 c. 750 d. 0.15 e. 0.024 f. 1500

1.29 Write each of the following as a standard number: a. 1.2 * 104 b. 8.25 * 10-2 c. 4 * 106 d. 5.8 * 10-3

1.30 Write each of the following as a standard number: a. 3.6 * 10-5 b. 8.75 * 104 c. 3 * 10-2 d. 2.12 * 105

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Concept Map 21

UPDATE Forensic Evidence Helps Solve the Crime

Using a variety of laboratory tests, Sarah finds ethylene glycol in Gloria’s blood. The quantitative tests indicate that Gloria had ingested 125 g of ethylene glycol. Sarah determines that the liquid in a glass found in Gloria’s home was ethylene glycol that had

been added to an alcoholic beverage. Ethylene glycol is a clear, sweet-tasting, thick liquid that is odorless and mixes with water. It is easy to obtain since it is used as antifreeze in automobiles and in brake fluid. Because the initial symptoms of ethylene glycol poisoning are similar to being intoxicated, the victim is often unaware of its presence.

If ingestion of ethylene glycol occurs, it can cause depression of the central nervous system, cardiovascular damage, and kidney failure. If discovered quickly, hemodialysis may be used to remove ethylene glycol from the blood. A toxic amount of ethylene glycol is 1.5 g of ethylene glycol/kg of body mass. Thus, 75 g could be fatal for a 50-kg (110-lb) person.

Sarah determines that fingerprints on the glass containing the ethylene glycol were those of Gloria’s

husband. This evidence along with the container of antifreeze found in the home led to the arrest and conviction of the husband for poisoning his wife.

Applications

1.31 Identify each of the following comments in the police report as an observation, a hypothesis, an experiment, or a conclusion: a. Gloria may have had a heart attack. b. Test results indicate that Gloria was poisoned. c. The liquid in the glass was analyzed. d. The antifreeze in the pantry was the same color as the

liquid in the glass.

1.32 Identify each of the following comments in the police report as an observation, a hypothesis, an experiment, or a conclusion: a. Gloria may have committed suicide. b. Sarah ran blood tests to identify any toxic substances. c. The temperature of Gloria’s body was 34 °C. d. The fingerprints found on the glass were determined to

be her husband’s.

1.33 A container was found in the home of the victim that contained 120 g of ethylene glycol in 450 g of liquid. What was the percentage of ethylene glycol? Express your answer to the ones place.

1.34 If the toxic quantity is 1.5 g of ethylene glycol per 1000 g of body mass, what percentage of ethylene glycol is fatal?

Substances

CHEMISTRY IN OUR LIVES

deals with uses the is learned by uses key math skills

Hypothesis

Experiments

Conclusion

Reading the Text

Practicing Problem Solving

Working with a Group

Self-Testing

Answering the Engage Questions

Trying It First

Chemicals

Scientific Method

Observations

Identifying Place Values

Using Positive and Negative Numbers

Calculating Percentages

Solving Equations

Converting Between Standard

Numbers and Scientific Notation

called starting with

that lead to

CONCEPT MAP

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22 CHAPTER 1 Chemistry in Our Lives

CHAPTER REVIEW

1.1 Chemistry and Chemicals LEARNING GOAL Define the term chemistry, and identify chemicals. • Chemistry is the study of the composition,

structure, properties, and reactions of matter. • A chemical always has the same compo-

sition and properties wherever it is found.

1.2 Scientific Method: Thinking Like a Scientist LEARNING GOAL Describe the scientific method. • The scientific method is a process of

explaining natural phenomena beginning with making observations, forming a hypothesis, and performing experiments.

• After repeated successful experiments, a conclusion may be drawn that the hypothesis is valid.

1.3 Studying and Learning Chemistry LEARNING GOAL Identify strategies that are effective for learning. Develop a study plan for learning chemistry. • A plan for learning chemistry utilizes

the features in the text that help develop a successful approach to learning chemistry.

• By using the Learning Goals, Reviews, Engage questions, Analyze the Problems, Try It First, Sample Problems, and Self Tests in the chapter, and Practice Problems at the end of each section, you can successfully learn the concepts of chemistry.

1.4 Key Math Skills for Chemistry LEARNING GOAL Review math concepts used in chemistry: place values, positive and negative numbers, percentages, solving equations, and interpreting graphs. • Solving chemistry problems involves a

number of math skills: identifying place values, using positive and negative numbers, calculating percentages, solving equations, and interpreting graphs.

1.5 Writing Numbers in Scientific Notation LEARNING GOAL Write a number in scientific notation. • A number written in

scientific notation has two parts, a coefficient and a power of 10.

• When a number greater than 1 is converted to scientific notation, the power of 10 is positive.

• When a number less than 1 is written in scientific notation, the power of 10 is negative.

1 * 105 hairs 8 * 10-6 m

chemical A substance that has the same composition and properties wherever it is found.

chemistry The study of the composition, structure, properties, and reactions of matter.

conclusion An explanation of an observation that has been validated by repeated experiments that support a hypothesis.

experiment A procedure that tests the validity of a hypothesis. hypothesis An unverified explanation of a natural phenomenon. observation Information determined by noting and recording a

natural phenomenon.

scientific method The process of making observations, proposing a hypothesis, and testing the hypothesis; after repeated experiments validate the hypothesis, a conclusion may be drawn as to its validity.

scientific notation A form of writing large and small numbers using a coefficient that is at least 1 but less than 10, followed by a power of 10.

KEY TERMS

The chapter section containing each Key Math Skill is shown in parentheses at the end of each heading.

Identifying Place Values (1.4) • The place value identifies the numerical value of each digit in a

number.

Example: Identify the place value for each of the digits in the number 456.78.

KEY MATH SKILLS

Answer: Digit Place Value

4 hundreds

5 tens

6 ones

7 tenths

8 hundredths

M01_TIMB8119_06_SE_C01.indd 22 11/29/18 9:24 AM

Key Math Skills 23

Using Positive and Negative Numbers in Calculations (1.4)

• A positive number is any number that is greater than zero and has a positive sign ( + ), which is often omitted. A negative number is any number that is less than zero and is written with a negative sign ( - ).

• When two positive numbers are added, multiplied, or divided, the answer is positive.

• When two negative numbers are multiplied or divided, the answer is positive. When two negative numbers are added, the answer is negative.

• When a positive and a negative number are multiplied or divided, the answer is negative.

• When a positive and a negative number are added, the smaller number is subtracted from the larger number, and the result has the same sign as the larger number.

• When two numbers are subtracted, change the sign of the number to be subtracted, then follow the rules for addition.

Example: Evaluate each of the following: a. - 8 - 14 = b. 6 * ( - 3) = Answer: a. - 22 b. - 18

Calculating Percentages (1.4) • A percentage is the part divided by the total (whole) multiplied

by 100%.

Example: A drawer contains 6 white socks and 18 black socks. What is the percentage of white socks?

Answer: 6 white socks 24 total socks

* 100% = 25% white socks

Solving Equations (1.4) An equation in chemistry often contains an unknown. To rearrange an equation to obtain the unknown factor by itself, you keep it balanced by performing matching mathematical operations on both sides of the equation.

• If you eliminate a number or symbol by subtracting, subtract that same number or symbol from both sides.

• If you eliminate a number or symbol by adding, add that same number or symbol to both sides.

• If you cancel a number or symbol by dividing, divide both sides by that same number or symbol.

• If you cancel a number or symbol by multiplying, multiply both sides by that same number or symbol.

Example: Solve the equation for a: 3a - 8 = 28 Answer: Add 8 to both sides 3a - 8 + 8 = 28 + 8

3a = 36

Divide both sides by 3 3a 3

= 36 3

a = 12 Check: 3(12) - 8 = 28

36 - 8 = 28 28 = 28

Your answer a = 12 is correct.

500

450

400

350

300

250

200

150

100

0 20 40 60 80 100

S ol

ub il

it y

(g s

ug ar

/1 00

m L

w at

er )

Temperature (°C)

Solubility of Sugar in Water versus Temperature

Interpreting Graphs (1.4)

• A graph represents the relationship between two variables. • The quantities are plotted along two perpendicular axes, which are

the x axis (horizontal) and y axis (vertical). • The title indicates the components of the x and y axes. • Numbers on the x and y axes show the range of values of the

variables. • The graph shows the relationship between the component on the

y axis and that on the x axis.

Example:

a. Does the amount of sugar that dissolves in 100 mL of water increase or decrease when the temperature increases?

b. How many grams of sugar dissolve in 100 mL of water at 70 °C?

c. At what temperature (°C) will 275 g of sugar dissolve in 100 mL of water?

Answer: a. increase b. 320 g c. 55 °C

Writing Numbers in Scientific Notation (1.5) • A number written in scientific notation consists of a coefficient

and a power of 10. A number is written in scientific notation by:

• Moving the decimal point to obtain a coefficient that is at least 1 but less than 10.

• Expressing the number of places moved as a power of 10. The power of 10 is positive if the decimal point is moved to the left, negative if the decimal point is moved to the right.

Example: Write the number 28 000 in scientific notation. Answer: Moving the decimal point four places to the left gives

a coefficient of 2.8 and a positive power of 10, 104. The number 28 000 written in scientific notation is 2.8 * 104.

M01_TIMB8119_06_SE_C01.indd 23 11/29/18 9:24 AM

24 CHAPTER 1 Chemistry in Our Lives

UNDERSTANDING THE CONCEPTS

The chapter sections to review are shown in parentheses at the end of each problem.

1.35 A “chemical-free” shampoo includes the following ingredients: water, cocamide, glycerin, and citric acid. Is the shampoo truly “chemical-free”? (1.1)

1.36 A “chemical-free” sunscreen includes the following ingredients: titanium dioxide, vitamin E, and vitamin C. Is the sunscreen truly “chemical-free”? (1.1)

1.37 According to Sherlock Holmes, “One must follow the rules of scientific inquiry, gathering, observing, and testing data, then formulating, modifying, and rejecting hypotheses, until only one remains.” Did Holmes use the scientific method? Why or why not? (1.2)

1.38 In A Scandal in Bohemia, Sherlock Holmes receives a mysterious note. He states, “I have no data yet. It is a capital mistake to theorize before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts.” What do you think Holmes meant? (1.2)

1.39 For each of the following, indicate if the answer has a positive or negative sign: (1.4)

a. Two negative numbers are added. b. A positive and negative number are multiplied.

1.40 For each of the following, indicate if the answer has a positive or negative sign: (1.4)

a. A negative number is subtracted from a positive number. b. Two negative numbers are divided.

Applications

1.41 Classify each of the following statements as an observation or a hypothesis: (1.2)

a. A patient breaks out in hives after receiving penicillin. b. Dinosaurs became extinct when a large meteorite struck

Earth and caused a huge dust cloud that severely decreased the amount of light reaching Earth.

c. A patient’s blood pressure was 110/75.

1.42 Classify each of the following statements as an observation or a hypothesis: (1.2)

a. Analysis of 10 ceramic dishes showed that four dishes contained lead levels that exceeded federal safety standards.

b. Marble statues undergo corrosion in acid rain. c. A child with a high fever and a rash may have chickenpox.

1.43 Select the correct phrase(s) to complete the following statement: If experimental results do not support your hypothesis, you should (1.2)

a. pretend that the experimental results support your hypothesis

b. modify your hypothesis c. do more experiments

1.44 Select the correct phrase(s) to complete the following statement: A hypothesis is supported when (1.2)

a. one experiment proves the hypothesis b. many experiments validate the hypothesis c. you think your hypothesis is correct

1.45 Identify each of the following as an observation, a hypothesis, an experiment, or a conclusion: (1.2)

a. During an assessment in the emergency room, a nurse writes that the patient has a resting pulse of 30 beats/min.

b. A nurse thinks that an incision from a recent surgery that is red and swollen is infected.

c. Repeated studies show that lowering sodium in the diet leads to a decrease in blood pressure.

1.46 Identify each of the following as an observation, a hypothesis, an experiment, or a conclusion: (1.2)

a. Drinking coffee at night keeps me awake. b. If I stop drinking coffee in the afternoon, I will be able to

sleep at night. c. I will try drinking coffee only in the morning.

1.47 Which of the following would be part of a successful study plan? (1.3)

a. skipping class and just reading the text b. working the Sample Problems as you go through a chapter

c. self-testing d. reading through the chapter, but working the problems later e. reading the assignment before class

1.48 Which of the following would be part of a successful study plan? (1.3)

a. studying all night before the exam b. highlighting important ideas in the text c. working problems in a notebook for easy reference d. forming a study group and discussing the problems together e. reviewing Key Math Skills and Core Chemistry Skills

1.49 Evaluate each of the following: (1.4)

a. 4 * ( - 8) = _______ b. - 12 - 48 = _______

c. - 168 - 4

= _______

1.50 Evaluate each of the following: (1.4)

a. - 95 - ( - 11) = _______ b. 152 - 19

= _______

c. 4 - 56 = _______ 1.51 A bag of gumdrops contains 16 orange gumdrops, 8 yellow

gumdrops, and 16 black gumdrops. (1.4)

a. What is the percentage of yellow gumdrops? Express your answer to the ones place.

b. What is the percentage of black gumdrops? Express your answer to the ones place.

1.52 On the first chemistry test, 12 students got As, 18 students got Bs, and 20 students got Cs. (1.4)

a. What is the percentage of students who received Bs? Express your answer to the ones place.

b. What is the percentage of students who received Cs? Express your answer to the ones place.

ADDITIONAL PRACTICE PROBLEMS

M01_TIMB8119_06_SE_C01.indd 24 11/29/18 9:24 AM

Answers to Engage Questions 25

1.53 Write each of the following in scientific notation: (1.5)

a. 120 000 b. 0.000 000 34 c. 0.066 d. 2700

1.54 Write each of the following in scientific notation: (1.5)

a. 0.0042 b. 310 c. 890 000 000 d. 0.000 000 056

1.55 Write each of the following as a standard number: (1.5)

a. 2.6 * 10-5 b. 6.5 * 102 c. 3.7 * 10-1 d. 5.3 * 105

1.56 Write each of the following as a standard number: (1.5)

a. 7.2 * 10-2 b. 1.44 * 103 c. 4.8 * 10-4 d. 9.1 * 106

Applications

1.57 Identify each of the following as an observation, a hypothesis, an experiment, or a conclusion: (1.2)

a. A patient has a high fever and a rash on her back. b. A nurse tells a patient that her baby who gets sick after

drinking milk may be lactose intolerant. c. Numerous studies have shown that omega-3 fatty acids

lower triglyceride levels.

1.58 Identify each of the following as an observation, a hypothesis, an experiment, or a conclusion: (1.2)

a. Every spring, you have congestion and a runny nose. b. An overweight patient decides to exercise more to lose weight. c. Many research studies have linked obesity to heart disease.

ANSWERS TO ENGAGE QUESTIONS 1.8 Dividing both sides of the equation solves the equation for the

unknown V2.

1.9 A graph represents the relationship between two variables plotted in equal intervals along the x axis (horizontal) and the y axis (vertical).

1.10 The decimal point is moved five places to the left to give a coefficient that is at least 1 but less than 10 and a power of 10 that is a positive 5.

1.11 The decimal point is moved five places to the right to give a coefficient that is at least 1 but less than 10, and a power of 10 that is a negative 5.

1.12 Enter the coefficient, press the EE or EXP key, and enter the power of 10. To enter a negative power of 10, press the + / - key or the - key.

1.1 A chemical has the same composition and properties wherever it is found.

1.2 An experiment is done to find a relationship between the hypothesis and the observations.

1.3 Self-testing practices the retrieval of new information and helps connect to prior knowledge.

1.4 The value 0 is located in the first place to the right of the decimal point, which is the tenths place.

1.5 Addition of a negative number and a positive number gives an answer that has the sign of the larger value. Addition of two negative numbers gives a larger number with a negative sign.

1.6 The value of 100% represents exactly 100 units of the whole.

1.7 Subtracting 8 from both sides of the equation places like terms on the same sides.

The following problems are related to the topics in this chapter. However, they do not all follow the chapter order, and they require you to combine concepts and skills from several sections. These problems will help you increase your critical thinking skills and prepare for your next exam.

1.59 Classify each of the following as an observation, a hypothesis, an experiment, or a conclusion: (1.2)

a. The bicycle tire is flat. b. If I add air to the bicycle tire, it will expand to the proper size. c. When I added air to the bicycle tire, it was still flat. d. The bicycle tire has a leak in it.

1.60 Classify each of the following as an observation, a hypothesis, an experiment, or a conclusion: (1.2)

a. A big log in the fire does not burn well. b. If I chop the log into smaller wood pieces, it will burn better. c. The small wood pieces burn brighter and make a hotter fire. d. The small wood pieces are used up faster than burning the

big log.

1.61 Solve each of the following for x: (1.4)

a. 2x + 5 = 41 b. 5x 3

= 40

1.62 Solve each of the following for z: (1.4)

a. 3z - ( - 6) = 12 b. 4z

- 12 = - 8

Use the following graph for problems 1.63 and 1.64:

CHALLENGE PROBLEMS

0.35

0.30 0.25 0.20

S ol

ub il

it y

(g C

O 2/

10 0

g w

at er

) 0.15

0.10 0.05

0 0 10 20

Temperature (°C)

Solubility of Carbon Dioxide in Water versus Temperature

30 40 50 60

1.63 a. What does the title indicate about the graph? (1.4) b. What is measured on the vertical axis? c. What is the range of values on the vertical axis? d. Does the solubility of carbon dioxide increase or decrease

with an increase in temperature?

1.64 a. What is measured on the horizontal axis? (1.4) b. What is the range of values on the horizontal axis? c. What is the solubility of carbon dioxide in water at 25 °C? d. At what temperature does carbon dioxide have a solubility

of 0.20 g/100 g water?

M01_TIMB8119_06_SE_C01.indd 25 11/29/18 9:24 AM

26 CHAPTER 1 Chemistry in Our Lives

ANSWERS TO SELECTED PROBLEMS 1.25 a. This graph shows the relationship between body

temperature and time since death. b. temperature, in °C c. 20 °C to 44 °C d. decrease

1.27 a. 5.5 * 104 b. 4.8 * 102 c. 5 * 10-6 d. 1.4 * 10-4 e. 7.2 * 10-3 f. 6.7 * 105

1.29 a. 12 000 b. 0.0825 c. 4 000 000 d. 0.0058

1.31 a. hypothesis b. conclusion c. experiment d. observation

1.33 27% ethylene glycol

1.35 No. All of the ingredients are chemicals.

1.37 Yes. Sherlock’s investigation includes making observations (gathering data), formulating a hypothesis, testing the hypothesis, and modifying it until one of the hypotheses is validated.

1.39 a. negative b. negative

1.41 a. observation b. hypothesis c. observation

1.43 b and c

1.45 a. observation b. hypothesis c. conclusion

1.47 b, c, and e

1.49 a. - 32 b. - 60 c. 42 1.51 a. 20% b. 40%

1.53 a. 1.2 * 105 b. 3.4 * 10-7 c. 6.6 * 10-2 d. 2.7 * 103

1.55 a. 0.000 026 b. 650 c. 0.37 d. 530 000

1.57 a. observation b. hypothesis c. conclusion

1.59 a. observation b. hypothesis c. experiment d. conclusion

1.61 a. 18 b. 24

1.63 a. The graph shows the relationship between the solubility of carbon dioxide in water and temperature.

b. solubility of carbon dioxide (g CO2/100 g water) c. 0 to 0.35 g of CO2/100 g of water d. decrease

1.1 a. Chemistry is the study of the composition, structure, properties, and reactions of matter.

b. A chemical is a substance that has the same composition and properties wherever it is found.

1.3 Many chemicals are listed on a multivitamin bottle, such as vitamin A, vitamin B3, vitamin B12, vitamin C, and folic acid.

1.5 Typical items found in a drugstore, and some of the chemicals they contain, are as follows:

Antacid tablets: calcium carbonate, cellulose, starch, stearic acid, silicon dioxide

Mouthwash: water, alcohol, thymol, glycerol, sodium benzoate, benzoic acid

Cough suppressant: menthol, beta-carotene, sucrose, glucose

1.7 a. observation b. hypothesis c. experiment d. observation e. observation f. conclusion

1.9 a. observation b. hypothesis c. experiment d. experiment

1.11 There are several things you can do that will help you successfully learn chemistry: forming a study group, retesting, working Sample Problems before reading the given answer, working the Self Test items, answering Engage questions as you study, working Practice Problems and checking Answers, reading the assignment ahead of class, and keeping a problem notebook.

1.13 a, c, and e

1.15 a. thousandths b. ones c. tenths d. hundreds e. hundredths

1.17 a. 23 b. - 30 c. - 2 d. 54 e. - 12 1.19 a. The graph shows the relationship between the temperature

of a cup of tea and time. b. temperature, in °C c. 20 °C to 90 °C d. decrease

1.21 a. 9 b. 42 c. 14 d. 3

1.23 a. 84% b. 72% c. 30% d. 5%

M01_TIMB8119_06_SE_C01.indd 26 11/29/18 9:24 AM

27

During the past few months, Greg has had an increased number of headaches, dizzy spells, and nausea. He goes to his doctor’s office, where Sandra, the registered nurse, completes the initial part of his exam by recording several measurements: weight 164 lb, height 5 ft 8 in., temperature 37.2 °C, and blood pressure 155/95. Normal blood pressure is 120/80 or lower.

When Greg sees his doctor, he is diagnosed with high blood pressure, or hypertension. The doctor prescribes 80. mg of Inderal (propranolol), which is available in 40.-mg tablets. Inderal is a beta blocker, which relaxes the muscles of the heart. It is used to treat hypertension, angina (chest pain), arrhythmia, and migraine headaches.

Two weeks later, Greg visits his doctor again, who determines that Greg’s blood pressure is now 152/90. The doctor increases the dosage of Inderal to 160. mg. Sandra informs Greg that he needs to increase his daily dosage from two tablets to four tablets.

CAREER

Registered Nurse In addition to assisting physicians, registered nurses work to promote patient health and prevent and treat disease. They provide patient care and help patients cope with illness. They take measurements such as a patient’s weight, height, temperature, and blood pressure; make conversions; and calculate drug dosage rates. Registered nurses also maintain detailed medical records of patient symptoms and prescribed medications.

Chemistry and Measurements

2

A few weeks later, Greg complains to his doctor that he is feel- ing tired. He has a blood test to determine if his iron level is low. You can see the results of Greg’s blood serum iron level in the UPDATE Greg’s Visit with His Doctor, page 59, and determine if Greg should be given an iron supplement.

UPDATE Greg’s Visit with His Doctor

M02_TIMB8119_06_SE_C02.indd 27 11/27/18 11:38 AM

28 CHAPTER 2 Chemistry and Measurements

2.1 Units of Measurement LEARNING GOAL Write the names and abbreviations for the metric and SI units used in measurements of volume, length, mass, temperature, and time.

Think about your day. You probably took some measurements. Perhaps you checked your weight by stepping on a bathroom scale. If you made rice for dinner, you added two cups of water to one cup of rice. If you did not feel well, you may have taken your temperature. Whenever you take a measurement, you use a measuring device such as a scale, a measuring cup, or a thermometer.

Scientists and health professionals throughout the world use the metric system of measurement. It is also the common measuring system in all but a few countries in the world. The International System of Units (SI), or Système International, is the official system of measurement throughout the world except for the United States. In chemistry, we use metric units and SI units for volume, length, mass, temperature, and time, as listed in TABLE 2.1.

Measurement Metric SI

Volume liter (L) cubic meter (m3)

Length meter (m) meter (m)

Mass gram (g) kilogram (kg)

Temperature degree Celsius (°C) kelvin (K)

Time second (s) second (s)

TABLE 2.1 Units of Measurement and Their Abbreviations

Your weight on a bathroom scale is a measurement.

1.7 m (65 in.)

58.0 kg (128 lb)

2.1 km (1.3 mi) 12 kg (26 lb)

22 °C (72 °F)

There are many measurements in everyday life.

LOOKING AHEAD

2.1 Units of Measurement 28 2.2 Measured Numbers and

Significant Figures 31 2.3 Significant Figures in

Calculations 34 2.4 Prefixes and Equalities 39 2.5 Writing Conversion

Factors 42 2.6 Problem Solving Using

Unit Conversion 47 2.7 Density 53

FIGURE 2.1 In the metric system, volume is based on the liter.

946.4 mL = 1 qt 1000 mL = 1 L

Suppose you walked 1.3 mi to campus today, carrying a backpack that weighs 26 lb. The temperature was 72 °F. Perhaps you weigh 128 lb and your height is 65 in. These measurements and units may seem familiar to you because they are stated in the U.S. system of measurement. However, in chemistry, we use the metric system in making our measure- ments. Using the metric system, you walked 2.1 km to campus, carrying a backpack that has a mass of 12 kg, when the temperature was 22 °C. You have a mass of 58.0 kg and a height of 1.7 m.

Volume Volume (V ) is the amount of space a substance occupies. The metric unit for volume is the liter (L), which is slightly larger than a quart (qt). In a laboratory or a hospital, chemists work with metric units of volume that are smaller and more convenient, such as the milliliter (mL). There are 1000 mL in 1 L (see FIGURE 2.1). Some relationships between units for volume are

1 L = 1000 mL 1 L = 1.057 qt 946.4 mL = 1 qt

M02_TIMB8119_06_SE_C02.indd 28 11/27/18 11:38 AM

2.1 Units of Measurement 29

Length The metric and SI unit of length is the meter (m). The centimeter (cm), a smaller unit of length, is commonly used in chemistry and is about equal to the width of your little finger (see FIGURE 2.2). Some relationships between units for length are

1 m = 100 cm 1 m = 39.37 in. 1 m = 1.094 yd 2.54 cm = 1 in.

Mass The mass of an object is a measure of the quantity of material it contains. The SI unit of mass, the kilogram (kg), is used for larger masses such as body mass. In the metric system, the unit for mass is the gram (g), which is used for smaller masses. There are 1000 g in 1 kg. One pound (lb) is equal to 453.6 g. Some relationships between units for mass are

1 kg = 1000 g 1 kg = 2.205 lb 453.6 g = 1 lb

You may be more familiar with the term weight than with mass. Weight is a measure of the gravitational pull on an object. On Earth, an astronaut with a mass of 75.0 kg has a weight of 165 lb. On the Moon, where the gravitational pull is one-sixth that of Earth, the astronaut has a weight of 27.5 lb. However, the mass of the astronaut is the same as on Earth, 75.0 kg. Scientists measure mass rather than weight because mass does not depend on gravity. In a chemistry laboratory, an electronic balance is used to measure the mass in grams of a substance (see FIGURE 2.3).

Temperature Temperature tells us how hot something is, tells us how cold it is outside, or helps us deter- mine if we have a fever (see FIGURE 2.4). In the metric system, temperature is measured using Celsius temperature. On the Celsius (°C) temperature scale, water freezes at 0 °C and boils at 100 °C, whereas on the Fahrenheit (°F) scale, water freezes at 32 °F and boils at 212 °F. In the SI system, temperature is measured using the Kelvin (K) temperature scale on which the lowest possible temperature is 0 K, known as absolute zero. A unit on the Kelvin scale is called a kelvin (K) and is not written with a degree sign.

Time We typically measure time in units such as years (yr), days, hours (h), minutes (min), or seconds (s). Of these, the SI and metric unit of time is the second (s). The standard now used to determine a second is an atomic clock. Some relationships between units for time are

1 day = 24 h 1 h = 60 min 1 min = 60 s

FIGURE 2.2 Length in the metric system (SI) is based on the meter, which is slightly longer than a yard.

1 in. = 2.54 cm

Centimeters 1 2 3 4 5

1

10 20 30 40 50 60 70 80 90 100

12 36

1 ft 2 ft 3 ft

Meterstick

Yardstick

1 m = 39.37 in.

Inches24

FIGURE 2.3 On an electronic balance, the digital readout gives the mass of a nickel, which is 5.01 g.

FIGURE 2.4 A thermometer is used to determine temperature.

A stopwatch is used to measure the time of a race.

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30 CHAPTER 2 Chemistry and Measurements

2.2 Measured Numbers and Significant Figures LEARNING GOAL Identify a number as measured or exact; determine the number of significant figures in a measured number.

When you make a measurement, you use some type of measuring device. For example, you may use a meterstick to measure your height, a scale to check your weight, or a thermometer to take your temperature.

Measured Numbers Measured numbers are the numbers you obtain when you measure a quantity such as your height, weight, or temperature. Suppose you are going to measure the lengths of the objects in FIGURE 2.5. To report the length of the object, you observe the numerical values of the marked lines at the end of the object. Then you can estimate by visually dividing the space between the marked lines. This estimated value is the final digit in a measured number.

For example, in Figure 2.5a, the end of the object is between the marks of 4 cm and 5 cm, which means that the length is more than 4 cm but less than 5 cm. If you estimate that the end of the object is halfway between 4 cm and 5 cm, you would report its length as 4.5 cm. Another student might report the length of the same object as 4.4 cm because people do not estimate in the same way.

The metric ruler shown in Figure 2.5b is marked at every 0.1 cm. Now you can determine that the end of the object is between 4.5 cm and 4.6 cm. Perhaps you report its length as 4.55 cm, whereas another student reports its length as 4.56 cm. Both results are acceptable.

In Figure 2.5c, the end of the object appears to line up with the 3-cm mark. Because the end of the object is on the 3-cm mark, the estimated digit is 0, which means the measurement is reported as 3.0 cm.

Significant Figures In a measured number, the significant figures (SFs) are all the digits including the estimated digit. Nonzero numbers are always counted as significant figures. However, a zero may or may not be a significant figure depending on its position in a number. TABLE 2.2 gives the rules and examples of counting significant figures.

PRACTICE PROBLEMS Try Practice Problems 2.1 to 2.8

REVIEW Writing Numbers in Scientific

Notation (1.5)

CORE CHEMISTRY SKILL Counting Significant Figures

PRACTICE PROBLEMS Try Practice Problems 2.9 to 2.14

SAMPLE PROBLEM 2.1 Units of Measurement

TRY IT FIRST

On a typical day, a nurse encounters several situations involving measurement. State the name and type of measurement indicated by the units in each of the following:

a. A patient has a temperature of 38.5 °C. b. A physician orders 1.5 g of cefuroxime for injection. c. A physician orders 1 L of a sodium chloride solution to be given intravenously. d. A medication is to be given to a patient every 4 h.

SOLUTION

a. A degree Celsius is a unit of temperature. b. A gram is a unit of mass. c. A liter is a unit of volume. d. An hour is a unit of time.

SELF TEST 2.1

State the names and types of measurements for an infant that is 54.6 cm long with a mass of 5.2 kg.

PRACTICE PROBLEMS

2.1 Units of Measurement

2.1 Write the abbreviation for each of the following: a. gram b. degree Celsius c. liter d. pound e. second

2.2 Write the abbreviation for each of the following: a. kilogram b. kelvin c. quart d. meter e. centimeter

2.3 State the type of measurement in each of the following statements: a. I put 12 L of gasoline in my gas tank. b. My friend is 170 cm tall. c. Earth is 385 000 km away from the Moon. d. The horse won the race by 1.2 s.

2.4 State the type of measurement in each of the following statements: a. I rode my bicycle 15 km today. b. My dog weighs 12 kg. c. It is hot today. It is 30 °C. d. I added 2 L of water to my fish tank.

2.5 State the name of the unit and the type of measurement indicated for each of the following quantities:

a. 4.8 m b. 325 g c. 1.5 mL d. 4.8 * 102 s e. 28 °C

2.6 State the name of the unit and the type of measurement indicated for each of the following quantities:

a. 0.8 L b. 3.6 cm c. 4 kg d. 3.5 h e. 373 K

Applications

2.7 On a typical day, medical personnel may encounter several situations involving measurement. State the name and type of measurement indicated by the units in each of the following: a. The clotting time for a blood sample is 12 s. b. A premature baby weighs 2.0 kg. c. An antacid tablet contains 1.0 g of calcium carbonate. d. An infant has a temperature of 39.2 °C.

2.8 On a typical day, medical personnel may encounter several situations involving measurement. State the name and type of measurement indicated by the units in each of the following: a. During open-heart surgery, the temperature of a patient is

lowered to 29 °C. b. The circulation time of a red blood cell through the body

is 20 s. c. A patient with a persistent cough is given 10. mL of cough

syrup. d. The amount of iron in the red blood cells of the body

is 2.5 g.

ANSWER

A centimeter (cm) is a unit of length. A kilogram (kg) is a unit of mass.

M02_TIMB8119_06_SE_C02.indd 30 11/27/18 11:38 AM

2.2 Measured Numbers and Significant Figures 31

2.2 Measured Numbers and Significant Figures LEARNING GOAL Identify a number as measured or exact; determine the number of significant figures in a measured number.

When you make a measurement, you use some type of measuring device. For example, you may use a meterstick to measure your height, a scale to check your weight, or a thermometer to take your temperature.

Measured Numbers Measured numbers are the numbers you obtain when you measure a quantity such as your height, weight, or temperature. Suppose you are going to measure the lengths of the objects in FIGURE 2.5. To report the length of the object, you observe the numerical values of the marked lines at the end of the object. Then you can estimate by visually dividing the space between the marked lines. This estimated value is the final digit in a measured number.

For example, in Figure 2.5a, the end of the object is between the marks of 4 cm and 5 cm, which means that the length is more than 4 cm but less than 5 cm. If you estimate that the end of the object is halfway between 4 cm and 5 cm, you would report its length as 4.5 cm. Another student might report the length of the same object as 4.4 cm because people do not estimate in the same way.

The metric ruler shown in Figure 2.5b is marked at every 0.1 cm. Now you can determine that the end of the object is between 4.5 cm and 4.6 cm. Perhaps you report its length as 4.55 cm, whereas another student reports its length as 4.56 cm. Both results are acceptable.

In Figure 2.5c, the end of the object appears to line up with the 3-cm mark. Because the end of the object is on the 3-cm mark, the estimated digit is 0, which means the measurement is reported as 3.0 cm.

Significant Figures In a measured number, the significant figures (SFs) are all the digits including the estimated digit. Nonzero numbers are always counted as significant figures. However, a zero may or may not be a significant figure depending on its position in a number. TABLE 2.2 gives the rules and examples of counting significant figures.

PRACTICE PROBLEMS Try Practice Problems 2.1 to 2.8

REVIEW Writing Numbers in Scientific

Notation (1.5)

CORE CHEMISTRY SKILL Counting Significant Figures

PRACTICE PROBLEMS Try Practice Problems 2.9 to 2.14

FIGURE 2.5 The lengths of the rectangular objects are measured in centimeters.

10 cm

2 3 4 5

10 cm

2 3 4 5

10 cm

2 3 4 5

(a) 4.5 cm

(b) 4.55 cm

(c) 3.0 cm

Rule Measured Number Number of Significant Figures

1. A number is a significant figure if it is

a. a nonzero digit 4.5 g 122.35 m

2 5

b. a zero between nonzero digits 205 °C 5.008 kg

3 4

c. a zero at the end of a decimal number 50. L 16.00 mL

2 4

d. in the coefficient of a number written in scientific notation

4.8 * 105 m 5.70 * 10-3 g

2 3

2. A zero is not significant if it is

a. at the beginning of a decimal number 0.0004 s 0.075 cm

1 2

b. used as a placeholder in a large number without a decimal point

850 000 m 1 250 000 g

2 3

TABLE 2.2 Significant Figures in Measured Numbers

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32 CHAPTER 2 Chemistry and Measurements

Significant Zeros and Scientific Notation In this text, we will place a decimal point after a significant zero at the end of a number. For example, if a measurement is written as 500. g, the decimal point after the second zero indicates that both zeros are significant. To show this more clearly, we can write it as 5.00 * 102 g. When the first zero in the measurement 300 m is a significant zero, but the second zero is not, the measurement is written as 3.0 * 102 m. We will assume that all zeros at the end of large standard numbers without a decimal point are not significant. Therefore, we write 400 000 g as 4 * 105 g, which has only one significant figure.

ENGAGE 2.1 Why is the zero in the coefficient of 3.20 * 104 cm a significant figure?

PRACTICE PROBLEMS Try Practice Problems 2.15 to 2.18

Counted Numbers Defined Equalities

Items Metric System U.S. System

8 doughnuts 1 L = 1000 mL 1 ft = 12 in. 2 baseballs 1 m = 100 cm 1 qt = 4 cups 5 capsules 1 kg = 1000 g 1 lb = 16 oz

TABLE 2.3 Examples of Some Exact Numbers

Exact Numbers Exact numbers are those numbers obtained by counting items or using a definition that c ompares two units in the same measuring system. Suppose a friend asks you how many classes you are taking. You would answer by counting the number of classes in your schedule. Suppose you want to state the number of seconds in one minute. Without using any measuring device, you would give the definition: There are 60 s in 1 min. Exact numbers are not measured, do not have a limited number of significant figures, and do not affect the number of significant figures in a calculated answer. For more examples of exact numbers, see TABLE 2.3.

The number of baseballs is counted, which means 2 is an exact number.

SAMPLE PROBLEM 2.2 Significant Figures

TRY IT FIRST

Write the number of significant figures in each of the following measured numbers:

a. 0.000 250 m b. 70.040 g c. 1.020 * 106 L

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

measured number number of significant figures

Table 2.2, rules for significant figures

a. Nonzero digits and zeros at the end of a decimal number are significant. 0.000 250 m has three SFs (Rules 1a, 1c).

b. Nonzero digits, zeros between nonzero digits, or at end of a decimal number are significant. 70.040 g has five SFs (Rules 1a, 1b, 1c).

c. All digits in the coefficient of a number written in scientific notation are significant. 1.020 * 106 L has four SFs (Rule 1d).

SELF TEST 2.2

Write the number of significant figures in each of the following measured numbers:

a. 0.040 08 m b. 6.00 * 103 g ANSWER

a. four SFs (Rules 1a, 1b) b. three SFs (Rule 1d)

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2.2 Measured Numbers and Significant Figures 33

For example, a mass of 42.2 g and a length of 5.0 * 10-3 cm are measured numbers because they are obtained using measuring tools. There are three SFs in 42.2 g because all nonzero digits are always significant. There are two SFs in 5.0 * 10-3 cm because all the digits in the coefficient of a number written in scientific notation are significant. However, a quantity of three eggs is an exact number that is obtained by counting. In the equality 1 kg = 1000 g, the masses of 1 kg and 1000 g are both exact numbers because this is a definition in the metric system.

ENGAGE 2.2 Why is the 4 in 4 hats an exact number, whereas the 6.24 in 6.24 cm has three significant figures?

PRACTICE PROBLEMS Try Practice Problems 2.19 to 2.24

SAMPLE PROBLEM 2.3 Measured and Exact Numbers

TRY IT FIRST

Identify each of the following numbers as measured or exact, give the number of significant figures (SFs) in each of the measured numbers, and explain:

a. 0.170 L b. 4 knives c. 6.3 * 10-6 s d. 1 m = 100 cm

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

numbers identify exact or measured, number of SFs

Table 2.2, rules for SFs in measured numbers

STEP 1 A measured number is the result of a measurement. An exact number is obtained by counting or by definition.

a. measured b. exact, counted c. measured d. exact, definition

STEP 2 Significant figures are the nonzero digits and any zeros after nonzero digits in a decimal number. In numbers written in scientific notation, the numbers in the coefficient are all significant.

a. three SFs c. two SFs

SELF TEST 2.3

Identify each of the following numbers as measured or exact, and give the number of significant figures (SFs) in each of the measured numbers:

a. 0.020 80 kg b. 5.06 * 104 h c. 4 chemistry books ANSWER

a. measured; four SFs b. measured; three SFs c. exact

PRACTICE PROBLEMS

2.2 Measured Numbers and Significant Figures

2.9 Use the metric ruler to measure the length in each of the following, and write the measurement, the estimated digit, and the number of significant figures:

10 cm

2 3 4 5 10 cm

2 3 4 5 10 cm

2 3 4 5

(a) (b) (c)

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34 CHAPTER 2 Chemistry and Measurements

2.10 Determine the volume, in milliliters, of each liquid in the graduated cylinders by reading the lowest point of the curve of the liquid in the diagrams (a), (b), and (c) using the correct number of significant figures.

2.11 How many significant figures are in each of the following? a. 11.005 g b. 0.000 32 m c. 36 000 000 km d. 1.80 * 104 kg e. 0.8250 L f. 30.0 °C

2.12 How many significant figures are in each of the following? a. 20.60 mL b. 1036.48 kg c. 4.00 m d. 20 °C e. 60 800 000 g f. 5.0 * 10-3 L 2.13 In which of the following pairs do both numbers contain the

same number of significant figures? a. 11.0 m and 11.00 m b. 0.0250 m and 0.205 m c. 0.000 12 s and 12 000 s d. 250.0 L and 2.5 * 10-2 L

2.14 In which of the following pairs do both numbers contain the same number of significant figures? a. 0.005 75 g and 5.75 * 10-3 g b. 405 K and 405.0 K c. 150 000 s and 1.50 * 104 s d. 3.8 * 10-2 L and 3.80 * 105 L

2.15 Indicate if the zeros are significant in each of the following measurements:

a. 0.0038 m b. 5.04 cm c. 800. L d. 3.0 * 10-3 kg e. 85 000 g

2.16 Indicate if the zeros are significant in each of the following measurements:

a. 20.05 °C b. 5.00 m c. 0.000 02 g d. 120 000 yr e. 8.05 * 102 L 2.17 Write each of the following in scientific notation with two

significant figures: a. 5000 L b. 30 000 g c. 100 000 m d. 0.000 25 cm

2.18 Write each of the following in scientific notation with two significant figures:

a. 5 100 000 g b. 26 000 s c. 40 000 m d. 0.000 820 kg

2.19 Identify the numbers in each of the following statements as measured or exact: a. A patient has a mass of 67.5 kg. b. A patient is given 2 tablets of medication. c. In the metric system, 1 L is equal to 1000 mL. d. The distance from Denver, Colorado, to Houston, Texas, is

1720 km.

2.20 Identify the numbers in each of the following statements as measured or exact: a. There are 31 students in the laboratory. b. The oldest known flower lived 1.20 * 108 yr ago. c. The largest gem ever found, an aquamarine, has a mass of

104 kg. d. A laboratory test shows a blood cholesterol level of

184 mg/dL.

2.21 Identify the measured number(s), if any, in each of the following pairs of numbers: a. 3 hamburgers and 6 oz of hamburger b. 1 table and 4 chairs c. 0.75 lb of grapes and 350 g of butter d. 60 s = 1 min

2.22 Identify the exact number(s), if any, in each of the following pairs of numbers: a. 5 pizzas and 50.0 g of cheese b. 6 nickels and 16 g of nickel c. 3 onions and 3 lb of onions d. 5 miles and 5 cars

Applications

2.23 Identify each of the following as measured or exact, and give the number of significant figures (SFs) in each measured number: a. The mass of a neonate is 1.607 kg. b. The Daily Value (DV) for iodine for an infant is 130 mcg. c. There are 4.02 * 106 red blood cells in a blood sample. d. In November, 23 babies were born in a hospital.

2.24 Identify each of the following as measured or exact, and give the number of significant figures (SFs) in each measured number: a. An adult with the flu has a temperature of 103.5 °F. b. A blister (push-through) pack of prednisone contains

21 tablets. c. The time for a nerve impulse to travel from the feet to the

brain is 0.46 s. d. A brain contains 1.20 * 1010 neurons.

1

2

(a)

20

30

40

(b)

50

75

(c)

2.3 Significant Figures in Calculations LEARNING GOAL Give the correct number of significant figures for a calculated answer.

In the sciences, we measure many things: the length of a bacterium, the volume of a gas sample, the temperature of a reaction mixture, or the mass of iron in a sample. The number

REVIEW Identifying Place Values (1.4)

Using Positive and Negative Numbers in Calculations (1.4)

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2.3 Significant Figures in Calculations 35

of significant figures in measured numbers determines the number of significant figures in the calculated answer.

Using a calculator will help you perform calculations faster. However, calculators cannot think for you. It is up to you to enter the numbers correctly, press the correct function keys, and give the answer with the correct number of significant figures.

Rounding Off Suppose you decide to buy carpeting for a room that has a length of 5.52 m and a width of 3.58 m. To determine how much carpeting you need, you would calculate the area of the room by multiplying 5.52 times 3.58 on your calculator. The calculator shows the number 19.7616 in its display. Because each of the original measurements has only three signifi- cant figures, the calculator display (19.7616) should be rounded off to three significant figures, 19.8.

5.52 m * 3.58 m = 19.7616 = 19.8 m2

Therefore, you can order carpeting that will cover an area of 19.8 m2. Each time you use a calculator, it is important to look at the original measurements and

determine the number of significant figures that can be used for the answer. You can use the following rules to round off the numbers shown in a calculator display.

Rules for Rounding Off

1. If the first digit to be dropped is 4 or less, it and all following digits are simply dropped from the number.

2. If the first digit to be dropped is 5 or greater, the last retained digit of the number is increased by 1.

KEY MATH SKILL Rounding Off

ENGAGE 2.3 Why is 10.072 rounded off to three significant figures equal to 10.1?

A technician uses a calculator in the laboratory.

Number to Round Off Three Significant Figures Two Significant Figures

8.4234 8.42 (drop 34) 8.4 (drop 234)

14.780 14.8 (drop 80, increase the last retained digit by 1)

15 (drop 780, increase the last retained digit by 1)

3256 3260* (drop 6, increase the last retained digit by 1, add 0) (3.26 * 103)

3300* (drop 56, increase the last retained digit by 1, add 00) (3.3 * 103)

*The value of a large number is retained by using placeholder zeros to replace dropped digits.

TABLE 2.4 Rules for Rounding Off

SAMPLE PROBLEM 2.4 Rounding Off

TRY IT FIRST

Round off each of the following numbers to three significant figures: a. 35.7823 m b. 0.002 621 7 L c. 3.8268 * 103 g

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

measured numbers

three significant figures

Table 2.4, rules for rounding off

Three SFs

Three SFs

Calculator display

Final answer, rounded off to three SFs

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36 CHAPTER 2 Chemistry and Measurements

Multiplication and Division with Measured Numbers In multiplication or division, the final answer is written so that it has the same number of significant figures (SFs) as the measurement with the fewest SFs.

Perform the following operations with measured numbers:

2.8 * 67.40 34.8

=

When the problem has multiple steps, the numbers in the numerator are multiplied and then divided by each of the numbers in the denominator.

2.8 * 67.40 , 34.8 = 5.422988506 = 5.4

Because the calculator display has more digits than the significant figures in the measured numbers allow, we need to round it off. Using the measured number that has the smallest number (two) of significant figures, 2.8, we round off the calculator display to an answer with two SFs.

Adding Significant Zeros Sometimes, a calculator display gives a small whole number. For example, suppose the calculator display is 4, but you used measurements that have three significant numbers. Then two significant zeros are added to give 4.00 as the correct answer.

8.00 2.00 = 4. = 4.00

PRACTICE PROBLEMS Try Practice Problems 2.25 to 2.28

CORE CHEMISTRY SKILL Using Significant Figures in

Calculations

ENGAGE 2.4 Why is the answer for the multiplication of 0.3 * 52.6 written with one significant figure?

PRACTICE PROBLEMS Try Practice Problems 2.29 and 2.30

ENGAGE 2.5 Why does the calculator answer of 5 in part c. require two significant zeros to give an answer of 5.00?

Three SFs

Three SFs Calculator display

Final answer, two zeros added to give three SFs

STEP 1 Identify the digits to be dropped.

a. drop 823 b. drop 17 c. drop 68

STEP 2 Increase the last retained digit by 1 if the first digit dropped is 5 or greater. Keep the last retained digit if the first digit dropped is 4 or less.

a. Increase the last retained digit by 1 to give 35.8 m. b. Retain digits to give 0.002 62 L. c. Increase the last retained digit by 1 to give 3.83 * 103 g.

SELF TEST 2.4

Round off each of the numbers in Sample Problem 2.4 to two significant figures.

ANSWER

a. 36 m b. 0.0026 L c. 3.8 * 103 g

A calculator is helpful in working problems and doing calculations faster.

SAMPLE PROBLEM 2.5 Significant Figures in Multiplication and Division

TRY IT FIRST

Perform the following calculations with measured numbers, and give the answers with the correct number of significant figures:

a. 56.8 * 0.37 b. (2.075) (0.585)

(8.42) (0.0245) c.

25.0 5.00

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

multiplication and division

answer with the correct number of SFs

rules for multiplication/ division

Two SFs Four SFs Three SFs

Calculator display

Answer, rounded off to two SFs

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2.3 Significant Figures in Calculations 37

Addition and Subtraction with Measured Numbers In addition or subtraction, the final answer is written so that it has the same number of decimal places as the measurement having the fewest decimal places.

2.045

+ 34.1 36.145

36.1

When numbers are added or subtracted to give an answer ending in zero, the zero does not appear after the decimal point in the calculator display. For example, 14.5 g - 2.5 g = 12.0 g. However, if you do the subtraction on your calculator, the display shows 12. To write the correct answer, a significant zero is written after the decimal point.

Thousandths place

Calculator display

Answer, rounded off to the tenths place

ENGAGE 2.6 Why is the answer 57.7 correct for the addition of 55.2 and 2.506?

Multiplication and Division with Measured Numbers In multiplication or division, the final answer is written so that it has the same number of significant figures (SFs) as the measurement with the fewest SFs.

Perform the following operations with measured numbers:

2.8 * 67.40 34.8

=

When the problem has multiple steps, the numbers in the numerator are multiplied and then divided by each of the numbers in the denominator.

2.8 * 67.40 , 34.8 = 5.422988506 = 5.4

Because the calculator display has more digits than the significant figures in the measured numbers allow, we need to round it off. Using the measured number that has the smallest number (two) of significant figures, 2.8, we round off the calculator display to an answer with two SFs.

Adding Significant Zeros Sometimes, a calculator display gives a small whole number. For example, suppose the calculator display is 4, but you used measurements that have three significant numbers. Then two significant zeros are added to give 4.00 as the correct answer.

8.00 2.00 = 4. = 4.00

PRACTICE PROBLEMS Try Practice Problems 2.25 to 2.28

CORE CHEMISTRY SKILL Using Significant Figures in

Calculations

ENGAGE 2.4 Why is the answer for the multiplication of 0.3 * 52.6 written with one significant figure?

PRACTICE PROBLEMS Try Practice Problems 2.29 and 2.30

ENGAGE 2.5 Why does the calculator answer of 5 in part c. require two significant zeros to give an answer of 5.00?

Three SFs

Three SFs Calculator display

Final answer, two zeros added to give three SFs

STEP 1 Determine the number of significant figures in each measured number.

a. 56.8 * 0.37 b. (2.075) (0.585)

(8.42) (0.0245) c.

25.0 5.00

STEP 2 Perform the indicated operations on the calculator.

a. 21.016 b. 5.884313345 c. 5.

STEP 3 Round off (or add zeros) to give the same number of significant figures as the measurement having the fewest significant figures.

a. 21 b. 5.88 c. 5.00

SELF TEST 2.5

Perform the following calculations with measured numbers, and give the answers with the correct number of significant figures:

a. 45.26 * 0.010 88 b. 2.6 , 324 c. 4.0 * 8.00

16

Three SFs Two SFs Four SFs Three SFs

Three SFs Three SFs

Three SFs

Three SFs

Calculator display

Calculator display

Calculator display

ANSWER

a. 0.4924 b. 0.0080 or 8.0 * 10-3 c. 2.0

Tenths place (fewer decimal places)

SAMPLE PROBLEM 2.6 Decimal Places in Addition and Subtraction

TRY IT FIRST

Perform the following calculations with measured numbers, and give each answer with the correct number of decimal places:

a. 104 + 7.8 + 40 b. 153.247 - 14.82

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

addition and subtraction

answer with the correct number of decimal places

rules for addition/ subtraction

STEP 1 Determine the number of decimal places in each measured number.

a. 104 7.8

+ 40

Ones place

Tenths place

Tens place (fewest decimal places)

b. 153.247 - 14.82

Thousandths place

Hundredths place

(fewer decimal places)

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38 CHAPTER 2 Chemistry and Measurements

ANSWER

a. 83.70 b. 0.5 PRACTICE PROBLEMS

Try Practice Problems 2.31 and 2.32

STEP 2 Perform the indicated operations on the calculator.

a. 151.8 b. 138.427

STEP 3 Round off (or add zeros) to give the same number of decimal places as the measured number having the fewest decimal places.

a. 150 b. 138.43

SELF TEST 2.6

Perform the following operations with measured numbers, and give each answer with the correct number of decimal places:

a. 82.45 + 1.245 + 0.000 56 b. 4.259 - 3.8

Rounded off to the tens place Rounded off to the hundredths place

Calculator display

Calculator display

PRACTICE PROBLEMS

2.3 Significant Figures in Calculations

2.25 Round off each of the following calculator answers to three significant figures:

a. 1.854 kg b. 88.2038 L c. 0.004 738 265 cm d. 8807 m e. 1.832 * 105 s 2.26 Round off each of the calculator answers in problem 2.25 to two

significant figures.

2.27 Round off or add zeros to each of the following to give three significant figures:

a. 56.855 m b. 0.002 282 g c. 11 527 s d. 8.1 L

2.28 Round off or add zeros to each of the following to give two significant figures:

a. 3.2805 m b. 1.855 * 102 g c. 0.002 341 mL d. 2 L

2.29 Perform each of the following operations, and give each answer with the correct number of significant figures:

a. 45.7 * 0.034 b. 0.002 78 * 5

c. 34.56 1.25

d. (0.2465) (25)

1.78

e. (2.8 * 104) (5.05 * 10-6) f. (3.45 * 10-2) (1.8 * 105)

(8 * 103) 2.30 Perform each of the following operations, and give each answer

with the correct number of significant figures: a. 400 * 185

b. 2.40

(4) (125)

c. 0.825 * 3.6 * 5.1

d. (3.5) (0.261)

(8.24) (20.0)

e. (5 * 10-5) (1.05 * 104)

(8.24 * 10-8)

f. (4.25 * 102) (2.56 * 10-3)

(2.245 * 10-3) (56.5)

2.31 Perform each of the following operations, and give each answer with the correct number of decimal places: a. 45.48 + 8.057 b. 23.45 + 104.1 + 0.025 c. 145.675 - 24.2 d. 1.08 - 0.585 e. 2300 + 196.11 f. 145.111 - 22.9 + 34.49

2.32 Perform each of the following operations, and give each answer with the correct number of decimal places: a. 5.08 + 25.1 b. 85.66 + 104.10 + 0.025 c. 24.568 - 14.25 d. 0.2654 - 0.2585 e. 66.77 + 17 - 0.33 f. 460 - 33.77

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2.4 Prefixes and Equalities 39

2.4 Prefixes and Equalities LEARNING GOAL Use the numerical values of prefixes to write a metric equality.

The special feature of the metric system is that a prefix can be placed in front of any unit to increase or decrease its size by some factor of 10. For example, the prefixes milli and micro are used to make the smaller units, milligram (mg) and microgram (mg).

The U.S. Food and Drug Administration (FDA) has determined the Daily Values (DV) for nutrients for adults and children age 4 or older. Examples of these recommended Daily Values, some of which use prefixes, are listed in TABLE 2.5.

The prefix centi is like cents in a dollar. One cent would be a “centidollar” or 0.01 of a dollar. That also means that one dollar is the same as 100 cents. The prefix deci is like dimes in a dollar. One dime would be a “decidollar” or 0.1 of a dollar. That also means that one dollar is the same as 10 dimes. TABLE 2.6 lists some of the metric prefixes, their symbols, and their numerical values.

(DV) for Selected Nutrients

Nutrient Amount Recommended

Calcium 1.0 g

Copper 2 mg

Iodine 150 mg (150 mcg)

Iron 18 mg

Magnesium 400 mg

Niacin 20 mg

Phosphorus 800 mg

Potassium 3.5 g

Selenium 70. mg (70. mcg)

Sodium 2.4 g

Zinc 15 mg

TABLE 2.5 Daily Values

TABLE 2.6 Metric and SI Prefixes

Prefix Symbol Numerical Value Scientific Notation Equality

Prefixes That Increase the Size of the Unit

peta P 1 000 000 000 000 000 1015 1 Pg = 1 * 1015 g 1 g = 1 * 10-15 Pg

tera T 1 000 000 000 000 1012 1 Ts = 1 * 1012 s 1 s = 1 * 10-12 Ts

giga G 1 000 000 000 109 1 Gm = 1 * 109 m 1 m = 1 * 10-9 Gm

mega M 1 000 000 106 1 Mg = 1 * 106 g 1 g = 1 * 10-6 Mg

kilo k 1 000 103 1 km = 1 * 103 m 1 m = 1 * 10-3 km

Prefixes That Decrease the Size of the Unit

deci d 0.1 10-1 1 dL = 1 * 10-1 L 1 L = 10 dL

centi c 0.01 10-2 1 cm = 1 * 10-2 m 1 m = 100 cm

milli m 0.001 10-3 1 ms = 1 * 10-3 s 1 s = 1 * 103 ms

micro m* 0.000 001 10-6 1 mg = 1 * 10-6 g 1 g = 1 * 106 mg

nano n 0.000 000 001 10-9 1 nm = 1 * 10-9 m 1 m = 1 * 109 nm

pico p 0.000 000 000 001 10-12 1 ps = 1 * 10-12 s 1 s = 1 * 1012 ps

femto f 0.000 000 000 000 001 10-15 1 fs = 1 * 10-15 s 1 s = 1 * 1015 fs

*In medicine, the abbreviation mc is used for the prefix micro because the symbol m may be misread, which could result in a medication error. Thus, 1 mg would be written as 1 mcg.

The relationship of a prefix to a unit can be expressed by replacing the prefix with its numerical value. For example, when the prefix kilo in kilometer is replaced with its value of 1000, we find that a kilometer is equal to 1000 m. Other examples follow:

1 kilometer (1 km) = 1000 meters (1000 m = 103 m)

1 kiloliter (1 kL) = 1000 liters (1000 L = 103 L)

1 kilogram (1 kg) = 1000 grams (1000 g = 103 g)

CORE CHEMISTRY SKILL Using Prefixes

PRACTICE PROBLEMS Try Practice Problems 2.33 to 2.40

ENGAGE 2.7 Why is 60. mg of vitamin C the same as 0.060 g of vitamin C?

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40 CHAPTER 2 Chemistry and Measurements

Equalities Between Prefixes Equalities can also be written for two prefix units. Suppose you need an equality for milliliters and deciliters. From Table 2.6, using L as the base unit, we can write the equalities 1 L = 10 dL and 1 L = 1000 mL and then write an equality for dL and mL.

1 L = 10 dL = 1000 mL or 1 dL = 100 mL

An endoscope has a video camera with a width of 1 mm attached to the end of a thin cable.

SAMPLE PROBLEM 2.7 Prefixes and Equalities

TRY IT FIRST

The camera in an endoscope has a width of 1 mm. Complete each of the following equalities involving millimeters:

a. 1 m = mm b. 1 cm = mm

SOLUTION

a. 1 m = 1000 mm b. 1 cm = 10 mm

SELF TEST 2.7

a. What is the relationship between liters and centiliters? b. What is the relationship between millimeters and micrometers? ANSWER

a. 1 L = 100 cL b. 1 mm = 1000 mm (mcm)

Measuring Length An ophthalmologist may measure the diameter of the retina of an eye in centimeters (cm), whereas a surgeon may need to know the length of a nerve in millimeters (mm). When the prefix centi is used with the unit meter, it becomes centimeter, a length that is one-hundredth of a meter (0.01 m). When the prefix milli is used with the unit meter, it becomes millimeter, a length that is one-thousandth of a meter (0.001 m). There are 100 cm and 1000 mm in a meter.

If we compare the lengths of a millimeter and a centimeter, we find that 1 mm is 0.1 cm; there are 10 mm in 1 cm. These comparisons are examples of equalities, which show the relationship between two units that measure the same quantity. Examples of equalities between different metric units of length follow:

1 m = 100 cm = 1 * 102 cm 1 m = 1000 mm = 1 * 103 mm

1 cm = 10 mm = 1 * 101 mm Some metric units for length are compared in FIGURE 2.6.

First quantity

unit+Number unit+Number

Second quantity

1 m 100 cm=

This example of an equality shows the relationship between meters and centimeters.

FIGURE 2.6 The metric length of 1 m is the same length as 10 dm, 100 cm, or 1000 mm.

10 20 30 40 50 60 70 80 90 100

m dm cm mm

1 10

100 1000

1 2 3 4

1 mm 1 cm = 10 mm

1 dm

ENGAGE 2.8 Why can the relationship of centimeters and meters be written as 1 m = 100 cm or 0.01 m = 1 cm?

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2.4 Prefixes and Equalities 41

When you see 1 cm, you are reading about length; when you see 1 cm3 or 1 cc or 1 mL, you are reading about volume. A comparison of units of volume is illustrated in FIGURE 2.8.

Substance in Blood Normal Range

Albumin 3.5–5.4 g/dL

Ammonia 20970 mg/dL (mcg/dL)

Calcium 8.5–10.5 mg/dL

Cholesterol 105–250 mg/dL

Iron (male) 809160 mg/dL (mcg/dL)

Protein (total) 6.0–8.5 g/dL

TABLE 2.7 Some Normal Laboratory Test Values

FIGURE 2.7 A plastic intravenous fluid container contains 1000 mL.

ENGAGE 2.9 Why is the volume 3.1 mL equal to the volume of 3.1 cm3?

Measuring Volume Volumes smaller than 1 L are common in the health sciences. When a liter is divided into 10 equal portions, each portion is a deciliter (dL). There are 10 dL in 1 L. Laboratory results for bloodwork are often reported in mass per deciliter. TABLE 2.7 lists normal laboratory test values for some substances in the blood.

When a liter is divided into a thousand parts, each of the smaller volumes is a milliliter (mL). In a 1-L container of physiological saline, there are 1000 mL of solution (see FIGURE 2.7). Examples of equalities between different metric units of volume follow:

1 L = 10 dL = 1 * 101 dL 1 L = 1000 mL = 1 * 103 mL

1 dL = 100 mL = 1 * 102 mL 1 mL = 1000 mL (mcL) = 1 * 103 mL (mcL) The cubic centimeter (abbreviated as cm3 or cc) is the volume of a cube whose

dimensions are 1 cm on each side. A cubic centimeter has the same volume as a milliliter, and the units are often used interchangeably.

1 cm3 = 1 cc = 1 mL

1

2

3

4

5 mL

1 cm

10 cm–=-1 dm

Volume–=-1 cm–*–1 cm–*–1 cm =-1 cm3–=-1 mL

Volume–=-10 cm–*–10 cm–*–10 cm =-1000 cm3 =-1000 mL =-1 L

10 c

m

10 cm

1.0 mL–= =-1.0 cc1.0 cm3

A cube measuring 10 cm on each side has a volume of 1000 cm3 or 1 L; a cube measuring 1 cm on each side has a volume of 1 cm3 (cc) or 1 mL.

FIGURE 2.8 Measurement of volume

A syringe contains 1.0 mL, which is the same as 1.0 cm3 and 1.0 cc.

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42 CHAPTER 2 Chemistry and Measurements

Measuring Mass When you go to the doctor for a physical examination, your mass is recorded in kilograms, whereas the results of your laboratory tests are reported in grams, milligrams (mg), or micrograms (mg or mcg). A kilogram is equal to 1000 g. One gram represents the same mass as 1000 mg, and one mg equals 1000 mg (or 1000 mcg). Examples of equalities between different metric units of mass follow:

1 kg = 1000 g = 1 * 103 g 1 g = 1000 mg = 1 * 103 mg

1 mg = 1000 mg (mcg) = 1 * 103 mg (mcg)

PRACTICE PROBLEMS Try Practice Problems 2.41 to 2.44

PRACTICE PROBLEMS

2.4 Prefixes and Equalities

2.33 Write the abbreviation for each of the following units: a. milligram b. deciliter c. kilometer d. picogram

2.34 Write the abbreviation for each of the following units: a. gigagram b. megameter c. microliter d. femtosecond

2.35 Write the complete name for each of the following units: a. cL b. kg c. ms d. Pm

2.36 Write the complete name for each of the following units: a. dL b. Ts c. mcg d. pm

2.37 Write the numerical value for each of the following prefixes: a. centi b. tera c. milli d. deci

2.38 Write the numerical value for each of the following prefixes: a. giga b. micro c. mega d. nano

2.39 Use a prefix to write the name for each of the following: a. 0.1 g b. 10-6 g c. 1000 g d. 0.01 g

2.40 Use a prefix to write the name for each of the following: a. 109 m b. 106 m c. 0.001 m d. 10-12 m

2.41 Complete each of the following metric relationships: a. 1 m = cm b. 1 m = nm c. 1 mm = m d. 1 L = mL 2.42 Complete each of the following metric relationships: a. 1 Mg = g b. 1 mL = mL c. 1 g = kg d. 1 g = mg 2.43 For each of the following pairs, which is the larger unit? a. milligram or kilogram b. milliliter or microliter c. m or km d. kL or dL e. nanometer or picometer

2.44 For each of the following pairs, which is the smaller unit? a. mg or g b. centimeter or nanometer c. millimeter or micrometer d. mL or dL e. centigram or megagram

2.5 Writing Conversion Factors LEARNING GOAL Write a conversion factor for two units that describe the same quantity.

Many problems in chemistry and the health sciences require you to change from one unit to another unit. Suppose you worked 2.0 h on your homework, and someone asked you how many minutes that was. You would answer 120 min. You must have multiplied 2.0 h * 60 min/h because you knew the equality (1 h = 60 min) that related the two units. When you expressed 2.0 h as 120 min, you changed only the unit of measurement used to express the time. Any equality can be written as fractions called conversion factors with one of the quantities in the numerator and the other quantity in the denominator. Two conversion factors are always possible from any equality. Be sure to include the units when you write the conversion factors.

Two Conversion Factors for the Equality: 1 h = 60 min

Numerator Denominator

h h

60 min

1 h and

1 h 60 min

These factors are read as “60 minutes per 1 hour” and “1 hour per 60 minutes.” The term per means “divide.” Some common relationships are given in TABLE 2.8.

The numbers in any definition between two metric units or between two U.S. system units are exact. Because numbers in a definition are exact, they are not used to determine

REVIEW Calculating Percentages (1.4)

CORE CHEMISTRY SKILL Writing Conversion Factors from

Equalities

PRACTICE PROBLEMS Try Practice Problems 2.45 and 2.46

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2.5 Writing Conversion Factors 43

significant figures. For example, the equality of 1 g = 1000 mg is a definition, which means that both of the numbers 1 and 1000 are exact.

When an equality consists of a metric unit and a U.S. unit, one of the numbers in the equality is obtained by measurement and counts toward the significant figures in the answer. For example, the equality of 1 lb = 453.6 g is obtained by measuring the grams in exactly 1 lb. In this equality, the measured quantity 453.6 g has four significant figures, whereas the 1 is exact. An exception is the relationship of 1 in. = 2.54 cm, which has been defined as exact.

Metric Conversion Factors We can write two metric conversion factors for any of the metric relationships. For example, from the equality for meters and centimeters, we can write the following factors:

Metric Equality Conversion Factors

1 m = 100 cm 100 cm 1 m

and 1 m

100 cm

Both are proper conversion factors for the relationship; one is just the inverse of the other. The usefulness of conversion factors is enhanced by the fact that we can turn a conversion factor over and use its inverse. The numbers 100 and 1 in this equality, and its conversion factors are both exact numbers.

Metric–U.S. System Conversion Factors Suppose you need to convert from pounds, a unit in the U.S. system, to kilograms in the metric system. A relationship you could use is

1 kg = 2.205 lb

The corresponding conversion factors would be

2.205 lb 1 kg

and 1 kg

2.205 lb

FIGURE 2.9 illustrates the contents of some packaged foods in both U.S. and metric units.

ENGAGE 2.10 Why does the equality 1 day = 24 h have two conversion factors?

Quantity Metric (SI) U.S. Metric–U.S.

Length 1 km = 1000 m 1 ft = 12 in. 2.54 cm = 1 in. (exact) 1 m = 1000 mm 1 yd = 3 ft 1 m = 39.37 in.

1 cm = 10 mm 1 mi = 5280 ft 1 m = 1.094 yd

1 km = 0.6214 mi

Volume 1 L = 1000 mL 1 qt = 4 cups 946.4 mL = 1 qt 1 dL = 100 mL 1 qt = 2 pt 1 L = 1.057 qt

1 mL = 1 cm3 1 gal = 4 qt 473.2 mL = 1 pt 1 mL = 1 cc* 3.785 L = 1 gal

5 mL = 1 tsp*

15 mL = 1 T (tbsp)*

Mass 1 kg = 1000 g 1 lb = 16 oz 1 kg = 2.205 lb 1 g = 1000 mg 453.6 g = 1 lb

1 mg = 1000 mcg* 28.35 g = 1 oz

Time 1 h = 60 min 1 h = 60 min

1 min = 60 s 1 min = 60 s

*Used in medicine.

TABLE 2.8 Some Common Equalities

FIGURE 2.9 In the United States, the contents of many packaged foods are listed in both U.S. and metric units.

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44 CHAPTER 2 Chemistry and Measurements

Equalities and Conversion Factors Stated Within a Problem An equality may also be stated within a problem that applies only to that problem. For example, the speed of a car in kilometers per hour or the price of onions in dollars per pound would be specific relationships for that problem only. From each of the following statements, we can write an equality and two conversion factors, and identify each number as exact or give the number of significant figures.

The car was traveling at a speed of 85 km/h.

Equality Conversion Factors Significant Figures or Exact

1 h = 85 km 85 km 1 h

and 1 h

85 km The 85 km is measured: It has two significant

figures. The 1 h is exact.

The price of onions is $1.24 per pound.

Equality Conversion Factors Significant Figures or Exact

1 lb = $1.24 $1.24 1 lb

and 1 lb

$1.24

The $1.24 is measured: It has three significant figures. The 1 lb is exact.

Conversion Factors from Dosage Problems Equalities stated within dosage problems for medications can also be written as conversion factors. Keflex (cephalexin), an antibiotic used for respiratory and ear infections, is available in 250-mg capsules. Vitamin C, an antioxidant, is available in 500-mg tablets. These dosage relationships can be used to write equalities from which two conversion factors can be derived.

One capsule contains 250 mg of Keflex.

Equality Conversion Factors Significant Figures or Exact

1 capsule = 250 mg of Keflex

250 mg Keflex

1 capsule and

1 capsule

250 mg Keflex

The 250 mg is measured: It has two significant figures. The 1 capsule is exact.

One tablet contains 500 mg of vitamin C.

Equality Conversion Factors Significant Figures or Exact

1 tablet = 500 mg of vitamin C

500 mg vitamin C

1 tablet and

1 tablet 500 mg vitamin C

The 500 mg is measured: It has one significant figure. The 1 tablet is exact.

Conversion Factors from a Percent, ppm, and ppb A percent (%) is written as a conversion factor by choosing a unit and expressing the numerical relationship of the parts of this unit to 100 parts of the whole. For example, a person might have 18% body fat by mass. The percent can be written as 18 mass units of body fat in every 100 mass units of body mass. Different mass units such as grams (g), kilograms (kg), or pounds (lb) can be used, but both units in the factor must be the same.

Equality Conversion Factors Significant Figures or Exact

100 kg of body mass = 18 kg of body fat

18 kg body fat

100 kg body mass and

100 kg body mass

18 kg body fat

The 18 kg is measured: It has two significant figures. The 100 kg is exact.

ENGAGE 2.11 How is a percent used to write an equality and two conversion factors?

Keflex (cephalexin), used to treat respiratory infections, is available in 250-mg capsules.

The thickness of the skin fold at the abdomen is used to determine the percentage of body fat.

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2.5 Writing Conversion Factors 45

When scientists want to indicate very small ratios, they use numerical relationships called parts per million (ppm) or parts per billion (ppb). The ratio of parts per million is the same as the milligrams of a substance per kilogram (mg/kg). The ratio of parts per billion equals the micrograms per kilogram (mg/kg, mcg/kg).

Ratio Units

parts per million (ppm) milligrams per kilogram (mg/kg)

parts per billion (ppb) micrograms per kilogram (mg/kg, mcg/kg)

For example, the maximum amount of lead that is allowed by the FDA in glazed pottery bowls is 2 ppm, which is 2 mg/kg.

Equality Conversion Factors Significant Figures or Exact

1 kg of glaze = 2 mg of lead

2 mg lead

1 kg glaze and

1 kg glaze

2 mg lead

The 2 mg is measured: It has one significant figure. The 1 kg is exact.

Conversion Factors with Powers Sometimes we use a conversion factor that is squared or cubed. This is the case when we need to calculate an area or a volume.

Distance = length

Area = length * length = length2

Volume = length * length * length = length3

Suppose you want to write the equality and the conversion factors for the relationship between an area in square centimeters and in square meters. To square the equality 1 m = 100 cm, we square both the number and the unit on each side.

Equality: 1 m = 100 cm

Area: (1 m)2 = (100 cm)2 or 1 m2 = (100)2 cm2

From the new equality, we can write two conversion factors as follows:

Conversion factors: (100 cm)2

(1 m)2 and

(1 m)2

(100 cm)2

In the following example, we show that the equality 1 in. = 2.54 cm can be squared to give area or can be cubed to give volume. Both the number and the unit must be squared or cubed.

Measurement Equality Conversion Factors

Length 1 in. = 2.54 cm 2.54 cm 1 in.

and 1 in.

2.54 cm

Area (1 in.)2 = (2.54 cm)2

(1 in.)2 = (2.54)2 cm2 = 6.45 cm2 (2.54 cm)2

(1 in.)2 and

(1 in.)2

(2.54 cm)2

Volume (1 in.)3 = (2.54 cm)3

(1 in.)3 = (2.54)3 cm3 = 16.4 cm3 (2.54 cm)3

(1 in.)3 and

(1 in.)3

(2.54 cm)3

SAMPLE PROBLEM 2.8 Equalities and Conversion Factors in a Problem

TRY IT FIRST

Write the equality and two conversion factors, and identify each number as exact or give the number of significant figures for each of the following:

a. The medication that Greg takes for his high blood pressure contains 40. mg of propranolol in 1 tablet.

b. Cold-water fish such as salmon contains 1.9% omega-3 fatty acids by mass.

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46 CHAPTER 2 Chemistry and Measurements

c. The U.S. Environmental Protection Agency (EPA) has set the maximum level for mercury in tuna at 0.5 ppm.

d. The area of (1 cm)2 is the same as (10 mm)2.

SOLUTION

a. Equality Conversion Factors Significant Figures or Exact 1 tablet = 40. mg of

propranolol 40. mg propranolol

1 tablet

and

1 tablet 40. mg propranolol

The 40. mg is measured: It has two significant figures. The 1 tablet is exact.

b. Equality Conversion Factors Significant Figures or Exact 100 g of salmon = 1.9 g

of omega-3 fatty acids 1.9 g omega@3 fatty acids

100 g salmon

and 100 g salmon

1.9 g omega@3 fatty acids

The 1.9 g is measured: It has two significant figures. The 100 g is exact.

c. Equality Conversion Factors Significant Figures or Exact 1 kg of tuna = 0.5 mg of

mercury 0.5 mg mercury

1 kg tuna

and 1 kg tuna

0.5 mg mercury

The 0.5 mg is measured: It has one significant figure. The 1 kg is exact.

d. Equality Conversion Factors Significant Figures or Exact

(1 cm)2 = (10 mm)2 (10 mm)2

(1 cm)2 and

(1 cm)2

(10 mm)2 Both 1 cm and 10 mm are exact,

by definition.

SELF TEST 2.8

Write the equality and its corresponding conversion factors, and identify each number as exact or give the number of significant figures for each of the following:

a. Levsin (hyoscyamine), used to treat stomach and bladder problems, is available as drops with 0.125 mg of Levsin per 1 mL of solution.

b. The EPA has set the maximum level of cadmium in rice as 0.4 ppm.

Propranolol is used to lower high blood pressure.

Salmon contains high levels of omega-3 fatty acids.

The maximum amount of mercury allowed in tuna is 0.5 ppm.

ANSWER

a. 1 mL of solution = 0.125 mg of Levsin

0.125 mg Levsin

1 mL solution and

1 mL solution 0.125 mg Levsin

The 0.125 mg is measured: It has three significant figures. The 1 mL is exact.

b. 1 kg of rice = 0.4 mg of cadmium

0.4 mg cadmium

1 kg rice and

1 kg rice

0.4 mg cadmium

The 0.4 mg is measured: It has one significant figure. The 1 kg is exact. PRACTICE PROBLEMS

Try Practice Problems 2.47 to 2.56

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2.6 Problem Solving Using Unit Conversion 47

PRACTICE PROBLEMS

2.5 Writing Conversion Factors

2.45 Why can two conversion factors be written for an equality such as 1 m = 100 cm?

2.46 How can you check that you have written the correct conversion factors for an equality?

2.47 Write the equality and two conversion factors for each of the following pairs of units: a. milligrams and kilograms b. nanograms and grams c. liters and kiloliters d. seconds and milliseconds e. millimeters and decimeters f. cubic meters and cubic centimeters

2.48 Write the equality and two conversion factors for each of the following pairs of units: a. centimeters and inches b. kilometers and miles c. pounds and grams d. liters and deciliters e. grams and picograms f. square centimeters and square inches

2.49 Write the equality and two conversion factors, and identify the numbers as exact or give the number of significant figures for each of the following: a. One yard is 3 ft. b. One kilogram is 2.205 lb. c. A car goes 27 mi on 1 gal of gas. d. Sterling silver is 93% silver by mass. e. One minute is 60 s. f. One cubic meter is 1 * 106 cm3.

2.50 Write the equality and two conversion factors, and identify the numbers as exact or give the number of significant figures for each of the following: a. One liter is 1.057 qt. b. At the store, oranges are $1.29 per lb. c. There are 7 days in 1 week. d. One deciliter contains 100 mL. e. An 18-carat gold ring contains 75% gold by mass. f. One square foot is 144 in.2

2.51 Write the equality and two conversion factors, and identify the numbers as exact or give the number of significant figures for each of the following: a. A bee flies at an average speed of 3.5 m per second. b. The Daily Value (DV) for potassium is 3.5 g.

c. An automobile traveled 26.0 km on 1 L of gasoline. d. The pesticide level in plums was 29 ppb. e. Silicon makes up 28.2% by mass of Earth’s crust.

2.52 Write the equality and two conversion factors, and identify the numbers as exact or give the number of significant figures for each of the following: a. The Daily Value (DV) for iodine is 150 mcg. b. The nitrate level in well water was 32 ppm. c. Gold jewelry contains 58% gold by mass. d. The price of a liter of milk is $1.65. e. A metric ton is 1000 kg.

Applications

2.53 Write the equality and two conversion factors, and identify the numbers as exact or give the number of significant figures for each of the following: a. A calcium supplement contains 630 mg of calcium per

tablet. b. The Daily Value (DV) for vitamin C is 60 mg. c. The label on a bottle reads 50 mg of atenolol per tablet. d. A low-dose aspirin contains 81 mg of aspirin per tablet.

2.54 Write the equality and two conversion factors, and identify the numbers as exact or give the number of significant figures for each of the following: a. The label on a bottle reads 10 mg of furosemide per 1 mL. b. The Daily Value (DV) for selenium is 70. mcg. c. An IV of normal saline solution has a flow rate of 85 mL per

hour. d. One capsule of fish oil contains 360 mg of omega-3 fatty

acids.

2.55 Write an equality and two conversion factors for each of the following medications: a. 10 mg of Atarax per 5 mL of Atarax syrup b. 0.25 g of Lanoxin per 1 tablet of Lanoxin c. 300 mg of Motrin per 1 tablet of Motrin

2.56 Write an equality and two conversion factors for each of the following medications: a. 2.5 mg of Coumadin per 1 tablet of Coumadin b. 100 mg of Clozapine per 1 tablet of Clozapine c. 1.5 g of Cefuroxime per 1 mL of Cefuroxime

2.6 Problem Solving Using Unit Conversion LEARNING GOAL Use conversion factors to change from one unit to another.

The process of problem solving in chemistry often requires one or more conversion factors to change a given unit to the needed unit. For the problem, the unit of the given and the unit of the needed are identified. From there, the problem is set up with one or more conversion factors used to convert the given unit to the needed unit as seen in Sample Problem 2.9.

Given unit * one or more conversion factors = needed unit

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48 CHAPTER 2 Chemistry and Measurements

SAMPLE PROBLEM 2.9 Using Conversion Factors

TRY IT FIRST

Greg’s doctor has ordered a PET scan of his heart. In radiological imaging, dosages of pharmaceuticals are based on body mass. If Greg weighs 164 lb, what is his body mass in kilograms?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

164 lb kilograms U.S.–metric conversion factor

STEP 2 Write a plan to convert the given unit to the needed unit.

2.205 lb and

1 kg = 2.205 lb

1 kg

1 kg

2.205 lb

164 lb 74.4 kg=* 1 kg

2.205 lb Given Conversion factor

to cancel given unit

Unit for answer goes here

Answer

pounds kilograms U.S.– metric factor

STEP 3 State the equalities and conversion factors.

STEP 4 Set up the problem to cancel units and calculate the answer. Write the given, 164 lb, and multiply by the conversion factor that has lb in the denominator (bottom number) to cancel lb in the given.

ENGAGE 2.12 Why would you use the conversion

factor 1 lb

16 oz , not

16 oz 1 lb

, to convert

6.4 oz to pounds?

The given unit lb cancels out, and the needed unit kg is in the numerator. The unit you want in the final answer is the one that remains after all the other units have canceled out. This is a helpful way to check that you set up a problem properly.

lb * kg

lb = kg

The calculator display gives the numerical answer, which is adjusted to give a final answer with the proper number of significant figures (SFs). The value of 74.4 combined with the unit, kg, gives the final answer of 74.4 kg.

Unit needed for answer

2.205 2.205 7 4 . 3 7 6 4 1 7 2 3

1 164 74.4

Calculator display

Three SFs Four SFs

Exact

Three SFs (rounded off)

== =,164 * INTERACTIVE VIDEO

Conversion Factors

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2.6 Problem Solving Using Unit Conversion 49

SELF TEST 2.9

a. A total of 2500 mL of a boric acid antiseptic solution is prepared from boric acid concentrate. How many quarts of boric acid have been prepared?

b. A vial contains 65 mg of phenobarbital per 1 mL of solution. How many milliliters are needed for an order of 40. mg of phenobarbital?

PRACTICE PROBLEMS Try Practice Problems 2.57 to 2.62

ANSWER

a. 2.6 qt b. 0.62 mL

Using Two or More Conversion Factors In problem solving, two or more conversion factors are often needed to complete the change of units. In setting up these problems, one factor follows the other. Each factor is arranged to cancel the preceding unit until the needed unit is obtained. Once the problem is set up to cancel units properly, the calculations can be done without writing intermediate results. In this text, when two or more conversion factors are required, the final answer will be based on obtaining a final calculator display and rounding off (or adding zeros) to give the correct number of significant figures as shown in Sample Problem 2.10.

CORE CHEMISTRY SKILL Using Conversion Factors

ENGAGE 2.13 How are two conversion factors used in a problem setup?

SAMPLE PROBLEM 2.10 Using Two Conversion Factors

TRY IT FIRST

Greg has been diagnosed with diminished thyroid function. His doctor prescribes a dosage of 0.150 mg of Synthroid to be taken once a day. If tablets in stock contain 75 mcg of Synthroid, how many tablets are required to provide the prescribed medication?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

0.150 mg of Synthroid

number of tablets

metric conversion factor, clinical conversion factor

STEP 2 Write a plan to convert the given unit to the needed unit.

2 tablets= 1 tablet

75 mcg Synthroid *

Exact

Two SFs

0.150 mg Synthroid * 1000 mcg

1 mg

Three SFs Exact

Exact

milligrams number of tabletsmicrograms Metric factor

Clinical factor

STEP 3 State the equalities and conversion factors.

1 mg and

1 mg = 1000 mcg

1000 mcg

1000 mcg

1 mg

1 tablet and

1 tablet = 75 mcg of Synthroid

75 mcg Synthroid

75 mcg Synthroid

1 tablet

STEP 4 Set up the problem to cancel units and calculate the answer. The problem can be set up using the metric factor to cancel milligrams, and then the clinical factor to obtain the number of tablets as the final unit.

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50 CHAPTER 2 Chemistry and Measurements

ANSWER

a. 6 days b. 2.5 mL

SELF TEST 2.10

a. A bottle contains 120 mL of cough syrup. If one teaspoon (5 mL) is given four times a day, how many days will elapse before a refill is needed?

b. A patient is given a solution containing 0.625 g of calcium carbonate. If the calcium carbonate solution contains 1250 mg per 5 mL, how many milliliters of the solution were given to the patient?

One teaspoon of cough syrup is measured for a patient.

SAMPLE PROBLEM 2.11 Using a Percent as a Conversion Factor

TRY IT FIRST

A person who exercises regularly has 16% body fat by mass. If this person weighs 155 lb, what is the mass, in kilograms, of body fat?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

155 lb body weight kilograms of body fat

U.S.–metric conversion factor, percent conversion factor

STEP 2 Write a plan to convert the given unit to the needed unit.

Exercising regularly helps reduce body fat.

ANSWER

a. 160 g of fat b. 480 mg

2.205 lb body weight and

1 kg of body mass = 2.205 lb of body weight

1 kg body mass

1 kg body mass

2.205 lb body weight

16 kg body fat and

100 kg of body mass = 16 kg of body fat

100 kg body mass

100 kg body mass

16 kg body fat

pounds of body weight

kilograms of body mass

kilograms of body fat

U.S.–metric factor

Percent factor

155 lb body weight 11 kg of body fat* = 1 kg body mass

2.205 lb body weight

16 kg body fat

100 kg body mass *

Three SFs Four SFs Two SFsExact

Two SFsExact

STEP 3 State the equalities and conversion factors.

STEP 4 Set up the problem to cancel units and calculate the answer.

For problems that use several conversion factors, other plans may be possible. For exam- ple, for this problem, the pounds of body weight could be converted to pounds of body fat using the percent factor, which can be converted to kilograms of body fat using the metric–U.S. factor for kilograms and pounds.

SELF TEST 2.11

a. A package contains 2.4 lb of ground round. If it contains 15% fat, how many grams of fat are in the ground round?

b. A cream contains 4.0% (by mass) lidocaine to relieve back pain. If a patient uses 12 g of cream, how many milligrams of lidocaine are used?

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2.6 Problem Solving Using Unit Conversion 51

Conversion of Units in a Fraction Sometime units in an equality are written as a fraction that relates the two units. For example, we measure fuel consumption in a car in units of miles per gallon. Perhaps, the mileage for a car is 39.0 mi per gal (39.0 mi/gal). Someone driving a car in Europe would express the use of fuel in kilometers per liter (km/L). If we want to express mi/gal in units of km/L, we one use conversion factor to change miles to kilometers, and another to change gallons to liters as shown in Sample Problem 2.12.

SAMPLE PROBLEM 2.12 Conversion of Units in a Fraction

TRY IT FIRST

Your new car has a fuel consumption of 39.0 mi/gal. Convert this fuel consumption to km/L.

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

39.0 mi 1 gal

km L

U.S.–metric conversion factors, km/mi, L/gal

ANSWER

a. 3 * 10-5 oz/lb b. 110 mi/h PRACTICE PROBLEMS

Try Practice Problems 2.63 to 2.68

0.6214 mi =

1 km = 0.6214 mi

1 km

1 km

0.6214 mi

3.785 L =

1 gal = 3.785 L

1 gal

1 gal

3.785 L

miles gallon

kilometers gallon

kilometers liter

U.S.–metric factor

U.S.–metric factor

* = 1 km

0.6214 mi

39.0 mi

1 gal

1 gal

3.785 L

16.6 km

1 L *

Exact

Three SFs

Four SFs Exact

Three SFs

Four SFs

ExactExact

STEP 2 Write a plan to convert the given unit to the needed unit.

STEP 3 State the equalities and conversion factors.

STEP 4 Set up the problem to cancel units and calculate the answer.

The problem can be set up using the U.S.–metric factor to convert miles to kilometers, and another U.S.–metric factor to convert gallons to liters to obtain the fraction km/L.

SELF TEST 2.12

Convert each of the following fractions to the indicated units:

a. 2 mg/kg to oz/lb b. 50. m/s to mi/h

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52 CHAPTER 2 Chemistry and Measurements

Chemistry Link to Health Toxicology and Risk–Benefit Assessment

Each day, we make choices about what we do or what we eat, often without thinking about the risks associated with these choices. We are aware of the risks of cancer from smoking or the risks of lead poisoning, and we know there is a greater risk of having an accident if we cross a street where there is no light or crosswalk.

A basic concept of toxicology is the statement of Paracelsus that the dose is the difference between a poison and a cure. To evaluate the level of danger from various substances, natural or synthetic, a risk assessment is made by exposing laboratory animals to the sub- stances and monitoring the health effects. Often, doses very much greater than humans might ordinarily encounter are given to the test animals.

Many hazardous chemicals or substances have been identified by these tests. One measure of toxicity is the LD50, or lethal dose, which is the con- centration of the substance that causes death in 50% of the test animals. A dosage is typically measured in milligrams per kilogram (mg/kg) of body mass or micrograms per kilogram (mcg/kg) of body mass.

Other evaluations need to be made, but it is easy to compare LD50 values. Parathion, a pesticide, with an LD50 of 3 mg/kg, would be highly toxic. This means that 3 mg of parathion per kg of body mass would be fatal to half the test animals. Table salt (sodium chloride) with an LD50 of 3300 mg/kg would have a much lower toxicity. You would need to ingest a huge amount of salt before any toxic effect

would be observed. Although the risk to animals can be evaluated in the laboratory, it is more difficult to determine the impact in the envi- ronment since there is also a difference between continued exposure and a single, large dose of the substance.

TABLE 2.9 lists some LD50 values and compares substances in order of increasing toxicity.

Substance LD50 (mg/kg)

Table sugar 29 700

Boric acid 5140

Baking soda 4220

Table salt 3300

Ethanol 2080

Aspirin 1100

Codeine 800

Oxycodone 480

Caffeine 192

DDT 113

Cocaine (injected) 95

Dichlorvos (pesticide strips) 56

Ricin 30

Sodium cyanide 6

Parathion 3

TABLE 2.9 Some LD50 Values for Substances Tested in Rats

The LD50 of caffeine is 192 mg/kg.

PRACTICE PROBLEMS

2.6 Problem Solving Using Unit Conversion

2.57 Perform each of the following conversions using metric conversion factors:

a. 44.2 mL to liters b. 8.65 m to nanometers c. 5.2 * 108 g to megagrams d. 0.72 ks to milliseconds

2.58 Perform each of the following conversions using metric conversion factors:

a. 4.82 * 10-5 L to picoliters b. 575.2 dm to kilometers c. 5 * 10-4 kg to micrograms d. 6.4 * 1010 ps to seconds

2.59 Perform each of the following conversions using metric and U.S. conversion factors:

a. 3.428 lb to kilograms b. 1.6 m to inches c. 4.2 L to quarts d. 0.672 ft to millimeters

2.60 Perform each of the following conversions using metric and U.S. conversion factors:

a. 0.21 lb to grams b. 11.6 in. to centimeters c. 0.15 qt to milliliters d. 35.41 kg to pounds

2.61 Use metric conversion factors to solve each of the following problems: a. If a student is 175 cm tall, how tall is the student in

meters? b. A cooler has a volume of 5000 mL. What is the capacity of

the cooler in liters? c. A hummingbird has a mass of 0.0055 kg. What is the mass,

in grams, of the hummingbird? d. A balloon has a volume of 3500 cm3. What is the volume in

liters?

2.62 Use metric conversion factors to solve each of the following problems: a. The Daily Value (DV) for phosphorus is 800 mg. How many

grams of phosphorus are recommended? b. A glass of orange juice contains 3.2 dL of juice. How many

milliliters of orange juice are in the glass? c. A package of chocolate instant pudding contains 2840 mg of

sodium. How many grams of sodium are in the pudding? d. A jar contains 0.29 kg of olives. How many grams of olives

are in the jar?

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2.7 Density 53

2.63 Solve each of the following problems using one or more conversion factors: a. A container holds 0.500 qt of liquid. How many milliliters of

lemonade will it hold? b. What is the mass, in kilograms, of a person who weighs 175 lb? c. An athlete has 15% body fat by mass. What is the weight of

fat, in pounds, of a 74-kg athlete? d. A plant fertilizer contains 15% nitrogen (N) by mass. In a

container of soluble plant food, there are 10.0 oz of fertilizer. How many grams of nitrogen are in the container?

2.64 Solve each of the following problems using one or more conversion factors: a. Wine is 12% alcohol by volume. How many milliliters of

alcohol are in a 0.750-L bottle of wine? b. Blueberry high-fiber muffins contain 51% dietary fiber by

mass. If a package with a net weight of 12 oz contains six muffins, how many grams of fiber are in each muffin?

c. A jar of crunchy peanut butter contains 1.43 kg of peanut butter. If you use 8.0% of the peanut butter for a sandwich, how many ounces of peanut butter did you take out of the container?

d. In a candy factory, the nutty chocolate bars contain 22.0% pecans by mass. If 5.0 kg of pecans were used for candy last Tuesday, how many pounds of nutty chocolate bars were made?

Applications

2.65 Using conversion factors, solve each of the following problems: a. You have used 250 L of distilled water for a dialysis patient.

How many gallons of water is that? b. A patient needs 0.024 g of a sulfa drug. There are 8-mg

tablets in stock. How many tablets should be given? c. The daily dose of ampicillin for the treatment of an ear

infection is 115 mg/kg of body mass. What is the daily dose for a 34-lb child?

d. You need 4.0 oz of a steroid ointment. How many grams of ointment does the pharmacist need to prepare?

2.66 Using conversion factors, solve each of the following problems: a. The physician has ordered 1.0 g of tetracycline to be given

every six hours to a patient. If your stock on hand is 500-mg tablets, how many will you need for one day’s treatment?

b. An intramuscular medication is given at 5.00 mg/kg of body mass. What is the dose for a 180-lb patient?

c. A physician has ordered 0.50 mg of atropine, intramuscularly. If atropine were available as 0.10 mg/mL of solution, how many milliliters would you need to give?

d. During surgery, a patient receives 5.0 pt of plasma. How many milliliters of plasma were given?

2.67 Using conversion factors, solve each of the following problems: a. A nurse practitioner prepares 500. mL of an IV of normal

saline solution to be delivered at a rate of 80. mL/h. What is the infusion time, in hours, to deliver 500. mL?

b. A nurse practitioner orders Medrol to be given 1.5 mg/kg of body mass. i. What is the dosage of Medrol in g/lb of body weight?

ii. If a child weighs 72.6 lb and the available stock of Medrol is 20. mg/mL, how many milliliters does the nurse administer to the child?

c. A dosage for a patient is 25.0 mg of a drug/kg of body mass. i. What is that dosage in oz of drug/lb of body weight?

ii. What is the dose, in ounces, required for a 164-lb patient?

2.68 Using conversion factors, solve each of the following problems: a. A nurse practitioner prepares an injection of promethazine,

an antihistamine used to treat allergic rhinitis. If the stock bottle is labeled 25 mg/mL and the order is a dose of 12.5 mg, how many milliliters will the nurse draw up in the syringe?

b. An order for ampicillin is 25 mg/kg of body mass. i. What is the dosage of ampicillin in g/lb of body

weight? ii. If stock on hand is 250 mg/capsule, how many capsules

should be given to a 67-lb child? c. The dosage of an asthma drug is 5 mg/kg of body mass.

i. What is that dosage in g/lb of body weight? ii. What is the dose, in milligrams, required for a 48-lb

child?

2.7 Density LEARNING GOAL Calculate the density of a substance; use the density to calculate the mass or volume of a substance.

The mass and volume of any object can be measured. If we compare the mass of the object to its volume, we obtain a relationship called density.

Density = mass of substance

volume of substance

Every substance has a unique density, which distinguishes it from other substances. For example, lead has a density of 11.3 g/mL, whereas cork has a density of 0.26 g/mL. From these densities, we can predict if these substances will sink or float in water. If an object is less dense than a liquid, the object floats when placed in the liquid. If a substance, such as cork, is less dense than water, it will float. However, a lead object sinks because its density is greater than that of water (see FIGURE 2.10).

ENGAGE 2.14 If a piece of iron sinks in water, how does its density compare to that of water?

Agricultural fertilizers applied to a field provide nitrogen for plant growth.

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54 CHAPTER 2 Chemistry and Measurements

FIGURE 2.10 Objects that sink in water are more dense than water; objects that float are less dense.

Cork (D = 0.26 g/mL)

Ice (D = 0.92 g/mL)

Water (D = 1.000 g/mL)

Aluminum (D = 2.70 g/mL)

Lead (D = 11.3 g/mL)

Solids (at 25 °C)

Density (g/cm3)

Liquids (at 25 °C)

Density (g/mL)

Gases (at 0 °C)

Density (g/L)

Cork  0.26 Gasoline  0.74 Hydrogen 0.090

Body fat  0.909 Ethanol  0.79 Helium 0.179

Ice (at 0 °C)  0.92 Olive oil  0.92 Methane 0.714

Muscle  1.06 Water (at 4 °C) 1.000 Neon 0.902

Sugar  1.59 Urine 1.003 - 1.030 Nitrogen 1.25 Bone  1.80 Plasma (blood)  1.03 Air (dry) 1.29

Salt (NaCl)  2.16 Milk  1.04 Oxygen 1.43

Aluminum  2.70 Blood  1.06 Carbon dioxide 1.96

Iron  7.86 Mercury 13.6

Copper  8.92

Silver 10.5

Lead 11.3

Gold 19.3

TABLE 2.10 Densities of Some Common Substances

Calculating Density We can calculate the density of a substance from its mass and volume as shown in Sample Problem 2.13.

SAMPLE PROBLEM 2.13 Calculating Density

TRY IT FIRST

High-density lipoprotein (HDL) is a type of cholesterol, sometimes called “good cholesterol,” that is measured in a routine blood test. If a 0.258-g sample of HDL has a volume of 0.215 mL, what is the density, in grams per milliliter, of the HDL sample?

Density is used in chemistry in many ways. If we calculate the density of a pure metal as 10.5 g/mL, then we could identify it as silver, but not gold or aluminum. Metals such as gold and silver have higher densities, whereas gases have low densities. In the metric sys- tem, the densities of solids and liquids are usually expressed as grams per cubic centimeter (g/cm3) or grams per milliliter (g/mL). The densities of gases are usually stated as grams per liter (g/L). TABLE 2.10 gives the densities of some common substances.

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2.7 Density 55

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

0.258 g of HDL, 0.215 mL

density (g/mL) of HDL density expression

STEP 2 Write the density expression.

ANSWER

a. 1.05 g/mL b. 1.54 g/cm3

Density 0.258 g

0.215 mL

1.20 g

1 mL

Three SFs

Three SFs

= = 1.20 g/mL

Three SFs

=

mass of substance

volume of substance Density =

STEP 3 Express mass in grams and volume in milliliters.

Mass of HDL sample = 0.258 g

Volume of HDL sample = 0.215 mL

STEP 4 Substitute mass and volume into the density expression and calculate the density.

SELF TEST 2.13

a. Low-density lipoprotein (LDL), sometimes called “bad cholesterol,” is also measured in a routine blood test. If a 0.380-g sample of LDL has a volume of 0.362 mL, what is the density, in grams per milliliter, of the LDL sample?

b. Osteoporosis is a condition in which bone deteriorates to cause a decreased bone mass. If a bone sample has a mass of 2.15 g and a volume of 1.40 cm3, what is its density in grams per cubic centimeter?

Density of Solids Using Volume Displacement The volume of a solid can be determined by volume displacement. When a solid is com- pletely submerged in water, it displaces a volume that is equal to the volume of the solid. In FIGURE 2.11, the water level rises from 35.5 mL to 45.0 mL after the zinc object is added. This means that 9.5 mL of water is displaced and that the volume of the object is 9.5 mL.

The density of the zinc is calculated using volume displacement as follows:

Density = 68.60 g Zn

9.5 mL = 7.2 g/mL

Problem Solving Using Density Density can be used as a conversion factor. For example, if the volume and the density of a sample are known, the mass in grams of the sample can be calculated as shown in Sample Problem 2.14.

Two SFs

Four SFs

Two SFs

CORE CHEMISTRY SKILL Using Density as a Conversion

Factor

PRACTICE PROBLEMS Try Practice Problems 2.69 to 2.72

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56 CHAPTER 2 Chemistry and Measurements

FIGURE 2.11 Calculation of the density of a solid

Submerged zinc object

45.0 mL

Volume increase

35.5 mL

The mass of a zinc object is determined.

The volume of the zinc object is determined by volume displacement.

SAMPLE PROBLEM 2.14 Problem Solving Using Density

TRY IT FIRST

Greg has a blood volume of 5.9 qt. If the density of blood is 1.06 g/mL, what is the mass, in grams, of Greg’s blood?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

5.9 qt of blood grams of blood U.S.–metric conversion factor, density conversion factor

Chemistry Link to Health Bone Density

Our bones’ density is a measure of their health and strength. Our bones are constantly gaining and losing calcium, magnesium, and phosphate. In childhood, bones form at a faster rate than they break down. As we age, bone breakdown occurs more rapidly than new bone forms. As bone loss increases, bones begin to thin, causing a decrease in mass and density. Thinner bones lack strength, which increases the risk of fracture. Hormonal changes, disease, and certain medications can also contribute to the bone thinning. Eventually, a condition of severe bone thinning known as osteoporosis, may occur. Scanning electron micrographs (SEMs) show (a) normal bone and (b) bone with osteoporosis due to loss of bone minerals.

Bone density is often determined by passing low-dose X-rays through the narrow part at the top of the femur (hip) and the spine (c). These locations are where fractures are more likely to occur, especially as we age. Bones with high density will block more of the X-rays compared to bones that are less dense. The results of a bone density test are compared to a healthy young adult as well as to other people of the same age.

Recommendations to improve bone strength include calcium and vitamin D supplements. Weight-bearing exercise such as walking and lifting weights can also improve muscle strength, which in turn increases bone strength.

(a) Normal bone (b) Bone with osteoporosis (c) Viewing a low-dose X-ray of the spine

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2.7 Density 57

1 pt of blood contains 473.2 mL.

ANSWER

a. 1.5 kg b. 1.21 L

946.4 mL and

1 qt = 946.4 mL

1 qt

1 qt

946.4 mL

1 mL blood and

1 mL of blood = 1.06 g of blood

1.06 g blood

1.06 g blood

1 mL blood

quarts gramsmilliliters U.S.–metric factor

Density factor

5.9 qt blood 5900 g of blood* = 1 qt

1.06 g blood

1 mL blood *

Two SFs Exact Two SFsExact

Three SFsFour SFs

946.4 mL

STEP 2 Write a plan to calculate the needed quantity.

STEP 3 Write the equalities and their conversion factors including density.

STEP 4 Set up the problem to calculate the needed quantity.

SELF TEST 2.14

a. During surgery, a patient receives 3.0 pt of blood. How many kilograms of blood (density = 1.06 g/mL) were needed for the transfusion?

b. A woman receives 1280 g of type A blood. If the blood has a density of 1.06 g/mL, how many liters of blood did she receive?

PRACTICE PROBLEMS Try Practice Problems 2.73 to 2.78

Specific Gravity Specific gravity (sp gr) is a relationship between the density of a substance and the density of water. Specific gravity is calculated by dividing the density of a sample by the density of water, which is 1.000 g/mL at 4 °C. A substance with a specific gravity of 1.000 has the same density as water (1.000 g/mL).

Specific gravity = density of sample

density of water

Specific gravity is one of the few unitless values you will encounter in chemistry. The spe- cific gravity of urine helps evaluate the water balance in the body and the substances in the urine. In FIGURE 2.12, a hydrometer is used to measure the specific gravity of urine. The normal range of specific gravity for urine is 1.003 to 1.030. The specific gravity can decrease with type 2 diabetes and kidney disease. Increased specific gravity may occur with dehydration, kidney infection, and liver disease. In a clinic or hospital, a dipstick containing chemical pads is used to evaluate specific gravity.

FIGURE 2.12 A hydrometer is used to measure the specific gravity of urine, which, for adults, is 1.003 to 1.030.

1.000

1.010

A dipstick is used to measure the specific gravity of a urine sample.

PRACTICE PROBLEMS Try Practice Problems 2.79 and 2.80

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58 CHAPTER 2 Chemistry and Measurements

PRACTICE PROBLEMS

2.7 Density

2.69 Determine the density (g/mL) for each of the following: a. A 20.0-mL sample of a salt solution has a mass of 24.0 g. b. A cube of butter weighs 0.250 lb and has a volume of

130.3 mL. c. A gem has a mass of 4.50 g. When the gem is placed in a

graduated cylinder containing 12.00 mL of water, the water level rises to 13.45 mL.

d. A 3.00-mL sample of a medication has a mass of 3.85 g.

2.70 Determine the density (g/mL) for each of the following: a. The fluid in a car battery has a volume of 125 mL and a

mass of 155 g. b. A plastic material weighs 2.68 lb and has a volume of 3.5 L. c. A 4.000-mL urine sample from a person suffering from

diabetes mellitus has a mass of 4.004 g. d. A solid object has a mass of 1.65 lb and a volume of 170 mL.

2.71 What is the density (g/mL) of each of the following samples? a. A lightweight head on a golf club is made of titanium. The

volume of a sample of titanium is 114 cm3, and the mass is 514.1 g.

2.74 Use the density values in Table 2.10 to solve each of the following problems: a. A graduated cylinder contains 18.0 mL of water. What is the

new water level, in milliliters, after 35.6 g of silver metal is submerged in the water?

b. A thermometer containing 8.3 g of mercury has broken. What volume, in milliliters, of mercury spilled?

c. A fish tank holds 35 gal of water. How many kilograms of water are in the fish tank?

2.75 Use the density values in Table 2.10 to solve each of the following problems: a. What is the mass, in grams, of a cube of copper that has a

volume of 74.1 cm3? b. How many kilograms of gasoline fill a 12.0-gal gas tank? c. What is the volume, in cubic centimeters, of an ice cube that

has a mass of 27 g?

2.76 Use the density values in Table 2.10 to solve each of the following problems: a. If a bottle of olive oil contains 1.2 kg of olive oil, what is the

volume, in milliliters, of the olive oil? b. A cannonball made of iron has a volume of 115 cm3. What is

the mass, in kilograms, of the cannonball? c. A balloon filled with helium has a volume of 7.3 L. What is

the mass, in grams, of helium in the balloon?

2.77 In an old trunk, you find a piece of metal that you think may be aluminum, silver, or lead. You take it to a lab, where you find it has a mass of 217 g and a volume of 19.2 cm3. Using Table 2.10, what is the metal you found?

2.78 Suppose you have two 100-mL graduated cylinders. In each cylinder, there is 40.0 mL of water. You also have two cubes: one is lead, and the other is aluminum. Each cube measures 2.0 cm on each side. After you carefully lower each cube into the water of its own cylinder, what will the new water level be in each of the cylinders? Use Table 2.10 for density values.

Applications

2.79 Solve each of the following problems: a. A urine sample has a density of 1.030 g/mL. What is the

specific gravity of the sample? b. A 20.0-mL sample of a glucose IV solution has a mass of

20.6 g. What is the density of the glucose solution? c. The specific gravity of a vegetable oil is 0.92. What is the

mass, in grams, of 750 mL of vegetable oil? d. A bottle containing 325 g of cleaning solution is used to

clean hospital equipment. If the cleaning solution has a specific gravity of 0.850, what volume, in milliliters, of solution was used?

2.80 Solve each of the following problems: a. A glucose solution has a density of 1.02 g/mL. What is its

specific gravity? b. A 0.200-mL sample of very-low-density lipoprotein (VLDL)

has a mass of 190 mg. What is the density of the VLDL? c. Butter has a specific gravity of 0.86. What is the mass, in

grams, of 2.15 L of butter? d. A 5.000-mL urine sample has a mass of 5.025 g. If the

normal range for the specific gravity of urine is 1.003 to 1.030, would the specific gravity of this urine sample indicate that the patient could have type 2 diabetes?

115.25 g 182.48 g

Lightweight heads on golf clubs are made of titanium.

b. A syrup is added to an empty container with a mass of 115.25 g. When 0.100 pt of syrup is added, the total mass of the container and syrup is 182.48 g.

c. A block of aluminum metal has a volume of 3.15 L and a mass of 8.51 kg.

2.72 What is the density (g/mL) of each of the following samples? a. An ebony carving has a mass of 275 g and a volume of 207 cm3. b. A 14.3@cm3 sample of tin has a mass of 0.104 kg. c. A bottle of acetone (fingernail polish remover) contains

55.0 mL of acetone with a mass of 43.5 g.

2.73 Use the density values in Table 2.10 to solve each of the following problems: a. How many liters of ethanol contain 1.50 kg of ethanol? b. How many grams of mercury are present in a barometer that

holds 6.5 mL of mercury? c. A sculptor has prepared a mold for casting a silver figure.

The figure has a volume of 225 cm3. How many ounces of silver are needed in the preparation of the silver figure?

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Concept Map 59

UPDATE Greg’s Visit with His Doctor

On Greg’s visit to his doctor, he complains of feeling tired. Sandra, the registered nurse, with- draws 8.0 mL of blood, which is sent to the lab and tested for iron. When the iron level is low, a person may have fatigue and decreased immunity.

The normal range for serum iron in men is 80 to 160 mcg/dL. Greg’s iron test shows a blood serum iron level of 42 mcg/dL, which indicates that Greg has iron-deficiency anemia. His doctor orders an iron supplement. One tablet of the iron supple- ment contains 50 mg of iron.

Applications

2.81 a. Write an equality and two conversion factors for Greg’s serum iron level.

b. How many micrograms of iron were in the 8.0-mL sample of Greg’s blood?

2.82 a. Write an equality and two conversion factors for one tablet of the iron supplement.

b. How many grams of iron will Greg consume in one week, if he takes two tablets each day?

Each tablet contains 50 mg of iron, which is given for iron supplementation.

CONCEPT MAP

Measurements

Rounding Off AnswersDensity

Length (m) Metric Units

CHEMISTRY AND MEASUREMENTS

that requiregive

for measuring that change the size of

Measured Numbers

Mass (g)

Volume (L)

Temperature (°C, K)

Time (s)

Metric Units

Significant Figures

Equalities

to give

Adding Zeros

or

Conversion Factors

used for

Problem Solving

to change units in

Prefixes

in chemistry involve

Specific Gravity

and

have

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60 CHAPTER 2 Chemistry and Measurements

CHAPTER REVIEW

2.1 Units of Measurement LEARNING GOAL Write the names and abbreviations for the metric and SI units used in measurements of volume, length, mass, temperature, and time. • In science, physical quanti-

ties are described in units of the metric or International System of Units (SI).

• Some important units are liter (L) for volume, meter (m) for length, gram (g) and kilogram (kg) for mass, degree Celsius (°C) and kelvin (K) for temperature, and second (s) for time.

2.2 Measured Numbers and Significant Figures LEARNING GOAL Identify a number as measured or exact; determine the number of significant figures in a measured number. • A measured number is any number

obtained by using a measuring device.

• An exact number is obtained by counting items or from a definition; no measuring device is needed.

• Significant figures are the numbers reported in a measurement including the estimated digit.

• Zeros in front of a decimal number or at the end of a nondecimal number are not significant.

2.3 Significant Figures in Calculations LEARNING GOAL Give the correct number of significant figures for a calculated answer. • In multiplication and division, the

final answer is written so that it has the same number of significant figures as the measurement with the fewest significant figures.

• In addition and subtraction, the final answer is written so that it has the same number of decimal places as the measurement with the fewest decimal places.

2.4 Prefixes and Equalities LEARNING GOAL Use the numerical values of prefixes to write a metric equality. • A prefix placed in

front of a metric or SI unit changes the size of the unit by factors of 10.

• Prefixes such as centi, milli, and micro provide smaller units; prefixes such as kilo, mega, and tera provide larger units.

• An equality shows the relationship between two units that measure the same quantity of volume, length, mass, or time.

• Examples of metric equalities are 1 L = 1000 mL, 1 m = 100 cm, 1 kg = 1000 g, and 1 min = 60 s.

2.5 Writing Conversion Factors LEARNING GOAL Write a conversion factor for two units that describe the same quantity. • Conversion factors are used to express a

relationship in the form of a fraction. • Two conversion factors can be written

for any relationship in the metric or U.S. system.

• A percentage is written as a conversion factor by expressing matching units as the parts in 100 parts of the whole.

2.6 Problem Solving Using Unit Conversion LEARNING GOAL Use conversion factors to change from one unit to another. • Conversion factors are

useful when changing a quantity expressed in one unit to a quantity expressed in another unit.

• In the problem-solving process, a given unit is multiplied by one or more conversion factors that cancel units until the needed answer is obtained.

2.7 Density LEARNING GOAL Calculate the density of a substance; use the density to calculate the mass or volume of a substance. • The density of a substance is a ratio of

its mass to its volume, usually g/mL or g/cm3.

• The units of density can be used to write conversion factors that convert between the mass and volume of a substance.

• Specific gravity (sp gr) compares the density of a substance to the density of water, 1.000 g/mL.

10 cm

2 3 4 5

10 cm

2 3 4 5

(a) 4.5 cm

(b) 4.55 cm

10 cm = 1 dm

1 cm

10 c

m

10 cm

1 L = 1.057 qt 946.4 mL = 1 qt

M02_TIMB8119_06_SE_C02.indd 60 11/27/18 11:38 AM

Core Chemistry Skills 61

Celsius (°C) temperature scale A temperature scale on which water has a freezing point of 0 °C and a boiling point of 100 °C.

centimeter (cm) A unit of length in the metric system; there are 2.54 cm in 1 in.

conversion factor A ratio in which the numerator and denominator are quantities from an equality or given relationship. For exam- ple, the two conversion factors for the equality 1 kg = 2.205 lb are written as

2.205 lb 1 kg

and 1 kg

2.205 lb

cubic centimeter (cm3, cc) The volume of a cube that has 1-cm sides; 1 cm3 is equal to 1 mL.

density The relationship of the mass of an object to its volume expressed as grams per cubic centimeter (g/cm3), grams per milliliter (g/mL), or grams per liter (g/L).

equality A relationship between two units that measure the same quantity.

exact number A number obtained by counting or by definition. gram (g) The metric unit used in measurements of mass. International System of Units (SI) The official system of measure-

ment throughout the world, except for the United States, that modifies the metric system.

Kelvin (K) temperature scale A temperature scale on which the lowest possible temperature is 0 K.

kilogram (kg) A metric mass of 1000 g, equal to 2.205 lb. The kilogram is the SI standard unit of mass.

liter (L) The metric unit for volume that is slightly larger than a quart. mass A measure of the quantity of material in an object. measured number A number obtained when a quantity is determined

by using a measuring device. meter (m) The metric unit for length that is slightly longer than a

yard. The meter is the SI standard unit of length. metric system A system of measurement used by scientists and in

most countries of the world. milliliter (mL) A metric unit of volume equal to one-thousandth of a

liter (0.001 L). prefix The part of the name of a metric unit that precedes the base

unit and specifies the size of the measurement. All prefixes are related on a decimal scale.

second (s) A unit of time used in both the SI and metric systems. SI See International System of Units (SI). significant figures (SFs) The numbers recorded in a measurement. specific gravity (sp gr) A relationship between the density of a

substance and the density of water:

sp gr = density of sample

density of water

temperature An indicator of the hotness or coldness of an object. volume (V ) The amount of space occupied by a substance.

KEY TERMS

The chapter section containing each Key Math Skill is shown in parentheses at the end of each heading.

Rounding Off (2.3) Calculator displays are rounded off to give the correct number of significant figures.

• If the first digit to be dropped is 4 or less, then it and all following digits are simply dropped from the number.

• If the first digit to be dropped is 5 or greater, then the last retained digit of the number is increased by 1.

KEY MATH SKILLS

One or more significant zeros are added when the calculator display has fewer digits than the needed number of significant figures.

Example: Round off each of the following to three significant figures:

a. 3.608 92 L b. 0.003 870 298 m c. 6 g

Answer: a. 3.61 L b. 0.003 87 m c. 6.00 g

The chapter section containing each Core Chemistry Skill is shown in parentheses at the end of each heading.

Counting Significant Figures (2.2) The significant figures (SFs) are all the measured numbers including the last, estimated digit.

• All nonzero digits • Zeros between nonzero digits • Zeros after a nonzero digit in a decimal number • All digits in a coefficient of a number written in scientific notation

An exact number is obtained from counting or a definition and has no effect on the number of significant figures in the final answer.

CORE CHEMISTRY SKILLS

Example: State the number of significant figures in each of the following:

a. 0.003 045 mm b. 15 000 m c. 45.067 kg d. 5.30 * 103 g e. 2 cans of soda

Answer: a. four SFs b. two SFs c. five SFs d. three SFs e. exact

Using Significant Figures in Calculations (2.3)

• In multiplication or division, the final answer is written so that it has the same number of significant figures as the measurement with the fewest SFs.

M02_TIMB8119_06_SE_C02.indd 61 11/27/18 11:38 AM

62 CHAPTER 2 Chemistry and Measurements

• In addition or subtraction, the final answer is written so that it has the same number of decimal places as the measurement having the fewest decimal places.

Example: Perform the following calculations using measured numbers, and give answers with the correct number of SFs or decimal places:

a. 4.05 m * 0.6078 m b. 4.50 g

3.27 mL c. 0.758 g + 3.10 g d. 13.538 km - 8.6 km Answer: a. 2.46 m2 b. 1.38 g/mL c. 3.86 g d. 4.9 km

Using Prefixes (2.4) In the metric and SI systems of units, a prefix attached to any unit increases or decreases its size by some factor of 10.

• When the prefix centi is used with the unit meter, it becomes centimeter, a length that is one-hundredth of a meter (0.01 m).

• When the prefix milli is used with the unit meter, it becomes millimeter, a length that is one-thousandth of a meter (0.001 m).

Example: Complete each of the following metric relationships:

a. 1000 m = 1 m b. 0.01 g = 1 g Answer: a. 1000 m = 1 km b. 0.01 g = 1 cg

Writing Conversion Factors from Equalities (2.5)

• A conversion factor allows you to change from one unit to another. • Two conversion factors can be written for any equality in the

metric, U.S., or metric–U.S. systems of measurement. • Two conversion factors can be written for a relationship stated

within a problem.

Example: Write two conversion factors for the equality: 1 L = 1000 mL

Answer: 1000 mL

1 L and

1 L 1000 mL

Using Conversion Factors (2.6) In problem solving, conversion factors are used to cancel the given unit and to provide the needed unit for the answer.

• State the given and needed quantities. • Write a plan to convert the given unit to the needed unit. • State the equalities and conversion factors. • Set up the problem to cancel units and calculate the answer.

Example: A computer chip has a width of 0.75 in. What is the width in millimeters?

Answer: 0.75 in. * 2.54 cm

1 in. *

10 mm 1 cm

= 19 mm

Using Density as a Conversion Factor (2.7) Density is an equality of mass and volume for a substance, which is written as the density expression.

Density = mass of substance

volume of substance

Density is useful as a conversion factor to convert between mass and volume.

Example: The element tungsten used in light bulb filaments has a density of 19.3 g/cm3. What is the volume, in cubic centimeters, of 250 g of tungsten?

Answer: 250 g * 1 cm3

19.3 g = 13 cm3

UNDERSTANDING THE CONCEPTS

The chapter sections to review are shown in parentheses at the end of each problem.

2.83 In which of the following pairs do both numbers contain the same number of significant figures? (2.2) a. 2.0500 m and 0.0205 m b. 600.0 K and 60 K c. 0.000 75 s and 75 000 s d. 6.240 L and 6.240 * 10-2 L

2.84 In which of the following pairs do both numbers contain the same number of significant figures? (2.2) a. 3.44 * 10-3 g and 0.0344 g b. 0.0098 s and 9.8 * 104 s c. 6.8 * 103 m and 68 000 m d. 258.000 g and 2.58 * 10-2 g

2.85 Indicate if each of the following is an exact number or a measured number: (2.2)

a. number of legs b. height of table c. chairs at the table

2.86 Use the figure in problem 2.85 to indicate if each of the following is an exact number or a measured number: (2.2) a. area of tabletop b. plates on the table c. length of tabletop

10

0

70

60

50

2.88 State the temperature on the Celsius thermometer to the correct number of significant figures: (2.3)

2.87 State the temperature on the Celsius thermometer to the correct number of significant figures: (2.3)

M02_TIMB8119_06_SE_C02.indd 62 20/12/18 4:35 PM

Understanding the Concepts 63

2.92 What is the density of the solid object that is weighed and submerged in water? (2.7)

2.93 Consider the following solids. The solids A, B, and C represent aluminum (D = 2.70 g/cm3), gold (D = 19.3 g/cm3), and sil- ver (D = 10.5 g/cm3). If each has a mass of 10.0 g, what is the identity of each solid? (2.7)

2.94 A graduated cylinder contains three liquids A, B, and C, which have different densities and do not mix: mercury (D = 13.6 g/mL), vegetable oil (D = 0.92 g/mL), and water (D = 1.00 g/mL). Identify the liquids A, B, and C in the cylinder. (2.7)

2.95 The gray cube has a density of 4.5 g/cm3. Is the density of the green cube the same, lower than, or higher than that of the gray cube? (2.7)

2.96 The gray cube has a density of 4.5 g/cm3. Is the density of the green cube the same, lower than, or higher than that of the gray cube? (2.7)

2.89 The length of this rug is 38.4 in. and the width is 24.2 in. (2.3, 2.6)

a. What is the length of this rug, in centimeters? b. What is the width of this rug, in centimeters? c. How many significant figures are in the length

measurement? d. Calculate the area of the rug, in square centime-

ters, to the correct number of significant figures. (Area = Length * Width)

2.90 A shipping box has a length of 7.00 in., a width of 6.00 in., and a height of 4.00 in. (2.3, 2.6)

a. What is the length of the box, in centimeters? b. What is the width of the box, in centimeters? c. How many significant figures are in the width

measurement? d. Calculate the volume of the box, in cubic centimeters,

to the correct number of significant figures. (Volume = Length * Width * Height)

2.91 Each of the following diagrams represents a container of water and a cube. Some cubes float while others sink. Match diagrams 1, 2, 3, or 4 with one of the following descriptions and explain your choices: (2.7)

a. The cube has a greater density than water. b. The cube has a density that is 0.80 g/mL. c. The cube has a density that is one-half the density of water. d. The cube has the same density as water.

A

30

20

10

50

B

C

40

A B C

1 2 3 4

Solid Water

18.5 mL + 23.1 mL

8.24 g

M02_TIMB8119_06_SE_C02.indd 63 11/27/18 11:38 AM

64 CHAPTER 2 Chemistry and Measurements

2.97 Identify each of the following numbers as measured or exact, and write the number of significant figures (SFs) in each of the measured numbers: (2.2, 2.3) a. 5.82 g b. 1 L = 1000 mL c. 0.009 050 m d. 1.08 * 103 kg

2.98 Identify each of the following numbers as measured or exact, and write the number of significant figures (SFs) in each of the measured numbers: (2.2, 2.3)

a. 2.3055 cm b. 3.40 * 10-4 km c. 82 mg d. 2.54 cm = 1 in. 2.99 Round off or add zeros to the following calculated answers to

give a final answer with three significant figures: (2.2) a. 0.000 012 58 L b. 3.528 * 102 kg c. 125 111 m d. 34.9673 s

2.100 Round off or add zeros to the following calculated answers to give a final answer with three significant figures: (2.2)

a. 58.703 mL b. 3 * 10-3 s c. 0.010 826 g d. 1.7484 * 103 ms 2.101 Perform each of the following operations, and write answers

with the correct number of significant figures or decimal places: (2.2, 2.3)

a. 3.285 + 14.8 + 7.06 b. (1.20 * 10-4) * (2.234 * 106)

c. 46.7 * 23.22 90.10

d. 85.2 + 6.081 - 5.608 2.102 Perform each of the following operations, and write answers

with the correct numbers of significant figures or decimal places: (2.2, 2.3)

a. 3.7 + 4.888 b. (4.09 * 102) * (6.33 * 104) c. 3.376 * 23 * 0.123 d. 0.0467 + 23.32 - 22.8 2.103 A dessert contains 137.25 g of vanilla ice cream, 84 g of

fudge sauce, and 43.7 g of nuts. (2.3, 2.6) a. What is the total mass, in grams, of the dessert? a. What is the total weight, in pounds, of the dessert?

2.104 A fish company delivers 22 kg of salmon, 5.5 kg of crab, and 3.48 kg of oysters to your seafood restaurant. (2.3, 2.6) a. What is the total mass, in kilograms, of the seafood? a. What is the total number of pounds?

2.105 Perform each of the following conversions: (2.5) a. 3.47 mm to m b. 37 g to mg c. 873 L to kL d. 2.05 * 105 ns to s 2.106 Perform each of the following conversions: (2.5) a. 4.8 L to mL b. 77.8 g to kg c. 0.008 67 mg to mcg d. 4.6 * 109 mm to km 2.107 In France, grapes are 1.95 euros per kilogram. What is the

cost of grapes, in dollars per pound, if the exchange rate is 1.14 dollars/euro? (2.6)

2.108 In Mexico, avocados are 48 pesos per kilogram. What is the cost, in cents, of an avocado that weighs 0.45 lb if the exchange rate is 18 pesos to the dollar? (2.6)

2.109 Bill’s recipe for onion soup calls for 4.0 lb of thinly sliced onions. If an onion has an average mass of 115 g, how many onions does Bill need? (2.6)

2.110 The price of 1 lb of potatoes is $1.75. If all the potatoes sold today at the store bring in $1420, how many kilograms of potatoes did grocery shoppers buy? (2.6)

2.111 During a workout at the gym, you set the treadmill at a pace of 55.0 m/min. How many minutes will you walk if you cover a distance of 7500 ft? (2.6)

2.112 The distance between two cities is 1700 km. How long will it take, in hours, to drive from one city to the other if your average speed is 63 mi/h? (2.6)

2.113 Perform each of the following conversions: (2.5) a. 40.5 in.2 to ft2 b. 8.2 km2 to m2

c. 3.24 yd2 to cm2 d. 2.2 * 106 cm3 to m3

2.114 Perform each of the following conversions: (2.5) a. 0.254 ft2 to cm2 b. 2.26 in.3 to cm3

c. 0.22 cm2 to mm2 d. 7.8 m2 to ft2

2.115 Perform each of the following conversions: (2.6) a. 65 mi/h to ft/s b. $0.75/lb to cents/kg c. 1.63 g/mL to lb/gal d. 11 m/s to mi/h

2.116 Perform each of the following conversions: (2.6) a. 76 km/L to mi/gal b. 35 lb/in.2 to kg/cm2

c. 55 mi/h to m/s d. 2.5 mg/kg to oz/lb

2.117 The water level in a graduated cylinder initially at 215 mL rises to 285 mL after a piece of lead is submerged. What is the mass, in grams, of the lead (see Table 2.10)? (2.7)

2.118 A graduated cylinder contains 155 mL of water. A 15.0-g piece of iron and a 20.0-g piece of lead are added. What is the new water level, in milliliters, in the cylinder (see Table 2.10)? (2.7)

2.119 How many milliliters of gasoline have a mass of 1.2 kg (see Table 2.10)? (2.7)

2.120 What is the volume, in quarts, of 3.40 kg of ethanol (see Table 2.10)? (2.7)

Applications

2.121 The following nutrition information is listed on a box of crackers: (2.6)

Serving size 0.50 oz (6 crackers)

Fat 4 g per serving; Sodium 140 mg per serving

a. If the box has a net weight (contents only) of 8.0 oz, about how many crackers are in the box?

b. If you ate 10 crackers, how many ounces of fat did you consume?

c. How many servings of crackers in part a would it take to obtain the Daily Value (DV) for sodium, which is 2.4 g?

2.122 A dialysis unit requires 75 000 mL of distilled water. How many gallons of water are needed? (2.6)

2.123 To treat a bacterial infection, a doctor orders 4 tablets of amoxicillin per day for 10 days. If each tablet contains 250 mg of amoxicillin, how many ounces of the medication are given in 10 days? (2.6)

2.124 Celeste’s diet restricts her intake of protein to 24 g per day. If she eats 1.2 oz of protein, has she exceeded her protein limit for the day? (2.6)

2.125 A doctor orders 5.0 mL of phenobarbital elixir. If the phenobarbital elixir is available as 30. mg per 7.5 mL, how many milligrams is given to the patient? (2.6)

2.126 A doctor orders 2.0 mg of morphine. The vial of morphine on hand is 10. mg/mL. How many milliliters of morphine should you administer to the patient? (2.6)

ADDITIONAL PRACTICE PROBLEMS

M02_TIMB8119_06_SE_C02.indd 64 11/27/18 11:38 AM

Answers to Selected Problems 65

The following problems are related to the topics in this chapter. However, they do not all follow the chapter order, and may require you to combine concepts and skills from several sections. These problems will help you increase your critical thinking skills and prepare for your next exam.

2.127 A balance measures mass to 0.001 g. If you determine the mass of an object that weighs about 31 g, would you record the mass as 31 g, 31.1 g, 31.08 g, 31.075 g, or 31.0750? Explain your choice by writing two to three complete sentences that describe your thinking. (2.3)

2.128 When three students use the same meterstick to measure the length of a paper clip, they obtain results of 5.8 cm, 5.75 cm, and 5.76 cm. If the meterstick has millimeter markings, what are some reasons for the different values? (2.3)

2.129 A car travels at 55 mi/h and gets 11 km/L of gasoline. How many gallons of gasoline are needed for a 3.0-h trip? (2.6)

2.130 A sunscreen preparation contains 2.50% benzyl salicylate by mass. If a tube contains 4.0 oz of sunscreen, how many kilograms of benzyl salicylate are needed to manufacture 325 tubes of sunscreen? (2.6)

2.131 How many milliliters of olive oil have the same mass as 1.50 L of gasoline (see Table 2.10)? (2.7)

2.132 A 50.0-g silver object and a 50.0-g gold object are both added to 75.5 mL of water contained in a graduated cylinder. What is the new water level, in milliliters, in the cylinder (see Table 2.10)? (2.7)

Applications

2.133 a. An athlete with a body mass of 65 kg has 3.0% body fat. How many pounds of body fat does that person have? (2.6)

b. In liposuction, a doctor removes fat deposits from a person’s body. If body fat has a density of 0.909 g/cm3 and 3.0 L of fat is removed, how many pounds of fat were removed from the patient?

2.134 A mouthwash is 21.6% ethanol by mass. If each bottle contains 1.06 pt of mouthwash with a density of 0.876 g/mL, how many kilograms of ethanol are in 180 bottles of the mouthwash? (2.6, 2.7)

CHALLENGE PROBLEMS

ANSWERS TO ENGAGE QUESTIONS 2.7 Because the numbers in the equality 1 g = 1000 mg are exact,

60. mg with two significant figures is the same as 0.060 g. 60. mg * 1 g/1000 mg = 0.060 g (2 SFs)

2.8 When we divide both sides of the equality 1 m = 100 cm by 100, we get 0.01 m = 1 cm.

2.9 1 mL is the same volume as a cube with sides of 1 cm * 1 cm * 1 cm = 1 cm3.

2.10 The equality 1 day = 24 h can be written as two conversion factors, 24 h/1 day or 1 day/24 h.

2.11 The equality of a percent states the parts of a relationship to 100 parts of the whole, parts/100 parts * 100%. Two conversion factors can be written from this equality.

2.12 6.4 oz is multiplied by the conversion factor 1 lb/16 oz to cancel oz and give an answer in lb.

2.13 Two or more conversion factors can be used in a problem, as long as the units cancel to give the desired unit for the final answer.

2.14 Iron is more dense than water.

2.1 All the digits in the coefficient of a number written in scientific notation are significant.

2.2 The quantity 4 hats is obtained from counting. It is an exact number, whereas the value of 6.24 cm is obtained by measurement, which has three significant figures.

2.3 Because the first of the digits to be dropped (7) is 5 or greater, the last retained digit is increased by 1 to give 10.1.

2.4 The answer is rounded off to the same number of SFs as the measurement with the fewer SFs, which is one significant figure.

2.5 When the multiplication and division of the numbers with three significant figures give the whole number 5, two significant zeros are added to give the correct number of significant figures in the answer.

2.6 The answer from addition and subtraction is adjusted to the decimal place of the measurement with the fewest decimal places.

ANSWERS TO SELECTED PROBLEMS 2.9 a. 2.2 cm, two SFs b. 3.90 cm, three SFs c. 4.81 cm, three SFs

2.11 a. five SFs b. two SFs c. two SFs d. three SFs e. four SFs f. three SFs

2.13 b and c

2.15 a. not significant b. significant c. significant d. significant e. not significant

2.1 a. g b. °C c. L d. lb e. s

2.3 a. volume b. length c. length d. time

2.5 a. meter, length b. gram, mass c. milliliter, volume d. second, time e. degree Celsius, temperature

2.7 a. second, time b. kilogram, mass c. gram, mass d. degree Celsius, temperature

M02_TIMB8119_06_SE_C02.indd 65 11/27/18 11:38 AM

66 CHAPTER 2 Chemistry and Measurements

2.17 a. 5.0 * 103 L b. 3.0 * 104 g c. 1.0 * 105 m d. 2.5 * 10-4 cm 2.19 a. measured b. exact c. exact d. measured

2.21 a. 6 oz b. none c. 0.75 lb, 350 g d. none (definitions are exact)

2.23 a. measured, four SFs b. measured, two SFs c. measured, three SFs d. exact

2.25 a. 1.85 kg b. 88.2 L c. 0.004 74 cm d. 8810 m e. 1.83 * 105 s 2.27 a. 56.9 m b. 0.002 28 g c. 11 500 s (1.15 * 104 s) d. 8.10 L 2.29 a. 1.6 b. 0.01 c. 27.6 d. 3.5 e. 0.14 (1.4 * 10-1) f. 0.8 (8 * 10-1) 2.31 a. 53.54 b. 127.6 c. 121.5 d. 0.50 e. 2500 f. 156.7

2.33 a. mg b. dL c. km d. pg

2.35 a. centiliter b. kilogram c. millisecond d. petameter

2.37 a. 0.01 b. 1 * 1012 c. 0.001 d. 0.1

2.39 a. decigram b. microgram c. kilogram d. centigram

2.41 a. 100 cm b. 1 * 109 nm c. 0.001 m d. 1000 mL

2.43 a. kilogram b. milliliter c. km d. kL e. nanometer

2.45 A conversion factor can be inverted to give a second conversion factor.

2.47 a. 1 kg = 1 * 106 mg; 1 * 106 mg

1 kg and

1 kg

1 * 106 mg

b. 1 g = 1 * 109 ng; 1 * 109 ng

1 g and

1 g

1 * 109 ng

c. 1 kL = 1000 L; 1000 L 1 kL

and 1 kL

1000 L

d. 1 s = 1000 ms; 1000 ms

1 s and

1 s 1000 ms

d. 100 g of sterling = 93 g of silver;

93 g silver

100 g sterling and

100 g sterling

93 g silver

The 93 g is measured: It has two SFs. The 100 g is exact.

e. 1 min = 60 s; 60 s

1 min and

1 min 60 s

The 1 min and 60 s are both exact.

f. 1 m3 = 1 * 106 cm3; 1 * 106 cm3

1 m3 and

1 m3

1 * 106 cm3

The 1 m and 1 * 106 cm3 are both exact.

2.51 a. 1 s = 3.5 m; 3.5 m

1 s and

1 s 3.5 m

The 3.5 m is measured: It has two SFs. The 1 s is exact.

b. 1 day = 3.5 g of potassium;

3.5 g potassium

1 day and

1 day

3.5 g potassium

The 3.5 g is measured: It has two SFs. The 1 day is exact.

c. 1 L = 26.0 km; 26.0 km

1 L and

1 L 26.0 km

The 26.0 km is measured: It has three SFs. The 1 L is exact.

d. 1 kg of plums = 29 mcg of pesticide;

29 mcg pesticide

1 kg plums and

1 kg plums

29 mcg pesticide

The 29 mcg is measured: It has two SFs. The 1 kg is exact.

e. 100 g of crust = 28.2 g of silicon;

28.2 g silicon

100 g crust and

100 g crust

28.2 g silicon

The 28.2 g is measured: It has three SFs. The 100 g is exact.

2.53 a. 1 tablet = 630 mg of calcium;

630 mg calcium

1 tablet and

1 tablet 630 mg calcium

The 630 mg is measured: It has two SFs. The 1 tablet is exact.

b. 1 day = 60 mg of vitamin C;

60 mg vitamin C

1 day and

1 day

60 mg vitamin C

The 60 mg is measured: It has one SF. The 1 day is exact.

c. 1 tablet = 50 mg of atenolol;

50 mg atenolol

1 tablet and

1 tablet 50 mg atenolol

The 50 mg is measured: It has one SF. The 1 tablet is exact.

d. 1 tablet = 81 mg of aspirin;

81 mg aspirin

1 tablet and

1 tablet 81 mg aspirin

The 81 mg is measured: It has two SFs. The 1 tablet is exact.

e. 1 dm = 100 mm; 100 mm

1 dm and

1 dm 100 mm

f. (1 m)3 = (100 cm)3 = 106 cm3; (100 cm)3

(1 m)3 and

(1 m)3

(100 cm)3

2.49 a. 1 yd = 3 ft; 3 ft 1 yd

and 1 yd

3 ft

The 1 yd and 3 ft are both exact.

b. 1 kg = 2.205 lb; 2.205 lb

1 kg and

1 kg

2.205 lb

The 2.205 lb is measured: It has four SFs. The 1 kg is exact.

c. 1 gal = 27 mi; 27 mi 1 gal

and 1 gal

27 mi

The 27 mi is measured: It has two SFs. The 1 gal is exact.

M02_TIMB8119_06_SE_C02.indd 66 11/27/18 11:38 AM

Answers to Selected Problems 67

2.55 a. 5 mL of syrup = 10 mg of Atarax;

10 mg Atarax

5 mL syrup and

5 mL syrup

10 mg Atarax

b. 1 tablet = 0.25 g of Lanoxin;

0.25 g Lanoxin

1 tablet and

1 tablet 0.25 g Lanoxin

c. 1 tablet = 300 mg of Motrin;

300 mg Motrin

1 tablet and

1 tablet 300 mg Motrin

2.57 a. 0.0442 L b. 8.65 * 109 nm c. 5.2 * 102 Mg d. 7.2 * 105 ms 2.59 a. 1.555 kg b. 63 in. c. 4.4 qt d. 205 mm

2.61 a. 1.75 m b. 5 L c. 5.5 g d. 3.5 L

2.63 a. 473 mL b. 79.4 kg c. 24 lb d. 43 g

2.65 a. 66 gal b. 3 tablets c. 1800 mg d. 110 g

2.67 a. 6.3 h b. i. 6.8 * 10-4 g/lb ii. 2.5 mL c. i. 4.00 * 10-4 oz/lb ii. 0.0656 oz 2.69 a. 1.20 g/mL b. 0.870 g/mL c. 3.10 g/mL d. 1.28 g/mL

2.71 a. 4.51 g/mL b. 1.42 g/mL c. 2.70 g/mL

2.73 a. 1.9 L of ethanol b. 88 g of mercury c. 83.3 oz of silver

2.75 a. 661 g b. 34 kg c. 29 cm3

2.77 Because we calculate the density to be 11.3 g/cm3, we identify the metal as lead.

2.79 a. 1.030 b. 1.03 g/mL c. 690 g d. 382 mL

2.81 a. 1 dL of blood = 42 mcg of iron;

42 mcg iron

1 dL blood and

1 dL blood 42 mcg iron

b. 3.4 mcg of iron

2.83 c and d

2.85 a. exact b. measured c. exact

2.87 61.5 °C

2.89 a. 97.5 cm b. 61.5 cm c. three SFs d. 6.00 * 103 cm2

2.91 a. Diagram 3; a cube that has a greater density than the water will sink to the bottom.

b. Diagram 4; a cube with a density of 0.80 g/mL will be about four-fifths submerged in the water.

c. Diagram 1; a cube with a density that is one-half the density of water will be one-half submerged in the water.

d. Diagram 2; a cube with the same density as water will float just at the surface of the water.

2.93 A would be gold; it has the highest density (19.3 g/cm3) and the smallest volume. B would be silver; its density is intermediate (10.5 g/cm3) and the volume is intermediate. C would be aluminum; it has the lowest density (2.70 g/cm3) and the largest volume.

2.95 The green cube has the same volume as the gray cube. However, the green cube has a larger mass on the scale, which means that its mass/volume ratio is larger. Thus, the density of the green cube is higher than the density of the gray cube.

2.97 a. measured, three SFs b. exact c. measured, four SFs d. measured, three SFs

2.99 a. 0.000 012 6 L (1.26 * 10-5 L) b. 353 kg (3.53 * 102 kg) c. 125 000 m (1.25 * 105 m) d. 35.0 s

2.101 a. 25.1 b. 2.68 * 102 c. 12.0 d. 85.7 2.103 a. 265 g b. 0.584 lb

2.105 a. 0.003 47 m b. 37 000 mg c. 0.873 kL d. 2.05 * 10-4 s

2.107 $1.01 per lb

2.109 16 onions

2.111 42 min

2.113 a. 0.281 ft2 b. 8.2 * 106 m2 c. 2.71 * 104 cm2 d. 2.2 m3

2.115 a. 95 ft/s b. 170 cents/kg

c. 13.6 lb/gal d. 25 mi/h

2.117 790 g

2.119 1600 mL (1.6 * 103 mL) 2.121 a. 96 crackers b. 0.2 oz of fat c. 17 servings

2.123 0.35 oz

2.125 20. mg

2.127 You would record the mass as 31.075 g. Because the balance will weigh to the nearest 0.001 g, the mass value would be reported to 0.001 g.

2.129 6.4 gal

2.131 1200 mL

2.133 a. 4.3 lb of body fat b. 6.0 lb

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68

When Daniel sees Charles and his mother, they discuss a menu for weight loss. Charles is going to record his food intake and return to discuss his diet with Daniel. You can view the results in the UPDATE A Diet and Exercise Program, page 90, and calculate both the kilocalories that Charles consumes in one day and the weight that Charles has lost.

UPDATE A Diet and Exercise Program

Charles is 13 years old and overweight. His doctor is worried that Charles is at risk for type 2 diabetes and advises his mother to make an appointment with a dietitian. Daniel, a dietitian, explains to them that choosing the appropriate foods is important to living a healthy lifestyle, losing weight, and preventing or managing diabetes.

Daniel also explains that food contains potential or stored energy, and different foods contain different amounts of potential energy. For instance, carbohydrates contain 4 kcal/g (17 kJ/g), whereas fats contain 9 kcal/g (38 kJ/g). He then explains that diets high in fat require more exercise to burn the fats, as they contain more energy. When Daniel looks at Charles’s typical daily diet, he calculates that Charles obtains 2500 kcal in one day, which exceeds the recommendation of the American Heart Association of 1800 kcal for boys 9 to 13 years of age. Daniel encourages Charles and his mother to include whole grains, fruits, and vegetables in their diet instead of foods high in fat. They also discuss food labels and the fact that smaller serving sizes of healthy foods are necessary to lose weight. Daniel also recommends that Charles exercise at least 60 min every day. Before leaving, Charles and his mother make an appointment for the following week to look at a weight loss plan.

CAREER

Dietitian Dietitians specialize in helping individuals learn about good nutrition and the need for a balanced diet. This requires them to understand biochemical processes, the importance of vitamins and food labels, as well as the differences between carbohydrates, fats, and proteins in terms of their energy value and how they are metabolized. Dietitians work in a variety of environments, including hospitals, nursing homes, school cafeterias, and public health clinics. In these roles, they create specialized diets for individuals diagnosed with a specific disease or create meal plans for those in a nursing home.

Matter and Energy 3

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3.1 Classification of Matter 69

3.1 Classification of Matter LEARNING GOAL Classify examples of matter as pure substances or mixtures.

Matter is anything that has mass and occupies space. Matter is everywhere around us: the orange juice we had for breakfast, the water we put in the coffee maker, the plastic bag we put our sandwich in, our toothbrush and toothpaste, the oxygen we inhale, and the carbon dioxide we exhale. All of this material is matter. The different types of matter are classified by their composition.

Pure Substances: Elements and Compounds All matter is made of extremely small particles called atoms. Much of matter is made of atoms bonded together in definite arrangements called molecules. A pure substance is matter that consists of just one type of atom or one type of molecule. An element, the simplest type of a pure substance, is composed of only one type of atom such as silver, iron, or aluminum. Silver is composed of silver atoms, iron of iron atoms, and aluminum of aluminum atoms. A full list of the elements is found on the inside front cover of this text.

A compound is also a pure substance, but it consists of atoms of two or more elements always chemically combined in the same proportion. For example, in the compound water, there are two hydrogen atoms for every one oxygen atom, which is represented by the formula H2O. This means that water always has the same composi- tion of H2O. Another compound that consists of a chemical combination of hydrogen and oxygen is hydrogen peroxide. It has two hydrogen atoms for every two oxygen atoms and is represented by the formula H2O2. Thus, water (H2O) and hydrogen peroxide (H2O2) are different compounds even though they contain the same elements, hydrogen and oxygen.

ENGAGE 3.1 Why are elements and compounds both pure substances?

PRACTICE PROBLEMS Try Practice Problems 3.1 and 3.2

LOOKING AHEAD

3.1 Classification of Matter 69

3.2 States and Properties of Matter 72

3.3 Temperature 75 3.4 Energy 79 3.5 Specific Heat 82 3.6 Energy and Nutrition 87

Aluminum atom

An aluminum can consists of many atoms of aluminum.

H H O

Water molecule

A water molecule, H2O, consists of two atoms of hydrogen (white) for one atom of oxygen (red).

H

H O

O Hydroxide peroxide molecule

A hydroxide peroxide molecule, H2O2, consists of two atoms of hydrogen (white) for every two atoms of oxygen (red).

Mixtures In a mixture, two or more different substances are physically mixed. Much of the matter in our everyday lives consists of mixtures. The air we breathe is a mixture of mostly oxygen and nitrogen gases. The steel in buildings and railroad tracks is a mixture of iron, nickel, carbon, and chromium. The brass in doorknobs and musical instruments is a mixture of

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70 CHAPTER 3 Matter and Energy

copper and zinc (see FIGURE 3.1). Tea, coffee, and ocean water are mixtures too. The proportions of substances in a mixture are not consistent but can vary. For example, two sugar–water mixtures may look the same, but the one with the higher ratio of sugar to water would taste sweeter.

Types of Mixtures Mixtures are classified further as homogeneous or heterogeneous. In a homogeneous mixture, also called a solution, the composition is uniform throughout the sample. We cannot see the individual components, which appear as one state. Familiar examples of homoge- neous mixtures are air, which contains oxygen and nitrogen gases, and seawater, a solution of salt and water.

In a heterogeneous mixture, the components do not have a uniform composition throughout the sample. The components appear as two separate regions. A mixture of oil and water is heterogeneous because the oil f loats on the surface of the water. Other examples of heterogeneous mixtures are a cookie with raisins and orange juice with pulp.

ENGAGE 3.2 Why is a pizza heterogeneous, whereas vinegar is a homogeneous mixture?

Chemistry Link to Health Breathing Mixtures

The air we breathe is composed mostly of the gases oxygen (21%) and nitrogen (79%). The homogeneous breathing mixtures used by scuba divers differ from the air we breathe depending on the depth of the dive. Nitrox is a mixture of oxygen and nitrogen, but with more oxygen gas (up to 32%) and less nitrogen gas (68%) than air. A breathing mixture with less nitrogen gas decreases the risk of nitrogen narcosis associated with breathing regular air while diving. Heliox contains oxygen and helium, which is typically used for div- ing to more than 200 ft. By replacing nitrogen with helium, nitrogen narcosis does not occur. However, at dive depths over 300 ft, helium is associated with severe shaking and a drop in body temperature.

A breathing mixture used for dives over 400 ft is trimix, which contains oxygen, helium, and some nitrogen. The addition of some

nitrogen lessens the problem of shaking that comes with breathing high levels of helium. Heliox and trimix are used only by professional, military, or other highly trained divers.

In hospitals, heliox may be used as a treatment for respiratory disorders and lung constriction in adults and prema- ture infants. Heliox is less dense than air, which reduces the effort of breath- ing and helps distribute the oxygen gas to the tissues.

A nitrox mixture is used to fill scuba tanks.

FIGURE 3.1 Matter is organized by its components: elements, compounds, and mixtures.

Heterogeneous

Mixtures

CompoundsElements

Pure substances

Matter

Copper atoms Water molecules Brass (copper and zinc atoms)

Water molecules and copper atoms

Homogeneous

A cookie with raisins is a heterogeneous mixture.

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3.1 Classification of Matter 71

PRACTICE PROBLEMS Try Practice Problems 3.3 to 3.6

SAMPLE PROBLEM 3.1 Classifying Mixtures

TRY IT FIRST

Classify each of the following as a pure substance (element or compound) or a mixture (homogeneous or heterogeneous):

a. copper in wire b. chocolate-chip ice cream c. nitrox, a combination of oxygen and nitrogen used to fill scuba tanks

SOLUTION

a. Copper, which is only one type of matter, is a pure substance. It is an element. b. Chocolate-chip ice cream is two or more substances mixed together; it is a mixture.

Because the distribution of substances in the ice cream is not uniform, it is heterogeneous. c. Nitrox, consisting of nitrogen and oxygen, is a mixture. It has a uniform composition;

it is homogeneous.

SELF TEST 3.1

a. A salad dressing is prepared with oil, vinegar, and chunks of blue cheese. Is this a homogeneous or heterogeneous mixture?

b. A mouthwash used to reduce plaque and clean gums and teeth contains several ingredients, such as menthol, alcohol, hydrogen peroxide, and a flavoring. Is this a homogeneous or heterogeneous mixture?

ANSWER

a. heterogeneous mixture b. homogeneous mixture

PRACTICE PROBLEMS

3.1 Classification of Matter

3.1 Classify each of the following pure substances as an element or a compound:

a. a silicon (Si) chip b. hydrogen peroxide (H2O2) c. oxygen gas (O2) d. rust (Fe2O3) e. methane (CH4) in natural gas

3.2 Classify each of the following pure substances as an element or a compound:

a. helium gas (He) b. sulfur (S) c. sugar (C12H22O11) d. mercury (Hg) in a thermometer e. lye (NaOH)

3.3 Classify each of the following as a pure substance or a mixture: a. baking soda (NaHCO3) b. a blueberry muffin c. ice (H2O) d. zinc (Zn) e. trimix (oxygen, nitrogen, and helium) in a scuba tank

3.4 Classify each of the following as a pure substance or a mixture: a. a soft drink b. propane (C3H8)

c. a cheese sandwich d. an iron (Fe) nail e. salt substitute (KCl)

Applications

3.5 A dietitian includes one of the following mixtures in the lunch menu. Classify each of the following as homogeneous or heterogeneous:

a. vegetable soup b. tea c. fruit salad d. tea with ice and lemon slices

3.6 A dietitian includes one of the following mixtures in the lunch menu. Classify each of the following as homogeneous or heterogeneous:

a. nonfat milk b. spaghetti and meatballs c. peanut butter sandwich d. cranberry juice

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72 CHAPTER 3 Matter and Energy

A liquid has a definite volume, but not a definite shape. In a liquid, the particles move in random directions but are sufficiently attracted to each other to maintain a definite volume, although not a rigid structure. Thus, when water, oil, or vinegar is poured from one container to another, the liquid maintains its own volume but takes the shape of the new container.

A gas does not have a definite shape or volume. In a gas, the particles are far apart, have little attraction to each other, and move rapidly to fill the shape and volume of their container. When you inflate a bicycle tire, the air, which is a gas, fills the entire volume of the tire. The propane gas in a tank fills the entire volume of the tank.

ENGAGE 3.3 Why does a gas take both the shape and volume of its container?

H H O

Water as a liquid takes the shape of its container.

He

A gas takes the shape and volume of its container.

PRACTICE PROBLEMS Try Practice Problems 3.7 and 3.8

Characteristic Solid Liquid Gas

Shape Has a definite shape Takes the shape of the container Takes the shape of the container

Volume Has a definite volume Has a definite volume Fills the volume of the container

Arrangement of Particles Fixed, very close Random, close Random, far apart

Interaction Between Particles Very strong Strong Essentially none

Movement of Particles Vibrate in fixed positions Move slowly around each other Move rapidly, spread out

Examples Ice, salt, iron Water, oil, vinegar Water vapor, helium, air

TABLE 3.1 A Comparison of Solids, Liquids, and Gases

3.2 States and Properties of Matter LEARNING GOAL Identify the states and the physical and chemical properties of matter.

On Earth, matter exists in one of three physical forms called the states of matter: solids, liquids, and gases. Water is a familiar substance that we routinely observe in all three states. In the solid state, water can be an ice cube or a snowflake. It is a liquid when it comes out of a faucet or fills a pool. Water forms a gas, or vapor, when it evaporates from wet clothes or boils in a pan. A solid, such as a pebble or a baseball, has a definite shape and volume. You can probably recognize several solids within your reach right now such as books, pencils, or a computer mouse. In a solid, strong attractive forces hold the particles close together. The particles in a solid are arranged in such a rigid pattern, their only movement is to vibrate in fixed positions. For many solids, this rigid structure produces a crystal such as that seen in amethyst. TABLE 3.1 compares the three states of matter.

Si O

Amethyst, a solid, is a form of quartz that contains atoms of Si and O.

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3.2 States and Properties of Matter 73

Physical Properties and Physical Changes One way to describe matter is to observe its properties. For example, if you were asked to describe yourself, you might list characteristics such as your height and weight, the color of your eyes and skin, or the length, color, and texture of your hair.

Physical properties are those characteristics that can be observed or measured without affecting the identity of a substance. In chemistry, typical physical properties include the shape, color, melting point, boiling point, and physical state of a sub- stance. For example, some of the physical properties of a penny include its round shape, orange-red color (from copper), solid state, and shiny luster. TABLE 3.2 gives more examples of physical properties of copper, which is found in pennies, electrical wiring, and copper pans.

Water is a substance that is commonly found in all three states: solid, liquid, and gas. When matter undergoes a physical change, its state, size, or appearance will change, but its composition does not change. The solid state of water, snow or ice, has a different appear- ance than its liquid or gaseous state, but all three states are water. In a physical change, no new substances are produced.

Physical change is used to separate mixtures because there are no chemical inter- actions between the components. For example, different coins, such as nickels, dimes, and quarters, can be separated by size; iron particles mixed with sand can be picked up with a magnet; and water is separated from cooked spaghetti by using a strainer (see FIGURE 3.2).

In the chemistry laboratory, mixtures are separated by various methods. Solids are separated from liquids by filtration, which involves pouring a mixture through a filter paper, set in a funnel. The solid (residue) remains in the filter paper, and the filtered liquid (filtrate) moves through. In chromatography, different components of a liquid mixture separate as they move at different rates up the surface of a piece of chromatography paper.

Copper, used in cookware, is a good conductor of heat.

FIGURE 3.2 A mixture of spaghetti and water is separated using a strainer.

State at 25 °C Solid

Color Orange-red

Odor Odorless

Melting Point 1083 °C

Boiling Point 2567 °C

Luster Shiny

Conduction of Electricity

Excellent

Conduction of Heat Excellent

TABLE 3.2 Some Physical Properties of Copper

Chemical Properties and Chemical Changes Chemical properties describe the ability of a substance to change into a new substance. For example, the rusting or corrosion of a metal, such as iron, is a chemical property. When a chemical change takes place, the original substances, iron (Fe) and oxygen (O2), are converted into a new substance, rust (Fe2O3), which has different physical and chemical properties. Another example of a chemical change occurs when table salt, sodium chloride, is separated into sodium metal and chlorine gas, as seen in FIGURE 3.3. TABLE 3.3 sum- marizes physical and chemical properties and changes. TABLE 3.4 gives examples of some physical and chemical changes.

A mixture of a liquid and a solid is separated by filtration.

Different substances in a mixture of inks are separated as they travel at different rates up the surface of chromatography paper.

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74 CHAPTER 3 Matter and Energy

TABLE 3.3 Summary of Physical and Chemical Properties and Changes Physical Chemical

Property A characteristic of a substance: color, shape, odor, luster, size, melting point, or density.

A characteristic that indicates the ability of a substance to form another substance: paper can burn, iron can rust, silver can tarnish.

Change A change in a physical property that retains the identity of the substance: a change of state, a change in size, or a change in shape.

A change in which the original substance is converted to one or more new substances: paper burns, iron rusts, silver tarnishes.

TABLE 3.4 Examples of Some Physical and Chemical Changes Physical Changes Chemical Changes

Water boils to form water vapor. Water and cesium combine explosively.

Paper is cut into tiny pieces of confetti. Paper burns with a bright flame and produces heat, ashes, carbon dioxide, and water vapor.

Sugar dissolves in water to form a sugar solution.

Heating sugar forms a smooth, caramel- colored substance.

Iron has a melting point of 1538 °C. Iron, which is gray and shiny, combines with oxygen to form orange-red iron oxide (rust).

A chemical change occurs when sugar is heated, forming a caramelized topping for flan.

CORE CHEMISTRY SKILL Identifying Physical and Chemical

Changes

ENGAGE 3.5 Why is the melting point of iron, 1538 °C, a physical property, whereas the heating of iron with oxygen to form rust, Fe2O3, is a chemical property?

FIGURE 3.3 The decomposition of salt, NaCl, produces the elements sodium and chlorine.

Sodium metal

Sodium chloride

Chemical change

Chlorine gasand

+

ENGAGE 3.4 Why is the decomposition of salt a chemical change?

SAMPLE PROBLEM 3.2 Physical and Chemical Changes

TRY IT FIRST

Classify each of the following as a physical or chemical change:

a. A gold ingot is hammered to form gold leaf. b. Gasoline burns in air. c. Garlic is chopped into small pieces. d. Milk left in a warm room turns sour. e. A mixture of oil and water is separated.

A gold ingot is hammered to form gold leaf.

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3.3 Temperature 75

PRACTICE PROBLEMS

3.2 States and Properties of Matter

3.7 Indicate whether each of the following describes a gas, a liquid, or a solid: a. The breathing mixture in a scuba tank has no definite

volume or shape. b. The neon atoms in a lighting display do not interact with

each other. c. The particles in an ice cube are held in a rigid structure.

3.8 Indicate whether each of the following describes a gas, a liquid, or a solid: a. Lemonade has a definite volume but takes the shape of its

container. b. The particles in a tank of oxygen are very far apart. c. Helium occupies the entire volume of a balloon.

3.9 Describe each of the following as a physical or chemical property: a. Chromium is a steel-gray solid. b. Hydrogen reacts readily with oxygen. c. A patient has a temperature of 40.2 °C. d. Ammonia will corrode iron. e. Butane gas in an igniter burns in oxygen.

3.10 Describe each of the following as a physical or chemical property: a. Neon is a colorless gas at room temperature. b. Apple slices turn brown when they are exposed to air. c. Phosphorus will ignite when exposed to air. d. At room temperature, mercury is a liquid. e. Propane gas is compressed to a liquid for placement in a

small cylinder.

3.11 What type of change, physical or chemical, takes place in each of the following? a. Water vapor condenses to form rain. b. Cesium metal reacts explosively with water. c. Gold melts at 1064 °C. d. A puzzle is cut into 1000 pieces. e. Cheese is grated.

3.12 What type of change, physical or chemical, takes place in each of the following? a. Pie dough is rolled into thin pieces for a crust. b. A silver pin tarnishes in the air. c. A tree is cut into boards at a saw mill. d. Food is digested. e. A chocolate bar melts.

3.13 Describe each of the following properties for the element fluorine as physical or chemical: a. is highly reactive b. is a gas at room temperature c. has a pale, yellow color d. will explode in the presence of hydrogen e. has a melting point of - 220 °C

3.14 Describe each of the following properties for the element zirconium as physical or chemical: a. melts at 1852 °C b. is resistant to corrosion c. has a grayish white color d. ignites spontaneously in air when finely divided e. is a shiny metal

3.3 Temperature LEARNING GOAL Given a temperature, calculate the corresponding temperature on another scale.

Temperatures in science are measured and reported in units of degrees Celsius (°C). On the Celsius scale, the reference points are the freezing point of water, defined as 0 °C, and the boiling point, defined as 100 °C. In the United States, everyday temperatures are commonly reported in units of degrees Fahrenheit (°F). On the Fahrenheit scale, water freezes at 32 °F

REVIEW Using Positive and Negative

Numbers in Calculations (1.4)

Solving Equations (1.4)

Counting Significant Figures (2.2)

SOLUTION

a. physical change b. chemical change c. physical change d. chemical change e. physical change

SELF TEST 3.2

Classify each of the following as a physical or chemical change: a. Water freezes on a pond. b. Gas bubbles form when baking powder is placed in vinegar. c. A log is cut for firewood. d. Butter melts in a warm room.

ANSWER

a. physical change b. chemical change c. physical change d. physical change

PRACTICE PROBLEMS Try Practice Problems 3.9 to 3.14

INTERACTIVE VIDEO

Chemical vs. Physical Changes

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76 CHAPTER 3 Matter and Energy

and boils at 212 °F. A typical room temperature of 22 °C is the same as 72 °F. Normal human body temperature is 37.0 °C, which is the same temperature as 98.6 °F.

On the Celsius and Fahrenheit temperature scales, the temperature difference between freezing and boiling is divided into smaller units called degrees. On the Celsius scale, there are 100 degrees Celsius between the freezing and boiling points of water, whereas the Fahrenheit scale has 180 degrees Fahrenheit between the freezing and boiling points of water. That makes a degree Celsius almost twice the size of a degree Fahrenheit: 1 °C = 1.8 °F (see FIGURE 3.4).

180 degrees Fahrenheit = 100 degrees Celsius

180 degrees Fahrenheit

100 degrees Celsius =

1.8 °F 1 °C

We can write a temperature equation that relates a Fahrenheit temperature and its corresponding Celsius temperature.

TF = 1.8(TC) + 32

A digital ear thermometer is used to measure body temperature.

FIGURE 3.4 A comparison of the Fahrenheit, Celsius, and Kelvin temperature scales between the freezing and boiling points of water.

KelvinCelsiusFahrenheit

373 K

310. K

273 K

37.0 °C

0 °C

98.6 °F

32 °F

100 °C212 °F

100 kelvins

180 degrees

Fahrenheit

Boiling water

Freezing point of water

Normal body temperature

Boiling point of water

100 degrees Celsius

In the equation, the Celsius temperature is multiplied by 1.8 to change °C to °F; then 32 is added to adjust the freezing point from 0 °C to the Fahrenheit freezing point, 32 °F. The values, 1.8 and 32, used in the temperature equation are exact numbers and are not used to determine significant figures in the answer.

To convert from degrees Fahrenheit to degrees Celsius, the temperature equation is rearranged to solve for TC. First, we subtract 32 from both sides because we must apply the same operation to both sides of the equation.

TF - 32 = 1.8(TC) + 32 - 32

TF - 32 = 1.8(TC)

ENGAGE 3.6 Why is a degree Celsius a larger unit of temperature than a degree Fahrenheit?

CORE CHEMISTRY SKILL Converting Between Temperature

Scales

Adjusts freezing point

Changes °C to °F

Temperature equation to obtain degrees Fahrenheit

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3.3 Temperature 77

Second, we solve the equation for TC by dividing both sides by 1.8.

TF - 32 1.8

= 1.8(TC)

1.8

TC = TF - 32

1.8

Scientists have learned that the coldest temperature possible is - 273 °C (more precisely, - 273.15 °C). On the Kelvin scale, this temperature, called absolute zero, has the value of 0 K. Units on the Kelvin scale are called kelvins (K); no degree symbol is used. Because there are no lower temperatures than absolute zero, the Kelvin scale has no negative tem- perature values. Between the freezing point of water, 273 K, and the boiling point, 373 K, there are 100 kelvins, which makes a kelvin equal in size to a degree Celsius.

1 K = 1 °C We can write an equation that relates a Celsius temperature to its corresponding Kelvin

temperature by adding 273 to the Celsius temperature. TABLE 3.5 gives a comparison of some temperatures on the three scales.

TK = TC + 273 An antifreeze mixture in a car radiator will not freeze until the temperature drops to

- 37 °C. We can calculate the temperature of the antifreeze mixture in kelvins by adding 273 to the temperature in degrees Celsius.

TK = - 37 °C + 273 = 236 K

ENGAGE 3.7 Show that - 40. °C is the same temperature as - 40. °F.

Temperature equation to obtain kelvins

Temperature equation to obtain degrees Celsius

Example Fahrenheit (°F) Celsius (°C) Kelvin (K)

Sun 9937 5503 5776

A hot oven 450 232 505

Water boils 212 100 373

A high fever 104 40 313

Normal body temperature 98.6 37.0 310

Room temperature 70 21 294

Water freezes 32 0 273

A northern winter - 66 - 54 219 Nitrogen liquefies - 346 - 210 63 Absolute zero - 459 - 273 0

TABLE 3.5 A Comparison of Temperatures

SAMPLE PROBLEM 3.3 Calculating Temperature

TRY IT FIRST

A dermatologist uses cryogenic nitrogen at - 196 °C to remove skin lesions and some skin cancers. What is the temperature, in degrees Fahrenheit, of the nitrogen?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

- 196 °C T in degrees Fahrenheit temperature equation

STEP 2 Write a temperature equation.

TF = 1.8(TC) + 32

The low temperature of cryogenic nitrogen is used to destroy skin lesions.

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78 CHAPTER 3 Matter and Energy

STEP 3 Substitute in the known values and calculate the new temperature.

TF = 1.8(-196) + 32 1.8 is exact; 32 is exact

= - 353 + 32 = - 321 °F Answer to the ones place

SELF TEST 3.3

a. In the process of making ice cream, rock salt is added to crushed ice to chill the ice cream mixture to - 11 °C. What is that temperature in degrees Fahrenheit?

b. A hot tub reaches a temperature of 40.6 °C. What is that temperature in degrees Fahrenheit?

ANSWER

a. 12 °F b. 105 °F PRACTICE PROBLEMS

Try Practice Problems 3.15 to 3.18

TK = 45 + 273

Ones place

Ones place

= 318 K

Ones place

SAMPLE PROBLEM 3.4 Calculating Degrees Celsius and Kelvins

TRY IT FIRST

In a type of cancer treatment called thermotherapy, temperatures as high as 113 °F are used to destroy cancer cells. What is that temperature in degrees Celsius? In kelvins?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

113 °F T in degrees Celsius, kelvins temperature equations

STEP 2 Write a temperature equation.

TF - 32 1.8

TC = TK = TC + 273

32 is exact; 1.8 is exact

STEP 3 Substitute in the known values and calculate the new temperature.

(113 - 32) 1.8

TC =

Two SFs

Exact Two SFs

= 81 1.8

= 45 °C

Using the equation that converts degrees Celsius to kelvins, we substitute in degrees Celsius.

SELF TEST 3.4

a. A child has a temperature of 103.6 °F. What is this temperature on a Celsius thermometer? b. A child who fell through the ice on a lake has hypothermia with a core temperature of

90. °F. What is this temperature in degrees Celsius?

ANSWER

a. 39.8 °C b. 32 °C

PRACTICE PROBLEMS Try Practice Problems 3.19 and 3.20

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3.4 Energy 79

PRACTICE PROBLEMS

3.3 Temperature

3.15 Your friend who is visiting from Canada just took her tempera- ture. When she reads 99.8 °F, she becomes concerned that she is quite ill. How would you explain this temperature to your friend?

3.16 You have a friend who is using a recipe for flan from a Mexican cookbook. You notice that he set your oven temperature at 175 °F. What would you advise him to do?

3.17 Calculate the unknown temperature in each of the following: a. 37.0 °C = °F b. 65.3 °F = °C c. - 27 °C = K d. 62 °C = K e. 114 °F = °C

3.18 Calculate the unknown temperature in each of the following: a. 25 °C = °F b. 155 °C = °F

c. - 25 °F = °C d. 224 K = °C e. 145 °C = K

Applications

3.19 a. A patient with hyperthermia has a temperature of 106 °F. What does this read on a Celsius thermometer?

b. Because high fevers can cause convulsions in children, the doctor needs to be called if the child’s temperature goes over 40.0 °C. Should the doctor be called if a child has a tempera- ture of 103 °F?

3.20 a. Water is heated to 145 °F. What is the temperature of the hot water in degrees Celsius?

b. During extreme hypothermia, a child’s temperature dropped to 20.6 °C. What was his temperature in degrees Fahrenheit?

Chemistry Link to Health Variation in Body Temperature

Normal body temperature is considered to be 37.0 °C, although it var- ies throughout the day and from person to person. Oral temperatures of 36.1 °C are common in the morning and climb to a high of 37.2 °C between 6 p.m. and 10 p.m. Individuals who are involved in prolonged exercise may also experience elevated temperatures. Body temperatures of marathon runners can range from 39 °C to 41 °C as heat production during exercise exceeds the body’s ability to lose heat. Temperatures above 37.2 °C for a person at rest are usually an indication of illness.

Hyperthermia occurs at body temperatures above 41 °C. Sweat production stops, and the skin becomes hot and dry. The pulse rate is elevated, and respiration becomes weak and rapid. The person can become lethargic and lapse into a coma. High body temperatures can lead to convulsions, particularly in children, which may cause per- manent brain damage. Damage to internal organs is a major concern, and treatment, which must be immediate, may include immersing the person in an ice-water bath.

At the low temperature extreme of hypothermia, body tempera- ture can drop as low as 28.5 °C. The person may appear cold and pale and have an irregular heartbeat. Unconsciousness can occur if the body temperature drops below 26.7 °C. Respiration becomes slow and shallow, and oxygenation of the tissues decreases. Treatment

involves providing oxygen and increasing blood volume with glucose and saline fluids. Injecting warm fluids (37.0 °C) into the peritoneal cavity may restore the internal temperature.

107.6

105.8 Hyperthermia

Normal range

Hypothermia

Fever

Death

104.0

102.2

100.4

98.6

96.8

95.0

93.2

42.0

41.0

40.0

39.0

38.0

37.0

36.0

35.0

34.0

°C °F

3.4 Energy LEARNING GOAL Identify energy as potential or kinetic; convert between units of energy.

Almost everything you do involves energy. When you are running, walking, dancing, or thinking, you are using energy to do work. In fact, energy is defined as the ability to do work. Suppose you are climbing a steep hill and you become too tired to go on. At that moment, you do not have the energy to do any more work. Now suppose you sit down and have lunch. In a while, you will have obtained energy from the food, and you will be able to do more work and complete the climb.

REVIEW Rounding Off (2.3)

Using Significant Figures in Calculations (2.3)

Writing Conversion Factors from Equalities (2.5)

Using Conversion Factors (2.6)

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80 CHAPTER 3 Matter and Energy

Kinetic and Potential Energy Energy can be classified as either kinetic energy or potential energy. Kinetic energy is the energy of motion. Any object that is moving has kinetic energy. Potential energy is deter- mined by the position or composition of a substance. A boulder resting on top of a mountain has potential energy because of its location. If the boulder rolls down the mountain, the potential energy becomes kinetic energy. Water stored in a reservoir has potential energy. When the water goes over the dam and falls to the stream below, its potential energy is converted to kinetic energy. Foods and fossil fuels have potential energy. When you digest food or burn gasoline in your car, potential energy is converted to kinetic energy to do work.

Heat and Energy Heat is the energy associated with the motion of particles. An ice cube feels cold because heat f lows from your hand into the ice cube. The faster the particles move, the greater the heat or thermal energy of the substance. In the ice cube, the particles are moving very slowly. As heat is added, the motion of the particles in the ice cube increases. Eventually, the particles have enough energy to make the ice cube melt as it changes from a solid to a liquid.

Units of Energy The SI unit of energy and work is the joule (J) (pronounced “jewel”). The joule is a small amount of energy, so scientists often use the kilojoule (kJ), 1000 joules. To heat water for one cup of tea, you need about 75 000 J or 75 kJ of heat. TABLE 3.6 shows a comparison of energy in joules for several energy sources or uses.

You may be more familiar with the unit calorie (cal), from the Latin caloric, meaning “heat.” The calorie was originally defined as the amount of energy (heat) needed to raise the temperature of 1 g of water by 1 °C. Now, one calorie is defined as exactly 4.184 J. This equality can be written as two conversion factors:

1 cal = 4.184 J (exact) 4.184 J 1 cal

and 1 cal

4.184 J

One kilocalorie (kcal) is equal to 1000 calories, and one kilojoule (kJ) is equal to 1000 joules. The equalities and conversion factors follow:

1 kcal = 1000 cal 1000 cal 1 kcal

and 1 kcal

1000 cal

1 kJ = 1000 J 1000 J 1 kJ

and 1 kJ

1000 J

ENGAGE 3.8 Why does a book have more potential energy when it is on the top of a high table than when it is on the floor?

PRACTICE PROBLEMS Try Practice Problems 3.21 to 3.24

Water at the top of the dam has potential energy. When the water flows over the dam, potential energy is converted to kinetic energy.

TABLE 3.6 A Comparison of Energy for Various Resources and Uses

1027 Energy radiated by the Sun in 1 s (1026)

Energy in Joules

World reserves of fossil fuel (1023)

Energy consumption for 1 yr in the United States (1020)

Solar energy reaching Earth in 1 s (1017)

Energy use per person in 1 yr in the United States (1011)

Energy from 1 gal of gasoline (108)

Energy from one serving of pasta, a doughnut, or needed to bicycle for 1 h (106)

Energy used to sleep for 1 h (105)

1024

1021

1018

1015

1012

109

106

103

100

CORE CHEMISTRY SKILL Using Energy Units

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3.4 Energy 81

PRACTICE PROBLEMS Try Practice Problems 3.25 to 3.28

SAMPLE PROBLEM 3.5 Energy Units

TRY IT FIRST

A defibrillator gives a high-energy shock of 360 J. What is this quantity of energy in calories?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

360 J calories energy factor

STEP 2 Write a plan to convert the given unit to the needed unit.

and

1 cal = 4.184 J 1 cal

4.184 J 4.184 J 1 cal

joules calories Energy factor

STEP 3 State the equalities and conversion factors.

STEP 4 Set up the problem to calculate the needed quantity.

360 J 86 cal* = 1 cal

4.184 J

Two SFs Exact

Exact

Two SFs

SELF TEST 3.5

a. When 1.0 g of glucose is metabolized in the body, it produces 3900 cal. How many joules are produced?

b. A swimmer expends 855 kcal during practice. How many kilojoules did the swimmer expend?

ANSWER

a. 16 000 J b. 3580 kJ or 3.58 * 103 kJ

A defibrillator provides electrical energy to heart muscle to re-establish normal rhythm.

PRACTICE PROBLEMS

3.4 Energy

3.21 Discuss the changes in the potential and kinetic energy of a roller-coaster ride as the roller-coaster car descends from the top of the ramp.

3.22 Discuss the changes in the potential and kinetic energy of a ski jumper taking the elevator to the top of the jump and going down the ramp.

3.23 Indicate whether each of the following statements describes potential or kinetic energy:

a. water at the top of a waterfall b. kicking a ball

c. the energy in a lump of coal d. a skier at the top of a hill

3.24 Indicate whether each of the following statements describes potential or kinetic energy:

a. the energy in your food b. a tightly wound spring c. a car speeding down the freeway d. an earthquake

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82 CHAPTER 3 Matter and Energy

3.25 Convert each of the following energy units: a. 3500 cal to kcal b. 415 J to cal c. 28 cal to J d. 4.5 kJ to cal

3.26 Convert each of the following energy units: a. 8.1 kcal to cal b. 325 J to kJ c. 2550 cal to kJ d. 2.50 kcal to J

Applications

3.27 The energy needed to keep a 75-watt light bulb burning for 1.0 h is 270 kJ. Calculate the energy required to keep the light bulb burning for 3.0 h in each of the following energy units:

a. joules b. kilocalories

3.28 A person uses 750 kcal on a long walk. Calculate the energy used for the walk in each of the following energy units:

a. joules b. kilojoules

3.5 Specific Heat LEARNING GOAL Calculate the specific heat for a substance. Use specific heat to calculate heat loss or gain.

Every substance has its own characteristic ability to absorb heat. When you bake a potato, you place it in a hot oven. If you are cooking pasta, you add the pasta to boiling water. You already know that adding heat to water increases its temperature until it boils. Certain sub- stances must absorb more heat than others to reach a certain temperature.

The energy requirements for different substances are described in terms of a physi- cal property called specific heat. The specific heat (SH) for a substance is defined as the amount of heat (q) in calories (or joules) needed to change the temperature of exactly 1 g of a substance by exactly 1 °C. To calculate the specific heat for a substance, we measure the heat in calories or joules, the mass in grams, and the temperature change written as ∆T.

Symbols Used in Specific Heat Equation

Symbol Meaning Unit

SH specific heat cal (or J)

g °C

q heat calories (cal), joules (J)

m mass grams (g)

∆T temperature change

degrees Celsius (°C)

ENGAGE 3.9 Why will 1 g of silver have a greater increase of temperature than 1 g of iron when each absorbs 400 J of heat?

TABLE 3.7 Specific Heats for Some Substances

Substance cal/g °C J/g °C

Elements

Aluminum, Al(s) 0.214 0.897

Copper, Cu(s) 0.0920 0.385

Gold, Au(s) 0.0308 0.129

Iron, Fe(s) 0.108 0.452

Silver, Ag(s) 0.0562 0.235

Titanium, Ti(s) 0.125 0.523

Compounds

Ammonia, NH3(g) 0.488 2.04

Ethanol, C2H6O(l) 0.588 2.46

Sodium chloride, NaCl(s) 0.207 0.864

Water, H2O(l) 1.00 4.184

Water, H2O(s) 0.485 2.03

Specific heat (SH)

= q

mass ¢T =

cal (or J)

g °C

The specific heat for water is written using our definition of the calorie and joule.

SH for H2O(l) = 1.00 cal

g °C =

4.184 J g °C

If we look at TABLE 3.7, we see that 1 g of water requires 1.00 cal or 4.184 J to increase its temperature by 1 °C. The specific heat of water is about five times the specific heat of aluminum. Aluminum has a specific heat that is about twice that of copper. However, adding the same amount of heat (1.00 cal or 4.184 J) will raise the temperature of 1 g of aluminum

The high specific heat of water keeps temperatures more moderate in summer and winter.

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3.5 Specific Heat 83

by about 5 °C and 1 g of copper by about 10 °C. The low specific heats of aluminum and copper mean they transfer heat efficiently, which makes them useful in cookware.

The high specific heat of water has a major impact on the temperatures in a coastal city compared to an inland city. A large mass of water near a coastal city can absorb or release five times the energy absorbed or released by the same mass of rock near an inland city. This means that in the summer a body of water absorbs large quantities of heat, which cools a coastal city, and then in the winter that same body of water releases large quantities of heat, which provides warmer temperatures. A similar effect happens with our bodies, which contain 70% water by mass. Water in the body absorbs or releases large quantities of heat to maintain body temperature.

CORE CHEMISTRY SKILL Using the Heat Equation

Heat Equation When we know the specific heat of a substance, we can calculate the heat lost or gained by measuring the mass of the substance and the initial and final temperatures. We can substi- tute these measurements into the specific heat equation that is rearranged to solve for heat, which we call the heat equation.

SH = q

m * ∆T

m * ∆T * SH = q

m * ∆T * m * ∆T

We can write the heat equation as

q m ¢T SH= * *

PRACTICE PROBLEMS Try Practice Problems 3.29 to 3.32

CORE CHEMISTRY SKILL Calculating Specific Heat

SH = 57.0 J

35.6 g 12.5 °C

0.128 J

g °C =

Three SFs

Three SFs Three SFs Three SFs

SH = heat

mass ¢T =

q

m ¢T

STEP 3 Set up the problem to calculate the specific heat.

SELF TEST 3.6

a. What is the specific heat, in J/g °C, of sodium if 92.3 J is needed to raise the temperature of 3.00 g of sodium by 25.0 °C?

b. What is the specific heat, in J/g °C, of magnesium if 58.8 J is needed to raise the temperature of 4.65 g of sodium by 12.4 °C?

ANSWER

a. 1.23 J/g °C b. 1.02 J/g °C

SAMPLE PROBLEM 3.6 Calculating Specific Heat

TRY IT FIRST

What is the specific heat, in J/g °C, of lead if 57.0 J raises the temperature of 35.6 g of lead by 12.5 °C?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

q = 57.0 J, m = 35.6 g of lead, ∆T = 12.5 °C

specific heat of lead (J/g °C)

specific heat equation

STEP 2 Write the relationship for specific heat. The specific heat (SH) is calcu- lated by dividing the heat (q) by the mass (m) and by the temperature change (∆T).

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84 CHAPTER 3 Matter and Energy

The heat lost or gained, in calories or joules, is obtained when the units of grams and °C in the numerator cancel grams and °C in the denominator of specific heat in the heat equation.

cal = g * °C * cal

g °C

J = g * °C * J

g °C

A cooling cap lowers the body temperature to reduce the oxygen required by the tissues.

SAMPLE PROBLEM 3.7 Calculating Heat Loss

TRY IT FIRST

During surgery or when a patient has suffered a cardiac arrest or stroke, lowering the body temperature will reduce the amount of oxygen needed by the body. Some methods used to lower body temperature include cooled saline solution, cool water blankets, or cooling caps worn on the head. How many kilojoules are lost when the body temperature of a surgery patient with a blood volume of 5500 mL is cooled from 38.5 °C to 33.2 °C? (Assume that the specific heat and density of blood is the same as for water.)

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

5500 mL of blood = 5500 g of blood, cooled from 38.5 °C to 33.2 °C

kilojoules removed

heat equation, specific heat of water

STEP 2 Calculate the temperature change (�T ).

∆T = 38.5 °C - 33.2 °C = 5.3 °C

STEP 3 Write the heat equation and needed conversion factors.

120 kJq * == 4.184 J

Two SFs

Exact

Two SFs

5500 g 5.3 °C

Two SFs

g °C *

1 kJ

1000 J

ExactExact

Exact

*

q m ¢T SH= * *

STEP 4 Substitute in the given values and calculate the heat, making sure units cancel.

SELF TEST 3.7

a. Dental implants use titanium because it is strong, is nontoxic, and can fuse to bone. How many calories are needed to raise the temperature of 16 g of titanium from 36.0 °C to 41.3 °C (see Table 3.7)?

b. Some cooking pans have a layer of copper on the bottom. How many kilojoules are needed to raise the temperature of 125 g of copper from 22 °C to 325 °C (see Table 3.7)?

ANSWER

a. 11 cal b. 14.6 kJ

A dental implant contains the metal titanium that is attached to a crown.

=

and g °C

g °C 4.184 J

g °C 4.184 J

SHwater

4.184 J 1000 J and

1 kJ = 1000 J

1 kJ

1 kJ

1000 J

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3.5 Specific Heat 85

Another use of the heat equation is to calculate the mass, in grams, of a substance by rearranging the heat equation to solve for mass (m) as shown in Sample Problem 3.8.

ENGAGE 3.10 Rearrange the heat equation to solve for mass (m).

PRACTICE PROBLEMS Try Practice Problems 3.33 to 3.40

SAMPLE PROBLEM 3.8 Using the Heat Equation

TRY IT FIRST

When 655 J is added to a sample of ethanol, its temperature rises from 18.2 °C to 32.8 °C. What is the mass, in grams, of the ethanol sample (see Table 3.7)?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

q = 655 J, heated from 18.2 °C to 32.8 °C

grams of ethanol

heat equation

STEP 2 Calculate the temperature change (�T).

∆T = 32.8 °C - 18.2 °C = 14.6 °C

STEP 3 Write the heat equation.

q m ¢T SH= * *

q

¢T SH m =

STEP 4 Substitute in the given values and solve, making sure units cancel.

When the heat equation is rearranged for mass (m), the heat is divided by the tempera- ture change and the specific heat.

SELF TEST 3.8

a. When 8.81 kJ is absorbed by a piece of iron, its temperature rises from 15 °C to 122 °C. What is the mass, in grams, of the piece of iron (see Table 3.7)?

b. A silver bar with a mass of 284 g at an initial temperature of 25.3 °C absorbs 1.13 kJ of heat. What is the final temperature of the silver bar (see Table 3.7)?

ANSWER

a. 182 g b. 42.2 °C

655 J m

14.6 °C

= 18.2 g = 2.46 J

g °C

Three SFs

Three SFsThree SFs

Three SFs

Heat Exchange: Heat Gain Equals Heat Loss If a piece of metal is dropped into a container of cold water, the metal cools and the water warms until they are both at the same temperature. We assume that the heat lost by the metal is equal to the heat gained by the water. The heat equation allows us to calculate the heat gained by the water. Because the heat loss and heat gain are equal, we can use the heat equation again to calculate the specific heat of the metal.

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86 CHAPTER 3 Matter and Energy

PRACTICE PROBLEMS

3.5 Specific Heat

3.29 If the same amount of heat is supplied to samples of 10.0 g each of aluminum, iron, and copper, all at 15.0 °C, which sample would reach the highest temperature (see Table 3.7)?

3.30 Substances A and B are the same mass and at the same initial temperature. When the same amount of heat is added to each, the final temperature of A is 75 °C and B is 35 °C. What does this tell you about the specific heats of A and B?

3.31 Calculate the specific heat (J/g °C) for each of the following: a. a 13.5-g sample of zinc heated from 24.2 °C to 83.6 °C that

absorbs 312 J of heat b. a 48.2-g sample of a metal that absorbs 345 J with a

temperature increase from 35.0 °C to 57.9 °C

3.32 Calculate the specific heat (J/g °C) for each of the following: a. an 18.5-g sample of tin that absorbs 183 J of heat when its

temperature increases from 35.0 °C to 78.6 °C b. a 22.5-g sample of a metal that absorbs 645 J when its

temperature increases from 36.2 °C to 92.0 °C

3.33 Use the heat equation to calculate the energy, in joules, for each of the following (see Table 3.7): a. required to heat 25.0 g of water from 12.5 °C to 25.7 °C b. required to heat 38.0 g of copper from 122 °C to 246 °C

3.34 Use the heat equation to calculate the energy, in joules, for each of the following (see Table 3.7): a. required to heat 5.25 g of water from 5.5 °C to 64.8 °C b. required to heat 10.0 g of silver from 112 °C to 275 °C

SELF TEST 3.9

A 50.-g piece of zinc metal at 62.0 °C is dropped into 24 g of water at 29.3 °C. All the heat lost by the metal is used to heat the water, and the final temperature of the water and metal is 34.6 °C. Calculate the specific heat, in J/g °C, of zinc.

ANSWER

0.39 J g °C

SAMPLE PROBLEM 3.9 Heat Exchange

TRY IT FIRST

A 44-g piece of chromium metal at 58.3 °C is dropped into a beaker containing 20. g of water at 24.6 °C. All the heat lost by the metal heats the water. The final temperature of the water and metal is 31.0 °C. Calculate the specific heat, in J/g °C, of chromium.

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

44 g of metal, ∆T = 58.3 °C - 31.0 °C = 27.3 °C, 20. g of water,

∆T = 31.0 °C - 24.6 °C = 6.4 °C

specific heat of metal

heat gain equals heat loss, specific heat of water

STEP 2 Write the heat equation.

q m ¢T SH= * *

For water,

q = 20. g * 6.4 °C * 4.184 J

g °C = 540 J

Because the heat gained by water is equal to the heat lost by the metal, q for the metal is equal to 540 J.

STEP 3 Write the relationship for specific heat and calculate the specific heat. The specific heat (SH) is calculated by dividing the heat (q) by the mass (m) and by the temperature change (∆T).

Solving the heat equation for the specific heat of the metal:

SH = q

m ∆T =

540 J 44 g * 27.3 °C

= 0.45 J g °C

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3.6 Energy and Nutrition 87

Steel combustion chamber

Food sample

Water

Stirrer

Insulated container

Ignition wires Thermometer

Oxygen

FIGURE 3.5 Heat released from burning a food sample in a calorimeter is used to determine the energy value for the food.

3.6 Energy and Nutrition LEARNING GOAL Use the energy values to calculate the kilocalories (kcal) or kilojoules (kJ) for a food.

The food we eat provides energy to do work in the body, which includes the growth and repair of cells. Carbohydrates are the primary fuel for the body, but if the carbohydrate reserves are exhausted, fats and then proteins are used for energy.

For many years in the field of nutrition, the energy from food was measured as Calories or kilocalories. The nutritional unit Calorie, Cal (with an uppercase C), is the same as 1000 cal, or l kcal. The international unit, kilojoule (kJ), is becoming more prevalent. For example, a baked potato has an energy content of 100 Calories, which is 100 kcal or 440 kJ. A typical diet that provides 2100 Cal (kcal) is the same as an 8800 kJ diet.

1 Cal = 1 kcal = 1000 cal

1 Cal = 4.184 kJ = 4184 J

In the nutrition laboratory, foods are burned in a steel container called a calorimeter to determine their energy value (kcal/g or kJ/g) (see FIGURE 3.5). A measured amount of water is added to fill the area surrounding the combustion chamber. The food sample

PRACTICE PROBLEMS Try Practice Problems 3.41 and 3.42

3.35 Calculate the mass, in grams, for each of the following using Table 3.7: a. a sample of gold that absorbs 225 J to increase its

temperature from 15.0 °C to 47.0 °C b. a sample of iron that loses 8.40 kJ when its temperature

decreases from 168.0 °C to 82.0 °C

3.36 Calculate the mass, in grams, for each of the following using Table 3.7: a. a sample of water, that absorbs 8250 J when its temperature

increases from 18.4 °C to 92.6 °C b. a sample of silver that loses 3.22 kJ when its temperature

decreases from 145 °C to 24 °C

3.37 Calculate the change in temperature (∆T) for each of the following using Table 3.7: a. 20.0 g of iron that absorbs 1580 J b. 150.0 g of water that absorbs 7.10 kJ

3.38 Calculate the change in temperature (∆T) for each of the following using Table 3.7: a. 115 g of copper that loses 2.45 kJ b. 22.0 g of silver that loses 625 J

3.39 a. A 13.0-g piece of glass at 64.8 °C is dropped into 38 g of water at 22.9 °C. All the heat lost by the glass is used to heat the water, and the final temperature of the water and glass is 25.6 °C. Calculate the specific heat, in J/g °C, of the glass.

b. A 19.6-g piece of metal at 88.7 °C is dropped into 42 g of water at 24.2 °C. All the heat lost by the metal is used to heat the water, and the final temperature of the water and metal is 27.6 °C. Calculate the specific heat, in J/g °C, of the metal.

3.40 a. A 22.8-g piece of metal at 92.6 °C is dropped into 46 g of water at 24.7 °C. All the heat lost by the metal is used to heat the water, and the final temperature of the water and metal is 28.2 °C. Calculate the specific heat, in J/g °C, of the metal.

b. A 23.4-g piece of metal at 96.9 °C is dropped into 52 g of water at 23.3 °C. All the heat lost by the metal is used to heat the water, and the final temperature of the water and metal is 24.4 °C. Calculate the specific heat, in J/g °C, of the metal.

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88 CHAPTER 3 Matter and Energy

Serving Size 14 crackers (31g) Servings Per Container About 7

Amount Per Serving

Calories 130 Calories from Fat 40

% Daily Value*

Total Fat 4 g 6%

Saturated Fat 0.5 g 3%

Trans Fat 0 g

Polyunsaturated Fat 0.5%

Monounsaturated Fat 1.5 g

Cholesterol 0 mg 0%

Sodium 310 mg 13%

Total Carbohydrate 19 g 6%

Dietary Fiber Less than 1g 4%

Sugars 2 g

Proteins 2 g

Nutrition Facts

Snack Crackers Snack Crackers

The Nutrition Facts include the total Calories and the grams of carbohydrate, fat, and protein per serving.

SAMPLE PROBLEM 3.10 Calculating the Energy from a Food

TRY IT FIRST

While working on his diet log, Charles observed that the Nutrition Facts label for Snack Crackers states that one serving contains 19 g of carbohydrate, 4 g of fat, and 2 g of pro- tein. If Charles eats one serving of Snack Crackers, what is the energy, in kilocalories, from each food type and the total kilocalories? (Round off the kilocalories for each food type to the tens place.)

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

19 g of carbohydrate, 4 g of fat, 2 g of protein

total number of kilocalories

energy values

is burned, releasing heat that increases the temperature of the water. From the known mass of the food and water, as well as the measured temperature increase, the energy value for the food is calculated. We assume that the energy absorbed by the calorimeter is negligible.

Energy Values for Foods The energy values for food, listed in TABLE 3.8, are the kilocalories or kilojoules obtained from burning 1 g of carbohydrate, fat, or protein. Using these energy values, we can calcu- late the total energy for a food if the mass of each food type is known.

kilocalories = g * kcal

g kilojoules = g *

kJ g

On packaged food, the energy content is listed on the Nutrition Facts label, usually in terms of the number of Calories or kilojoules for one serving. The composition and energy content for some foods are given in TABLE 3.9. The total energy in kilocalories for each food type was calculated using energy values in kilocalories. Total energy in kilojoules was calculated using energy values in kilojoules. The energy for each food type was rounded off to the tens place.

TABLE 3.8 Typical Energy Values for the Three Food Types

Food Type kcal/g kJ/g

Carbohydrate 4 17

Fat 9 38

Protein 4 17

ENGAGE 3.11 What type of food provides the most energy per gram?

TABLE 3.9 Composition and Energy Content for Some Foods Food Carbohydrate (g) Fat (g) Protein (g) Energy (kcal, kJ)

Apple, 1 medium 15 0 0 60 kcal (260 kJ)

Banana, 1 medium 26 0 1 110 kcal (460 kJ)

Beef, ground, 3 oz 0 14 22 220 kcal (900 kJ)

Broccoli, 3 oz 4 0 3 30 kcal (120 kJ)

Carrots, 1 cup 11 0 2 50 kcal (220 kJ)

Chicken, no skin, 3 oz 0 3 20 110 kcal (450 kJ)

Egg, 1 large 0 6 6 70 kcal (330 kJ)

Milk, nonfat, 1 cup 12 0 9 90 kcal (350 kJ)

Potato, baked 23 0 3 100 kcal (440 kJ)

Salmon, 3 oz 0 5 16 110 kcal (460 kJ)

Steak, 3 oz 0 27 19 320 kcal (1350 kJ)

M03_TIMB8119_06_SE_C03.indd 88 11/27/18 11:43 AM

3.6 Energy and Nutrition 89

PRACTICE PROBLEMS Try Practice Problems 3.43 to 3.50

STEP 2 Use the energy value for each food type to calculate the kilocalories, rounded off to the tens place.

Food Type Mass Energy Value Energy

Carbohydrate 19 g * 4 kcal

1 g = 80 kcal

Fat 4 g * 9 kcal

1 g = 40 kcal

Protein 2 g * 4 kcal

1 g = 10 kcal

STEP 3 Add the energy for each food type to give the total energy from the food.

Total energy = 80 kcal + 40 kcal + 10 kcal = 130 kcal

SELF TEST 3.10

a. Using the mass of carbohydrate listed in the Nutrition Facts, calculate the energy, in kilojoules, from the carbohydrate in one serving of Snack Crackers. (Round off the kilojoules to the tens place.)

b. Using the energy calculations from Sample Problem 3.10, calculate the percentage of energy obtained from fat in one serving of Snack Crackers.

ANSWER

a. 320 kJ b. 31%

TABLE 3.10 Typical Energy Requirements for Adults

Gender Age Moderately Active kcal (kJ)

Highly Active kcal (kJ)

Female 19–30 2100 (8800) 2400 (10 000)

31–50 2000 (8400) 2200 (9200)

Male 19–30 2700 (11 300) 3000 (12 600)

31–50 2500 (10 500) 2900 (12 100)

One hour of swimming uses 2100 kJ of energy.

TABLE 3.11 Energy Expended by a 70.0-kg (154-lb) Adult

Activity Energy (kcal/h) Energy (kJ/h)

Sleeping 60 250

Sitting 100 420

Walking 200 840

Swimming 500 2100

Running 750 3100

Chemistry Link to Health Losing and Gaining Weight

The number of kilocalories or kilojoules needed in the daily diet of an adult depends on gender, age, and level of physical activity. Some typical levels of energy needs are given in TABLE 3.10.

A person gains weight when food intake exceeds energy output. The amount of food a person eats is regulated by the hunger center in the hypothalamus, which is located in the brain. Food intake is normally proportional to the nutrient stores in the body. If these nutrient stores are low, you feel hungry; if they are high, you do not feel like eating.

A person loses weight when food intake is less than energy out- put. Many diet products contain cellulose, which has no nutritive value but provides bulk and makes you feel full. Some diet drugs depress the hunger center and must be used with caution because

they excite the nervous system and can elevate blood pressure. Because muscular exercise is an important way to expend energy, an increase in daily exercise aids weight loss. TABLE 3.11 lists some activities and the amount of energy they require.

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90 CHAPTER 3 Matter and Energy

PRACTICE PROBLEMS

3.6 Energy and Nutrition

3.41 Calculate the kilocalories for each of the following: a. one stalk of celery that produces 125 kJ when burned in a

calorimeter b. a waffle that produces 870 kJ when burned in a calorimeter

3.42 Calculate the kilocalories for each of the following: a. one cup of popcorn that produces 131 kJ when burned in a

calorimeter b. a sample of butter that produces 23.4 kJ when burned in a

calorimeter

3.43 Using the energy values for foods (see Table 3.8), determine each of the following (round off the answer for each food type to the tens place): a. the total kilojoules for one cup of orange juice that contains

26 g of carbohydrate, no fat, and 2 g of protein b. the grams of carbohydrate in one apple if the apple has no fat

and no protein and provides 72 kcal of energy c. the kilocalories in one tablespoon of vegetable oil, which

contains 14 g of fat and no carbohydrate or protein d. the grams of fat in one avocado that has 410 kcal, 13 g of

carbohydrate, and 5 g of protein

3.44 Using the energy values for foods (see Table 3.8), determine each of the following (round off the answer for each food type to the tens place): a. the total kilojoules in two tablespoons of crunchy peanut butter

that contains 6 g of carbohydrate, 16 g of fat, and 7 g of protein b. the grams of protein in one cup of soup that has 110 kcal

with 9 g of carbohydrate and 6 g of fat c. the kilocalories in one can of cola if it has 40. g of

carbohydrate and no fat or protein d. the total kilocalories for one cup of vanilla ice cream that

contains 26 g of carbohydrate, 14 g of fat, and 4 g of protein

Applications

3.45 For dinner, Charles had one cup of clam chowder, which con- tains 16 g of carbohydrate, 12 g of fat, and 9 g of protein (see Table 3.8). How much energy, in kilocalories and kilojoules, is in the clam chowder? (Round off the answer for each food type to the tens place.)

3.46 For lunch, Charles consumed 3 oz of skinless chicken, 3 oz of broccoli, 1 medium apple, and 1 cup of nonfat milk (see Table 3.9). How many kilocalories did Charles obtain from the lunch?

3.47 A patient receives 3.2 L of intravenous (IV) glucose solution. If 100. mL of the solution contains 5.0 g of glucose (carbohydrate), how many kilocalories did the patient obtain from the glucose solution (see Table 3.8)?

3.48 A high-protein diet contains 70.0 g of carbohydrate, 5.0 g of fat, and 150 g of protein (see Table 3.8). How much energy, in kilocalories and kilojoules, does this diet provide? (Round off the answer for each food type to the tens place.)

3.49 When a 1.50-g sample of walnuts is burned in a calorimeter, the heat released increases the temperature of 450. g of water from 21.8 °C to 44.1 °C. What is the energy value (kJ/g) for the walnuts?

3.50 When a 1.2-g sample of chocolate-chip cookie is burned in a calorimeter, the heat released increases the temperature of 360 g of water from 23 °C to 40. °C. What is the energy value (kJ/g) for the chocolate-chip cookie?

UPDATE A Diet and Exercise Program

After Charles has been on his diet and exercise plan for a month, he and his mother meet again with the dietitian. Daniel looks at the diary of food intake and exercise, and weighs Charles to evaluate how well the diet is working. The following is what Charles ate in one day:

Breakfast 1 banana, 1 cup of nonfat milk, 1 egg

Lunch 1 cup of carrots, 3 oz of ground beef, 1 apple, 1 cup of nonfat milk

Dinner 6 oz of skinless chicken, 1 baked potato, 3 oz of broccoli, 1 cup of nonfat milk

Applications

3.51 Using energy contents from Table 3.9, determine each of the following: a. the total kilocalories for each meal b. the total kilocalories for one day c. If Charles consumes 1800 kcal per day, he will maintain

his weight. Would he lose weight on his new diet? d. If expending 3500 kcal is equal to a loss of 1.0 lb, how

many days will it take Charles to lose 5.0 lb?

3.52 a. During one week, Charles swam for a total of 2.5 h and walked for a total of 8.0 h. If Charles expends 340 kcal/h swimming and 160 kcal/h walking, how many total kilocalories did he expend for one week (see Table 3.11)?

b. For the amount of exercise that Charles did for one week in part a, if expending 3500 kcal is equal to a loss of 1.0 lb, how many pounds did he lose?

c. How many hours would Charles have to walk to lose 1.0 lb?

d. How many hours would Charles have to swim to lose 1.0 lb?

M03_TIMB8119_06_SE_C03.indd 90 11/27/18 11:43 AM

Chapter Review 91

MATTER AND ENERGY

Matter

Elements

Compounds

has states of

can be

or

or

or

Homogeneous

Heterogeneous

Particle MotionSolid Liquid Gas

Energy

affects

Heat

as

Calories or Joules

Temperature Change

Specific Heat

Heat Equation

Mass

using

Pure Substances Mixtures

that are that are measured in

CONCEPT MAP

CHAPTER REVIEW

3.1 Classification of Matter LEARNING GOAL Classify examples of matter as pure substances or mixtures. • Matter is anything that has mass and

occupies space. • Matter is classified as pure substances

or mixtures. • Pure substances, which are ele-

ments or compounds, have fixed compositions, and mixtures have variable compositions.

• The substances in mixtures can be separated using physical methods.

3.2 States and Properties of Matter LEARNING GOAL Identify the states and the physical and chemical properties of matter. • The three states of matter are solid,

liquid, and gas. • A physical property is a

characteristic of a substance that can be observed or measured without affecting the identity of the substance.

• A physical change occurs when physical properties change, but not the composition of the substance.

• A chemical property indicates the ability of a substance to change into another substance.

• A chemical change occurs when one or more substances react to form a new substance with new physical and chemical properties.

3.3 Temperature LEARNING GOAL Given a temperature, calculate the corresponding temperature on another scale. • In science, temperature is measured in

degrees Celsius (°C) or kelvins (K). • On the Celsius scale, there are 100 units

between the freezing point (0 °C) and the boiling point (100 °C) of water.

• On the Fahrenheit scale, there are 180 units between the freezing point (32 °F) and the boiling point (212 °F) of water. A Fahrenheit temperature is related to its Celsius temperature by the equation TF = 1.8(TC) + 32.

• The SI unit, kelvin, is related to the Celsius temperature by the equation TK = TC + 273.

3.4 Energy LEARNING GOAL Identify energy as potential or kinetic; convert between units of energy. • Energy is the ability to do work. • Potential energy is determined by

the position or by the composition of a substance; kinetic energy is the energy of motion.

• Common units of energy are the calorie (cal), kilocalorie (kcal), joule (J), and kilojoule (kJ).

• One calorie is equal to 4.184 J.

H H O

Aluminum atom 107.6

105.8 Hyperthermia

Normal range

Hypothermia

Fever

Death

104.0

102.2

100.4

98.6

96.8

95.0

93.2

42.0

41.0

40.0

39.0

38.0

37.0

36.0

35.0

34.0

°C °F

M03_TIMB8119_06_SE_C03.indd 91 11/27/18 11:43 AM

92 CHAPTER 3 Matter and Energy

3.5 Specific Heat LEARNING GOAL Calculate the specific heat for a substance. Use specific heat to calculate heat loss or gain. • Specific heat is the amount of

energy required to raise the temperature of exactly 1 g of a substance by exactly 1 °C.

• The heat lost or gained by a substance is determined by multiply- ing its mass, the temperature change, and its specific heat.

3.6 Energy and Nutrition LEARNING GOAL Use the energy values to calculate the kilocalories (kcal) or kilojoules (kJ) for a food. • The nutritional Calorie is the same

amount of energy as 1 kcal or 1000 calories.

• The energy of a food is the sum of kilocalories or kilojoules from carbohydrate, fat, and protein.

Serving Size 14 crackers (31g) Servings Per Container About 7

Amount Per Serving

Calories 130 Calories from Fat 40

Nutrition Facts

Snack Crackers Snack Crackers

calorie (cal) The amount of heat energy that raises the temperature of exactly 1 g of water by exactly 1 °C.

chemical change A change during which the original substance is converted into a new substance that has a different composition and new physical and chemical properties.

chemical properties The properties that indicate the ability of a sub- stance to change into a new substance.

compound A pure substance consisting of two or more elements, with a definite composition, that can be broken down into sim- pler substances only by chemical methods.

element A pure substance containing only one type of matter, which cannot be broken down by chemical methods.

energy The ability to do work. energy value The kilocalories (or kilojoules) obtained per gram of

the food types: carbohydrate, fat, and protein. gas A state of matter that does not have a definite shape or volume. heat The energy associated with the motion of particles in a

substance. heat equation A relationship that calculates heat (q) given the mass,

specific heat, and temperature change for a substance.

KEY TERMS

joule (J) The SI unit of heat energy; 4.184 J = 1 cal. kinetic energy The energy of moving particles. liquid A state of matter that takes the shape of its container but has a

definite volume. matter The material that makes up a substance and has mass and

occupies space. mixture The physical combination of two or more substances that are

not chemically combined. physical change A change in which the physical properties of a sub-

stance change but its identity stays the same. physical properties The properties that can be observed or measured

without affecting the identity of a substance. potential energy A type of energy related to position or composition

of a substance. pure substance A type of matter that has a definite composition. solid A state of matter that has its own shape and volume. specific heat (SH) A quantity of heat that changes the temperature of

exactly 1 g of a substance by exactly 1 °C. states of matter Three forms of matter: solid, liquid, and gas.

The chapter section containing each Core Chemistry Skill is shown in parentheses at the end of each heading.

Identifying Physical and Chemical Changes (3.2) • Physical properties can be observed or measured without changing

the identity of a substance. • Chemical properties describe the ability of a substance to change

into a new substance. • When matter undergoes a physical change, its state or its appear-

ance changes, but its composition remains the same. • When a chemical change takes place, the original substance is

converted into a new substance, which has different physical and chemical properties.

Example: Classify each of the following as a physical or chemical property:

a. Helium in a balloon is a gas. b. Methane, in natural gas, burns. c. Hydrogen sulfide smells like rotten eggs.

Answer: a. A gas is a state of matter, which makes it a physical property.

b. When methane burns, it changes to different substances with new properties, which is a chemical property.

c. The odor of hydrogen sulfide is a physical property.

CORE CHEMISTRY SKILLS

Converting Between Temperature Scales (3.3) • The temperature equation TF = 1.8(TC) + 32 is used to convert

from Celsius to Fahrenheit and can be rearranged to convert from Fahrenheit to Celsius.

• The temperature equation TK = TC + 273 is used to convert from Celsius to Kelvin and can be rearranged to convert from Kelvin to Celsius.

Example: Convert 75.0 °C to degrees Fahrenheit.

Answer: TF = 1.8(TC) + 32 TF = 1.8(75.0) + 32 = 135 + 32

= 167 °F Example: Convert 355 K to degrees Celsius.

Answer: TK = TC + 273 To solve the equation for TC, subtract 273 from both sides. TK - 273 = TC + 273 - 273

TC = TK - 273 TC = 355 - 273

= 82 °C

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Using Energy Units (3.4) • Equalities for energy units include 1 cal = 4.184 J,

1 kcal = 1000 cal, and 1 kJ = 1000 J. • Each equality for energy units can be written as two conversion

factors: 4.184 J 1 cal

and 1 cal

4.184 J

1000 cal 1 kcal

and 1 kcal

1000 cal

1000 J 1 kJ

and 1 kJ

1000 J

• The energy unit conversion factors are used to cancel given units of energy and to obtain the needed unit of energy.

Example: Convert 45 000 J to kilocalories.

Answer: Using the conversion factors above, we start with the given 45 000 J and convert it to kilocalories.

45 000 J * 1 cal

4.184 J *

1 kcal 1000 cal

= 11 kcal

Calculating Specific Heat (3.5) • Specific heat (SH) is the amount of heat (q) that raises the tempera-

ture of 1 g of a substance by 1 °C.

SH = q

m * ∆T =

cal (or J)

g °C

• To calculate specific heat, the heat lost or gained is divided by the mass of the substance and the change in temperature (∆T).

Example: Calculate the specific heat, in J/g °C, for a 4.0-g sample of tin that absorbs 63 J when heated from 125 °C to 197 °C.

Answer: q = 63 J, m = 4.0 g, ∆T = 197 °C - 125 °C = 72 °C

SH = q

m * ∆T =

63 J 4.0 g * 72 °C

= 0.22 J g °C

Using the Heat Equation (3.5) • The quantity of heat absorbed or lost by a substance is calculated

using the heat equation.

q = m * ∆T * SH • Heat, in calories, is obtained when the specific heat of a substance

in cal/g °C is used. • Heat, in joules, is obtained when the specific heat of a substance in

J/g °C is used. • To cancel, the unit grams is used for mass, and the unit °C is used

for temperature change.

Example: How many joules are required to heat 5.25 g of titanium from 85.5 °C to 132.5 °C?

Answer: m = 5.25 g, ∆T = 132.5 °C - 85.5 °C = 47.0 °C, SH for titanium = 0.523 J/g °C

The known values are substituted into the heat equation making sure units cancel.

q = m * ∆T * SH = 5.25 g * 47.0 °C * 0.523 J

g °C = 129 J

UNDERSTANDING THE CONCEPTS

The chapter sections to review are shown in parentheses at the end of each problem.

3.53 Identify each of the following as an element, a compound, or a mixture. Explain your choice. (3.1)

3.55 Classify each of the following as a homogeneous or heterogeneous mixture: (3.1)

a. lemon-flavored water b. stuffed mushrooms c. eye drops

3.56 Classify each of the following as a homogeneous or heterogeneous mixture: (3.1)

a. ketchup b. hard-boiled egg c. tortilla soup

3.54 Identify each of the following as a homogeneous or hetero- geneous mixture. Explain your choice. (3.1) 3.57 State the temperature on the Celsius thermometer and convert

to Fahrenheit. (3.3)

50

40

30

Understanding the Concepts 93

a. b.

c.

a. b.

c.

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94 CHAPTER 3 Matter and Energy

a. Using Table 3.8, calculate the total kilocalories for each food type in this meal. (Round off the kilocalories to the tens place.)

b. Determine the total kilocalories for the meal. c. Using Table 3.11, determine the number of hours of sleep

needed to burn off the kilocalories in this meal. d. Using Table 3.11, determine the number of hours of running

needed to burn off the kilocalories in this meal.

3.64 Your friend, who has a mass of 70.0 kg, has a slice of pizza, a cola soft drink, and ice cream. (3.6)

Item Carbohydrate (g) Fat (g) Protein (g)

Pizza 29 10. 13

Cola 51 0 0

Ice cream 44 28 8

a. Using Table 3.8, calculate the total kilocalories for each food type in this meal. (Round off the kilocalories to the tens place.)

b. Determine the total kilocalories for the meal. c. Using Table 3.11, determine the number of hours of sitting

needed to burn off the kilocalories in this meal. d. Using Table 3.11, determine the number of hours of swim-

ming needed to burn off the kilocalories in this meal.

3.58 State the temperature on the Celsius thermometer and convert to Fahrenheit. (3.3)

46

45

44

Compost produced from decayed plant material is used to enrich the soil.

3.65 Classify each of the following as an element, a compound, or a mixture: (3.1) a. carbon in pencils b. carbon monoxide (CO) in automobile exhaust c. orange juice

3.66 Classify each of the following as an element, a compound, or a mixture: (3.1) a. neon gas in lights b. a salad dressing of oil and vinegar c. sodium hypochlorite (NaClO) in bleach

3.67 Classify each of the following mixtures as homogeneous or heterogeneous: (3.1) a. hot fudge sundae b. herbal tea c. vegetable oil

3.68 Classify each of the following mixtures as homogeneous or heterogeneous: (3.1) a. water and sand b. mustard c. blue ink

ADDITIONAL PRACTICE PROBLEMS

3.59 Compost can be made at home from grass clippings, kitchen scraps, and dry leaves. As microbes break down organic matter, heat is generated, and the compost can reach a temperature of 155 °F, which kills most pathogens. What is this temperature in degrees Celsius? In kelvins? (3.3)

3.60 After a week, biochemical reactions in compost slow, and the temperature drops to 45 °C. The dark brown organic-rich mix- ture is ready for use in the garden. What is this temperature in degrees Fahrenheit? In kelvins? (3.3)

3.61 Calculate the energy, in joules, to heat two cubes (gold and alu- minum), each with a volume of 10.0 cm3, from 15 °C to 25 °C. Refer to Tables 2.10 and 3.7. (3.5)

3.62 Calculate the energy, in joules, to heat two cubes (silver and copper), each with a volume of 10.0 cm3, from 15 °C to 25 °C. Refer to Tables 2.10 and 3.7. (3.5)

Applications

3.63 A 70.0-kg person had a quarter-pound cheeseburger, french fries, and a chocolate shake. (3.6)

Item Carbohydrate (g) Fat (g) Protein (g)

Cheeseburger 46 40. 47

French fries 47 16 4

Chocolate shake 76 10. 10.

M03_TIMB8119_06_SE_C03.indd 94 11/27/18 11:43 AM

On a sunny day, the sand gets hot but the water stays cool.

3.69 Identify each of the following as solid, liquid, or gas: (3.2) a. vitamin tablets in a bottle b. helium in a balloon c. milk in a bottle d. the air you breathe e. charcoal briquettes on a barbecue

3.70 Identify each of the following as solid, liquid, or gas: (3.2) a. popcorn in a bag b. water in a garden hose c. a computer mouse d. air in a tire e. hot tea in a teacup

3.71 Identify each of the following as a physical or chemical property: (3.2) a. Gold is shiny. b. Gold melts at 1064 °C. c. Gold is a good conductor of electricity. d. When gold reacts with sulfur, a black sulfide compound

forms.

3.72 Identify each of the following as a physical or chemical property: (3.2) a. A candle is 10 cm high and 2 cm in diameter. b. A candle burns. c. The wax of a candle softens on a hot day. d. A candle is blue.

3.73 Identify each of the following as a physical or chemical change: (3.2) a. A plant grows a new leaf. b. Chocolate is melted for a dessert. c. Wood is chopped for the fireplace. d. Wood burns in a woodstove.

3.74 Identify each of the following as a physical or chemical change: (3.2) a. Aspirin tablets are broken in half. b. Carrots are grated for use in a salad. c. Malt undergoes fermentation to make beer. d. A copper pipe reacts with air and turns green.

3.75 Calculate each of the following temperatures in degrees Celsius and kelvins: (3.3) a. The highest recorded temperature in the continental United

States was 134 °F in Death Valley, California, on July 10, 1913.

b. The lowest recorded temperature in the continental United States was - 69.7 °F in Rogers Pass, Montana, on January 20, 1954.

3.76 Calculate each of the following temperatures in kelvins and degrees Fahrenheit: (3.3)

a. The highest recorded temperature in the world was 58.0 °C in El Azizia, Libya, on September 13, 1922.

b. The lowest recorded temperature in the world was - 89.2 °C in Vostok, Antarctica, on July 21, 1983.

3.77 What is - 15 °F in degrees Celsius and in kelvins? (3.3) 3.78 What is 56 °F in degrees Celsius and in kelvins? (3.3)

3.79 On a hot day, the beach sand gets hot but the water stays cool. Would you predict that the specific heat of sand is higher or lower than that of water? Explain. (3.5)

3.80 On a hot sunny day, you get out of the swimming pool and sit in a metal chair, which is very hot. Would you predict that the specific heat of the metal is higher or lower than that of water? Explain. (3.5)

3.81 A 0.50-g sample of vegetable oil is placed in a calorimeter. When the sample is burned, 18.9 kJ is given off. What is the energy value (kcal/g) for the oil? (3.6)

3.82 A 1.3-g sample of rice is placed in a calorimeter. When the sample is burned, 22 kJ is given off. What is the energy value (kcal/g) for the rice? (3.6)

Applications

3.83 A hot-water bottle for a patient contains 725 g of water at 65 °C. If the water cools to body temperature (37 °C), how many kilojoules of heat could be transferred to sore muscles? (3.5)

3.84 The highest recorded body temperature that a person has survived is 46.5 °C. Calculate that temperature in degrees Fahrenheit and in kelvins. (3.3)

3.85 If you want to lose 1 lb of “body fat,” which is 15% water, how many kilocalories do you need to expend? (3.6)

3.86 A young patient drinks whole milk as part of her diet. Calculate the total kilocalories if the glass of milk contains 12 g of carbohydrate, 9 g of fat, and 9 g of protein. (Round off the answer for each food type to the tens place.) (3.6)

Challenge Problems 95

The following problems are related to the topics in this chapter. However, they do not all follow the chapter order, and they require you to combine concepts and skills from several sections. These problems will help you increase your critical thinking skills and prepare for your next exam.

3.87 When a 0.80-g sample of olive oil is burned in a calorimeter, the heat released increases the temperature of 450 g of water from 22.7 °C to 38.8 °C. What is the energy value (kcal/g) for the olive oil? (3.5, 3.6)

3.88 When 1.0 g of gasoline burns, it releases 11 kcal. The density of gasoline is 0.74 g/mL. (3.4)

a. How many megajoules are released when 1.0 gal of gasoline burns?

b. If a television requires 150 kJ/h to run, how many hours can the television run on the energy provided by 1.0 gal of gasoline?

3.89 A 70.-g piece of nickel metal at 54.0 °C is placed in 25 g of water at 21.9 °C. All the heat lost by the metal is used to heat

CHALLENGE PROBLEMS

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96 CHAPTER 3 Matter and Energy

ANSWERS TO SELECTED PROBLEMS 3.25 a. 3.5 kcal b. 99.2 cal c. 120 J d. 1100 cal

3.27 a. 8.1 * 105 J b. 190 kcal 3.29 Copper, which has the lowest specific heat, would reach the

highest temperature.

3.31 a. 0.389 J/g °C b. 0.313 J/g °C

3.33 a. 1380 J b. 1810 J

3.35 a. 54.5 g b. 216 g

3.37 a. 175 °C b. 11.3 °C

3.39 a. 0.84 J/g °C b. 0.50 J/g °C

3.41 a. 29.9 kcal b. 208 kcal

3.43 a. 470 kJ b. 18 g c. 130 kcal d. 38 g

3.45 210 kcal, 880 kJ

3.47 640 kcal

3.49 28.0 kJ/g

3.51 a. breakfast 270 kcal; lunch 420 kcal; dinner 440 kcal b. 1130 kcal total c. Yes. Charles should be losing weight. d. 26 days

3.53 a. compound, the particles have a 2:1 ratio of atoms b. mixture, has two different kinds of particles c. element, has a single kind of atom

3.55 a. homogeneous b. heterogeneous c. homogeneous

3.57 41.5 °C, 106.7 °F

3.59 68.3 °C, 341 K

3.1 a. element b. compound c. element d. compound e. compound

3.3 a. pure substance b. mixture c. pure substance d. pure substance e. mixture

3.5 a. heterogeneous b. homogeneous c. heterogeneous d. heterogeneous

3.7 a. gas b. gas c. solid

3.9 a. physical b. chemical c. physical d. chemical e. chemical

3.11 a. physical b. chemical c. physical d. physical e. physical

3.13 a. chemical b. physical c. physical d. chemical e. physical

3.15 In the United States, we still use the Fahrenheit temperature scale. In °F, normal body temperature is 98.6. On the Celsius scale, her temperature would be 37.7 °C, a mild fever.

3.17 a. 98.6 °F b. 18.5 °C c. 246 K d. 335 K e. 46 °C

3.19 a. 41 °C b. No. The temperature is equivalent to 39 °C.

3.21 When the roller-coaster car stops at the top of the ramp, it has its maximum potential energy and a minimum of kinetic energy. As it descends, potential energy changes to kinetic energy. At the bottom, all the energy is kinetic.

3.23 a. potential b. kinetic c. potential d. potential

ANSWERS TO ENGAGE QUESTIONS 3.6 A degree Celsius is larger: It is 1.8 times the size of a degree

Fahrenheit.

3.7 TC = TF - 32

1.8 =

- 40 - 32 1.8

= - 72 1.8

= - 40. °C

3.8 A book on a high table has more potential energy because of its position above the floor.

3.9 Because silver has a smaller specific heat than iron, 1 g of silver will reach a higher temperature when it absorbs 400 J of heat than 1 g of iron.

3.10 m = q

∆T * SH 3.11 Fat has the most energy per gram compared to carbohydrate or

protein.

3.1 Both elements and compounds have a definite composition of one kind of atom or molecule.

3.2 A pizza is heterogeneous because the food items do not have a uniform distribution, whereas in vinegar, the components have a uniform distribution.

3.3 The components in a gas are moving rapidly in all directions until they take the shape and volume of the container.

3.4 The products of the decomposition of salt are new substances with different properties, which means that a chemical change has taken place.

3.5 The boiling point is a characteristic of iron, a physical property. Heating iron to form Fe2O3, rust, shows the ability of iron to from a new substance, which is a chemical property.

the water, and the final temperature of the water and metal is 29.2 °C. What is the specific heat, in J/g °C, of nickel? (3.5)

3.90 A 125-g piece of metal is heated to 288 °C and dropped into 85.0 g of water at 12.0 °C. All the heat lost by the metal is used to heat the water, and the metal and water come to a final tem- perature of 24.0 °C. What is the specific heat, in J/g °C, of the metal? (3.5)

3.91 A metal is thought to be titanium or aluminum. When 4.7 g of the metal absorbs 11 J, its temperature rises by 4.5 °C. (3.5)

a. What is the specific heat, in J/g °C, of the metal? b. Would you identify the metal as titanium or aluminum

(see Table 3.7)?

3.92 A metal is thought to be copper or gold. When 18 g of the metal absorbs 58 cal, its temperature rises by 35 °C. (3.5)

a. What is the specific heat, in cal/g °C, of the metal? b. Would you identify the metal as copper or gold (see

Table 3.7)?

M03_TIMB8119_06_SE_C03.indd 96 11/27/18 11:43 AM

3.61 gold, 250 J or 60. cal; aluminum, 240 J or 58 cal

3.63 a. carbohydrate, 680 kcal; fat, 590 kcal; protein, 240 kcal b. 1510 kcal c. 3 * 101 h d. 2.0 h

3.65 a. element b. compound c. mixture

3.67 a. heterogeneous b. homogeneous c. homogeneous

3.69 a. solid b. gas c. liquid d. gas e. solid

3.71 a. physical b. physical c. physical d. chemical

3.73 a. chemical b. physical c. physical d. chemical

3.75 a. 56.7 °C, 330. K b. - 56.5 °C, 217 K 3.77 - 26 °C, 247 K 3.79 The same amount of heat causes a greater temperature change

in the sand than in the water; thus the sand must have a lower specific heat than that of water.

3.81 9.0 kcal/g

3.83 85 kJ

3.85 3500 kcal

3.87 9.0 kcal/g

3.89 0.44 J/g °C

3.91 a. 0.52 J/g °C b. titanium

Answers to Selected Problems 97

M03_TIMB8119_06_SE_C03.indd 97 11/27/18 11:43 AM

98

COMBINING IDEAS from Chapters 1 to 3

CI.1 Gold, one of the most sought-after metals in the world, has a density of 19.3 g/cm3, a melting point of 1064 °C, and a specific heat of 0.129 J/g °C. A gold nugget found in Alaska in 1998 weighed 20.17 lb. (2.4, 2.6, 2.7, 3.3, 3.5)

a. In which sample (A or B) does the water have its own shape?

b. Which diagram (1 or 2 or 3) represents the arrangement of particles in water sample A?

c. Which diagram (1 or 2 or 3) represents the arrangement of particles in water sample B?

Gold nuggets, also called native gold, can be found in streams and mines.

a. How many significant figures are in the measurement of the weight of the nugget?

b. What is the mass of the nugget in kilograms? c. If the nugget were pure gold, what would its volume be in

cubic centimeters? d. What is the melting point of gold in degrees Fahrenheit and

kelvins? e. How many kilocalories are required to raise the temperature

of the nugget from 500. °C to 1064 °C? f. If the price of gold is $42.06 per gram, what is the nugget

worth in dollars?

CI.2 The mileage for a motorcycle with a fuel-tank capacity of 22 L is 35 mi/gal. (2.5, 2.6, 2.7, 3.4) a. How long a trip, in kilome-

ters, can be made on one full tank of gasoline?

b. If the price of gasoline is $2.82 per gallon, what would be the cost of fuel for the trip?

c. If the average speed dur- ing the trip is 44 mi/h, how many hours will it take to reach the destination?

d. If the density of gasoline is 0.74 g/mL, what is the mass, in grams, of the fuel in the tank?

e. When 1.00 g of gasoline burns, 47 kJ of energy is released. How many kilojoules are produced when the fuel in one full tank is burned?

CI.3 Answer the following for the water samples A and B shown in the diagrams: (3.1, 3.2, 3.3, 3.5)

A B

An energy bar contains carbohydrate, fat, and protein.

When 1.00 g of gasoline burns, 47 kJ of energy is released.

1 2 3

Answer the following for diagrams 1, 2, and 3: d. The state of matter indicated in diagram 1 is a _______;

in diagram 2, it is a _______; and in diagram 3, it is a _______.

e. The motion of the particles is slowest in diagram _______. f. The arrangement of particles is farthest apart in diagram

_______. g. The particles fill the volume of the container in diagram

_______.

CI.4 The label of an energy bar with a mass of 68 g lists the nutrition facts as 39 g of carbohydrate, 5 g of fat, and 10. g of protein. (2.5, 2.6, 3.4, 3.6)

a. Using the energy values for carbohydrates, fats, and pro- teins (see Table 3.8), what are the total kilocalories for the energy bar? (Round off the answer for each food type to the tens place.)

b. What are the kilojoules for the energy bar? (Round off the answer for each food type to the tens place.)

c. If you obtain 160 kJ, how many grams of the energy bar did you eat?

d. If you are walking and using energy at a rate of 840 kJ/h, how many minutes will you need to walk to expend the energy from two energy bars?

M03_TIMB8119_06_SE_C03.indd 98 11/27/18 11:43 AM

CI.6 A hot tub is filled with 450 gal of water. (2.5, 2.6, 2.7, 3.3, 3.4, 3.5)

CI.5 In one box of nails weighing 0.250 lb, there are 75 iron nails. The density of iron is 7.86 g/cm3. The specific heat of iron is 0.452 J/g °C. The melting point of iron is 1535 °C. (2.5, 2.6, 2.7, 3.4, 3.5)

Nails made of iron have a density of 7.86 g/cm3.

a. What is the volume, in cubic centimeters, of the iron nails in the box?

b. If 30 nails are added to a graduated cylinder containing 17.6 mL of water, what is the new level of water, in milliliters, in the cylinder?

c. How much heat, in joules, must be added to the nails in the box to raise their temperature from 16 °C to 125 °C?

a. What is the volume of water, in liters, in the tub? b. What is the mass, in kilograms, of water in the tub? c. How many kilocalories are needed to heat the water from

62 °F to 105 °F? d. If the hot-tub heater provides 5900 kJ/min, how long, in

minutes, will it take to heat the water in the hot tub from 62 °F to 105 °F?

A hot tub filled with water is heated to 105 °F.

ANSWERS d. solid; liquid; gas e. diagram 1 f. diagram 3 g. diagram 3

CI.5 a. 14.4 cm3

b. 23.4 mL c. 5590 J

CI.1 a. four SFs b. 9.17 kg c. 475 cm3

d. 1947 °F; 1337 K e. 159 kcal f. $386 000

CI.3 a. B b. A is represented by diagram 2. c. B is represented by diagram 1.

Combining Ideas from Chapters 1 to 3 99

M03_TIMB8119_06_SE_C03.indd 99 11/27/18 11:43 AM

100

John prepares for the next growing season by deciding how much of each crop to plant and where on his farm. The quality of the soil, including the pH, the amount of moisture, and the nutrient content in the soil helps John make his decisions. After performing chemical tests on the soil, John determines that several of his fields need additional fertilizer before the crops can be planted. John considers several different types of fertilizers, as each supplies different nutrients to the soil. Plants need three elements for growth: potassium, nitrogen, and phosphorus. Potassium (K on the periodic table) is a metal, whereas nitrogen (N) and phosphorus (P) are nonmetals. Fertilizers may also contain several other elements including calcium (Ca), magnesium (Mg), and sulfur (S). John applies a fertilizer containing a mixture of all of these elements to his soil and plans to recheck the soil nutrient content in a few days.

CAREER

Farmer Farming involves much more than growing crops and raising animals. Farmers must understand how to perform chemical tests and how to apply fertilizer to soil and pesticides or herbicides to crops. Pesticides are chemicals used to kill insects that could destroy the crop, whereas herbicides are chemicals used to kill weeds that would compete with the crops for water and nutrients. Farmers must understand how to use these chemicals safely and effectively. In using this information, farmers are able to grow crops that produce a higher yield, greater nutritional value, and better taste.

Atoms and Elements 4

Last year, John noticed that the potatoes from one of his fields had brown spots on the leaves and were undersized. Last week, he obtained soil samples from that field to check the nutrient levels. You can see how John improved the soil and crop production in the UPDATE Improving Crop Production, page 117, and learn what kind of fertilizer John used and the quantity applied.

UPDATE Improving Crop Production

M04_TIMB8119_06_SE_C04.indd 100 11/27/18 11:45 AM

4.1 Elements and Symbols 101

4.1 Elements and Symbols LEARNING GOAL Given the name of an element, write its correct symbol; from the symbol, write the correct name.

All matter is composed of elements, of which there are 118 different kinds. Of these, 88 elements occur naturally and make up all the substances in our world. Many elements are already familiar to you. Perhaps you use aluminum in the form of foil or drink soft drinks from aluminum cans. You may have a ring or necklace made of gold, silver, or perhaps platinum. If you play tennis or golf, then you may have noticed that your racket or clubs may be made from the elements titanium or carbon. In our bodies, calcium and phosphorus form the structure of bones and teeth, iron and copper are needed in the formation of red blood cells, and iodine is required for thyroid function.

Elements are pure substances from which all other things are built. Elements cannot be broken down into simpler substances. Over the centuries, elements have been named for planets, mythological figures, colors, minerals, geographic locations, and famous people. Some sources of names of elements are listed in TABLE 4.1. A complete list of all the ele- ments and their symbols are found on the inside front cover of this text.

ENGAGE 4.1 What are the names and symbols of some elements that you encounter every day?

LOOKING AHEAD

4.1 Elements and Symbols 101

4.2 The Periodic Table 103 4.3 The Atom 108 4.4 Atomic Number and Mass

Number 111 4.5 Isotopes and Atomic

Mass 113

TABLE 4.1 Some Elements, Symbols, and Source of Names Element Symbol Source of Name

Uranium U The planet Uranus

Titanium Ti Titans (mythology)

Chlorine Cl Chloros: “greenish yellow” (Greek)

Iodine I Ioeides: “violet” (Greek)

Magnesium Mg Magnesia, a mineral

Tennessine Ts Tennessee

Curium Cm Marie and Pierre Curie

Copernicium Cn Nicolaus Copernicus

Chemical Symbols Chemical symbols are one- or two-letter abbreviations for the names of the elements. Only the first letter of an element’s symbol is capitalized. If the symbol has a second letter, it is lower- case so that we know when a different element is indicated. If two letters are capitalized, they

Aluminum (Al)

TABLE 4.2 Names and Symbols of Some Common Elements Name* Symbol Name* Symbol Name* Symbol

Aluminum Al Gallium Ga Oxygen O

Argon Ar Gold (aurum) Au Phosphorus P

Arsenic As Helium He Platinum Pt

Barium Ba Hydrogen H Potassium (kalium) K

Boron B Iodine I Radium Ra

Bromine Br Iron (ferrum) Fe Silicon Si

Cadmium Cd Lead (plumbum) Pb Silver (argentum) Ag

Calcium Ca Lithium Li Sodium (natrium) Na

Carbon C Magnesium Mg Strontium Sr

Chlorine Cl Manganese Mn Sulfur S

Chromium Cr Mercury (hydrargyrum)

Hg Tin (stannum) Sn

Cobalt Co Neon Ne Titanium Ti

Copper (cuprum) Cu Nickel Ni Uranium U

Fluorine F Nitrogen N Zinc Zn *Names given in parentheses are ancient Latin or Greek words from which the symbols are derived.

Gold (Au)

Carbon (C)

M04_TIMB8119_06_SE_C04.indd 101 11/27/18 11:45 AM

102 CHAPTER 4 Atoms and Elements

represent the symbols of two different elements. For example, the element cobalt has the sym- bol Co. However, the two capital letters CO specify two elements, carbon (C) and oxygen (O).

Although most of the symbols use letters from the current names, some are derived from their ancient names. For example, Na, the symbol for sodium, comes from the Latin word natrium. The symbol for silver, Ag, is derived from the Latin name argentum. TABLE 4.2 lists the names and symbols of some common elements. Learning their names and symbols will greatly help your learning of chemistry.

PRACTICE PROBLEMS

4.1 Elements and Symbols

4.1 Write the symbols for the following elements: a. copper b. platinum c. calcium d. manganese e. iron f. barium g. lead h. strontium

4.2 Write the symbols for the following elements: a. oxygen b. lithium c. uranium d. titanium e. hydrogen f. chromium g. tin h. gold

4.3 Determine if each of the following symbols is correct. If incorrect, write the correct symbol.

a. silver, Si b. silicon, SI c. antimony, SB d. mercury, Hg e. fluorine, Fl

4.4 Determine if each of the following symbols is correct. If incorrect, write the correct symbol.

a. potassium, P b. sodium, Na c. chlorine, Ch d. neon, N e. nickel, Ni

Applications

4.5 Write the name for the symbol of each of the following elements found in the body:

a. C b. Cl c. I d. Se e. N f. S g. Zn h. Co

4.6 Write the name for the symbol of each of the following elements found in the body:

a. V b. P c. Na d. As e. Ca f. Mo g. Mg h. Si

4.7 Write the names for the elements in each of the following formulas of compounds used in medicine: a. table salt, NaCl b. plaster casts, CaSO4 c. Demerol, C15H22ClNO2 d. treatment of bipolar disorder, Li2CO3

4.8 Write the names for the elements in each of the following formulas of compounds used in medicine: a. salt substitute, KCl b. dental cement, Zn3(PO4)2 c. antacid, Mg(OH)2 d. contrast agent for X-ray, BaSO4

Silver (Ag)

Sulfur (S)

PRACTICE PROBLEMS Try Practice Problems 4.1 to 4.8

SAMPLE PROBLEM 4.1 Names and Symbols of Chemical Elements

TRY IT FIRST

Complete the following table with the correct name or symbol for each element:

Name Symbol

nickel

nitrogen

Zn

K

SOLUTION

Name Symbol

nickel Ni

nitrogen N

zinc Zn

potassium K

SELF TEST 4.1

Write the chemical symbols for the elements silicon, selenium, sodium, and seaborgium.

ANSWER

Si, Se, Na, and Sg

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4.2 The Periodic Table 103

4.2 The Periodic Table LEARNING GOAL Use the periodic table to identify the group and the period of an element; identify an element as a metal, a nonmetal, or a metalloid.

By the late 1800s, scientists recognized that certain elements looked alike and behaved in similar ways. In 1869, a Russian chemist, Dmitri Mendeleev (1834–1907), arranged the 60 elements known at that time into groups with similar properties and placed them in order of increasing atomic masses. The current arrangement of the 118 elements we know about today is called the periodic table (see FIGURE 4.1).

Representative elements

Transition elements

†Actinides

1 Group

1A 2

Group 2A

3 3B

4 4B

5 5B

6 6B

7 7B

8 9 8B

10 11 1B

12 2B

*Lanthanides

1

3 4

11 12

19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

2

5 6 7 8 9 10

13 14 15 16 17 18

37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54

86858483828180797877767574737257*5655

87 88 89† 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118

58 59 60 61 62 63 64 65 66 67 68 69 70 71

90 91 92 93 94 95 96 97 98 99 100 101 102 103

H

Li Be

Na Mg

K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn

He

B C N O F Ne

Al Si P S Cl Ar

Ga Ge As Se Br Kr

Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe

Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn

Fr Ra Ac Rf Db Sg Bh Hs Mt

Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu

Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr

Ds Rg Cn Fl LvNh

Period numbers

Group numbers

1

2

3

4

5

6

7

13 Group

3A

14 Group

4A

15 Group

5A

16 Group

6A

17 Group

7A

18 Group

8A

Metals Metalloids Nonmetals

Mc Ts Og

Periodic Table of Elements

Alkali metals

Noble gases

Alkaline earth

metals

Halogens

FIGURE 4.1 On the periodic table, groups are the elements arranged as vertical columns, and periods are the elements in each horizontal row.

Groups and Periods Each vertical column on the periodic table contains a group (or family) of elements that have similar properties. A group number is written at the top of each vertical column (group) in the periodic table. For many years, the representative elements have had group numbers 1A to 8A. In the center of the periodic table is a block of elements known as the transition elements, which have numbers followed by the letter “B.” A newer system assigns numbers 1 to 18 to the groups going left to right across the periodic table. Because both systems are in use, they are shown on the periodic table and are included in our discus- sions of elements and group numbers.

Each horizontal row in the periodic table is a period. The periods are counted down from the top of the table as Periods 1 to 7. The first period contains two elements: hydrogen (H) and helium (He). The second period contains eight elements: lithium (Li), beryllium (Be), boron (B), carbon (C), nitrogen (N), oxygen (O), fluorine (F), and neon (Ne). The third period also contains eight elements beginning with sodium (Na) and ending with argon (Ar). The fourth period, which begins with potassium (K), and the fifth period, which begins with rubidium (Rb), have 18 elements each. The sixth period, which begins with cesium (Cs), has 32 elements. The seventh period contains 32 elements, for a total of 118 elements. The elements called the lanthanides (atomic numbers 57 through 71) and the actinides (atomic numbers 89 through 103) are part of periods 6 and 7, but are usually moved to the bottom of the periodic table to allow them to fit on a page.

ENGAGE 4.2 What is the name and symbol of the element in Group 1A (1), Period 4?

M04_TIMB8119_06_SE_C04.indd 103 11/27/18 11:45 AM

104 CHAPTER 4 Atoms and Elements

No common names

1 1A

2 2A

13 3A

14 4A

15 5A

16 6A

17 7A

18 8A

Representative elements

Transition elements

H al

og en

s

A lk

al in

e ea

rt h

m et

al s

A lk

al i

m et

al s

N ob

le g

as es

FIGURE 4.2 Certain groups on the periodic table have common names.

Names of Groups Several groups in the periodic table have special names (see FIGURE 4.2). Group 1A (1) elements—lithium (Li), sodium (Na), potassium (K), rubidium (Rb), cesium (Cs), and francium (Fr)—are a family of elements known as the alkali metals (see FIGURE 4.3). The elements within this group are soft, shiny metals that are good conductors of heat and electricity and have relatively low melting points. Alkali metals react vigorously with water and form white products when they combine with oxygen.

Although hydrogen (H) is at the top of Group 1A (1), it is not an alkali metal and has very different properties than the rest of the elements in this group. Thus, hydrogen is not included in the alkali metals.

The alkaline earth metals are found in Group 2A (2). They include the elements beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra). The alkaline earth metals are shiny metals like those in Group 1A (1), but they are not as reactive.

The halogens are found on the right side of the periodic table in Group 7A (17). They include the elements fluorine (F), chlorine (Cl), bromine (Br), iodine (I), astatine (At), and tennessine (Ts) (see FIGURE 4.4). The halogens, especially fluorine and chlorine, are highly reactive and form compounds with most of the elements.

The noble gases are found in Group 8A (18). They include helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), radon (Rn), and oganesson (Og). They are quite unreactive and are seldom found in combination with other elements.

Metals, Nonmetals, and Metalloids Another feature of the periodic table is the heavy zigzag line that separates the elements into the metals and the nonmetals. Except for hydrogen, the metals are to the left of the line with the nonmetals to the right.

In general, most metals are shiny solids, such as copper (Cu), gold (Au), and silver (Ag). Metals can be shaped into wires (ductile) or hammered into a flat sheet (malleable). Metals are good conductors of heat and electricity. They usually melt at higher temperatures than nonmetals. All the metals are solids at room temperature, except for mercury (Hg), which is a liquid.

Nonmetals are not especially shiny, ductile, or malleable, and they are often poor conductors of heat and electricity. They typically have low melting points and low densi- ties. Some examples of nonmetals are hydrogen (H), carbon (C), nitrogen (N), oxygen (O), chlorine (Cl), and sulfur (S).

Except for aluminum and oganesson, the elements located along the heavy line are metalloids: B, Si, Ge, As, Sb, Te, Po, At, and Ts. Metalloids are elements that exhibit some properties that are typical of the metals and other properties that are characteristic of the nonmetals. For example, they are better conductors of heat and electricity than the

ENGAGE 4.3 What properties do the alkali metals have in common?

Group 1A (1)

Li

Na

K

3

11

19

37

55

87

Rb

Cs

Fr

Lithium (Li)

Sodium (Na)

Potassium (K)

FIGURE 4.3 Lithium (Li), sodium (Na), and potassium (K) are alkali metals from Group 1A (1).

Group 7A (17)

F

Cl

Br

9

17

35

53

85

I

At 117

Ts Chlorine

(Cl2) Bromine

(Br2 ) Iodine

(I2 )

FIGURE 4.4 Chlorine (Cl2), bromine (Br2), and iodine (I2) are halogens from Group 7A (17).

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4.2 The Periodic Table 105

nonmetals, but not as good as the metals. The metalloids are semiconductors because they can be modified to function as conductors or insulators. TABLE 4.3 compares some characteristics of silver, a metal, with those of antimony, a metalloid, and sulfur, a nonmetal.

TABLE 4.3 Some Characteristics of a Metal, a Metalloid, and a Nonmetal Silver (Ag) Antimony (Sb) Sulfur (S)

Type Metal Metalloid Nonmetal

Luster Shiny Blue-gray, shiny Dull, yellow

Ductility Extremely ductile Brittle Brittle

Malleability Can be hammered into sheets

Shatters when hammered

Shatters when hammered

Conductor Good conductor Poor conductor Poor conductor

Uses Manufacture of coins, jewelry, tableware

Hardens lead, colors glass and plastics

Manufacture of gunpowder, rubber, fungicides

Density 10.5 g/cm3 6.7 g/cm3 2.1 g/cm3

Melting Point 962 °C 630 °C 113 °C

Sulfur

Antimony

Silver

Silver is a metal, antimony is a metalloid, and sulfur is a nonmetal.

32 33

5

14

51 52

8584

B

Si

Ge As

Sb Te

Po At 117

Ts

Metalloids

Nonmetals

Metals

The metalloids on the zigzag line exhibit characteristics of both metals and nonmetals.

Strontium provides the red color in fireworks.

ENGAGE 4.4 What elements are in Group 7A (17)?

PRACTICE PROBLEMS Try Practice Problems 4.9 to 4.18

SAMPLE PROBLEM 4.2 Metals, Nonmetals, and Metalloids

TRY IT FIRST

Use the periodic table to classify each of the following elements by its group and period, group name (if any), and as a metal, a nonmetal, or a metalloid:

a. Na, important in nerve impulses, regulates blood pressure b. I, needed to produce thyroid hormones c. Si, needed for tendons and ligaments

SOLUTION

a. Na (sodium), Group 1A (1), Period 3, is an alkali metal. b. I (iodine), Group 7A (17), Period 5, halogen, is a nonmetal. c. Si (silicon), Group 4A (14), Period 3, is a metalloid.

SELF TEST 4.2

Strontium is an element that gives a brilliant red color to fireworks.

a. In what group is strontium found? b. What is the name of this chemical family? c. In what period is strontium found? d. Is strontium a metal, a nonmetal, or a metalloid?

ANSWER

a. Group 2A (2) b. alkaline earth metals c. Period 5 d. a metal

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106 CHAPTER 4 Atoms and Elements

TABLE 4.4 Typical Amounts of Essential Elements in a 60.-kg Adult Element Quantity Function

Major Elements

Oxygen (O) 39 kg Building block of biomolecules and water (H2O)

Carbon (C) 11 kg Building block of organic and biomolecules

Hydrogen (H) 6 kg Component of biomolecules, water (H2O), regulates pH of body fluids, stomach acid (HCl)

Nitrogen (N) 2 kg Component of proteins and nucleic acids

Macrominerals

Calcium (Ca) 1000 g Needed for bones and teeth, muscle contraction, nerve impulses

Phosphorus (P) 600 g Needed for bones and teeth, nucleic acids

Potassium (K) 120 g Most common positive ion (K+) in cells, muscle contraction, nerve impulses

Chlorine (Cl) 100 g Most common negative ion (Cl-) in fluids outside cells, stomach acid (HCl)

Sulfur (S) 86 g Component of proteins, vitamin B1, insulin

Sodium (Na) 60 g Most common positive ion (Na+) in fluids outside cells, involved in water balance, muscle contraction, nerve impulses

Magnesium (Mg) 36 g Component of bones, required for metabolic reactions

Chemistry Link to Health Elements Essential to Health

Of all the elements, only about 20 are essential for the well-being and survival of the human body. Of those, four elements—oxygen, carbon, hydrogen, and nitrogen—which are representative elements in Period 1 and Period 2 on the periodic table—make up 96% of our body mass. Most of the food in our daily diet provides these elements to maintain a healthy body. These elements are found in carbohy- drates, fats, and proteins. Most of the hydrogen and oxygen is found in water, which makes up 55 to 60% of our body mass.

The macrominerals—Ca, P, K, Cl, S, Na, and Mg—are located in Period 3 and Period 4 of the periodic table. They are involved in the formation of bones and teeth, maintenance of heart and blood ves- sels, muscle contraction, nerve impulses, acid–base balance of body fluids, and regulation of cellular metabolism. The macrominerals are

present in lower amounts than the major elements, so that smaller amounts are required in our daily diets.

The other essential elements, called microminerals or trace elements, are mostly transition elements in Period 4 along with Si in Period 3 and Mo and I in Period 5. They are present in the human body in very small amounts, some less than 100 mg. In recent years, the ability to detect small amounts has improved so that research- ers can more easily identify the roles of trace elements. Some trace elements such as arsenic, chromium, and selenium are toxic at high levels in the body but are still required by the body. Other elements, such as tin and nickel, are thought to be essential, but their metabolic role has not yet been determined. Some examples and the amounts present in a 60.-kg person are listed in TABLE 4.4.

114

1 Group

1A 2

Group 2A

3 3B

4 4B

5 5B

6 6B

7 7B

8 9 8B

10 11 1B

12 2B

111

1

3 4

11 12

19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

2

5 6 7 8 9 10

13 14 15 16 17 18

37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54

86858483828180797877767574737257*5655

87 88 89† 104 105 106 107 108 109 110

H

Li Be

Na Mg

K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn

He

B C N O F Ne

Al Si P S Cl Ar

Ga Ge As Se Br Kr

Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe

Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn

Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg 112

Cn Fl Lv 113

Nh

1

2

3

4

5

6

7

13 Group

3A

14 Group

4A

15 Group

5A

16 Group

6A

17 Group

7A

18 Group

8A

115

Mc 116 117

Ts 118

Og

Major elements in the human body Macrominerals Microminerals (trace elements)

Elements essential to health include the major elements H, C, N, O (red), the macrominerals (blue), and the microminerals (purple).

M04_TIMB8119_06_SE_C04.indd 106 11/27/18 11:45 AM

4.2 The Periodic Table 107

PRACTICE PROBLEMS

4.2 The Periodic Table

4.9 Identify the group or period number described by each of the following:

a. contains C, N, and O b. begins with helium c. contains the alkali metals d. ends with neon

4.10 Identify the group or period number described by each of the following:

a. contains Na, K, and Rb b. begins with Be c. contains the noble gases d. contains B, N, and F

4.11 Give the symbol of the element described by each of the following: a. Group 4A (14), Period 2 b. the noble gas in Period 1

c. the alkali metal in Period 3 d. Group 2A (2), Period 4 e. Group 3A (13), Period 3

4.12 Give the symbol of the element described by each of the following: a. the alkaline earth metal in Period 2 b. Group 5A (15), Period 3 c. the noble gas in Period 4 d. the halogen in Period 5 e. Group 4A (14), Period 4

4.13 Identify each of the following elements as a metal, a nonmetal, or a metalloid: a. barium

b. sulfur c. a shiny element d. an element that is a gas at room temperature

e. located in Group 8A (18) f. bromine g. boron h. silver

4.14 Identify each of the following elements as a metal, a nonmetal, or a metalloid:

a. located in Group 2A (2) b. a good conductor of electricity c. fluorine d. bismuth e. an element that is not shiny f. scandium g. polonium h. tin

Applications

4.15 Using Table 4.4, identify the function of each of the following elements in the body and classify each as an alkali metal, an alkaline earth metal, a transition element, or a halogen:

a. Ca b. Fe c. K d. Cl e. V f. Cr

4.16 Using Table 4.4, identify the function of each of the following elements in the body, and classify each as an alkali metal, an alkaline earth metal, a transition element, or a halogen:

a. Mg b. Cu c. I d. Na e. Co f. Mn

4.17 Identify each of the elements in problem 4.15 as a metal, a nonmetal, or a metalloid.

4.18 Identify each of the elements in problem 4.16 as a metal, a nonmetal, or a metalloid.

Element Quantity Function

Microminerals (Trace Elements)

Iron (Fe) 3600 mg Component of oxygen carrier hemoglobin

Silicon (Si) 3000 mg Needed for growth and maintenance of bones and teeth, tendons and ligaments, hair and skin

Zinc (Zn) 2000 mg Needed for metabolic reactions in cells, DNA synthesis, growth of bones, teeth, connective tissue, immune system

Copper (Cu) 240 mg Needed for blood vessels, blood pressure, immune system

Manganese (Mn) 60 mg Needed for growth of bones, blood clotting, metabolic reactions

Iodine (I) 20 mg Needed for proper thyroid function

Molybdenum (Mo) 12 mg Needed to process Fe and N from food

Arsenic (As) 3 mg Needed for growth and reproduction

Chromium (Cr) 3 mg Needed for maintenance of blood sugar levels, synthesis of biomolecules

Cobalt (Co) 3 mg Component of vitamin B12, red blood cells

Selenium (Se) 2 mg Needed for the immune system, heart, and pancreas

Vanadium (V) 2 mg Needed in the formation of bones and teeth, and the extraction of energy from food

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108 CHAPTER 4 Atoms and Elements

4.3 The Atom LEARNING GOAL Describe the electrical charge of a proton, neutron, and electron and identify their location within an atom.

All the elements listed on the periodic table are made up of atoms. An atom is the smallest particle of an element that retains the characteristics of that element. Imagine that you are tearing a piece of aluminum foil into smaller and smaller pieces. Now imagine that you have a microscopic piece so small that it cannot be divided any further. Then you would have a single atom of aluminum.

The concept of the atom is relatively recent. Although the Greek philosophers in 500 b.c.e. reasoned that everything must contain minute particles they called atomos, the idea of atoms did not become a scientific theory until 1808. Then, John Dalton (1766–1844) developed an atomic theory that proposed that atoms were responsible for the combinations of elements found in compounds.

Dalton’s Atomic Theory

1. All matter is made up of tiny particles called atoms. 2. All atoms of a given element are the same and different from atoms of other elements. 3. Atoms of two or more different elements combine to form compounds. A particular

compound is always made up of the same kinds of atoms and always has the same number of each kind of atom.

4. A chemical reaction involves the rearrangement, separation, or combination of atoms. Atoms are neither created nor destroyed during a chemical reaction.

Dalton’s atomic theory formed the basis of current atomic theory, although we have modi- fied some of Dalton’s statements. We now know that atoms of the same element are not completely identical to each other and consist of even smaller particles. However, an atom is still the smallest particle that retains the properties of an element.

Although atoms are the building blocks of everything we see around us, we cannot see an atom or even a billion atoms with the naked eye. However, when billions and billions of atoms are packed together, the characteristics of each atom are added to those of the next until we can see the characteristics we associate with the element. For example, a small piece of the element gold consists of many, many gold atoms. A special kind of microscope called a scanning tunneling microscope (STM) produces images of individual atoms (see FIGURE 4.5).

Electrical Charges in an Atom By the end of the 1800s, experiments showed that atoms were not solid spheres but were composed of smaller bits of matter called subatomic particles, three of which are the proton, neutron, and electron. Two of these subatomic particles were discovered because they have electrical charges.

An electrical charge can be positive or negative. Experiments show that like charges repel or push away from each other. When you brush your hair on a dry day, electrical charges that are alike build up on the brush and in your hair. As a result, your hair flies away from the brush. However, opposite or unlike charges attract. The crackle of clothes taken from the clothes dryer indicates the presence of electrical charges. The clinginess of the clothing results from the attraction of opposite, unlike, charges (see FIGURE 4.6).

Structure of the Atom In 1897, J. J. Thomson (1856–1940), an English physicist, applied electricity to electrodes sealed in a glass tube, which produced streams of small particles called cathode rays. Because these rays were attracted to a positively charged electrode, Thomson realized that the particles in the rays must be negatively charged. In further experiments, these particles called electrons were found to be much smaller than the atom and to have extremely small masses. Because atoms are neutral, scientists soon discovered that atoms contained positively charged particles called protons that were much heavier than the electrons.

ENGAGE 4.5 What does Dalton’s atomic theory say about the atoms in a compound?

Aluminum foil consists of atoms of aluminum.

FIGURE 4.5 Images of gold atoms magnified 16 million times by a scanning tunneling microscope.

FIGURE 4.6 Like charges repel and unlike charges attract.

Positive charges repel

Negative charges repel

Unlike charges attract

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4.3 The Atom 109

Positive electrode

Cathode ray tube

+

+

-

-

Electron beam

Negatively charged cathode rays (electrons) are attracted to the positive electrode.

Thomson proposed a model for the atom in which the electrons were distributed in a positively charged cloud. Thomson’s model of the atom became know as the “plum-pudding” model because electrons reminded scientists at that time of “plums” in an English dessert, plum pudding. In 1911, Ernest Rutherford (1871–1937) worked with Thomson to test this model. In Rutherford’s experiment, positively charged particles were aimed at a thin sheet of gold foil (see FIGURE 4.7). If the Thomson model were correct, the particles would travel in straight paths through the gold foil. Rutherford was greatly surprised to find that some of the particles were deflected as they passed through the gold foil, and a few particles were deflected so much that they went back in the opposite direction. According to Rutherford, it was as though he had shot a cannonball at a piece of tissue paper, and it bounced back at him.

Negatively charged electrons

Positively charged cloud

Thomson’s “plum-pudding” model describes the electrons in an atom as scattered throughout a positively charged cloud.

ENGAGE 4.6 Why are some particles deflected, whereas most pass through the gold foil undeflected?

From his gold-foil experiments, Rutherford realized that the protons must be con- tained in a small, positively charged region at the center of the atom, which he called the nucleus. He proposed that the electrons in the atom occupy the space surrounding the nucleus through which most of the particles traveled undisturbed. Only the particles that came near this dense, positive center were def lected. The size of the nucleus is extremely small compared with the overall size of the atom. If an atom were the size of a football stadium, the nucleus would be about the size of a golf ball placed in the center of the field.

Later, when scientists realized that the nucleus was heavier than the mass of the protons, they looked for another subatomic particle. Eventually, they discovered that the nucleus also contained a particle that is neutral, which they called a neutron.

Mass of the Atom All the subatomic particles are extremely small compared with the things you see around you (see FIGURE 4.8). One proton has a mass of 1.67 * 10-24 g, and the neutron is about the same. However, the electron has a mass 9.11 * 10-28 g, which is much less than the mass of either a proton or neutron. Because the masses of subatomic particles are so small, chemists use a very small unit of mass called an atomic mass unit (amu). An amu is defined

INTERACTIVE VIDEO

Rutherford’s Gold-Foil Experiment

FIGURE 4.7 Rutherford's gold-foil experiment

Source of particles

Beam of particles

Lead container

Fluorescent screen

Thin gold foil

Deflected particles Positively charged nucleus

Atoms of gold

Undeflected particles

Positive particles are aimed at a piece of gold foil. Positive particles that come very close to atomic nuclei are deflected from their straight path.

Positive particles are aimed at a piece of gold foil. Positive particles that come very close to atomic nuclei are deflected from their straight path.

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110 CHAPTER 4 Atoms and Elements

Atomic diameter L 10-10 m

Nuclear diameter L 10-15 m

Lithium atoms Lithium atom Lithium nucleusLithium metal

Electron (-) Proton (+)Neutron

Nucleus (+)

FIGURE 4.8 In an atom, the protons and neutrons that make up almost all the mass are packed into the tiny volume of the nucleus. The rapidly moving electrons (negative charge) surround the nucleus and account for the large volume of the atom.

as one-twelfth of the mass of a carbon atom, which has a nucleus containing six protons and six neutrons. In biology, the atomic mass unit is called a Dalton (Da) in honor of John Dalton. On the amu scale, the proton and neutron each have a mass of about 1 amu. Because the electron mass is so small, it is usually ignored in atomic mass calculations. TABLE 4.5 summarizes some information about the subatomic particles in an atom.

ENGAGE 4.7 Why can we say that the atom is mostly empty space?

Proton (+)

Neutron (no charge)

Nucleus Electron (-)

+ + +

- -

-

The nucleus of a typical lithium atom contains three protons and four neutrons.

PRACTICE PROBLEMS Try Practice Problems 4.19 to 4.26

TABLE 4.5 Subatomic Particles in the Atom Particle Symbol Charge Mass (amu) Location in Atom

Proton p or p+ 1 + 1.007 Nucleus

Neutron n or n0 0 1.008 Nucleus

Electron e- 1 - 0.000 55 Outside nucleus

PRACTICE PROBLEMS

4.3 The Atom

4.19 Identify each of the following as describing either a proton, a neutron, or an electron:

a. has the smallest mass b. has a 1 + charge c. is found outside the nucleus d. is electrically neutral

4.20 Identify each of the following as describing either a proton, a neutron, or an electron: a. has a mass about the same as a proton b. is found in the nucleus c. is attracted to the protons d. has a 1 - charge

4.21 What did Rutherford determine about the structure of the atom from his gold-foil experiment?

4.22 How did Thomson determine that the electrons have a negative charge?

4.23 Is each of the following statements true or false? a. A proton and an electron have opposite charges. b. The nucleus contains most of the mass of an atom. c. Electrons repel each other. d. A proton is attracted to a neutron.

4.24 Is each of the following statements true or false? a. A proton is attracted to an electron. b. A neutron has twice the mass of a proton. c. Neutrons repel each other. d. Electrons and neutrons have opposite charges.

4.25 On a dry day, your hair flies apart when you brush it. How would you explain this?

4.26 Sometimes clothes cling together when removed from a dryer. What kinds of charges are on the clothes?

M04_TIMB8119_06_SE_C04.indd 110 11/27/18 11:45 AM

4.4 Atomic Number and Mass Number 111

4.4 Atomic Number and Mass Number LEARNING GOAL Given the atomic number and the mass number of an atom, state the number of protons, neutrons, and electrons.

All the atoms of the same element always have the same number of protons. This feature distinguishes atoms of one element from atoms of all the other elements.

Atomic Number The atomic number of an element is equal to the number of protons in every atom of that element. The atomic number is the whole number that appears above the symbol of each element on the periodic table.

Atomic number = number of protons in an atom

The periodic table on the inside front cover of this text shows the elements in order of atomic number from 1 to 118. We can use an atomic number to identify the number of protons in an atom of any element. For example, a lithium atom, with atomic number 3, has 3 protons. Any atom with 3 protons is always a lithium atom. In the same way, we determine that a carbon atom, with atomic number 6, has 6 protons. Any atom with 6 protons is carbon.

An atom is electrically neutral. That means that the number of protons in an atom is equal to the number of electrons, which gives every atom an overall charge of zero. Thus, the atomic number also gives the number of electrons in a neutral atom.

REVIEW Using Positive and Negative

Numbers in Calculations (1.4)

CORE CHEMISTRY SKILL Counting Protons and Neutrons

ENGAGE 4.8 Why does every atom of barium have 56 protons and 56 electrons?

ENGAGE 4.9 Which subatomic particles in an atom determine its mass number?

Lithium 3 protons

Carbon 6 protons

Li 3

C 6

Atomic number

Proton

Neutron Neutron

Proton

Electron Electron

All atoms of lithium contain three protons and three electrons.

All atoms of carbon contain six protons and six electrons.

Mass Number We now know that the protons and neutrons determine the mass of the nucleus. Thus, for a single atom, we assign a mass number, which is the total number of protons and neutrons in its nucleus. It is important to understand that the term mass number tells you how many protons and neutrons there are in the nucleus of an atom. Because these are the main subatomic particles that determine the mass of an atom, the count of protons and neutrons is called the “mass” number. The mass number does not appear on the periodic table because it applies to single atoms only.

Mass number = number of protons + number of neutrons

For example, the nucleus of an oxygen atom that contains 8 protons and 8 neutrons has a mass number of 16. An atom of iron that contains 26 protons and 32 neutrons has a mass number of 58.

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112 CHAPTER 4 Atoms and Elements

If we are given the mass number of an atom and its atomic number, we can calculate the number of neutrons in its nucleus.

Number of neutrons in a nucleus = mass number - number of protons

For example, if we are given an atom of chlorine (atomic number 17) that has a mass number of 37, we can calculate the number of neutrons in its nucleus.

Number of neutrons = 37 (mass number) - 17 (protons) = 20 neutrons

TABLE 4.6 illustrates the relationships between atomic number, mass number, and the number of protons, neutrons, and electrons in examples of single atoms for different elements.

ENGAGE 4.10 How many neutrons are in an atom of tin that has a mass number of 102?

PRACTICE PROBLEMS Try Practice Problems 4.27 to 4.34

TABLE 4.6 Composition of Some Atoms of Different Elements

Element Symbol Atomic Number

Mass Number

Number of Protons

Number of Neutrons

Number of Electrons

Hydrogen H 1 1 1 0 1

Nitrogen N 7 14 7 7 7

Oxygen O 8 16 8 8 8

Chlorine Cl 17 37 17 20 17

Iron Fe 26 58 26 32 26

Gold Au 79 197 79 118 79

TRY IT FIRST

Zinc, a micromineral, is needed for metabolic reactions in cells, DNA synthesis, the growth of bone, teeth, and connective tissue, and the proper functioning of the immune system. For an atom of zinc that has a mass number of 68, determine the number of:

a. protons b. neutrons c. electrons

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

zinc (Zn), mass number 68

number of protons, neutrons, electrons

periodic table, atomic number

a. Zinc (Zn), with an atomic number of 30, has 30 protons. b. The number of neutrons in this atom is found by subtracting the number of protons

(atomic number) from the mass number.

Mass number - atomic number = number of neutrons 68 - 30 = 38

c. Because a zinc atom is neutral, the number of electrons is equal to the number of protons. A zinc atom has 30 electrons.

SELF TEST 4.3

a. How many neutrons are in the nucleus of a bromine atom that has a mass number of 80? b. What is the mass number of a cesium atom that has 71 neutrons?

ANSWER

a. 45 b. 126

SAMPLE PROBLEM 4.3 Calculating Numbers of Protons, Neutrons, and Electrons

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4.5 Isotopes and Atomic Mass 113

PRACTICE PROBLEMS

4.4 Atomic Number and Mass Number

4.27 Would you use the atomic number, mass number, or both to determine each of the following? a. number of protons in an atom b. number of neutrons in an atom c. number of particles in the nucleus d. number of electrons in a neutral atom

4.28 Identify the type of subatomic particles described by each of the following: a. atomic number b. mass number c. mass number - atomic number d. mass number + atomic number

4.29 Write the names and symbols for the elements with the following atomic numbers:

a. 3 b. 9 c. 20 d. 30 e. 10 f. 14 g. 53 h. 8

4.30 Write the names and symbols for the elements with the following atomic numbers:

a. 1 b. 11 c. 19 d. 82 e. 35 f. 47 g. 15 h. 2

4.31 How many protons and electrons are there in a neutral atom of each of the following elements?

a. argon b. manganese c. iodine d. cadmium

4.32 How many protons and electrons are there in a neutral atom of each of the following elements?

a. carbon b. fluorine c. tin d. nickel

Applications

4.33 Complete the following table for atoms of essential elements in the body:

Name of the Element Symbol

Atomic Number

Mass Number

Number of Protons

Number of Neutrons

Number of Electrons

Zn 66

12 12

Potassium 20

16 15

56 26

4.34 Complete the following table for atoms of essential elements in the body:

Name of the Element Symbol

Atomic Number

Mass Number

Number of Protons

Number of Neutrons

Number of Electrons

N 15

Calcium 42

53 72

14 16

29 65

4.5 Isotopes and Atomic Mass LEARNING GOAL Determine the number of protons, neutrons, and electrons in one or more of the isotopes of an element; calculate the atomic mass of an element using the percentage abundance and mass of its naturally occurring isotopes.

We have seen that all atoms of the same element have the same number of protons and elec- trons. However, the atoms of any one element are not entirely identical because the atoms of most elements have different numbers of neutrons.

Atoms and Isotopes Isotopes are atoms of the same element that have the same atomic number but different numbers of neutrons. For example, all atoms of the element magnesium (Mg) have an atomic number of 12. Thus every magnesium atom always has 12 protons. However, some naturally occurring magnesium atoms have 12 neutrons, others have 13 neutrons, and still others have 14 neutrons. The different numbers of neutrons give the magnesium atoms dif- ferent mass numbers but do not change their chemical behavior.

To distinguish between the different isotopes of an element, we write an atomic symbol for a particular isotope that indicates the mass number in the upper left corner and the atomic number in the lower left corner.

REVIEW Calculating Percentages (1.4)

CORE CHEMISTRY SKILL Writing Atomic Symbols for

Isotopes

Symbol of element

Mass number

Atomic number 12 Mg 24

Atomic symbol for an isotope of magnesium, Mg-24.

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114 CHAPTER 4 Atoms and Elements

Atoms of Mg

Isotopes of Mg

12 Mg 24

12 Mg 25

12 Mg 26

The nuclei of three naturally occurring magnesium isotopes have the same number of protons but different numbers of neutrons.

An isotope may be referred to by its name or symbol, followed by its mass number, such as magnesium-24 or Mg-24. Magnesium has three naturally occurring isotopes, as shown in TABLE 4.7. In a large sample of naturally occurring magnesium atoms, each type of isotope can be present as a specific percentage. For example, the Mg-24 isotope makes up almost 80% of the total sample, whereas Mg-25 and Mg-26 each make up only about 10% of the total number of magnesium atoms.

PRACTICE PROBLEMS Try Practice Problems 4.35 to 4.38

TABLE 4.7 Isotopes of Magnesium

Atomic Symbol 12 24Mg 12

25Mg 12 26Mg

Name Mg-24 Mg-25 Mg-26

Number of Protons 12 12 12

Number of Electrons 12 12 12

Mass Number 24 25 26

Number of Neutrons 12 13 14

Mass of Isotope (amu) 23.99 24.99 25.98

Percentage Abundance 78.70 10.13 11.17

ENGAGE 4.11 What is different and what is the same for an atom of Sn-112 and an atom of Sn-126?

TRY IT FIRST

Chromium, needed for maintenance of blood sugar levels, has four naturally occurring isotopes. Calculate the number of protons and number of neutrons in each of the fol- lowing isotopes:

a. 24 50Cr b. 24

52Cr c. 24 53Cr d. 24

54Cr

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

atomic symbols for Cr isotopes

number of protons, number of neutrons

atomic number

In the atomic symbol, the mass number is shown in the upper left corner of the symbol, and the atomic number is shown in the lower left corner of the symbol. Thus, each isotope of Cr, atomic number 24, has 24 protons. The number of neutrons is found by subtracting the number of protons (24) from the mass number of each isotope.

Atomic Symbol

Atomic Number

Mass Number

Number of Protons

Number of Neutrons

a. 24 50Cr 24 50 24 26 (50 - 24)

b. 24 52Cr 24 52 24 28 (52 - 24)

c. 24 53Cr 24 53 24 29 (53 - 24)

d. 24 54Cr 24 54 24 30 (54 - 24)

SELF TEST 4.4

a. Vanadium is a micromineral needed in the formation of bones and teeth. Write the atomic symbol for the isotope of vanadium that has 27 neutrons.

b. Molybdenum, a micromineral needed to process Fe and N from food, has seven naturally occurring isotopes. Write the atomic symbols for two isotopes, Mo-97 and Mo-98.

c. Calculate the number of protons and number of neutrons in an atom of Mo-97.

ANSWER

a. 23 50V b. 42

97Mo, 42 98Mo c. 42 protons and 55 neutrons

SAMPLE PROBLEM 4.4 Identifying Protons and Neutrons in Isotopes

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4.5 Isotopes and Atomic Mass 115

Atomic Mass In laboratory work, a chemist generally uses samples with many atoms that contain all the different atoms or isotopes of an element. Because each isotope has a different mass, chemists have calculated an atomic mass for an “average atom,” which is a weighted average of the masses of all the naturally occurring isotopes of that element. On the periodic table, the atomic mass is the number including decimal places that is given below the symbol of each element. Most elements consist of two or more isotopes, which is one reason that the atomic masses on the periodic table are seldom whole numbers.

Weighted Average Analogy To understand how the atomic mass as a weighted average for a group of isotopes is calcu- lated, we will use an analogy of bowling balls with different weights. Suppose that a bowl- ing alley has ordered 5 bowling balls that weigh 8 lb each and 20 bowling balls that weigh 14 lb each. In this group of bowling balls, there are more 14-lb balls than 8-lb balls. The abundance of the 14-lb balls is 80.% (20/25), and the abundance of the 8-lb balls is 20.% (5/25). Now we can calculate a weighted average for the “average” bowling ball using the weight and percentage abundance of the two types of bowling balls.

Item Weight (lb) Percentage Abundance

Weight from Each Type

14-lb bowling ball 14 * 80. 100 = 11.2 lb

8-lb bowling ball 8 * 20. 100 = 1.6 lb

Weighted average mass of a bowling ball = 12.8 lb “Atomic mass” of a bowling ball = 12.8 lb

Calculating Atomic Mass To calculate the atomic mass of an element, we need to know the percentage abundance and the mass of each isotope, which must be determined experimentally. For example, a large sample of naturally occurring chlorine atoms is experimentally determined to contain 75.76% of 17

35Cl atoms and 24.24% of 17 37Cl atoms. The 17

35Cl isotope has a mass of 34.97 amu,

and the 17 37Cl isotope has a mass of 36.97 amu.

Atomic mass of Cl = mass of 17 35Cl * 17

35Cl%

100% + mass of 1737Cl *

17 37Cl%

100% amu from 1735Cl amu from 1737Cl

Atomic Symbol Mass (amu) Percentage Abundance

Contribution to the Atomic Mass

17 35Cl 34.97 * 75.76

100 = 26.49 amu

17 37Cl 36.97 * 24.24

100 = 8.962 amu

Weighted average mass of Cl = 35.45 amu Atomic mass of Cl = 35.45 amu

The atomic mass of 35.45 amu is the weighted average mass of a sample of Cl atoms, although no individual Cl atom actually has this mass. An atomic mass of 35.45, which is closer to the mass number of Cl-35, indicates there is a higher percentage of 17

35Cl atoms in the chlorine sample. In fact, there are about three atoms of 17

35Cl for every one atom of 17 37Cl

in a sample of chlorine atoms. TABLE 4.8 lists the naturally occurring isotopes of some selected elements and their

atomic masses.

ENGAGE 4.12 What is the difference between mass number and atomic mass?

CORE CHEMISTRY SKILL Calculating Atomic Mass

The weighted average of 8-lb and 14-lb bowling balls is calculated using their percentage abundance.

35.45

17

Cl 35 17

Cl

24.24%

37 17

75.76%

Cl 17 protons

Symbol for chlorine

Atomic mass 35.45 amu

Chlorine, with two naturally occurring isotopes, has an atomic mass of 35.45 amu.

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116 CHAPTER 4 Atoms and Elements

TABLE 4.8 The Atomic Mass of Some Elements

Element Naturally Occurring Isotopes

Atomic Mass (weighted average)

Lithium 3 6Li, 3

7Li 6.941 amu

Carbon 6 12C, 6

13C, 6 14C 12.01 amu

Oxygen 8 16O, 8

17O, 8 18O 16.00 amu

Fluorine 9 19F 19.00 amu

Sulfur 16 32S, 16

33S, 16 34S, 16

36S 32.07 amu

Potassium 19 39K, 19

40K, 19 41K 39.10 amu

Copper 29 63Cu, 29

65Cu 63.55 amu

78.70% 10.13% 11.17%

Mg2412

Mg2512 Mg2612

Mg 24.31

12

Magnesium, with three naturally occurring isotopes, has an atomic mass of 24.31 amu.

SAMPLE PROBLEM 4.5 Calculating Atomic Mass

TRY IT FIRST

Magnesium is a macromineral needed in the contraction of muscles and metabolic reac- tions. Using Table 4.7, calculate the atomic mass for magnesium using the weighted average mass method.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

percentage abundance, atomic mass

atomic mass of Mg weighted average mass

STEP 1 Multiply the mass of each isotope by its percentage abundance divided by 100.

Atomic Symbol Mass (amu) Percentage Abundance Contribution to the Atomic Mass

12 24Mg 23.99 * 78.70

100 = 18.88 amu

12 25Mg 24.99 * 10.13

100 = 2.531 amu

12 26Mg 25.98 * 11.17

100 = 2.902 amu

STEP 2 Add the contribution of each isotope to obtain the atomic mass.

Atomic mass of Mg = 18.88 amu + 2.531 amu + 2.902 amu = 24.31 amu (weighted average mass)

SELF TEST 4.5

There are two naturally occurring isotopes of boron. The isotope 5 10B has a mass of 10.01 amu

with an abundance of 19.80%, and the isotope 5 11B has a mass of 11.01 amu with an abundance

of 80.20%. Calculate the atomic mass for boron using the weighted average mass method.

ANSWER

10.81 amu

PRACTICE PROBLEMS

4.5 Isotopes and Atomic Mass

4.35 What are the number of protons, neutrons, and electrons in the following isotopes?

a. 38 89Sr b. 24

52Cr c. 16 34S d. 35

81Br

4.36 What are the number of protons, neutrons, and electrons in the following isotopes?

a. 1 2H b. 7

14N c. 14 26Si d. 30

70Zn

PRACTICE PROBLEMS Try Practice Problems 4.39 to 4.48

INTERACTIVE VIDEO

Isotopes and Atomic Mass

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Plants need potassium (K) for many metabolic processes, including the regulation of growth, protein synthesis, photosynthesis, and ionic balance. Potassium- deficient potato plants show purple or brown spots and reduced plant, root, and seed growth. John notices that the

leaves of his last crop of potatoes have brown spots, the potatoes are undersized, and the crop yield is low.

Tests on soil samples show that the potassium levels are below 100 ppm, indicating that the potato plants need supplemental potassium. John applies a fertilizer containing potassium chloride (KCl). To obtain the correct amount of potassium, John calculates that he needs to apply 170 kg of fertilizer per hectare.

Applications

4.49 a. What is the group number and name of the group that contains potassium?

b. Is potassium a metal, a nonmetal, or a metalloid? c. How many protons are in an atom of potassium?

A mineral deficiency of potassium causes brown spots on potato leaves.

UPDATE Improving Crop Production

d. Potassium has three naturally occurring isotopes. They are K-39, K-40, and K-41. Using the atomic mass of potassium, determine which isotope of potassium is the most abundant.

4.50 a. How many neutrons are in K-41? b. If John’s potato field has an area of 34.5 hectares, how

many pounds of fertilizer does John need to use? c. Potassium has three naturally occurring isotopes.

They are K-39 (93.26%, 38.964 amu), K-40 (0.0117%, 39.964 amu), and K-41 (6.73%, 40.962 amu). Calculate the atomic mass for potassium from the naturally occurring isotopes and percentage abundance using the weighted average mass method.

4.37 Write the atomic symbol for the isotope with each of the following characteristics: a. 15 protons and 16 neutrons b. 35 protons and 45 neutrons c. 50 electrons and 72 neutrons d. a chlorine atom with 18 neutrons e. a mercury atom with 122 neutrons

4.38 Write the atomic symbol for the isotope with each of the following characteristics: a. an oxygen atom with 10 neutrons b. 4 protons and 5 neutrons c. 25 electrons and 28 neutrons d. a mass number of 24 and 13 neutrons e. a nickel atom with 32 neutrons

4.39 Argon has three naturally occurring isotopes, with mass numbers 36, 38, and 40. a. Write the atomic symbol for each of these atoms. b. How are these isotopes alike? c. How are they different? d. Why is the atomic mass of argon listed on the periodic table

not a whole number? e. Which isotope is the most abundant in a sample of argon?

4.40 Strontium has four naturally occurring isotopes, with mass numbers 84, 86, 87, and 88. a. Write the atomic symbol for each of these atoms. b. How are these isotopes alike? c. How are they different?

d. Why is the atomic mass of strontium listed on the periodic table not a whole number?

e. Which isotope is the most abundant in a sample of strontium?

4.41 What is the difference between the mass of an isotope and the atomic mass of an element?

4.42 What is the difference between the mass number and the atomic mass of an element?

4.43 Copper consists of two isotopes, 29 63Cu and 29

65Cu. If the atomic mass for copper on the periodic table is 63.55, are there more atoms of 29

63Cu or 29 65Cu in a sample of copper?

4.44 There are two naturally occurring isotopes of iridium: Ir-191 and Ir-193. Use the atomic mass of iridium listed on the periodic table to identify the more abundant isotope.

4.45 Cadmium consists of eight naturally occurring isotopes. Do you expect any of the isotopes to have the atomic mass listed on the periodic table for cadmium? Explain.

4.46 Zinc consists of five naturally occurring isotopes: 30 64Zn, 30

66Zn,

30 67Zn, 30

68Zn, and 30 70Zn. None of these isotopes has the atomic

mass of 65.41 listed for zinc on the periodic table. Explain.

4.47 Two isotopes of gallium are naturally occurring, with 31 69Ga at

60.11% (68.93 amu) and 31 71Ga at 39.89% (70.92 amu). Calculate

the atomic mass for gallium using the weighted average mass method.

4.48 Two isotopes of rubidium occur naturally, with 37 85Rb at 72.17%

(84.91 amu) and 37 87Rb at 27.83% (86.91 amu). Calculate the atomic

mass for rubidium using the weighted average mass method.

Update 117

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118 CHAPTER 4 Atoms and Elements

CONCEPT MAP

are

ATOMS

Elements

Metalloids

Metals

Nonmetals

PeriodsGroups

Atomic Number

Atomic Mass

Nucleus

Mass Number

Isotopes

Subatomic Particles

Neutrons ElectronsProtons

by

determine

that give

make up the

that has a

that differs in

Chemical Symbols

Periodic Table

arranged in the

that have

havemake up

which are

CHAPTER REVIEW

4.1 Elements and Symbols LEARNING GOAL Given the name of an element, write its correct symbol; from the symbol, write the correct name. • Elements are the primary

substances of matter. • Chemical symbols are one- or two-

letter abbreviations of the names of the elements.

4.2 The Periodic Table LEARNING GOAL Use the periodic table to identify the group and the period of an element; identify an element as a metal, a nonmetal, or a metalloid. • The periodic table is an arrange-

ment of the elements by increasing atomic number.

• A horizontal row is called a period. • A vertical column on the periodic

table containing elements with similar properties is called a group.

• Elements in Group 1A (1) are called the alkali metals; Group 2A (2), the alkaline earth metals; Group 7A (17), the halogens; and Group 8A (18), the noble gases.

• On the periodic table, metals are located on the left of the heavy zigzag line, and nonmetals are to the right of the heavy zigzag line.

• Except for aluminum and tennessine, elements located along the heavy zigzag line are called metalloids.

4.3 The Atom LEARNING GOAL Describe the electrical charge of a proton, neutron, and electron and identify their location within an atom. • An atom is the smallest

particle that retains the char- acteristics of an element.

• Atoms are composed of three types of subatomic particles.

• Protons have a positive charge ( + ), electrons carry a negative charge ( - ), and neutrons are electrically neutral.

• The protons and neutrons are found in the tiny, dense nucleus; electrons are located outside the nucleus.

Electrons (-)

Nucleus (+) 32 33

5

14

51 52

85

117

84

B

Si

Ge As

Sb Te

Po At

Ts

Metalloids

Nonmetals

Metals

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4.4 Atomic Number and Mass Number LEARNING GOAL Given the atomic number and the mass number of an atom, state the number of protons, neutrons, and electrons. • The atomic number gives the

number of protons in all the atoms of the same element.

• In a neutral atom, the number of protons and electrons is equal. • The mass number is the total number of protons and neutrons

in an atom.

4.5 Isotopes and Atomic Mass LEARNING GOAL Determine the number of protons, neutrons, and electrons in one or more of the isotopes of an element; calculate the atomic mass of an element using the percentage abundance and mass of its naturally occurring isotopes. • Atoms that have the same number

of protons but different numbers of neutrons are called isotopes. • The atomic mass of an element is the weighted average mass of all

the isotopes in a naturally occurring sample of that element.

Lithium 3 protons

Li 3

Proton Neutron

Electron

Isotopes of Mg

Atoms of Mg

12 Mg 24

12 Mg 25

12 Mg 26

Core Chemistry Skills 119

alkali metal An element in Group 1A (1), except hydrogen, that is a soft, shiny metal with one electron in its outermost energy level.

alkaline earth metal An element in Group 2A (2) that has two electrons in its outermost energy level.

atom The smallest particle of an element that retains the characteristics of the element.

atomic mass The weighted average mass of all the naturally occurring isotopes of an element.

atomic mass unit (amu) A small mass unit used to describe the mass of extremely small particles such as atoms and subatomic particles; 1 amu is equal to one-twelfth the mass of a 6

12C atom . atomic number A number that is equal to the number of protons in

an atom. atomic symbol An abbreviation used to indicate the mass number

and atomic number of an isotope. chemical symbol An abbreviation that represents the name of an

element. electron A negatively charged subatomic particle having a minute

mass that is usually ignored in mass calculations; its symbol is e-. group A vertical column in the periodic table that contains elements

having similar physical and chemical properties. group number A number that appears at the top of each vertical

column (group) in the periodic table and indicates the number of electrons in the outermost energy level.

halogen An element in Group 7A (17)—fluorine, chlorine, bromine, iodine, astatine, and tennessine—that has seven electrons in its outermost energy level.

isotope An atom that differs only in mass number from another atom of the same element. Isotopes have the same atomic number (number of protons), but different numbers of neutrons.

mass number The total number of protons and neutrons in the nucleus of an atom.

metal An element that is shiny, malleable, ductile, and a good conductor of heat and electricity. The metals are located to the left of the heavy zigzag line on the periodic table.

metalloid An element with properties of both metals and nonmetals located along the heavy zigzag line on the periodic table.

neutron A neutral subatomic particle having a mass of about 1 amu and found in the nucleus of an atom; its symbol is n or n0.

noble gas An element in Group 8A (18) of the periodic table, generally unreactive and seldom found in combination with other elements, that has eight electrons (helium has two electrons) in its outermost energy level.

nonmetal An element with little or no luster that is a poor conductor of heat and electricity. The nonmetals are located to the right of the heavy zigzag line on the periodic table.

nucleus The compact, extremely dense center of an atom, containing the protons and neutrons of the atom.

period A horizontal row of elements in the periodic table. periodic table An arrangement of elements by increasing atomic

number such that elements having similar chemical behavior are grouped in vertical columns.

proton A positively charged subatomic particle having a mass of about 1 amu and found in the nucleus of an atom; its symbol is p or p+.

representative element An element in the first two columns on the left of the periodic table and the last six columns on the right that has a group number of 1A through 8A or 1, 2, and 13 through 18.

subatomic particle A particle within an atom; protons, neutrons, and electrons are subatomic particles.

transition element An element in the center of the periodic table that is designated with the letter “B” or the group number of 3 through 12.

KEY TERMS

The chapter section containing each Core Chemistry Skill is shown in parentheses at the end of each heading.

Counting Protons and Neutrons (4.4) • The atomic number of an element is equal to the number of protons

in every atom of that element. The atomic number is the whole number that appears above the symbol of each element on the periodic table.

Atomic number = number of protons in an atom

CORE CHEMISTRY SKILLS • Because atoms are neutral, the number of electrons is equal to the

number of protons. Thus, the atomic number gives the number of electrons.

• The mass number is the total number of protons and neutrons in the nucleus of an atom. Mass number = number of protons + number of neutrons

• The number of neutrons is calculated from the mass number and atomic number.

Number of neutrons = mass number - number of protons

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120 CHAPTER 4 Atoms and Elements

Example: Calculate the number of protons, neutrons, and electrons in a krypton atom with a mass number of 80.

Answer:

Element Atomic Number

Mass Number

Number of Protons

Number of Neutrons

Number of Electrons

Kr 36 80 equal to atomic number 36

equal to mass number - number of protons 80 - 36 = 44

equal to number of protons 36

Writing Atomic Symbols for Isotopes (4.5) • Isotopes are atoms of the same element that have the same atomic

number but different numbers of neutrons. • An atomic symbol is written for a particular isotope, with its mass

number (protons and neutrons) shown in the upper left corner and its atomic number (protons) shown in the lower left corner.

Example: Calculate the number of protons and neutrons in the cadmium isotope 48

112Cd. Answer:

Atomic Symbol

Atomic Number

Mass Number

Number of Protons

Number of Neutrons

48 112Cd number in

lower left corner 48

number in upper left corner 112

equal to atomic number 48

equal to mass number - number of protons 112 - 48 = 64

Calculating Atomic Mass (4.5) • The atomic mass is calculated from the masses and percentage

abundances of the isotopes of an element.

Atomic mass = %/100 * isotope (1) + %/100 * isotope (2) c

Example: Calculate the atomic mass for rhenium with two naturally occurring isotopes: 37.40% Re-185 (184.95 amu) and 62.60% Re-187 (186.96 amu).

Symbol of element

Mass number

Atomic number 12 Mg 24 Answer:

atomic mass Re = 37.40/100 * 184.95 amu + 62.60/100 * 186.96 amu = 69.17 amu + 117.0 amu = 186.2 amu

UNDERSTANDING THE CONCEPTS

The chapter sections to review are shown in parentheses at the end of each problem.

4.51 According to Dalton’s atomic theory, which of the following are true or false? If false, correct the statement to make it true. (4.3)

a. Atoms of an element are identical to atoms of other elements.

b. Every element is made of atoms. c. Atoms of different elements combine to form compounds. d. In a chemical reaction, some atoms disappear and new

atoms appear.

4.52 Use Rutherford’s gold-foil experiment to answer each of the following: (4.3)

a. What did Rutherford expect to happen when he aimed particles at the gold foil?

b. How did the results differ from what he expected? c. How did he use the results to propose a model of the atom?

4.53 Match the subatomic particles (1 to 3) to each of the descriptions below. (4.4)

1. protons 2. neutrons 3. electrons

a. atomic mass b. atomic number c. positive charge d. negative charge e. mass number - atomic number

4.54 Match the subatomic particles (1 to 3) to each of the descriptions below. (4.4)

1. protons 2. neutrons 3. electrons

a. mass number b. surround the nucleus

c. in the nucleus d. charge of 0 e. equal to number of electrons

4.55 Consider the following atoms in which X represents the chemical symbol of the element: (4.5)

8 16X 9

16X 10 18X 8

17X 8 18X

a. What atoms have the same number of protons? b. Which atoms are isotopes? Of what element? c. Which atoms have the same mass number?

4.56 Consider the following atoms in which X represents the chemical symbol of the element: (4.5)

47 124X 49

116X 50 116X 50

124X 48 116X

a. What atoms have the same number of protons? b. Which atoms are isotopes? Of what element? c. Which atoms have the same mass number?

4.57 For each representation of a nucleus A through E, write the atomic symbol, and identify which are isotopes. (4.5)

A B C D E

Neutron

Proton

4.58 Identify the element represented by each nucleus A through E in problem 4.57 as a metal, a nonmetal, or a metalloid. (4.2)

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Applications

4.59 Complete the following table for three of the naturally occurring isotopes of germanium, which is a metalloid used in semiconductors: (4.5)

Atomic Symbol

32 70Ge 32

73Ge 32 76Ge

Atomic Number

Mass Number

Number of Protons

Number of Neutrons

Number of Electrons

ADDITIONAL PRACTICE PROBLEMS

4.61 Complete the following statements: (4.2, 4.4) a. The atomic number gives the number of __________ in the

nucleus. b. In an atom, the number of electrons is equal to the number

of __________. c. Sodium and potassium are examples of elements called

__________.

4.62 Complete the following statements: (4.2, 4.4) a. The number of protons and neutrons in an atom is the

__________ number. b. The elements in Group 7A (17) are called the __________. c. Elements that are shiny and conduct heat are called

__________.

4.63 Write the chemical symbol and group and period number for each of the following elements: (4.2)

a. bromine b. argon c. lithium d. radium

4.64 Write the chemical symbol and group and period number for each of the following elements: (4.2)

a. radon b. tin c. carbon d. magnesium

4.65 The following trace elements have been found to be crucial to the functions of the body. Indicate each as a metal, a nonmetal, or a metalloid. (4.2)

a. molybdenum b. vanadium c. silicon d. iodine

4.66 The following trace elements have been found to be crucial to the functions of the body. Indicate each as a metal, a nonmetal, or a metalloid. (4.2)

a. copper b. selenium c. arsenic d. chromium

4.67 Indicate if each of the following statements is true or false: (4.3) a. The proton is a negatively charged particle. b. The neutron is 2000 times as heavy as a proton. c. The atomic mass unit is based on a carbon atom with six

protons and six neutrons. d. The nucleus is the largest part of the atom. e. The electrons are located outside the nucleus.

4.68 Indicate if each of the following statements is true or false: (4.3) a. The neutron is electrically neutral. b. Most of the mass of an atom is because of the protons and

neutrons. c. The charge of an electron is equal, but opposite, to the

charge of a neutron. d. The proton and the electron have about the same mass. e. The mass number is the number of protons.

4.69 For the following atoms, give the number of protons, neutrons, and electrons: (4.4, 4.5)

a. 48 114Cd b. 43

98Tc

c. 79 199Au d. 86

222Rn

e. 54 136Xe

4.70 For the following atoms, give the number of protons, neutrons, and electrons: (4.4, 4.5)

a. 80 202Hg b. 53

127I

c. 35 75Br d. 55

133Cs

e. 78 195Pt

Computer chips consist primarily of the element silicon.

4.60 Complete the following table for the three naturally occurring isotopes of silicon, the major component in computer chips: (4.5)

Additional Practice Problems 121

Atomic Symbol

14 28Si 14

29Si 14 30Si

Atomic Number

Mass Number

Number of Protons

Number of Neutrons

Number of Electrons

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122 CHAPTER 4 Atoms and Elements

4.71 Complete the following table: (4.4, 4.5)

Name of the Element

Atomic Symbol

Number of Protons

Number of Neutrons

Number of Electrons

34 80Se

28 34

Magnesium 14

88 228Ra

4.72 Complete the following table: (4.4, 4.5)

Name of the Element

Atomic Symbol

Number of Protons

Number of Neutrons

Number of Electrons

Potassium 22

23 51V

48 64

Barium 82

4.73 Provide the following: (4.2, 4.4) a. the atomic number and symbol of the lightest alkali metal b. the atomic number and symbol of the heaviest noble gas c. the atomic mass and symbol of the alkaline earth metal in

Period 3 d. the atomic mass and symbol of the halogen with the fewest

electrons

4.74 Provide the following: (4.2, 4.4) a. the atomic number and symbol of the heaviest metalloid

in Group 4A (14) b. the atomic number and symbol of the element in

Group 5A (15), Period 6 c. the atomic mass and symbol of the alkali metal in Period 4 d. the metalloid in Group 3A (13)

4.75 Write the names and symbols of the elements with the following atomic numbers: (4.4)

a. 35 b. 57 c. 52 d. 33 e. 50 f. 55

4.76 Write the names and symbols of the elements with the following atomic numbers: (4.4)

a. 22 b. 49 c. 26 d. 54 e. 78 f. 83

4.77 How many protons and electrons are there in a neutral atom of each of the following elements? (4.4)

a. Pt b. phosphorus c. Sr d. Co e. uranium

4.78 How many protons and electrons are there in a neutral atom of each of the following elements? (4.4)

a. chromium b. Cs c. copper d. chlorine e. Ba

4.79 Write the atomic symbol for each of the following: (4.5) a. an atom with 22 protons and 26 neutrons b. an atom with 12 protons and 14 neutrons c. a gallium atom with a mass number of 71 d. an atom with 30 electrons and 40 neutrons

4.80 Write the atomic symbol for each of the following: (4.5) a. an aluminum atom with 14 neutrons b. an atom with atomic number 26 and 32 neutrons c. a lead atom with 125 neutrons d. an atom with a mass number of 72 and atomic number 33

4.81 Lead consists of four naturally occurring isotopes. Calculate the atomic mass for lead using the weighted average mass method. (4.5)

Isotope Mass (amu) Abundance (%)

82 204Pb 204.0 1.40

82 206Pb 206.0 24.10

82 207Pb 207.0 22.10

82 208Pb 208.0 52.40

4.82 Indium (In) has two naturally occurring isotopes: In-113 and In-115. In-113 has a 4.30% abundance and a mass of 112.9 amu, and In-115 has a 95.70% abundance and a mass of 114.9 amu. Calculate the atomic mass for indium using the weighted average mass method. (4.5)

The following problems are related to the topics in this chapter. However, they do not all follow the chapter order, and they require you to combine concepts and skills from several sections. These problems will help you increase your critical thinking skills and prepare for your next exam.

4.83 The most abundant isotope of lead is 82 208Pb. (4.4, 4.5)

a. How many protons, neutrons, and electrons are in 82 208Pb?

b. What is the atomic symbol of another isotope of lead with 132 neutrons?

c. What is the name and symbol of an atom with the same mass number as in part b and 131 neutrons?

4.84 The most abundant isotope of silver is 47 107Ag. (4.4, 4.5)

a. How many protons, neutrons, and electrons are in 47 107Ag?

b. What is the symbol of another isotope of silver with 62 neutrons?

c. What is the name and symbol of an atom with the same mass number as in part b and 61 neutrons?

4.85 The most abundant isotope of gold is Au-197. (4.5) a. How many protons, neutrons, and electrons are in this isotope? b. What is the atomic symbol of another isotope of gold with

116 neutrons? c. What is the atomic symbol of an atom with an atomic

number of 78 and 116 neutrons?

4.86 The most abundant isotope of calcium is Ca-40. (4.5) a. How many protons, neutrons, and electrons are in this isotope? b. What is the atomic symbol of another isotope of calcium

with 24 neutrons? c. What is the atomic symbol of an atom with an atomic

number of 21 and 24 neutrons?

4.87 Silicon has three naturally occurring isotopes: Si-28 that has a percentage abundance of 92.23% and a mass of 27.977 amu, Si-29 that has an abundance of 4.68% and a mass of 28.976 amu, and Si-30 that has a percentage abundance of 3.09% and a mass of 29.974 amu. Calculate the atomic mass for silicon using the weighted average mass method. (4.5)

CHALLENGE PROBLEMS

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ANSWERS TO ENGAGE QUESTIONS 4.8 Because atoms are neutral, 56 positively charged protons of

every barium atom are balanced by 56 negatively charged electrons, which gives the atom an overall charge of zero.

4.9 The number of protons and neutrons determines the mass number of an atom.

4.10 An atom of tin with a mass number of 102 has 50 protons and 52 neutrons.

4.11 Atoms of Sn-112 and Sn-126 both have 50 protons; atoms of Sn-112 have 62 neutrons and atoms of Sn-126 have 76 neutrons.

4.12 The mass number is the number of protons and neutrons in an atom. The atomic mass is the weighted average of the masses of all the naturally occurring isotopes of an element.

4.1 Some elements we may encounter everyday are aluminum (Al), calcium (Ca), silver (Ag), carbon (C), and oxygen (O).

4.2 potassium, K

4.3 shiny, soft, good conductors of heat, and electricity, have relatively low melting points, react vigorously with water, and form white products when they combine with oxygen

4.4 fluorine, chlorine, bromine, iodine, astatine, and tennessine

4.5 Dalton said that compounds are always made up of the same kind and number of atoms.

4.6 Particles passing through the gold foil are deflected only if they go very close to the positively charged nucleus.

4.7 The neutrons and protons are contained in the very small nucleus with the electrons occupying the very large volume of the atom, which is mostly empty space.

4.88 Antimony (Sb) has two naturally occurring isotopes: Sb-121 that has a percentage abundance of 57.21% and a mass of 120.90 amu, and Sb-123 that has a percentage abundance of 42.79% and a mass of 122.90 amu. Calculate the atomic mass for antimony using the weighted average mass method. (4.5)

4.89 There are four naturally occurring isotopes of strontium: (4.4, 4.5)

38 84Sr, 38

86Sr, 38 87Sr, and 38

88Sr

a. How many protons, neutrons, and electrons are in Sr-87? b. What is the most abundant isotope in a strontium sample?

c. How many neutrons are in Sr-84? d. Why don’t any of the isotopes of strontium have the atomic

mass of 87.62 amu listed on the periodic table?

4.90 There are four naturally occurring isotopes of iron: (4.4, 4.5)

26 54Fe, 26

56Fe, 26 57Fe, and 26

58Fe

a. How many protons, neutrons, and electrons are in 26 58Fe?

b. What is the most abundant isotope in an iron sample? c. How many neutrons are in 26

57Fe? d. Why don’t any of the isotopes of iron have the atomic mass

of 55.85 amu listed on the periodic table?

ANSWERS TO SELECTED PROBLEMS 4.21 Rutherford determined that an atom contains a small, compact

nucleus that is positively charged.

4.23 a. true b. true c. true d. false. A proton is attracted to an electron. 4.25 In the process of brushing hair, strands of hair become charged

with like charges that repel each other.

4.27 a. atomic number b. both c. mass number d. atomic number

4.29 a. lithium, Li b. fluorine, F c. calcium, Ca d. zinc, Zn e. neon, Ne f. silicon, Si g. iodine, I h. oxygen, O

4.31 a. 18 protons and 18 electrons b. 25 protons and 25 electrons c. 53 protons and 53 electrons d. 48 protons and 48 electrons

4.33

Name of the Element Symbol

Atomic Number

Mass Number

Number of Protons

Number of Neutrons

Number of Electrons

Zinc Zn 30 66 30 36 30

Magnesium Mg 12 24 12 12 12

Potassium K 19 39 19 20 19

Sulfur S 16 31 16 15 16

Iron Fe 26 56 26 30 26

4.1 a. Cu b. Pt c. Ca d. Mn e. Fe f. Ba g. Pb h. Sr

4.3 a. Ag b. Si c. Sb d. correct e. F

4.5 a. carbon b. chlorine c. iodine d. selenium e. nitrogen f. sulfur g. zinc h. cobalt

4.7 a. sodium, chlorine b. calcium, sulfur, oxygen c. carbon, hydrogen, chlorine, nitrogen, oxygen d. lithium, carbon, oxygen

4.9 a. Period 2 b. Group 8A (18) c. Group 1A (1) d. Period 2

4.11 a. C b. He c. Na d. Ca e. Al

4.13 a. metal b. nonmetal c. metal d. nonmetal e. nonmetal f. nonmetal g. metalloid h. metal

4.15 a. needed for bones and teeth, muscle contraction, nerve impulses; alkaline earth metal

b. component of hemoglobin; transition element c. muscle contraction, nerve impulses; alkali metal d. found in fluids outside cells; halogen e. needed in the formation of bones and teeth, and the extrac-

tion of energy from food; transition element f. needed for maintenance of blood sugar levels; transition

element

4.17 a. metal b. metal c. metal d. nonmetal e. metal f. metal

4.19 a. electron b. proton c. electron d. neutron

Answers to Selected Problems 123

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124 CHAPTER 4 Atoms and Elements

4.35 a. 38 protons, 51 neutrons, 38 electrons b. 24 protons, 28 neutrons, 24 electrons c. 16 protons, 18 neutrons, 16 electrons d. 35 protons, 46 neutrons, 35 electrons

4.37 a. 15 31P b. 35

80Br c. 50 122Sn d. 17

35Cl e. 80 202Hg

4.39 a. 18 36Ar 18

38Ar 18 40Ar

b. They all have the same number of protons and electrons. c. They have different numbers of neutrons, which gives them

different mass numbers. d. The atomic mass of Ar listed on the periodic table is the

weighted average atomic mass of all the naturally occurring isotopes.

e. The isotope Ar-40 is most abundant because its mass is closest to the atomic mass of Ar on the periodic table.

4.41 The mass of an isotope is the mass of an individual atom; the atomic mass of an element is the weighted average of the masses of all the naturally occurring isotopes of that element.

4.43 Because the atomic mass of copper is closer to 63 amu, there are more atoms of 29

63Cu.

4.45 No. The atomic mass is the weighted average of the masses of the eight naturally occurring isotopes of cadmium.

4.47 69.72 amu

4.49 a. Group 1A (1), alkali metals b. metal c. 19 protons d. K-39

4.51 a. false. All atoms of a given element are different from atoms of other elements.

b. true c. true d. false. In a chemical reaction, atoms are neither created nor

destroyed.

4.53 a. 1 + 2 b. 1 c. 1 d. 3 e. 2 4.55 a. 8

16X, 8 17X, 8

18X all have eight protons.

b. 8 16X, 8

17X, 8 18X are all isotopes of oxygen.

c. 8 16X and 9

16X have mass numbers of 16, and 10 18X and 8

18X have mass numbers of 18.

4.57 a. 4 9Be b. 5

11B c. 6 13C

d. 5 10B e. 6

12C

B and D are isotopes of boron; C and E are isotopes of carbon.

4.59 Atomic Symbol

32 70Ge 32

73Ge 32 76Ge

Atomic Number 32 32 32

Mass Number 70 73 76

Number of Protons 32 32 32

Number of Neutrons 38 41 44

Number of Electrons 32 32 32

4.61 a. protons b. protons c. alkali metals

4.63 a. Group 7A (17), Period 4 b. Group 8A (18), Period 3 c. Group 1A (1), Period 2 d. Group 2A (2), Period 7

4.65 a. metal b. metal c. metalloid d. nonmetal

4.67 a. false b. false c. true d. false e. true

4.69 a. 48 protons, 66 neutrons, 48 electrons b. 43 protons, 55 neutrons, 43 electrons c. 79 protons, 120 neutrons, 79 electrons d. 86 protons, 136 neutrons, 86 electrons e. 54 protons, 82 neutrons, 54 electrons

4.71 Name of the Element

Atomic Symbol

Number of Protons

Number of Neutrons

Number of Electrons

Selenium 34 80Se 34 46 34

Nickel 28 62Ni 28 34 28

Magnesium 12 26Mg 12 14 12

Radium 88 228Ra 88 140 88

4.73 a. 3, Li b. 118, Og c. 24.31 amu, Mg d. 19.00 amu, F

4.75 a. bromine, Br b. lanthanum, La c. tellurium, Te d. arsenic, As e. tin, Sn f. cesium, Cs

4.77 a. 78 protons, 78 electrons b. 15 protons, 15 electrons c. 38 protons, 38 electrons d. 27 protons, 27 electrons e. 92 protons, 92 electrons

4.79 a. 22 48Ti b. 12

26Mg c. 31 71Ga d. 30

70Zn

4.81 207.3 amu

4.83 a. 82 protons, 126 neutrons, 82 electrons b. 82

214Pb

c. 83 214Bi

4.85 a. 79 protons, 118 neutrons, 79 electrons b. 79

195Au c. 78

194Pt

4.87 28.09 amu

4.89 a. 38 protons, 49 neutrons, 38 electrons b. 38

88Sr c. 46 neutrons d. The atomic mass on the periodic table is the average of all

the naturally occurring isotopes.

M04_TIMB8119_06_SE_C04.indd 124 11/27/18 11:45 AM

125

Robert and Jennifer work in the materials science department of a research laboratory. As materials engineers, they develop and test various materials that are used in the manufacturing of consumer goods like computer chips, television sets, golf clubs, and snow skis. These engineers create new materials, which are needed for mechanical, chemical, and electrical industries. Often materials engineers study materials at an atomic level to learn how to improve the characteristics of materials.

In their research, Robert and Jennifer work with some of the elements in Groups 3A (13), 4A (14), and 5A (15) of the periodic table. These elements, such as silicon, have properties that make them good semiconductors. A microchip requires growing a single crystal of a semiconductor such as pure silicon. Microchips are manufactured for use in computers, cell phones, satellites, televisions, calculators, GPS, and many other devices. Robert and Jennifer are working on materials that may be used to develop more complex microchips that lead to new applications.

CAREER

Materials Engineer Materials engineers work with metals, ceramics, plastics, semiconductors, and composites to develop new or improved products such as computer chips, aircraft material, and tennis racquets.

Engineers use mathematics and the principles of chemistry and other sciences to solve technical problems. They may also work in the development and testing of new materials. Materials engineers typically have bachelor’s degrees in mechanical, electrical, or chemical engineering, or related fields.

Electronic Structure of Atoms and Periodic Trends

Last week, Robert and Jennifer added impurities to silicon to evaluate changes in its properties as a semiconductor. You can see how Robert and Jennifer used the elements indium, in Group 3A (13), and tellurium, in Group 6A (16), to try to make new computer chips in the UPDATE Developing New Materials for Computer Chips, page 148, and learn some of the properties of indium and tellurium.

UPDATE Developing New Materials for Computer Chips

5

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126 CHAPTER 5 Electronic Structure of Atoms and Periodic Trends

LOOKING AHEAD

5.1 Electromagnetic Radiation 126

5.2 Atomic Spectra and Energy Levels 129

5.3 Sublevels and Orbitals 131 5.4 Orbital Diagrams

and Electron Configurations 135

5.5 Electron Configurations and the Periodic Table 139

5.6 Trends in Periodic Properties 143

The wavelength is the distance between adjacent crests or troughs in a wave.

5.1 Electromagnetic Radiation LEARNING GOAL Compare the wavelength, frequency, and energy of electromagnetic radiation.

When we listen to a radio, use a microwave oven, turn on a light, see the colors of a rainbow, or have an X-ray taken, we are experiencing various forms of electromagnetic radiation. All of these types of electromagnetic radiation, including light, consist of particles that move as waves of energy.

Wavelength and Frequency You are probably familiar with the action of waves in the ocean. When you go to a beach, you notice that the water in each wave rises and falls as the wave comes in to shore. The highest point on the wave is called a crest, whereas the lowest point is a trough. On a calm day, there might be long distances between crests or troughs. However, if there is a storm with a lot of energy, the crests or troughs are much closer together.

The waves of electromagnetic radiation also have crests and troughs. The wavelength (symbol l, lambda) is the distance from a crest or trough in the wave to the next crest or trough in the wave (see FIGURE 5.1). In some types of radiation, the crests or troughs are far apart, while in others, they are close together.

REVIEW Writing Numbers in Scientific

Notation (1.5)

Converting Between Standard Numbers and Scientific Notation (1.5)

Using Prefixes (2.4)

ENGAGE 5.1 How does the wavelength of red light compare to that of blue light?

ENGAGE 5.2 How many seconds will it take light to travel from Rome to Los Angeles, a distance of 10 200 km?

The frequency (symbol n, nu) is the number of times the crests of a wave pass a point in 1 s. All electromagnetic radiation travels at the speed of light (c), which is equal to 3.00 * 108 m/s. Mathematically, the wave equation expresses the relationship of the speed of light (m/s) to wavelength (m) and frequency (s-1).

c = ln Wave equation Speed of light (c) = 3.00 * 108 m/s = wavelength (l) * frequency (n)

The speed of light is about a million times faster than the speed of sound, which is the reason we see lightning before we hear thunder during a storm.

Electromagnetic Spectrum The electromagnetic spectrum is an arrangement of different types of electromagnetic radiation from longest wavelength to shortest wavelength. According to the wave equation, as the wavelength decreases, the frequency increases. Or as the wavelength increases, the frequency decreases. This type of relationship is called an inverse relationship.

Scientists have shown that the energy of electromagnetic radiation is directly related to frequency, which means that energy is inversely related to the wavelength. Thus, as the wavelength of radiation increases, the frequency and, therefore, the energy decrease.

At one end of the electromagnetic spectrum are radiations with long wavelengths such as radio waves that are used for AM and FM radio bands, cellular phones, and TV signals. The wavelength of a typical AM radio wave can be as long as a football field. Microwaves

PRACTICE PROBLEMS Try Practice Problems 5.1 to 5.8

FIGURE 5.1 Wavelengths for light

Light

Light source

Blue light

Red light

l

l

l

l

CrestTrough

Light passing through a prism is separated into a spectrum of colors we see in a rainbow.

The wavelength is the distance from a crest or trough on a wave to the next crest or trough.

M05_TIMB8119_06_SE_C05.indd 126 11/29/18 9:25 AM

5.1 Electromagnetic Radiation 127

FIGURE 5.2 The electromagnetic spectrum shows the arrangement of wavelengths of electromagnetic radiation. The visible portion consists of wavelengths from 700 nm to 400 nm.

104 105 106

400700(nm)

107 108 109 1010 1011 1012 1013 1014 1015 1016 1017 1018 1019

105 104 103 102 10 1 10-1 10-2 10-3 10-4 10-5 10-6 10-7 10-8 10-9 10-10 10-11

Visible light

Wavelength (m)

Power lines

AM/FM Radio

Cell phone

Microwave

Heat

Medical X-rays

Ultraviolet

Television Satellite

Frequency (s-1)

Gamma rays

X-raysUltravioletInfraredMicrowaveRadio waves

Low energy High energy

Energy Increases

have shorter wavelengths and higher frequencies than radio waves. Infrared radiation (IR) is responsible for the heat we feel from sunlight and the infrared lamps used to warm food in restaurants. When we change the volume or the station on a TV set, we use a remote control to send infrared impulses to the receiver in the TV. Wireless technology uses radiation with higher frequencies than infrared to connect many electronic devices, including mobile and cell phones and laptops (see FIGURE 5.2).

Visible light, with wavelengths from 700 to 400 nm, is the only radiation our eyes can detect. Red light has the longest wavelength at 700 nm; orange is about 600 nm; green is about 500 nm; and violet at 400 nm has the shortest wavelength of visible light. We see objects as different colors because the objects reflect only certain wavelengths, which are absorbed by our eyes.

Ultraviolet (UV) light has shorter wavelengths and higher frequencies than violet light of the visible range. The UV radiation in sunlight can cause serious sunburn, which may lead to skin cancer. While some UV light from the Sun is blocked by the ozone layer, the cosmetic industry has developed sunscreens to prevent the absorption of UV light by the skin. X-rays have shorter wavelengths than ultraviolet light, which means they have some of the highest frequencies. X-rays can pass through soft substances but not metals or bone, which allow us to see images of the bones and teeth in the body.

ENGAGE 5.3 Which type of electromagnetic radiation has lower frequency, ultraviolet or infrared?

PRACTICE PROBLEMS Try Practice Problems 5.9 to 5.14

Microwaves with wavelengths of about 1 cm heat water molecules in food.

Infrared radiation from heat lamps keeps food at proper serving temperature.

SAMPLE PROBLEM 5.1 The Electromagnetic Spectrum

TRY IT FIRST

Arrange the following in order of decreasing wavelengths: X-rays, ultraviolet light, FM radio waves, and microwaves.

SOLUTION

The electromagnetic radiation with the longest wavelength is FM radio waves, then micro- waves, followed by ultraviolet light, and then X-rays, which have the shortest wavelengths.

SELF TEST 5.1

Visible light contains colors from red to violet.

a. Does red light or violet light have the shorter wavelength? b. Does red light or violet light have the lower frequency?

ANSWER

a. violet light b. red light

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128 CHAPTER 5 Electronic Structure of Atoms and Periodic Trends

Chemistry Link to Health Biological Reactions to UV Light

Our everyday life depends on sunlight, but exposure to sunlight can have damaging effects on living cells, and too much exposure can even cause their death. Light energy, especially ultraviolet (UV), excites electrons and may lead to unwanted chemical reactions. The list of damaging effects of sunlight includes sunburn; wrinkling; premature aging of the skin; changes in the DNA of the cells, which can lead to skin cancers; inflammation of the eyes; and perhaps cata- racts. Some drugs, like the acne medications Accutane and Retin-A, as well as antibiotics, diuretics, sulfonamides, and estrogen, make the skin extremely sensitive to light.

Phototherapy uses light to treat certain skin conditions, including psoriasis, eczema, and dermatitis. In the treatment of psoriasis, for example, oral drugs are given to make the skin more photosensitive;

then exposure to UV radiation follows. Low-energy radiation (blue light) with wavelengths from 390 to 470 nm is used to treat babies with neonatal jaundice, which converts high levels of bilirubin to water-

soluble compounds that are excreted from the body. Sun- light is also a factor in stimu- lating the immune system.

In a disorder called sea- sonal affective disorder or SAD, people experience mood swings and depression during the winter. Some research sug- gests that SAD is the result of a decrease in serotonin, or an increase in melatonin, when there are fewer hours of sun- light. One treatment for SAD is therapy using bright light provided by a lamp called a light box. A daily exposure to blue light (460 nm) for 30 to 60 min seems to reduce symp- toms of SAD.

Phototherapy is used to treat babies with neonatal jaundice.

A light box is used to provide light, which reduces symptoms of SAD.

PRACTICE PROBLEMS

5.1 Electromagnetic Radiation

5.1 What is meant by the wavelength of UV light?

5.2 How are the wavelength and frequency of light related?

5.3 What is the difference between “white” light and blue or red light?

5.4 Why can we use X-rays, but not radio waves or microwaves, to give an image of bones and teeth?

Applications

5.5 Ultraviolet radiation (UVB) used to treat psoriasis has a wavelength of 3 * 10-7 m, whereas infrared radiation used in

a thermal imaging camera has a wavelength of 1 * 10-5 m. Which has a higher frequency? Which has a higher energy?

5.6 AM radio waves have a frequency of 8 * 105 s-1, whereas TV Channel 2 has a frequency of 6 * 107 s-1. Which has a shorter wavelength? Which has a lower energy?

5.7 If orange light has a wavelength of 6.3 * 10-5 cm, what is its wavelength in meters and nanometers?

5.8 A wavelength of 850 nm is used for fiber-optic transmission. What is its wavelength in meters?

5.9 Which type of electromagnetic radiation, cell phones, AM radio, or infrared light has the longest wavelengths?

5.10 Of radio waves, infrared light, and UV light, which has the shortest wavelengths?

5.11 Place the following types of electromagnetic radiation in order of increasing wavelengths: the blue color in a rainbow, X-rays, and microwaves from an oven.

5.12 Place the following types of electromagnetic radiation in order of decreasing wavelengths: FM radio station, red color in neon lights, and cell phone.

5.13 Place the following types of electromagnetic radiation in order of increasing frequencies: TV signal, X-rays, and microwaves.

5.14 Place the following types of electromagnetic radiation in order of decreasing frequencies: AM music station, UV radiation from the Sun, and infrared radiation from a heat lamp.

A thermal image of the hand of a patient with multiple sclerosis shows the warm areas in the palm of the hand as red and cool areas as blue in the fingers where blood flow is poor.

M05_TIMB8119_06_SE_C05.indd 128 11/29/18 9:25 AM

5.2 Atomic Spectra and Energy Levels 129

5.2 Atomic Spectra and Energy Levels LEARNING GOAL Explain how atomic spectra correlate with the energy levels in atoms.

When the white light from the Sun or a light bulb is passed through a prism or raindrops, it produces a continuous spectrum, like a rainbow. When atoms of elements are heated, they also produce light. At night, you may have seen the yellow color of sodium streetlamps or the red color of neon lights.

Photons The light emitted from a streetlamp or by atoms that are heated is a stream of particles called photons. A photon is a packet of energy with both particle and wave characteristics that travels at the speed of light. High-energy photons have short wavelengths, whereas low-energy photons have long wavelengths.

Photons play an important role in our modern world, particularly in the use of lasers, which use a narrow range of wavelengths. For example, lasers use photons of a single frequency to scan bar codes on labels when we buy groceries. In hospitals, high-energy photons from X-rays and gamma rays are used in scans for diagnosis or for radiation treatments to reach tumors within the tissues without damaging the sur- rounding tissue.

Atomic Spectra When the light emitted from heated elements is passed through a prism, it does not produce a continuous spectrum. Instead, an atomic spectrum is produced that consists of lines of different colors separated by dark areas (see FIGURE 5.3). This separation of colors indicates that only certain wavelengths of light are produced when an element is heated, which gives each element a unique atomic spectrum.

ENGAGE 5.4 Why does the light emitted by heated elements form atomic spectra whereas white light produces a continuous spectrum?

A rainbow forms when light passes through water droplets.

In gamma knife radiation therapy, gamma photons are used to kill cancer cells or reduce tumor size.

FIGURE 5.3 In an atomic spectrum, light from a heated element separates into distinct lines.

Sr

Strontium atomic spectrum

Light from a heated element passes through a prism

Prism Film

Barium atomic spectrum

Light

Ba Light

Electron Energy Levels Scientists have now determined that the lines in atomic spectra are associated with changes in the energies of the electrons. In an atom, each electron has a specific energy known as its energy level, which is assigned a value called the principal quantum number (n), (n = 1, n = 2, c ). Generally, electrons in the lower energy levels are closer to the nucleus, while electrons in the higher energy levels are farther away. The energy of an

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130 CHAPTER 5 Electronic Structure of Atoms and Periodic Trends

FIGURE 5.4 Changes in energy levels

Energy absorbed

Energy absorbed

Low-energy photons

High-energy photons

When electrons lose energy, photons with specific energies are emitted.

When electrons absorb a specific amount of energy, they move to a higher energy level.

SAMPLE PROBLEM 5.2 Changes in Energy Levels

TRY IT FIRST

a. When does an electron move to a higher energy level? b. When an electron drops to a lower energy level, how is energy lost?

SOLUTION

a. An electron moves to a higher energy level when it absorbs an amount of energy equal to the difference between energy levels.

b. Energy, equal to the difference between energy levels, is emitted as a photon when an electron drops to a lower energy level.

electron is quantized, which means that the energy of an electron can only have specific energy values, but cannot have values between these values.

Principal Quantum Number (n)

1 6 2 6 3 6 4 6 5 6 6 6 7 Lowest h Highest energy energy

All the electrons with the same energy are grouped in the same energy level. As an analogy, we can think of the energy levels of an atom as similar to the shelves in a bookcase. The first shelf is the lowest energy level; the second shelf is the second energy level; and so on. If we are arranging books on the shelves, it would take less energy to fill the bottom shelf first, and then the second shelf, and so on. However, we could never get any book to stay in the space between any of the shelves. Similarly, the energy of an electron must be at a specific energy level, and not between.

Unlike standard bookcases, however, there is a large difference between the energy of the first and second energy levels, but then the higher energy levels are closer together. Another difference is that the lower electron energy levels hold fewer electrons than the higher energy levels.

Changes in Energy Levels When an atom absorbs energy, an electron moves to a higher energy level called an excited state. An electron can change from one energy level to a higher energy level only if it absorbs the energy equal to the difference in energy levels. However, an electron in the excited state is less stable and falls to a lower-energy state by emitting a photon of a specific energy. If the energy emitted is in the visible range, we see one of the colors of visible light (see FIGURE 5.4). The yellow color of sodium streetlights and the red color of neon lights are examples of electrons emitting energy in the visible color range.

ENGAGE 5.5 What causes electrons to move to higher energy levels?

THE

RED BA

DG E

OF

COU RAG

E

by

Step hen

Cra ne

n = 5

n = 4

n = 3

n = 2

n = 1

Nucleus

E ne

rg y

In cr

ea se

s

An electron can have only the energy of one of the energy levels in an atom.

Colors are produced when electricity excites electrons in noble gases.

M05_TIMB8119_06_SE_C05.indd 130 11/29/18 9:25 AM

5.3 Sublevels and Orbitals 131

Energy Level (n) 1 2 3 4

2n2 2(1)2 2(2)2 2(3)2 2(4)2

Maximum Number of Electrons 2 8 18 32

SELF TEST 5.2

a. Why did scientists propose that electrons occupy specific energy levels in an atom? b. How does the wavelength of a low-energy photon compare to that of a high-energy

photon?

ANSWER

a. Because the atomic spectra of elements consist of discrete, separated lines, scientists concluded that electrons occupy only certain energy levels in the atom.

b. The wavelength of a low-energy photon would be longer than the wavelength of a high- energy photon.

PRACTICE PROBLEMS

5.2 Atomic Spectra and Energy Levels

5.15 What feature of an atomic spectrum indicates that the energy emitted by heating an element is not continuous?

5.16 How can we explain the distinct lines that appear in an atomic spectrum?

5.17 Electrons can jump to higher energy levels when they ___________ (absorb/emit) a photon.

5.18 Electrons drop to lower energy levels when they ___________ (absorb/emit) a photon.

5.19 Identify the photon in each pair with the greater energy. a. green light or yellow light b. red light or blue light

5.20 Identify the photon in each pair with the greater energy. a. orange light or violet light b. infrared light or ultraviolet light

5.3 Sublevels and Orbitals LEARNING GOAL Describe the sublevels and orbitals for the electrons in an atom.

We have seen that the protons and neutrons are contained in the small, dense nucleus of an atom. However, it is the electrons within the atoms that determine the physical and chemical properties of the elements. Therefore, we will look at the arrangement of electrons within the large volume of space surrounding the nucleus.

There is a limit to the number of electrons allowed in each energy level. Only a few electrons can occupy the lower energy levels, while more electrons can be accommodated in higher energy levels. The maximum number of electrons allowed in any energy level is calculated using the formula 2n2 (two times the square of the principal quantum number). TABLE 5.1 shows the maximum number of electrons allowed in the first four energy levels.

TABLE 5.1 Maximum Number of Electrons Allowed in Energy Levels 1 to 4

PRACTICE PROBLEMS Try Practice Problems 5.15 to 5.20

Sublevels Each of the energy levels consists of one or more sublevels, in which electrons with identical energy are found. The sublevels are identified by the letters s, p, d, and f. The number of sublevels within an energy level is equal to the principal quantum number, n. For example, the first energy level (n = 1) has only one sublevel, 1s. The second energy level (n = 2) has two sublevels, 2s and 2p. The third energy level (n = 3) has three sublevels, 3s, 3p, and 3d. The fourth energy level (n = 4) has four sublevels, 4s, 4p, 4d, and 4f. Energy levels n = 5, n = 6, and n = 7 also have as many sublevels as the value of n, but only s, p, d, and f sublevels are needed to hold the electrons in atoms of the 118 known elements (see FIGURE 5.5).

ENGAGE 5.6 How many sublevels are in energy level n = 5?

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132 CHAPTER 5 Electronic Structure of Atoms and Periodic Trends

FIGURE 5.5 The number of sublevels in an energy level is the same as the principal quantum number, n. n = 4

s p d f

n = 3

n = 2

n = 1

4

3

2

1

Types of SublevelsNumber of Sublevels

Energy Level

Within each energy level, the s sublevel has the lowest energy. If there are additional sublevels, the p sublevel has the next lowest energy, then the d sublevel, and finally the f sublevel.

Order of Increasing Energy of Sublevels in an Energy Level s 6 p 6 d 6 f Lowest

h Highest

energy energy

Orbitals There is no way to know the exact location of an electron in an atom. Instead, scientists describe the location of an electron in terms of probability. The orbital is the three- dimensional volume in which electrons have the highest probability of being found.

As an analogy, imagine that you draw a circle with a 100-m radius around your chemistry classroom. There is a high probability of finding you within that circle when your chemistry class is in session. But once in a while, you may be outside that circle because you were sick or your car did not start.

Shapes of Orbitals Each type of orbital has a unique three-dimensional shape. Electrons in an s orbital are most likely found in a region with a spherical shape. Imagine that you take a picture of the location of an electron in an s orbital every second for an hour. When all these pictures are overlaid, the result, called a probability density, would look like the electron cloud shown in FIGURE 5.6. For convenience, we draw this electron cloud as a sphere called an s orbital. There is one s orbital for every energy level starting with n = 1. For example, in the first, second, and third energy levels, there are s orbitals designated as 1s, 2s, and 3s. As the principal quantum number increases, there is an increase in the size of the s orbitals, although the shape is the same (see FIGURE 5.7).

The orbitals occupied by p, d, and f electrons have three-dimensional shapes different from those of the s electrons. There are three p orbitals, starting with n = 2. Each p orbital has two lobes like a balloon tied in the middle. The three p orbitals are arranged in three perpendicular directions, along the x, y, and z axes around the nucleus (see FIGURE 5.8). As with s orbitals, the shape of p orbitals is the same, but the volume increases at higher energy levels.

z

y

x

FIGURE 5.6 The electron cloud of an s orbital is spherical, which indicates a higher probability of finding an s electron near the nucleus.

s orbitals

2s

1s

3s

FIGURE 5.7 The sizes of the s orbitals increase because they contain electrons at higher energy levels.

ENGAGE 5.7 What are some similarities and differences of the p orbitals in the n = 3 energy level?

FIGURE 5.8 Shapes of p orbitals

2pz

z

x y

z

x y

p orbitals Combined p orbitals around the nucleus

2px

2px

z

x y

2py

2py

z

x y

2pz

The p orbitals in the same energy level are arranged on their axes around the nucleus.

The electron cloud of a p orbital shows the highest probability of finding a p electron.

M05_TIMB8119_06_SE_C05.indd 132 11/29/18 9:25 AM

5.3 Sublevels and Orbitals 133

In summary, the n = 2 energy level, which has 2s and 2p sublevels, consists of one s orbital and three p orbitals.

Energy level n = 2 2p sublevel

2s sublevel 2s orbital

2p orbitals Energy level n = 2 consists of one 2s orbital and three 2p orbitals.

Energy level n = 3 consists of three sublevels s, p, and d. The d sublevels contain five d orbitals (see FIGURE 5.9).

Energy level n = 4 consists of four sublevels s, p, d, and f. In the f sublevel, there are seven f orbitals. The shapes of f orbitals are complex, and we have not included them in this text.

ENGAGE 5.8 How many orbitals are there in the 5d sublevel?

d

z

x

y

z

x

y

dxy

z

x y

x

dxz

z

y

dx2- y2 z2

x

y

dyz

z

FIGURE 5.9 Four of the five d orbitals consist of four lobes that are aligned along or between different axes. One d orbital consists of two lobes and a doughnut-shaped ring around its center.

SAMPLE PROBLEM 5.3 Energy Levels, Sublevels, and Orbitals

TRY IT FIRST

Indicate the type and number of orbitals in each of the following energy levels or sublevels:

a. 3p sublevel b. n = 2 c. n = 3 d. 4d sublevel

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

energy level, sublevel orbitals n, orbitals in s, p, d, f

a. The 3p sublevel contains three 3p orbitals. b. The n = 2 energy level consists of one 2s and three 2p orbitals. c. The n = 3 energy level consists of one 3s, three 3p, and five 3d orbitals. d. The 4d sublevel contains five 4d orbitals.

SELF TEST 5.3

a. What is similar and what is different for 1s, 2s, and 3s orbitals? b. How many orbitals would be in the 4f sublevel?

ANSWER

a. The 1s, 2s, and 3s orbitals are all spherical, but they increase in volume because the electron is most likely to be found farther from the nucleus for higher energy levels.

b. The 4f sublevel has seven orbitals. PRACTICE PROBLEMS

Try Practice Problems 5.21 to 5.24

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134 CHAPTER 5 Electronic Structure of Atoms and Periodic Trends

Energy Level (n)

Number of Sublevels

Type of Sublevel

Number of Orbitals

Maximum Number of Electrons Total Electrons

1 1 1s 1 2 2

2 2 2s 1 2

2p 3 6 8

3 3 3s 1 2

3p 3 6

3d 5 10 18

4 4 4s 1 2

4p 3 6

4d 5 10

4f 7 14 32

TABLE 5.2 Electron Capacity in Sublevels for Energy Levels 1 to 4

Orbital Capacity and Electron Spin The Pauli exclusion principle states that each orbital can hold a maximum of two electrons. According to a model for electron behavior, an electron is seen as spinning on its axis, which generates a magnetic field. When two electrons are in the same orbital, they will repel each other unless their magnetic fields cancel. This happens only when the two electrons spin in opposite directions. We can represent the spins of the electrons in the same orbital with one arrow pointing up and the other pointing down.

Number of Electrons in Sublevels There is a maximum number of electrons that can occupy each sublevel. An s sublevel holds one or two electrons. Because each p orbital can hold up to two electrons, the three p orbitals in a p sublevel can accommodate six electrons. A d sublevel with five d orbitals can hold a maximum of 10 electrons. With seven f orbitals, an f sublevel can hold up to 14 electrons.

As mentioned earlier, higher energy levels such as n = 5, 6, and 7 would have 5, 6, and 7 sublevels, but those beyond sublevel f are not utilized by the atoms of the elements known today. The total number of electrons in all the sublevels adds up to give the electrons allowed in an energy level. The number of sublevels, the number of orbitals, and the maximum number of electrons for energy levels 1 to 4 are shown in TABLE 5.2.

Electron spinning counterclockwise

Electron spinning clockwise

Opposite spins of electrons in an orbital

An orbital can hold up to two electrons with opposite spins.

PRACTICE PROBLEMS Try Practice Problems 5.25 to 5.28

PRACTICE PROBLEMS

5.3 Sublevels and Orbitals

5.21 Describe the shape of each of the following orbitals: a. 1s b. 2p c. 5s

5.22 Describe the shape of each of the following orbitals: a. 3p b. 6s c. 4p

5.23 Match statements 1 to 3 with a to d: 1. They have the same shape. 2. The maximum number of electrons is the same. 3. They are in the same energy level.

a. 1s and 2s orbitals b. 3s and 3p sublevels c. 3p and 4p sublevels d. three 3p orbitals

5.24 Match statements 1 to 3 with a to d: 1. They have the same shape. 2. The maximum number of electrons is the same. 3. They are in the same energy level.

a. 5s and 6s orbitals b. 3p and 4p orbitals c. 3s and 4s sublevels d. 2s and 2p orbitals

5.25 Indicate the number of each in the following: a. orbitals in the 3d sublevel b. sublevels in the n = 1 energy level c. orbitals in the 6s sublevel d. orbitals in the n = 3 energy level 5.26 Indicate the number of each in the following: a. orbitals in the n = 2 energy level b. sublevels in the n = 4 energy level c. orbitals in the 5f sublevel d. orbitals in the 6p sublevel

5.27 Indicate the maximum number of electrons in the following: a. 2p orbital b. 3p sublevel c. n = 4 energy level d. 5d sublevel 5.28 Indicate the maximum number of electrons in the following: a. 3s sublevel b. 4p orbital c. n = 3 energy level d. 4f sublevel

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5.4 Orbital Diagrams and Electron Configurations 135

5.4 Orbital Diagrams and Electron Configurations LEARNING GOAL Draw the orbital diagram and write the electron configuration for an element.

We can now look at how electrons are arranged in the orbitals within an atom. An orbital diagram shows the placement of the electrons in the orbitals in order of increasing energy (see FIGURE 5.10). In this energy diagram, we see that the electrons in the 1s orbital have the lowest energy level. The energy level is higher for the 2s orbital and is even higher for the 2p orbitals.

ENGAGE 5.10 When do we need to use Hund’s rule in drawing orbital diagrams?

1s 2s 2p

Orbital diagram for carbon

Half-filled Empty

E ne

rg y

In cr

ea se

s

1s

4f

5f

5d

6p

6s

7p

7s

6d

4d

3d

5s

4s

3s

2s

5p

4p

3p

2p

FIGURE 5.10 The orbitals in an atom fill in order of increasing energy beginning with 1s.

ENGAGE 5.9 Why do 3d orbitals fill after the 4s orbital?

We begin our discussion of electron configuration by using orbital diagrams in which boxes represent the orbitals. Any orbital can have a maximum of two electrons.

To draw an orbital diagram, the lowest energy orbitals are filled first. For example, we can draw the orbital diagram for carbon. The atomic number of carbon is 6, which means that a carbon atom has six electrons. The first two electrons go into the 1s orbital; the next two electrons go into the 2s orbital. In the orbital diagram, the two electrons in the 1s and 2s orbitals are shown with opposite spins; the first arrow is up and the second is down. The last two electrons in carbon begin to fill the 2p sublevel, which has the next lowest energy. However, there are three 2p orbitals of equal energy. Because the negatively charged electrons repel each other, they are placed in separate 2p orbitals. Hund’s rule states that the lowest energy state is attained when electrons are placed singly in available orbitals of the same sublevel with the same spin. With few exceptions (which will be noted later in this chapter), lower energy sublevels are filled first, and then the “building” of electrons continues to the next lowest energy sublevel that is available until all the electrons are placed.

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136 CHAPTER 5 Electronic Structure of Atoms and Periodic Trends

CORE CHEMISTRY SKILL Writing Electron Configurations

Electron Configurations Chemists use a notation called the electron configuration to indicate the placement of the electrons of an atom in order of increasing energy. As in the orbital diagrams, an electron configuration is written with the lowest energy sublevel first, followed by the next lowest energy sublevel. The number of electrons in each sublevel is shown as a superscript.

1s22s22p2

Number of electrons

Read as “one s two, two s two, two p two”

Electron Configuration for Carbon Type of orbital

Period 1: Hydrogen and Helium We will draw orbital diagrams and write the corresponding electron configurations for the elements H and He in Period 1. The 1s orbital (which is the 1s sublevel) is written first because it has the lowest energy. Hydrogen has one electron in the 1s sublevel; helium has two. In the orbital diagram, the electrons for helium are shown as arrows pointing in opposite directions.

Period 2: Lithium to Neon Period 2 begins with lithium, which has three electrons. The first two electrons fill the 1s orbital, whereas the third electron goes into the 2s orbital, the sublevel with the next lowest energy. In beryllium, another electron is added to complete the 2s orbital. The next six electrons are used to fill the 2p orbitals. The electrons are added to separate p orbitals (Hund’s rule) from boron to nitrogen, which gives three half-filled 2p orbitals.

From oxygen to neon, the remaining three electrons are paired up using opposite spins to complete the 2p sublevel. In writing the electron configurations for the elements in Period 2, begin with the 1s orbital followed by the 2s and then the 2p orbitals.

An electron configuration can also be written in an abbreviated configuration. The electron configuration of the preceding noble gas is replaced by writing its element symbol inside square brackets. For example, the electron configuration for lithium, 1s22s1, can be abbreviated as [He]2s1 where [He] replaces 1s2.

1H 1s1

1s2

Atomic NumberElement

Orbital Diagram

Electron Configuration

1s

2He

3Li 1s22s1

1s22s22p1

1s22s2

Atomic NumberElement Orbital Diagram

Electron Configuration

1s 2s

5B

4Be

[He]2s1

Abbreviated Electron Configuration

[He]2s22p1

[He]2s2

2p

1s22s22p26C [He]2s22p2

1s22s22p37N [He]2s22p3

1s22s22p48O [He]2s22p4

1s22s22p59F [He]2s 22p5

1s22s22p610Ne [He]2s 22p6

Unpaired electrons

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5.4 Orbital Diagrams and Electron Configurations 137

2p2s1s

2p2s1s

2p2s1s

SAMPLE PROBLEM 5.4 Drawing Orbital Diagrams

TRY IT FIRST

Nitrogen is an element that is used in the formation of amino acids, proteins, and nucleic acids. Draw the orbital diagram for nitrogen.

SOLUTION

STEP 1 Draw boxes to represent the orbitals. Nitrogen has atomic number 7, which means it has seven electrons. For the orbital diagram, we draw boxes to r epresent the 1s, 2s, and 2p orbitals.

2p2s1s

Orbital diagram for nitrogen (N)

STEP 2 Place a pair of electrons with opposite spins in each filled orbital. First, we place a pair of electrons with opposite spins in both the 1s and 2s orbitals.

STEP 3 Place the remaining electrons in the last occupied sublevel in separate orbitals. Then we place the three remaining electrons in three separate 2p orbitals with arrows drawn in the same direction.

SELF TEST 5.4

Draw the orbital diagram for f luorine, which is used to make nonstick coatings for cookware.

ANSWER

Period 3: Sodium to Argon In Period 3, electrons enter the orbitals of the 3s and 3p sublevels, but not the 3d sublevel. We notice that the elements sodium to argon, which are directly below the elements lithium to neon in Period 2, have a similar pattern of filling their s and p orbitals. In sodium and magnesium, one and two electrons go into the 3s orbital. The electrons for aluminum, silicon, and phosphorus go into separate 3p orbitals. The remaining electrons in sulfur, chlorine, and argon are paired up (opposite spins) with the electrons already in the 3p orbitals. For the abbreviated electron configurations of Period 3, the symbol [Ne] replaces 1s22s22p6.

Fluorine is used to make nonstick coatings for cookware.

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138 CHAPTER 5 Electronic Structure of Atoms and Periodic Trends

PRACTICE PROBLEMS Try Practice Problems 5.29 to 5.40

SAMPLE PROBLEM 5.5 Electron Configurations

TRY IT FIRST

Silicon is the basis of semiconductors. Write the complete and abbreviated electron con- figurations for silicon.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

silicon (Si) electron configuration, abbreviated electron configuration

periodic table, atomic number, order of filling orbitals

STEP 1 State the number of electrons from the atomic number on the periodic table. Silicon has an atomic number of 14, which means it has 14 electrons.

STEP 2 Write the number of electrons for each orbital in order of increasing energy until filling is complete.

1s22s22p63s23p2 Electron configuration for Si

STEP 3 Write an abbreviated electron configuration by replacing the configuration of the preceding noble gas with its symbol.

[Ne]3s23p2 Abbreviated electron configuration for Si

SELF TEST 5.5

Write the complete and abbreviated electron configurations for each of the following:

a. sulfur, a macromineral in proteins, vitamin B1, and insulin b. vanadium, a micromineral needed for the extraction of energy from foods

ANSWER

a. 1s22s22p63s23p4, [Ne]3s23p4

b. 1s22s22p63s23p64s23d3, [Ar]4s23d3

Sulfur atoms are part of the structure of human insulin.

11Na 1s22s22p63s1

1s22s22p63s2

1s22s22p63s23p1

1s22s22p63s23p2

1s22s22p63s23p3

1s22s22p63s23p4

1s22s22p63s23p5

1s22s22p63s23p6

Atomic NumberElement

Orbital Diagram (3s and 3p orbitals only)

Electron Configuration

3s 3p

13Al

12Mg

[Ne]3s1

Abbreviated Electron Configuration

[Ne]3s23p1

[Ne]3s2

14Si [Ne]3s23p2

15P [Ne]3s23p3

16S [Ne]3s23p4

17Cl [Ne]3s23p5

18Ar [Ne]3s23p6

[Ne]

[Ne]

[Ne]

[Ne]

[Ne]

[Ne]

[Ne]

[Ne]

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5.5 Electron Configurations and the Periodic Table 139

PRACTICE PROBLEMS

5.4 Orbital Diagrams and Electron Configurations

5.29 Compare the terms electron configuration and abbreviated electron configuration.

5.30 Compare the terms orbital diagram and electron configuration.

5.31 Draw the orbital diagram for each of the following: a. boron b. aluminum c. phosphorus d. argon

5.32 Draw the orbital diagram for each of the following: a. chlorine b. sulfur c. magnesium d. beryllium

5.33 Write the complete electron configuration for each of the following:

a. nickel b. sodium c. lithium d. titanium

5.34 Write the complete electron configuration for each of the following:

a. nitrogen b. chlorine c. strontium d. neon

5.35 Write the abbreviated electron configuration for each of the following:

a. tin b. cadmium c. selenium d. fluorine

5.36 Write the abbreviated electron configuration for each of the following:

a. barium b. oxygen c. manganese d. arsenic

5.37 Give the symbol of the element with each of the following electron or abbreviated electron configurations:

a. 1s22s1 b. 1s22s22p63s23p64s23d2

c. [Ar]4s23d104p2 d. [Ne]3s23p1

5.38 Give the symbol of the element with each of the following electron or abbreviated electron configurations:

a. 1s22s22p4 b. [Ne]3s2

c. 1s22s22p63s23p6 d. [He]2s22p5

5.39 Give the symbol of the element that meets the following conditions:

a. has three electrons in the n = 3 energy level b. has two 2p electrons c. completes the 3p sublevel d. completes the 2s sublevel

5.40 Give the symbol of the element that meets the following conditions:

a. has five electrons in the 3p sublevel b. has four 2p electrons c. completes the 3s sublevel d. has one electron in the 3s sublevel

5.5 Electron Configurations and the Periodic Table

LEARNING GOAL Write the electron configuration for an atom using the sublevel blocks on the periodic table.

Up to now, we have written electron configurations using their energy level diagrams. As configurations involve more energy levels, this can become tedious. However, the electron configurations of the elements are related to their position on the periodic table. Different sections or blocks within the periodic table correspond to the s, p, d, and f sublevels (see FIGURE 5.11). Therefore, we can “build” the electron configurations of atoms by reading the periodic table in order of increasing atomic number.

ENGAGE 5.11 How do you use the periodic table to determine the number of electrons in the 1s, 2s, and 2p sublevels of neon?

Period Number

2

1

3

4

5

6

7

2s

1s

3s

4s

5s

6s

7s

3d

4d

5d

4f

5f

6d

2p

3p

4p

5p

6p

7p

s Block

d Block

p Block

f Block

FIGURE 5.11 Electron configurations follow the order of occupied sublevels on the periodic table.

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140 CHAPTER 5 Electronic Structure of Atoms and Periodic Trends

Blocks on the Periodic Table 1. The s block includes hydrogen and helium as well as the elements in Group 1A (1)

and Group 2A (2). This means that the final one or two electrons in the elements of the s block are located in an s orbital. The period number indicates the particular s orbital that is filling: 1s, 2s, and so on.

2. The p block consists of the elements in Group 3A (13) to Group 8A (18). There are six p block elements in each period because three p orbitals can hold up to six electrons. The period number indicates the particular p sublevel that is filling: 2p, 3p, and so on.

3. The d block, containing the transition elements, first appears after calcium (atomic number 20). There are 10 elements in each period of the d block because five d orbitals can hold up to 10 electrons. The particular d sublevel is one less (n - 1) than the period number. For example, in Period 4, the d block is the 3d sublevel. In Period 5, the d block is the 4d sublevel.

4. The f block, the inner transition elements, are the two rows at the bottom of the periodic table. There are 14 elements in each f block because seven f orbitals can hold up to 14 electrons. Elements that have atomic numbers higher than 57 (La) have electrons in the 4f block. The particular f sublevel is two less (n - 2) than the period number. For example, in Period 6, the f block is the 4f sublevel. In Period 7, the f block is the 5f sublevel.

Writing Electron Configurations Using Sublevel Blocks Now we can write electron configurations using the sublevel blocks on the periodic table. As before, each configuration begins at H. But now we move across the table from left to right, writing down each sublevel block we come to until we reach the element for which we are writing an electron configuration. We will show how to write the electron configuration for chlorine (atomic number 17) from the sublevel blocks on the periodic table in Sample Problem 5.6.

ENGAGE 5.12 On the periodic table, what are the group numbers that make up the s block? the p block? the d block?

CORE CHEMISTRY SKILL Using the Periodic Table to

Write Electron Configurations

SAMPLE PROBLEM 5.6 Using the Sublevel Blocks to Write Electron Configurations

TRY IT FIRST Use the sublevel blocks on the periodic table to write the electron configuration for chlorine.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

element chlorine electron configuration sublevel blocks

STEP 1 Locate the element on the periodic table. Chlorine (atomic number 17) is in Period 3 and Group 7A (17).

STEP 2 Write the filled sublevels in order, going across each period.

Period Sublevel Block Filling Sublevel Block Notation (filled)

1 1s (H S He) 1s2

2 2s (Li S Be) and 2p (B S Ne) 2s22p6

3 3s (Na S Mg) 3s2

STEP 3 Complete the configuration by counting the electrons in the last occupied sublevel block. Because chlorine is the fifth element in the 3p block, there are five electrons in the 3p sublevel.

Period Sublevel Block Filling Sublevel Block Notation

3 3p (Al S Cl) 3p5

The electron configuration for chlorine (Cl) is: 1s22s22p63s23p5.

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5.5 Electron Configurations and the Periodic Table 141

SELF TEST 5.6

Use the sublevel blocks on the periodic table to write the electron configuration for each of the following:

a. argon b. cobalt

ANSWER

a. 1s22s22p63s23p6 b. 1s22s22p63s23p64s23d7

Electron Configurations for Period 4 and Above Up to Period 4, the filling of the sublevels has progressed in order. However, if we look at the sublevel blocks in Period 4, we see that the 4s sublevel fills before the 3d sublevel. This occurs because the electrons in the 4s sublevel have slightly lower energy than the electrons in the 3d sublevel. This order occurs again in Period 5 when the 5s sublevel fills before the 4d sublevel, in Period 6 when the 6s fills before the 5d, and in Period 7 when the 7s fills before the 6d.

At the beginning of Period 4, the electrons in potassium (19) and calcium (20) go into the 4s sublevel. In scandium, the next electron added after the 4s sublevel is filled goes into the 3d block. The 3d block continues to fill until it is complete with 10 electrons at zinc (30). Once the 3d block is complete, the next six electrons, gallium to krypton, go into the 4p block.

ENGAGE 5.13 What sublevel blocks are used to add electrons to the elements from K to Zn?

Element Atomic Number Electron Configuration Abbreviated Electron Configuration

4s Block

K 19 1s22s22p63s23p64s1 [Ar]4s1

Ca 20 1s22s22p63s23p64s2 [Ar]4s2

3d Block

Sc 21 1s22s22p63s23p64s23d1 [Ar]4s23d1

Ti 22 1s22s22p63s23p64s23d2 [Ar]4s23d2

V 23 1s22s22p63s23p64s23d3 [Ar]4s23d3

Cr* 24 1s22s22p63s23p64s13d5 [Ar]4s13d5 half-filled d sublevel is stable

Mn 25 1s22s22p63s23p64s23d5 [Ar]4s23d5

Fe 26 1s22s22p63s23p64s23d6 [Ar]4s23d6

Co 27 1s22s22p63s23p64s23d7 [Ar]4s23d7

Ni 28 1s22s22p63s23p64s23d8 [Ar]4s23d8

Cu* 29 1s22s22p63s23p64s13d10 [Ar]4s13d10 filled d sublevel is stable

Zn 30 1s22s22p63s23p64s23d10 [Ar]4s23d10

4p Block

Ga 31 1s22s22p63s23p64s23d104p1 [Ar]4s23d104p1

Ge 32 1s22s22p63s23p64s23d104p2 [Ar]4s23d104p2

As 33 1s22s22p63s23p64s23d104p3 [Ar]4s23d104p3

Se 34 1s22s22p63s23p64s23d104p4 [Ar]4s23d104p4

Br 35 1s22s22p63s23p64s23d104p5 [Ar]4s23d104p5

Kr 36 1s22s22p63s23p64s23d104p6 [Ar]4s23d104p6

*Exceptions to the order of filling.

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142 CHAPTER 5 Electronic Structure of Atoms and Periodic Trends

In Sample Problem 5.7, we use sublevel blocks on the periodic table to write the electron configuration for selenium (atomic number 34).

Some dietary sources of iodine include seafood, eggs, and milk.

SAMPLE PROBLEM 5.7 Using Sublevel Blocks to Write Electron Configurations

TRY IT FIRST Selenium is used in making glass and in pigments. Use the sublevel blocks on the periodic table to write the electron configuration for selenium.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

selenium electron configuration sublevel blocks

STEP 1 Locate the element on the periodic table. Selenium is in Period 4 and Group 6A (16).

STEP 2 Write the filled sublevels in order, going across each period.

Period Sublevel Block Filling Sublevel Block Notation (filled)

1 1s (H S He) 1s2

2 2s (Li S Be) and 2p (B S Ne) 2s22p6

3 3s (Na S Mg) and 3p (Al S Ar) 3s23p6

4 4s (K S Ca) and 3d (Sc S Zn) 4s23d10

STEP 3 Complete the configuration by counting the electrons in the last occupied sublevel block. Because selenium is the fourth element in the 4p block, there are four electrons in the 4p sublevel.

Period Sublevel Block Filling Sublevel Block Notation

4 4p (Ga S Se) 4p4

The electron configuration for selenium (Se) is: 1s22s22p63s23p64s23d104p4.

SELF TEST 5.7

a. Iodine is a micromineral needed for thyroid function. Use the sublevel blocks on the periodic table to write the electron configuration for iodine.

b. Manganese is a micromineral needed for metabolic function. Use the sublevel blocks on the periodic table to write the electron configuration for manganese.

ANSWER

a. 1s22s22p63s23p64s23d104p65s24d105p5 b. 1s22s22p63s23p64s23d5

Some Exceptions in Sublevel Block Order Within the filling of the 3d sublevel, exceptions occur for chromium and copper. In Cr and Cu, the 3d sublevel is close to being a half-filled or filled sublevel, which is particularly stable. Thus, the electron configuration for chromium has only one electron in the 4s and five electrons in the 3d sublevel to give the added stability of a half-filled d sublevel. This is shown in the abbreviated orbital diagram for chromium:

3d (half-filled)4s

[Ar]

4p Abbreviated orbital diagram for chromium

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5.6 Trends in Periodic Properties 143

3d (filled)4s

[Ar]

4p Abbreviated orbital diagram for copper

A similar exception occurs when copper achieves a stable, filled 3d sublevel with 10 electrons and only one electron in the 4s orbital. This is shown in the abbreviated orbital diagram for copper:

After the 4s and 3d sublevels are completed, the 4p sublevel fills as expected from gallium to krypton, the noble gas that completes Period 4. There are also exceptions in filling for the higher d and f electron sublevels, some caused by the added stability of half-filled shells and others where the cause is not known.

PRACTICE PROBLEMS Try Practice Problems 5.41 to 5.50

PRACTICE PROBLEMS

5.5 Electron Configurations and the Periodic Table

5.41 Use the sublevel blocks on the periodic table to write a complete electron configuration for an atom of each of the following:

a. arsenic b. iron c. tin d. krypton

5.42 Use the sublevel blocks on the periodic table to write a complete electron configuration for an atom of each of the following:

a. calcium b. nickel c. gallium d. cadmium

5.43 Use the sublevel blocks on the periodic table to write an abbreviated electron configuration for an atom of each of the following:

a. titanium b. bromine c. barium d. lead

5.44 Use the sublevel blocks on the periodic table to write an abbreviated electron configuration for an atom of each of the following:

a. vanadium b. palladium c. zinc d. cesium

5.45 Use the periodic table to give the symbol of the element with each of the following electron configurations:

a. 1s22s22p63s23p3 b. 1s22s22p63s23p64s23d7

c. [Ar]4s23d10 d. [Xe]6s24f 145d106p3

5.46 Use the periodic table to give the symbol of the element with each of the following electron configurations:

a. 1s22s22p63s23p64s23d8 b. [Kr]5s24d105p4

c. 1s22s22p63s23p64s23d104p2 d. [Ar]4s23d104p5

5.47 Use the periodic table to give the symbol of the element that meets the following conditions:

a. has three electrons in the n = 4 energy level b. has three 2p electrons c. completes the 5p sublevel d. has two electrons in the 4d sublevel

5.48 Use the periodic table to give the symbol of the element that meets the following conditions:

a. has five electrons in the n = 3 energy level b. has one electron in the 6p sublevel c. completes the 7s sublevel d. has four 5p electrons

5.49 Use the periodic table to give the number of electrons in the indicated sublevels for the following:

a. 3d in zinc b. 2p in sodium c. 4p in arsenic d. 5s in rubidium

5.50 Use the periodic table to give the number of electrons in the indicated sublevels for the following:

a. 3d in manganese b. 5p in antimony c. 6p in lead d. 3s in magnesium

5.6 Trends in Periodic Properties LEARNING GOAL Use the electron configurations of elements to explain the trends in periodic properties.

The electron configurations of atoms are an important factor in the physical and chemical properties of the elements and in the properties of the compounds that they form. In this section, we will look at the valence electrons in atoms, the trends in atomic size, ionization energy, and metallic character. Going across a period, there is a pattern of regular change in these properties from one group to the next. Known as periodic properties, each property increases or decreases across a period, and then the trend is repeated in each successive period. We can use the seasonal changes in temperatures as an analogy for periodic properties. In the winter, temperatures are cold and become warmer in the spring. By summer, the outdoor temperatures are hot but begin to cool in the fall. By winter, we expect cold temperatures again as the pattern of decreasing and increasing temperatures repeats for another year.

SPRING SUMMER AUTUMN WINTER

The change in temperature with the seasons is a periodic property.

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144 CHAPTER 5 Electronic Structure of Atoms and Periodic Trends

1A (1) 2A (2) 3A (13) 4A (14) 5A (15) 6A (16) 7A (17) 8A (18)

1

H

1s1

2

He

1s2

3

Li

2s1

4

Be

2s2

5

B

2s22p1

6

C

2s22p2

7

N

2s22p3

8

O

2s22p4

9

F

2s22p5

10

Ne

2s22p6

11

Na

3s1

12

Mg

3s2

13

Al

3s23p1

14

Si

3s23p2

15

P

3s23p3

16

S

3s23p4

17

Cl

3s23p5

18

Ar

3s23p6

19

K

4s1

20

Ca

4s2

31

Ga

4s24p1

32

Ge

4s24p2

33

As

4s24p3

34

Se

4s24p4

35

Br

4s24p5

36

Kr

4s24p6

Group Number and Valence Electrons The chemical properties of representative elements are mostly due to the valence electrons, which are the electrons in the outermost energy level. These valence electrons occupy the s and p sublevels with the highest principal quantum number n. The group numbers indicate the number of valence (outer) electrons for the elements in each vertical column. For example, the elements in Group 1A (1), such as lithium, sodium, and potassium, all have one electron in an s orbital. Looking at the sublevel block, we can represent the valence electron in the alkali metals of Group 1A (1) as ns1. All the elements in Group 2A (2), the alkaline earth metals, have two valence electrons, ns2. The halogens in Group 7A (17) have seven valence electrons, ns2np5.

We can see the repetition of the outermost s and p electrons for the representative ele- ments for Periods 1 to 4 in TABLE 5.3. Helium is included in Group 8A (18) because it is a noble gas, but it has only two electrons in its complete energy level.

ENGAGE 5.14 What do the group numbers indicate about the electron configurations of the elements in Groups 1A (1) to 8A (18)?

PRACTICE PROBLEMS Try Practice Problems 5.51 to 5.58

CORE CHEMISTRY SKILL Identifying Trends in Periodic

Properties

TABLE 5.3 Valence Electron Configuration for Representative Elements in Periods 1 to 4

SAMPLE PROBLEM 5.8 Using Group Numbers

TRY IT FIRST

Using the periodic table, write the period, the group number, the number of valence electrons, and the valence electron configuration for each of the following:

a. calcium b. iodine c. lead

SOLUTION

The valence electrons are the outermost s and p electrons. Although there may be electrons in the d or f sublevels, they are not valence electrons.

a. Calcium is in Period 4, Group 2A (2), has two valence electrons, and has a valence electron configuration of 4s2.

b. Iodine is in Period 5, Group 7A (17), has seven valence electrons, and has a valence electron configuration of 5s25p5.

c. Lead is in Period 6, Group 4A (14), has four valence electrons, and has a valence electron configuration of 6s26p2.

SELF TEST 5.8

Using the periodic table, write the period, the group number, the number of valence electrons, and the valence electron configuration for each of the following:

a. sulfur b. strontium

ANSWER

a. Sulfur is in Period 3, Group 6A (16), has six valence electrons, and a 3s23p4 valence electron configuration.

b. Strontium is in Period 5, Group 2A (2), has two valence electrons, and a 5s2 valence electron configuration.

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5.6 Trends in Periodic Properties 145

Atomic Size The atomic size of an atom is determined by the distance of the valence electrons from the nucleus. For each group of representative elements, the atomic size increases going from the top to the bottom because the outermost electrons in each energy level are farther from the nucleus. For example, in Group 1A (1), Li has a valence electron in energy level 2; Na has a valence electron in energy level 3; and K has a valence electron in energy level 4. This means that a K atom is larger than a Na atom and a Na atom is larger than a Li atom (see FIGURE 5.12).

ENGAGE 5.15 Why is a phosphorus atom larger than a nitrogen atom but smaller than a silicon atom?

Atomic Size Decreases

A to

m ic

S iz

e In

cr ea

se s

1

2

3

4

5

P er

io d

1A (1)

2A (2)

3A (13)

4A (14)

5A (15)

6A (16)

7A (17)

8A (18)

3B (3)

4B (4)

5B (5)

6B (6)

7B (7) (9) (10)

1B (11)

2B (12)(8)

Group

H

Li Be

Na Mg

K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn

He

B C N O F Ne

Al Si P S Cl Ar

Ga Ge As Se Br Kr

Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe

8B

FIGURE 5.12 The atomic size increases going down a group but decreases going from left to right across a period.

SAMPLE PROBLEM 5.9 Sizes of Atoms

TRY IT FIRST Identify the smaller atom in each of the following pairs:

a. N or F b. K or Kr c. Ca or Sr

SOLUTION

a. The F atom has a greater positive charge on the nucleus, which pulls electrons closer, and makes the F atom smaller than the N atom. Atomic size decreases going from left to right across a period.

b. The Kr atom has a greater positive charge on the nucleus, which pulls electrons closer, and makes the Kr atom smaller than the K atom. Atomic size decreases going from left to right across a period.

c. The outer electrons in the Ca atom are closer to the nucleus than in the Sr atom, which makes the Ca atom smaller than the Sr atom. Atomic size increases going down a group.

The atomic size of representative elements is affected by the attractive forces of the protons in the nucleus on the electrons in the outermost level. For the elements going across a period, the increase in the number of protons in the nucleus increases the positive charge of the nucleus. As a result, the electrons are pulled closer to the nucleus, which means that the atomic size of representative elements decreases going from left to right across a period.

The size of atoms of transition elements within the same period changes only slightly because electrons are filling d orbitals rather than the outermost energy level. Because the increase in nuclear charge is canceled by an increase in d electrons, the attraction of the valence electrons by the nucleus remains about the same. Because there is little change in the nuclear attraction for the valence electrons, the atomic size remains relatively constant for the transition elements.

M05_TIMB8119_06_SE_C05.indd 145 11/29/18 9:26 AM

146 CHAPTER 5 Electronic Structure of Atoms and Periodic Trends

Ionization Energy In an atom, negatively charged electrons are attracted to the positive charge of the protons in the nucleus. Therefore, energy is required to remove an electron from an atom. The ionization energy is the energy needed to remove one electron from an atom in the gaseous (g) state. When an electron is removed from a neutral atom, a cation with a 1 + charge is formed.

ENGAGE 5.17 Why would Cs have a lower ionization energy than K?

2p2s1s

Na (g) + energy (ionization) Na+(g) + e-

3s 2p2s1s 3s

+

The attraction of a nucleus for the outermost electrons decreases as those electrons are farther from the nucleus. Thus the ionization energy decreases going down a group (see FIGURE 5.13). However, going across a period from left to right, the positive charge of the nucleus increases because there is an increase in the number of protons. Thus the ionization energy increases going from left to right across the periodic table.

In summary, the ionization energy is low for the metals and high for the nonmetals. The high ionization energies of the noble gases indicate that their electron configurations are especially stable.

PRACTICE PROBLEMS Try Practice Problems 5.63 to 5.66

FIGURE 5.13 As the distance from the nucleus to the valence electron in Li, Na, and K atoms increases in Group 1A (1), the ionization energy decreases and less energy is required to remove the valence electron.

Io ni

za ti

on E

ne rg

y D

ec re

as es

Na atom

Li atom

K atom

Distance between the nucleus and valence electron

ENGAGE 5.16 When a magnesium atom ionizes to form a magnesium ion Mg2+, which electrons are lost?

SAMPLE PROBLEM 5.10 Ionization Energy

TRY IT FIRST Indicate the element in each set that has the higher ionization energy and explain your choice.

a. S or Se b. Mg or Cl c. F, N, or C

SOLUTION

a. S. In S, an electron is removed from a sublevel closer to the nucleus, which requires a higher ionization energy for S compared with Se.

b. Cl. The increased nuclear charge of Cl increases the attraction for the valence electrons, which requires a higher ionization energy for Cl compared to Mg.

c. F. The increased nuclear charge of F increases the attraction for the valence electrons, which requires a higher ionization energy for F compared to C or N.

SELF TEST 5.10

a. Arrange Sr, I, and Sn in order of increasing ionization energy. b. Arrange Ba, Mg, and Ca in order of decreasing ionization energy.

ANSWER

a. Sr, Sn, I. Ionization energy increases going from left to right across a period. b. Ionization energy decreases going down a group: Mg, Ca, Ba.

Ionization Energy Increases

Io ni

za ti

on E

ne rg

y D

ec re

as es

Ionization energy decreases going down a group and increases going from left to right across a period.

SELF TEST 5.9

Identify the largest atom in each of the following: a. P, As, or Se b. Mg, Si, or S

ANSWER

a. As b. Mg PRACTICE PROBLEMS

Try Practice Problems 5.59 to 5.62

M05_TIMB8119_06_SE_C05.indd 146 11/29/18 9:26 AM

5.6 Trends in Periodic Properties 147

Metallic Character An element that has metallic character is an element that loses valence electrons easily. Metallic character is more prevalent in the elements on the left side of the periodic table (metals) and decreases going from left to right across a period. The elements on the right side of the periodic table (nonmetals) do not easily lose electrons, which means they are less metallic. Most of the metalloids between the metals and nonmetals tend to lose electrons, but not as easily as the metals. Thus, in Period 3, sodium, which loses electrons most easily, would be the most metallic. Going across from left to right in Period 3, metallic character decreases to argon, which has the least metallic character.

For elements in the same group of representative elements, metallic character increases going from top to bottom. Atoms at the bottom of any group have more electron levels, which makes it easier to lose electrons. Thus, the elements at the bottom of a group on the periodic table have lower ionization energy and are more metallic compared to the elements at the top.

A summary of the trends in periodic properties we have discussed is given in TABLE 5.4. ENGAGE 5.18

Why is magnesium more metallic than aluminum?

PRACTICE PROBLEMS Try Practice Problems 5.67 to 5.74

TABLE 5.4 Summary of Trends in Periodic Properties of Representative Elements

Periodic Property Top to Bottom Within a Group Left to Right Across a Period

Valence Electrons Remains the same Increases

Atomic Size Increases because there is an increase in the number of energy levels

Decreases as the number of protons increases, which strengthens the attraction of the nucleus for the valence electrons, and pulls them closer to the nucleus

Ionization Energy Decreases because the valence electrons are easier to remove when they are farther from the nucleus

Increases as the number of protons increases, which strengthens the attraction of the nucleus for the valence electrons, and more energy is needed to remove a valence electron

Metallic Character Increases because the valence electrons are easier to remove when they are farther from the nucleus

Decreases as the number of protons increases, which strengthens the attraction of the nucleus for the valence electrons, and makes it more difficult to remove a valence electron

PRACTICE PROBLEMS

5.6 Trends in Periodic Properties

5.51 What do the group numbers from 1A (1) to 8A (18) for the elements indicate about electron configurations of those elements?

5.52 What is similar and what is different about the valence electrons of the elements in a group?

5.53 Write the group number using both A/B and 1 to 18 notations for elements that have the following outer electron configuration: a. 2s2 b. 3s23p3 c. 4s23d5 d. 5s24d105p4

5.54 Write the group number using both A/B and 1 to 18 notations for elements that have the following outer electron configuration:

a. 4s23d104p5 b. 4s1 c. 4s23d8 d. 5s24d105p2

5.55 Write the valence electron configuration for each of the following: a. alkali metals b. Group 4A (14) c. Group 7A (17) d. Group 5A (15)

5.56 Write the valence electron configuration for each of the following: a. halogens b. Group 6A (16) c. Group 13 d. alkaline earth metals

5.57 Indicate the number of valence electrons in each of the following: a. aluminum b. Group 5A (15) c. barium d. F, Cl, Br, and I

5.58 Indicate the number of valence electrons in each of the following: a. Li, Na, K, Rb, and Cs b. Se c. C, Si, Ge, Sn, and Pb d. Group 8A (18)

5.59 Select the larger atom in each pair. a. Na or Cl b. Na or Rb c. Na or Mg d. Rb or I

5.60 Select the larger atom in each pair. a. S or Ar b. S or O c. S or K d. S or Mg

Metallic Character Decreases

M et

al li

c C

ha ra

ct er

I nc

re as

es

Metallic character increases going down a group and decreases going from left to right across a period.

M05_TIMB8119_06_SE_C05.indd 147 11/29/18 9:26 AM

148 CHAPTER 5 Electronic Structure of Atoms and Periodic Trends

5.61 Place the elements in each set in order of decreasing atomic size. a. Al, Si, Mg b. Cl, I, Br c. Sb, Sr, I d. P, Si, Na

5.62 Place the elements in each set in order of decreasing atomic size. a. Cl, S, P b. Ge, Si, C c. Ba, Ca, Sr d. S, O, Se

5.63 Select the element in each pair with the higher ionization energy. a. Br or I b. Mg or Sr c. Si or P d. I or Xe

5.64 Select the element in each pair with the higher ionization energy. a. O or Ne b. K or Br c. Ca or Ba d. N or Ne

5.65 Arrange each set of elements in order of increasing ionization energy.

a. F, Cl, Br b. Na, Cl, Al c. Na, K, Cs d. As, Ca, Br

5.66 Arrange each set of elements in order of increasing ionization energy.

a. O, N, C b. S, P, Cl c. P, As, N d. Al, Si, P

5.67 Place the following in order of decreasing metallic character:

Br, Ge, Ca, Ga

5.68 Place the following in order of increasing metallic character:

Na, P, Al, Ar

5.69 Fill in each of the following blanks using higher or lower, more or less: Sr has a ______________ ionization energy and is ______________ metallic than Sb.

5.70 Fill in each of the following blanks using higher or lower, more or less: N has a _____________ ionization energy and is _____________ metallic than As.

5.71 Complete each of the following statements a to d using 1, 2, or 3: 1. decreases 2. increases 3. remains the same

Going down Group 6A (16), a. the ionization energy ____________ b. the atomic size ____________ c. the metallic character ___________ d. the number of valence electrons ___________

5.72 Complete each of the following statements a to d using 1, 2, or 3: 1. decreases 2. increases 3. remains the same

Going from left to right across Period 3, a. the ionization energy ____________ b. the atomic size ____________ c. the metallic character ____________ d. the number of valence electrons ____________

5.73 Which statements completed with a to e will be true and which will be false?

An atom of N compared to an atom of Li has a larger (greater) a. atomic size b. ionization energy c. number of protons d. metallic character e. number of valence electrons

5.74 Which statements completed with a to e will be true and which will be false?

An atom of C compared to an atom of Sn has a larger (greater) a. atomic size b. ionization energy c. number of protons d. metallic character e. number of valence electrons

UPDATE Developing New Materials for Computer Chips

As part of a new research project, Jennifer and Robert are assigned to investigate the properties of indium (In) and tellurium (Te). Both these elements are used to make com- puter chips. As a result of their research, Jennifer and

Robert hope to develop new computer chips that will be faster and less expensive to manufacture.

5.75 a. What is the atomic number of In? b. How many electrons are in an atom of In? c. Use the sublevel blocks on the periodic table to write

the electron configuration and abbreviated electron configuration for In.

d. Which is larger, an atom of indium or an atom of iodine? e. Which has a higher ionization energy, an atom of indium

or an atom of iodine?

5.76 a. What is the atomic number of Te? b. How many electrons are in an atom of Te? c. Use the sublevel blocks on the periodic table to write

the electron configuration and abbreviated electron configuration for Te.

d. Which is smaller, an atom of selenium or an atom of tellurium?

e. Which has a lower ionization energy, an atom of selenium or an atom of tellurium?

M05_TIMB8119_06_SE_C05.indd 148 11/29/18 9:26 AM

z

y

x

CONCEPT MAP

ELECTRONIC STRUCTURE OF ATOMS AND PERIODIC TRENDS

Elements

PeriodsGroups

Atomic Number

Sublevels Orbitals

ElectronsProtons

by

that determine

Metallic Character

Atomic Size

Ionization Energy

Energy Levels

Valence Electrons

Chemical Symbols

Periodic Table

Periodic Trends

Group Number

Electron Configurations

Orbital Diagrams

such as

in

have containing

shown as that determine

arranged in the

that have

havemake up have

CHAPTER REVIEW

5.1 Electromagnetic Radiation LEARNING GOAL Compare the wavelength, frequency, and energy of electromagnetic radiation. • Electromagnetic radiation

such as radio waves and vis- ible light is energy that travels at the speed of light.

• Each particular type of radiation has a specific wavelength and frequency.

• A wavelength (symbol l, lambda) is the distance between a crest or trough in a wave and the next crest or trough on that wave.

• The frequency (symbol n, nu) is the number of waves that pass a certain point in 1 s.

• All electromagnetic radiation travels at the speed of light (c), which is 3.00 * 108 m/s.

• Mathematically, the relationship of the speed of light, wavelength, and frequency is expressed as c = ln.

• Long-wavelength radiation has low frequencies, while short- wavelength radiation has high frequencies.

• Radiation with a high frequency has high energy.

5.2 Atomic Spectra and Energy Levels LEARNING GOAL Explain how atomic spectra correlate with the energy levels in atoms. • The atomic spectra of elements

are related to the specific energy levels occupied by electrons.

• Light consists of photons, which are particles of a specific energy.

• When an electron absorbs a photon of a particular energy, it attains a higher energy level. When an electron drops to a lower energy level, a photon of a particular energy is emitted.

• Each element has its own unique atomic spectrum.

5.3 Sublevels and Orbitals LEARNING GOAL Describe the sublevels and orbitals for the electrons in an atom. • An orbital is a region around the

nucleus where an electron with a specific energy is most likely to be found.

Chapter Review 149

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150 CHAPTER 5 Electronic Structure of Atoms and Periodic Trends

atomic size The distance between the outermost electrons and the nucleus.

atomic spectrum A series of lines specific for each element produced by photons emitted by electrons dropping to lower energy levels.

d block The 10 elements in Groups 3B (3) to 2B (12) in which electrons fill the five d orbitals.

electromagnetic radiation Forms of energy such as visible light, microwaves, radio waves, infrared, ultraviolet light, and X-rays that travel as waves at the speed of light.

electromagnetic spectrum The arrangement of types of radiation from long wavelengths to short wavelengths.

electron configuration A list of the number of electrons in each sublevel within an atom, arranged by increasing energy.

energy level A group of electrons with similar energy. f block The 14 elements in the rows at the bottom of the periodic

table in which electrons fill the seven 4f and 5f orbitals. frequency The number of times the crests of a wave pass a point in 1 s. ionization energy The energy needed to remove the least tightly

bound electron from the outermost energy level of an atom. metallic character A measure of how easily an element loses a

valence electron.

orbital The region around the nucleus of an atom where electrons of certain energy are most likely to be found: s orbitals are spherical; p orbitals have two lobes.

orbital diagram A diagram that shows the distribution of electrons in the orbitals of the energy levels.

p block The elements in Groups 3A (13) to 8A (18) in which electrons fill the p orbitals.

photon A packet of energy that has both particle and wave characteristics and travels at the speed of light.

principal quantum number (n) The number (n = 1, n = 2, c) assigned to an energy level.

s block The elements in Groups 1A (1) and 2A (2) in which electrons fill the s orbitals.

sublevel A group of orbitals of equal energy within energy levels. The number of sublevels in each energy level is the same as the principal quantum number (n).

valence electrons The electrons in the highest energy level of an atom.

wavelength The distance between adjacent crests or troughs in a wave.

KEY TERMS

• Each orbital holds a maximum of two electrons, which must have opposite spins.

• In each energy level (n), electrons occupy orbitals of identical energy within sublevels.

• An s sublevel contains one s orbital, a p sublevel contains three p orbitals, a d sublevel contains five d orbitals, and an f sublevel contains seven f orbitals.

• Each type of orbital has a unique shape.

5.4 Orbital Diagrams and Electron Configurations LEARNING GOAL Draw the orbital diagram and write the electron configuration for an element. • Within a sublevel, electrons enter orbitals in the same

energy level one at a time until all the orbitals are half-filled.

• Additional electrons enter with opposite spins until the orbitals in that sublevel are filled with two electrons each.

• The orbital diagram for an element such as silicon shows the orbitals that are occupied by paired and unpaired electrons:

5.6 Trends in Periodic Properties LEARNING GOAL Use the electron configurations of elements to explain the trends in periodic properties. • The properties of elements are

related to the valence electrons of the atoms.

• With only a few exceptions, each group of elements has the same arrangement of valence electrons differing only in the energy level.

• The size of an atom increases going down a group and decreases going from left to right across a period.

• The energy required to remove a valence electron is the ionization energy, which decreases going down a group, and increases going from left to right across a period.

• The metallic character of an element increases going down a group and decreases going from left to right across a period.

1s1

1s2

1s

2p1s 2s 3p3s

• The electron configuration for an element such as silicon shows the number of electrons in each sublevel: 1s22s22p63s23p2.

• An abbreviated electron configuration for an element such as silicon places the symbol of a noble gas in brackets to represent the filled sublevels: [Ne]3s23p2.

5.5 Electron Configurations and the Periodic Table LEARNING GOAL Write the electron configuration for an atom using the sublevel blocks on the periodic table. • The periodic table consists of s, p, d, and f sublevel blocks. • An electron configuration can be written following the order of the

sublevel blocks on the periodic table.

Na atom

Li atom

Distance between the nucleus and valence electron

K atom

Period Number

2

1

3

4

5

6

7

2s

1s

3s

4s

5s

6s

7s

3d

4d

5d

4f

5f

6d

2p

3p

4p

5p

6p

7p

s Block

d Block

p Block

f Block

• Beginning with 1s, an electron configuration is obtained by writing the sublevel blocks in order going across each period on the periodic table until the element is reached.

M05_TIMB8119_06_SE_C05.indd 150 11/29/18 9:26 AM

The chapter section containing each Core Chemistry Skill is shown in parentheses at the end of each heading.

Writing Electron Configurations (5.4) • The electron configuration for an atom specifies the energy levels

and sublevels occupied by the electrons of an atom. • An electron configuration is written starting with the lowest energy

sublevel, followed by the next lowest energy sublevel. • The number of electrons in each sublevel is shown as a superscript.

Example: Write the electron configuration for palladium. Answer: Palladium has atomic number 46, which means it has

46 protons and 46 electrons.

1s22s22p63s23p64s23d104p65s24d8

Using the Periodic Table to Write Electron Configurations (5.5) • An electron configuration corresponds to the location of an ele-

ment on the periodic table, where different blocks within the peri- odic table are identified as the s, p, d, and f sublevels.

Example: Use the periodic table to write the electron configuration for sulfur.

Answer: Sulfur (atomic number 16) is in Group 6A (16) and Period 3.

Period Sublevel Block Filling Sublevel Block Notation

1 1s (H S He) 1s2

2 2s (Li S Be) and 2p (B S Ne) 2s22p6

3 3s (Na S Mg) 3s2

3p (Al S S) 3p4

The electron configuration for sulfur (S) is: 1s22s22p63s23p4.

CORE CHEMISTRY SKILLS

Identifying Trends in Periodic Properties (5.6) • The size of an atom increases going down a group and decreases

going from left to right across a period. • The ionization energy decreases going down a group and increases

going from left to right across a period. • The metallic character of an element increases going down a group

and decreases going from left to right across a period.

Example: For Mg, P, and Cl, identify which has the a. largest atomic size b. highest ionization energy c. greatest metallic character

Answer: a. Mg b. Cl c. Mg

UNDERSTANDING THE CONCEPTS

The chapter sections to review are shown in parentheses at the end of each problem.

Use the following diagram for problems 5.77 and 5.78:

5.78 Select diagram A, B, or C that (5.1) a. has the highest energy b. has the lowest energy c. would represent blue light d. would represent red lightA.

B.

C.

5.77 Select diagram A, B, or C that (5.1) a. has the longest wavelength b. has the shortest wavelength c. has the highest frequency d. has the lowest frequency

b. c.

5.80 Match the following with s or p orbitals: (5.3) a. two lobes b. spherical shape c. found in n = 1 d. found in n = 3

Understanding the Concepts 151

5.79 Match the following with an s or p orbital: (5.3)

a.

M05_TIMB8119_06_SE_C05.indd 151 11/29/18 9:26 AM

152 CHAPTER 5 Electronic Structure of Atoms and Periodic Trends

A B C D

A B C D

5.81 Indicate whether or not each of the following orbital diagrams is possible and explain. When possible, indicate the element it represents. (5.4) a.

b. 2p 3s1s 2s

2p2s

[He]

5.82 Indicate whether or not each of the following abbreviated orbital diagrams is possible and explain. When possible, indi- cate the element it represents. (5.4) a.

b.

2p2s1s 3s

[Ar]

3d4s

5.83 Match the spheres A through D with atoms of Li, Na, K, and Rb. (5.6)

5.84 Match the spheres A through D with atoms of K, Ge, Ca, and Kr. (5.6)

5.85 What is the difference between a continuous spectrum and an atomic spectrum? (5.1)

5.86 Why does a neon sign give off red light? (5.1)

5.87 What is the Pauli exclusion principle? (5.3)

5.88 Why would there be five unpaired electrons in a d sublevel but no paired electrons? (5.3)

5.89 Which of the following orbitals are possible in an atom: 4p, 2d, 3f, and 5f ? (5.3)

5.90 Which of the following orbitals are possible in an atom: 1p, 4f, 6s, and 4d ? (5.3)

5.91 a. What electron sublevel starts to fill after completion of the 3s sublevel? (5.4)

b. What electron sublevel starts to fill after completion of the 4p sublevel?

c. What electron sublevel starts to fill after completion of the 3d sublevel?

d. What electron sublevel starts to fill after completion of the 3p sublevel?

5.92 a. What electron sublevel starts to fill after completion of the 5s sublevel? (5.4)

b. What electron sublevel starts to fill after completion of the 4d sublevel?

c. What electron sublevel starts to fill after completion of the 4f sublevel?

d. What electron sublevel starts to fill after completion of the 5p sublevel?

5.93 a. How many 3d electrons are in Fe? (5.4) b. How many 5p electrons are in Ba? c. How many 4d electrons are in I? d. How many 7s electrons are in Ra?

5.94 a. How many 4d electrons are in Cd? (5.4) b. How many 4p electrons are in Br? c. How many 6p electrons are in Bi? d. How many 4s electrons are in Zn?

5.95 Write the abbreviated electron configuration and group number for each of the following elements: (5.4)

a. Si b. Se c. Mn d. Sb

5.96 Write the abbreviated electron configuration and group number for each of the following elements: (5.4)

a. Br b. Rh c. Tc d. Ra

5.97 What do the elements Ca, Sr, and Ba have in common in terms of their electron configuration? Where are they located on the periodic table? (5.4, 5.5)

5.98 What do the elements O, S, and Se have in common in terms of their electron configuration? Where are they located on the periodic table? (5.4, 5.5)

5.99 Name the element that corresponds to each of the following: (5.4, 5.5, 5.6)

a. 1s22s22p63s23p3

b. alkali metal with the smallest atomic size c. [Kr]5s24d10

d. Group 5A (15) element with the highest ionization energy e. Period 3 element with the largest atomic size

5.100 Name the element that corresponds to each of the following: (5.4, 5.5, 5.6)

a. 1s22s22p63s23p64s13d5

b. [Xe]6s24f 145d106p5

c. halogen with the highest ionization energy d. Group 2A (2) element with the lowest ionization energy e. Period 4 element with the smallest atomic size

5.101 Why is the ionization energy of Ca higher than that of K but lower than that of Mg? (5.6)

5.102 Why is the ionization energy of Br lower than that of Cl but higher than that of Se? (5.6)

5.103 Select the more metallic element in each pair. (5.6) a. As or Sb b. Sn or Sb c. Cl or P d. O or P

ADDITIONAL PRACTICE PROBLEMS

M05_TIMB8119_06_SE_C05.indd 152 11/29/18 9:26 AM

Answers to Engage Questions 153

5.104 Select the more metallic element in each pair. (5.6) a. Sn or As b. Cl or I c. Ca or Ba d. Ba or Hg

5.105 Of the elements Na, P, Cl, and F, which (5.6) a. is a metal? b. is in Group 5A (15)? c. has the highest ionization energy? d. loses an electron most easily? e. is found in Group 7A (17), Period 3?

5.106 Of the elements K, Ca, Br, and Kr, which (5.6) a. is a halogen? b. has the smallest atomic size? c. has the lowest ionization energy? d. requires the most energy to remove an electron? e. is found in Group 2A (2), Period 4?

5.107 Consider three elements with the following abbreviated electron configurations: (5.4, 5.5, 5.6) X = [Ar]4s2 Y = [Ne]3s23p4 Z = [Ar]4s23d104p4

a. Identify each element as a metal, nonmetal, or metalloid. b. Which element has the largest atomic size? c. Which element has the highest ionization energy? d. Which element has the smallest atomic size?

5.108 Consider three elements with the following abbreviated electron configurations: (5.4, 5.5, 5.6) X = [Ar]4s23d5 Y = [Ar]4s23d104p1 Z = [Ar]4s23d104p6

a. Identify each element as a metal, nonmetal, or metalloid. b. Which element has the smallest atomic size? c. Which element has the highest ionization energy? d. Which element has a half-filled sublevel?

The following problems are related to the topics in this chapter. However, they do not all follow the chapter order, and they require you to combine concepts and skills from several sections. These problems will help you increase your critical thinking skills and prepare for your next exam.

5.109 How do scientists explain the colored lines observed in the spectra of heated atoms? (5.2)

5.110 Even though H has only one electron, there are many lines in the atomic spectrum of H. Explain. (5.2)

5.111 What is meant by an energy level, a sublevel, and an orbital? (5.3)

5.112 In some periodic tables, H is placed in Group 1A (1). In other periodic tables, H is also placed in Group 7A (17). Why? (5.4, 5.5)

5.113 Compare F, S, and Cl in terms of atomic size and ionization energy. (5.6)

5.114 Compare K, Mg, and Ca in terms of atomic size and ionization energy. (5.6)

5.115 Give the symbol of the element that has the (5.6) a. smallest atomic size in Group 6A (16) b. smallest atomic size in Period 3 c. highest ionization energy in Group 3A (13) d. lowest ionization energy in Period 3 e. abbreviated electron configuration [Kr]5s24d6

5.116 Give the symbol of the element that has the (5.6) a. largest atomic size in Period 5 b. largest atomic size in Group 2A (2) c. highest ionization energy in Group 8A (18) d. lowest ionization energy in Period 2 e. abbreviated electron configuration [Kr]5s24d105p2

CHALLENGE PROBLEMS

ANSWERS TO ENGAGE QUESTIONS 5.11 In neon, the 1s block and the 2s block each have two electrons.

Neon is located at the end of Period 2; thus, neon has six electrons in the 2p sublevel.

5.12 The s block contains elements with group numbers 1 (1A) and 2 (2A), the p block contains elements with group numbers 13 (3A) through 18 (8A), and the d block contains elements with group numbers 3 (3B) through 12 (2B).

5.13 For elements K to Zn, the sublevel blocks used are 4s and 3d.

5.14 The groups 1A to 8A indicate the number of s and p valence electrons for the elements in those groups.

5.15 Atomic size increases going from the top to the bottom of each group, which makes a phosphorus atom larger than a nitrogen atom. Atomic size also decreases going from left to right across a period as the number of protons in the nucleus increases. Thus, a phosphorus atom with more protons is smaller than a silicon atom.

5.16 In Cs, the outermost electron is farther from the nucleus, which means less energy is required to remove that electron.

5.17 The ionization of Mg to Mg2+ occurs with the loss of the two outermost electrons in the 3s sublevel.

5.18 The metallic character decreases going from left to right within a period. Thus, magnesium is more metallic than aluminum.

5.1 The wavelength (distance from crest to crest) of red light is longer than that of blue light.

5.2 10 200 km * 1000 m/1 km * 1 s/3.00 * 108 m = 0.034 s 5.3 Infrared radiation has a lower frequency than ultraviolet light.

5.4 When electrons in a heated element absorb energy, they move to higher energy levels, then emit that energy by falling to lower energy levels. Only specific energies are emitted, not continuous energies.

5.5 Electrons absorb a specific amount of energy to move between their current energy level and a higher energy level.

5.6 Energy level n = 5 has five sublevels. 5.7 The p orbitals in the n = 3 energy level have the same shape

and energy. However, they are oriented in different directions along the x, y, and z axes.

5.8 There are 5 orbitals in the 5d sublevel.

5.9 Because the 4s sublevel has a lower energy than the 3d sublevel, the 4s energy level fills before the 3d sublevel.

5.10 Hund’s rule states there is less repulsion between electrons when they occupy orbitals of equal energy singly with the same spin. Thus, in an electron configuration, electrons are added singly to the orbitals in a sublevel, and then additional electrons for the same sublevel double up.

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154 CHAPTER 5 Electronic Structure of Atoms and Periodic Trends

1s 2s 2p

3s2s1s 3p2p

3s2p2s1s 3p

3s 3p2p2s1s

ANSWERS TO SELECTED PROBLEMS 5.39 a. Al b. C c. Ar d. Be

5.41 a. 1s22s22p63s23p64s23d104p3 b. 1s22s22p63s23p64s23d6

c. 1s22s22p63s23p64s23d104p65s24d105p2

d. 1s22s22p63s23p64s23d104p6

5.43 a. [Ar]4s23d2 b. [Ar]4s23d104p5

c. [Xe]6s2 d. [Xe]6s24f 145d106p2

5.45 a. P b. Co c. Zn d. Bi

5.47 a. Ga b. N c. Xe d. Zr

5.49 a. 10 b. 6 c. 3 d. 1

5.51 The group numbers 1A to 8A indicate the number of valence electrons from 1 to 8.

5.53 a. 2A (2) b. 5A (15) c. 7B (7) d. 6A (16)

5.55 a. ns1 b. ns2np2 c. ns2np5 d. ns2np3

5.57 a. 3 b. 5 c. 2 d. 7

5.59 a. Na b. Rb c. Na d. Rb

5.61 a. Mg, Al, Si b. I, Br, Cl c. Sr, Sb, I d. Na, Si, P

5.63 a. Br b. Mg c. P d. Xe

5.65 a. Br, Cl, F b. Na, Al, Cl c. Cs, K, Na d. Ca, As, Br

5.67 Ca, Ga, Ge, Br

5.69 lower, more

5.71 a. 1. decreases b. 2. increases c. 2. increases d. 3. remains the same 5.73 a. false b. true c. true d. false e. true

5.75 a. 49 b. 49 c. 1s22s22p63s23p64s23d104p65s24d105p1; [Kr]5s24d105p1

d. indium e. iodine

5.77 a. C has the longest wavelength. b. A has the shortest wavelength. c. A has the highest frequency. d. C has the lowest frequency.

5.79 a. p b. s c. p

5.81 a. This is possible. This element is magnesium. b. Not possible. The 2p sublevel would fill before the 3s, and

only two electrons are allowed in an s orbital.

5.83 Li is D, Na is A, K is C, and Rb is B.

5.85 A continuous spectrum from white light contains wavelengths of all energies. Atomic spectra are line spectra in which a series of lines corresponds to energy emitted when electrons drop from a higher energy level to a lower level.

5.87 The Pauli exclusion principle states that two electrons in the same orbital must have opposite spins.

5.89 A 4p orbital is possible because the n = 4 energy level has four sublevels, including a p sublevel. A 2d orbital is not possible because the n = 2 energy level has only s and p sublevels. There are no 3f orbitals because only s, p, and d sublevels are allowed for the n = 3 energy level. A 5f sublevel is possible in the n = 5 energy level because five sublevels are allowed, including an f sublevel.

5.1 The wavelength of UV light is the distance between crests of the wave.

5.3 White light has all the colors of the spectrum, including red and blue light.

5.5 Ultraviolet radiation has a higher frequency and higher energy.

5.7 6.3 * 10-7 m; 630 nm 5.9 AM radio has a longer wavelength than cell phones or infrared.

5.11 Order of increasing wavelengths: X-rays, blue light, microwaves

5.13 Order of increasing frequencies: TV, microwaves, X-rays

5.15 Atomic spectra consist of a series of lines separated by dark sections, indicating that the energy emitted by the elements is not continuous.

5.17 absorb

5.19 a. green light b. blue light

5.21 a. spherical b. two lobes c. spherical

5.23 a. 1 and 2 b. 3 c. 1 and 2 d. 1, 2, and 3

5.25 a. There are five orbitals in the 3d sublevel. b. There is one sublevel in the n = 1 energy level. c. There is one orbital in the 6s sublevel. d. There are nine orbitals in the n = 3 energy level. 5.27 a. There is a maximum of two electrons in a 2p orbital. b. There is a maximum of six electrons in the 3p sublevel. c. There is a maximum of 32 electrons in the n = 4 energy

level. d. There is a maximum of 10 electrons in the 5d sublevel.

5.29 The electron configuration shows the number of electrons in each sublevel of an atom. The abbreviated electron configuration uses the symbol of the preceding noble gas to show completed sublevels.

5.31 a.

b.

c.

d.

5.33 a. 1s22s22p63p64s23d8

b. 1s22s22p63s1

c. 1s22s1

d. 1s22s22p63s23p64s23d2

5.35 a. [Kr]5s24d105p2

b. [Kr]5s24d10

c. [Ar]4s23d104p4

d. [He]2s22p5

5.37 a. Li b. Ti c. Ge d. Al

M05_TIMB8119_06_SE_C05.indd 154 11/29/18 9:26 AM

Answers to Selected Problems 155

5.91 a. 3p b. 5s c. 4p d. 4s

5.93 a. 6 b. 6 c. 10 d. 2

5.95 a. [Ne]3s23p2; Group 4A (14) b. [Ar]4s23d104p4; Group 6A (16) c. [Ar]4s23d5; Group 7B (7) d. [Kr]5s24d105p3; Group 5A (15)

5.97 Ca, Sr, and Ba all have two valence electrons, ns2, which place them in Group 2A (2).

5.99 a. phosphorus b. lithium (H is a nonmetal) c. cadmium d. nitrogen e. sodium

5.101 Calcium has a greater number of protons than K. The least tightly bound electron in Ca is farther from the nucleus than in Mg and needs less energy to remove.

5.103 a. Sb b. Sn c. P d. P

5.105 a. Na b. P c. F d. Na e. Cl

5.107 a. X is a metal; Y and Z are nonmetals. b. X has the largest atomic size. c. Y has the highest ionization energy. d. Y has the smallest atomic size.

5.109 The series of lines separated by dark sections in atomic spectra indicate that the energy emitted by the elements is not continuous and that electrons are moving between discrete energy levels.

5.111 The energy level contains all the electrons with similar energy. A sublevel contains electrons with the same energy, while an orbital is the region around the nucleus where electrons of a certain energy are most likely to be found.

5.113 S has a larger atomic size than Cl; Cl is larger than F: S 7 Cl 7 F. F has a higher ionization energy than Cl; Cl has a higher ionization energy than S: F 7 Cl 7 S.

5.115 a. O b. Ar c. B d. Na e. Ru

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156

Richard’s doctor recommends that Richard take a low-dose aspirin (81 mg) every day to decrease the chance of a heart attack or stroke. Richard is concerned about taking aspirin and asks Dr. Chavez, a pharmacist working at a local pharmacy, about the effects of aspirin. Dr. Chavez, explains that aspirin is acetylsalicylic acid and has the chemical formula C9H8O4. Aspirin is a molecular compound, often referred to as an organic molecule because it contains the nonmetal carbon (C) combined with the nonmetals hydrogen (H) and oxygen (O). She explains that aspirin is used to relieve minor pains, to reduce inflammation and fever, and to slow blood clotting. Aspirin is one of several nonsteroidal anti-inflammatory drugs (NSAIDs) that reduce pain and fever by blocking the formation of prostaglandins, which are chemical messengers that transmit pain signals to the brain and cause fever. Some potential side effects of aspirin may include heartburn, upset stomach, nausea, and an increased risk of a stomach ulcer.

CAREER

Pharmacist Pharmacists work in hospitals, pharmacies, clinics, and long-term care facilities where they are responsible for the preparation and distribution of pharmaceutical medications based on a doctor’s orders. They obtain the proper medication and also calculate, measure, and label the patients’ medication. Pharmacists advise clients and health care practitioners on the selection of both prescription and over-the-counter drugs, proper dosages, and geriatric considerations, as well as possible side effects and interactions. They may also administer vaccinations; prepare sterile intravenous solutions; and advise clients about health, diet, and home medical equipment. Pharmacists also prepare insurance claims and create and maintain patient profiles.

Ionic and Molecular Compounds

Two weeks later, Richard returns to the pharmacy. Using his

pharmacist's recommendations, he purchases Epsom salts for a sore toe, an antacid for an upset stomach, and an iron supple- ment. You can see the chemical formulas of these medications in the UPDATE Compounds at the Pharmacy, page 176.

UPDATE Compounds at the Pharmacy

6

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6.1 Ions: Transfer of Electrons 157

6.1 Ions: Transfer of Electrons LEARNING GOAL Write the symbols for the simple ions of the representative elements.

Most of the elements, except the noble gases, are found in nature combined as compounds. The noble gases are so stable that they form compounds only under extreme conditions. One explanation for the stability of noble gases is that they have a filled valence electron energy level.

Compounds form when electrons are transferred or shared between atoms to give stable electron configurations to the atoms. In the formation of either an ionic bond or a covalent bond, atoms lose, gain, or share valence electrons to acquire an octet of eight valence electrons. This tendency of atoms to attain a stable electron configuration is known as the octet rule and provides a key to our understanding of the ways in which atoms bond and form compounds. A few elements achieve the stability of helium with two valence electrons. However, the octet rule is not used with transition elements.

Ionic bonds occur when the valence electrons of atoms of a metal are transferred to atoms of nonmetals. For example, sodium atoms lose electrons and chlorine atoms gain electrons to form the ionic compound NaCl. Covalent bonds form when atoms of nonmetals share valence electrons. In the molecular compounds H2O and C3H8, atoms share electrons (see TABLE 6.1).

REVIEW Using Positive and Negative

Numbers in Calculations (1.4)

Writing Electron Configurations (4.7)

NmM

Nm -

M+

Transfer of electrons

M is a metal Nm is a nonmetal

Nm Nm

Nm Nm

Sharing electrons

Covalent bondIonic bond

LOOKING AHEAD

6.1 Ions: Transfer of Electrons 157

6.2 Ionic Compounds 161 6.3 Naming and Writing

Ionic Formulas 164 6.4 Polyatomic Ions 168 6.5 Molecular Compounds:

Sharing Electrons 172

Positive Ions: Loss of Electrons In ionic bonding, ions, which have electrical charges, form when atoms lose or gain electrons to form a stable electron configuration. Because the ionization energies of metals of Groups 1A (1), 2A (2), and 3A (13) are low, metal atoms readily lose their valence electrons. In doing so, they form ions with positive charges. A metal atom obtains the same electron configuration as its nearest noble gas (usually eight valence electrons). For example, when a sodium atom loses its single valence electron, the remaining electrons have a stable configuration. By losing an electron, sodium has 10 negatively charged electrons instead of 11.

Loss of one valence

electron

Sodium atom Na

Name

Protons

Electrons

Electron Configuration

e-

11 p+ 11 p+

11 e- 10 e-

Sodium ion Na+

1s22s22p63s1 1s22s22p6

TABLE 6.1 Types of Particles and Bonds in Compounds Type Ionic Compounds Molecular Compounds

Particles Ions Molecules

Bonds Ionic Covalent

Examples Na+ Cl- ions H2O molecules C3H8 molecules

Na+ Cl- ions H2O molecules C3H8 moleculesNa

+ Cl- ions H2O molecules C3H8 moleculesNa + Cl- ions H2O molecules C3H8 molecules

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158 CHAPTER 6 Ionic and Molecular Compounds

Because there are still 11 positively charged protons in its nucleus, the sodium atom becomes a sodium ion, which is no longer neutral. It is now a sodium ion with a positive electrical charge (1 + ), called an ionic charge. The ionic charge of 1 + is written in the upper right corner Na+, where the 1 is understood. A metal ion is named by its element name. Thus, Na+ is named the sodium ion. The sodium ion is smaller than the sodium atom because the ion has lost its outermost electron from the third energy level. A positively charged ion of a metal is called a cation (pronounced cat-eye-un).

Ionic charge = Charge of protons + Charge of electrons 1 + = (11 + ) + (10 - ) Magnesium, a metal in Group 2A (2), obtains a stable electron configuration by

losing two valence electrons to form a magnesium ion with a 2 + ionic charge, Mg2+. The magnesium ion is smaller than the magnesium atom because the outermost electrons in the third energy level were removed. The octet in the magnesium ion is made up of electrons that fill its second energy level.

ENGAGE 6.1 A cesium ion has 55 protons and 54 electrons. Why does the cesium ion have a 1 + charge?

Gain of one valence

electron

Chlorine atom Cl e-

17 p+ 17 p+

17 e-

Chloride ion Cl-

1s22s22p63s23p5 1s22s22p63s23p6

18 e-

Name

Protons

Electrons

Electron Configuration

Loss of two valence

electrons

Magnesium atom

e-

12 p+ 12 p+

12 e- 10 e -

Magnesium ion

e-

1s22s22p63s2 1s22s22p6

Mg2+ Name

Protons

Electrons

Electron Configuration

Mg

Negative Ions: Gain of Electrons The ionization energy of a nonmetal atom in Groups 5A (15), 6A (16), or 7A (17) is high, which means that nonmetal atoms do not easily lose their valence electrons. Thus, a nonmetal atom gains one or more valence electrons to obtain a stable electron configuration. By gaining electrons, a nonmetal atom forms a negatively charged ion. For example, an atom of chlorine with seven valence electrons gains one electron to form an octet. Because it now has 18 electrons and 17 protons in its nucleus, chlorine is no longer a neutral atom. It is a chloride ion with an ionic charge of 1 - , which is written as Cl-, with the 1 understood. A negatively charged ion, called an anion (pronounced an-eye-un), is named by using the first syllable of its element name followed by ide. The chloride ion is larger than the chlorine atom because the ion has an additional electron, which completes its outermost energy level.

Ionic charge 1 -

= Charge of protons = (17 + )

+ Charge of electrons + (18 - )

CORE CHEMISTRY SKILL Writing Positive and Negative Ions

ENGAGE 6.2 Why does Li form a positive ion, Li+, whereas Br forms a negative ion, Br-?

PRACTICE PROBLEMS Try Practice Problems 6.1 to 6.4

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6.1 Ions: Transfer of Electrons 159

TABLE 6.2 lists the names of some important metal and nonmetal ions.

TABLE 6.2 Formulas and Names of Some Common Ions Metals Nonmetals

Group Number Cation Name of Cation Group Number Anion Name of Anion

1A (1) Li+ Lithium 5A (15) N3- Nitride

Na+ Sodium P3- Phosphide

K+ Potassium 6A (16) O2- Oxide

2A (2) Mg2+ Magnesium S2- Sulfide

Ca2+ Calcium 7A (17) F - Fluoride

Ba2+ Barium Cl- Chloride

3A (13) Al3+ Aluminum Br- Bromide

I- Iodide

Ionic Charges from Group Numbers In ionic compounds, representative elements usually lose or gain electrons to give a stable electron arrangement like their nearest noble gas. We can use the group numbers in the periodic table to determine the charges for the ions of the representative elements. The elements in Group 1A (1) lose one electron to form ions with a 1 + charge. The elements in Group 2A (2) lose two electrons to form ions with a 2 + charge. The elements in Group 3A (13) lose three electrons to form ions with a 3 + charge. In this text, we do not use the group numbers of the transition elements to determine their ionic charges.

In ionic compounds, the elements in Group 7A (17) gain one electron to form ions with a 1 - charge. The elements in Group 6A (16) gain two electrons to form ions with a 2 - charge. The elements in Group 5A (15) gain three electrons to form ions with a 3 - charge.

The nonmetals of Group 4A (14) do not typically form ions. However, the metals Sn and Pb in Group 4A (14) lose electrons to form positive ions. TABLE 6.3 lists the ionic charges for some common monatomic ions of representative elements.

Metals Lose Valence

Electrons

Nonmetals Gain Valence

Electrons Noble Gas

1A (1)

2A (2)

3A (13)

5A (15)

6A (16)

7A (17)

He

Ne

Ar

Kr

Xe

Li+

Na+

K+

Rb+

Cs+

Mg2+ Al3+

Ca2+

N3-

P3-

O2-

S2-

Se2-Sr2+

Ba2+

F-

Cl-

Br-

I-

TABLE 6.3 Examples of Monatomic Ions and Their Nearest Noble Gases

PRACTICE PROBLEMS Try Practice Problems 6.5 to 6.8

SAMPLE PROBLEM 6.1 Ions

TRY IT FIRST

a. Write the symbol and name for the ion that has 7 protons and 10 electrons. b. Write the symbol and name for the ion that has 20 protons and 18 electrons.

SOLUTION

a. The element with 7 protons is nitrogen. In an ion of nitrogen with 10 electrons, the ionic charge would be 3 - , [(7 + ) + (10 - ) = 3 - ]. The ion, written as N3-, is the nitride ion.

b. The element with 20 protons is calcium. In an ion of calcium with 18 electrons, the ionic charge would be 2 + , [(20 + ) + (18 - ) = 2 + ]. The ion, written as Ca2+, is the calcium ion.

SELF TEST 6.1

How many protons and electrons are in each of the following ions?

a. Sr2+ b. Cl- c. As3+

ANSWER

a. 38 protons, 36 electrons b. 17 protons, 18 electrons c. 33 protons, 30 electrons

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160 CHAPTER 6 Ionic and Molecular Compounds

PRACTICE PROBLEMS

6.1 Ions: Transfer of Electrons

6.1 State the number of electrons that must be lost by atoms of each of the following to achieve a stable electron configuration:

a. Li b. Ca c. Ga d. Cs e. Ba

6.2 State the number of electrons that must be gained by atoms of each of the following to achieve a stable electron configuration:

a. Cl b. Se c. N d. I e. S

6.3 State the number of electrons lost or gained when the following elements form ions:

a. Sr b. P c. Group 7A (17) d. Na e. Br

6.4 State the number of electrons lost or gained when the following elements form ions:

a. O b. Group 2A (2) c. F d. K e. Rb

6.5 Write the symbols for the ions with the following number of protons and electrons:

a. 3 protons, 2 electrons b. 9 protons, 10 electrons c. 12 protons, 10 electrons d. 27 protons, 24 electrons

6.6 Write the symbols for the ions with the following number of protons and electrons:

a. 8 protons, 10 electrons b. 19 protons, 18 electrons c. 35 protons, 36 electrons d. 50 protons, 46 electrons

6.7 State the number of protons and electrons in each of the following: a. Cu2+ b. Se2- c. Br- d. Fe2+

6.8 State the number of protons and electrons in each of the following: a. S2- b. Ni2+ c. Au3+ d. Ag +

6.9 Write the symbol for the ion of each of the following: a. chlorine b. cesium c. nitrogen d. radium

6.10 Write the symbol for the ion of each of the following: a. fluorine b. barium c. sodium d. iodine

6.11 Write the names for each of the following ions: a. Li+ b. Ca2+ c. Ga3+ d. P3-

6.12 Write the names for each of the following ions: a. Rb+ b. Sr2+ c. S2- d. F -

Applications 6.13 State the number of protons and electrons in each of the

following ions: a. O2-, used to build biomolecules and water b. K+, most prevalent positive ion in cells; needed for muscle

contraction, nerve impulses c. I-, needed for thyroid function d. Ca2+, needed for bones and teeth

6.14 State the number of protons and electrons in each of the following ions:

a. P3-, needed for bones and teeth b. F -, used to strengthen tooth enamel c. Mg2+, needed for bones and teeth d. Na+, most prevalent positive ion in extracellular fluid

PRACTICE PROBLEMS Try Practice Problems 6.9 to 6.14

ENGAGE 6.3 Why do all the atoms in Group 2A (2) form ions with 2 + charges?

SAMPLE PROBLEM 6.2 Writing Symbols for Ions

TRY IT FIRST

Consider the elements aluminum and oxygen.

a. Identify each as a metal or a nonmetal. b. State the number of valence electrons for each. c. State the number of electrons that must be lost or gained for each to achieve an octet. d. Write the symbol, including its ionic charge, and the name for each ion.

SOLUTION

Aluminum Oxygen

a. metal nonmetal

b. 3 valence electrons 6 valence electrons

c. loses 3 e- gains 2 e-

d. Al3+, [(13 + ) + (10 - )] = 3 + , aluminum ion O2-, [(8 + ) + (10 - )] = 2 - , oxide ion

SELF TEST 6.2

Consider the elements sulfur and potassium.

a. Identify each as a metal or a nonmetal. b. State the number of valence electrons for each. c. State the number of electrons that must be lost or gained for each to achieve an octet. d. Write the symbol, including its ionic charge, and the name for each ion.

ANSWER

a. Sulfur is a nonmetal; potassium is a metal. b. Sulfur has six valence electrons; potassium has one valence electron. c. Sulfur will gain two electrons; potassium will lose one electron. d. S2-, sulfide ion; K+, potassium ion

M06_TIMB8119_06_SE_C06.indd 160 11/30/18 7:58 AM

6.2 Ionic Compounds 161

Chemistry Link to Health Some Important Ions in the Body

Several ions in body f luids have important physiological and metabolic functions. Some are listed in TABLE 6.4.

Foods such as bananas, milk, cheese, and potatoes provide the body with ions that are important in regulating body functions.

TABLE 6.4 Ions in the Body Ion Occurrence Function Source Result of Too Little Result of Too Much

Na+ Principal cation outside the cell

Regulates and controls body fluids

Salt, cheese, pickles Hyponatremia, anxiety, diarrhea, circulatory failure, decrease in body fluid

Hypernatremia, little urine, thirst, edema

K+ Principal cation inside the cell

Regulates body fluids and cellular functions

Bananas, orange juice, milk, prunes, potatoes

Hypokalemia (hypopotassemia), lethargy, muscle weakness, failure of neurological impulses

Hyperkalemia (hyperpotassemia), irritability, nausea, little urine, cardiac arrest

Ca2+ Cation outside the cell; 90% of calcium in the body in bones

Major cation of bones; needed for muscle contraction

Milk, yogurt, cheese, greens, spinach

Hypocalcemia, tingling fingertips, muscle cramps, osteoporosis

Hypercalcemia, relaxed muscles, kidney stones, deep bone pain

Mg2+ Cation outside the cell; 50% of magnesium in the body in bones

Essential for certain enzymes, muscles, nerve control

Widely distributed (part of chlorophyll of all green plants), nuts, whole grains

Disorientation, hypertension, tremors, slow pulse

Drowsiness

Cl- Principal anion outside the cell

Gastric juice, regulates body fluids

Salt Same as for Na+ Same as for Na+

Milk, cheese, bananas, cereal, and potatoes provide ions for the body.

6.2 Ionic Compounds LEARNING GOAL Using charge balance, write the correct formula for an ionic compound.

We use ionic compounds such as salt, NaCl, and baking soda, NaHCO3, every day. Milk of magnesia, Mg(OH)2, or calcium carbonate, CaCO3, may be taken to settle an upset stomach. If you take a mineral supplement, iron may be present as iron(II) sulfate, FeSO4, iodine as potassium iodide, KI, and manganese as manganese(II) sulfate, MnSO4. Some sunscreens contain zinc oxide, ZnO, while tin(II) fluoride, SnF2, in toothpaste provides fluoride to help prevent tooth decay. Gemstones are ionic compounds that are cut and polished to make jewelry. For example, sapphires and rubies are aluminum oxide, Al2O3. Impurities of chromium ions make rubies red, and iron and titanium ions make sapphires blue.

Properties of Ionic Compounds In an ionic compound, one or more electrons are transferred from metals to nonmetals, which form positive and negative ions. The attraction between these ions is called an ionic bond.

The physical and chemical properties of an ionic compound such as NaCl are very different from those of the original elements. For example, the original elements of NaCl were sodium, a soft, shiny metal, and chlorine, a yellow-green poisonous gas. However, when

Rubies and sapphires are the ionic compound aluminum oxide, with chromium ions in rubies, and titanium and iron ions in sapphires.

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162 CHAPTER 6 Ionic and Molecular Compounds

they react and form positive and negative ions, they produce NaCl, which is ordinary table salt, a hard, white, crystalline substance that is important in our diet.

In a crystal of NaCl, the larger Cl- ions are arranged in a three-dimensional structure in which the smaller Na+ ions occupy the spaces between the Cl- ions (see FIGURE 6.1). In this crystal, every Na+ ion is surrounded by six Cl- ions, and every Cl- ion is surrounded by six Na+ ions. Thus, there are many strong attractions between the positive and negative ions, which account for the high melting points of ionic compounds. For example, the melting point of NaCl is 801 °C. At room temperature, ionic compounds are solids.

ENGAGE 6.4 What is the type of bonding between Na+ and Cl- ions in NaCl?

PRACTICE PROBLEMS Try Practice Problems 6.15 and 6.16

+

Sodium metal Chlorine gasand

Sodium chloride

Cl-

Na+FIGURE 6.1 The elements sodium and chlorine react to form the ionic compound sodium chloride, the compound that makes up table salt. The magnification of NaCl crystals shows the arrangement of Na+ and Cl- ions.

Chemical Formulas of Ionic Compounds The chemical formula of a compound represents the symbols and subscripts in the lowest whole-number ratio of the atoms or ions. In the formula of an ionic compound, the sum of the ionic charges in the formula is always zero. Thus, the total amount of positive charge is equal to the total amount of negative charge. For example, to achieve a stable electron con- figuration, one Na atom (metal) loses its one valence electron to form Na+, and one Cl atom (nonmetal) gains one electron to form a Cl- ion. The formula NaCl indicates that the com- pound has charge balance because there is one sodium ion, Na+, for every chloride ion, Cl-. The ionic charges are not shown in the formula of the compound.

Loses 1 e- Gains 1 e- Na+ Cl-

1(1+) + 1(1-) = 0 NaCl, sodium chloride

Cl -

Na+ClNa

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6.2 Ionic Compounds 163

Subscripts in Formulas Consider a compound of magnesium and chlorine. To achieve a stable electron configura- tion, one Mg atom (metal) loses its two valence electrons to form Mg2+. Two Cl atoms (nonmetals) each gain one electron to form two Cl- ions. The two Cl- ions are needed to balance the positive charge of Mg2+. This gives the formula MgCl2, magnesium chloride, in which the subscript 2 shows that two Cl- ions are needed for charge balance.

Loses 2 e- Each gains 1 e- Mg2+ 2Cl-

1(2+) + 2(1-) = 0 MgCl2, magnesium chloride

Cl

Cl Cl

Cl

Mg Mg2+

-

-

Writing Ionic Formulas from Ionic Charges The subscripts in the formula of an ionic compound represent the number of positive and negative ions that give an overall charge of zero. Thus, we can now write a formula directly from the ionic charges of the positive and negative ions. Suppose we wish to write the formula for the ionic compound containing Na+ and S2- ions. To balance the ionic charge of the S2- ion, we will need to place two Na+ ions in the formula. This gives the formula Na2S, which has an overall charge of zero. In the formula of an ionic compound, the cation is written first, followed by the anion. Appropriate subscripts are used to show the number of each of the ions. This formula is the lowest ratio of ions in the ionic compound. This lowest ratio of ions is called a formula unit.

CORE CHEMISTRY SKILL Writing Ionic Formulas

Each loses 1 e- Gains 2 e- 2Na+ S2-

2(1+) + 1(2-) = 0 Na2S, sodium sulfide

Na

Na

Na+

Na+

S S 2 -

SAMPLE PROBLEM 6.3 Writing Formulas from Ionic Charges

TRY IT FIRST

Write the symbols for the ions, and the correct formula for the ionic compound formed when lithium and nitrogen react.

SOLUTION

Lithium, which is a metal in Group 1A (1), forms Li+; nitrogen, which is a nonmetal in Group 5A (15), forms N3-. The charge of 3 - is balanced by three Li+ ions.

3(1 + ) + 1(3 - ) = 0

Writing the cation (positive ion) first and the anion (negative ion) second gives the formula Li3N.

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164 CHAPTER 6 Ionic and Molecular Compounds

PRACTICE PROBLEMS Try Practice Problems 6.17 to 6.20

SELF TEST 6.3

Write the symbols for the ions, and the correct formula for the ionic compound that would form when each of the following react:

a. calcium and oxygen b. magnesium and phosphorus

ANSWER

a. Ca2+, O2-, CaO b. Mg2+, P3-, Mg3P2

PRACTICE PROBLEMS

6.2 Ionic Compounds

6.15 Which of the following pairs of elements are likely to form an ionic compound?

a. lithium and chlorine b. oxygen and bromine c. potassium and oxygen d. sodium and neon e. cesium and magnesium f. nitrogen and fluorine

6.16 Which of the following pairs of elements are likely to form an ionic compound?

a. helium and oxygen b. magnesium and chlorine c. chlorine and bromine d. potassium and sulfur e. sodium and potassium f. nitrogen and iodine

6.17 Write the correct ionic formula for the compound formed between each of the following pairs of ions:

a. Na+ and O2- b. Al3+ and Br- c. Ba2+ and N3-

d. Mg2+ and F - e. Al3+ and S2-

6.18 Write the correct ionic formula for the compound formed between each of the following pairs of ions:

a. Al3+ and Cl- b. Ca2+ and S2-

c. Li+ and S2- d. Rb+ and P3-

e. Cs+ and I-

6.19 Write the symbols for the ions, and the correct formula for the ionic compound formed by each of the following:

a. potassium and sulfur b. sodium and nitrogen c. aluminum and iodine d. gallium and oxygen

6.20 Write the symbols for the ions, and the correct formula for the ionic compound formed by each of the following:

a. calcium and chlorine b. rubidium and bromine c. sodium and phosphorus d. magnesium and oxygen

6.3 Naming and Writing Ionic Formulas LEARNING GOAL Given the formula of an ionic compound, write the correct name; given the name of an ionic compound, write the correct formula.

In the name of an ionic compound made up of two elements, the name of the metal ion, which is written first, is the same as its element name. The name of the nonmetal ion is obtained by using the first syllable of its element name followed by ide. In the name of any ionic compound, a space separates the name of the cation from the name of the anion. Subscripts are not used; they are understood because of the charge balance of the ions in the compound (see TABLE 6.5).

CORE CHEMISTRY SKILL Naming Ionic Compounds

ENGAGE 6.5 How are the positive and negative ions named in an ionic compound?

TABLE 6.5 Names of Some Ionic Compounds Compound Metal Ion Nonmetal Ion Name

KI K+

Potassium I-

Iodide Potassium iodide

MgBr2 Mg2+

Magnesium Br-

Bromide Magnesium bromide

Al2O3 Al3+

Aluminum O2-

Oxide Aluminum oxide

REVIEW Solving Equations (1.4)

SAMPLE PROBLEM 6.4 Naming Ionic Compounds

TRY IT FIRST

Write the name for the ionic compound Mg3N2.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

Mg3N2 name cation, anion

Iodized salt contains KI to prevent iodine deficiency.

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6.3 Naming and Writing Ionic Formulas 165

Metals with Variable Charges We have seen that the charge of an ion of a representative element can be obtained from its group number. However, we cannot determine the charge of a transition element because it typically forms two or more positive ions. The transition elements lose electrons, but they are lost from the highest energy level and sometimes from a lower energy level as well. This is also true for metals of representative elements in Groups 4A (14) and 5A (15), such as Pb, Sn, and Bi.

In some ionic compounds, iron is in the Fe2+ form, but in other compounds, it has the Fe3+ form. Copper also forms two different ions, Cu+ and Cu2+. When a metal can form two or more types of ions, it has variable charge. Then we cannot predict the ionic charge from the group number.

For metals that form two or more ions, a naming system is used to identify the particular cation. A Roman numeral that is equal to the ionic charge is placed in parentheses immediately after the name of the metal. For example, Fe2+ is iron(II), and Fe3+ is iron(III). TABLE 6.6 lists the ions of some metals that produce more than one ion.

The transition elements form more than one positive ion except for zinc (Zn2+), cadmium (Cd2+), and silver (Ag +), which form only one ion. Thus, no Roman numerals are used with zinc, cadmium, and silver when naming their cations in ionic compounds. Metals in Groups 4A (14) and 5A (15) also form more than one type of positive ion. For example, lead and tin in Group 4A (14) form cations with charges of 2 + and 4 + , and bismuth in Group 5A (15) forms cations with charges of 3 + and 5 + .

Determination of Variable Charge When you name an ionic compound, you need to determine if the metal is a representative element or a transition element. If it is a transition element, except for zinc, cadmium, or silver, you will need to use its ionic charge as a Roman numeral as part of its name. The calculation of ionic charge depends on the negative charge of the anions in the formula. For example, if we want to name the ionic compound CuCl2, we use charge balance to determine the charge of the copper cation. Because there are two chloride ions, each with a 1 - charge, the total negative charge is 2 - . To balance this 2 - charge, the copper ion must have a charge of 2 + , or Cu2+:

CuCl2 Cu? Cl-

Cl-

1(?) + 2(1 - ) = 0 ? = 2 +

To indicate the 2 + charge for the copper ion Cu2+, we place the Roman numeral (II) immediately after copper when naming this compound: copper(II) chloride. Some ions and their location on the periodic table are seen in FIGURE 6.2.

ENGAGE 6.6 Why is a Roman numeral placed after the name of the cations of most transition elements?

PRACTICE PROBLEMS Try Practice Problems 6.21 and 6.22

STEP 1 Identify the cation and anion. The cation is Mg2+ and the anion is N3-.

STEP 2 Name the cation by its element name. The cation Mg2+ is magnesium.

STEP 3 Name the anion by using the first syllable of its element name followed by ide. The anion N3- is nitride.

STEP 4 Write the name for the cation first and the name for the anion second. magnesium nitride

SELF TEST 6.4

Write the name for each of the following ionic compounds: a. Ga2S3 b. Li3P

ANSWER

a. gallium sulfide b. lithium phosphide

TABLE 6.6 Some Metals That Form More Than One Positive Ion

Element Positive Ions Name of Ion

Bismuth Bi3+

Bi5+ Bismuth(III) Bismuth(V)

Chromium Cr2+

Cr3+ Chromium(II) Chromium(III)

Cobalt Co2+

Co3+ Cobalt(II) Cobalt(III)

Copper Cu+

Cu2+ Copper(I) Copper(II)

Gold Au+

Au3+ Gold(I) Gold(III)

Iron Fe2+

Fe3+ Iron(II) Iron(III)

Lead Pb2+

Pb4+ Lead(II) Lead(IV)

Manganese Mn2+

Mn3+ Manganese(II) Manganese(III)

Mercury Hg2 2+

Hg2+ Mercury(I)* Mercury(II)

Nickel Ni2+

Ni3+ Nickel(II) Nickel(III)

Tin Sn2+

Sn4+ Tin(II) Tin(IV)

*Mercury(I) ions form an ion pair with a 2 + charge.

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166 CHAPTER 6 Ionic and Molecular Compounds

FIGURE 6.2 In ionic compounds, metals form positive ions, and nonmetals form negative ions.

1 1A

2 2A

3 3B

4 4B

5 5B

6 6B

7 7B

8 9 8B

10 11 1B

12 2B

H+

Li+

Na+ Mg2+

K+ Ca2+ Cr 2+

Cr3+ Mn2+

Mn3+ Fe2+

Fe3+ Co2+

Co3+ Ni2+

Ni3+ Cu+

Cu2+ Zn2+

N3- O2- F-

Al3+ P3- S2- Cl-

Br-

Rb+ Sr2+ Ag+ Cd2+ Sn 2+

Sn4+ I-

Cs+ Ba2+ Au +

Au3+ Hg2

2+

Hg2+ Pb2+

Pb4+

13 3A

14 4A

15 5A

16 6A

17 7A

18 8A

Metals Metalloids Nonmetals

Bi3+

Bi5+

TABLE 6.7 Some Ionic Compounds of Metals That Form Two Compounds

Compound Systematic Name

FeCl2 Iron(II) chloride

Fe2O3 Iron(III) oxide

Cu3P Copper(I) phosphide

CrBr2 Chromium(II) bromide

SnCl2 Tin(II) chloride

PbS2 Lead(IV) sulfide

BiF3 Bismuth(III) fluoride

The growth of barnacles is prevented by using a paint with Cu2O on the bottom of a boat.

TABLE 6.7 lists the names of some ionic compounds in which the transition elements and metals from Groups 4A (14) and 5A (15) have more than one positive ion.

SAMPLE PROBLEM 6.5 Naming Ionic Compounds with Variable Charge Metal Ions

TRY IT FIRST

Antifouling paint contains Cu2O, which prevents the growth of barnacles and algae on the bottoms of boats. What is the name of Cu2O?

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

Cu2O name cation, anion, charge balance

STEP 1 Determine the charge of the cation from the anion.

Cation Anion

Element copper (Cu) oxygen (O)

Group transition element 6A (16)

Ion Cu? O2-

Calculation of Cation Charge 2Cu? + 2 - = 0 2Cu? = 2 +

2Cu?

2 =

2 + 2

= 1 +

Ion Cu+ O2-

STEP 2 Name the cation by its element name, and use a Roman numeral in parentheses for the charge. copper(I)

STEP 3 Name the anion by using the first syllable of its element name followed by ide. oxide

STEP 4 Write the name for the cation first and the name for the anion second. copper(I) oxide

ENGAGE 6.7 What are the ions formed by copper and oxygen in ionic compounds?

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6.3 Naming and Writing Ionic Formulas 167

PRACTICE PROBLEMS

6.3 Naming and Writing Ionic Formulas

6.21 Write the name for each of the following ionic compounds: a. AlF3 b. CaCl2 c. Na2O d. Mg3P2 e. KI f. BaF2 6.22 Write the name for each of the following ionic compounds: a. MgCl2 b. K3P c. Li2S d. CsF e. MgO f. SrBr2

6.23 Write the name for each of the following ions (include the Roman numeral when necessary):

a. Fe2+ b. Cu2+ c. Zn2+

d. Pb4+ e. Cr3+ f. Mn2+

6.24 Write the name for each of the following ions (include the Roman numeral when necessary):

a. Ag + b. Cu+ c. Bi3+

d. Sn2+ e. Au3+ f. Ni2+

PRACTICE PROBLEMS Try Practice Problems 6.23 to 6.28

SELF TEST 6.5

Write the name for each of the following compounds: a. Mn2S3 b. SnF4

ANSWER

a. Mn2S3 with the variable charge metal ion Mn 3+ is named manganese(III) sulfide.

b. SnF4 with the variable charge metal ion Sn 4+ is named tin(IV) fluoride.

Writing Formulas from the Name of an Ionic Compound The formula for an ionic compound is written from the first part of the name that describes the metal ion, including its charge, and the second part of the name that specifies the non- metal ion. Subscripts are added, as needed, to balance the charge. The steps for writing a formula from the name of an ionic compound are shown in Sample Problem 6.6.

INTERACTIVE VIDEO

Naming and Writing Ionic Formulas

PRACTICE PROBLEMS Try Practice Problems 6.29 to 6.34

SAMPLE PROBLEM 6.6 Writing Formulas for Ionic Compounds

TRY IT FIRST

Write the correct formula for iron(III) chloride.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

iron(III) chloride formula cation, anion, charge balance

STEP 1 Identify the cation and anion.

Type of Ion Cation Anion

Name iron(III) chloride

Symbol of Ion Fe3+ Cl-

STEP 2 Balance the charges. The charge of 3 + is balanced by three Cl- ions. 1(3 + ) + 3(1 - ) = 0

STEP 3 Write the formula, cation first, using subscripts from the charge balance. FeCl3

SELF TEST 6.6

Write the correct formula for each of the following paint pigments: a. chrome green, chromium(III) oxide b. titanium white, titanium(IV) oxide

ANSWER

a. Cr2O3 b. TiO2

The pigment chrome green contains chromium(III) oxide.

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168 CHAPTER 6 Ionic and Molecular Compounds

6.25 Write the name for each of the following ionic compounds: a. SnCl2 b. FeO c. Cu2S d. CuS e. CdBr2 f. HgCl2 6.26 Write the name for each of the following ionic compounds: a. Ag3P b. PbS c. SnO2 d. MnCl3 e. Bi2O3 f. CoCl2 6.27 Write the symbol for the cation in each of the following ionic

compounds: a. AuCl3 b. Fe2O3 c. PbI4 d. AlP

6.28 Write the symbol for the cation in each of the following ionic compounds:

a. FeCl2 b. CrO c. Ni2S3 d. SnCl2 6.29 Write the formula for each of the following ionic compounds: a. magnesium chloride b. sodium sulfide c. copper(I) oxide d. zinc phosphide e. gold(III) nitride f. cobalt(III) fluoride

6.30 Write the formula for each of the following ionic compounds: a. nickel(III) oxide b. barium fluoride c. tin(IV) chloride d. silver sulfide e. bismuth(V) chloride f. potassium nitride

6.31 Write the formula for each of the following ionic compounds: a. cobalt(III) chloride b. lead(IV) oxide c. silver iodide d. calcium nitride e. copper(I) phosphide f. chromium(II) chloride

6.32 Write the formula for each of the following ionic compounds: a. zinc bromide b. iron(III) sulfide c. manganese(IV) oxide d. chromium(III) iodide e. lithium nitride f. gold(I) oxide

Applications 6.33 The following compounds contain ions that are required in

small amounts by the body. Write the formula for each. a. potassium phosphide b. copper(II) chloride c. iron(III) bromide d. magnesium oxide

6.34 The following compounds contain ions that are required in small amounts by the body. Write the formula for each.

a. calcium chloride b. nickel(II) iodide c. manganese(II) oxide d. zinc nitride

6.4 Polyatomic Ions LEARNING GOAL Write the name and formula for an ionic compound containing a polyatomic ion.

An ionic compound may also contain a polyatomic ion as one of its cations or anions. A polyatomic ion is a group of covalently bonded atoms that has an overall ionic charge. Most polyatomic ions consist of a nonmetal such as phosphorus, sulfur, carbon, or nitrogen covalently bonded to oxygen atoms.

Almost all the polyatomic ions are anions with charges 1 - , 2 - , or 3 - . Only one common polyatomic ion, NH4

+, has a positive charge. Some models of common polyatomic ions are shown in FIGURE 6.3.

FIGURE 6.3 Many products contain polyatomic ions, which are groups of atoms that have an ionic charge.

SO4 2-

Sulfate ion Ca2+

Plaster cast CaSO4

NO3 -

Nitrate ion NH4

+

Ammonium ion

Fertilizer NH4NO3

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6.4 Polyatomic Ions 169

Names of Polyatomic Ions The names of the most common polyatomic ions, which end in ate, are shown in bold in TABLE 6.8. When a related ion has one less O atom, the ite ending is used. For the same nonmetal, the ate ion and the ite ion have the same charge. For example, the sulfate ion is SO4

2-, and the sulfite ion, which has one less oxygen atom, is SO3 2-.

Formula Charge Name SO4

2- 2 - sulfate SO3

2- 2 - sulfite

Phosphate and phosphite ions each have a 3 - charge.

Formula Charge Name PO4

3- 3 - phosphate PO3

3- 3 - phosphite

The formula of hydrogen carbonate, or bicarbonate, is written by placing a hydrogen in front of the polyatomic ion formula for carbonate (CO3

2-), and the charge is decreased from 2 - to 1 - to give HCO3 -.

Formula Charge Name CO3

2- 2 - carbonate HCO3

- 1 - hydrogen carbonate The halogens form four different polyatomic ions with oxygen. The prefix per is used

for one more O than the ate ion and the prefix hypo for one less O than the ite ion.

Formula Charge Name ClO4

- 1 - perchlorate ClO3

- 1 - chlorate ClO2

- 1 - chlorite ClO- 1 - hypochlorite Recognizing these prefixes and endings will help you identify polyatomic ions in the

name of a compound. The hydroxide ion (OH-) and cyanide ion (CN-) are exceptions to this naming pattern.

ENGAGE 6.8 How do the number of O atoms in a phosphate ion compare to the number of O atoms in a phosphite ion?

PRACTICE PROBLEMS Try Practice Problems 6.35 to 6.38

TABLE 6.8 Names and Formulas of Some Common Polyatomic Ions Nonmetal Formula of Ion* Name of Ion

Hydrogen OH- Hydroxide

Nitrogen NH4 +

NO3 −

NO2 -

Ammonium Nitrate Nitrite

Chlorine ClO4 -

ClO3 −

ClO2 -

ClO-

Perchlorate Chlorate Chlorite Hypochlorite

Carbon CO3 2−

HCO3 -

CN-

C2H3O2 -

Carbonate Hydrogen carbonate (or bicarbonate) Cyanide Acetate

Sulfur SO4 2−

HSO4 -

SO3 2-

HSO3 -

Sulfate Hydrogen sulfate (or bisulfate) Sulfite Hydrogen sulfite (or bisulfite)

Phosphorus PO4 3−

HPO4 2-

H2PO4 -

PO3 3-

Phosphate Hydrogen phosphate Dihydrogen phosphate Phosphite

*Formulas and names in bold type indicate the most common polyatomic ion for that element.

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170 CHAPTER 6 Ionic and Molecular Compounds

ENGAGE 6.9 Why does the formula for magnesium nitrate contain two NO3

- ions?

Sodium chlorite is used as a disinfectant in contact lens cleaning solutions, mouthwashes, and as a medication to treat amyotrophic lateral sclerosis (ALS).

Aluminum hydroxide is an antacid used to treat acid indigestion.

Writing Formulas for Compounds Containing Polyatomic Ions No polyatomic ion exists by itself. Like any ion, a polyatomic ion must be associated with ions of opposite charge. The bonding between polyatomic ions and other ions is one of electrical attraction.

To write correct formulas for compounds containing polyatomic ions, we follow the same rules of charge balance that we used for writing the formulas for simple ionic compounds. The total negative and positive charges must equal zero. For example, consider the formula for a compound containing sodium ions and chlorite ions. The ions are written as

Na+ ClO2 -

Sodium ion Chlorite ion

(1 + ) + (1 - ) = 0 Because one ion of each balances the charge, the formula is written as

NaClO2 Sodium chlorite

When more than one polyatomic ion is needed for charge balance, parentheses are used to enclose the formula of the ion. A subscript is written outside the right parenthesis of the polyatomic ion to indicate the number needed for charge balance. Consider the formula for magnesium nitrate. The ions in this compound are the magnesium ion and the nitrate ion, a polyatomic ion.

Mg2+ NO3 -

Magnesium ion Nitrate ion

To balance the positive charge of 2 + of the magnesium ion, two nitrate ions are needed. In the formula of the compound, parentheses are placed around the nitrate ion, and the sub- script 2 is written outside the right parenthesis.

Magnesium nitrate

Parentheses enclose the formula of the nitrate

ion

Subscript outside the parenthesis

indicates the use of two nitrate

ions

NO3 -

NO3 -

Mg2+ Mg(NO3)2

(2+) + 2(1-) = 0

SAMPLE PROBLEM 6.7 Writing Formulas Containing Polyatomic Ions

TRY IT FIRST

Antacid tablets contain aluminum hydroxide, which treats acid indigestion and heartburn. Write the formula for aluminum hydroxide.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

aluminum hydroxide

formula cation, polyatomic anion, charge balance

STEP 1 Identify the cation and polyatomic ion (anion).

Cation Polyatomic Ion (anion)

aluminum hydroxide

Al3+ OH-

STEP 2 Balance the charges. The charge of 3 + is balanced by three OH- ions.

1(3 + ) + 3(1 - ) = 0

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6.4 Polyatomic Ions 171

STEP 3 Write the formula, cation first, using the subscripts from charge balance. The formula for the compound is written by enclosing the formula of the hydroxide ion, OH-, in parentheses and writing the subscript 3 outside the right parenthesis.

Al(OH)3

SELF TEST 6.7

Write the formula for a compound containing the following: a. ammonium ions and phosphate ions b. iron(III) ions and bicarbonate ions

ANSWER

a. (NH4)3PO4 b. Fe(HCO3)3

Naming Ionic Compounds Containing Polyatomic Ions When naming ionic compounds containing polyatomic ions, we first write the positive ion, usually a metal, and then we write the name for the polyatomic ion. It is important that you learn to recognize the polyatomic ion in the formula and name it correctly.

SAMPLE PROBLEM 6.8 Naming Compounds Containing Polyatomic Ions

TRY IT FIRST

Name the following ionic compounds: a. Cu(NO3)2, used in fireworks b. KClO3, used to produce oxygen in aircraft and space stations

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

formula name cation, polyatomic ion

TABLE 6.9 Some Ionic Compounds That Contain Polyatomic Ions Formula Name Medical Use

AlPO4 Aluminum phosphate Antacid

Al2(SO4)3 Aluminum sulfate Used as pH stabilizer in paper industry

BaSO4 Barium sulfate Contrast medium for X-rays

CaCO3 Calcium carbonate Antacid, calcium supplement

Ca3(PO4)2 Calcium phosphate Calcium dietary supplement

Cr2(SO4)3 Chromium(III) sulfate Green pigment for paints and inks

MgSO4 Magnesium sulfate Epsom salts, used to sooth sore muscles

K2CO3 Potassium carbonate Alkalizer, diuretic

AgNO3 Silver nitrate Topical anti-infective

Na2CO3 Sodium carbonate Manufacture of glass

NaHCO3 Sodium hydrogen carbonate or sodium bicarbonate

Antacid, used to make baking powder

Zn3(PO4)2 Zinc phosphate Dental cement

SO4 2-Mg2+

A solution of Epsom salts, magnesium sulfate, MgSO4, may be used to soothe sore muscles.

Na2SO4

Na2 SO4 Sodium sulfate Iron(III) phosphate Aluminum carbonate

FePO4

Fe PO4

Al2(CO3)3

Al2( )3CO3

TABLE 6.9 lists the formulas and names of some ionic compounds that include polyatomic ions and also gives their uses.

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172 CHAPTER 6 Ionic and Molecular Compounds

PRACTICE PROBLEMS Try Practice Problems 6.39 to 6.46

Formula Cation Anion Name of Cation Name of Anion Name of Compound

a. Cu(NO3)2 Cu 2+ NO3

- Copper(II) ion Nitrate ion Copper(II) nitrate

b. KClO3 K + ClO3

- Potassium ion Chlorate ion Potassium chlorate

SELF TEST 6.8

Name each of the following compounds: a. Co3(PO4)2, a pigment called cobalt violet b. NaHCO3, corrects pH imbalance

ANSWER

a. cobalt(II) phosphate b. sodium hydrogen carbonate or sodium bicarbonate

PRACTICE PROBLEMS

6.4 Polyatomic Ions

6.35 Write the formula including the charge for each of the following polyatomic ions:

a. hydrogen carbonate (bicarbonate) b. ammonium c. phosphite d. chlorate

6.36 Write the formula including the charge for each of the following polyatomic ions:

a. nitrite b. sulfite c. hydroxide d. acetate

6.37 Name the following polyatomic ions: a. SO4

2- b. CO3 2- c. HSO3

- d. NO3 -

6.38 Name the following polyatomic ions: a. OH- b. PO4

3- c. CN- d. NO2 -

6.39 Complete the following table with the formula and name of the compound that forms between each pair of ions:

NO2 − CO3

2− HSO4 − PO4

3−

Li+

Cu2+

Ba2+

6.40 Complete the following table with the formula and name of the compound that forms between each pair of ions:

NO3 − HCO3

− SO3 2− HPO4

2−

NH4 +

Al3+

Pb4+

6.41 Write the correct formula for the following ionic compounds: a. barium hydroxide b. sodium hydrogen sulfate c. iron(II) nitrite d. zinc phosphate e. iron(III) carbonate

6.42 Write the correct formula for the following ionic compounds: a. aluminum chlorate b. ammonium oxide c. magnesium bicarbonate d. sodium nitrite e. copper(I) sulfate

6.43 Write the formula for the polyatomic ion and name each of the following compounds:

a. Na2CO3 b. (NH4)2S c. Ca(OH)2 d. Sn(NO2)2 6.44 Write the formula for the polyatomic ion and name each of the

following compounds: a. MnCO3 b. Au2SO4 c. Ca3(PO4)2 d. Fe(HCO3)3

Applications 6.45 Name each of the following ionic compounds: a. Zn(C2H3O2)2, cold remedy b. Mg3(PO4)2, antacid c. NH4Cl, expectorant d. Sr(NO3)2, produces red color in fireworks e. NaNO2, meat preservative

6.46 Name each of the following ionic compounds: a. Li2CO3, antidepressant b. MgSO4, Epsom salts c. NaClO, disinfectant d. Na3PO4, laxative e. Ba(OH)2, component of antacids

6.5 Molecular Compounds: Sharing Electrons LEARNING GOAL Given the formula of a molecular compound, write its correct name; given the name of a molecular compound, write its formula.

A molecular compound consists of atoms of two or more nonmetals that share one or more valence electrons. The atoms are held together by covalent bonds that form a molecule. There are many more molecular compounds than there are ionic ones. For example, water (H2O) and carbon dioxide (CO2) are both molecular compounds. Molecular compounds

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6.5 Molecular Compounds: Sharing Electrons 173

consist of molecules, which are discrete groups of atoms in a definite proportion. A molecule of water (H2O) consists of two atoms of hydrogen and one atom of oxygen. When you have iced tea, perhaps you add molecules of sugar (C12H22O11), which is a molecular compound. Other familiar molecular compounds include propane (C3H8), alcohol (C2H6O), the antibiotic amoxicillin (C16H19N3O5S), and the antidepressant Prozac (C17H18F3NO).

Names and Formulas of Molecular Compounds When naming a molecular compound, the first nonmetal in the formula is named by its element name; the second nonmetal is named using the first syllable of its element name, followed by ide. When a subscript indicates two or more atoms of an element, a prefix is shown in front of its name. TABLE 6.10 lists prefixes used in naming molecular compounds.

The names of molecular compounds need prefixes because several different compounds can be formed from the same two nonmetals. For example, carbon and oxygen can form two different compounds, carbon monoxide, CO, and carbon dioxide, CO2, in which the number of atoms of oxygen in each compound is indicated by the prefixes mono or di in their names.

When the vowels o and o or a and o appear together, the first vowel is omitted, as in carbon monoxide. In the name of a molecular compound, the prefix mono is usually omitted, as in NO, nitrogen oxide. Traditionally, however, CO is named carbon monoxide. TABLE 6.11 lists the formulas, names, and commercial uses of some molecular compounds.

CORE CHEMISTRY SKILL Writing the Names and Formulas

for Molecular Compounds

TABLE 6.10 Prefixes Used in Naming Molecular Compounds

1 mono 6 hexa

2 di 7 hepta

3 tri 8 octa

4 tetra 9 nona

5 penta 10 deca

TABLE 6.11 Some Common Molecular Compounds Formula Name Commercial Uses

CO2 Carbon dioxide Fire extinguishers, dry ice, propellant in aerosols, carbonation of beverages

CS2 Carbon disulfide Manufacture of rayon

NO Nitrogen oxide Stabilizer, biochemical messenger in cells

N2O Dinitrogen oxide Inhalation anesthetic, “laughing gas”

SF6 Sulfur hexafluoride Electrical circuits

SO2 Sulfur dioxide Preserving fruits, vegetables; disinfectant in breweries; bleaching textiles

SO3 Sulfur trioxide Manufacture of explosives

SAMPLE PROBLEM 6.9 Naming Molecular Compounds

TRY IT FIRST

Name the molecular compound NCl3.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

NCl3 name prefixes

STEP 1 Name the first nonmetal by its element name. In NCl3, the first nonmetal (N) is nitrogen.

STEP 2 Name the second nonmetal by using the first syllable of its element name followed by ide. The second nonmetal (Cl) is chloride.

STEP 3 Add prefixes to indicate the number of atoms (subscripts). Because there is one nitrogen atom, no prefix is needed. The subscript 3 for the Cl atoms is shown as the prefix tri. The name of NCl3 is nitrogen trichloride.

ENGAGE 6.10 Why are prefixes used to name molecular compounds?

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174 CHAPTER 6 Ionic and Molecular Compounds

SELF TEST 6.9

Name each of the following molecular compounds: a. SiBr4 b. Br2O c. S3N2

ANSWER

a. silicon tetrabromide b. dibromine oxide c. trisulfur dinitride PRACTICE PROBLEMS

Try Practice Problems 6.47 to 6.50

Writing Formulas from the Names of Molecular Compounds In the name of a molecular compound, the names of two nonmetals are given along with prefixes for the number of atoms of each. To write the formula from the name, we use the symbol for each element and a subscript if a prefix indicates two or more atoms.

SAMPLE PROBLEM 6.10 Writing Formulas for Molecular Compounds

TRY IT FIRST

Write the formula for the molecular compound diboron trioxide.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

diboron trioxide formula subscripts from prefixes

STEP 1 Write the symbols in the order of the elements in the name.

Name of Element boron oxygen

Symbol of Element B O

STEP 2 Write any prefixes as subscripts. The prefix di in diboron indicates that there are two atoms of boron, shown as a subscript 2 in the formula. The prefix tri in trioxide indicates that there are three atoms of oxygen, shown as a subscript 3 in the formula. B2O3

SELF TEST 6.10

Write the formula for each of the following molecular compounds: a. iodine pentafluoride b. carbon diselenide

ANSWER

a. IF5 b. CSe2

PRACTICE PROBLEMS Try Practice Problems 6.51 to 6.54

Summary of Naming Ionic and Molecular Compounds We have now examined strategies for naming ionic and molecular compounds. In general, compounds having two elements are named by stating the first element name followed by the name of the second element with an ide ending. If the first element is a metal, the compound is usually ionic; if the first element is a nonmetal, the compound is usually molecular. For ionic compounds, it is necessary to determine whether the metal can form more than one type of positive ion; if so, a Roman numeral following the name of the metal indicates the particular ionic charge. One exception is the ammonium ion, NH4

+, which is also written first as a positively charged polyatomic ion. Ionic compounds having three or more elements include some type of polyatomic ion. They are named by ionic rules but have an ate or ite ending when the polyatomic ion has a negative charge.

In naming molecular compounds having two elements, prefixes are necessary to indicate two or more atoms of each nonmetal as shown in that particular formula (see FIGURE 6.4).

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6.5 Molecular Compounds: Sharing Electrons 175

SAMPLE PROBLEM 6.11 Naming Ionic and Molecular Compounds

TRY IT FIRST

Identify each of the following compounds as ionic or molecular and write its name: a. NiSO4 b. SO3

SOLUTION

a. NiSO4, consisting of a cation of a transition element and a polyatomic ion SO4 2-, is an

ionic compound. As a transition element, Ni forms more than one type of ion. In this formula, the 2 - charge of SO4 2- is balanced by one nickel ion, Ni2+. In the name, a Roman numeral written after the metal name, nickel(II), specifies the 2 + charge. The anion SO4

2- is a polyatomic ion named sulfate. The compound is named nickel(II) sulfate.

b. SO3 consists of two nonmetals, which indicates that it is a molecular compound. The first element S is sulfur (no prefix is needed). The second element O, oxide, has subscript 3, which requires a prefix tri in the name. The compound is named sulfur trioxide.

SELF TEST 6.11

Name each of the following compounds: a. IF7 b. Fe(NO3)3

ANSWER

a. iodine heptafluoride b. iron(III) nitrate

ENGAGE 6.11 How do you know that sodium phosphate is an ionic compound and diphosphorus pentoxide a molecular compound?

PRACTICE PROBLEMS Try Practice Problems 6.55 and 6.56

FIGURE 6.4 A flowchart illustrates naming for ionic and molecular compounds.

Flowchart for Naming Chemical Compounds

Ionic (metal and nonmetal) Molecular (two nonmetals)

Metal or ammonium (cation)

Nonmetal (anion)

First nonmetal

Second nonmetal

Forms only one positive

ion

Forms more than one

positive ion

Monatomic ion

Polyatomic ion

Use the element

name followed by a Roman

numeral equal to the charge

Use the name of the

element or ammonium

Use the first syllable

of the element name, followed

by ide

Use the name of the polyatomic

ion

Use a prefix to match

subscript before the

element name

Use a prefix to match subscript

before the element name

and end with ide

ENGAGE 6.12 When are the names of some metal ions followed by a Roman numeral in the name of a compound?

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176 CHAPTER 6 Ionic and Molecular Compounds

UPDATE Compounds at the Pharmacy

Richard returns to the pharmacy to talk to Dr. Chavez about a way to treat his sore toe. She recommends soaking his foot in a solution of Epsom salts, which is magnesium sulfate. Richard also asks Dr. Chavez to recommend an antacid for his upset stomach

and an iron supplement. She suggests an antacid that contains calcium carbonate and aluminum hydroxide, and iron(II) sulfate as an iron supplement. Richard also picks up

toothpaste containing tin(II) fluoride, and carbonated water,

which contains carbon dioxide.

Applications 6.57 Write the correct formula for each of the following

compounds: a. magnesium sulfate b. tin(II) fluoride c. aluminum hydroxide

6.58 Write the correct formula for each of the following compounds:

a. iron(II) sulfate b. calcium carbonate c. carbon dioxide

PRACTICE PROBLEMS

6.5 Molecular Compounds: Sharing Electrons

6.47 Name each of the following molecular compounds: a. PBr3 b. Cl2O c. CBr4 d. HF e. NF3 6.48 Name each of the following molecular compounds: a. IF3 b. P2O5 c. SiO2 d. PCl3 e. CO

6.49 Name each of the following molecular compounds: a. N2O3 b. Si2Br6 c. P4S3 d. PCl5 e. SeF6 6.50 Name each of the following molecular compounds: a. SiF4 b. IBr3 c. CO2 d. N2F2 e. N2S3 6.51 Write the formula for each of the following molecular

compounds: a. carbon tetrachloride b. carbon monoxide c. phosphorus trifluoride d. dinitrogen tetroxide

6.52 Write the formula for each of the following molecular compounds:

a. sulfur dioxide b. silicon tetrachloride c. iodine trifluoride d. dinitrogen oxide

6.53 Write the formula for each of the following molecular compounds:

a. oxygen difluoride b. boron trichloride c. dinitrogen trioxide d. sulfur hexafluoride

6.54 Write the formula for each of the following molecular compounds:

a. sulfur dibromide b. carbon disulfide c. tetraphosphorus hexoxide d. dinitrogen pentoxide

Applications

6.55 Name each of the following ionic or molecular compounds: a. Al2(SO4)3, antiperspirant b. Zn3(PO4)2, dental cement c. N2O , “laughing gas,” inhaled anesthetic d. Mg(OH)2, laxative

6.56 Name each of the following ionic or molecular compounds: a. (NH4)3PO4, used as a fertilizer b. BaSO4, used to create images of the esophagus and stomach c. NO, vasodilator d. Cu(OH)2, fungicide

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Chapter Review 177

6.3 Naming and Writing Ionic Formulas LEARNING GOAL Given the formula of an ionic compound, write the correct name; given the name of an ionic compound, write the correct formula. • In naming ionic compounds, the positive ion is given first followed

by the name of the negative ion. • The names of ionic compounds containing two elements end with

ide. • Except for Ag, Cd, and Zn, transition elements form cations with

two or more ionic charges. • The charge of the cation is determined from the total negative

charge in the formula and included as a Roman numeral immediately following the name of the metal that has a variable charge.

6.4 Polyatomic Ions LEARNING GOAL Write the name and formula for an ionic compound containing a polyatomic ion. • A polyatomic ion is a covalently bonded

group of atoms with an electrical charge; for example, the carbonate ion has the formula CO3

2-.

CONCEPT MAP

IONIC AND MOLECULAR COMPOUNDS

Ionic Compounds Molecular Compounds

Ionic Bonds

contain

between

NonmetalsMetals

Charge Balance

use

to write subscripts for the

Chemical Formula

that form

Negative IonsPositive Ions

contain

between

Nonmetals

Covalent Bonds

that form

CHAPTER REVIEW

6.1 Ions: Transfer of Electrons LEARNING GOAL Write the symbols for the simple ions of the representative elements. • The stability of the noble gases is associated with

a stable electron configuration in the outermost energy level.

• With the exception of helium, which has two electrons, noble gases have eight valence electrons, which is an octet.

• Atoms of elements in Groups 1A to 7A (1, 2, 13 to 17) achieve stability by losing, gaining, or sharing their valence electrons in the formation of compounds.

• Metals of the representative elements lose valence electrons to form positively charged ions (cations): Groups 1A (1), 1 + , 2A (2), 2 + , and 3A (13), 3 + .

• When reacting with metals, nonmetals gain electrons to form octets and form negatively charged ions (anions): Groups 5A (15), 3 - , 6A (16), 2 - , and 7A (17), 1 - .

6.2 Ionic Compounds LEARNING GOAL Using charge balance, write the correct formula for an ionic compound. • The total positive and negative ionic charge is

balanced in the formula of an ionic compound. • Charge balance in a formula is achieved by

using subscripts after each symbol so that the overall charge is zero.

NmM

Nm -

M+

Transfer of electrons

Ionic bond

Sodium chloride

Cl- Na+

1 1A

2 2A

3 3B

4 4B

5 5B

6 6B

7 7B

8 9 8B

10 11 1B

12 2B

H+

Li+

Na+ Mg2+

K+ Ca2+ Cr 2+

Cr3+ Mn2+

Mn3+ Fe2+

Fe3+ Co2+

Co3+ Ni2+

Ni3+ Cu+

Cu2+ Zn2+

N3- O2- F-

Al3+ P3- S2- Cl-

Br-

Rb+ Sr2+ Ag+ Cd2+

Bi3+

Bi5+

Sn2+

Sn4+ I-

Cs+ Ba2+ Au+

Au3+ Hg2

2+

Hg2+ Pb2+

Pb4+

13 3A

14 4A

15 5A

16 6A

17 7A

18 8A

Metals Metalloids Nonmetals

SO4 2-

Sulfate ion

Ca2+

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178 CHAPTER 6 Ionic and Molecular Compounds

• Most polyatomic ions have names that end with ate or ite. • Most polyatomic ions contain a nonmetal and one or more oxygen

atoms. • The ammonium ion, NH4

+, is a positive polyatomic ion. • When more than one polyatomic ion is used for charge balance,

parentheses enclose the formula of the polyatomic ion.

6.5 Molecular Compounds: Sharing Electrons LEARNING GOAL Given the formula of a molecular compound, write its correct name; given the name of a molecular compound, write its formula.

• The first nonmetal in a molecular compound uses its element name; the second nonmetal uses the first syllable of its element name followed by ide.

• The name of a molecular compound with two different atoms uses prefixes to indicate the subscripts in the formula.

1 mono

2 di

3 tri

4 tetra

5 penta • In a covalent bond, atoms of nonmetals

share valence electrons such that each atom has a stable electron configuration.

The chapter section containing each Core Chemistry Skill is shown in parentheses at the end of each heading.

Writing Positive and Negative Ions (6.1) • In the formation of an ionic bond, atoms of a metal lose and atoms

of a nonmetal gain valence electrons to acquire a stable electron configuration, usually eight valence electrons.

• This tendency of atoms to attain a stable electron configuration is known as the octet rule.

Example: State the number of electrons lost or gained by atoms and the ion formed for each of the following to obtain a stable electron configuration:

a. Br b. Ca c. S

Answer: a. Br atoms gain one electron to achieve a stable electron configuration, Br9.

b. Ca atoms lose two electrons to achieve a stable electron configuration, Ca2+.

c. S atoms gain two electrons to achieve a stable electron configuration, S29.

Writing Ionic Formulas (6.2) • The chemical formula of a compound represents the lowest whole-

number ratio of the atoms or ions. • In the chemical formula of an ionic compound, the sum of the

positive and negative charges is always zero. • In a chemical formula of an ionic compound, the total positive

charge is equal to the total negative charge.

CORE CHEMISTRY SKILLS Example: Write the formula for magnesium phosphide. Answer: Magnesium phosphide is an ionic compound that contains the

ions Mg2+ and P3-. Using charge balance, we determine the number(s) of each type of ion. 3(2 + ) + 2(3 - ) = 0 3Mg2+ and 2P3- give the formula Mg3P2.

Naming Ionic Compounds (6.3) • In the name of an ionic compound made up of two elements, the

name of the metal ion, which is written first, is the same as its element name.

• For metals that form two or more ions, a Roman numeral that is equal to the ionic charge is placed in parentheses immediately after the name of the metal.

• The name of a nonmetal ion is obtained by using the first syllable of its element name followed by ide.

• The name of a polyatomic ion ends in ate or ite.

Example: What is the name of PbSO4? Answer: This compound contains the SO4

2- ion, which has a 2 - charge. For charge balance, the positive ion must have a charge of 2 + . Pb? + (2 - ) = 0 Pb = 2 + Because lead can form two different positive ions, a Roman numeral (II) is used in the name of the compound: lead(II) sulfate.

anion A negatively charged ion such as Cl-, O2-, or SO4 2-.

cation A positively charged ion such as Na+, Mg2+, Al3+, or NH4 +.

chemical formula The group of symbols and subscripts that repre- sents the atoms or ions in a compound.

covalent bond A sharing of valence electrons by atoms. ion An atom or group of atoms having an electrical charge because

of a loss or gain of electrons. ionic bond The attraction between a positive ion and a negative ion

when electrons are transferred from a metal to a nonmetal. ionic charge The difference between the number of protons (positive)

and the number of electrons (negative) written in the upper right corner of the symbol for the element or polyatomic ion.

ionic compound A compound of positive and negative ions held together by ionic bonds.

molecular compound A combination of atoms in which stable  electron configurations are attained by sharing electrons.

molecule The smallest unit of two or more atoms held together by covalent bonds.

octet rule Elements in Groups 1A to 7A (1, 2, 13 to 17) react with other elements by forming ionic or covalent bonds to produce a stable electron configuration.

polyatomic ion A group of covalently bonded nonmetal atoms that has an overall electrical charge.

KEY TERMS

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Writing the Names and Formulas for Molecular Compounds (6.5) • When naming a molecular compound, the first nonmetal in the

formula is named by its element name; the second nonmetal is named using the first syllable of its element name followed by ide.

• When a subscript indicates two or more atoms of an element, a prefix is shown in front of its name.

Example: Name the molecular compound BrF5. Answer: Two nonmetals share electrons and form a molecular

compound. Br (first nonmetal) is bromine; F (second nonmetal) is fluoride. In the name for a molecular compound, prefixes indicate the subscripts in the formulas. The subscript 1 is understood for Br. The subscript 5 for fluoride is written with the prefix penta. The name is bromine pentafluoride.

UNDERSTANDING THE CONCEPTS

The chapter sections to review are shown in parentheses at the end of each problem.

6.59 a. How does the octet rule explain the formation of a magnesium ion? (6.1)

b. What noble gas has the same electron configuration as the magnesium ion?

c. Why are Group 1A (1) and Group 2A (2) elements found in many compounds, but not Group 8A (18) elements?

6.60 a. How does the octet rule explain the formation of a chloride ion? (6.1)

b. What noble gas has the same electron configuration as the chloride ion?

c. Why are Group 7A (17) elements found in many compounds, but not Group 8A (18) elements?

6.61 Identify each of the following atoms or ions: (6.1)

Electron Configurations Cation Anion Formula of Compound

Name of Compound

1s22s22p63s2 1s22s22p3

1s22s22p63s23p64s1 1s22s22p4

1s22s22p63s23p1 1s22s22p63s23p5

6.66 Using each of the following electron configurations, write the formulas for the cation and anion that form, the formula for the compound they form, and its name. (6.2, 6.3)

Electron Configurations Cation Anion Formula of Compound

Name of Compound

1s22s22p63s1 1s22s22p5

1s22s22p63s23p64s2 1s22s22p63s23p4

1s22s1 1s22s22p63s23p3

7 p+

8 n

10 e-

3 p+

4 n

2 e-

1 p+

0 n

0 e-

3 p+

4 n

3 e-

A B C D

6.62 Identify each of the following atoms or ions: (6.1)

15 p+

16 n

18 e-

8 p+

8 n

8 e-

30 p+

35 n

28 e-

26 p+

28 n

23 e-

DCBA

6.63 Element X has one valence electron and element Y has 6 valence electrons.

a. What are the group numbers of X and Y? b. Will a compound of X and Y be ionic or molecular? c. What ions would be formed by X and Y? d. What would be the formula of a compound of X and Y? e. What would be the formula of a compound of X and sulfur?

f. What would be the formula of a compound of Y and chlorine?

g. Is the compound in part f ionic or molecular?

6.64 Element X has two valence electrons and element Y has 5 valence electrons.

a. What are the group numbers of X and Y? b. Will a compound of X and Y be ionic or molecular? c. What ions would be formed by X and Y? d. What would be the formula of a compound of X and Y? e. What would be the formula of a compound of X and sulfur? f. What would be the formula of a compound of Y and

chlorine? g. Is the compound in part f ionic or molecular?

6.65 Using each of the following electron configurations, write the formulas for the cation and anion that form, the formula for the compound they form, and its name. (6.2, 6.3)

ADDITIONAL PRACTICE PROBLEMS

6.67 Write the name for each of the following ions: (6.1) a. N3- b. Mg2+

c. O2- d. Al3+

6.68 Write the name for each of the following ions: (6.1) a. K+ b. Na+

c. Ba2+ d. Cl-

6.69 Consider an ion with the symbol X2+ formed from a representative element. (6.1, 6.2, 6.3)

a. What is the group number of the element? b. If X is in Period 3, what is the element? c. What is the formula of the compound formed from X and

the nitride ion?

Additional Practice Problems 179

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180 CHAPTER 6 Ionic and Molecular Compounds

6.70 Consider an ion with the symbol Y3- formed from a representative element. (6.1, 6.2, 6.3)

a. What is the group number of the element? b. If Y is in Period 3, what is the element? c. What is the formula of the compound formed from the

barium ion and Y?

6.71 One of the ions of tin is tin(IV). (6.1, 6.2, 6.3, 6.4) a. What is the symbol for this ion? b. How many protons and electrons are in the ion? c. What is the formula of tin(IV) oxide? d. What is the formula of tin(IV) phosphate?

6.72 One of the ions of gold is gold(III). (6.1, 6.2, 6.3, 6.4) a. What is the symbol for this ion? b. How many protons and electrons are in the ion? c. What is the formula of gold(III) sulfate? d. What is the formula of gold(III) nitrate?

6.73 Write the formula for each of the following ionic compounds: (6.2, 6.3)

a. tin(II) sulfide b. lead(IV) oxide c. silver chloride d. calcium nitride e. copper(I) phosphide f. chromium(II) bromide

6.74 Write the formula for each of the following ionic compounds: (6.2, 6.3)

a. nickel(III) oxide b. iron(III) sulfide c. lead(II) sulfate d. chromium(III) iodide e. lithium nitride f. gold(I) oxide

6.75 Name each of the following molecular compounds: (6.5) a. NCl3 b. N2S3 c. N2O d. IF e. BF3 f. P2O5

6.76 Name each of the following molecular compounds: (6.5) a. CF4 b. SF6 c. BrCl d. N2O4 e. SO2 f. CS2 6.77 Write the formula for each of the following molecular

compounds: (6.5) a. carbon sulfide b. diphosphorus pentoxide c. dihydrogen sulfide d. sulfur dichloride

6.78 Write the formula for each of the following molecular compounds: (6.5)

a. silicon dioxide b. carbon tetrabromide c. diphosphorus tetraiodide d. dinitrogen trioxide

6.79 Classify each of the following as ionic or molecular, and write its name: (6.3, 6.5)

a. FeCl3 b. Na2SO4 c. NO2 d. Rb2S e. PF5 f. CF4 6.80 Classify each of the following as ionic or molecular, and write

its name: (6.3, 6.5) a. Al2(CO3)3 b. ClF5 c. BCl3 d. Mg3N2 e. ClO2 f. CrPO4

6.81 Write the formula for each of the following: (6.3, 6.4, 6.5) a. tin(II) carbonate b. lithium phosphide c. silicon tetrachloride d. manganese(III) oxide e. tetraphosphorus triselenide f. calcium bromide

6.82 Write the formula for each of the following: (6.3, 6.4, 6.5) a. sodium carbonate b. nitrogen dioxide c. aluminum nitrate d. copper(I) nitride e. potassium phosphate f. cobalt(III) sulfate

The following problems are related to the topics in this chapter. However, they do not all follow the chapter order, and they require you to combine concepts and skills from several sections. These problems will help you increase your critical thinking skills and prepare for your next exam.

6.83 Complete the following table for atoms or ions: (6.1)

Atom or Ion Number of Protons

Number of Electrons

Electrons Lost/Gained

K+

12 p+ 10 e-

8 p+ 2 e- gained

10 e- 3 e- lost

6.84 Complete the following table for atoms or ions: (6.1)

Atom or Ion Number of Protons

Number of Electrons

Electrons Lost/Gained

30 p+ 2 e- lost

36 p+ 36 e-

16 p+ 2 e- gained

46 e- 4 e- lost

6.85 Identify the group number in the periodic table of X, a represen- tative element, in each of the following ionic compounds: (6.2)

a. XCl3 b. Al2X3 c. XCO3

6.86 Identify the group number in the periodic table of X, a represen- tative element, in each of the following ionic compounds: (6.2)

a. X2O3 b. X2SO3 c. Na3X

6.87 Classify each of the following as ionic or molecular, and name each: (6.2, 6.3, 6.4, 6.5)

a. Li2HPO4 b. ClF3 c. Mg(ClO2)2 d. NF3 e. Ca(HSO4)2 f. KClO4 g. Au2(SO3)3

6.88 Classify each of the following as ionic or molecular, and name each: (6.2, 6.3, 6.4, 6.5)

a. FePO3 b. Cl2O7 c. Ca3(PO4)2 d. PCl3 e. Al(ClO2)3 f. Pb(C2H3O2)2 g. MgCO3

CHALLENGE PROBLEMS

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ANSWERS TO ENGAGE QUESTIONS 6.8 In a phosphate ion, there are four O atoms; in a phosphite ion,

there are three O atoms.

6.9 The 2 + charge of the magnesium ion must be balanced by the charges of two nitrate (NO3

-) ions.

6.10 Because two nonmetals may form more than one molecular compound, prefixes are used.

6.11 Sodium phosphate is a combination of a metal ion and a polyatomic ion; it is an ionic compound. Diphosphorus pentoxide is a combination of two nonmetals; it is a molecular compound.

6.12 When a transition element that forms more than one positive ion is present in an ionic compound, a Roman numeral is used.

6.1 The net charge of the cesium ion with 55 protons (55 + ) and 54 electrons (54 - ) is 1 + .

6.2 A Li atom loses one electron to form a stable Li+ ion, whereas a Br atom gains one electron to form a stable Br- ion.

6.3 The atoms in Group 2A (2) lose 2 electrons to form stable 2 + ions.

6.4 The bonding between Na+ and Cl9 ions in NaCl is ionic.

6.5 In an ionic compound, the positive ion has the same name as the atom, and the negative ion uses the first syllable of the atom’s name followed by ide.

6.6 Because most transition elements have more than one positive ion, a Roman numeral is used.

6.7 In ionic compounds, copper can form two different ions, Cu+ and Cu2+. Oxygen forms the ion O29.

ANSWERS TO SELECTED PROBLEMS 6.39

NO2 − CO3

2− HSO4 − PO4

3−

Li+ LiNO2 Lithium

nitrite

Li2CO3 Lithium

carbonate

LiHSO4 Lithium

hydrogen sulfate

Li3PO4 Lithium

phosphate

Cu2+ Cu(NO2)2 Copper(II)

nitrite

CuCO3 Copper(II)

carbonate

Cu(HSO4)2 Copper(II)

hydrogen sulfate

Cu3(PO4)2 Copper(II)

phosphate

Ba2+ Ba(NO2)2 Barium

nitrite

BaCO3 Barium

carbonate

Ba(HSO4)2 Barium

hydrogen sulfate

Ba3(PO4)2 Barium

phosphate

6.41 a. Ba(OH)2 b. NaHSO4 c. Fe(NO2)2 d. Zn3(PO4)2 e. Fe2(CO3)3 6.43 a. CO3

2-, sodium carbonate b. NH4

+, ammonium sulfide c. OH-, calcium hydroxide d. NO2

-, tin(II) nitrite

6.45 a. zinc acetate b. magnesium phosphate c. ammonium chloride d. strontium nitrate e. sodium nitrite

6.47 a. phosphorus tribromide b. dichlorine oxide c. carbon tetrabromide d. hydrogen fluoride e. nitrogen trifluoride

6.49 a. dinitrogen trioxide b. disilicon hexabromide c. tetraphosphorus trisulfide d. phosphorus pentachloride e. selenium hexafluoride

6.51 a. CCl4 b. CO c. PF3 d. N2O4 6.53 a. OF2 b. BCl3 c. N2O3 d. SF6

6.1 a. 1 b. 2 c. 3 d. 1 e. 2

6.3 a. 2 e- lost b. 3 e- gained c. 1 e- gained d. 1 e- lost e. 1 e- gained

6.5 a. Li+ b. F - c. Mg2+ d. Co3+

6.7 a. 29 protons, 27 electrons b. 34 protons, 36 electrons c. 35 protons, 36 electrons d. 26 protons, 24 electrons

6.9 a. Cl- b. Cs+ c. N3- d. Ra2+

6.11 a. lithium b. calcium c. gallium d. phosphide

6.13 a. 8 protons, 10 electrons b. 19 protons, 18 electrons c. 53 protons, 54 electrons d. 20 protons, 18 electrons

6.15 a and c

6.17 a. Na2O b. AlBr3 c. Ba3N2 d. MgF2 e. Al2S3 6.19 a. K+ and S2-, K2S b. Na

+ and N3-, Na3N c. Al3+ and I-, AlI3 d. Ga

3+ and O2-, Ga2O3 6.21 a. aluminum fluoride b. calcium chloride c. sodium oxide d. magnesium phosphide e. potassium iodide f. barium fluoride

6.23 a. iron(II) b. copper(II) c. zinc d. lead(IV) e. chromium(III) f. manganese(II)

6.25 a. tin(II) chloride b. iron(II) oxide c. copper(I) sulfide d. copper(II) sulfide e. cadmium bromide f. mercury(II) chloride

6.27 a. Au3+ b. Fe3+ c. Pb4+ d. Al3+

6.29 a. MgCl2 b. Na2S c. Cu2O d. Zn3P2 e. AuN f. CoF3 6.31 a. CoCl3 b. PbO2 c. AgI d. Ca3N2 e. Cu3P f. CrCl2 6.33 a. K3P b. CuCl2 c. FeBr3 d. MgO

6.35 a. HCO3 - b. NH4

+ c. PO3 3- d. ClO3

-

6.37 a. sulfate b. carbonate c. hydrogen sulfite (bisulfite) d. nitrate

Answers to Selected Problems 181

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182 CHAPTER 6 Ionic and Molecular Compounds

6.55 a. aluminum sulfate b. zinc phosphate c. dinitrogen oxide d. magnesium hydroxide

6.57 a. MgSO4 b. SnF2 c. Al(OH)3 6.59 a. By losing two valence electrons from the third energy level,

magnesium achieves an octet in the second energy level. b. The magnesium ion Mg2+ has the same electron

configuration as Ne (1s22s22p6). c. Group 1A (1) and 2A (2) elements achieve octets by losing

electrons to form compounds. Group 8A (18) elements are stable with octets (or two electrons for helium).

6.61 a. P3- ion b. O atom c. Zn2+ ion d. Fe3+ ion

6.63 a. X = Group 1A (1), Y = Group 6A (16) b. ionic c. X+ and Y2-

d. X2Y e. X2S f. YCl2 g. molecular

6.65

Electron Configurations Cation Anion Formula of Compound

Name of Compound

1s22s22p63s2 1s22s22p3 Mg2+ N3- Mg3N2 Magnesium nitride

1s22s22p63s23p64s1 1s22s22p4 K+ O2- K2O Potassium oxide

1s22s22p63s23p1 1s22s22p63s23p5 Al3+ Cl- AlCl3 Aluminum chloride

6.67 a. nitride b. magnesium c. oxide d. aluminum

6.69 a. 2A (2) b. Mg c. X3N2 6.71 a. Sn4+

b. 50 protons and 46 electrons c. SnO2 d. Sn3(PO4)4

6.73 a. SnS b. PbO2 c. AgCl d. Ca3N2 e. Cu3P f. CrBr2 6.75 a. nitrogen trichloride b. dinitrogen trisulfide c. dinitrogen oxide d. iodine fluoride e. boron trifluoride f. diphosphorus pentoxide

6.77 a. CS b. P2O5 c. H2S d. SCl2 6.79 a. ionic, iron(III) chloride b. ionic, sodium sulfate c. molecular, nitrogen dioxide d. ionic, rubidium sulfide e. molecular, phosphorus pentafluoride f. molecular, carbon tetrafluoride

6.81 a. SnCO3 b. Li3P c. SiCl4 d. Mn2O3 e. P4Se3 f. CaBr2 6.83

Atom or Ion Number of Protons

Number of Electrons

Electrons Lost/ Gained

K+ 19 p+ 18 e- 1 e- lost

Mg2+ 12 p+ 10 e- 2 e- lost

O2- 8 p+ 10 e- 2 e- gained

Al3+ 13 p+ 10 e- 3 e- lost

6.85 a. Group 3A (13) b. Group 6A (16) c. Group 2A (2)

6.87 a. ionic, lithium hydrogen phosphate b. molecular, chlorine trifluoride c. ionic, magnesium chlorite d. molecular, nitrogen trifluoride e. ionic, calcium bisulfate or calcium hydrogen sulfate f. ionic, potassium perchlorate g. ionic, gold(III) sulfite

M06_TIMB8119_06_SE_C06.indd 182 11/30/18 7:58 AM

183

Max a six-year-old dog, is listless, drinking large amounts of water, and not eating his food. His owner takes Max to his veterinarian, Dr. Evans, for an examination. Dr. Evans weighs Max and obtains a blood sample for a blood chemistry profile, which determines the overall health, detects any metabolic disorders, and measures the concentration of electrolytes.

The results of the lab tests for Max indicate that the white blood count is elevated, which may indicate an infection or inflammation. His electrolytes, which are also indicators of good health, were all in the normal ranges. The electrolyte chloride (Cl-) was in the normal range of 0.106 to 0.118 mol/L, and the electrolyte sodium (Na+) has a normal range of 0.144 to 0.160 mol/L. Potassium (K+), another important electrolyte, has a normal range of 0.0035 to 0.0058 mol/L.

CAREER

Veterinarian Veterinarians care for domesticated pets such as dogs, cats, rats, and birds. Some veterinarians specialize in the treatment of large animals, such as horses and cattle. Veterinarians interact with pet owners as they give advice about feeding, behavior, and breeding. In the assessment of an ill animal, they record the animal’s symptoms and medical history including dietary intake, medications, eating habits, weight, and any signs of disease.

To diagnose health problems, veterinarians perform laboratory tests on animals including a complete blood count and urinalysis. They also obtain tissue and blood samples, as well as vaccinate against distemper, rabies, and other diseases. Veterinarians prescribe medication if an animal has an infection or illness. If an animal is injured, a veterinarian treats wounds, and sets fractures. A veterinarian also performs surgery such as neutering and spaying, provides dental cleanings, removes tumors, and euthanizes animals.

After Dr. Evans obtained the lab results from Max’s tests, she determined that Max had a low-grade infection. You can see how Dr. Evans treated Max for the infection in the UPDATE Prescription for Max, page 204, and learn about the compounds that Max was given.

Chemical Quantities 7

UPDATE Prescriptions for Max

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184 CHAPTER 7 Chemical Quantities

7.1 The Mole LEARNING GOAL Use Avogadro’s number to calculate the number of particles in a given number of moles. Calculate the number of moles of an element in a given number of moles of a compound.

At the grocery store, you buy eggs by the dozen or soda by the case. In an office-supply store, pencils are ordered by the gross and paper by the ream. Common terms such as dozen, case, gross, and ream are used to count the number of items present. For example, when you buy a dozen eggs, you know you will get 12 eggs in the carton.

REVIEW Writing Numbers in Scientific

Notation (1.5)

Counting Significant Figures (2.2)

Using Significant Figures in Calculations (2.3)

Writing Conversion Factors from Equalities (2.5)

Using Conversion Factors (2.6)

LOOKING AHEAD

7.1 The Mole 184 7.2 Molar Mass 188 7.3 Calculations Using

Molar Mass 190 7.4 Mass Percent

Composition 194 7.5 Empirical Formulas 197 7.6 Molecular Formulas 201

Avogadro’s Number In chemistry, particles such as atoms, molecules, and ions are counted by the mole (abbre- viated mol in calculations), which contains 6.022 * 1023 items. This value, known as Avogadro’s number, is a very big number because atoms are so small that it takes an extremely large number of atoms to provide a sufficient amount to weigh and use in chemical reactions. Avogadro’s number is named for Amedeo Avogadro (1776–1856), an Italian physicist.

Avogadro’s Number

6.022 * 1023 = 602 200 000 000 000 000 000 000

One mole of any element always contains Avogadro’s number of atoms. For example, 1 mol of carbon contains 6.022 * 1023 carbon atoms; 1 mol of aluminum contains 6.022 * 1023 aluminum atoms; 1 mol of sulfur contains 6.022 * 1023 sulfur atoms.

1 mol of an element = 6.022 * 1023 atoms of that element

Avogadro’s number tells us that 1 mol of a compound contains 6.022 * 1023 of the particular type of particles that make up that compound. One mole of a molecular com- pound contains Avogadro’s number of molecules. For example, 1 mol of CO2 contains 6.022 * 1023 molecules of CO2. For an ionic compound, 1 mol contains Avogadro’s num- ber of formula units, which are the groups of ions represented by its formula. For the ionic formula, NaCl, 1 mol contains 6.022 * 1023 formula units of NaCl (Na+, Cl-). TABLE 7.1 gives examples of the number of particles in some 1-mol quantities.

CORE CHEMISTRY SKILL Converting Particles to Moles

144 pencils = 1 gross

500 sheets = 1 ream

12 eggs = 1 dozen

Collections of items include dozen, gross, ream, and mole.

TABLE 7.1 Number of Particles in 1-Mol Quantities Substance Number and Type of Particles

1 mol of Al 6.022 * 1023 atoms of Al

1 mol of Fe 6.022 * 1023 atoms of Fe

1 mol of water (H2O) 6.022 * 1023 molecules of H2O

1 mol of vitamin C (C6H8O6) 6.022 * 1023 molecules of vitamin C

1 mol of NaCl 6.022 * 1023 formula units of NaCl

1 mol of K3PO4 6.022 * 1023 formula units of K3PO4

One mole of sulfur contains 6.022 * 1023 sulfur atoms.

M07_TIMB8119_06_SE_C07.indd 184 11/30/18 8:03 AM

7.1 The Mole 185

Using Avogadro’s Number as a Conversion Factor We use Avogadro’s number as a conversion factor to convert between the moles of a substance and the number of particles it contains.

6.022 * 1023 particles 1 mol

and 1 mol

6.022 * 1023 particles

For example, we use Avogadro’s number to convert 4.00 mol of iron to atoms of iron.

4.00 mol Fe * 6.022 * 1023 atoms Fe

1 mol Fe = 2.41 * 1024 atoms of Fe

Avogadro’s number as a conversion factor

We also use Avogadro’s number to convert 3.01 * 1024 molecules of CO2 to moles of CO2.

3.01 * 1024 molecules CO2 * 1 mol CO2

6.022 * 1023 molecules CO2 = 5.00 mol of CO2

Avogadro’s number as a conversion factor

In calculations that convert between moles and particles, the number of moles will be a small number compared to the number of atoms or molecules, which will be large as shown in Sample Problem 7.1.

ENGAGE 7.1 Why is 0.20 mol of aluminum a small number, but the number 1.2 * 1023 atoms of aluminum in 0.20 mol is a large number?

SAMPLE PROBLEM 7.1 Calculating the Number of Molecules

TRY IT FIRST

How many molecules are present in 1.75 mol of carbon dioxide?

CO2 molecules

The solid form of carbon dioxide is known as “dry ice.”

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

1.75 mol of CO2 molecules of CO2 Avogadro’s number

STEP 2 Write a plan to convert moles to particles.

STEP 3 Use Avogadro’s number to write conversion factors.

1 mol of CO2 = 6.022 * 1023 molecules of CO2 6.022 * 1023 molecules CO2

1 mol CO2 and

1 mol CO2

6.022 * 1023 molecules CO2

Avogadro’s numbermoles of CO2 molecules of CO2

M07_TIMB8119_06_SE_C07.indd 185 11/30/18 8:03 AM

186 CHAPTER 7 Chemical Quantities

Moles of Elements in a Chemical Compound We have seen that the subscripts in a chemical formula indicate the number of atoms of each type of element in a compound. For example, aspirin, C9H8O4, is a drug used to reduce pain and inflammation in the body. Using the subscripts in the chemical formula of aspirin shows that there are 9 carbon atoms, 8 hydrogen atoms, and 4 oxygen atoms. The subscripts of the formula of aspirin, C9H8O4, also tell us the number of moles of each element in 1 mol of aspirin: 9 mol of C atoms, 8 mol of H atoms, and 4 mol of O atoms.

ENGAGE 7.2 Why does 1 mol of Zn(C2H3O2)2, a dietary supplement, contain 1 mol of Zn, 4 mol of C, 6 mol of H, and 4 mol of O?

Number of atoms in 1 molecule Oxygen (O)Hydrogen (H)Carbon (C)

Aspirin C9H8O4

Every molecule of aspirin C9H8O4 contains 9 C atoms, 8 H atoms, and 4 O atoms.

C9H8O4

Atoms in 1 molecule 9 atoms of C

The chemical formula subscripts specify the

8 atoms of H 4 atoms of O Carbon Hydrogen Oxygen

Moles of each element in 1 mol 9 mol of C 8 mol of H 4 mol of O

Using a Chemical Formula to Write Conversion Factors Using the subscripts from the formula, C9H8O4, we can write the conversion factors for each of the elements in 1 mol of aspirin:

9 mol C 1 mol C9H8O4

8 mol H 1 mol C9H8O4

4 mol O 1 mol C9H8O4

1 mol C9H8O4 9 mol C

1 mol C9H8O4 8 mol H

1 mol C9H8O4 4 mol O

STEP 4 Set up the problem to calculate the number of particles.

6.022 * 1023 molecules CO2 1 mol CO2

= 1.05 * 1024 molecules of CO2 1.75 mol CO2 *

SELF TEST 7.1

a. How many moles of water, H2O, contain 2.60 * 1023 molecules of water? b. How many molecules are in 2.65 mol of dinitrogen oxide, N2O, which is an anesthetic

called “laughing gas”?

ANSWER

a. 0.432 mol of H2O b. 1.60 * 1024 molecules of N2O PRACTICE PROBLEMS

Try Practice Problems 7.1 to 7.4

M07_TIMB8119_06_SE_C07.indd 186 11/30/18 8:03 AM

7.1 The Mole 187

PRACTICE PROBLEMS

7.1 The Mole 7.1 What is a mole?

7.2 What is Avogadro’s number?

7.3 Calculate each of the following: a. number of C atoms in 0.500 mol of C b. number of SO2 molecules in 1.28 mol of SO2 c. moles of Fe in 5.22 * 1022 atoms of Fe d. moles of C2H6O in 8.50 * 1024 molecules of C2H6O 7.4 Calculate each of the following: a. number of Li atoms in 4.5 mol of Li b. number of CO2 molecules in 0.0180 mol of CO2 c. moles of Cu in 7.8 * 1021 atoms of Cu d. moles of C2H6 in 3.75 * 1023 molecules of C2H6 7.5 Calculate each of the following quantities in 2.00 mol of H3PO4: a. moles of H b. moles of O c. atoms of P d. atoms of O

7.6 Calculate each of the following quantities in 0.185 mol of C6H14O: a. moles of C b. moles of O c. atoms of H d. atoms of C

Applications

7.7 Quinine, C20H24N2O2, is a component of tonic water and bitter lemon.

a. How many moles of H are in 1.5 mol of quinine? b. How many moles of C are in 5.0 mol of quinine? c. How many moles of N are in 0.020 mol of quinine?

7.8 Aluminum sulfate, Al2(SO4)3, is used in some antiperspirants. a. How many moles of O are present in 3.0 mol of Al2(SO4)3? b. How many moles of aluminum ions (Al3+) are present in

0.40 mol of Al2(SO4)3? c. How many moles of sulfate ions (SO4

2-) are present in 1.5 mol of Al2(SO4)3?

SAMPLE PROBLEM 7.2 Calculating the Moles of an Element in a Compound

TRY IT FIRST

How many moles of carbon are present in 1.50 mol of aspirin, C9H8O4?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

1.50 mol of aspirin, C9H8O4 moles of C subscripts in formula

STEP 2 Write a plan to convert moles of a compound to moles of an element.

Subscriptmoles of C9H8O4 moles of C

STEP 3 Write the equalities and conversion factors using subscripts.

9 mol C and

1 mol of C9H8O4 = 9 mol of C

9 mol C 1 mol C9H8O4

1 mol C9H8O4

STEP 4 Set up the problem to calculate the moles of an element.

13.5 mol of C* = 9 mol C

1.50 mol C9H8O4 1 mol C9H8O4

SELF TEST 7.2

a. How many moles of aspirin, C9H8O4, contain 0.480 mol of O? b. How many moles of C are in 0.125 mol of aspirin?

ANSWER

a. 0.120 mol of aspirin b. 1.13 mol of C PRACTICE PROBLEMS

Try Practice Problems 7.5 to 7.10

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188 CHAPTER 7 Chemical Quantities

7.9 Naproxen is used to treat pain and inflammation caused by arthritis. Naproxen has a formula of C14H14O3.

a. How many moles of C are present in 2.30 mol of naproxen? b. How many moles of H are present in 0.444 mol of

naproxen? c. How many moles of O are present in 0.0765 mol of naproxen?

7.10 Benadryl is an over-the-counter drug used to treat allergy symptoms. The formula of Benadryl is C17H21NO.

a. How many moles of C are present in 0.733 mol of Benadryl? b. How many moles of H are present in 2.20 mol of Benadryl? c. How many moles of N are present in 1.54 mol of Benadryl?

7.2 Molar Mass LEARNING GOAL Given the chemical formula of a substance, calculate its molar mass.

A single atom or molecule is much too small to weigh, even on the most accurate balance. In fact, it takes a huge number of atoms or molecules to make enough of a substance for you to see. An amount of water that contains Avogadro’s number of water molecules is only a few sips. However, in the laboratory, we can use a balance to weigh out Avogadro’s number of particles for 1 mol of substance.

For any element, the quantity called molar mass is the quantity in grams that equals the atomic mass of that element. We are counting 6.022 * 1023 atoms of an element when we weigh out the number of grams equal to its molar mass. For example, carbon has an atomic mass of 12.01 on the periodic table. This means 1 mol of carbon atoms has a mass of 12.01 g. Then to obtain 1 mol of carbon atoms, we would need to weigh out 12.01 g of carbon. Thus, the molar mass of carbon is found by looking at its atomic mass on the periodic table.

118

Og

6.022 * 1023 atoms of C

1 mol of C atoms

12.01 g of C atoms

114

1

3 4

11 12

19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

2

5 6 7 8 9 10

13 14 15 16 17 18

37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54

86858483828180797877767574737257*5655

87 88 89† 104 105 106 107 108 109 110 111 112

H

Li Be

Na Mg

K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn

He

B C N O F Ne

Al Si P S Cl Ar

Ga Ge As Se Br Kr

Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe

Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn

Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg Cn Fl Lv

107.9

47

Ag 12.01

6

C 32.07

16

S 1 mol of

silver atoms has a mass of 107.9 g

1 mol of carbon atoms

has a mass of 12.01 g

1 mol of sulfur atoms has a mass of 32.07 g

113

Nh 115

Mc 116 117

Ts

The molar mass of an element is obtained from the periodic table.

Molar Mass of a Compound To determine the molar mass of a compound, multiply the molar mass of each element by its subscript in the formula and add the results as shown in Sample Problem 7.3. In this text, we round the molar mass of an element to the hundredths place (0.01) or use at least four significant figures for calculations. FIGURE 7.1 shows some 1-mol quantities of substances.

CORE CHEMISTRY SKILL Calculating Molar Mass

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7.2 Molar Mass 189

Lithium carbonate is used to treat bipolar disorder.

FIGURE 7.1 Each beaker contains a 1-mol sample of an element or compound.

6.941 g Li 1 mol Li

16.00 g O 1 mol O

12.01 g C 1 mol C

STEP 2 Multiply each molar mass by the number of moles (subscript) in the formula.

Grams from 2 mol of Li:

6.941 g Li * = 13.88 g of Li2 mol Li

1 mol Li

Grams from 1 mol of C:

12.01 g C * = 12.01 g of C1 mol C

1 mol C

Grams from 3 mol of O:

16.00 g O * = 48.00 g of O3 mol O

1 mol O

SAMPLE PROBLEM 7.3 Calculating Molar Mass

TRY IT FIRST

Calculate the molar mass for lithium carbonate, Li2CO3, used to treat bipolar disorder.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

formula Li2CO3 molar mass of Li2CO3 periodic table

STEP 1 Obtain the molar mass of each element.

ENGAGE 7.3 How is the molar mass for K2Cr2O7 obtained?

(342.3 g)(294.2 g)(58.44 g)(55.85 g)

S Fe NaCl

1-Mol Quantities

K2Cr2O7 C12H22O11 (32.07 g)

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190 CHAPTER 7 Chemical Quantities

PRACTICE PROBLEMS

7.2 Molar Mass 7.11 Calculate the molar mass for each of the following: a. Cl2 b. C3H6O3 c. Mg3(PO4)2 7.12 Calculate the molar mass for each of the following: a. O2 b. KH2PO4 c. Fe(ClO4)3 7.13 Calculate the molar mass for each of the following: a. AlF3 b. C2H4Cl2 c. SnF2 7.14 Calculate the molar mass for each of the following: a. C4H8O4 b. Ga2(CO3)3 c. KBrO4

Applications

7.15 Calculate the molar mass for each of the following: a. KCl, salt substitute b. Fe2O3, rust c. C19H20FNO3, Paxil, antidepressant

7.16 Calculate the molar mass for each of the following: a. FeSO4, iron supplement b. Al2O3, absorbent and abrasive c. C7H5NO3S, saccharin, artificial sweetener

7.17 Calculate the molar mass for each of the following: a. Al2(SO4)3, antiperspirant b. KC4H5O6, cream of tartar c. C16H19N3O5S, amoxicillin, antibiotic

7.18 Calculate the molar mass for each of the following: a. C3H8O, rubbing alcohol b. (NH4)2CO3, baking powder c. Zn(C2H3O2)2, zinc supplement

7.19 Calculate the molar mass for each of the following: a. C8H9NO2, acetaminophen used in Tylenol b. Ca3(C6H5O7)2, calcium supplement c. C17H18FN3O3, Cipro, used to treat a range of bacterial

infections

7.20 Calculate the molar mass for each of the following: a. CaSO4, calcium sulfate, used to make casts to protect broken

bones b. C6H12N2O4Pt, Carboplatin, used in chemotherapy c. C12H18O, Propofol, used to induce anesthesia during surgery

7.3 Calculations Using Molar Mass LEARNING GOAL Use molar mass to convert between grams and moles.

The molar mass of an element is one of the most useful conversion factors in chemistry because it converts moles of a substance to grams, or grams to moles. For example, 1 mol of silver has a mass of 107.9 g. To express molar mass of Ag as an equality, we write

1 mol of Ag = 107.9 g of Ag

From this equality for the molar mass, two conversion factors can be written as

107.9 g Ag

1 mol Ag and

1 mol Ag

107.9 g Ag

Sample Problem 7.4 shows how the molar mass of silver is used as a conversion factor.

CORE CHEMISTRY SKILL Using Molar Mass as a Conversion

Factor

STEP 3 Calculate the molar mass by adding the masses of the elements.

2 mol of Li = 13.88 g of Li 1 mol of C = 12.01 g of C 3 mol of O = 48.00 g of O Molar mass of Li2CO3 = 73.89 g

SELF TEST 7.3

Calculate the molar mass for each of the following: a. salicylic acid, C7H6O3, used to treat skin conditions such as acne, psoriasis, and dandruff b. vitamin D3, C27H44O, a vitamin that prevents rickets

ANSWER

a. 138.12 g b. 384.6 g PRACTICE PROBLEMS

Try Practice Problems 7.11 to 7.20

M07_TIMB8119_06_SE_C07.indd 190 11/30/18 8:03 AM

7.3 Calculations Using Molar Mass 191

Silver metal is used to make jewelry.

Writing Conversion Factors for the Molar Mass of a Compound The conversion factors for a compound are also written from the molar mass. For example, the molar mass of the compound H2O is written

1 mol of H2O = 18.02 g of H2O

From this equality, conversion factors for the molar mass of H2O are written as

18.02 g H2O

1 mol H2O and

1 mol H2O

18.02 g H2O

We can now change from moles to grams, or grams to moles, using the conversion factors derived from the molar mass of a compound shown in Sample Problem 7.5. (Remember, you must determine the molar mass first.)

STEP 3 Determine the molar mass and write conversion factors.

Molar massmoles of Ag grams of Ag

STEP 4 Set up the problem to convert moles to grams.

107.9 g Ag and

1 mol of Ag = 107.9 g of Ag

107.9 g Ag1 mol Ag

1 mol Ag

* 107.9 g Ag

0.750 mol Ag = 80.9 g of Ag 1 mol Ag

SELF TEST 7.4

a. A dentist orders 24.4 g of gold (Au) to prepare dental crowns and fillings. Calculate the number of moles of gold in the order.

b. Pencil lead contains graphite, which is the element carbon. Calculate the number of grams in 0.520 mol of carbon.

ANSWER

a. 0.124 mol of Au b. 6.25 g of C

SAMPLE PROBLEM 7.4 Converting Between Moles and Grams

TRY IT FIRST

Silver metal is used in the manufacture of tableware, mirrors, jewelry, and dental alloys. If the design for a piece of jewelry requires 0.750 mol of silver, how many grams of silver are needed?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

0.750 mol of Ag grams of Ag molar mass

STEP 2 Write a plan to convert moles to grams.

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192 CHAPTER 7 Chemical Quantities

FIGURE 7.2 The moles of a compound are related to its mass in grams by molar mass, to the number of molecules (or formula units) by Avogadro’s number, and to the moles of each element by the subscripts in the formula.

Grams of compound

Molar Mass (g/mol) Moles of

compound

Formula Subscripts

Formula Subscripts

Avogadro’s Number Molecules (or

formula units)

Mass Moles Particles

Grams of element

Molar Mass (g/mol) Moles of

element

Avogadro’s Number Atoms

(or ions)

PRACTICE PROBLEMS Try Practice Problems 7.21 to 7.30

FIGURE 7.2 gives a summary of the calculations to show the connections between the moles of a compound, its mass in grams, the number of molecules (or formula units if ionic), and the moles and atoms of each element in that compound in the following flowchart:

ENGAGE 7.5 Why are there more grams of chlorine than grams of fluorine in 1 mol of Freon-12, CCl2F2?

ENGAGE 7.4 Which conversion steps are needed to calculate the number of atoms of H in 5.00 g of CH4?

SAMPLE PROBLEM 7.5 Converting Between Mass and Moles of a Compound

TRY IT FIRST

A salt shaker contains 73.7 g of NaCl. How many moles of NaCl are present?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

73.7 g of NaCl moles of NaCl molar mass

STEP 2 Write a plan to convert grams to moles.

grams of NaCl moles of NaCl Molar mass

STEP 3 Determine the molar mass and write conversion factors.

(1 * 22.99) + (1 * 35.45) = 58.44 g/mol

STEP 4 Set up the problem to convert grams to moles.

58.44 g NaCl and

1 mol of NaCl = 58.44 g of NaCl

58.44 g NaCl1 mol NaCl

1 mol NaCl

58.44 g NaCl 1.26 mol of NaCl*

1 mol NaCl 73.7 g NaCl =

SELF TEST 7.5

a. One tablet of an antacid contains 680. mg of CaCO3. How many moles of CaCO3 are present?

b. A different brand of antacid contains 6.9 * 10-3 mol of Mg(OH)2. How many grams of Mg(OH)2 are present?

ANSWER

a. 0.006 79 or 6.79 * 10-3 mol of CaCO3 b. 0.40 g of Mg(OH)2

Table salt is sodium chloride, NaCl.

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7.3 Calculations Using Molar Mass 193

FIGURE 7.2 gives a summary of the calculations to show the connections between the moles of a compound, its mass in grams, the number of molecules (or formula units if ionic), and the moles and atoms of each element in that compound in the following flowchart:

ENGAGE 7.5 Why are there more grams of chlorine than grams of fluorine in 1 mol of Freon-12, CCl2F2?

ENGAGE 7.4 Which conversion steps are needed to calculate the number of atoms of H in 5.00 g of CH4?

We can now convert the mass in grams of a compound to the mass of one of the elements in a compound as shown in Sample Problem 7.6.

PRACTICE PROBLEMS Try Practice Problems 7.31 to 7.36

SAMPLE PROBLEM 7.6 Converting Between Grams of Compound and Grams of Element

TRY IT FIRST

Hot packs are used to reduce muscle aches, inflammation, and muscle spasms. A hot pack consists of a bag of water and an inner bag containing 10.2 g of CaCl2. When the bag is broken, the CaCl2 dissolves in the water and heat is released. How many grams of Cl are in the CaCl2 in the inner bag?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

10.2 g of CaCl2 grams of Cl molar mass, subscript for Cl

STEP 2 Write a plan to convert grams of a compound to grams of an element.

STEP 3 Write the equalities and conversion factors for molar mass and mole factors.

Molar massgrams of CaCl2 moles of CaCl2 Mole factor Molar mass grams of Clmoles of Cl

Molar massgrams of CaCl2 moles of CaCl2 Mole factor Molar mass grams of Clmoles of Cl

110.98 g CaCl2 1 mol CaCl2

and

1 mol of CaCl2 = 110.98 g of CaCl2 1 mol CaCl2

110.98 g CaCl2

1 mol CaCl2 2 mol Cl

and

1 mol of CaCl2 = 2 mol of Cl 2 mol Cl

1 mol CaCl2

35.45 g Cl

1 mol Cl and

1 mol of Cl = 35.45 g of Cl 1 mol Cl

35.45 g Cl

110.98 g CaCl2 1 mol CaCl2

and

1 mol of CaCl2 = 110.98 g of CaCl2 1 mol CaCl2

110.98 g CaCl2

1 mol CaCl2 2 mol Cl

and

1 mol of CaCl2 = 2 mol of Cl 2 mol Cl

1 mol CaCl2

35.45 g Cl

1 mol Cl and

1 mol of Cl = 35.45 g of Cl 1 mol Cl

35.45 g Cl

STEP 4 Set up the problem to convert grams of a compound to grams of an element.

110.98 g CaCl2

1 mol CaCl210.2 g CaCl2 * 1 mol CaCl2

2 mol Cl *

1 mol Cl

35.45 g Cl * = 6.52 g of Cl

SELF TEST 7.6

a. Tin(II) fluoride (SnF2) is added to toothpaste to strengthen tooth enamel. How many grams of tin(II) fluoride contain 1.46 g of F?

b. Sodium nitrite, NaNO2, is used to give a longer storage life to ham and hot dogs. A package of ten hot dogs contains 110 milligrams of sodium nitrite. How many milligrams of N are in the package?

ANSWER

a. 6.02 g of SnF2 b. 22 mg of N

In some hot packs, heat is generated when CaCl2 dissolves in water.

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194 CHAPTER 7 Chemical Quantities

PRACTICE PROBLEMS

7.3 Calculations Using Molar Mass 7.21 Calculate the mass, in grams, for each of the following: a. 1.50 mol of Na b. 2.80 mol of Ca c. 0.125 mol of CO2 d. 0.0485 mol of Na2CO3 e. 7.14 * 102 mol of PCl3 7.22 Calculate the mass, in grams, for each of the following: a. 5.12 mol of Al b. 0.75 mol of Cu c. 3.52 mol of MgBr2 d. 0.145 mol of C2H6O e. 2.08 mol of (NH4)2SO4 7.23 Calculate the mass, in grams, in 0.150 mol of each of the

following: a. Ne b. I2 c. Na2O d. Ca(NO3)2 e. C6H14 7.24 Calculate the mass, in grams, in 2.28 mol of each of the

following: a. Pd b. SO3 c. C3H6O3 d. Mg(HCO3)2 e. SF6 7.25 Calculate the number of moles in each of the following: a. 82.0 g of Ag b. 0.288 g of C c. 15.0 g of ammonia, NH3 d. 7.25 g of CH4 e. 245 g of Fe2O3 7.26 Calculate the number of moles in each of the following: a. 85.2 g of Ni b. 144 g of K c. 6.4 g of H2O d. 308 g of BaSO4 e. 252.8 g of fructose, C6H12O6 7.27 Calculate the number of moles in 25.0 g of each of the

following: a. He b. O2 c. Al(OH)3 d. Ga2S3 e. C4H10, butane

7.28 Calculate the number of moles in 4.00 g of each of the following:

a. Au b. SnO2 c. CS2 d. Ca3N2 e. C6H8O6, vitamin C

7.29 Calculate the mass, in grams, of C in each of the following: a. 0.688 g of CO2 b. 275 g of C3H6 c. 1.84 g of C2H6O d. 73.4 g of C8H16O2 7.30 Calculate the mass, in grams, of N in each of the following: a. 0.82 g of NaNO3 b. 40.0 g of (NH4)3P c. 0.464 g of N2H4 d. 1.46 g of N4O6

Applications

7.31 Propane gas, C3H8, is used as a fuel for many barbecues. a. How many moles of H are in 34.0 g of propane? b. How many grams of C are in 1.50 mol of propane? c. How many grams of C are in 34.0 g of propane? d. How many grams of H are in 0.254 g of propane?

7.32 Allyl sulfide, (C3H5)2S, gives garlic, onions, and leeks their characteristic odor.

The characteristic odor of garlic is due to a sulfur- containing compound.

a. How many moles of S are in 23.2 g of allyl sulfide? b. How many grams of H are in 0.75 mol of allyl sulfide? c. How many grams of C are in 4.66 g of allyl sulfide? d. How many grams of S are in 15.0 g of allyl sulfide?

7.33 a. The compound MgSO4, Epsom salts, is used to soothe sore feet and muscles. How many grams will you need to prepare a bath containing 5.00 mol of Epsom salts?

b. Potassium iodide, KI, is used as an expectorant. How many grams of KI are in 0.450 mol of potassium iodide?

7.34 a. Cyclopropane, C3H6, is an anesthetic given by inhalation. How many grams are in 0.25 mol of cyclopropane?

b. The sedative Demerol hydrochloride has the formula C15H22ClNO2. How many grams are in 0.025 mol of Demerol hydrochloride?

7.35 Dinitrogen oxide (or nitrous oxide), N2O, also known as laughing gas, is widely used as an anesthetic in dentistry.

a. How many grams of the compound are in 1.50 mol of dinitrogen oxide?

b. How many moles of the compound are in 34.0 g of dinitrogen oxide?

c. How many grams of N are in 34.0 g of dinitrogen oxide?

7.36 Chloroform, CHCl3, was formerly used as an anesthetic but its use was discontinued due to respiratory and cardiac failure.

a. How many grams of the compound are in 0.122 mol of chloroform?

b. How many moles of the compound are in 15.6 g of chloroform?

c. How many grams of Cl are in 26.7 g of chloroform?

7.4 Mass Percent Composition LEARNING GOAL Given the formula of a compound, calculate the mass percent composition.

Because the atoms of the elements in a compound are combined in a definite mole ratio, they are also combined in a definite proportion by mass. When we know the mass of an element in the mass of a sample of a compound, we can calculate its mass percent composition or mass percent, which is the mass of an element divided by the total mass of the compound and multiplied by 100%. For example, we can calculate the mass percent of N if we find from experiment that 7.64 g of N are present in 12.0 g of N2O, “laughing gas,” which is used as an anesthetic for surgery and in dentistry.

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7.4 Mass Percent Composition 195

From the grams of N and the grams of N2O, we calculate the mass percent of nitrogen as follows:

Mass percent of an element = mass of an element

total mass of the compound * 100%

Mass percent of N = 7.64 g N

12.0 g N2O * 100% = 63.7% N

Mass Percent Composition Using Molar Mass We can also calculate mass percent composition of a compound from its molar mass. Then the total mass of each element is divided by the molar mass of the compound and multiplied by 100%.

Mass percent composition = mass of each element

molar mass of the compound * 100%

CORE CHEMISTRY SKILL Calculating Mass Percent

Composition

SAMPLE PROBLEM 7.7 Calculating Mass Percent Composition from Molar Mass

TRY IT FIRST

The odor of pears is due to the compound propyl acetate, which has a formula of C5H10O2. What is the mass percent composition of propyl acetate?

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

C5H10O2 mass percent composition: %C, %H, %O periodic table

STEP 1 Determine the total mass of each element in the molar mass of a formula.

5 mol C * 12.01 g C

1 mol C = 60.05 g of C

10 mol H * 1.008 g H

1 mol H = 10.08 g of H

2 mol O * 16.00 g O

1 mol O = 32.00 g of O

Molar mass of C5H10O2 = 102.13 g of C5H10O2

STEP 2 Divide the total mass of each element by the molar mass and multiply by 100%.

Mass % C = 60.05 g C

102.13 g C5H10O2 * 100% = 58.80% C

Mass % H = 10.08 g H

102.13 g C5H10O2 * 100% = 9.870% H

Mass % O = 32.00 g O

102.13 g C5H10O2 * 100% = 31.33% O

The total mass percent for all the elements in the compound should equal 100%. In some cases, because of rounding off, the sum of the mass percents may not total exactly 100%.

58.80% C + 9.870% H + 31.33% O = 100.00%

The odor of pears is due to propyl acetate, C5H10O2.

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196 CHAPTER 7 Chemical Quantities

PRACTICE PROBLEMS

7.4 Mass Percent Composition 7.37 Calculate the mass percent composition of each of the

following: a. 4.68 g of Si and 5.32 g of O b. 5.72 g of C and 1.28 g of H c. 16.1 of Na, 22.5 g of S, and 11.3 g of O d. 6.22 g of C, 1.04 g of H, and 4.14 g of O

7.38 Calculate the mass percent composition of each of the following:

a. 0.890 g of Ba and 1.04 g of Br b. 3.82 g of K and 1.18 g of I c. 3.29 of N, 0.946 g of H, and 3.76 g of S d. 4.14 g of C, 0.695 g of H, and 2.76 g of O

PRACTICE PROBLEMS Try Practice Problems 7.37 to 7.42

SELF TEST 7.7

a. Ethylene glycol, C2H6O2, used as automobile antifreeze, is a sweet-tasting liquid, which is toxic to humans and animals. What is the mass percent composition of ethylene glycol?

b. Iron(II) sulfate, FeSO4, is a dietary supplement. What is the mass percent composition of iron(II) sulfate?

ANSWER

a. 38.70% C; 9.744% H; 51.56% O b. 36.76% Fe; 21.11% S; 42.13% O

Chemistry Link to the Environment Fertilizers

Every year in the spring, homeowners and farmers add fertilizers to the soil to produce greener lawns and larger crops. Plants require several nutrients, but the major ones are nitrogen, phosphorus, and potassium. Nitrogen promotes green growth, phosphorus promotes strong root development for strong plants and abundant f lowers, and potassium helps plants defend against diseases and weather extremes. The numbers on a package of fertilizer give the percent- ages each of N, P, and K by mass. For example, the set of numbers 30–3–4 describes a fertilizer that contains 30% N, 3% P, and 4% K.

The major nutrient, nitrogen, is present in huge quantities as N2 in the atmosphere, but plants cannot utilize nitrogen in this form. Bacteria in the soil convert atmospheric N2 to usable forms by nitrogen fixation. To provide additional nitrogen to plants, several types of nitrogen-containing chemicals, including ammonia, nitrates, and ammonium compounds, are added to soil. The nitrates are absorbed directly, but ammonia and ammonium salts are first converted to nitrates by soil bacteria.

The percent nitrogen depends on the type of nitrogen compound used in the fertilizer. The percent nitrogen by mass in each type is calculated using mass percent composition.

Type of Fertilizer Percent Nitrogen by Mass

NH3 14.01 g N

17.03 g NH3 * 100% = 82.27% N

NH4NO3 28.02 g N

80.05 g NH4NO3 * 100% = 35.00% N

(NH4)2HPO4 28.02 g N

132.06 g (NH4)2HPO4 * 100% = 21.22% N

(NH4)2SO4 28.02 g N

132.15 g (NH4)2SO4 * 100% = 21.20% N

The choice of a fertilizer depends on its use and convenience. A fertilizer can be prepared as crystals or a powder, in a liquid solu- tion, or as a gas such as ammonia. The ammonia and ammonium fertilizers are water soluble and quick-acting. Other forms may be made to slow-release by enclosing water-soluble ammonium salts in a thin plastic coating. The most commonly used fertilizer is NH4NO3 because it is easy to apply and has a high percent of N by mass.

The label on a bag of fertilizer states the percentages of N, P, and K.

Nitrogen (N) 30% Phosphorus (P) 3% Potassium (K) 4%

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7.5 Empirical Formulas 197

7.5 Empirical Formulas LEARNING GOAL From the mass percent composition, calculate the empirical formula for a compound.

Up to now, the formulas you have seen have been molecular formulas, which are the actual formulas of compounds. If we write a formula that represents the lowest whole- number ratio of the atoms in a compound, it is called the simplest or empirical formula. For example, the compound benzene, with molecular formula C6H6, has the empirical formula CH. Some molecular formulas and their empirical formulas are shown in TABLE 7.2.

Applications

7.39 Calculate the mass percent composition of each of the following:

a. MgF2, magnesium fluoride b. Ca(OH)2, calcium hydroxide c. C4H8O4, erythrose, a carbohydrate d. (NH4)3PO4, ammonium phosphate, fertilizer e. C17H19NO3, morphine, a painkiller

7.40 Calculate the mass percent composition of each of the following:

a. CaCl2, calcium chloride b. Na2Cr2O7, sodium dichromate c. C2H3Cl3, trichloroethane, cleaning solvent d. Ca3(PO4)2, calcium phosphate, found in bone and teeth e. C18H36O2, stearic acid, from animal fat

7.41 Calculate the mass percent of N in each of the following: a. N2O5, dinitrogen pentoxide b. NH4Cl, expectorant in cough medicine c. C2H8N2, dimethylhydrazine, rocket fuel d. C9H15N5O, Rogaine, stimulates hair growth e. C14H22N2O, lidocaine, local anesthetic

7.42 Calculate the mass percent of S in each of the following: a. Na2SO4, sodium sulfate b. Al2S3, aluminum sulfide c. SO3, sulfur trioxide d. C2H6SO, dimethylsulfoxide, topical anti-inflammatory e. C10H10N4O2S, sulfadiazine, antibacterial

TABLE 7.2 Examples of Molecular and Empirical Formulas Name Molecular (actual formula) Empirical (simplest formula)

Acetylene C2H2 CH

Benzene C6H6 CH

Ammonia NH3 NH3 Hydrazine N2H4 NH2 Ribose C5H10O5 CH2O

Glucose C6H12O6 CH2O

The empirical formula of a compound is determined by converting the number of grams of each element to moles and finding the lowest whole-number ratio to use as subscripts, as shown in Sample Problem 7.8.

CORE CHEMISTRY SKILL Calculating an Empirical Formula

SAMPLE PROBLEM 7.8 Calculating an Empirical Formula

TRY IT FIRST

A compound of iron and chlorine is used to purify water in water-treatment plants. What is the empirical formula of the compound if experimental analysis shows that a sample of the compound contains 6.87 g of iron and 13.1 g of chlorine?

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

6.87 g of Fe, 13.1 g of Cl empirical formula molar mass

Water-treatment plants use chemicals to purify sewage.

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198 CHAPTER 7 Chemical Quantities

The mass percent composition for any amount of methane is always the same.

Mass percent composition of methane, CH4

Composition, in grams, for 50.0 g of methane, CH4

Composition, in grams, for 100.0 g of methane, CH4

25.1% Hydrogen

25.1 g of H12.6 g

of H

37.4 g of C

74.9% Carbon

74.9 g of C

STEP 1 Calculate the moles of each element.

6.87 g Fe * 1 mol Fe

55.85 g Fe = 0.123 mol of Fe

13.1 g Cl * 1 mol Cl

35.45 g Cl = 0.370 mol of Cl

STEP 2 Divide by the smallest number of moles. The smallest number of moles is 0.123.

0.123 mol Fe

0.123 = 1.00 mol of Fe

0.370 mol Cl

0.123 = 3.01 mol of Cl

STEP 3 Use the lowest whole-number ratio of moles as subscripts. The relation- ship of moles of Fe to moles of Cl is 1 to 3, which we obtain by rounding off 3.01 to 3.

Fe1.00Cl3.01 h Fe1Cl3, written as FeCl3 Empirical formula

SELF TEST 7.8

a. Phosphine is a highly toxic compound used for pest and rodent control. If a sample of phosphine contains 0.456 g of P and 0.0440 g of H, what is its empirical formula?

b. A compound of titanium and oxygen is used to make white pigment. If a sample of this compound contains 1.21 g of Ti and 0.806 g of O, what is its empirical formula?

ANSWER

a. PH3 b. TiO2

Often, the mass percent of each element in a compound is given. The mass percent composition is true for any quantity of the compound. For example, methane, CH4, always has a mass percent composition of 74.9% C and 25.1% H. Thus, if we assume that we have a sample of 100.0 g of the compound, we can determine the grams of each element in that sample and calculate the empirical formula as shown in Sample Problem 7.9.

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7.5 Empirical Formulas 199

SAMPLE PROBLEM 7.9 Calculating an Empirical Formula from the Mass Percent Composition

TRY IT FIRST

Tetrachloroethene is a colorless liquid used for dry cleaning. What is the empirical formula of tetrachloroethene if its mass percent composition is 14.5% C and 85.5% Cl?

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

14.5% C, 85.5% Cl empirical formula molar mass

STEP 1 Calculate the moles of each element. In a sample size of 100. g of this compound, there are 14.5 g of C and 85.5 g of Cl.

14.5 g C * 1 mol C

12.01 g C = 1.21 mol of C

85.5 g Cl * 1 mol Cl

35.45 g Cl = 2.41 mol of Cl

STEP 2 Divide by the smallest number of moles. The smallest number of moles is 1.21.

1.21 mol C

1.21 = 1.00 mol of C

2.41 mol Cl

1.21 = 1.99 mol of Cl

STEP 3 Use the lowest whole-number ratio of moles as subscripts.

C1.00Cl1.99 h C1Cl2 = CCl2

SELF TEST 7.9

a. Sulfate of potash is the common name of a compound used in fertilizers to supply potas- sium and sulfur. What is the empirical formula of this compound if it has a mass percent composition of 44.9% K, 18.4% S, and 36.7% O?

b. Acrylonitrile is a compound used to make a variety of polymers. What is the empirical formula of acrylonitrile if it has a mass percent composition of 67.9% C, 5.7% H, and 26.4% N?

ANSWER

a. K2SO4 b. C3H3N Sulfate of potash supplies sulfur and potassium.

Converting Decimal Numbers to Whole Numbers Sometimes the result of dividing by the smallest number of moles gives a decimal instead of a whole number. Decimal values that are very close to whole numbers can be rounded off. For example, 2.04 rounds off to 2 and 6.98 rounds off to 7. However, a decimal that is greater than 0.1 or less than 0.9 should not be rounded off. Instead, we multiply by a small integer to obtain a whole number. Some multipliers that are typically used are listed in TABLE 7.3.

ENGAGE 7.6 If we obtain a ratio of 1.0 mol of P and 2.5 mol of O, why do we multiply by 2 to obtain an empirical formula of P2O5?

TABLE 7.3 Some Multipliers That Convert Decimals to Whole-Number Subscripts

Decimal Multiplier Example Whole Number

0.20 5 1.20 * 5 = 6 0.25 4 2.25 * 4 = 9 0.33 3 1.33 * 3 = 4 0.50 2 2.50 * 2 = 5 0.67 3 1.67 * 3 = 5

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200 CHAPTER 7 Chemical Quantities

Let us suppose the numbers of moles we obtain give subscripts in the ratio of C1.00H2.33O0.99. While 0.99 rounds off to 1, we cannot round off 2.33. If we multiply 2.33 * 2, we obtain 4.66, which is still not a whole number. If we multiply 2.33 by 3, the answer is 6.99, which rounds off to 7. To complete the empirical formula, all the other subscripts must be multiplied by 3.

C(1.00 * 3)H(2.33 * 3)O(0.99 * 3) = C3.00H6.99O2.97 h C3H7O3

PRACTICE PROBLEMS Try Practice Problems 7.43 to 7.46

Citrus fruits are a good source of vitamin C.

SAMPLE PROBLEM 7.10 Calculating an Empirical Formula Using Multipliers

TRY IT FIRST

Ascorbic acid (vitamin C), found in citrus fruits and vegetables, is important in metabolic reactions in the body, in the synthesis of collagen, and in the prevention of scurvy. If the mass percent composition of ascorbic acid is 40.9% C, 4.58% H, and 54.5% O, what is the empirical formula of ascorbic acid?

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

40.9% C, 4.58% H, 54.5% O empirical formula molar mass

STEP 1 Calculate the moles of each element. In a sample size of 100. g, there are 40.9 g of C, 4.58 g of H, and 54.5 g of O.

40.9 g C * 1 mol C

12.01 g C = 3.41 mol of C

4.58 g H * 1 mol H

1.008 g H = 4.54 mol of H

54.5 g O * 1 mol O

16.00 g O = 3.41 mol of O

STEP 2 Divide by the smallest number of moles. The smallest number of moles is 3.41.

3.41 mol C

3.41 = 1.00 mol of C

4.54 mol H

3.41 = 1.33 mol of H

3.41 mol O

3.41 = 1.00 mol of O

STEP 3 Use the lowest whole-number ratio of moles as subscripts. As calculated, the ratio of moles gives the formula

C1.00H1.33O1.00 Because the subscript for H has a decimal that is greater than 0.1 and less than 0.9, it should not be rounded off. Then, we multiply each of the subscripts by 3 to obtain a whole number for H, which is 4. Thus, the empirical formula of ascorbic acid is C3H4O3.

C(1.00 * 3)H(1.33 * 3)O(1.00 * 3) = C3.00H3.99O3.00 h C3H4O3

SELF TEST 7.10

a. Glyoxylic acid is used by plants and bacteria to convert fats into glucose. What is the empirical formula of glyoxylic acid if it has a mass percent composition of 32.5% C, 2.70% H, and 64.8% O?

b. A compound of boron and oxygen is used to make glass that is resistant to thermal shocks. If this compound contains 31.1% B and 68.9% O, what is its empirical formula?

ANSWER

a. C2H2O3 b. B2O3

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7.6 Molecular Formulas 201

7.6 Molecular Formulas LEARNING GOAL Determine the molecular formula of a substance from the empirical formula and molar mass.

An empirical formula represents the lowest whole-number ratio of atoms in a compound. How- ever, empirical formulas do not necessarily represent the actual number of atoms in a molecule. A molecular formula is related to the empirical formula by a small integer such as 1, 2, or 3.

Molecular formula = small integer * empirical formula

For example, in TABLE 7.4, we see several different compounds that have the same empiri- cal formula, CH2O. The molecular formulas are related to the empirical formulas by small whole numbers (integers). The same relationship is true for the molar mass and empirical formula mass. The molar mass of each of the different compounds is related to the mass of the empirical formula (30.03 g) by the same small integer.

PRACTICE PROBLEMS

7.5 Empirical Formulas 7.43 Calculate the empirical formula for each of the following: a. 3.57 g of N and 2.04 g of O b. 7.00 g of C and 1.75 g of H c. 0.175 g of H, 2.44 g of N, and 8.38 g of O d. 2.06 g of Ca, 2.66 g of Cr, and 3.28 g of O

7.44 Calculate the empirical formula for each of the following: a. 2.90 g of Ag and 0.430 g of S b. 2.22 g of Na and 0.774 g of O c. 2.11 g of Na, 0.0900 g of H, 2.94 g of S, and 5.86 g of O d. 5.52 g of K, 1.45 g of P, and 3.00 g of O

7.45 Calculate the empirical formula for each of the following: a. 70.9% K and 29.1% S b. 55.0% Ga and 45.0% F

c. 69.6% Mn and 30.4% O d. 18.8% Li, 16.3% C, and 64.9% O e. 51.7% C, 6.95% H, and 41.3% O

7.46 Calculate the empirical formula for each of the following: a. 55.5% Ca and 44.5% S b. 78.3% Ba and 21.7% F c. 76.0% Zn and 24.0% P d. 29.1% Na, 40.6% S, and 30.3% O e. 19.8% C, 2.20% H, and 78.0% Cl

PRACTICE PROBLEMS Try Practice Problems 7.47 and 7.48

TABLE 7.4 Comparing the Molar Mass of Some Compounds with the Empirical Formula of CH2O

Compound Empirical Formula

Molecular Formula

Molar Mass (g)

Integer : Empirical Formula

Integer : Empirical Formula Mass

Acetaldehyde CH2O CH2O 30.03 1(CH2O) 1 * 30.03 Acetic acid CH2O C2H4O2 60.06 2(CH2O) 2 * 30.03 Lactic acid CH2O C3H6O3 90.09 3(CH2O) 3 * 30.03 Erythrose CH2O C4H8O4 120.12 4(CH2O) 4 * 30.03 Ribose CH2O C5H10O5 150.15 5(CH2O) 5 * 30.03

Small integer

Molecular formula

Empirical formula

Molar mass

Empirical formula mass

Relating Empirical and Molecular Formulas Once we determine the empirical formula, we can calculate the empirical formula mass in grams. If we are given the molar mass of the compound, we can calculate the value of the small integer.

Small integer = molar mass of compound

empirical formula mass

ENGAGE 7.7 Why do the molecular formulas C2H4O2 and C5H10O5 both have an empirical formula of CH2O?

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202 CHAPTER 7 Chemical Quantities

For example, when the molar mass of ribose is divided by the empirical formula mass, the integer is 5.

Small integer = molar mass of ribose

empirical formula mass of CH2O =

150.15 g

30.03 g = 5

Multiplying each of the subscripts in the empirical formula (CH2O) by 5 gives the molecular formula of ribose, C5H10O5.

5 * empirical formula (CH2O) = molecular formula (C5H10O5)

Calculating a Molecular Formula Earlier, in Sample Problem 7.10, we determined that the empirical formula of ascorbic acid (vitamin C) was C3H4O3. From the molar mass for ascorbic acid, which is experimentally determined to be 176.12 g, we can calculate its molecular formula as follows:

The mass of the empirical formula C3H4O3 is obtained in the same way as molar mass.

Empirical formula = 3 mol of C + 4 mol of H + 3 mol of O

Empirical formula mass = (3 * 12.01 g) + (4 * 1.008 g) + (3 * 16.00 g)

= 88.06 g

Small integer = molar mass of ascorbic acid

empirical formula mass of C3H4O3 =

176.12 g

88.06 g = 2

Multiplying all the subscripts in the empirical formula of ascorbic acid by 2 gives its molecular formula.

C(3 * 2)H(4 * 2)O(3 * 2) = C6H8O6 Molecular formula

CORE CHEMISTRY SKILL Calculating a Molecular Formula

SAMPLE PROBLEM 7.11 Determination of a Molecular Formula

TRY IT FIRST

Melamine, which is used to make plastic items such as dishes and toys, contains 28.57% C, 4.80% H, and 66.64% N. If the experimental molar mass is 125 g, what is the molecular formula of melamine?

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

28.57% C, 4.80% H, 66.64% N, molar mass 125 g

molecular formula empirical formula mass

STEP 1 Obtain the empirical formula and calculate the empirical formula mass. In a sample size of 100. g of this compound, there are 28.57 g of C, 4.80 g of H, and 66.64 g of N.

28.57 g C * 1 mol C

12.01 g C = 2.38 mol of C

4.80 g H * 1 mol H

1.008 g H = 4.76 mol of H

66.64 g N * 1 mol N

14.01 g N = 4.76 mol of N

Divide the moles of each element by the smallest number of moles, 2.38, to obtain the subscripts of each element in the formula.

2.38 mol C

2.38 = 1.00 mol of C

4.76 mol H

2.38 = 2.00 mol of H

4.76 mol N

2.38 = 2.00 mol of N

Some brightly colored dishes are made of melamine.

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7.6 Molecular Formulas 203

Using these values as subscripts, C1.00H2.00N2.00, we write the empirical formula for melamine as CH2N2.

Now we calculate the molar mass for this empirical formula as follows:

Empirical formula = 1 mol of C + 2 mol of H + 2 mol of N

Empirical formula mass = (1 * 12.01) + (2 * 1.008) + (2 * 14.01) = 42.05 g

STEP 2 Divide the molar mass by the empirical formula mass to obtain a small integer.

Small integer = molar mass of melamine

empirical formula mass of CH2N2 =

125 g

42.05 g = 2.97

STEP 3 Multiply the empirical formula by the small integer to obtain the molecular formula. Because the experimental molar mass is close to 3 times the empirical formula mass, the subscripts in the empirical formula are multiplied by 3 to give the molecular formula.

C(1 * 3)H(2 * 3)N(2 * 3) = C3H6N6 Molecular formula

SELF TEST 7.11

a. The insecticide lindane has a mass percent composition of 24.78% C, 2.08% H, and 73.14% Cl. If its experimental molar mass is 290 g, what is the molecular formula?

b. Difluoroethane, used as a refrigerant, has a mass percent composition of 36.37% C, 6.10% H, and 57.53% F. If its experimental molar mass is 65 g, what is the molecular formula?

ANSWER

a. C6H6Cl6 b. C2H4F2

PRACTICE PROBLEMS Try Practice Problems 7.49 to 7.58

PRACTICE PROBLEMS

7.6 Molecular Formulas

Applications

7.47 Write the empirical formula for each of the following substances:

a. H2O2, peroxide b. C18H12, chrysene, used in the manufacture of dyes c. C10H16O2, chrysanthemic acid, in pyrethrum flowers d. C9H18N6, altretamine, anticancer medication e. C2H4N2O2, oxamide, fertilizer

7.48 Write the empirical formula for each of the following substances:

a. C6H6O3, pyrogallol, developer in photography b. C6H12O6, galactose, carbohydrate c. C8H6O4, terephthalic acid, used in the manufacture of plastic

bottles d. C6Cl6, hexachlorobenzene, fungicide e. C24H16O12, laccaic acid, crimson dye

7.49 The carbohydrate fructose found in honey and fruits has an empirical formula of CH2O. If the experimental molar mass of fructose is 180 g, what is its molecular formula?

7.50 Caffeine has an empirical formula of C4H5N2O. If it has an experimental molar mass of 194 g, what is the molecular formula of caffeine?

7.51 Benzene and acetylene have the same empirical formula, CH. However, benzene has an experimental molar mass of 78 g, and acetylene has an experimental molar mass of 26 g. What are the molecular formulas of benzene and acetylene?

7.52 Glyoxal, used in textiles; maleic acid, used to retard oxida- tion of fats and oils; and aconitic acid, a plasticizer, all have the same empirical formula, CHO. However, the experimental molar masses are glyoxal 58 g, maleic acid 117 g, and aconitic acid 174 g. What are the molecular formulas of glyoxal, maleic acid, and aconitic acid?

Coffee beans are a source of caffeine.

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204 CHAPTER 7 Chemical Quantities

After completing the examina- tion and lab tests, Dr. Evans, the veterinarian, prescribes medications that contain clavu- lanic acid and phenylpropanol- amine for Max.

7.53 Mevalonic acid is involved in the biosynthesis of cholesterol. Mevalonic acid contains 48.64% C, 8.16% H, and 43.20% O. If mevalonic acid has an experimental molar mass of 148 g, what is its molecular formula?

7.54 Chloral hydrate, a sedative, contains 14.52% C, 1.83% H, 64.30% Cl, and 19.35% O. If it has an experimental molar mass of 165 g, what is the molecular formula of chloral hydrate?

7.55 Vanillic acid contains 57.14% C, 4.80% H, and 38.06% O, and has an experimental molar mass of 168 g. What is the molecular formula of vanillic acid?

7.56 Lactic acid, the substance that builds up in muscles during exercise, has a mass percent composition of 40.0% C, 6.71% H,

and 53.3% O, and an experimental molar mass of 90. g. What is the molecular formula of lactic acid?

7.57 A sample of nicotine, a poisonous compound found in tobacco leaves, contains 74.0% C, 8.70% H, and 17.3% N. If the experi- mental molar mass of nicotine is 162 g, what is its molecular formula?

7.58 Adenine, a nitrogen-containing compound found in DNA and RNA, contains 44.5% C, 3.70% H, and 51.8% N. If adenine has an experimental molar mass of 135 g, what is its molecular formula?

7.60 Phenylpropanolamine (PPA) is used to treat urinary tract infections. The molecular formula of PPA is C9H13NO.

a. What is the molar mass of PPA? b. What is the mass percent of N in PPA? c. Max weighs 12 kg. If the dose of PPA is 2.0 mg/kg, how

many moles of PPA are given?

UPDATE Prescriptions for Max

Applications

7.59 Clavulanic acid has a molecular formula of C8H9NO5. a. What is the molar mass of clavulanic acid? b. What is the mass percent of C in clavulanic acid? c. Max weighs 12 kg. If the dose of clavulanic acid is

2.5 mg/kg, how many moles of clavulanic acid are given?

CONCEPT MAP

Avogadro’s Number

(6.022 : 1023)

CHEMICAL QUANTITIES

Moles to Grams Grams to Moles

that convert

Atoms

Atomic Mass

Periodic Table

have an

on the

Moles

Molar Masses

contain

and have

used to calculate

Mass Percent Composition

Empirical Formula

Molecular Formula

give molar ratios for

and to find

M07_TIMB8119_06_SE_C07.indd 204 11/30/18 8:03 AM

Core Chemistry Skills 205

CHAPTER REVIEW

7.1 The Mole LEARNING GOAL Use Avogadro’s number to calculate the number of particles in a given number of moles. Calculate the number of moles of an element in a given number of moles of a compound. • One mole of an element contains 6.022 * 1023 atoms. • One mole of a compound contains 6.022 * 1023 molecules or

formula units.

7.2 Molar Mass LEARNING GOAL Given the chemical formula of a substance, calculate its molar mass. • The molar mass (g/mol) of a substance is the

mass in grams equal numerically to its atomic mass, or the sum of the atomic masses, which have been multiplied by their subscripts in a formula.

7.3 Calculations Using Molar Mass LEARNING GOAL Use molar mass to convert between grams and moles. • The molar mass is used as a

conversion factor to change a quantity in grams to moles or to change a given number of moles to grams.

7.4 Mass Percent Composition LEARNING GOAL Given the formula of a compound, calculate the mass percent composition. • The mass percent composition is

obtained by dividing the mass in grams of each element in a compound by the mass of that compound.

7.5 Empirical Formulas LEARNING GOAL From the mass percent composition, calculate the empirical formula for a compound. • The empirical formula is calculated

by determining the lowest whole- number mole ratio from the grams of the elements present in a sample.

• If mole ratios for an empirical formula are not all whole numbers, multiply all values by an integer to give whole numbers.

7.6 Molecular Formulas LEARNING GOAL Determine the molecular formula of a substance from the empirical formula and molar mass. • A molecular formula is equal to, or a

multiple of, the empirical formula. • The experimental molar mass, which must be known, is divided by

the mass of the empirical formula to obtain the small integer used to convert the empirical formula to the molecular formula.

Avogadro’s number The number of items in a mole, equal to 6.022 * 1023.

empirical formula The simplest or smallest whole-number ratio of the atoms in a formula.

mass percent composition The percent by mass of the elements in a formula.

molar mass The mass in grams of 1 mol of an element equal numerically to its atomic mass. The molar mass of a compound

is equal to the sum of the atomic masses of the elements in the formula.

mole A group of atoms, molecules, or formula units that contains 6.022 * 1023 of these items.

molecular formula The actual formula that gives the number of atoms of each type of element in the compound.

KEY TERMS

The chapter section containing each Core Chemistry Skill is shown in parentheses at the end of each heading.

Converting Particles to Moles (7.1) • In chemistry, atoms, molecules, and ions are counted by the mole

(abbreviated mol in calculations), a unit that contains 6.022 * 1023 items, which is Avogadro’s number.

• For example, 1 mol of carbon contains 6.022 * 1023 atoms of carbon and 1 mol of H2O contains 6.022 * 1023 molecules of H2O.

• Avogadro’s number is used to convert between particles and moles.

Example: How many moles of nickel contain 2.45 * 1024 Ni atoms?

CORE CHEMISTRY SKILLS Answer:

2.45 * 1024 Ni atoms * 1 mol Ni

6.022 * 1023 Ni atoms = 4.07 mol of Ni

Calculating Molar Mass (7.2) • The molar mass of an element is its mass in grams equal numeri-

cally to its atomic mass. • The molar mass of a compound is the sum of the molar mass of

each element in its chemical formula multiplied by its subscript in the formula.

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206 CHAPTER 7 Chemical Quantities

Example: Calculate the molar mass for pinene, C10H16, a component of pine tree sap.

Calculating an Empirical Formula (7.5)

• The empirical formula or simplest formula represents the lowest whole-number ratio of the atoms and therefore moles of elements in a compound.

• For example, dinitrogen tetroxide, N2O4, has the empirical formula NO2.

• To calculate the empirical formula, the grams of each element are converted to moles and divided by the smallest number of moles to obtain the lowest whole-number ratio.

Example: Calculate the empirical formula for a compound that contains 3.28 g of Cr and 6.72 g of Cl.

Answer: Convert the grams of each element to moles.

3.28 g Cr * 1 mol Cr

52.00 g Cr = 0.0631 mol of Cr

6.72 g Cl * 1 mol Cl

35.45 g Cl = 0.190 mol of Cl

Divide by the smallest number of moles (0.0631) to obtain the empirical formula.

0.0631 mol Cr

0.0631 = 1.00 mol of Cr

0.190 mol Cl

0.0631 = 3.01 mol of Cl

Write the empirical formula using the whole-number ratios of moles. Cr1.00Cl3.01 h CrCl3

Calculating a Molecular Formula (7.6) • A molecular formula is related to the empirical formula by a small

whole number (integer) such as 1, 2, or 3.

Molecular formula = small integer * empirical formula

• If the empirical formula mass and the molar mass are known for a compound, an integer can be calculated by dividing the molar mass by the empirical formula mass.

Small integer = molar mass of compound

empirical formula mass

Example: Cymene, a component in oil of thyme, has an empirical formula C5H7 and an experimental molar mass of 135 g. What is the molecular formula of cymene?

Answer: Empirical formula = 5 mol of C + 7 mol of H Empirical formula mass = (5 * 12.01 g) + (7 * 1.008 g)

= 67.11 g

Molar mass of cymene

Empirical formula mass of C5H7 =

135 g

67.11 g = 2.01 (round off to 2)

The molecular formula of cymene is calculated by multiply- ing each of the subscripts in the empirical formula by 2.

C(5 * 2)H(7 * 2) = C10H14

Pinene is a component of pine sap.

Answer: 10 mol C * 12.01 g C

1 mol C = 120.1 g of C

16 mol H * 1.008 g H

1 mol H = 16.13 g of H

Molar mass of C10H16 = 136.2 g

Using Molar Mass as a Conversion Factor (7.3) • Molar mass is used as a conversion factor to convert between the

moles and grams of a substance.

Example: The frame of a bicycle contains 6500 g of aluminum. How many moles of aluminum are in the bicycle frame?

A racing bicycle has an aluminum frame.

Answer: Equality: 1 mol of Al = 26.98 g of Al

Conversion Factors: 26.98 g Al

1 mol Al and

1 mol Al 26.98 g Al

6500 g Al * 1 mol Al

26.98 g Al = 240 mol of Al

Calculating Mass Percent Composition (7.4) • The mass of a compound contains a definite proportion by mass of

its elements. • The mass percent of an element in a compound is calculated by

dividing the mass of that element by the mass of the compound.

Mass percent of an element = mass of an element

mass of the compound * 100%

Example: Dinitrogen tetroxide, N2O4, is used in liquid fuels for rockets. If it has a molar mass of 92.02 g, what is the mass percent of nitrogen?

Answer: Mass % N = 28.02 g N

92.02 g N2O4 * 100% = 30.45% N

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Additional Practice Problems 207

2.

ADDITIONAL PRACTICE PROBLEMS

7.65 Calculate the mass, in grams, of O in each of the following: (7.3) a. 50.0 g of H2O b. 17.5 g of CO2 c. 48 g of C7H6O2 7.66 Calculate the mass, in grams, of Cu in each of the following: (7.3) a. 4.92 g of CuCO3 b. 0.654 g of Cu2S c. 89 g of Cu3(PO4)2 7.67 Calculate the mass percent composition for each of the

following compounds: (7.4) a. 3.85 g of Ca and 3.65 g of F b. 0.389 g of Na and 0.271 g of O c. 12.4 of K, 17.4 g of Mn, and 20.3 g of O

7.68 Calculate the mass percent composition for each of the following compounds: (7.4)

a. 0.457 g of C and 0.043 g of H b. 3.65 g of Na, 2.54 g of S, and 3.81 g of O c. 0.907 g of Na, 1.40 g of Cl, and 1.89 g of O

7.69 Calculate the mass percent composition for each of the following compounds: (7.4)

a. K2CrO4 b. Al(HCO3)3 c. C6H12O6 7.70 Calculate the mass percent composition for each of the

following compounds: (7.4) a. CaCO3 b. NaC2H3O2 c. Ba(NO3)2 7.71 A mixture contains 0.250 mol of Mn2O3 and 20.0 g of MnO2.

(7.1, 7.2, 7.3) a. How many moles of O are present in the mixture? b. How many grams of Mn are in the mixture?

7.72 A mixture contains 4.00 * 1023 molecules of PCl3 and 0.250 mol of PCl5. (7.1, 7.2, 7.3)

a. How many grams of Cl are present in the mixture? b. How many moles of P are in the mixture?

UNDERSTANDING THE CONCEPTS

The chapter sections to review are shown in parentheses at the end of each problem.

1.

Dandruff shampoo contains dipyrithione.

Ibuprofen is an anti-inflammatory drug.

1. 2.

7.61 Using the models of the molecules (black = C, white = H, yellow = S, green = Cl), determine each of the following for models of compounds 1 and 2: (7.2, 7.4, 7.5, 7.6)

a. molecular formula b. empirical formula c. molar mass d. mass percent composition

7.62 Using the models of the molecules (black = C, white = H, yellow = S, red = O), determine each of the following for models of compounds 1 and 2: (7.2, 7.4, 7.5, 7.6)

a. molecular formula b. empirical formula c. molar mass d. mass percent composition

Applications

7.63 A dandruff shampoo contains dipyrithione, C10H8N2O2S2, an antibacterial and antifungal agent. (7.1, 7.2, 7.3, 7.4, 7.5)

a. What is the empirical formula of dipyrithione? b. What is the molar mass of dipyrithione? c. What is the mass percent of O in dipyrithione? d. How many grams of C are in 25.0 g of dipyrithione? e. How many moles of dipyrithione are in 25.0 g of

dipyrithione?

7.64 Ibuprofen, the anti-inflammatory drug in Advil, has the formula C13H18O2. (7.1, 7.2, 7.3, 7.4, 7.5)

a. What is the empirical formula of ibuprofen? b. What is the molar mass of ibuprofen? c. What is the mass percent of O in ibuprofen? d. How many grams of C are in 0.425 g of ibuprofen? e. How many moles of ibuprofen are in 2.45 g of ibuprofen?

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208 CHAPTER 7 Chemical Quantities

c. How many grams of O are in 7.50 g of aspirin? d. How many molecules of aspirin contain 2.50 g of H?

7.82 Ammonium sulfate, (NH4)2SO4, is used in fertilizers. (7.1, 7.2, 7.3, 7.4)

a. What is the mass percent composition of (NH4)2SO4? b. How many atoms of H are in 0.75 mol of (NH4)2SO4? c. How many grams of O are in 4.50 * 1023 formula units

of (NH4)2SO4? d. What mass of (NH4)2SO4 contains 2.50 g of S?

7.83 Write the empirical formula for each of the following: (7.5) a. C5H5N5, adenine, found in RNA and DNA b. FeC2O4, iron(II) oxalate, photographic developer c. C16H16N4, stilbamidine, antibiotic for animals d. C6H14N2O2, lysine, amino acid needed for growth 7.84 Write the empirical formula for each of the following: (7.5) a. C12H24N2O4, carisoprodol, skeletal muscle relaxant b. C10H10O5, opianic acid, used to synthesize a drug to treat

tuberculosis c. CrCl3, chromium(III) chloride, used in chrome plating d. C16H16N2O2, lysergic acid, controlled substance from ergot

7.85 Oleic acid, a component of olive oil, is 76.54% C, 12.13% H, and 11.33% O. The experimental value of the molar mass is 282 g. (7.1, 7.2, 7.3, 7.4, 7.5, 7.6)

a. What is the molecular formula of oleic acid? b. If oleic acid has a density of 0.895 g/mL, how many

molecules of oleic acid are in 3.00 mL of oleic acid?

7.86 Iron pyrite, commonly known as “fool’s gold,” is 46.5% Fe and 53.5% S. (7.1, 7.2, 7.3, 7.4, 7.5, 7.6)

a. If the empirical formula and the molecular formula are the same, what is the molecular formula of the compound?

b. If the crystal contains 4.85 g of iron, how many grams of S are in the crystal?

7.73 Calculate the empirical formula for each of the following compounds: (7.5)

a. 2.20 g of S and 7.81 g of F b. 6.35 g of Ag, 0.825 g of N, and 2.83 g of O c. 43.6% P and 56.4% O d. 22.1% Al, 25.4% P, and 52.5% O

7.74 Calculate the empirical formula for each of the following compounds: (7.5)

a. 5.13 g of Cr and 2.37 g of O b. 2.82 g of K, 0.870 g of C, and 2.31 g of O c. 61.0% Sn and 39.0% F d. 25.9% N and 74.1% O

7.75 Succinic acid is 40.7% C, 5.12% H, and 54.2% O. If it has an experimental molar mass of 118 g, what are the empirical and molecular formulas? (7.4, 7.5, 7.6)

7.76 A compound is 70.6% Hg, 12.5% Cl, and 16.9% O. If it has an experimental molar mass of 568 g, what are the empirical and molecular formulas? (7.4, 7.5, 7.6)

7.77 A sample of a compound contains 1.65 * 1023 atoms of C, 0.552 g of H, and 4.39 g of O. If 1 mol of the compound contains 4 mol of O, what is the molecular formula and molar mass of the compound? (7.4, 7.5, 7.6)

7.78 What is the molecular formula of a compound if 0.500 mol of the compound contains 0.500 mol of Sr, 1.81 * 1024 atoms of O, and 35.5 g of Cl? (7.4, 7.5, 7.6)

Applications 7.79 Calculate the molar mass for each of the following: (7.2) a. ZnSO4, zinc sulfate, zinc supplement b. Ca(IO3)2, calcium iodate, iodine source in table salt c. C5H8NNaO4, monosodium glutamate, flavor enhancer d. C6H12O2, isoamyl formate, used to make artificial fruit syrups

7.80 Calculate the molar mass for each of the following: (7.2) a. MgCO3, magnesium carbonate, an antacid b. Au(OH)3, gold(III) hydroxide, used in gold plating c. C18H34O2, oleic acid, from olive oil d. C21H26O5, prednisone, anti-inflammatory

7.81 Aspirin, C9H8O4, is used to reduce inflammation and reduce fever. (7.1, 7.2, 7.3, 7.4)

a. What is the mass percent composition of aspirin? b. How many moles of aspirin contain 5.0 * 1024 atoms of C?

The following problems are related to the topics in this chapter. However, they do not all follow the chapter order, and they require you to combine concepts and skills from several sections. These problems will help you increase your critical thinking skills and prepare for your next exam.

Applications

7.87 A toothpaste contains 0.240% by mass sodium fluoride used to prevent tooth decay and 5.0% by mass KNO3, which decreases pain and sensitivity. One tube contains 119 g of toothpaste. (7.1, 7.2, 7.3, 7.4)

a. How many moles of NaF are in the tube of toothpaste? b. How many fluoride ions, F -, are in the tube of toothpaste? c. How many grams of sodium ion, Na+, are in the tube of

toothpaste? d. How many KNO3 formula units are in the tube of

toothpaste?

CHALLENGE PROBLEMS

Iron pyrite is commonly known as “fool’s gold.”

7.88 A gold bar is 5.50 cm long, 3.10 cm wide, and 0.300 cm thick. (7.1, 7.2, 7.3, 7.4, 7.5, 7.6)

A gold bar consists of gold atoms.

a. If gold has a density of 19.3 g/cm3, what is the mass of the gold bar?

b. How many atoms of gold are in the bar? c. When the same mass of gold combines with oxygen, the

oxide product has a mass of 111 g. How many moles of O are combined with the gold?

d. What is the molecular formula of the oxide product if it is the same as the empirical formula?

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Answers to Selected Problems 209

7.90 Sorbic acid, an inhibitor of mold in cheese, has a mass percent composition of 64.27% C, 7.19% H, and 28.54% O. If sorbic acid has an experimental molar mass of 112 g, what is its molecular formula? (7.4, 7.5, 7.6)

7.89 Iron(III) chromate, a yellow powder used as a pigment in paints, contains 24.3% Fe, 33.9% Cr, and 41.8% O. If it has an experimental molar mass of 460 g, what are its empirical and molecular formulas? (7.4, 7.5, 7.6)

Iron(III) chromate is a yellow pigment used in paints.

Cheese contains sorbic acid, which prevents the growth of mold.

ANSWERS TO ENGAGE QUESTIONS 7.5 In 1 mol of Freon-12, there are 2 mol of fluorine and 2 mol of

chlorine. Because the molar mass of chlorine is almost double the molar mass of fluorine, there are more grams of chlorine.

7.6 The atoms in a compound must be whole numbers, so multiplying both by 2 gives the whole-number ratio of 2 to 5.

7.7 By dividing the molecular formula C2H4O2 by 2 and the molecular formula C5H10O5 by 5, we see that they both have the empirical formula CH2O.

7.1 There are a large number of atoms in 1 mol: 6.022 * 1023. A small number of moles will contain a large number of atoms.

7.2 The formula of the dietary supplement zinc acetate, Zn(C2H3O2)2, shows that 1 mol contains 1 mol of Zn, 2 * 2 or 4 mol of C, 3 * 2 or 6 mol of H, and 2 * 2 or 4 mol of O.

7.3 The molar mass obtained by multiplying the molar mass of each element by its subscript in the formula. For K2Cr2O7, the molar mass is 2 * 39.10 g + 2 * 52.00 g + 7 * 16.00 g or 294.2 g.

7.4 To calculate the number of atoms of H in 5.00 g of CH4, first the moles of CH4 are calculated, then the moles of H are calculated, and finally the atoms of H are calculated.

ANSWERS TO SELECTED PROBLEMS 7.25 a. 0.760 mol of Ag b. 0.0240 mol of C c. 0.881 mol of NH3 d. 0.452 mol of CH4 e. 1.53 mol of Fe2O3

7.27 a. 6.25 mol of He b. 0.781 mol of O2 c. 0.321 mol of Al(OH)3 d. 0.106 mol of Ga2S3 e. 0.430 mol of C4H10 7.29 a. 0.188 g of C b. 235 g of C c. 0.959 g of C d. 48.9 g of C

7.31 a. 6.17 mol of H b. 54.0 g of C c. 27.8 g of C d. 0.0465 g or 4.65 * 10-2 g of H 7.33 a. 602 g b. 74.7 g

7.35 a. 66.0 g of N2O b. 0.772 mol of N2O c. 21.6 g of N

7.37 a. 46.8% Si; 53.2% O b. 81.7% C; 18.3% H c. 32.3% Na; 45.1% S; 22.6% O d. 54.6% C; 9.12% H; 36.3% O

7.39 a. 39.01% Mg; 60.99% F b. 54.09% Ca; 43.18% O; 2.721% H c. 40.00% C; 6.714% H; 53.29% O d. 28.19% N; 8.115% H; 20.77% P; 42.92% O e. 71.55% C; 6.710% H; 4.909% N; 16.82% O

7.1 One mole contains 6.022 * 1023 atoms of an element, molecules of a molecular substance, or formula units of an ionic substance.

7.3 a. 3.01 * 1023 atoms of C b. 7.71 * 1023 molecules of SO2 c. 0.0867 mol of Fe d. 14.1 mol of C2H6O

7.5 a. 6.00 mol of H b. 8.00 mol of O c. 1.20 * 1024 atoms of P d. 4.82 * 1024 atoms of O 7.7 a. 36 mol of H b. 1.0 * 102 mol of C c. 0.040 mol of N

7.9 a. 32.2 mol of C b. 6.22 mol of H c. 0.230 mol of O

7.11 a. 70.90 g b. 90.08 g c. 262.9 g

7.13 a. 83.98 g b. 98.95 g c. 156.7 g

7.15 a. 74.55 g b. 159.7 g c. 329.4 g

7.17 a. 342.2 g b. 188.18 g c. 365.5 g

7.19 a. 151.16 g b. 498.4 g c. 331.4 g

7.21 a. 34.5 g b. 112 g c. 5.50 g d. 5.14 g e. 9.80 * 104 g 7.23 a. 3.03 g b. 38.1 g c. 9.30 g d. 24.6 g e. 12.9 g

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210 CHAPTER 7 Chemical Quantities

7.41 a. 25.94% N b. 26.19% N c. 46.62% N d. 33.47% N e. 11.96% N

7.43 a. N2O b. CH3 c. HNO3 d. CaCrO4 7.45 a. K2S b. GaF3 c. Mn2O3 d. Li2CO3 e. C5H8O3 7.47 a. HO b. C3H2 c. C5H8O d. C3H6N2 e. CH2NO

7.49 C6H12O6 7.51 benzene C6H6; acetylene C2H2 7.53 C6H12O4 7.55 C8H8O4 7.57 C10H14N2 7.59 a. 199.16 g b. 48.24% C c. 1.5 * 10-4 mol 7.61 1. a. S2Cl2 b. SCl c. 135.04 g d. 47.50% S; 52.50% Cl

2. a. C6H6 b. CH c. 78.11 g d. 92.25% C; 7.743% H

7.63 a. C5H4NOS b. 252.3 g c. 12.68% O d. 11.9 g of C e. 0.0991 mol of dipyrithione

7.65 a. 44.4 g b. 12.7 g c. 13 g

7.67 a. 51.3% Ca; 48.7% F b. 58.9% Na; 41.1% O c. 24.8% K; 34.7% Mn; 40.5% O

7.69 a. 40.27% K; 26.78% Cr; 32.96% O b. 12.85% Al; 1.440% H; 17.16% C; 68.57% O c. 40.00% C; 6.716% H; 53.29% O

7.71 a. 1.210 mol of O b. 40.1 g of Mn

7.73 a. SF6 b. AgNO3 c. P2O5 d. AlPO4 7.75 The empirical formula is C2H3O2; the molecular formula is

C4H6O4.

7.77 The molecular formula is C4H8O4; the molar mass is 120.10 g.

7.79 a. 161.48 g b. 389.9 g c. 169.11 g d. 116.16 g

7.81 a. 59.99% C; 4.475% H; 35.52% O b. 0.92 mol of aspirin c. 2.66 g of O d. 1.87 * 1023 molecules of aspirin 7.83 a. CHN b. FeC2O4 c. C4H4N d. C3H7NO

7.85 a. C18H34O2 b. 5.73 * 1021 molecules of oleic acid 7.87 a. 0.00680 mol of NaF b. 4.10 * 1021 F - ions c. 0.156 g of Na+

d. 3.5 * 1022 formula units of KNO3 7.89 The empirical formula is Fe2Cr3O12; the molecular formula is

Fe2Cr3O12.

M07_TIMB8119_06_SE_C07.indd 210 11/30/18 8:03 AM

211

a. What is the empirical formula of oxalic acid? b. If oxalic acid has an experimental molar mass of 90. g,

what is its molecular formula? c. Using the LD50, how many grams of oxalic acid would be

toxic for a 160-lb person? d. How many kilograms of rhubarb leaves would the person

in part c need to eat to reach the toxic level of oxalic acid?

CI.10 The active ingredient in an antacid tablet is calcium carbonate. One tablet contains 500. mg of calcium carbonate. (2.7, 6.3, 7.2, 7.3, 7.4)

a. What is the chemical formula of calcium carbonate? b. What is the molar mass of calcium carbonate? c. What is the mass percent composition of calcium

carbonate? d. How many moles of calcium carbonate are in 12 antacid

tablets? e. If a person takes two antacid tablets, how many grams of

calcium are obtained? f. If the Daily Value (DV) for Ca2+ to maintain bone strength

in older women is 1500 mg, how many antacid tablets are needed each day?

CI.11 Oseltamivir, C16H28N2O4, is a drug that is used to treat influ- enza. The preparation of oseltamivir begins with the extraction of shikimic acid from the seedpods of star anise. From 2.6 g of star anise, 0.13 g of shikimic acid can be obtained and used to produce one capsule containing 75 mg of oseltamivir. The usual adult dosage for treatment of influenza is two capsules of oseltamivir daily for 5 days. (2.7, 6.5, 7.2, 7.3, 7.4, 7.5, 7.6)

CI.7 For parts a to f, consider the loss of electrons by atoms of the element X, and a gain of electrons by atoms of the element Y. Element X is in Group 2A (2), Period 3, and Y is in Group 7A (17), Period 3. (4.2, 5.4, 5.5, 6.2, 6.3)

COMBINING IDEAS from Chapters 4 to 7

Sterling silver is 92.5% silver by mass.

Rhubarb leaves are a source of oxalic acid.

The spice called star anise is a plant source of shikimic acid.

+++ +

X Y Y

a. Which element is a metal, X or Y? b. Which element is a nonmetal, X or Y? c. What are the ionic charges of X and Y? d. Write the electron configurations for the atoms X and Y. e. Write the formula and name for the ionic compound

formed by the ions of X and Y.

CI.8 A sterling silver bracelet, which is 92.5% silver by mass, has a volume of 25.6 cm3 and a density of 10.2 g/cm3. (2.6, 2.8, 4.4, 7.1, 7.5, 7.6)

a. What is the mass, in kilograms, of the bracelet? b. How many atoms of silver are in the bracelet? c. Determine the number of protons and neutrons in each

of the two stable isotopes of silver:

47 107Ag 47

109Ag

d. When silver combines with oxygen, a compound forms that contains 93.10% Ag by mass. What are the name and the molecular formula of the oxide product if the molecu- lar formula is the same as the empirical formula?

CI.9 Oxalic acid, a compound found in plants and vegetables such as rhubarb, has a mass percent composition of 26.7% C, 2.24% H, and 71.1% O. Oxalic acid can interfere with respiration and cause kidney or bladder stones. If a large quantity of rhubarb leaves is ingested, the oxalic acid can be toxic. The lethal dose (LD50) in rats for oxalic acid is 375 mg/kg. Rhubarb leaves contain about 0.5% by mass of oxalic acid. (2.7, 7.4, 7.5, 7.6)

Shikimic acid is the basis for the antiviral drug oseltamivir.

a. What is the empirical formula of oseltamivir? b. What is the mass percent composition of oseltamivir? c. If black spheres are carbon atoms, white spheres are

hydrogen atoms, and red spheres are oxygen atoms, what is the molecular formula of shikimic acid?

d. How many moles of shikimic acid are contained in 130 g of shikimic acid?

e. How many capsules containing 75 mg of oseltamivir could be produced from 155 g of star anise?

f. How many grams of C are in one dose (75 mg) of oseltamivir?

g. How many kilograms of oseltamivir would be needed to treat all the people in a city with a population of 500 000 people if each person consumes two oseltamivir capsules a day for 5 days?

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212 CHAPTER 7 Chemical Quantities

CI.11 a. C8H14NO2 b. 61.52% C; 9.033% H; 8.969% N; 20.49% O c. C7H10O5 d. 0.75 mol of shikimic acid e. 59 capsules f. 0.046 g of C g. 4 * 102 kg

CI.7 a. X is a metal; elements in Group 2A (2) are metals. b. Y is a nonmetal; elements in Group 7A (17) are nonmetals. c. X2+, Y-

d. X = 1s22s22p63s2 Y = 1s22s22p63s23p5

e. MgCl2, magnesium chloride

CI.9 a. CHO2 b. C2H2O4 c. 27 g of oxalic acid d. 5 kg of rhubarb

CI.12 The compound butyric acid gives rancid butter its characteris- tic odor. (2.7, 2.8, 6.5, 7.1, 7.2, 7.3, 7.4, 7.5)

a. If black spheres are carbon atoms, white spheres are hydrogen atoms, and red spheres are oxygen atoms, what is the molecular formula of butyric acid?

b. What is the empirical formula of butyric acid? c. What is the mass percent composition of butyric acid? d. How many grams of C are in 0.850 g of butyric acid? e. How many grams of butyric acid contain 3.28 * 1023

O atoms? f. Butyric acid has a density of 0.959 g/mL at 20 °C. How

many moles of butyric acid are contained in 1.56 mL of butyric acid?

Butyric acid produces the characteristic odor of rancid butter.

ANSWERS

M07_TIMB8119_06_SE_C07.indd 212 11/30/18 8:03 AM

213

Natalie was recently diagnosed with mild pulmonary emphysema due to secondhand cigarette smoke. She was referred to Angela, an exercise physiologist, who assesses Natalie’s condition by connecting her to an electrocardiogram (ECG or EKG), a pulse oximeter, and a blood pressure cuff. The ECG records the electrical activity of Natalie’s heart, which is used to measure her heart rate and rhythm and detect the possible presence of heart damage. The pulse oximeter measures her pulse and the saturation level of oxygen (the percentage of hemoglobin that is saturated with O2) in her arterial blood. The blood pressure cuff determines the pressure exerted by the heart in pumping her blood.

To determine possible heart disease, Natalie has an exercise stress test on a treadmill to measure how her heart rate and blood pressure respond to exertion by walking faster as Angela increases the treadmill’s slope. Angela measures the heart rate and blood pressure first at rest and then on the treadmill. Angela uses a face mask to collect expired air and measures Natalie’s maximal volume of oxygen uptake, or VO2 max.

CAREER

Exercise Physiologist Exercise physiologists work with athletes as well as patients diagnosed with diabetes, heart disease, pulmonary disease, or other chronic disabilities or diseases. Patients who have one of these diseases are often prescribed exercise as a form of treatment, and they are referred to an exercise physiologist. The exercise physiologist evaluates the patient’s overall health and then creates a customized exercise program for that individual. The program for an athlete might focus on reducing the number of injuries, whereas a program for a cardiac patient would focus on strengthening the heart muscles. The exercise physiologist also monitors the patient for improvement and determines if the exercise is helping to reduce or reverse the progression of the disease.

Chemical Reactions

Natalie’s test results indicate that her blood oxygen level is below normal. You can view Natalie’s test results and lung function diagnosis in the UPDATE Improving Natalie’s Overall Fitness, page 232. After reviewing the test results, Angela teaches Natalie how to improve her respiration and her overall fitness.

UPDATE Improving Natalie’s Overall Fitness

8

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214 CHAPTER 8 Chemical Reactions

LOOKING AHEAD

8.1 Equations for Chemical Reactions 214

8.2 Balancing a Chemical Equation 217

8.3 Types of Chemical Reactions 223

8.4 Oxidation–Reduction Reactions 228

FIGURE 8.1 The tarnishing of silver is a chemical change that produces a new substance with new properties.

Ag2SAg

8.1 Equations for Chemical Reactions LEARNING GOAL Identify a balanced chemical equation; determine the number of atoms in the reactants and products.

Chemical reactions occur everywhere. The fuel in our cars burns with oxygen to make the car move and run the air conditioner. When we cook our food or bleach our hair, chemical reactions take place. In our bodies, chemical reactions convert food into molecules that build muscles and move them. In the leaves of trees and plants, carbon dioxide and water are converted into carbohydrates. Some chemical reactions are simple, whereas others are quite complex. However, they can all be written with chemical equations that chemists use to describe chemical reactions. In every chemical reaction, the atoms in the reacting substances, called reactants, are rearranged to give new substances called products.

A chemical change occurs when a substance is converted into one or more new substances. For example, when silver tarnishes, the shiny silver metal (Ag) reacts with sulfur (S) to become the dull, black substance we call tarnish (Ag2S) (see FIGURE 8.1).

A chemical reaction always involves chemical change because atoms of the reacting substances form new combinations with new properties. For example, a chemical reaction takes place when a piece of iron (Fe) combines with oxygen (O2) in the air to produce a new substance, rust (Fe2O3), which has a reddish-brown color. During a chemical change, new properties become visible, which are an indication that a chemical reaction has taken place (see TABLE 8.1).

REVIEW Writing Ionic Formulas (6.2)

Naming Ionic Compounds (6.3)

Writing the Names and Formulas for Molecular Compounds (6.5)

ENGAGE 8. 1 Why is the formation of rust a chemical change?

Iron nails change color when they react with oxygen to form rust.

Writing a Chemical Equation When you build a model airplane, prepare a new recipe, or mix a medication, you follow a set of directions. These directions tell you what materials to use and the products you will obtain. In chemistry, a chemical equation tells us the materials we need and the products that will form.

TABLE 8.1 Types of Evidence of a Chemical Reaction

1. Change in color

Fe Fe2O3

2. Formation of a gas (bubbles)

Bubbles (gas) form when CaCO3 reacts with acid.

3. Formation of a solid (precipitate)

A yellow solid forms when potassium iodide is added to lead nitrate.

4. Heat (or a flame) produced or heat absorbed

Methane gas burns in the air with a hot flame.

M08_TIMB8119_06_SE_C08.indd 214 11/27/18 11:52 AM

8.1 Equations for Chemical Reactions 215

Suppose you work in a bicycle shop, assembling wheels and frames into bicycles. You could represent this process by a simple equation:

Equation: 2 Wheels + 1 Frame 1 Bicycle

Reactants Product

Reactants Product

O2(g) CO2(g) ¢+C(s)Equation:

In a chemical equation, the formulas of the reactants are written on the left of the arrow and the formulas of the products on the right.

When you burn charcoal in a grill, the carbon in the charcoal combines with oxygen to form carbon dioxide. We can represent this reaction by a chemical equation.

In a chemical equation, the formulas of the reactants are written on the left of the arrow and the formulas of the products on the right. When there are two or more formulas on the same side, they are separated by plus ( + ) signs. The chemical equation for burning carbon is balanced because there is one carbon atom and two oxygen atoms in both the reactants and the products.

Generally, each formula in an equation is followed by an abbreviation, in parentheses, that gives the physical state of the substance: solid (s), liquid (l), or gas (g). If a substance is dissolved in water, it is in an aqueous (aq) solution. The delta sign (∆) indicates that heat was used to start the reaction. TABLE 8.2 summarizes some of the symbols used in equations.

Identifying a Balanced Chemical Equation When a chemical reaction takes place, the bonds between the atoms of the reactants are bro- ken, and new bonds are formed to give the products. All atoms are conserved, which means that atoms cannot be gained, lost, or changed into other types of atoms. Every chemical reaction must be written as a balanced equation, which shows the same number of atoms for each element in the reactants and in the products.

Now consider the balanced reaction in which hydrogen reacts with oxygen to form water, written as follows:

H2(g) + O2(g) h H2O(g) Not balanced

In the balanced equation, there are whole numbers called coefficients in front of the formulas. On the reactant side, the coefficient of 2 in front of the H2 formula represents two molecules of hydrogen, which is 4 atoms of H. A coefficient of 1 is understood for O2, which gives 2 atoms of O. On the product side, the coefficient of 2 in front of the H2O formula

TABLE 8.2 Some Symbols Used in Writing Equations

Symbol Meaning

+

(l)

(g)

(aq)

¢

(s)

Separates two or more formulas

Reacts to form products

Solid

Liquid

Gas

Aqueous

Reactants are heated

+

2H2O(g)2H2(g) O2(g)

+

+ Balanced

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216 CHAPTER 8 Chemical Reactions

represents 2 molecules of water. Because the coefficient of 2 multiplies all the atoms in H2O, there are 4 hydrogen atoms and 2 oxygen atoms in the products. Because there are the same number of hydrogen atoms and oxygen atoms in the reactants as in the products, we know that the equation is balanced. This illustrates the Law of Conservation of Matter, which states that matter cannot be created or destroyed during a chemical reaction.

PRACTICE PROBLEMS Try Practice Problems 8.1 to 8.6

H OHO

Reactant atoms Product atoms= SAMPLE PROBLEM 8.1 Number of Atoms in Balanced

Chemical Equations

TRY IT FIRST

Indicate the number of each type of atom in the following balanced chemical equation:

Fe2S3(s) + 6HCl(aq) h 2FeCl3(aq) + 3H2S(g)

Reactants Products

Fe

S

H

Cl

SOLUTION

The total number of atoms in each formula is obtained by multiplying the coefficient by each subscript in the chemical formula.

Reactants Products

Fe 2 (1 * 2) 2 (2 * 1) S 3 (1 * 3) 3 (3 * 1) H 6 (6 * 1) 6 (3 * 2) Cl 6 (6 * 1) 6 (2 * 3)

SELF TEST 8.1

State the number of atoms of each element in the reactants and in the products for each of the following:

a. When ethane, C2H6, burns in oxygen, the products are carbon dioxide and water. The balanced equation is written as

2C2H6(g) + 7O2(g) h ∆

4CO2(g) + 6H2O(g)

b. Iron(III) oxide reacts with carbon monoxide to produce iron and carbon dioxide. The balanced chemical equation is written as

Fe2O3(s) + 3CO(g) h 2Fe(s) + 3CO2(g)

ANSWER

a. In both the reactants and products, there are 4 C atoms, 12 H atoms, and 14 O atoms. b. In both the reactants and products, there are 2 Fe atoms, 6 O atoms, and 3 C atoms.

ENGAGE 8. 2 How can we determine that a chemical equation is balanced?

PRACTICE PROBLEMS

8.1 Equations for Chemical Reactions

8. 1 State the number of atoms of oxygen in the reactants and in the products for each of the following equations: a. 3NO2(g) + H2O(l) h NO(g) + 2HNO3(aq) b. 5C(s) + 2SO2(g) h CS2(g) + 4CO(g) c. 2C2H2(g) + 5O2(g) h

∆ 4CO2(g) + 2H2O(g)

d. N2H4(g) + 2H2O2(g) h N2(g) + 4H2O(g)

8. 2 State the number of atoms of oxygen in the reactants and in the products for each of the following equations: a. CH4(g) + 2O2(g) h

∆ CO2(g) + 2H2O(g)

b. 4P(s) + 5O2(g) h P4O10(s) c. 4NH3(g) + 6NO(g) h 5N2(g) + 6H2O(g) d. 6CO2(g) + 6H2O(l) h C6H12O6(aq) + 6O2(g)

Glucose

M08_TIMB8119_06_SE_C08.indd 216 11/27/18 11:52 AM

8.2 Balancing a Chemical Equation 217

8.3 Determine whether each of the following equations is balanced or not balanced: a. S(s) + O2(g) h SO3(g) b. Al(s) + Cl2(g) h AlCl3(s) c. H2SO4(aq) + 2NaOH(aq) h 2H2O(l) + Na2SO4(aq) d. C3H8(g) + 5O2(g) h

∆ 3CO2(g) + 4H2O(g)

8.4 Determine whether each of the following equations is balanced or not balanced: a. PCl3(s) + Cl2(g) h PCl5(s) b. CO(g) + 2H2(g) h CH4O(g) c. 2KClO3(s) h 2KCl(s) + O2(g) d. Mg(s) + N2(g) h Mg3N2(s)

8.5 All of the following are balanced equations. State the number of atoms of each element in the reactants and in the products. a. 2Na(s) + Cl2(g) h 2NaCl(s) b. PCl3(l) + 3H2(g) h PH3(g) + 3HCl(g) c. P4O10(s) + 6H2O(l) h 4H3PO4(aq)

8.6 All of the following are balanced equations. State the number of atoms of each element in the reactants and in the products. a. 2N2(g) + 3O2(g) h 2N2O3(g) b. Al2O3(s) + 6HCl(aq) h 3H2O(l) + 2AlCl3(aq) c. C5H12(l) + 8O2(g) h

∆ 5CO2(g) + 6H2O(g)

8.2 Balancing a Chemical Equation LEARNING GOAL Write a balanced chemical equation from the formulas of the reactants and products for a chemical reaction.

The chemical reaction that occurs in the flame of a gas burner you use in the laboratory or a gas cooktop is the reaction of methane gas, CH4, and oxygen to produce carbon dioxide and water. We now show the process of writing and balancing the chemical equation for this reaction in Sample Problem 8.2.

CORE CHEMISTRY SKILL Balancing a Chemical Equation

SAMPLE PROBLEM 8. 2 Writing and Balancing a Chemical Equation

TRY IT FIRST

The chemical reaction of methane gas (CH4) and oxygen gas (O2) produces the gases carbon dioxide (CO2) and water (H2O). Write a balanced chemical equation for this reaction.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

reactants, products balanced equation equal numbers of atoms in reactants and products

STEP 1 Write an equation using the correct formulas for the reactants and products.

CH4(g) + O2(g) h ∆

CO2(g) + H2O(g)

O2CH4 H2OCO2

In a balanced chemical equation, the number of each type of atom must be the same in the reactants and in the products.

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218 CHAPTER 8 Chemical Reactions

STEP 2 Count the atoms of each element in the reactants and products. When we count the atoms on the reactant side and the atoms on the product side, we see that there are more H atoms in the reactants and more O atoms in the products.

CH4(g) + O2(g) h ∆

CO2(g) + H2O(g)

Reactants Products

1 C atom 1 C atom Balanced

4 H atoms 2 H atoms Not balanced

2 O atoms 3 O atoms Not balanced

STEP 3 Use coefficients to balance each element. We will start by balancing the H atoms in CH4 because it has the most atoms. By placing a coefficient of 2 in front of the formula for H2O, a total of 4 H atoms in the products is obtained. Only use coefficients to balance an equation. Do not change any of the subscripts: This would alter the chemical formula of a reactant or product.

CH4(g) + O2(g) h ∆

CO2(g) + 2H2O(g)

Reactants Products

1 C atom 1 C atom Balanced

4 H atoms 4 H atoms Balanced

2 O atoms 4 O atoms Not balanced

We can balance the O atoms on the reactant side by placing a coefficient of 2 in front of the formula O2. There are now 4 O atoms in both the reactants and products.

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) Balanced

+ +

¢

¢

STEP 4 Check the final equation to confirm it is balanced.

CH4(g) + 2O2(g) h ∆

CO2(g) + 2H2O(g) The equation is balanced.

Reactants Products

1 C atom 1 C atom Balanced

4 H atoms 4 H atoms Balanced

4 O atoms 4 O atoms Balanced

In a balanced chemical equation, the coefficients must be the lowest possible whole numbers. Suppose you had obtained the following for the balanced equation:

2CH4(g) + 4O2(g) h ∆

2CO2(g) + 4H2O(g) Incorrect

Although there are equal numbers of atoms on both sides of the equation, this is not written correctly. To obtain coefficients that are the lowest whole numbers, we divide all the coefficients by 2.

SELF TEST 8. 2

Balance each of the following chemical equations:

a. Al(s) + Cl2(g) h AlCl3(s) b. Cr2O3(s) + CCl4(l) h CrCl3(s) + COCl2(g)

ANSWER

a. 2Al(s) + 3Cl2(g) h 2AlCl3(s) b. Cr2O3(s) + 3CCl4(l) h 2CrCl3(s) + 3COCl2(g)

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8.2 Balancing a Chemical Equation 219

Whole-Number Coefficients Sometimes the coefficients of the compounds in an equation need to be increased to give whole numbers for the coefficients. Then we need to adjust the coefficients and count the total atoms on both sides once again, as shown in Sample Problem 8.3.

SAMPLE PROBLEM 8. 3 Balancing a Chemical Equation with Whole-Number Coefficients

TRY IT FIRST

Acetylene, C2H2, is used to produce high temperatures for welding by reacting it with O2 to produce CO2 and H2O. All of the compounds are gases. Write a balanced chemical equation with whole-number coefficients for this reaction.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

reactants, products balanced equation equal numbers of atoms in reactants and products

STEP 1 Write an equation using the correct formulas of the reactants and products.

C2H2(g) + O2(g) h ∆

CO2(g) + H2O(g) Not balanced Acetylene

STEP 2 Count the atoms of each element in the reactants and products. When we count the atoms on the reactant side and the product side, we see that there are more C atoms in the reactants and more O atoms in the products.

C2H2(g) + O2(g) h ∆

CO2(g) + H2O(g)

Reactants Products

2 C atoms 1 C atom Not balanced

2 H atoms 2 H atoms Balanced

2 O atoms 3 O atoms Not balanced

STEP 3 Use coefficients to balance each element. We start by balancing the C in C2H2 by placing a coefficient of 2 in front of the formula for CO2. When we recheck all the atoms, there are 2 C atoms and 2 H atoms in both the reactants and products.

C2H2(g) + O2(g) h ∆

2CO2(g) + H2O(g)

Reactants Products

2 C atoms 2 C atoms Balanced

2 H atoms 2 H atoms Balanced

2 O atoms 5 O atoms Not balanced

To balance the O atoms, we place a coefficient of 5/2 in front of the formula for O2, which gives a total of 5 O atoms on each side of the equation.

C2H2(g) + 5 2

O2(g) h ∆

2CO2(g) + H2O(g)

Reactants Products

2 C atoms 2 C atoms Balanced

2 H atoms 2 H atoms Balanced

5 O atoms 5 O atoms Balanced

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220 CHAPTER 8 Chemical Reactions

Now the equation is balanced for atoms, but the coefficient for O2 is a fraction. To obtain all whole-number coefficients, we multiply all the coefficients by 2.

2C2H2(g) + 5O2(g) h ∆

4CO2(g) + 2H2O(g)

STEP 4 Check the final equation to confirm it is balanced. A check of the total number of atoms indicates that the equation is now balanced with whole-number coefficients.

2C2H2(g) + 5O2(g) h ∆

4CO2(g) + 2H2O(g)

Reactants Products

4 C atoms 4 C atoms Balanced

4 H atoms 4 H atoms Balanced

10 O atoms 10 O atoms Balanced

SELF TEST 8.3

Balance each of the following chemical equations: a. NO2(g) + H2(g) h NH3(g) + H2O(g) b. Fe(aq) + F2(g) h FeF3(s)

ANSWER

a. 2NO2(g) + 7H2(g) h 2NH3(g) + 4H2O(g) b. 2Fe(aq) + 3F2(g) h 2FeF3(s)

Equations with Polyatomic Ions Sometimes an equation contains the same polyatomic ion in both the reactants and the products. Then we can balance the polyatomic ions as a group on both sides of the equation, as shown in Sample Problem 8.4.

SAMPLE PROBLEM 8.4 Balancing Chemical Equations with Polyatomic Ions

TRY IT FIRST

Balance the following chemical equation:

Na3PO4(aq) + MgCl2(aq) h Mg3(PO4)2(s) + NaCl(aq)

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

reactants, products balanced equation equal numbers of atoms in reactants and products

STEP 1 Write an equation using the correct formulas for the reactants and products.

Na3PO4(aq) + MgCl2(aq) h Mg3(PO4)2(s) + NaCl(aq) Not balanced

STEP 2 Count the atoms of each element in the reactants and products. When we count the number of ions in the reactants and products, we find that the equation is not balanced. In this equation, we can balance the phosphate ion as a group of atoms because it appears on both sides of the equation.

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8.2 Balancing a Chemical Equation 221

Na3PO4(aq) + MgCl2(aq) h Mg3(PO4)2(s) + NaCl(aq)

Reactants Products

3 Na+ 1 Na+ Not balanced

1 PO4 3− 2 PO4

3− Not balanced

1 Mg2+ 3 Mg2+ Not balanced

2 Cl- 1 Cl- Not balanced

STEP 3 Use coefficients to balance each element. We begin with the formula that has the highest subscript values, which in this equation is Mg3(PO4)2. The subscript 3 in Mg3(PO4)2 is used as a coefficient for MgCl2 to balance magnesium. The subscript 2 in Mg3(PO4)2 is used as a coefficient for Na3PO4 to balance the phosphate ion.

2Na3PO4(aq) + 3MgCl2(aq) h Mg3(PO4)2(s) + NaCl(aq)

Reactants Products

6 Na+ 1 Na+ Not balanced

2 PO4 3− 2 PO4

3− Balanced

3 Mg2+ 3 Mg2+ Balanced

6 Cl- 1 Cl- Not balanced

In the reactants and products, we see that the sodium and chloride ions are not yet balanced. A coefficient of 6 is placed in front of the NaCl to balance the equation.

2Na3PO4(aq) + 3MgCl2(aq) h Mg3(PO4)2(s) + 6NaCl(aq)

STEP 4 Check the final equation to confirm it is balanced. A check of the total number of ions confirms the equation is balanced. A coefficient of 1 is understood and not usually written.

2Na3PO4(aq) + 3MgCl2(aq) h Mg3(PO4)2(s) + 6NaCl(aq) Balanced

Reactants Products

6 Na+ 6 Na+ Balanced

2 PO4 3- 2 PO4

3- Balanced

3 Mg2+ 3 Mg2+ Balanced

6 Cl- 6 Cl- Balanced

Mg2+ Na+Cl- PO4 3-

ENGAGE 8. 3 What is the evidence for a chemical reaction when Na3PO4(aq) and MgCl2(aq) are mixed?

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222 CHAPTER 8 Chemical Reactions

PRACTICE PROBLEMS

8. 2 Balancing a Chemical Equation

8.7 Balance each of the following chemical equations: a. N2(g) + O2(g) h NO(g) b. HgO(s) h

∆ Hg(l) + O2(g)

c. Fe(s) + O2(g) h Fe2O3(s) d. Na(s) + Cl2(g) h NaCl(s) e. Cu2O(s) + O2(g) h CuO(s)

8.8 Balance each of the following chemical equations: a. Ca(s) + Br2(l) h CaBr2(s) b. P4(s) + O2(g) h P4O10(s) c. C4H8(g) + O2(g) h

∆ CO2(g) + H2O(g)

d. HNO3(aq) + Ca(OH)2(aq) h H2O(l) + Ca(NO3)2(aq) e. Fe2O3(s) + C(s) h Fe(s) + CO(g)

8.9 Balance each of the following chemical equations: a. Mg(s) + AgNO3(aq) h Ag(s) + Mg(NO3)2(aq) b. CuCO3(s) h CuO(s) + CO2(g) c. Al(s) + CuSO4(aq) h Cu(s) + Al2(SO4)3(aq) d. Pb(NO3)2(aq) + NaCl(aq) h PbCl2(s) + NaNO3(aq) e. HCl(aq) + Al(s) h H2(g) + AlCl3(aq)

8.10 Balance each of the following chemical equations: a. HNO3(aq) + Zn(s) h H2(g) + Zn(NO3)2(aq) b. H2SO4(aq) + Al(s) h H2(g) + Al2(SO4)3(aq) c. K2SO4(aq) + BaCl2(aq) h BaSO4(s) + KCl(aq) d. CaCO3(s) h CaO(s) + CO2(g) e. AlCl3(aq) + KOH(aq) h Al(OH)3(s) + KCl(aq)

8.11 Balance each of the following chemical equations: a. Fe2O3(s) + CO(g) h Fe(s) + CO2(g) b. Li3N(s) h Li(s) + N2(g) c. HBr(aq) + Al(s) h H2(g) + AlBr3(aq) d. Ba(OH)2(aq) + Na3PO4(aq)

h Ba3(PO4)2(s) + NaOH(aq) e. As4S6(s) + O2(g) h As4O6(s) + SO2(g)

8.12 Balance each of the following chemical equations: a. K(s) + H2O(l) h H2(g) + KOH(aq) b. Cr(s) + S8(s) h Cr2S3(s) c. BCl3(s) + H2O(l) h H3BO3(aq) + HCl(aq) d. H2SO4(aq) + Fe(OH)3(s) h H2O(l) + Fe2(SO4)3(aq) e. BaCl2(aq) + Na3PO4(aq) h Ba3(PO4)2(s) + NaCl(aq)

8.13 Write a balanced equation using the correct formulas and include conditions (s, l, g, or aq) for each of the following chemical reactions: a. Lithium metal reacts with liquid water to form hydrogen gas

and aqueous lithium hydroxide. b. Solid phosphorus reacts with chlorine gas to form solid

phosphorus pentachloride.

c. Solid iron(II) oxide reacts with carbon monoxide gas to form solid iron and carbon dioxide gas.

d. Liquid pentene (C5H10) burns in oxygen gas to form carbon dioxide gas and water vapor.

e. Hydrogen sulfide gas and solid iron(III) chloride react to form solid iron(III) sulfide and hydrogen chloride gas.

8.14 Write a balanced equation using the correct formulas, and include conditions (s, l, g, or aq) for each of the following chemical reactions: a. Solid sodium carbonate decomposes to produce solid sodium

oxide and carbon dioxide gas. b. Nitrogen oxide gas reacts with carbon monoxide gas to

produce nitrogen gas and carbon dioxide gas. c. Iron metal reacts with solid sulfur to produce solid iron(III)

sulfide. d. Solid calcium reacts with nitrogen gas to produce solid

calcium nitride. e. In the Apollo lunar module, hydrazine gas, N2H4, reacts

with dinitrogen tetroxide gas to produce gaseous nitrogen and water vapor.

Applications

8.15 Dinitrogen oxide, also known as laughing gas, is a sweet-tasting gas used in dentistry. When solid ammonium nitrate is heated, the products are gaseous water and gaseous dinitrogen oxide. Write the balanced chemical equation for the reaction.

8.16 When ethanol C2H6O(aq) is consumed, it reacts with oxygen gas in the body to produce gaseous carbon dioxide and liquid water. Write the balanced chemical equation for the reaction.

8.17 In the body, the amino acid alanine C3H7NO2(aq) reacts with oxygen gas to produce gaseous carbon dioxide, liquid water, and urea, CH4N2O(aq). Write the balanced chemical equation for the reaction.

8.18 In the body, the amino acid asparagine C4H8N2O3(aq) reacts with oxygen gas to produce gaseous carbon dioxide, liquid water, and urea, CH4N2O(aq). Write the balanced chemical equation for the reaction.

SELF TEST 8. 4

Balance each of the following chemical equations: a. Pb(NO3)2(aq) + AlBr3(aq) h PbBr2(s) + Al(NO3)3(aq) b. HNO3(aq) + Fe2(SO4)3(aq) h H2SO4(aq) + Fe(NO3)3(aq)

ANSWER

a. 3Pb(NO3)2(aq) + 2AlBr3(aq) h 3PbBr2(s) + 2Al(NO3)3(aq) b. 6HNO3(aq) + Fe2(SO4)3(aq) h 3H2SO4(aq) + 2Fe(NO3)3(aq)

PRACTICE PROBLEMS Try Practice Problems 8.7 to 8.18

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8.3 Types of Chemical Reactions 223

8.3 Types of Chemical Reactions LEARNING GOAL Identify a chemical reaction as a combination, decomposition, single replacement, double replacement, or combustion.

A great number of chemical reactions occur in nature, in biological systems, and in the laboratory. However, there are some general patterns that help us classify reactions into five general types.

Combination Reactions In a combination reaction, two or more elements or compounds bond to form one product. For example, sulfur and oxygen combine to form the product sulfur dioxide.

CORE CHEMISTRY SKILL Classifying Types of Chemical

Reactions

ENGAGE 8.4 What happens to the atoms in the reactants in a combination reaction?

2Mg(s) Magnesium

+ O2(g) Oxygen

Mg Mg Mg Mg Mg2+

O2

Mg2+

O2- O2-

2MgO(s) Magnesium oxide

¢

FIGURE 8. 2 In a combination reaction, two or more substances combine to form one substance as product.

In other examples of combination reactions, elements or compounds combine to form a single product.

N2(g) + 3H2(g) h 2NH3(g) Cu(s) + S(s) h CuS(s) MgO(s) + CO2(g) h MgCO3(s)

Decomposition Reactions In a decomposition reaction, a reactant splits into two or more simpler products. For example, when mercury(II) oxide is heated, the compound breaks apart into mercury atoms and oxygen (see FIGURE 8.3).

2HgO(s) h ∆

2Hg(l) + O2(g) +

A reactant

splits into

two or more products

Decomposition

A AB B

Two or more reactants

combine to yield

a single product

+ AA BB

Combination

O2(g) SO2(g)+

+

S(s)

In FIGURE 8.2, the elements magnesium and oxygen combine to form a single product, which is the ionic compound magnesium oxide formed from Mg2+ and O2- ions.

2Mg(s) + O2(g) h 2MgO(s)

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224 CHAPTER 8 Chemical Reactions

FIGURE 8.3 In a decomposition reaction, one reactant breaks down into two or more products.

+ O2(g) Oxygen

2Hg(l) Mercury

2HgO(s) Mercury(II) oxide

Hg2+ Hg

Hg

HgHg

Hg Hg

Hg2+

O2- O2- O2

¢

In another example of a decomposition reaction, when calcium carbonate is heated, it breaks apart into simpler compounds of calcium oxide and carbon dioxide.

CaCO3(s) h ∆

CaO(s) + CO2(g)

Replacement Reactions In a replacement reaction, elements in a compound are replaced by other elements. In a single replacement reaction, a reacting element switches place with an element in the other reacting compound.

In the single replacement reaction shown in FIGURE 8. 4, zinc replaces hydrogen in hydrochloric acid, HCl(aq).

Zn(s) + 2HCl(aq) h H2(g) + ZnCl2(aq)

FIGURE 8.4 In a single replacement reaction, an atom or ion replaces an atom or ion in a compound.

H2(g) Hydrogen

ZnCl2(aq) Zinc chloride

Zn Cl-

Cl-H+ H2

Zn2+

++ 2HCl(aq) Hydrochloric acid

Zn(s) Zinc

ENGAGE 8.6 What changes in the formulas of the reactants identify this equation as a single replacement?

ENGAGE 8.5 How do the differences in the reactants and products classify this as a decomposition reaction?

+ One element replaces another element

+

Single replacement

A CB BC A

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8.3 Types of Chemical Reactions 225

In another single replacement reaction, chlorine replaces bromine in the compound potassium bromide.

Cl2(g) + 2KBr(s) h 2KCl(s) + Br2(l)

In a double replacement reaction, the positive ions in the reacting compounds switch places.

In the reaction shown in FIGURE 8. 5, barium ions change places with sodium ions in the reactants to form sodium chloride and a white solid precipitate of barium sulfate. The formulas of the products depend on the charges of the ions.

BaCl2(aq) + Na2SO4(aq) h BaSO4(s) + 2NaCl(aq)

When sodium hydroxide and hydrochloric acid (HCl) react, sodium and hydrogen ions switch places, forming water and sodium chloride.

HCl(aq) + NaOH(aq) h H2O(l) + NaCl(aq)

FIGURE 8.5 In a double replacement reaction, the positive ions in the reactants replace each other.

Ba2+

2NaCl(aq) Sodium chloride

BaSO4(s) Barium sulfate

++ BaCl2(aq) Barium chloride

Na2SO4(aq) Sodium sulfate

SO4 2-

Cl-Na+

ENGAGE 8.7 How do the formulas of the reactants change in a double replacement reaction?

Combustion Reactions The burning of a candle and the burning of fuel in the engine of a car are examples of combustion reactions. In a combustion reaction, a carbon-containing compound, usu- ally a fuel, burns in oxygen from the air to produce carbon dioxide (CO2), water (H2O), and energy in the form of heat or a f lame. For example, methane gas (CH4) undergoes combustion when used to cook our food on a gas cooktop and to heat our homes. In the equation for the combustion of methane, each element in the fuel (CH4) forms a compound with oxygen.

CH4(g) + 2O2(g) h ∆

CO2(g) + 2H2O(g) + energy Methane

The balanced equation for the combustion of propane (C3H8) is

C3H8(g) + 5O2(g) h ∆

3CO2(g) + 4H2O(g) + energy

In a combustion reaction, a candle burns by reacting with the oxygen in the air.

+ Two elements replace each other

+

Double replacement

ADA DB BC C

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226 CHAPTER 8 Chemical Reactions

TABLE 8.3 Summary of Reaction Types

Reaction Type Example

Combination A + B h AB Ca(s) + Cl2(g) h CaCl2(s) Decomposition AB h A + B Fe2S3(s) h 2Fe(s) + 3S(s) Single Replacement A + BC h AC + B Cu(s) + 2AgNO3(aq) h 2Ag(s) + Cu(NO3)2(aq) Double Replacement AB + CD h AD + CB BaCl2(aq) + K2SO4(aq) h BaSO4(s) + 2KCl(aq) Combustion

CXHY + ZO2(g) h ∆

XCO2(g) + Y 2

H2O(g) + energy CH4(g) + 2O2(g) h ∆

CO2(g) + 2H2O(g) + energy

Propane is the fuel used in portable heaters and gas barbecues. Gasoline, a mixture of liquid hydrocarbons, is the fuel that powers our cars, lawn mowers, and snow blowers.

TABLE 8.3 summarizes the reaction types and gives examples.

SAMPLE PROBLEM 8.5 Identifying Reaction Types

TRY IT FIRST

Classify each of the following as a combination, decomposition, single replacement, double replacement, or combustion reaction: a. 2Fe2O3(s) + 3C(s) h 3CO2(g) + 4Fe(s) b. 2KClO3(s) h

∆ 2KCl(s) + 3O2(g)

c. 2C3H6(g) + 9O2(g) h ∆

6CO2(g) + 6H2O(g) + energy

SOLUTION

a. When a C atom replaces Fe in Fe2O3 to form the compound CO2 and Fe atoms, the reaction is a single replacement.

b. When one reactant breaks down to produce two products, the reaction is decomposition. c. The reaction of a carbon compound with oxygen to produce carbon dioxide, water, and

energy makes this a combustion reaction.

SELF TEST 8.5

a. Nitrogen gas (N2) and oxygen gas (O2) react to form nitrogen dioxide gas. Write the balanced chemical equation using the correct chemical formulas for the reactants and product, and identify the reaction type.

b. Strontium metal and aqueous HCl react to give hydrogen gas and aqueous strontium chloride. Write the balanced chemical equation using the correct chemical formulas for the reactants and products, and identify the reaction type.

ANSWER

a. N2(g) + 2O2(g) h 2NO2(g) Combination b. 2HCl(aq) + Sr(s) h H2(g) + SrCl2(aq) Single replacement

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8.3 Types of Chemical Reactions 227

PRACTICE PROBLEMS Try Practice Problems 8.19 to 8. 24

Chemistry Link to Health Incomplete Combustion: Toxicity of Carbon Monoxide

When a propane heater, fireplace, or woodstove is used in a closed room, there must be adequate ventilation. If the supply of oxygen is limited, incomplete combustion from burning gas, oil, or wood produces carbon monoxide. The incomplete combustion of methane in natural gas is written

2CH4(g) + 3O2(g) h ∆

2CO(g) + 4H2O(g) + energy Limited Carbon oxygen monoxide supply

Carbon monoxide (CO) is a colorless, odorless, poisonous gas. When inhaled, CO passes into the bloodstream, where it attaches

to hemoglobin, which reduces the amount of oxygen (O2) reaching the cells. As a result, a person can experience a reduction in exercise capability, visual perception, and manual dexterity.

Hemoglobin is the protein that transports O2 in the blood. When the amount of hemoglobin bound to CO (COHb) is about 10%, a person may experience shortness of breath, mild headache, and drowsiness. Heavy smokers can have levels of COHb in their blood as high as 9%. When as much as 30% of the hemoglobin is bound to CO, a person may experience more severe symptoms, including dizziness, mental confusion, severe headache, and nausea. If 50% or more of the hemoglobin is bound to CO, a person could become unconscious and die if not treated immediately with oxygen.

SAMPLE PROBLEM 8.6 Writing an Equation for Combustion

TRY IT FIRST

A portable burner is fueled with butane gas, C4H10. Write the reactants and products for the complete combustion of butane, and balance the equation.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

C4H10 balanced equation for combustion

reactants: carbon compound + O2, products: CO2 + H2O

In a combustion reaction, butane gas reacts with the gas O2 to form the gases CO2, H2O, and energy. We write the unbalanced equation as

C4H10(g) + O2(g) h ∆

CO2(g) + H2O(g) + energy

We can begin by using the subscripts in C4H10 to balance the C atoms in CO2 and the H atoms in H2O. However, this gives a total of 13 O atoms in the products. This is balanced by placing a coefficient of 13/2 in front of the formula for O2.

C4H10(g) + 13 2

O2(g) h ∆

4CO2(g) + 5H2O(g) + energy 8

13 O atoms

To obtain a whole number coefficient for O2, we need to multiply all the coefficients by 2.

2C4H10(g) + 13O2(g) h ∆

8CO2(g) + 10H2O(g) + energy

SELF TEST 8.6

a. Ethene, used to ripen fruit, has the formula C2H4. Write a balanced chemical equation for the complete combustion of ethene.

b. Heptane, C7H16, is used in tests to determine the octane rating of gasoline. Write a balanced chemical equation for the complete combustion of heptane.

ANSWER

a. C2H4(g) + 3O2(g) h ∆

2CO2(g) + 2H2O(g) + energy

b. C7H16(l) + 11O2(g) h ∆

7CO2(g) + 8H2O(g) + energy

When camping, a butane cartridge provides fuel for a portable burner.

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228 CHAPTER 8 Chemical Reactions

PRACTICE PROBLEMS

8.3 Types of Chemical Reactions

8. 19 Classify each of the following as a combination, decomposition, single replacement, double replacement, or combustion reaction: a. 2Al2O3(s) h

∆ 4Al(s) + 3O2(g)

b. Br2(l) + BaI2(s) h BaBr2(s) + I2(g) c. 2C2H2(g) + 5O2(g) h

∆ 4CO2(g) + 2H2O(g)

d. BaCl2(aq) + K2CO3(aq) h BaCO3(s) + 2KCl(aq) e. Pb(s) + O2(g) h PbO2(s)

8.20 Classify each of the following as a combination, decomposition, single replacement, double replacement, or combustion reaction: a. H2(g) + Br2(l) h 2HBr(g) b. AgNO3(aq) + NaCl(aq) h AgCl(s) + NaNO3(aq) c. 2H2O2(aq) h 2H2O(l) + O2(g) d. Zn(s) + CuCl2(aq) h Cu(s) + ZnCl2(aq) e. C5H8(g) + 7O2(g) h

∆ 5CO2(g) + 4H2O(g)

8.21 Classify each of the following as a combination, decomposition, single replacement, double replacement, or combustion reaction: a. 4Fe(s) + 3O2(g) h 2Fe2O3(s) b. Mg(s) + 2AgNO3(aq) h 2Ag(s) + Mg(NO3)2(aq) c. CuCO3(s) h

∆ CuO(s) + CO2(g)

d. Al2(SO4)3(aq) + 6KOH(aq) h 2Al(OH)3(s) + 3K2SO4(aq)

e. C4H8(g) + 6O2(g) h ∆

4CO2(g) + 4H2O(g)

8.22 Classify each of the following as a combination, decomposition, single replacement, double replacement, or combustion reaction: a. CuO(s) + 2HCl(aq) h CuCl2(aq) + H2O(l) b. 2Al(s) + 3Br2(l) h 2AlBr3(s) c. C6H12(l) + 9O2(g) h

∆ 6CO2(g) + 6H2O(g)

d. Fe2O3(s) + 3C(s) h 2Fe(s) + 3CO(g) e. C6H12O6(aq) h 2C2H6O(aq) + 2CO2(g)

8.23 Using Table 8.3, predict the products that would result from each of the following reactions and balance: a. combination: Mg(s) + Cl2(g) h b. decomposition: HBr(g) h c. single replacement: Mg(s) + Zn(NO3)2(aq) h d. double replacement: K2S(aq) + Pb(NO3)2(aq) h e. combustion: C2H6(g) + O2(g) h

∆

8.24 Using Table 8.3, predict the products that would result from each of the following reactions and balance: a. combination: Ca(s) + O2(g) h b. combustion: C6H6(l) + O2(g) h

∆

c. decomposition: PbO2(s) h ∆

d. single replacement: KI(s) + Cl2(g) h e. double replacement: CuCl2(aq) + Na2S(aq) h

8.4 Oxidation–Reduction Reactions LEARNING GOAL Define the terms oxidation and reduction; identify the reactants oxidized and reduced.

Perhaps you have never heard of an oxidation and reduction reaction. However, this type of reaction has many important applications in your everyday life. When you see a rusty nail, tarnish on a silver spoon, or corrosion on metal, you are observing oxidation.

4Fe(s) + 3O2(g) h 2Fe2O3(s) Fe is oxidized Rust

When we turn the lights on in our automobiles, an oxidation–reduction reaction within the car battery provides the electricity. On a cold, wintry day, we might build a fire. As the wood burns, oxygen combines with carbon and hydrogen to produce carbon dioxide, water, and heat. In the previous section, we called this a combustion reaction, but it is also an oxidation–reduction reaction. When we eat foods with starches in them, the starches break down to give glucose, which is oxidized in our cells to give us energy along with carbon dioxide and water. Every breath we take provides oxygen to carry out oxidation in our cells.

C6H12O6(aq) + 6O2(g) h 6CO2(g) + 6H2O(l) + energy Glucose

Oxidation–Reduction Reactions In an oxidation–reduction reaction (redox), electrons are transferred from one substance to another. If one substance loses electrons, another substance must gain electrons. Oxidation is defined as the loss of electrons; reduction is the gain of electrons.

Rust forms when the oxygen in the air reacts with iron.

A oxidized

B reduced

A B

Reduction (gain of electron)

Oxidation (loss of electron)

e-

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8.4 Oxidation–Reduction Reactions 229

One way to remember these definitions is to use the following acronym:

OIL RIG Oxidation Is Loss of electrons Reduction Is Gain of electrons

In general, atoms of metals lose electrons to form positive ions, whereas nonmetals gain electrons to form negative ions. Now we can say that metals are oxidized and nonmetals are reduced.

The green color that appears on copper surfaces from weathering, known as patina, is a mixture of CuCO3 and CuO. We can now look at the oxidation and reduction reactions that take place when copper metal reacts with oxygen in the air to produce copper(II) oxide.

2Cu(s) + O2(g) h 2CuO(s)

The element Cu in the reactants has a charge of 0, but in the CuO product, it is present as Cu2+, which has a 2 + charge. Because the Cu atom lost two electrons, the Cu was oxidized in the reaction.

Cu0(s) h Cu2+(s) + 2 e- Oxidation: loss of electrons by Cu

At the same time, the element O in the reactants has a charge of 0, but in the CuO product, it is present as O2-, which has a 2 - charge. Because the O atom has gained two electrons, the O was reduced in the reaction.

O2 0(g) + 4 e- h 2O2-(s) Reduction: gain of electrons by O

Thus, the overall equation for the formation of CuO involves an oxidation and a reduc- tion that occur simultaneously. In every oxidation and reduction, the number of electrons lost must be equal to the number of electrons gained. Therefore, we multiply the oxidation reaction of Cu by 2. Canceling the 4 e- on each side, we obtain the overall oxidation– reduction equation for the formation of CuO.

2Cu(s) h 2Cu2+(s) + 4 e- Oxidation

O2(g) + 4 e- h 2O2-(s) Reduction

2Cu(s) + O2(g) h 2CuO(s) Oxidation–reduction equation

As we see in the next reaction between zinc and copper(II) sulfate, there is always an oxidation with every reduction (see FIGURE 8.6).

Zn(s) + CuSO4(aq) h ZnSO4(aq) + Cu(s)

We rewrite the equation to show the atoms and ions.

Zn(s) + Cu2+(aq) + SO4 2-(aq) h Zn2+(aq) + SO4 2-(aq) + Cu(s)

In this reaction, Zn atoms lose two electrons to form Zn2+, which indicates that Zn is oxidized. At the same time, Cu2+ gains two electrons, which indicates that Cu is reduced. The SO4

2- ions are spectator ions, which are present in both the reactants and products and do not change.

Zn(s) h Zn2+(aq) + 2 e- Oxidation of Zn

Cu2+(aq) + 2 e- h Cu(s) Reduction of Cu2+

In this single replacement reaction, zinc was oxidized and copper(II) ion was reduced.

Oxidation and Reduction in Biological Systems Oxidation may also involve the addition of oxygen or the loss of hydrogen, and reduction may involve the loss of oxygen or the gain of hydrogen. In the cells of the body, oxidation of organic (carbon) compounds involves the transfer of hydrogen atoms (H), which are composed of electrons and protons. For example, the oxidation of a typical biochemical

CORE CHEMISTRY SKILL Identifying Oxidized and Reduced

Substances

PRACTICE PROBLEMS Try Practice Problems 8. 25 to 8. 28

The green patina on copper is due to oxidation.

Oxidation: Lose e-

Reduction: Gain e-

Reduced

Na

Ca

2Br-

Fe2+

Oxidized

Na+ + e-

Ca2+ + 2 e-

Br2 + 2 e -

Fe3+ + e-

Oxidation is a loss of electrons; reduction is a gain of electrons.

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230 CHAPTER 8 Chemical Reactions

FIGURE 8.6 In this single replacement reaction, Zn(s) is oxidized to Zn2+(aq) when it provides two electrons to reduce Cu2+(aq) to Cu(s): Zn(s) + Cu2+(aq) h Zn2+(aq) + Cu(s)

Cu2+ ion Zn0 Zn0 Zn0

Zn0 Zn2+

Solution of Cu2+ and SO4

2 -

Zn strip

Cu(s)

Zn2+ ion

Cu2+ Cu0

Cu02 e -

molecule can involve the transfer of two hydrogen atoms (or 2H+ and 2 e-) to a hydrogen ion acceptor such as the coenzyme FAD (flavin adenine dinucleotide). The coenzyme is reduced to FADH2.

Reduction (gain of 2H)

Oxidation (loss of 2H)

2H in biological molecule

Coenzyme FAD

Coenzyme FADH2

Oxidized biological molecule

In many biochemical oxidation–reduction reactions, the transfer of hydrogen atoms is necessary for the production of energy in the cells. For example, methyl alcohol (CH4O), a poisonous substance, is metabolized in the body by the following reactions:

CH4O h CH2O + 2H Oxidation: loss of H atoms Methyl Formaldehyde alcohol

The formaldehyde can be oxidized further, this time by the addition of oxygen, to produce formic acid.

2CH2O + O2 h 2CH2O2 Oxidation: addition of O atoms Formaldehyde Formic acid

Finally, formic acid is oxidized to carbon dioxide and water.

2CH2O2 + O2 h 2CO2 + 2H2O Oxidation: addition of O atoms Formic acid

The intermediate products of the oxidation of methyl alcohol are quite toxic, causing blindness and possibly death as they interfere with key reactions in the cells of the body.

In summary, we find that the particular definition of oxidation and reduction we use depends on the process that occurs in the reaction. All these definitions are summarized in TABLE 8.4. Oxidation always involves a loss of electrons, but it may also be seen as an addition of oxygen or the loss of hydrogen atoms. A reduction always involves a gain of electrons and may also be seen as the loss of oxygen or the gain of hydrogen.

TABLE 8.4 Characteristics of Oxidation and Reduction

Always Involves May Involve

Oxidation

Loss of electrons Addition of oxygen

Loss of hydrogen

Reduction

Gain of electrons Loss of oxygen

Gain of hydrogen

ENGAGE 8.8 How can you determine that Cu2+ is reduced in this reaction?

M08_TIMB8119_06_SE_C08.indd 230 11/27/18 11:52 AM

8.4 Oxidation–Reduction Reactions 231

SAMPLE PROBLEM 8.7 Oxidation and Reduction

TRY IT FIRST

For each of the following reactions, determine the number of electrons lost or gained and identify as an oxidation or a reduction:

a. Mg2+(aq) h Mg(s) b. Fe(s) h Fe3+(aq)

SOLUTION

a. Mg2+(aq) + 2 e- h Mg(s) Reduction b. Fe(s) h Fe3+(aq) + 3 e- Oxidation

SELF TEST 8.7

Identify each of the following as an oxidation or a reduction: a. Co3+(aq) + e- h Co2+(aq) b. 2I-(aq) h I2(s) + 2 e-

ANSWER

a. reduction b. oxidation

PRACTICE PROBLEMS Try Practice Problems 8.29 to 8.32

PRACTICE PROBLEMS

8.4 Oxidation–Reduction Reactions

8.25 Identify each of the following as an oxidation or a reduction: a. Na+(aq) + e- h Na(s) b. Ni(s) h Ni2+(aq) + 2 e-

c. Cr3+(aq) + 3 e- h Cr(s) d. 2H+(aq) + 2 e- h H2(g)

8.26 Identify each of the following as an oxidation or a reduction: a. O2(g) + 4 e- h 2O2-(aq) b. Ag(s) h Ag +(aq) + e-

c. Fe3+(aq) + e- h Fe2+(aq) d. 2Br-(aq) h Br2(l) + 2 e-

8.27 In each of the following, identify the reactant that is oxidized and the reactant that is reduced: a. Zn(s) + Cl2(g) h ZnCl2(s) b. Cl2(g) + 2NaBr(aq) h 2NaCl(aq) + Br2(l) c. 2PbO(s) h 2Pb(s) + O2(g) d. 2Fe3+(aq) + Sn2+(aq) h 2Fe2+(aq) + Sn4+(aq)

8.28 In each of the following, identify the reactant that is oxidized and the reactant that is reduced: a. 2Li(s) + F2(g) h 2LiF(s) b. Cl2(g) + 2KI(aq) h I2(s) + 2KCl(aq) c. 2Al(s) + 3Sn2+(aq) h 2Al3+(aq) + 3Sn(s) d. Fe(s) + CuSO4(aq) h Cu(s) + FeSO4(aq)

Applications

8.29 In the mitochondria of human cells, energy is provided by the oxidation and reduction reactions of the iron ions in the cyto- chromes in electron transport. Identify each of the following as an oxidation or a reduction:

a. Fe3+ + e- h Fe2+ b. Fe2+ h Fe3+ + e-

8.30 Chlorine (Cl2) is a strong germicide used to disinfect drinking water and to kill microbes in swimming pools. If the product is Cl-, was the elemental chlorine oxidized or reduced?

8.31 When linoleic acid, an unsaturated fatty acid, reacts with hydrogen, it forms a saturated fatty acid. Is linoleic acid oxidized or reduced in the hydrogenation reaction?

C18H32O2 + 2H2 h C18H36O2 Linoleic acid

8.32 In one of the reactions in the citric acid cycle, which provides energy, succinic acid is converted to fumaric acid.

C4H6O4 h C4H4O4 + 2H Succinic acid Fumaric acid

The reaction is accompanied by a coenzyme, flavin adenine dinucleotide (FAD).

FAD + 2H h FADH2 a. Is succinic acid oxidized or reduced? b. Is FAD oxidized or reduced? c. Why would the two reactions occur together?

M08_TIMB8119_06_SE_C08.indd 231 11/27/18 11:52 AM

232 CHAPTER 8 Chemical Reactions

UPDATE Improving Natalie’s Overall Fitness

Natalie’s test results indicate that she has a blood oxygen level of 89%. The normal values for pulse oximeter readings are 95% to 100%, which means that Natalie’s O2 saturation is low. Thus, Natalie does not have an adequate amount of O2 in her blood. This may be the reason she has noticed a

shortness of breath and a dry cough. Her doctor diagnosed her with interstitial lung disease, which is scarring of the tissue of the lungs.

Angela teaches Natalie to inhale and exhale slower and deeper to fill the lungs with more air and thus more oxygen. Angela also develops a workout program with the goal of increasing Natalie’s overall fitness level. During the exercises, Angela continues to monitor Natalie’s heart rate, blood O2 level, and blood pressure to ensure that Natalie is exercising at a level that will enable her to become stronger without breaking down muscle due to a lack of oxygen.

Applications

8.33 a. During cellular respiration, aqueous glucose (C6H12O6) in the cells undergoes combustion. Write and balance the chemical equation for reaction of glucose.

b. In plants, carbon dioxide gas and liquid water are con- verted to aqueous glucose (C6H12O6) and oxygen gas. Write and balance the chemical equation for production of glucose in plants.

8.34 Aqueous fatty acids undergo reaction with oxygen gas and form gaseous carbon dioxide and liquid water when utilized for energy in the body. a. Write and balance the equation for the combustion of the

fatty acid capric acid (C10H20O2). b. Write and balance the equation for the combustion of the

fatty acid myristic acid (C14H28O2).

Low-intensity exercises are used at the beginning of Natalie’s exercise program.

CONCEPT MAP

are balanced with

to give

and those with

are also

Reactants and Products

Combination, Decomposition,

Single Replacement, Double Replacement,

and Combustion

Coefficients

Equal Numbers of Atoms on Each Side

Oxidation–Reduction Reactions

Loss or Gain of Electrons

are classified as

CHEMICAL REACTIONS

show a chemical change between

M08_TIMB8119_06_SE_C08.indd 232 11/27/18 11:52 AM

Core Chemistry Skills 233

CHAPTER REVIEW

8.1 Equations for Chemical Reactions LEARNING GOAL Identify a balanced chemical equation; determine the number of atoms in the reactants and products. • A chemical change occurs when the

atoms of the initial substances rearrange to form new substances. • A chemical equation shows the formulas of the substances that

react on the left side of a reaction arrow and the products that form on the right side of the reaction arrow.

8.2 Balancing a Chemical Equation LEARNING GOAL Write a balanced chemical equation from the formulas of the reactants and products for a chemical reaction. • A chemical equation

is balanced by writing coefficients, small whole numbers, in front of formulas to equalize the atoms of each of the elements in the reactants and the products.

8.3 Types of Chemical Reactions LEARNING GOAL Identify a chemical reaction as a combination, decomposition, single replacement, double replacement, or combustion. • Many chemical reactions can be organized by reaction type: com-

bination, decomposition, single replacement, double replacement, or combustion.

8.4 Oxidation–Reduction Reactions LEARNING GOAL Define the terms oxidation and reduction; identify the reactants oxidized and reduced. • When electrons are transferred

in a reaction, it is an oxidation– reduction reaction.

• One reactant loses electrons, and another reactant gains electrons. • Overall, the number of electrons lost and gained is equal.

+

2H2O(g)2H2(g) O2(g)

+

+ Balanced

A oxidized

B reduced

A B

Reduction (gain of electron)

Oxidation (loss of electron)

e-

+ One element replaces another element

+

Single replacement

A CB BC A

O2CH4 H2OCO2

balanced equation The final form of a chemical equation that shows the same number of atoms of each element in the reactants and products.

chemical equation A shorthand way to represent a chemical reaction using chemical formulas to indicate the reactants and products and coefficients to show reacting ratios.

coefficients Whole numbers placed in front of the formulas to bal- ance the number of atoms or moles of atoms of each element on both sides of an equation.

combination reaction A chemical reaction in which reactants com- bine to form a single product.

combustion reaction A chemical reaction in which a fuel containing carbon reacts with oxygen to produce CO2, H2O, and energy.

decomposition reaction A reaction in which a single reactant splits into two or more simpler substances.

KEY TERMS

double replacement reaction A reaction in which the positive ions in the reacting compounds exchange places.

oxidation The loss of electrons by a substance. Biological oxidation may involve the addition of oxygen or the loss of hydrogen.

oxidation–reduction reaction A reaction in which the oxidation of one reactant is always accompanied by the reduction of another reactant.

products The substances formed as a result of a chemical reaction. reactants The initial substances that undergo change in a chemical

reaction. reduction The gain of electrons by a substance. Biological reduction

may involve the loss of oxygen or the gain of hydrogen. single replacement reaction A reaction in which an element

replaces a different element in a compound.

The chapter section containing each Core Chemistry Skill is shown in parentheses at the end of each heading.

Balancing a Chemical Equation (8.2) • In a balanced chemical equation, whole numbers called coeffi-

cients multiply each of the atoms in the chemical formulas so that the number of each type of atom in the reactants is equal to the number of the same type of atom in the products.

Example: Balance the following chemical equation:

SnCl4(s) + H2O(l) h Sn(OH)4(s) + HCl(aq) Not balanced

Answer: When we count the atoms on the reactant side and the prod- uct side, we see that there are more Cl atoms in the reactants and more O and H atoms in the products.

CORE CHEMISTRY SKILLS

To balance the equation, we need to use coefficients in front of the formulas containing the Cl atoms, H atoms, and O atoms.

• Place a 4 in front of the formula HCl to give a total of 8 H atoms and 4 Cl atoms in the products.

SnCl4(s) + H2O(l) h Sn(OH)4(s) + 4HCl(aq)

• Place a 4 in front of the formula H2O to give 8 H atoms and 4 O atoms in the reactants.

SnCl4(s) + 4H2O(l) h Sn(OH)4(s) + 4HCl(aq)

• The total number of Sn (1), Cl (4), H (8), and O (4) atoms is now equal on both sides of the equation. Thus, this equation is balanced.

SnCl4(s) + 4H2O(l) h Sn(OH)4(s) + 4HCl(aq)

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234 CHAPTER 8 Chemical Reactions

Classifying Types of Chemical Reactions (8.3) • Chemical reactions are classified by identifying general patterns in

their equations. • In a combination reaction, two or more elements or compounds

bond to form one product. • In a decomposition reaction, a single reactant splits into two or

more products. • In a single replacement reaction, an uncombined element takes the

place of an element in a compound. • In a double replacement reaction, the positive ions in the reacting

compounds switch places. • In a combustion reaction, a fuel containing carbon and hydrogen

reacts with oxygen from the air to produce carbon dioxide (CO2), water (H2O), and energy.

Example: Classify the type of the following reaction:

2Al(s) + Fe2O3(s) h ∆

Al2O3(s) + 2Fe(l)

Answer: The iron in iron(III) oxide is replaced by aluminum, which makes this a single replacement reaction.

Identifying Oxidized and Reduced Substances (8.4)

• In an oxidation–reduction reaction (abbreviated redox), one reactant is oxidized when it loses electrons, and another reactant is reduced when it gains electrons.

• Oxidation is the loss of electrons; reduction is the gain of electrons.

Example: For the following redox reaction, identify the reactant that is oxidized, and the reactant that is reduced:

Fe(s) + Cu2+(aq) h Fe2+(aq) + Cu(s) Answer: Fe0(s) h Fe2+(aq) + 2 e-

Fe loses electrons; it is oxidized.

Cu2+(aq) + 2 e- h Cu0(s) Cu2+ gains electrons; it is reduced.

UNDERSTANDING THE CONCEPTS

The chapter sections to review are shown in parentheses at the end of each problem.

8.35 Balance each of the following by adding coefficients, and identify the type of reaction for each: (8.1, 8.2, 8.3)

+

b.

a.

+

Reactants Products

8.36 Balance each of the following by adding coefficients, and identify the type of reaction for each: (8.1, 8.2, 8.3)

8.37 If red spheres represent oxygen atoms, blue spheres represent nitrogen atoms, and all the molecules are gases, (8.1, 8.2, 8.3)

Reactants Products

a. write the formula for each of the reactants and products. b. write a balanced equation for the reaction. c. indicate the type of reaction as combination, decomposition,

single replacement, double replacement, or combustion.

Reactants Products

a. write the formula for each of the reactants and products. b. write a balanced equation for the reaction. c. indicate the type of reaction as combination, decomposition,

single replacement, double replacement, or combustion.

8.39 If blue spheres represent nitrogen atoms, purple spheres represent iodine atoms, the reacting molecules are solid, and the products are gases, (8.1, 8.2, 8.3)

a. write the formula for each of the reactants and products. b. write a balanced equation for the reaction. c. indicate the type of reaction as combination, decomposition,

single replacement, double replacement, or combustion.

+ +

+b.

a.

8. 38 If purple spheres represent iodine atoms, white spheres represent hydrogen atoms, and all the molecules are gases, (8.1, 8.2, 8.3)

M08_TIMB8119_06_SE_C08.indd 234 11/27/18 11:52 AM

Additional Practice Problems 235

a. write the formula for each of the reactants and products. b. write a balanced equation for the reaction. c. indicate the type of reaction as combination, decomposition,

single replacement, double replacement, or combustion.

8.42 If blue spheres represent nitrogen atoms, purple spheres rep- resent iodine atoms, the reacting molecules are gases, and the products are solid, (8.1, 8.2, 8.3)

8.40 If green spheres represent chlorine atoms, yellow-green spheres represent fluorine atoms, white spheres represent hydrogen atoms, and all the molecules are gases, (8.1, 8.2, 8.3)

Reactants Products

a. write the formula for each of the reactants and products. b. write a balanced equation for the reaction. c. indicate the type of reaction as combination, decomposition,

single replacement, double replacement, or combustion.

8.41 If green spheres represent chlorine atoms, red spheres represent oxygen atoms, and all the molecules are gases, (8.1, 8.2, 8.3)

Reactants Products

a. write the formula for each of the reactants and products. b. write a balanced equation for the reaction. c. indicate the type of reaction as combination, decomposition,

single replacement, double replacement, or combustion.

Reactants Products

8.43 Identify the type of reaction for each of the following as combination, decomposition, single replacement, double replacement, or combustion: (8.3) a. A metal and a nonmetal form an ionic compound. b. A compound of hydrogen and carbon reacts with oxygen to

produce carbon dioxide and water. c. Heating magnesium carbonate produces magnesium oxide

and carbon dioxide. d. Zinc replaces copper in Cu(NO3)2.

8.44 Identify the type of reaction for each of the following as combination, decomposition, single replacement, double replacement, or combustion: (8.3) a. A compound breaks apart into its elements. b. Copper and bromine form copper(II) bromide. c. Iron(II) sulfite breaks down to iron(II) oxide and sulfur

dioxide. d. Silver ion from AgNO3(aq) forms a solid with bromide ion

from KBr(aq).

8.45 Balance each of the following chemical equations, and identify the type of reaction: (8.1, 8.2, 8.3) a. NH3(g) + HCl(g) h NH4Cl(s) b. C4H8(g) + O2(g) h

∆ CO2(g) + H2O(g)

c. Sb(s) + Cl2(g) h SbCl3(s) d. NI3(s) h N2(g) + I2(g) e. KBr(aq) + Cl2(aq) h KCl(aq) + Br2(l) f. H2SO4(aq) + Fe(s) h H2(g) + Fe2(SO4)3(aq) g. Al2(SO4)3(aq) + NaOH(aq) h

Al(OH)3(s) + Na2SO4(aq)

8.46 Balance each of the following chemical equations, and identify the type of reaction: (8.1, 8.2, 8.3) a. Si3N4(s) h Si(s) + N2(g) b. Mg(s) + N2(g) h Mg3N2(s) c. H3PO4(aq) + Al(s) h H2(g) + AlPO4(aq) d. C3H4(g) + O2(g) h

∆ CO2(g) + H2O(g)

e. Cr2O3(s) + H2(g) h Cr(s) + H2O(g) f. Al(s) + Cl2(g) h AlCl3(s) g. MgCl2(aq) + AgNO3(aq) h AgCl(s) + Mg(NO3)2(aq)

8.47 Predict the products and write a balanced equation for each of the following: (8.1, 8.2, 8.3) a. single replacement:

HCl(aq) + Zn(s) h _____ + _____

b. decomposition: BaCO3(s) h ∆ _____ + _____

c. double replacement: HCl(aq) + NaOH(aq) h _____ + _____

d. combination: Al(s) + F2(g) h _____ 8.48 Predict the products and write a balanced equation for each of

the following: (8.1, 8.2, 8.3)

a. decomposition: NaCl(s) iih Electricity

_____ + _____ b. combination: Ca(s) + Br2(l) h c. combustion: C2H4(g) + O2(g) h

∆ _____ + _____ d. double replacement:

NiCl2(aq) + NaOH(aq) h _____ + _____

ADDITIONAL PRACTICE PROBLEMS

M08_TIMB8119_06_SE_C08.indd 235 11/27/18 11:53 AM

236 CHAPTER 8 Chemical Reactions

8.51 In each of the following reactions, identify the reactant that is oxidized and the reactant that is reduced: (8.4) a. N2(g) + 2O2(g) h 2NO2(g) b. CO(g) + 3H2(g) h CH4(g) + H2O(g) c. Mg(s) + Br2(l) h MgBr2(s)

8.52 In each of the following reactions, identify the reactant that is oxidized and the reactant that is reduced: (8.4) a. 2Al(s) + 3F2(g) h 2AlF3(s) b. ZnO(s) + H2(g) h Zn(s) + H2O(g) c. 2CuS(s) + 3O2(g) h 2CuO(s) + 2SO2(g)

8.49 Write a balanced equation for each of the following reactions and identify the type of reaction: (8.1, 8.2, 8.3) a. Sodium metal reacts with oxygen gas to form solid sodium

oxide. b. Aqueous sodium chloride and aqueous silver nitrate react to

form solid silver chloride and aqueous sodium nitrate. c. Gasohol is a fuel that contains liquid ethanol, C2H6O, which

burns in oxygen gas to form two gases, carbon dioxide and water.

8.50 Write a balanced equation for each of the following reactions and identify the type of reaction: (8.1, 8.2, 8.3) a. Solid potassium chlorate is heated to form solid potassium

chloride and oxygen gas. b. Carbon monoxide gas and oxygen gas combine to form

carbon dioxide gas. c. Ethene gas, C2H4, reacts with chlorine gas, Cl2, to form

dichloroethane, C2H4Cl2.

The following problems are related to the topics in this chapter. However, they do not all follow the chapter order, and they require you to combine concepts and skills from several sections. These problems will help you increase your critical thinking skills and prepare for your next exam.

8.53 Balance each of the following chemical equations, and identify the type of reaction: (8.1, 8.2, 8.3)

a. K2O(s) + H2O(g) h KOH(s) b. C8H18(l) + O2(g) h

∆ CO2(g) + H2O(g)

c. Fe(OH)3(s) h Fe2O3(s) + H2O(g) d. CuS(s) + HCl(aq) h CuCl2(aq) + H2S(g)

8.54 Balance each of the following chemical equations, and identify the type of reaction: (8.1, 8.2, 8.3)

a. TiCl4(s) + Mg(s) h MgCl2(s) + Ti(s) b. P4O10(s) + H2O(g) h

∆ H3PO4(aq)

c. KClO3(s) h ∆

KCl(s) + O2(g) d. C3H6(g) + O2(g) h

∆ CO2(g) + H2O(g)

8.55 Complete and balance each of the following chemical equations: (8.1, 8.2, 8.3)

a. single replacement: Fe3O4(s) + H2(g) h b. combustion: C4H10(g) + O2(g) h

∆

c. combination: Al(s) + O2(g) h d. double replacement: NaOH(aq) + ZnSO4(aq) h

8. 56 Complete and balance each of the following chemical equations: (8.1, 8.2, 8.3)

a. decomposition: HgO(s) h ∆

b. double replacement: BaCl2(aq) + AgNO3(aq) h c. single replacement: Ca(s) + AlCl3(s) h d. combination: Mg(s) + N2(g) h

8. 57 Write the correct formulas for the reactants and products, the balanced equation for each of the following reaction descriptions, and identify each type of reaction: (8.1, 8.2, 8.3)

a. An aqueous solution of lead(II) nitrate is mixed with aqueous sodium phosphate to produce solid lead(II) phosphate and aqueous sodium nitrate.

b. Gallium metal heated in oxygen gas forms solid gallium(III) oxide.

c. When solid sodium nitrate is heated, solid sodium nitrite and oxygen gas are produced.

8.58 Write the correct formulas for the reactants and products, the balanced equation for each of the following reaction descriptions, and identify each type of reaction: (8.1, 8.2, 8.3)

a. Solid bismuth(III) oxide and solid carbon react to form bismuth metal and carbon monoxide gas.

b. Solid sodium bicarbonate is heated and forms solid sodium carbonate, gaseous carbon dioxide, and liquid water.

c. Liquid hexane, C6H14, reacts with oxygen gas to form two gaseous products: carbon dioxide and liquid water.

8. 59 In the following diagram, if blue spheres are the element X and yellow spheres are the element Y: (8.1, 8.2, 8.3)

CHALLENGE PROBLEMS

Reactants Products

a. Write the formula for each of the reactants and products. b. Write a balanced equation for the reaction. c. Indicate the type of reaction as combination, decomposition,

single replacement, double replacement, or combustion.

8. 60 In the following diagram, if red spheres are the element A, white spheres are the element B, and green spheres are the element C: (8.1, 8.2, 8.3)

Reactants Products

a. Write the formula for each of the reactants and products. b. Write a balanced equation for the reaction. c. Indicate the type of reaction as combination, decomposition,

single replacement, double replacement, or combustion.

M08_TIMB8119_06_SE_C08.indd 236 11/27/18 11:53 AM

Answers to Selected Problems 237

ANSWERS TO ENGAGE QUESTIONS 8. 5 In a decomposition reaction, a reactant splits into two or more

simpler products.

8. 6 In a single replacement reaction, one element in the reactants switches places with an element in the other reacting compound.

8. 7 In a double replacement reaction, the positive ions in the reactants switch places.

8. 8 When Cu2+ gains two electrons to form Cu0, it is a reduction.

8. 1 A change in color in the formation of rust indicates a chemical change.

8. 2 A chemical equation is balanced when there are the same num- bers of atoms of each element in the reactants and the products, and the coefficients are the lowest possible whole numbers.

8. 3 The formation of a white solid (precipitate) in the products, which has a different formula from the reactants, is evidence of a chemical reaction.

8. 4 In a combination reaction, the atoms from the reactants combine to form one product.

ANSWERS TO SELECTED PROBLEMS 8. 23 a. Mg(s) + Cl2(g) h MgCl2(s)

b. 2HBr(g) h H2(g) + Br2(l) c. Mg(s) + Zn(NO3)2(aq) h Zn(s) + Mg(NO3)2(aq) d. K2S(aq) + Pb(NO3)2(aq) h PbS(s) + 2KNO3(aq) e. 2C2H6(g) + 7O2(g) h

∆ 4CO2(g) + 6H2O(g)

8. 25 a. reduction b. oxidation c. reduction d. reduction

8. 27 a. Zn is oxidized; Cl2 is reduced. b. The Br- in NaBr is oxidized; Cl2 is reduced. c. The O2- in PbO is oxidized; the Pb2+ in PbO is reduced. d. Sn2+ is oxidized; Fe3+ is reduced.

8. 29 a. reduction b. oxidation

8. 31 Linoleic acid gains hydrogen atoms and is reduced.

8. 33 a. C6H12O6(aq) + 6O2(g) h 6CO2(g) + 6H2O(l) b. 6CO2(g) + 6H2O(l) h C6H12O6(aq) + 6O2(g)

8. 35 a. 1,1,2 combination b. 2,2,1 decomposition

8. 37 a. reactants NO and O2; product NO2 b. 2NO(g) + O2(g) h 2NO2(g) c. combination

8. 39 a. reactant NI3; products N2 and I2 b. 2NI3(s) h N2(g) + 3I2(g) c. decomposition

8.41 a. reactants Cl2 and O2; product OCl2 b. 2Cl2(g) + O2(g) h 2OCl2(g) c. combination

8. 43 a. combination b. combustion c. decomposition d. single replacement

8. 45 a. NH3(g) + HCl(g) h NH4Cl(s) combination b. C4H8(g) + 6O2(g) h

∆ 4CO2(g) + 4H2O(g) combustion

c. 2Sb(s) + 3Cl2(g) h 2SbCl3(s) combination d. 2NI3(s) h N2(g) + 3I2(g) decomposition e. 2KBr(aq) + Cl2(aq) h 2KCl(aq) + Br2(l)

single replacement f. 3H2SO4(aq) + 2Fe(s) h 3H2(g) + Fe2(SO4)3(aq)

single replacement g. Al2(SO4)3(aq) + 6NaOH(aq) h 2Al(OH)3(s)

+ 3Na2SO4(aq) double replacement 8. 47 a. 2HCl(aq) + Zn(s) h H2(g) + ZnCl2(aq)

b. BaCO3(s) h ∆

BaO(s) + CO2(g) c. HCl(aq) + NaOH(aq) h H2O(l) + NaCl(aq) d. 2Al(s) + 3F2(g) h 2AlF3(s)

8.1 a. reactants/products: 7 O atoms b. reactants/products: 4 O atoms c. reactants/products: 10 O atoms d. reactants/products: 4 O atoms

8.3 a. not balanced b. not balanced c. balanced d. balanced

8.5 a. 2 Na atoms, 2 Cl atoms b. 1 P atom, 3 Cl atoms, 6 H atoms c. 4 P atoms, 16 O atoms, 12 H atoms

8.7 a. N2(g) + O2(g) h 2NO(g) b. 2HgO(s) h

∆ 2Hg(l) + O2(g)

c. 4Fe(s) + 3O2(g) h 2Fe2O3(s) d. 2Na(s) + Cl2(g) h 2NaCl(s) e. 2Cu2O(s) + O2(g) h 4CuO(s)

8.9 a. Mg(s) + 2AgNO3(aq) h 2Ag(s) + Mg(NO3)2(aq) b. CuCO3(s) h CuO(s) + CO2(g) c. 2Al(s) + 3CuSO4(aq) h 3Cu(s) + Al2(SO4)3(aq) d. Pb(NO3)2(aq) + 2NaCl(aq) h PbCl2(s) + 2NaNO3(aq) e. 6HCl(aq) + 2Al(s) h 3H2(g) + 2AlCl3(aq)

8.11 a. Fe2O3(s) + 3CO(g) h 2Fe(s) + 3CO2(g) b. 2Li3N(s) h 6Li(s) + N2(g) c. 6HBr(aq) + 2Al(s) h 3H2(g) + 2AlBr3(aq) d. 3Ba(OH)2(aq) + 2Na3PO4(aq) h

Ba3(PO4)2(s) + 6NaOH(aq) e. As4S6(s) + 9O2(g) h As4O6(s) + 6SO2(g)

8.13 a. 2Li(s) + 2H2O(l) h H2(g) + 2LiOH(aq) b. 2P(s) + 5Cl2(g) h 2PCl5(s) c. FeO(s) + CO(g) h Fe(s) + CO2(g) d. 2C5H10(l) + 15O2(g) h

∆ 10CO2(g) + 10H2O(g)

e. 3H2S(g) + 2FeCl3(s) h Fe2S3(s) + 6HCl(g)

8.15 NH4NO3(s) h ∆

2H2O(g) + N2O(g) 8.17 2C3H7NO2(aq) + 6O2(g) h

5CO2(g) + 5H2O(l) + CH4N2O(aq) 8.19 a. decomposition b. single replacement c. combustion d. double replacement e. combination

8.21 a. combination b. single replacement c. decomposition d. double replacement e. combustion

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238 CHAPTER 8 Chemical Reactions

8. 55 a. Fe3O4(s) + 4H2(g) h 3Fe(s) + 4H2O(g) b. 2C4H10(g) + 13O2(g) h

∆ 8CO2(g) + 10H2O(g)

c. 4Al(s) + 3O2(g) h 2Al2O3(s) d. 2NaOH(aq) + ZnSO4(aq) h

Zn(OH)2(s) + Na2SO4(aq) 8. 57 a. 3Pb(NO3)2(aq) + 2Na3PO4(aq) h

Pb3(PO4)2(s) + 6NaNO3(aq) double replacement b. 4Ga(s) + 3O2(g) h

∆ 2Ga2O3(s) combination

c. 2NaNO3(s) h ∆

2NaNO2(s) + O2(g) decomposition 8. 59 a. reactants: X and Y2; product: XY3

b. 2X + 3Y2 h 2XY3 c. combination

8. 49 a. 4Na(s) + O2(g) h 2Na2O(s) combination b. NaCl(aq) + AgNO3(aq) h AgCl(s) + NaNO3(aq)

double replacement c. C2H6O(l) + 3O2(g) h

∆ 2CO2(g) + 3H2O(g) combustion

8. 51 a. N2 is oxidized; O2 is reduced. b. H2 is oxidized; the C (in CO) is reduced. c. Mg is oxidized; Br2 is reduced.

8. 53 a. K2O(s) + H2O(g) h 2KOH(s) combination b. 2C8H18(l) + 25O2(g) h

∆ 16CO2(g) + 18H2O(g)

combustion c. 2Fe(OH)3(s) h Fe2O3(s) + 3H2O(g) decomposition d. CuS(s) + 2HCl(aq) h CuCl2(aq) + H2S(g)

double replacement

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239

Lance, an environmental scientist, collects soil and water samples at a nearby farm to test for the presence and concentration of insecticides and pharmaceuticals. Farmers use insecticides to increase food production and pharmaceuticals to treat and prevent animal-related diseases. Due to the common use of these chemicals, they may pass into the soil and water supply, potentially contaminating the environment and causing health problems.

Recently, a farmer treated his cotton and bean fields with an insecticide called Sevin (carbaryl), which is an acetylcholinesterase inhibitor that acts on the central nervous system. A few days later, Lance collects samples of soil and water and detects small amounts of Sevin. In humans, Sevin can cause headaches, nausea, and paralysis of the respiratory system. Because Sevin is very soluble, it is important that it does not contaminate the water supply. Lance advises the farmer to decrease the amount of Sevin he uses on his crops. Lance informs the farmer that he will return in a week to retest the soil and water for Sevin.

CAREER

Environmental Scientist Environmental science is a multidisciplinary field that combines chemistry, biology, ecology, and geology to study environmental problems. Environmental scientists monitor environmental pollution to protect the health of the public. By using specialized equipment, environmental scientists measure pollution levels in soil, air, and water, as well as noise and radiation levels. They can specialize in a specific area, such as air quality or hazardous and solid waste. For instance, air- quality experts monitor indoor air for allergens, mold, and toxins; they measure outdoor air pollutants created by businesses, vehicles, and agriculture. Since environmental scientists obtain samples containing potentially hazardous materials, they must be knowledgeable about safety protocols and wear personal protective equipment. They may also recommend methods to diminish various pollutants and may assist in cleanup and remediation efforts.

Chemical Quantities in Reactions

9

UPDATE Testing Water Samples for Insecticides

Last week, Lance's measurements indicated that the insecticide level exceeded government guidelines. This week, Lance is at the farm rechecking the levels of insecticides in the soil and water, which he finds are much lower and now within government guidelines. You can view the information in the UPDATE Testing Water Samples for Insecticides, page 260.

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240 CHAPTER 9 Chemical Quantities in Reactions

9.1 Conservation of Mass LEARNING GOAL Calculate the total mass of reactants and the total mass of products in a balanced chemical equation.

In any chemical reaction, the total amount of matter in the reactants is equal to the total amount of matter in the products. Thus, the total mass of all the reactants must be equal to the total mass of all the products. This is known as the Law of Conservation of Mass, which states that there is no change in the total mass of the substances reacting in a balanced chemical reaction. Thus, no material is lost or gained as original substances are changed to new substances. For example, tarnish (Ag2S) forms when silver reacts with sulfur to form silver sulfide.

2Ag(s) + S(s) h Ag2S(s)

2Ag(s) + S(s) Ag2S(s)

Mass of reactants = Mass of product

In the chemical reaction of Ag and S, the mass of the reactants is the same as the mass of the product, Ag2S.

LOOKING AHEAD

9.1 Conservation of Mass 240

9.2 Mole Relationships in Chemical Equations 241

9.3 Mass Calculations for Chemical Reactions 245

9.4 Limiting Reactants 247 9.5 Percent Yield 252 9.6 Energy in Chemical

Reactions 254

In this reaction, the number of silver atoms that reacts is twice the number of sulfur atoms. When 200 silver atoms react, 100 sulfur atoms are required. However, in the actual chemical reaction, many more atoms of both silver and sulfur would react. If we are dealing with molar amounts, then the coefficients in the equation can be interpreted in terms of moles. Thus, 2 mol of silver reacts with 1 mol of sulfur to produce 1 mol of Ag2S. Because the molar mass of each can be determined, the moles of Ag, S, and Ag2S can also be stated in terms of mass in grams of each. Thus, 215.8 g of Ag and 32.1 g of S react to form 247.9 g of Ag2S. The total mass of the reactants (247.9 g) is equal to the mass of the product (247.9 g). The various ways in which a chemical equation can be interpreted are seen in TABLE 9.1.

ENGAGE 9.1 Why is the total mass of the reactants equal to the total mass of the products?

Reactants Products

Equation 2Ag(s) + S(s) h Ag2S(s) Atoms/Formula Units 2 Ag atoms + 1 S atom h 1 Ag2S formula unit

200 Ag atoms + 100 S atoms h 100 Ag2S formula units Avogadro’s Number of Atoms

2(6.022 * 1023) Ag atoms

+ 1(6.022 * 1023) S atoms

h 1(6.022 * 1023) Ag2S formula units

Moles 2 mol of Ag + 1 mol of S h 1 mol of Ag2S Mass (g) 2(107.9 g) of Ag + 1(32.1 g) of S h 1(247.9 g) of Ag2S Total Mass (g) 247.9 g h 247.9 g

TABLE 9.1 Information Available from a Balanced Equation

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9.2 Mole Relationships in Chemical Equations 241

PRACTICE PROBLEMS

9.1 Conservation of Mass

9.1 Calculate the total mass of the reactants and the products for each of the following equations:

a. 2SO2(g) + O2(g) h 2SO3(g) b. 4P(s) + 5O2(g) h 2P2O5(s)

9.2 Calculate the total mass of the reactants and the products for each of the following equations:

a. 2Al(s) + 3Cl2(g) h 2AlCl3(s) b. 4HCl(g) + O2(g) h 2Cl2(g) + 2H2O(g)

9.2 Mole Relationships in Chemical Equations LEARNING GOAL Use a mole–mole factor from a balanced chemical equation to calculate the number of moles of another substance in the reaction.

When iron reacts with sulfur, the product is iron(III) sulfide.

2Fe(s) + 3S(s) h Fe2S3(s)

From the balanced equation, we see that 2 mol of iron reacts with 3 mol of sulfur to form 1 mol of iron(III) sulfide. Actually, any amount of iron or sulfur may be used, but the ratio

SAMPLE PROBLEM 9.1 Conservation of Mass

TRY IT FIRST

The combustion of methane (CH4) with oxygen produces carbon dioxide, water, and energy. Calculate the total mass of the reactants and the products for the following equa- tion when 1 mol of CH4 reacts:

CH4(g) + 2O2(g) h ∆

CO2(g) + 2H2O(g)

SOLUTION

Interpreting the coefficients in the equation as the number of moles of each substance and multiplying by its molar mass gives the total mass of reactants and products. The quantities of moles are exact because the coefficients in the balanced equation are exact.

Reactants Products

Equation CH4(g) + 2O2(g) h ∆

CO2(g) + 2H2O(g) Moles 1 mol of CH4 + 2 mol of O2 h 1 mol of CO2 + 2 mol of H2O Mass 16.04 g of CH4 + 64.00 g of O2

(++++++)++++++* h 44.01 g of CO2 + 36.03 g of H2O

(+++++++)++++++* Total Mass 80.04 g of reactants = 80.04 g of products

SELF TEST 9.1

Calculate the total mass of the reactants and the products for each of the following equations:

a. 4K(s) + O2(g) h 2K2O(s) b. 6HCl(aq) + 2Al(s) h 3H2(g) + 2AlCl3(aq)

ANSWER

a. 188.40 g of reactants and 188.40 g of products b. 272.71 g of reactants and 272.71 g of products

PRACTICE PROBLEMS Try Practice Problems 9.1 and 9.2

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242 CHAPTER 9 Chemical Quantities in Reactions

+

Iron (Fe) 2Fe(s) +

Sulfur (S) 3S(s)

Iron(III) sulfide (Fe2S3) Fe2S3(s)

In the chemical reaction of Fe and S, the mass of the reactants is the same as the mass of the product, Fe2S3.

of iron reacting with sulfur will always be the same. From the coefficients, we can write mole–mole factors between reactants and between reactants and products. The coefficients used in the mole–mole factors are exact numbers; they do not limit the number of significant figures.

Fe and S: 2 mol Fe 3 mol S

and 3 mol S 2 mol Fe

Fe and Fe2S3: 2 mol Fe

1 mol Fe2S3 and

1 mol Fe2S3 2 mol Fe

S and Fe2S3: 3 mol S

1 mol Fe2S3 and

1 mol Fe2S3 3 mol S

Using Mole–Mole Factors in Calculations Whenever you prepare a recipe, adjust an engine for the proper mixture of fuel and air, or prepare medicines in a pharmaceutical laboratory, you need to know the proper amounts of reactants to use and how much of the product will form. Now that we have written all the possible conversion factors for the balanced equation 2Fe(s) + 3S(s) h Fe2S3(s), we will use those mole–mole factors in a chemical calculation in Sample Problem 9.2.

PRACTICE PROBLEMS Try Practice Problems 9.3 to 9.6

CORE CHEMISTRY SKILL Using Mole–Mole Factors

SAMPLE PROBLEM 9.2 Calculating Moles of a Reactant

TRY IT FIRST

In the chemical reaction of iron and sulfur, how many moles of sulfur are needed to react with 1.42 mol of iron?

2Fe(s) + 3S(s) h Fe2S3(s)

SOLUTION

STEP 1 State the given and needed quantities (moles).

ANALYZE THE PROBLEM

Given Need Connect

1.42 mol of Fe moles of S mole–mole factor

Equation

2Fe(s) + 3S(s) h Fe2S3(s)

STEP 2 Write a plan to convert the given to the needed quantity (moles).

moles of Fe moles of S Mole–mole factor

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9.2 Mole Relationships in Chemical Equations 243

STEP 3 Use coefficients to write mole–mole factors.

2 mol of Fe = 3 mol of S 2 mol Fe 3 mol S

and 3 mol S 2 mol Fe

STEP 4 Set up the problem to give the needed quantity (moles).

1.42 mol Fe * 3 mol S 2 mol Fe

Exact

Exact

Three SFs Three SFs

= 2.13 mol of S

SELF TEST 9.2

Using the equation in Sample Problem 9.2, calculate each of the following: a. moles of iron needed to react with 2.75 mol of sulfur b. moles of iron(III) sulfide produced by the reaction of 0.758 mol of sulfur

ANSWER

a. 1.83 mol of iron b. 0.253 mol of iron(III) sulfide

Propane fuel reacts with O2 in the air to produce CO2, H2O, and energy.

moles of C3H8 moles of CO2 Mole–mole factor

STEP 3 Use coefficients to write mole–mole factors.

1 mol of C3H8 = 3 mol of CO2 1 mol C3H8 3 mol CO2

and 3 mol CO2 1 mol C3H8

SAMPLE PROBLEM 9.3 Calculating Moles of a Product

TRY IT FIRST

Propane gas (C3H8), a fuel used in camp stoves, soldering torches, and specially equipped automobiles, reacts with oxygen to produce carbon dioxide, water, and energy. How many moles of CO2 can be produced when 2.25 mol of C3H8 reacts?

C3H8(g) + 5O2(g) h ∆

3CO2(g) + 4H2O(g)

SOLUTION

STEP 1 State the given and needed quantities (moles).

ANALYZE THE PROBLEM

Given Need Connect

2.25 mol of C3H8 moles of CO2 mole–mole factor

Equation

C3H8(g) + 5O2(g) h ∆

3CO2(g) + 4H2O(g)

STEP 2 Write a plan to convert the given to the needed quantity (moles).

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244 CHAPTER 9 Chemical Quantities in Reactions

PRACTICE PROBLEMS

9.2 Mole Relationships in Chemical Equations

9.3 Write all of the mole–mole factors for each of the following chemical equations:

a. 2SO2(g) + O2(g) h 2SO3(g) b. 4P(s) + 5O2(g) h 2P2O5(s) 9.4 Write all of the mole–mole factors for each of the following

chemical equations: a. 2Al(s) + 3Cl2(g) h 2AlCl3(s) b. 4HCl(g) + O2(g) h 2Cl2(g) + 2H2O(g) 9.5 For the chemical equations in problem 9.3, write the problem

setup with the correct mole–mole factor needed to find each of the following:

a. moles of SO3 from the moles of SO2 b. moles of O2 needed to react with moles of P

9.6 For the chemical equations in problem 9.4, write the problem setup with the correct mole–mole factor needed to find each of the following:

a. moles of AlCl3 from the moles of Cl2 b. moles of O2 needed to react with moles of HCl

9.7 The chemical reaction of hydrogen with oxygen produces water.

2H2(g) + O2(g) h 2H2O(g)

a. How many moles of O2 are required to react with 2.6 mol of H2?

b. How many moles of H2 are needed to react with 5.0 mol of O2?

c. How many moles of H2O form when 2.5 mol of O2 reacts?

9.8 Ammonia is produced by the chemical reaction of nitrogen and hydrogen.

N2(g) + 3H2(g) h 2NH3(g) Ammonia

a. How many moles of H2 are needed to react with 1.8 mol of N2? b. How many moles of N2 reacted if 0.60 mol of NH3 is

produced? c. How many moles of NH3 are produced when 1.4 mol of H2

reacts?

9.9 Carbon disulfide and carbon monoxide are produced when carbon is heated with sulfur dioxide.

5C(s) + 2SO2(g) h ∆

CS2(l) + 4CO(g)

a. How many moles of C are needed to react with 0.500 mol of SO2?

b. How many moles of CO are produced when 1.2 mol of C reacts?

c. How many moles of SO2 are required to produce 0.50 mol of CS2?

d. How many moles of CS2 are produced when 2.5 mol of C reacts?

9.10 In the acetylene torch, acetylene gas burns in oxygen to produce carbon dioxide, water, and energy.

2C2H2(g) + 5O2(g) h ∆

4CO2(g) + 2H2O(g) Acetylene

a. How many moles of O2 are needed to react with 2.40 mol of C2H2?

b. How many moles of CO2 are produced when 3.5 mol of C2H2 reacts?

c. How many moles of C2H2 are required to produce 0.50 mol of H2O?

d. How many moles of CO2 are produced when 0.100 mol of O2 reacts?

SELF TEST 9.3

Using the equation in Sample Problem 9.3, calculate the moles of oxygen that must react to a. produce 0.756 mol of water b. react with 0.243 mol of C3H8

ANSWER

a. 0.945 mol of O2 b. 1.22 mol of O2

PRACTICE PROBLEMS Try Practice Problems 9.7 to 9.10

STEP 4 Set up the problem to give the needed quantity (moles).

2.25 mol C3H8 * 3 mol CO2 1 mol C3H8

Exact

Exact

Three SFs Three SFs

= 6.75 mol of CO2

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9.3 Mass Calculations for Chemical Reactions 245

9.3 Mass Calculations for Chemical Reactions LEARNING GOAL Given the mass in grams of a substance in a chemical reaction, calculate the mass in grams of another substance in the reaction.

When we have the balanced equation for a chemical reaction, we can use the mass of one of the substances (A) in the reaction to calculate the mass of another substance (B) in the reaction. However, the calculations require us to convert the mass of A to moles of A using the molar mass of A. Then we use the mole–mole factor that links substance A to substance B, which we obtain from the coefficients in the balanced equation. This mole–mole fac- tor (B/A) will convert the moles of A to moles of B. Then the molar mass of B is used to calculate the grams of substance B.

Substance A

grams of A Molar mass A

Molar mass B

moles of A

Substance B

moles of B Mole–mole factor B/A

grams of B

A mixture of acetylene and oxygen undergoes combustion during the welding of metals.

REVIEW Counting Significant Figures (2.2)

Using Significant Figures in Calculations (2.3)

Writing Conversion Factors from Equalities (2.5)

Using Conversion Factors (2.6)

Calculating Molar Mass (7.2)

Using Molar Mass as a Conversion Factor (7.3)

CORE CHEMISTRY SKILL Converting Grams to Grams

SAMPLE PROBLEM 9.4 Calculating Mass of a Product

TRY IT FIRST

When acetylene (C2H2) burns in oxygen, high temperatures are produced that are used for welding metals.

2C2H2(g) + 5O2(g) h ∆

4CO2(g) + 2H2O(g)

How many grams of CO2 are produced when 54.6 g of C2H2 is burned?

SOLUTION

STEP 1 State the given and needed quantities (grams).

ANALYZE THE PROBLEM

Given Need Connect

54.6 g of C2H2 grams of CO2 molar masses, mole–mole factor

Equation

2C2H2(g) + 5O2(g) h ∆

4CO2(g) + 2H2O(g)

STEP 2 Write a plan to convert the given to the needed quantity (grams).

grams of C2H2 moles of C2H2 moles of CO2 Molar mass

Mole–mole factor

Molar mass

grams of CO2

STEP 3 Use coefficients to write mole–mole factors; write molar masses.

1 mol of C2H2 = 26.04 g of C2H2 26.04 g C2H2 1 mol C2H2

and 1 mol C2H2

26.04 g C2H2

1 mol of CO2 = 44.01 g of CO2 44.01 g CO2 1 mol CO2

and 1 mol CO2

44.01 g CO2

2 mol of C2H2 = 4 mol of CO2 2 mol C2H2 4 mol CO2

and 4 mol CO2 2 mol C2H2

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246 CHAPTER 9 Chemical Quantities in Reactions

STEP 4 Set up the problem to give the needed quantity (grams).

54.6 g C2H2 = 185 g of CO2 1 mol C2H2

26.04 g C2H2

Four SFs

Exact

4 mol CO2 44.01 g CO2 1 mol CO22 mol C2H2

Exact

Exact

Exact

Four SFs

Three SFs Three SFs

* **

SELF TEST 9.4

Using the equation in Sample Problem 9.4, calculate each of the following: a. the grams of CO2 produced when 25.0 g of O2 reacts b. the grams of C2H2 needed when 65.0 g of H2O are produced

ANSWER

a. 27.5 g of CO2 b. 93.9 g of C2H2

ENGAGE 9.2 Why is a mole–mole factor needed in the problem setup that converts grams of a reactant to grams of a product?

SAMPLE PROBLEM 9.5 Calculating Mass of a Reactant

TRY IT FIRST

The fuel heptane (C7H16) is designated as the zero point in the octane rating of gasoline. Heptane is an undesirable compound in gasoline because it burns rapidly and causes engine knocking. How many grams of O2 are required to react with 22.5 g of C7H16?

C7H16(l) + 11O2(g) h ∆

7CO2(g) + 8H2O(g)

SOLUTION

STEP 1 State the given and needed quantities (grams).

ANALYZE THE PROBLEM

Given Need Connect

22.5 g of C7H16 grams of O2 molar masses, mole–mole factor

Equation

C7H16(l) + 11O2(g) h ∆

7CO2(g) + 8H2O(g)

STEP 2 Write a plan to convert the given to the needed quantity (grams).

grams of C7H16 moles of C7H16 moles of O2 Molar mass

Mole–mole factor

Molar mass

grams of O2

STEP 3 Use coefficients to write mole–mole factors; write molar masses.

1 mol of C7H16 = 100.2 g of C7H16 100.2 g C7H16 1 mol C7H16

and 1 mol C7H16

100.2 g C7H16

1 mol of O2 = 32.00 g of O2 32.00 g O2 1 mol O2

and 1 mol O2

32.00 g O2

1 mol of C7H16 = 11 mol of O2 1 mol C7H16

11 mol O2 and

11 mol O2 1 mol C7H16

STEP 4 Set up the problem to give the needed quantity (grams).

* * *22.5 g C7H16 1 mol C7H16

100.2 g C7H16

11 mol O2 1 mol C7H16

32.00 g O2 1 mol O2

= 79.1 g of O2

Four SFsThree SFs

Four SFs

Three SFsExact

Exact Exact

Exact

The octane rating is a measure of how well a fuel burns in an engine.

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9.4 Limiting Reactants 247

SELF TEST 9.5

Using the equation in Sample Problem 9.5, calculate each of the following: a. the grams of C7H16 that are needed to produce 15.0 g of H2O b. the grams of CO2 produced from the reaction of 25.0 g of C7H16

ANSWER

a. 10.4 g of C7H16 b. 76.9 g of CO2

PRACTICE PROBLEMS Try Practice Problems 9.11 to 9.18

PRACTICE PROBLEMS

9.3 Mass Calculations for Chemical Reactions

9.11 Sodium reacts with oxygen to produce sodium oxide.

4Na(s) + O2(g) h 2Na2O(s) a. How many grams of Na2O are produced when 57.5 g of Na

reacts? b. If you have 18.0 g of Na, how many grams of O2 are required

for reaction? c. How many grams of O2 are needed in a reaction that

produces 75.0 g of Na2O?

9.12 Nitrogen reacts with hydrogen to produce ammonia.

N2(g) + 3H2(g) h 2NH3(g) a. If you have 3.64 g of H2, how many grams of NH3 can be

produced? b. How many grams of H2 are needed to react with 2.80 g

of N2? c. How many grams of NH3 can be produced from 12.0 g

of H2?

9.13 Ammonia and oxygen react to form nitrogen and water.

4NH3(g) + 3O2(g) h 2N2(g) + 6H2O(g) a. How many grams of O2 are needed to react with 13.6 g

of NH3? b. How many grams of N2 can be produced when 6.50 g of O2

reacts? c. How many grams of H2O are formed from the reaction

of 34.0 g of NH3?

9.14 Iron(III) oxide reacts with carbon to give iron and carbon monoxide.

Fe2O3(s) + 3C(s) h 2Fe(s) + 3CO(g) a. How many grams of C are required to react with 16.5 g

of Fe2O3? b. How many grams of CO are produced when 36.0 g of C

reacts? c. How many grams of Fe can be produced when 6.00 g

of Fe2O3 reacts?

9.15 Nitrogen dioxide and water react to produce nitric acid, HNO3, and nitrogen oxide.

3NO2(g) + H2O(l) h 2HNO3(aq) + NO(g) a. How many grams of H2O are required to react with 28.0 g

of NO2? b. How many grams of NO are produced from 15.8 g of H2O? c. How many grams of HNO3 are produced from 8.25 g

of NO2?

9.16 Calcium cyanamide, CaCN2, reacts with water to form calcium carbonate and ammonia.

CaCN2(s) + 3H2O(l) h CaCO3(s) + 2NH3(g) a. How many grams of H2O are needed to react with 75.0 g

of CaCN2? b. How many grams of NH3 are produced from 5.24 g

of CaCN2? c. How many grams of CaCO3 form if 155 g of H2O reacts?

9.17 When solid lead(II) sulfide reacts with oxygen gas, the products are solid lead(II) oxide and sulfur dioxide gas.

a. Write the balanced chemical equation for the reaction. b. How many grams of oxygen are required to react with 29.9 g

of lead(II) sulfide? c. How many grams of sulfur dioxide can be produced when

65.0 g of lead(II) sulfide reacts? d. How many grams of lead(II) sulfide are used to produce

128 g of lead(II) oxide?

9.18 When the gases dihydrogen sulfide and oxygen react, they form the gases sulfur dioxide and water vapor.

a. Write the balanced chemical equation for the reaction. b. How many grams of oxygen are required to react with 2.50 g

of dihydrogen sulfide? c. How many grams of sulfur dioxide can be produced when

38.5 g of oxygen reacts? d. How many grams of oxygen are required to produce 55.8 g

of water vapor?

9.4 Limiting Reactants LEARNING GOAL Identify a limiting reactant when given the quantities of two reactants; calculate the amount of product formed from the limiting reactant.

When we make peanut butter sandwiches for lunch, we need 2 slices of bread and 1 tablespoon of peanut butter for each sandwich. As an equation, we could write:

2 slices of bread + 1 tablespoon of peanut butter h 1 peanut butter sandwich

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248 CHAPTER 9 Chemical Quantities in Reactions

If we have 8 slices of bread and a full jar of peanut butter, we will run out of bread after we make 4 peanut butter sandwiches. We cannot make any more sandwiches once the bread is used up, even though there is a lot of peanut butter left in the jar. The number of slices of bread has limited the number of sandwiches we can make.

On a different day, we might have 8 slices of bread but only a tablespoon of peanut butter left in the peanut butter jar. We will run out of peanut butter after we make just 1 peanut butter sandwich and have 6 slices of bread left over. The small amount of peanut butter available has limited the number of sandwiches we can make.

1 peanut butter sandwich

6 slices of bread left over

1 tablespoon of peanut butter

++

make8 slices of bread

8 slices of bread 1 jar of peanut butter

peanut butter left over

4 peanut butter sandwiches

+

+

+

+

+ make

+

SAMPLE PROBLEM 9.6 Moles of Product from a Limiting Reactant

TRY IT FIRST

Carbon monoxide and hydrogen are used to produce methanol (CH4O). The balanced chemical reaction is

CO(g) + 2H2(g) h CH4O(g)

If 3.00 mol of CO and 5.00 mol of H2 are the initial reactants, what is the limiting reactant, and how many moles of methanol can be produced?

Calculating Moles of Product from a Limiting Reactant In a similar way, the reactants in a chemical reaction do not always combine in quantities that allow each to be used up at exactly the same time. In many reactions, there is a limiting reac- tant that determines the amount of product that can be formed. When we know the quantities of the reactants of a chemical reaction, we calculate the amount of product that is possible from each reactant if it were completely consumed. We are looking for the limiting reactant, which is the one that runs out first, producing the smaller amount of product.

CORE CHEMISTRY SKILL Calculating Quantity of Product

from a Limiting Reactant

The reactant that is completely used up is the limiting reactant. The reactant that does not completely react and is left over is called the excess reactant.

ENGAGE 9.3 For a picnic you have 10 spoons, 8 forks, and 6 knives. If each person requires 1 spoon, 1 fork, and 1 knife, why can you only serve 6 people at the picnic?

Bread Peanut Butter Sandwiches Limiting Reactant Excess Reactant

1 loaf (20 slices) 1 tablespoon 1 peanut butter bread

4 slices 1 full jar 2 bread peanut butter

8 slices 1 full jar 4 bread peanut butter

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9.4 Limiting Reactants 249

SOLUTION

STEP 1 State the given and needed quantity (moles).

ANALYZE THE PROBLEM

Given Need Connect

3.00 mol of CO, 5.00 mol of H2

limiting reactant, moles of CH4O produced

mole–mole factors

Equation

CO(g) + 2H2(g) h CH4O(g)

STEP 2 Write a plan to convert the quantity (moles) of each reactant to quantity (moles) of product.

moles of CO moles of CH4O Mole–mole factor

moles of H2 moles of CH4O Mole–mole factor

STEP 3 Use coefficients to write mole–mole factors.

1 mol CH4O 2 mol H2

1 mol CH4O2 mol H2

2 mol of H2 = 1 mol of CH4O

and and 1 mol CH4O 1 mol CO

1 mol CO 1 mol CH4O 1 mol of CO = 1 mol of CH4O

STEP 4 Calculate the quantity (moles) of product from each reactant, and select the smaller quantity (moles) as the limiting reactant.

Moles of CH4O (product) from CO:

= 3.00 mol of CH4O3.00 mol CO * 1 mol CH4O

1 mol CO

Exact

ExactThree SFs Three SFs

Moles of CH4O (product) from H2:

= 2.50 mol of CH4O 1 mol CH4O

2 mol H2 5.00 mol H2 Smaller amount

of product

Limiting reactant

*

Exact

Three SFs Exact

Three SFs

PRACTICE PROBLEMS Try Practice Problems 9.19 to 9.22

The smaller amount, 2.50 mol of CH4O, is the maximum amount of methanol that can be produced from the limiting reactant, H2, because it is completely consumed.

SELF TEST 9.6

a. If an initial mixture of reactants for Sample Problem 9.6 contains 4.00 mol of CO and 4.00 mol of H2, what is the limiting reactant, and how many moles of methanol can be produced?

b. If an initial mixture of reactants for Sample Problem 9.6 contains 1.50 mol of CO and 6.00 mol of H2, what is the limiting reactant, and how many moles of methanol can be produced?

ANSWER

a. H2 is the limiting reactant; 2.00 mol of methanol can be produced. b. CO is the limiting reactant; 1.50 mol of methanol can be produced.

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250 CHAPTER 9 Chemical Quantities in Reactions

Calculating Mass of Product from a Limiting Reactant The quantities of the reactants can also be given in grams. The calculations to identify the limiting reactant are the same as before, but the grams of each reactant must first be converted to moles, then to moles of product, and finally to grams of product. Then select the smaller mass of product, which is from complete use of the limiting reactant. This calculation is shown in Sample Problem 9.7.

A ceramic brake disc in a car withstands temperatures of 1400 °C.

SAMPLE PROBLEM 9.7 Mass of Product from a Limiting Reactant

TRY IT FIRST

When silicon dioxide (sand) and carbon are heated, the products are silicon carbide, SiC, and carbon monoxide. Silicon carbide is a ceramic material that tolerates extreme tem- peratures and is used as an abrasive and in the brake discs of cars. How many grams of CO are formed from 70.0 g of SiO2 and 50.0 g of C?

SiO2(s) + 3C(s) h ∆

SiC(s) + 2CO(g)

SOLUTION

STEP 1 State the given and needed quantity (grams).

ANALYZE THE PROBLEM

Given Need Connect

70.0 g of SiO2, 50.0 g of C

grams of CO from limiting reactant

molar masses, mole–mole factors

Equation

SiO2(s) + 3C(s) h ∆

SiC(s) + 2CO(g)

STEP 2 Write a plan to convert the quantity (grams) of each reactant to quantity (grams) of product.

grams of SiO2

moles of SiO2

moles of CO

Molar mass

Mole–mole factor

Molar mass

grams of CO

grams of C

moles of C

moles of CO

Molar mass

Mole–mole factor

Molar mass

grams of CO

STEP 3 Use coefficients to write mole–mole factors; write molar masses.

1 mol of SiO2 = 60.09 g of SiO2 1 mol SiO2

60.09 g SiO2 and andand

60.09 g SiO2 1 mol SiO2

1 mol of C = 12.01 g of C 1 mol C

12.01 g C 12.01 g C 1 mol C

1 mol of CO = 28.01 g of CO 1 mol CO

28.01 g CO 28.01 g CO 1 mol CO

2 mol of CO = 1 mol of SiO2 2 mol CO

1 mol SiO2

1 mol SiO2 2 mol CO

and

2 mol of CO = 3 mol of C 2 mol CO 3 mol C

3 mol C 2 mol CO

and

1 mol of SiO2 = 60.09 g of SiO2 1 mol SiO2

60.09 g SiO2 and andand

60.09 g SiO2 1 mol SiO2

1 mol of C = 12.01 g of C 1 mol C

12.01 g C 12.01 g C 1 mol C

1 mol of CO = 28.01 g of CO 1 mol CO

28.01 g CO 28.01 g CO 1 mol CO

2 mol of CO = 1 mol of SiO2 2 mol CO

1 mol SiO2

1 mol SiO2 2 mol CO

and

2 mol of CO = 3 mol of C 2 mol CO 3 mol C

3 mol C 2 mol CO

and

M09_TIMB8119_06_SE_C09.indd 250 11/29/18 10:01 AM

9.4 Limiting Reactants 251

STEP 4 Calculate the quantity (grams) of product from each reactant, and select the smaller quantity (grams) as the limiting reactant.

Grams of CO (product) from SiO2:

Limiting reactant Smaller amount

of product

* * *70.0 g SiO2 1 mol SiO2

60.09 g SiO2

2 mol CO

1 mol SiO2

28.01 g CO

1 mol CO = 65.3 g of CO

Four SFs

Three SFs Four SFs Three SFs

Exact

Exact Exact

Exact

50.0 g C 1 mol C

12.01 g C 3 mol C

2 mol CO

1 mol CO

28.01 g CO 77.7 g of CO* * * =

Exact

ExactExact Four SFs

ExactFour SFs

Three SFs Three SFs Grams of CO (product) from C: Limiting reactant

Smaller amount of product

* * *70.0 g SiO2 1 mol SiO2

60.09 g SiO2

2 mol CO

1 mol SiO2

28.01 g CO

1 mol CO = 65.3 g of CO

Four SFs

Three SFs Four SFs Three SFs

Exact

Exact Exact

Exact

50.0 g C 1 mol C

12.01 g C 3 mol C

2 mol CO

1 mol CO

28.01 g CO 77.7 g of CO* * * =

Exact

ExactExact Four SFs

ExactFour SFs

Three SFs Three SFs

PRACTICE PROBLEMS Try Practice Problems 9.23 to 9.26

The smaller amount, 65.3 g of CO, is the most CO that can be produced. This also means that the SiO2 is the limiting reactant.

SELF TEST 9.7

Hydrogen sulfide burns with oxygen to give sulfur dioxide and water.

2H2S(g) + 3O2(g) h ∆

2SO2(g) + 2H2O(g)

a. How many grams of SO2 are formed from the reaction of 8.52 g of H2S and 9.60 g of O2?

b. How many grams of H2O are formed from the reaction of 15.0 g of H2S and 25.0 g of O2?

ANSWER

a. 12.8 g of SO2 b. 7.93 g of H2O

PRACTICE PROBLEMS

9.4 Limiting Reactants

9.19 A taxi company has 10 taxis. a. On a certain day, only eight taxi drivers show up for work.

How many taxis can be used to pick up passengers? b. On another day, 10 taxi drivers show up for work but three

taxis are in the repair shop. How many taxis can be driven?

9.20 A clock maker has 15 clock faces. Each clock requires one face and two hands.

a. If the clock maker has 42 hands, how many clocks can be produced?

b. If the clock maker has only eight hands, how many clocks can be produced?

9.21 Nitrogen and hydrogen react to form ammonia.

N2(g) + 3H2(g) h 2NH3(g) Determine the limiting reactant in each of the following mixtures of reactants:

a. 3.0 mol of N2 and 5.0 mol of H2 b. 8.0 mol of N2 and 4.0 mol of H2 c. 3.0 mol of N2 and 12.0 mol of H2

9.22 Iron and oxygen react to form iron(III) oxide.

4Fe(s) + 3O2(g) h 2Fe2O3(s) Determine the limiting reactant in each of the following mixtures of reactants:

a. 2.0 mol of Fe and 6.0 mol of O2 b. 5.0 mol of Fe and 4.0 mol of O2 c. 16.0 mol of Fe and 20.0 mol of O2 9.23 For each of the following reactions, 20.0 g of each reactant is

present initially. Determine the limiting reactant, and calculate the grams of product in parentheses that would be produced.

a. 2Al(s) + 3Cl2(g) h 2AlCl3(s) (AlCl3) b. 4NH3(g) + 5O2(g) h 4NO(g) + 6H2O(g) (H2O) c. CS2(g) + 3O2(g) h

∆ CO2(g) + 2SO2(g) (SO2)

9.24 For each of the following reactions, 20.0 g of each reactant is present initially. Determine the limiting reactant, and calculate the grams of product in parentheses that would be produced.

a. 4Al(s) + 3O2(g) h 2Al2O3(s) (Al2O3) b. 3NO2(g) + H2O(l) h 2HNO3(aq) + NO(g) (HNO3) c. C2H6O(l) + 3O2(g) h

∆ 2CO2(g) + 3H2O(g) (H2O)

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252 CHAPTER 9 Chemical Quantities in Reactions

9.25 For each of the following reactions, calculate the grams of indi- cated product when 25.0 g of the first reactant and 40.0 g of the second reactant are used:

a. 2SO2(g) + O2(g) h 2SO3(g) (SO3) b. 3Fe(s) + 4H2O(l) h Fe3O4(s) + 4H2(g) (Fe3O4) c. C7H16(l) + 11O2(g) h

∆ 7CO2(g) + 8H2O(g) (CO2)

9.26 For each of the following reactions, calculate the grams of indi- cated product when 15.0 g of the first reactant and 10.0 g of the second reactant are used:

a. 4Li(s) + O2(g) h 2Li2O(s) (Li2O) b. Fe2O3(s) + 3H2(g) h 2Fe(s) + 3H2O(l) (Fe) c. Al2S3(s) + 6H2O(l) h 2Al(OH)3(aq) + 3H2S(g) (H2S)

9.5 Percent Yield LEARNING GOAL Given the actual quantity of product, determine the percent yield for a chemical reaction.

In our problems up to now, we assumed that all of the reactants changed completely to product. Thus, we have calculated the amount of product as the maximum quantity possible, or 100%. While this would be an ideal situation, it does not usually happen. As we carry out a reaction and transfer products from one container to another, some product is usually lost. In the lab as well as commercially, the starting materials may not be completely pure, and side reactions may use some of the reactants to give unwanted products. Thus, 100% of the desired product is not actually obtained.

When we do a chemical reaction in the laboratory, we measure out specific quantities of the reactants. We calculate the theoretical yield for the reaction, which is the amount of product (100%) we would expect if all the reactants were converted to the desired prod- uct. When the reaction ends, we collect and measure the mass of the product, which is the actual yield for the product. Because some product is usually lost, the actual yield is less than the theoretical yield. Using the actual yield and the theoretical yield for a product, we can calculate the percent yield.

Percent yield (%) = actual yield

theoretical yield * 100%

REVIEW Calculating Percentages (1.4)

ENGAGE 9.4 For your chemistry class party, you have prepared cookie dough to make five dozen cookies. However, 12 of the cookies burned and you threw them away. Why is the percent yield of cookies for the chemistry party only 80%?

CORE CHEMISTRY SKILL Calculating Percent Yield

SAMPLE PROBLEM 9.8 Calculating Percent Yield

TRY IT FIRST

On spacecraft, LiOH can be used to absorb exhaled CO2 from breathing air to form LiHCO3.

LiOH(s) + CO2(g) h LiHCO3(s)

What is the percent yield of LiHCO3 for the reaction if 50.0 g of LiOH gives 72.8 g of LiHCO3?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

50.0 g of LiOH, 72.8 g of LiHCO3 (actual yield)

percent yield of LiHCO3

molar masses, mole–mole factor, percent yield expression

Equation

LiOH(s) + CO2(g) h LiHCO3(s)

STEP 2 Write a plan to calculate the theoretical yield and the percent yield.

On spacecraft, LiOH in canisters can be used to remove CO2 from the air.

grams of LiOH moles of LiOH moles of LiHCO3 Molar mass

Mole–mole factor

Molar mass

grams of LiHCO3 Theoretical yield

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9.5 Percent Yield 253

STEP 4 Calculate the percent yield by dividing the actual yield (given) by the theoretical yield and multiplying the result by 100%.

1 mol of LiOH = 23.95 g of LiOH 1 mol of LiHCO3 = 67.96 g of LiHCO3 1 mol LiOH

23.95 g LiOH1 mol LiOH 23.95 g LiOH

and 67.96 g LiHCO3 1 mol LiHCO3

and

1 mol of LiHCO3 = 1 mol of LiOH 1 mol LiHCO3

1 mol LiOH 1 mol LiHCO3

1 mol LiOH and

1 mol LiHCO3 67.96 g LiHCO3

Percent yield (%) = actual yield of LiHCO3

theoretical yield of LiHCO3 * 100%

* * *50.0 g LiOH 23.95 g LiOH

1 mol LiOH 1 mol LiHCO3

1 mol LiOH

67.96 g LiHCO3 1 mol LiHCO3

= 142 g of LiHCO3

Three SFs Four SFs

Exact Exact

Exact

Four SFs

Exact

Three SFs

Calculation of percent yield:

actual yield (given)

theoretical yield (calculated) * 100% =

72.8 g LiHCO3 142 g LiHCO3

* 100% = 51.3%

A percent yield of 51.3% means that 72.8 g of the theoretical amount of 142 g of LiHCO3 was actually produced by the reaction.

SELF TEST 9.8

Using the equation in Sample Problem 9.8, calculate each of the following: a. the percent yield of LiHCO3 if 8.00 g of CO2 produces 10.5 g of LiHCO3 b. the percent yield of LiHCO3 if 35.0 g of LiOH produces 76.6 g of LiHCO3

ANSWER

a. 84.7% b. 77.1%

Three SFs

Three SFs Three SFs

PRACTICE PROBLEMS Try Practice Problems 9.27 to 9.32

PRACTICE PROBLEMS

9.5 Percent Yield 9.27 Carbon disulfide is produced by the reaction of carbon and

sulfur dioxide.

5C(s) + 2SO2(g) h CS2(g) + 4CO(g) a. What is the percent yield of carbon disulfide if the reaction

of 40.0 g of carbon produces 36.0 g of carbon disulfide? b. What is the percent yield of carbon disulfide if the reaction

of 32.0 g of sulfur dioxide produces 12.0 g of carbon disulfide?

9.28 Iron(III) oxide reacts with carbon monoxide to produce iron and carbon dioxide.

Fe2O3(s) + 3CO(g) h 2Fe(s) + 3CO2(g)

a. What is the percent yield of iron if the reaction of 65.0 g of iron(III) oxide produces 15.0 g of iron?

b. What is the percent yield of carbon dioxide if the reaction of 75.0 g of carbon monoxide produces 85.0 g of carbon dioxide?

9.29 Aluminum reacts with oxygen to produce aluminum oxide.

4Al(s) + 3O2(g) h 2Al2O3(s)

Calculate the mass of Al2O3 that can be produced if the reaction of 50.0 g of aluminum and sufficient oxygen has a 75.0% yield.

1 mol of LiOH = 23.95 g of LiOH 1 mol of LiHCO3 = 67.96 g of LiHCO3 1 mol LiOH

23.95 g LiOH1 mol LiOH 23.95 g LiOH

and 67.96 g LiHCO3 1 mol LiHCO3

and

1 mol of LiHCO3 = 1 mol of LiOH 1 mol LiHCO3

1 mol LiOH 1 mol LiHCO3

1 mol LiOH and

1 mol LiHCO3 67.96 g LiHCO3

STEP 3 Use coefficients to write mole–mole factors; write molar masses.

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254 CHAPTER 9 Chemical Quantities in Reactions

9.6 Energy in Chemical Reactions LEARNING GOAL Given the heat of reaction, calculate the loss or gain of heat for an exothermic or endothermic reaction.

Almost every chemical reaction involves a loss or gain of energy. To discuss energy change for a reaction, we look at the energy of the reactants before the reaction and the energy of the products after the reaction.

Energy Units for Chemical Reactions The SI unit for energy is the joule (J). Often, the unit of kilojoules (kJ) is used to show the energy change in a reaction.

1 kilojoule (kJ) = 1000 joules (J)

Heat of Reaction The heat of reaction is the amount of heat absorbed or released during a reaction that takes place at constant pressure. A change of energy occurs as reactants interact, bonds break apart, and products form. We determine a heat of reaction, symbol ∆H, as the difference in the energy of the products and the reactants.

∆H = Hproducts - Hreactants

Exothermic Reactions In an exothermic reaction (exo means “out”), the energy of the products is lower than that of the reactants. This means that heat is released along with the products that form. Let us look at the equation for the exothermic reaction in which 185 kJ of heat is released when 1 mol of hydrogen and 1 mol of chlorine react to form 2 mol of hydrogen chloride. For an exothermic reaction, the heat of reaction can be written as one of the products. It can also be written as a ∆H value with a negative sign ( - ).

REVIEW Using Energy Units (3.4)

9.30 Propane (C3H8) burns in oxygen to produce carbon dioxide and water.

C3H8(g) + 5O2(g) h ∆

3CO2(g) + 4H2O(g)

Calculate the mass of CO2 that can be produced if the reaction of 45.0 g of propane and sufficient oxygen has a 60.0% yield.

9.31 When 30.0 g of carbon is heated with silicon dioxide, 28.2 g of carbon monoxide is produced. What is the percent yield of carbon monoxide for this reaction?

SiO2(s) + 3C(s) h ∆

SiC(s) + 2CO(g)

9.32 When 56.6 g of calcium is reacted with nitrogen gas, 32.4 g of calcium nitride is produced. What is the percent yield of calcium nitride for this reaction?

3Ca(s) + N2(g) h Ca3N2(s)

E ne

rg y

Reactants

Products

Energy is released

Progress of Reaction

H2 Cl2+

HCl HCl+ In an exothermic reaction, the energy of the products is lower than the energy of the reactants.

Exothermic, Heat Released Heat Is a Product

H2(g) + Cl2(g) h 2HCl(g) + 185 kJ ∆H = - 185 kJ Negative sign

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9.6 Energy in Chemical Reactions 255

Endothermic Reactions In an endothermic reaction (endo means “within”), the energy of the products is higher than that of the reactants. Heat is required to convert the reactants to products. Let us look at the equation for the endothermic reaction in which 180 kJ of heat is needed to convert 1 mol of nitrogen and 1 mol of oxygen to 2 mol of nitrogen oxide. For an endothermic reaction, the heat of reaction can be written as one of the reactants. It can also be written as a ∆H value with a positive sign ( + ).

E ne

rg y

Reactants

Products

Energy is absorbed

Progress of Reaction

N2 O2+

NO NO+

In an endothermic reaction, the energy of the products is higher than the energy of the reactants.

Endothermic, Heat Absorbed Heat Is a Reactant

N2(g) + O2(g) + 180 kJ h 2NO(g) ∆H = + 180 kJ Positive sign

Reaction Energy Change Heat in the Equation Sign of �H

Exothermic Heat released Product side Negative sign ( - ) Endothermic Heat absorbed Reactant side Positive sign ( + )

PRACTICE PROBLEMS Try Practice Problems 9.33 to 9.38

SAMPLE PROBLEM 9.9 Exothermic and Endothermic Reactions

TRY IT FIRST

In the reaction of 1 mol of solid carbon with 1 mol of oxygen gas, the energy of the carbon dioxide gas produced is 394 kJ lower than the energy of the reactants.

a. Is the reaction exothermic or endothermic? b. Write the balanced chemical equation for the reaction, including the heat of reaction. c. Is the sign of ∆H positive or negative?

SOLUTION

a. When the products have a lower energy than the reactants, the reaction is exothermic. b. C(s) + O2(g) h CO2(g) + 394 kJ c. negative

SELF TEST 9.9

In the reaction of 1 mol of solid carbon with 2 mol of solid sulfur, the energy of the liquid carbon disulfide produced is 92.0 kJ higher than the energy of the reactants.

a. Is the reaction exothermic or endothermic? b. Write the balanced chemical equation for the reaction, including the heat of reaction. c. Is the sign of ∆H positive or negative?

ANSWER

a. When the products have a higher energy than the reactants, the reaction is endothermic. b. C(s) + 2S(s) + 92.0 kJ h CS2(l) c. positive

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256 CHAPTER 9 Chemical Quantities in Reactions

Calculations of Heat in Reactions The value of ∆H refers to the heat change, in kilojoules, for each substance in the balanced equation for the reaction. Consider the following decomposition reaction:

2H2O(l) h 2H2(g) + O2(g) ∆ H = + 572 kJ 2H2O(l) + 572 kJ h 2H2(g) + O2(g)

For this reaction, 572 kJ are absorbed by 2 mol of H2O to produce 2 mol of H2 and 1 mol of O2. We can write two heat conversion factors for each substance in this reaction as:

SAMPLE PROBLEM 9.10 Calculating Heat in a Reaction

TRY IT FIRST

How much heat, in kilojoules, is released when nitrogen and hydrogen react to form 50.0 g of ammonia?

N2(g) + 3H2(g) h 2NH3(g) ∆H = - 92.2 kJ

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

50.0 g of NH3 heat released, in kilojoules

molar mass, ∆H = - 92.2 kJ

Equation

N2(g) + 3H2(g) h 2NH3(g)

STEP 2 Write a plan using the heat of reaction and any molar mass needed.

CORE CHEMISTRY SKILL Using the Heat of Reaction

grams of NH3 moles of NH3 kilojoules Molar mass

Heat of reaction

STEP 3 Write the conversion factors including heat of reaction.

1 mol of NH3 = 17.03 g of NH3 17.03 g NH3 1 mol NH3

1 mol NH3 17.03 g NH3

and

2 mol of NH3 = -92.2 kJ -92.2 kJ

2 mol NH3 and

-92.2 kJ 2 mol NH3

STEP 4 Set up the problem to calculate the heat.

* *50.0 g NH3 17.03 g NH3

1 mol NH3 2 mol NH3

-92.2 kJ = -135 kJ

Four SFs Exact

Exact Three SFs

Three SFsThree SFs

+ 572 kJ 2 mol H2O

and 2 mol H2O

+ 572 kJ

+ 572 kJ 2 mol H2

and 2 mol H2 + 572 kJ

+ 572 kJ 1 mol O2

and 1 mol O2 + 572 kJ

ENGAGE 9.5 Why is there a single heat of reaction whereas there are two or more heat conversion factors possible for the reaction?

Suppose in this reaction that 9.00 g of H2O undergoes reaction. We can calculate the kilojoules of heat absorbed as

9.00 g H2O * 1 mol H2O

18.02 g H2O *

+ 572 kJ 2 mol H2O

= + 143 kJ

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9.6 Energy in Chemical Reactions 257

PRACTICE PROBLEMS Try Practice Problems 9.39 and 9.40

SELF TEST 9.10

Mercury(II) oxide decomposes to mercury and oxygen.

2HgO(s) h 2Hg(l) + O2(g) ∆H = + 182 kJ

a. Is the reaction exothermic or endothermic? b. How many kilojoules are needed when 25.0 g of mercury(II) oxide reacts?

ANSWER

a. endothermic b. 20.5 kJ

Chemistry Link to Health Cold Packs and Hot Packs

In a hospital, at a first-aid station, or at an athletic event, an instant cold pack may be used to reduce swelling from an injury, remove heat from inflammation, or decrease capillary size to lessen the effect of hemorrhaging. Inside the plastic container of a cold pack, there is a compartment containing solid ammonium nitrate (NH4NO3) that is separated from a compartment containing water.

The pack is activated when it is hit or squeezed hard enough to break the walls between the compartments and cause the ammonium nitrate to mix with the water (shown as H2O over the reaction arrow). In an endothermic process, 1 mol of NH4NO3 that dissolves absorbs 26 kJ of heat. The temperature drops to about 4 to 5 °C to give a cold pack that is ready to use.

Endothermic Reaction in a Cold Pack

NH4NO3(s) + 26 kJ h H2O

NH4NO3(aq)

Hot packs are used to relax muscles, lessen aches and cramps, and increase circulation by expanding capillary size. Constructed in the same way as cold packs, a hot pack contains a salt such as CaCl2. When 1 mol of CaCl2 dissolves in water, 82 kJ are released as heat. The temperature increases as much as 66 °C to give a hot pack that is ready to use.

Exothermic Reaction in a Hot Pack

CaCl2(s) h H2O

CaCl2(aq) + 82 kJ Cold packs use an endothermic reaction.

Hess’s Law According to Hess’s law, heat can be absorbed or released in a single chemical reaction or in several steps. When there are two or more steps in the reaction, the overall energy change is the sum of the energy changes of those steps, provided they all occur at the same temperature.

Steps in solving problems involving Hess’s law:

1. If you reverse a chemical equation, you must also reverse the sign of ∆H. 2. If a chemical equation is multiplied by some factor, then ∆H must be multiplied by

the same factor.

We can see how the energy change for a specific reaction is the sum of two or more reactions in Sample Problem 9.11.

SAMPLE PROBLEM 9.11 Hess’s Law and Calculating Heat of Reaction

TRY IT FIRST

Calculate the ∆H value for the following chemical reaction:

C(s) + 2H2O(g) h CO2(g) + 2H2(g)

First equation C(s) + O2(g) h CO2(g) ∆H = - 394 kJ

Second equation 2H2(g) + O2(g) h 2H2O(g) ∆H = - 484 kJ

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258 CHAPTER 9 Chemical Quantities in Reactions

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

overall reaction, first and second equations

heat of reaction combine equations, total heats of reactions

STEP 1 Arrange the given equations to place reactants on the left and products on the right. Use the first equation as written with C(s) as a reactant and CO2(g) as a product.

C(s) + O2(g) h CO2(g) ∆H = - 394 kJ

Reverse the second equation to give H2O(g) as a reactant and H2(g) as a product, and change the sign of the ∆H.

2H2O(g) h 2H2(g) + O2(g) ∆H = + 484 kJ Changed sign

STEP 2 If an equation is multiplied to balance coefficients, multiply the �H by the same number. Now that C(s) and 2H2O(g) are reactants, and CO2(g) and 2H2(g) are products, no further changes are needed.

STEP 3 Combine the equations, cancel any substances that are common to both sides, and add the �H values.

C(s) + O2(g) h CO2(g) ∆H = - 394 kJ

2H2O(g) h 2H2(g) + O2(g) ∆H = + 484 kJ

C(s) + 2H2O(g) h CO2(g) + 2H2(g) ∆H = + 90. kJ

SELF TEST 9.11

Calculate the ∆H value for the following chemical reaction:

2NO(g) + O2(g) h N2O4(g)

First equation N2O4 (g) h 2NO2(g) ∆H = + 57.2 kJ

Second equation 2NO(g) + O2(g) h 2NO2(g) ∆H = - 114.0 kJ

ANSWER ∆H = - 171.2 kJ

PRACTICE PROBLEMS Try Practice Problems 9.41 to 9.44

SAMPLE PROBLEM 9.12 Calculating Heat of Reaction

TRY IT FIRST

Calculate the ∆H value for the following chemical reaction:

CS2(l) + 3O2(g) h CO2(g) + 2SO2(g)

First equation C(s) + O2(g) h CO2(g) ∆H = - 394 kJ

Second equation S(s) + O2(g) h SO2(g) ∆H = - 297 kJ

Third equation C(s) + 2S(s) h CS2(l) ∆H = + 88 kJ

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

overall reaction, first, second, third equations

heat of reaction combine equations, total heats of reactions

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9.6 Energy in Chemical Reactions 259

STEP 1 Arrange the given equations to place reactants on the left and products on the right. Use the first equation as written with CO2(g) as a product.

C(s) + O2(g) h CO2(g) ∆H = - 394 kJ

Use the second equation as written with SO2(g) as a product.

S(s) + O2(g) h SO2(g) ∆H = - 297 kJ

Reverse the third equation to give CS2(l) as a reactant, and change the sign of the ∆H.

CS2(l) h C(s) + 2S(s) ∆H = - 88 kJ

STEP 2 If an equation is multiplied to balance coefficients, multiply the �H by the same number. Obtain 2SO2(g) on the product side by multiplying the second equation and the ∆H by 2.

2S(s) + 2O2(g) h 2SO2(g) ∆H = - 594 kJ

STEP 3 Combine the equations, cancel any substances that are common to both sides, and add the �H values. The C(s) and 2S(s) cancel.

C(s) + O2(g) h CO2(g) ∆H = - 394 kJ

2S(s) + 2O2(g) h 2SO2(g) ∆H = - 594 kJ

CS2(l) h C(s) + 2S(s) ∆H = - 88 kJ

CS2(l) + 3O2(g) h CO2(g) + 2SO2(g) ∆H = - 1076 kJ

SELF TEST 9.12

Calculate the ∆H value for the following chemical reaction:

C(s) + 2H2(g) h CH4(g)

First equation C(s) + O2(g) h CO2(g) ∆H = - 394 kJ

Second equation 2H2(g) + O2(g) h 2H2O(g) ∆H = - 484 kJ

Third equation CH4(g) + 2O2(g) h CO2(g) + 2H2O(g) ∆H = - 802 kJ

ANSWER

- 76 kJ

PRACTICE PROBLEMS

9.6 Energy in Chemical Reactions

9.33 In an exothermic reaction, is the energy of the products higher or lower than that of the reactants?

9.34 In an endothermic reaction, is the energy of the products higher or lower than that of the reactants?

9.35 Classify each of the following as exothermic or endothermic: a. A reaction releases 550 kJ. b. The energy level of the products is higher than that of the

reactants. c. The metabolism of glucose in the body provides energy.

9.36 Classify each of the following as exothermic or endothermic: a. The energy level of the products is lower than that of the

reactants. b. In the body, the synthesis of proteins requires energy. c. A reaction absorbs 125 kJ.

9.37 Classify each of the following as exothermic or endothermic, and give the ∆H for each:

a. CH4(g) + 2O2(g) h ∆

CO2(g) + 2H2O(g) + 802 kJ b. Ca(OH)2(s) + 65.3 kJ h CaO(s) + H2O(l) c. 2Al(s) + Fe2O3(s) h Al2O3(s) + 2Fe(l) + 850 kJ

The thermite reaction of aluminum and iron(III) oxide produces very high temperatures used to cut or weld railroad tracks.

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260 CHAPTER 9 Chemical Quantities in Reactions

9.38 Classify each of the following as exothermic or endothermic, and give the ∆H for each:

a. C3H8(g) + 5O2(g) h ∆

3CO2(g) + 4H2O(g) + 2220 kJ b. 2Na(s) + Cl2(g) h 2NaCl(s) + 819 kJ c. PCl5(g) + 67 kJ h PCl3(g) + Cl2(g) 9.39 a. How many kilojoules are released when 125 g of Cl2 reacts

with silicon?

Si(s) + 2Cl2(g) h SiCl4(g) ∆H = - 657 kJ

b. How many kilojoules are absorbed when 278 g of PCl5 reacts?

PCl5(g) h PCl3(g) + Cl2(g) ∆H = + 67 kJ

9.40 a. How many kilojoules are released when 75.0 g of CH4O reacts?

2CH4O(l) + 3O2(g) h 2CO2(g) + 4H2O(l) ∆H = - 726 kJ

b. How many kilojoules are absorbed when 315 g of Ca(OH)2 reacts?

Ca(OH)2(s) h CaO(s) + H2O(l) ∆H = + 65.3 kJ

9.41 Calculate the energy change for the reaction N2(g) + 2O2(g) h 2NO2(g) from the following: N2(g) + O2(g) h 2NO(g) ∆H = + 180 kJ 2NO2(g) h 2NO(g) + O2(g) ∆H = + 114 kJ

9.42 Calculate the energy change for the reaction Fe2O3(s) + CO(g) h 2FeO(s) + CO2(g) from the following:

Fe2O3(s) + 3CO(g) h 2Fe(s) + 3CO2(g) ∆H = - 23.4 kJ

FeO(s) + CO(g) h Fe(s) + CO2(g) ∆H = - 10.9 kJ

9.43 Calculate the energy change for the reaction S(s) + O2(g) h SO2(g) from the following: 2S(s) + 3O2(g) h 2SO3(g) ∆H = - 792 kJ 2SO2(g) + O2(g) h 2SO3(g) ∆H = - 198 kJ

9.44 Calculate the energy change for the reaction 3C(s) + O2(g) h C3O2(g) from the following: 2CO(g) + C(s) h C3O2(g) ∆H = + 127 kJ 2C(s) + O2(g) h 2CO(g) ∆H = - 222 kJ

UPDATE Testing Water Samples for Insecticides

One of the problems that Lance monitors is water pollution by insecticides. These insecticides are made by organic synthesis, in which smaller molecules are combined to form larger molecules, in a stepwise fashion.

b. If 100. g of naphthol and 100. g of phosgene react, what is the theoretical yield of C11H7O2Cl?

c. If the actual yield of C11H7O2Cl in part b is 115 g, what is the percent yield of C11H7O2Cl?

9.46 Another widely used insecticide is carbofuran (Furadan), an extremely toxic compound. In one step in the synthesis of carbofuran, the reaction shown is used.

C6H6O2 + C4H7Cl h C10H12O2 + HCl a. How many grams of C6H6O2 are needed to produce

3.8 * 103 g of C10H12O2? b. If 67.0 g of C6H6O2 and 51.0 g of C4H7Cl react, what is

the theoretical yield of C10H12O2? c. If the actual yield of C10H12O2 in part b is 85.7 g, what

is the percent yield of C10H12O2?

Applications

9.45 In one step in the synthesis of the insecticide carbaryl (Sevin), naphthol reacts with phosgene as shown.

C10H8O + COCl2 h C11H7O2Cl + HCl Naphthol Phosgene

a. How many kilograms of C11H7O2Cl form from 2.2 * 102 kg of naphthol?

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CONCEPT MAP

Theoretical YieldLimiting Reactant

Percent Yield Hess’s Law

Heat Is Released Heat Is Absorbed

Heat of Reaction ¢H

Mole–Mole Factors

Molar MassCoefficients

Balanced Chemical Equation

Grams of Reactants or Products

Exothermic Reaction

Endothermic Reaction

CHEMICAL QUANTITIES IN REACTIONS

use a

to calculate that indicates an

when

to give

whento calculateto calculate

that indicates an

use havehave

that give ratios called

CHAPTER REVIEW

9.1 Conservation of Mass LEARNING GOAL Calculate the total mass of reactants and the total mass of products in a balanced chemical equation.

9.3 Mass Calculations for Chemical Reactions LEARNING GOAL Given the mass in grams of a substance in a chemical reaction, calculate the mass in grams of another substance in the reaction. • In calculations using equations, the molar

masses of the substances and their mole–mole factors are used to change the number of grams of one substance to the corresponding grams of a different substance.

9.4 Limiting Reactants LEARNING GOAL Identify a limiting reactant when given the quantities of two reactants; calculate the amount of product formed from the limiting reactant. • A limiting reactant is the reactant

that produces the smaller amount of product while the other reactant is left over.

• When the masses of two reactants are given, the mass of a product is calculated from the limiting reactant.

2Ag(s) + S(s) Ag2S(s)

Mass of reactants = Mass of product

• In a balanced equation, the total mass of the reactants is equal to the total mass of the products.

9.2 Mole Relationships in Chemical Equations LEARNING GOAL Use a mole–mole factor from a balanced chemical equation to calculate the number of moles of another substance in the reaction.

Fe and S: 2 mol Fe 3 mol S

and 3 mol S 2 mol Fe

• The coefficients in an equation describing the relationship between the moles of any two components are used to write mole–mole factors.

• When the number of moles for one substance is known, a mole–mole factor is used to find the moles of a different substance in the reaction.

Chapter Review 261

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262 CHAPTER 9 Chemical Quantities in Reactions

9.5 Percent Yield LEARNING GOAL Given the actual quantity of product, determine the percent yield for a chemical reaction. • The percent yield for a

chemical reaction indicates the percent of product actually produced during a reaction.

• The percent yield is calculated by dividing the actual yield in grams of a product by the theoretical yield in grams and multiply- ing by 100%.

9.6 Energy in Chemical Reactions LEARNING GOAL Given the heat of reaction, calculate the loss or gain of heat for an exothermic or endothermic reaction. • In chemical reactions, the heat

of reaction (∆H) is the energy difference between the products and the reactants.

• In an exothermic reaction, the energy of the products is lower than that of the reactants. Heat is released, and ∆H is negative.

• In an endothermic reaction, the energy of the products is higher than that of the reactants; heat is absorbed, and ∆H is positive.

• Hess’s law states that heat can be absorbed or released in a single step or in several steps.

H2 Cl2+

HCl HCl+

E ne

rg y

Reactants

Products

Energy is released

Progress of Reaction

actual yield The actual amount of product produced by a reaction. endothermic reaction A reaction in which the energy of the products

is higher than that of the reactants. exothermic reaction A reaction in which the energy of the products

is lower than that of the reactants. heat of reaction The heat (symbol ∆H) absorbed or released when a

reaction takes place at constant pressure. Hess’s law Heat can be absorbed or released in a single chemical

reaction or in several steps. Law of Conservation of Mass In a chemical reaction, the total mass

of the reactants is equal to the total mass of the products; matter is neither lost nor gained.

KEY TERMS

limiting reactant The reactant used up during a chemical reaction, which limits the amount of product that can form.

mole–mole factor A conversion factor that relates the number of moles of two compounds in an equation derived from their coefficients.

percent yield The ratio of the actual yield for a reaction to the theo- retical yield possible for the reaction.

theoretical yield The maximum amount of product that a reaction can produce from a given amount of reactant.

The chapter section containing each Core Chemistry Skill is shown in parentheses at the end of each heading.

Using Mole–Mole Factors (9.2) Consider the balanced chemical equation

4Na(s) + O2(g) h 2Na2O(s)

• The coefficients in a balanced chemical equation represent the moles of reactants and the moles of products. Thus, 4 mol of Na react with 1 mol of O2 to form 2 mol of Na2O.

• From the coefficients, mole–mole factors can be written for any two substances as follows:

Na and O2 4 mol Na 1 mol O2

and 1 mol O2 4 mol Na

Na and Na2O 4 mol Na

2 mol Na2O and

2 mol Na2O

4 mol Na

O2 and Na2O 2 mol Na2O

1 mol O2 and

1 mol O2 2 mol Na2O

• A mole–mole factor is used to convert the number of moles of one substance in the reaction to the number of moles of another substance in the reaction.

CORE CHEMISTRY SKILLS

Answer: Given Need Connect

3.5 mol of Na2O moles of Na mole–mole factor

3.5 mol Na2O * 4 mol Na

2 mol Na2O = 7.0 mol of Na

Exact

Two SFs Exact Two SFs

Converting Grams to Grams (9.3) When we have the balanced chemical equation for a reaction, we can use the mass of substance A and then calculate the mass of substance B. The process is as follows:

• Use the molar mass of A to convert the mass, in grams, of A to moles of A.

• Use the mole–mole factor that converts moles of A to moles of B. • Use the molar mass of B to calculate the mass, in grams, of B. Molar Mole–mole Molar mass A factor mass B

Example: How many moles of sodium are needed to produce 3.5 mol of sodium oxide?

grams of A h mole of A h moles of B h grams of B

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Calculating Quantity of Product from a Limiting Reactant (9.4) Often in reactions, the reactants are not present in quantities that allow both reactants to be completely used up. Then one of the reactants, called the limiting reactant, determines the maximum amount of product that can form.

• To determine the limiting reactant, we calculate the amount of product that is possible from each reactant.

• The limiting reactant is the one that produces the smaller amount of product.

Example: If 12.5 g of S reacts with 17.2 g of O2, what is the limiting reactant and the mass, in grams, of SO3 produced?

2S(s) + 3O2(g) h 2SO3(g)

Answer: 3.50 g CH4 * 1 mol CH4

16.04 g CH4 *

- 802 kJ 1 mol CH4

Exact

Three SFs

Three SFs

Four SFs Exact

Three SFs

Therefore, 175 kJ are released.

• The percent yield is calculated from the actual yield divided by the theoretical yield and multiplied by 100%.

Percent yield (%) = actual yield

theoretical yield * 100%

Calculating Percent Yield (9.5)

• The theoretical yield for a reaction is the amount of product (100%) formed if all the reactants were converted to desired product.

• The actual yield for the reaction is the mass, in grams, of the product obtained at the end of the experiment. Because some product is usually lost, the actual yield is less than the theoretical yield.

Example: How many grams of O2 are needed to completely react with 14.6 g of Na?

4Na(s) + O2(g) h 2Na2O(s)

Core Chemistry Skills 263

Answer:

Three SFs

Exact

Exact

Four SFs

Three SFs Four SFs

Exact

Exact

14.6 g Na * 1 mol Na

22.99 g Na *

1 mol O2 4 mol Na

* 32.00 g O2 1 mol O2

= 5.08 g of O2

Four SFs

Exact

Answer: Mass of SO3 from S:

12.5 g S * 1 mol S

32.07 g S *

2 mol SO3 2 mol S

* 80.07 g SO3 1 mol SO3

Exact Exact

Four SFsThree SFs Exact

Three SFs = 31.2 g of SO3

Example: If 22.6 g of Al reacts completely with O2, and 37.8 g of Al2O3 is obtained, what is the percent yield of Al2O3 for the reaction?

4Al(s) + 3O2(g) h 2Al2O3(s)

Using the Heat of Reaction (9.6)

• The heat of reaction is the amount of heat, usually in kJ, that is absorbed or released during a reaction.

• The heat of reaction or energy change, symbol ∆H, is the difference in the energy of the products and the reactants.

∆H = Hproducts - Hreactants • In an exothermic reaction, the energy of the products is lower than

that of the reactants. This means that heat is released along with the products that form. The sign for the heat of reaction, ∆H, is negative.

• In an endothermic reaction, the energy of the products is higher than that of the reactants. The heat is required to convert the reac- tants to products. The sign for the heat of reaction, ∆H, is positive.

Example: How many kilojoules are released when 3.50 g of CH4 undergoes combustion?

CH4(g) + 2O2(g) h CO2(g) + 2H2O(g) ∆H = - 802 kJ

Answer: Calculation of theoretical yield:

Percent yield (%) = actual yield (given)

theoretical yield (calculated) * 100%

= 37.8 g Al2O3 42.7 g Al2O3

* 100% = 88.5%

Three SFs

Three SFsThree SFs

= 28.7 g of SO3 Three SFs

Mass of SO3 from O2:

17.2 g O2 * 1 mol O2

32.00 g O2 *

2 mol SO3 3 mol O2

* 80.07 g SO3 1 mol SO3

Exact

ExactFour SFsThree SFs

Four SFs

Exact

Exact

Therefore, O2, is the limiting reactant, and 28.7 g of SO3 can be produced.

Exact

Exact Five SFs

Three SFs Four SFs Exact

Exact

Three SFs

22.6 g Al * 1 mol Al

26.98 g Al *

2 mol Al2O3 4 mol Al

* 101.96 g Al2O3

1 mol Al2O3

= 42.7 g of Al2O3 Theoretical yield

Calculation of percent yield:

= - 175 kJ

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264 CHAPTER 9 Chemical Quantities in Reactions

Products

or

A B

Reactants

Reactants Products

UNDERSTANDING THE CONCEPTS

The chapter sections to review are shown in parentheses at the end of each problem.

9.47 If red spheres represent oxygen atoms, blue spheres represent nitrogen atoms, and all the molecules are gases, (9.2, 9.4)

Reactants Products

a. write a balanced equation for the reaction. b. identify the limiting reactant.

9.48 If green spheres represent chlorine atoms, yellow-green spheres represent fluorine atoms, white spheres represent hydrogen atoms, and all the molecules are gases, (9.2, 9.4)

a. write a balanced equation for the reaction. b. identify the diagram that shows the products.

9.50 If purple spheres represent iodine atoms, white spheres represent hydrogen atoms, and all the molecules are gases, (9.2, 9.4)

a. write a balanced equation for the reaction. b. identify the limiting reactant.

9.49 If blue spheres represent nitrogen atoms, white spheres represent hydrogen atoms, and all the molecules are gases, (9.2, 9.4)

Products

or or

Reactants

A B C

a. write a balanced equation for the reaction. b. identify the diagram that shows the products.

9.51 If blue spheres represent nitrogen atoms, purple spheres represent iodine atoms, and the reacting molecules are solid, and the products are gases, (9.2, 9.4, 9.5)

Reactants Actual products

a. write a balanced equation for the reaction. b. from the diagram of the actual products that result,

calculate the percent yield for the reaction.

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Reactants Actual products

a. write a balanced equation for the reaction. b. identify the limiting reactant. c. from the diagram of the actual products that result, calculate

the percent yield for the reaction.

9.53 Use the balanced chemical equation to complete the table: (9.1, 9.2)

2FeS(s) + 3O2(g) h 2FeO(s) + 2SO2(g)

FeS O2 FeO SO2

2.0 mol _______ mol _______ mol _______ mol

_______ mol _______ mol _______ mol 4.6 mol

9.54 Use the balanced chemical equation to complete the table: (9.1, 9.2)

2C2H6(g) + 7O2(g) h ∆

4CO2(g) + 6H2O(g)

C2H6 O2 CO2 H2O

_______ mol _______ mol 3.2 mol _______ mol

_______ mol 2.8 mol _______ mol _______ mol

9.55 When ammonia (NH3) reacts with fluorine (F2), the products are dinitrogen tetrafluoride and hydrogen fluoride. (9.2, 9.3)

2NH3(g) + 5F2(g) h N2F4(g) + 6HF(g)

a. How many moles of each reactant are needed to produce 4.00 mol of HF?

b. How many grams of F2 are required to react with 25.5 g of NH3?

c. How many grams of N2F4 can be produced when 3.40 g of NH3 reacts?

9.56 Gasohol is a fuel that contains ethanol (C2H6O) that burns in oxygen (O2) to give carbon dioxide and water. (9.2, 9.3)

C2H6O(g) + 3O2(g) h ∆

2CO2(g) + 3H2O(g)

a. How many moles of O2 are needed to completely react with 4.0 mol of C2H6O?

b. If a car produces 88 g of CO2, how many grams of O2 are used up in the reaction?

c. If you add 125 g of C2H6O to your fuel, how many grams of CO2 can be produced from the ethanol?

9.57 When hydrogen peroxide (H2O2) is used in rocket fuels, it produces water and oxygen. (9.2, 9.3)

2H2O2(l) h 2H2O(l) + O2(g)

a. How many moles of H2O2 are needed to produce 3.00 mol of H2O?

b. How many grams of H2O2 are required to produce 36.5 g of O2?

c. How many grams of H2O can be produced when 12.2 g of H2O2 reacts?

9.58 Propane gas (C3H8) reacts with oxygen to produce carbon dioxide and water. (9.2, 9.3)

C3H8(g) + 5O2(g) h ∆

3CO2(g) + 4H2O(g)

a. How many moles of H2O form when 5.00 mol of C3H8 completely reacts?

b. How many grams of CO2 are produced from 18.5 g of O2? c. How many grams of H2O can be produced when 46.3 g

of C3H8 reacts?

9.59 When 12.8 g of Na and 10.2 g of Cl2 react, what is the mass, in grams, of NaCl that is produced? (9.2, 9.3, 9.4)

2Na(s) + Cl2(g) h 2NaCl(s)

9.60 If 35.8 g of CH4 and 75.5 g of S react, how many grams of H2S are produced? (9.2, 9.3, 9.4)

CH4(g) + 4S(g) h CS2(g) + 2H2S(g) 9.61 Pentane gas (C5H12) reacts with oxygen to produce carbon

dioxide and water. (9.2, 9.3, 9.4)

C5H12(g) + 8O2(g) h ∆

5CO2(g) + 6H2O(g) a. How many moles of C5H12 must react to produce 4.00 mol

of H2O? b. How many grams of CO2 are produced from 32.0 g of O2? c. How many grams of CO2 are formed if 44.5 g of C5H12 is

mixed with 108 g of O2?

9.62 When nitrogen dioxide (NO2) from car exhaust combines with water in the air, it forms nitrogen oxide and nitric acid (HNO3), which causes acid rain. (9.2, 9.3, 9.4)

3NO2(g) + H2O(l) h NO(g) + 2HNO3(aq) a. How many moles of NO2 are needed to react with

0.250 mol of H2O? b. How many grams of HNO3 are produced when 60.0 g

of NO2 completely reacts? c. How many grams of HNO3 can be produced if 225 g

of NO2 is mixed with 55.2 g of H2O?

9.63 The acetylene (C2H2) used in welders’ torches, burns according to the following equation: (9.2, 9.3, 9.5)

2C2H2(g) + 5O2(g) h ∆

4CO2(g) + 2H2O(g) a. What is the theoretical yield, in grams, of CO2, if 22.0 g

of C2H2 completely reacts? b. If the actual yield in part a is 64.0 g of CO2, what is the

percent yield of CO2 for the reaction?

9.64 The equation for the decomposition of potassium chlorate is written as (9.2, 9.3, 9.5)

2KClO3(s) h ∆

2KCl(s) + 3O2(g) a. When 46.0 g of KClO3 is completely decomposed, what

is the theoretical yield, in grams, of O2? b. If the actual yield in part a is 12.1 g of O2, what is the

percent yield of O2?

9.65 When 28.0 g of acetylene reacts with hydrogen, 24.5 g of ethane is produced. What is the percent yield of C2H6 for the reaction? (9.2, 9.3, 9.5)

C2H2(g) + 2H2(g) h Pt

C2H6(g)

ADDITIONAL PRACTICE PROBLEMS

Additional Practice Problems 265

9.52 If green spheres represent chlorine atoms, red spheres represent oxygen atoms, and all the molecules are gases, (9.2, 9.4, 9.5)

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266 CHAPTER 9 Chemical Quantities in Reactions

a. How many kilojoules are required to form 3.00 g of NO? b. How many grams of NO form when 66.0 kJ are absorbed?

9.70 The equation for the reaction of iron and oxygen gas to form rust (Fe2O3) is written as (9.2, 9.6)

4Fe(s) + 3O2(g) h 2Fe2O3(s) ∆H = - 1.7 * 103 kJ a. How many kilojoules are released when 2.00 g of Fe reacts? b. How many grams of Fe2O3 form when 150 kJ are released?

Applications

9.71 Each of the following is a reaction that occurs in the cells of the body. Identify each as exothermic or endothermic. (9.6)

a. Succinyl CoA + H2O h succinate + CoA + 37 kJ b. GDP + Pi + 34 kJ h GTP + H2O 9.72 Each of the following is a reaction that occurs in the cells of

the body. Identify each as exothermic or endothermic. (9.6)

a. Phosphocreatine + H2O h creatine + Pi + 42.7 kJ b. Fructose@6@phosphate + Pi + 16 kJ h

fructose@1,6@bisphosphate

9.66 When 50.0 g of iron(III) oxide reacts with carbon monoxide, 32.8 g of iron is produced. What is the percent yield of Fe for the reaction? (9.2, 9.3, 9.5)

Fe2O3(s) + 3CO(g) h 2Fe(s) + 3CO2(g) 9.67 Nitrogen and hydrogen combine to form ammonia. (9.2, 9.3,

9.4, 9.5)

N2(g) + 3H2(g) h 2NH3(g) a. If 50.0 g of N2 is mixed with 20.0 g of H2, what is the

theoretical yield, in grams, of NH3? b. If the reaction in part a has a percent yield of 62.0%, what

is the actual yield, in grams, of NH3?

9.68 Sodium and nitrogen combine to form sodium nitride. (9.2, 9.3, 9.4, 9.5)

6Na(s) + N2(g) h 2Na3N(s) a. If 80.0 g of Na is mixed with 20.0 g of nitrogen gas, what is

the theoretical yield, in grams, of Na3N? b. If the reaction in part a has a percent yield of 75.0%, what

is the actual yield, in grams, of Na3N?

9.69 The equation for the reaction of nitrogen and oxygen to form nitrogen oxide is written as (9.2, 9.6)

N2(g) + O2(g) h 2NO(g) ∆H = + 90.2 kJ

The following problems are related to the topics in this chapter. However, they do not all follow the chapter order, and they require you to combine concepts and skills from several sections. These problems will help you increase your critical thinking skills and prepare for your next exam.

9.73 Use the balanced chemical equation to complete the table: (9.1, 9.2, 9.3)

2FeS(s) + 3O2(g) h 2FeO(s) + 2SO2(g)

FeS O2 FeO SO2

26 g _______ g _______ g _______ g

_______ g _______ g _______ g 7.94 g

9.74 Use the balanced chemical equation to complete the table: (9.1, 9.2, 9.3)

2C2H6(g) + 7O2(g) h ∆

4CO2(g) + 6H2O(g)

C2H6 O2 CO2 H2O

_______ g _______ g 39 g _______ g

_______ g 73.3 g _______ g _______ g

9.75 Chromium and oxygen combine to form chromium(III) oxide. (9.2, 9.3, 9.4, 9.5)

4Cr(s) + 3O2(g) h 2Cr2O3(s) a. How many moles of O2 react with 4.50 mol of Cr? b. How many grams of Cr2O3 are produced when 24.8 g of

Cr reacts? c. When 26.0 g of Cr reacts with 8.00 g of O2, how many

grams of Cr2O3 can form? d. If 74.0 g of Cr and 62.0 g of O2 are mixed, and 87.3 g of

Cr2O3 is actually obtained, what is the percent yield of Cr2O3 for the reaction?

CHALLENGE PROBLEMS

9.76 Aluminum and chlorine combine to form aluminum chloride. (9.2, 9.3, 9.4, 9.5)

2Al(s) + 3Cl2(g) h 2AlCl3(s) a. How many moles of Cl2 are needed to react with 4.50 mol

of Al? b. How many grams of AlCl3 are produced when 50.2 g of Al

reacts? c. When 13.5 g of Al reacts with 8.00 g of Cl2, how many

grams of AlCl3 can form? d. If 45.0 g of Al and 62.0 g of Cl2 are mixed, and 66.5 g of

AlCl3 is actually obtained, what is the percent yield of AlCl3 for the reaction?

9.77 The combustion of propyne (C3H4) releases heat when it burns according to the following equation: (9.2, 9.3, 9.5)

C3H4(g) + 4O2(g) h ∆

3CO2(g) + 2H2O(g)

a. How many moles of O2 are needed to react completely with 0.225 mol of C3H4?

b. How many grams of H2O are produced from the complete reaction of 64.0 g of O2?

c. How many grams of CO2 are produced from the complete reaction of 78.0 g of C3H4?

d. If the reaction in part c produces 186 g of CO2, what is the percent yield of CO2 for the reaction?

9.78 Butane gas (C4H10) burns according to the following equation: (9.2, 9.3, 9.5)

2C4H10(g) + 13O2(g) h ∆

8CO2(g) + 10H2O(g)

a. How many moles of H2O are produced from the complete reaction of 2.50 mol of C4H10?

b. How many grams of O2 are needed to react completely with 22.5 g of C4H10?

c. How many grams of CO2 are produced from the complete reaction of 55.0 g of C4H10?

d. If the reaction in part c produces 145 g of CO2, what is the percent yield of CO2 for the reaction?

M09_TIMB8119_06_SE_C09.indd 266 11/29/18 10:01 AM

a. How many grams of O2 are needed to react with 15.0 g of glycine?

b. How many grams of urea are produced from 15.0 g of glycine? c. How many grams of CO2 are produced from 15.0 g of

glycine?

9.84 In the body, ethanol (C2H6O) reacts according to the following equation: (9.3)

C2H6O(aq) + 3O2(g) h 2CO2(g) + 3H2O(l) Ethanol

a. How many grams of O2 are needed to react with 8.40 g of ethanol?

b. How many grams of H2O are produced from 8.40 g of ethanol?

c. How many grams of CO2 are produced from 8.40 g of ethanol?

9.85 In photosynthesis, glucose (C6H12O6) and O2 are produced from CO2 and H2O. Glucose from starches is the major fuel for the body. (9.3, 9.6)

6CO2(g) + 6H2O(l) + 680 kcal h C6H12O6(aq) + 6O2(g) Glucose

a. Is the reaction exothermic or endothermic? b. How many grams of glucose are produced from 18.0 g

of CO2? c. How much heat, in kilojoules, is needed to produce 25.0 g

of C6H12O6?

9.86 Ethanol (C2H6O) reacts in the body using the following equation: (9.3, 9.5)

C2H6O(aq) + 3O2(g) h 2CO2(g) + 3H2O(l) + 327 kcal Ethanol

a. Is the reaction exothermic or endothermic? b. How much heat, in kilocalories, is produced when 5.00 g

of ethanol reacts with O2 gas? c. How many grams of ethanol react if 1250 kJ are produced?

9.79 Sulfur trioxide decomposes to sulfur and oxygen. (9.2, 9.6)

2SO3(g) h 2S(s) + 3O2(g) ∆H = + 790 kJ a. Is the reaction exothermic or endothermic? b. How many kilojoules are required when 1.5 mol of SO3

reacts? c. How many kilojoules are required when 150 g of O2 is

formed?

9.80 When hydrogen peroxide (H2O2) is used in rocket fuels, it produces water, oxygen, and heat. (9.2, 9.6)

2H2O2(l) h 2H2O(l) + O2(g) ∆H = - 196 kJ a. Is the reaction exothermic or endothermic? b. How many kilojoules are released when 2.50 mol of H2O2

reacts? c. How many kilojoules are released when 275 g of O2 is

produced?

9.81 Calculate the energy change for the reaction NH4Cl(s) h NH3(g) + HCl(g) from the following: (9.6) H2(g) + Cl2(g) h 2HCl(g) ∆H = - 184 kJ N2(g) + 4H2(g) + Cl2(g) h 2NH4Cl(s) ∆H = - 631 kJ N2(g) + 3H2(g) h 2NH3(g) ∆H = - 296 kJ

9.82 Calculate the energy change for the reaction Mg(s) + N2(g) + 3O2(g) h Mg(NO3)2(s) from the following: (9.6)

8Mg(s) + Mg(NO3)2(s) h Mg3N2(s) + 6MgO(s) ∆H = - 3281 kJ Mg3N2(s) h 3Mg(s) + N2(g) ∆H = + 461 kJ 2MgO(s) h 2Mg(s) + O2(g) ∆H = + 1204 kJ

Applications

9.83 In the body, the amino acid glycine (C2H5NO2) reacts according to the following equation: (9.3)

2C2H5NO2(aq) + 3O2(g) h Glycine 3CO2(g) + 3H2O(l) + CH4N2O(aq) Urea

9.1 The law of conservation of mass states that the total mass of reactants must be equal to the total mass of products in a chemical reaction. No material is lost or gained as original substances are changed to new substances.

9.2 The mole–mole factor relates the moles of reactants and prod- ucts in a reaction. The grams must first be changed to moles and then the mole–mole factor is used.

9.3 When the 6 knives are used, no more people can be served. The spoons and forks are excess.

ANSWERS TO ENGAGE QUESTIONS

9.4 You prepared enough dough to make five dozen, or 60, cookies. If 12 were burned, 48 were made.

48/60 * 100% = 80% yield 9.5 The heat of reaction can be related to each of the reactants and

products in a reaction, making many heat conversion factors possible.

Answers to Selected Problems 267

ANSWERS TO SELECTED PROBLEMS 9.1 a. 160.14 g of reactants = 160.14 g of products b. 283.88 g of reactants = 283.88 g of products

9.3 a. 2 mol SO2 1 mol O2

and 1 mol O2

2 mol SO2 2 mol SO2 2 mol SO3

and 2 mol SO3 2 mol SO2

2 mol SO3 1 mol O2

and 1 mol O2

2 mol SO3

b. 4 mol P

5 mol O2 and

5 mol O2 4 mol P

4 mol P 2 mol P2O5

and 2 mol P2O5

4 mol P

5 mol O2 2 mol P2O5

and 2 mol P2O5 5 mol O2

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268 CHAPTER 9 Chemical Quantities in Reactions

9.47 a. 2NO(g) + O2(g) h 2NO2(g) b. NO is the limiting reactant.

9.49 a. N2(g) + 3H2(g) h 2NH3(g) b. A

9.51 a. 2NI3(g) h N2(g) + 3I2(g) b. 67%

9.53

9.5 a. mol SO2 * 2 mol SO3 2 mol SO2

= mol of SO3

b. mol P * 5 mol O2 4 mol P

= mol of O2

9.7 a. 1.3 mol of O2 b. 10. mol of H2 c. 5.0 mol of H2O

9.9 a. 1.25 mol of C b. 0.96 mol of CO c. 1.0 mol of SO2 d. 0.50 mol of CS2 9.11 a. 77.5 g of Na2O b. 6.26 g of O2 c. 19.4 g of O2 9.13 a. 19.2 g of O2 b. 3.79 g of N2 c. 54.0 g of H2O

9.15 a. 3.66 g of H2O b. 26.3 g of NO c. 7.53 g of HNO3 9.17 a. 2PbS(s) + 3O2(g) h 2PbO(s) + 2SO2(g) b. 6.00 g of O2 c. 17.4 g of SO2 d. 137 g of PbS

9.19 a. Eight taxis can be used to pick up passengers. b. Seven taxis can be driven.

9.21 a. 5.0 mol of H2 b. 4.0 mol of H2 c. 3.0 mol of N2 9.23 a. Cl2 is the limiting reactant, which would produce 25.1 g of

AlCl3. b. O2 is the limiting reactant, which would produce 13.5 g of

H2O. c. O2 is the limiting reactant, which would produce 26.7 g of

SO2.

9.25 a. 31.2 g of SO3 b. 34.6 g of Fe3O4 c. 35.0 g of CO2 9.27 a. 71.0% b. 63.2%

9.29 70.9 g of Al2O3 9.31 60.5%

9.33 In exothermic reactions, the energy of the products is lower than that of the reactants.

9.35 a. exothermic b. endothermic c. exothermic

9.37 a. Heat is released, exothermic, ∆H = - 802 kJ b. Heat is absorbed, endothermic, ∆H = + 65.3 kJ c. Heat is released, exothermic, ∆H = - 850 kJ 9.39 a. 579 kJ b. 89 kJ

9.41 + 66 kJ 9.43 - 297 kJ 9.45 a. 3.2 * 102 kg b. 143 g c. 80.4%

FeS O2 FeO SO2

2.0 mol 3.0 mol 2.0 mol 2.0 mol

4.6 mol 6.9 mol 4.6 mol 4.6 mol

9.55 a. 1.33 mol of NH3 and 3.33 mol of F2 b. 142 g of F2 c. 10.4 g of N2F4 9.57 a. 3.00 mol of H2O2 b. 77.6 g of H2O2 c. 6.46 g of H2O

9.59 16.8 g of NaCl

9.61 a. 0.667 mol of C5H12 b. 27.5 g of CO2 c. 92.8 g of CO2 9.63 a. 74.4 g of CO2 b. 86.0%

9.65 75.9%

9.67 a. 60.8 g of NH3 b. 37.7 g of NH3 9.69 a. 4.51 kJ b. 43.9 g of NO

9.71 a. exothermic b. endothermic

9.73 FeS O2 FeO SO2

26 g 14 g 21 g 19 g

10.9 g 5.95 g 8.90 g 7.94 g

9.75 a. 3.38 mol of O2 b. 36.2 g of Cr2O3 c. 25.3 g of Cr2O3 d. 80.8%

9.77 a. 0.900 mol of O2 b. 18.0 g of O2 c. 257 g of CO2 d. 72.4%

9.79 a. endothermic b. 590 kJ c. 1200 kJ

9.81 76 kJ

9.83 a. 9.59 g of O2 b. 6.00 g of urea c. 13.2 g of CO2 9.85 a. endothermic b. 12.3 g of glucose c. 390 kJ

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269

UPDATE Histologist Stains Tissue with Dye

When Lisa obtains the tissue sample, she will freeze it to obtain thin slices of tissue for staining. You can learn more about how Lisa prepares the tissue from Bill's surgery for the pathologist by reading the UPDATE Histologist Stains Tissue with Dye, page 298.

Bill has been diagnosed with basal cell carcinoma, the most common form of skin cancer. He has an appointment to undergo Mohs surgery, a specialized procedure to remove the cancerous growth found on his shoulder. The surgeon begins by removing the abnormal growth, in addition to a thin layer of surrounding (margin) tissue, which he sends to Lisa, a histologist. Lisa prepares the tissue sample to be viewed by a pathologist. Tissue preparation requires Lisa to cut the tissue into a very thin section (normally about 0.001 cm), which is then mounted onto a microscope slide.

Lisa then treats the tissue with a dye to stain the cells, as this enables the pathologist to view any abnormal cells more easily. The pathologist examines the tissue sample and reports back to the surgeon. If no cancerous cells are detected, no more tissue will be removed. If cancer cells are still present, the surgeon will take more of the surrounding tissue.

CAREER

Histologist Histologists study the microscopic makeup of tissues, cells, and bodily fluids in order to detect and identify the presence of a specific disease. They determine blood types and the concentrations of drugs and other substances in the blood. Sample preparation is a critical component of a histologist’s job, as they prepare tissue samples from humans, animals, and plants. The tissue samples are cut into extremely thin sections, which are then mounted and stained using various chemical dyes. The dyes provide contrast for the cells to be viewed and help highlight any abnormalities that may exist. Utilization of various dyes requires the histologist to be familiar with solution preparation and the handling of potentially hazardous chemicals.

Bonding and Properties of Solids and Liquids

10

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270 CHAPTER 10 Bonding and Properties of Solids and Liquids

10.1 Lewis Structures for Molecules and Polyatomic Ions LEARNING GOAL Draw the Lewis structures for molecular compounds and polyatomic ions with single and multiple bonds.

Now we can investigate more complex chemical bonds and how they contribute to the structure of a molecule or a polyatomic ion. We will diagram the sharing of valence electrons in molecules and polyatomic ions, and determine the presence of single and multiple bonds.

Lewis Symbols for Atoms A Lewis symbol is a convenient way to represent the valence electrons, which are shown as dots placed on the sides, top, or bottom of the symbol for the element. One to four valence electrons are arranged as single dots. When there are five to eight electrons, one or more electrons are paired. Any of the following would be an acceptable Lewis symbol for mag- nesium, which has two valence electrons:

Lewis Symbols for Magnesium

Lewis symbols for selected elements are given in TABLE 10.1.

REVIEW Writing Electron Configurations (5.4)

Using the Periodic Table to Write Electron Configurations (5.5)

Writing Ionic Formulas (6.2)

LOOKING AHEAD

10.1 Lewis Structures for Molecules and Polyatomic Ions 270

10.2 Resonance Structures 276 10.3 Shapes of Molecules and

Polyatomic Ions (VSEPR Theory) 279

10.4 Electronegativity and Bond Polarity 283

10.5 Polarity of Molecules 287 10.6 Intermolecular Forces

Between Atoms or Molecules 288

10.7 Changes of State 291

SAMPLE PROBLEM 10.1 Drawing Lewis Symbols

TRY IT FIRST

Draw the Lewis symbol for each of the following: a. bromine b. aluminum

SOLUTION

a. The Lewis symbol for bromine, which is in Group 7A (17), has seven valence electrons. Thus, three pairs of dots and one single dot are drawn on the sides of the Br symbol.

Br

b. The Lewis symbol for aluminum, which is in Group 3A (13), has three valence electrons. They are drawn as single dots on the sides of the Al symbol.

Al

CORE CHEMISTRY SKILL Drawing Lewis Symbols

Number of Valence Electrons Increases

H He *

Li Be C NB O F Ne

Na Mg Si PAl S Cl Ar

K Ca Ge AsGa Se Br Kr

Mg Mg Mg Mg Mg Mg

TABLE 10.1 Lewis Symbols for Selected Elements in Periods 1 to 4 Group Number

Number of Valence Electrons

1A (1)

1

2A (2)

2

3A (13)

3

4A (14)

4

5A (15)

5

6A (16)

6

7A (17)

7

8A (18)

8

Lewis Symbol

Number of Valence Electrons Increases

H He *

Li Be C NB O F Ne

Na Mg Si PAl S Cl Ar

K Ca Ge AsGa Se Br Kr

Mg Mg Mg Mg Mg Mg

*Helium (He) is stable with two valence electrons.

ENGAGE 10.1 Why do nitrogen, phosphorus, and arsenic have five dots in their Lewis symbols?

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10.1 Lewis Structures for Molecules and Polyatomic Ions 271

FIGURE 10.1 A covalent bond forms as H atoms move close together to share electrons.

H

H

H H

H

H

Distance Between Nuclei Decreases

Far apart; no attractions

Attractions pull atoms closer

H2 molecule

E ne

rg y

In cr

ea se

s SELF TEST 10.1

Draw the Lewis symbol for each of the following:

a. phosphorus, a macromineral needed for bones and teeth b. selenium, a micromineral needed for the immune system

ANSWER

a. P Se b. P Se

Lewis Structures for Ionic Compounds An ionic compound forms when valence electrons from a metal are transferred to a nonmetal. For example, the transfer of one valence electron from sodium to chlorine gives octets to both. The loss of an electron by sodium produces a sodium ion, and the gain of an electron by chlorine produces a chloride ion. We can use the Lewis symbols to show this. The Lewis structure of the compound NaCl shows the chloride ion in brackets with its negative charge in the upper right corner.

Na Na+ -

Cl Cl+

Lewis Structures for Molecular Compounds The simplest molecule is hydrogen, H2. When two H atoms are far apart, there is no attrac- tion between them. As the H atoms move closer, the positive charge of each nucleus attracts the electron of the other atom. This attraction, which is greater than the repulsion between the valence electrons, pulls the H atoms closer until they share a pair of valence electrons (see FIGURE 10.1). The result is called a covalent bond, in which the shared electrons give the stable electron configuration of He to each of the H atoms. When the H atoms form H2, they are more stable than two individual H atoms.

ENGAGE 10.2 What determines the attraction between two H atoms?

H2 N2 O2 F2

Cl2

Br2

I2

The elements hydrogen, nitrogen, oxygen, fluorine, chlorine, bromine, and iodine exist as diatomic molecules.

A molecule is represented by a Lewis structure in which the valence electrons of all the atoms are arranged to give octets, except for hydrogen, which has two electrons. The shared electrons, or bonding pairs, are shown as two dots or a single line between atoms. The nonbonding pairs of electrons, or lone pairs, are placed on the outside. For example, a fluorine molecule, F2, consists of two fluorine atoms, which are in Group 7A (17), each with seven valence electrons. In the Lewis structure for the F2 molecule, each F atom achieves an octet by sharing its unpaired valence electron.

CORE CHEMISTRY SKILL Drawing Lewis Structures

Lewis structure

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272 CHAPTER 10 Bonding and Properties of Solids and Liquids

TABLE 10.2 Elements That Exist as Diatomic Molecules

Diatomic Molecule Name

H2 Hydrogen

N2 Nitrogen

O2 Oxygen

F2 Fluorine

Cl2 Chlorine

Br2 Bromine

I2 Iodine

TABLE 10.3 Typical Bonding Patterns of Some Nonmetals 1A (1) 3A (13) 4A (14) 5A (15) 6A (16) 7A (17)

*H 1 bond

*B

3 bonds

C

4 bonds

N

3 bonds

O

2 bonds

F

1 bond

Si

4 bonds

P

3 bonds

S

2 bonds

Cl, Br, I

1 bond

*H and B do not form octets. H atoms share one electron pair; B atoms share three electron pairs for a set of six electrons.

TABLE 10.4 Representations for Some Molecular Compounds

CH4 Methane molecule

NH3 Ammonia molecule

H2O Water molecule

Lewis Structures

H

H H H H H H H HOC N

H

H H H H H H H HOC N

H

H H H

HHHC HH ON

Ball-and-Stick Models

Space-Filling Models

+ =

Electrons to share

A shared pair of electrons

Lewis structures

A covalent bond

Space-filling model of

fluorine molecule

Ball-and-stick model of fluorine

molecule

F2

A lone pair A bonding pair

F F F FFF

Hydrogen (H2) and fluorine (F2) are examples of nonmetal elements whose natural state is diatomic; that is, they contain two like atoms. The elements that exist as diatomic molecules are listed in TABLE 10.2.

Sharing Electrons Between Atoms of Different Elements The number of electrons that a nonmetal atom shares and the number of covalent bonds it forms are usually equal to the number of electrons it needs to achieve a stable electron configuration. TABLE 10.3 gives the most typical bonding patterns for some nonmetals.

Drawing Lewis Structures To draw the Lewis structure for CH4, we first draw the Lewis symbols for carbon and hydrogen.

C H

Then we determine the number of valence electrons needed for carbon and hydrogen. When a carbon atom shares its four electrons with four hydrogen atoms, carbon obtains an octet, and each hydrogen atom is complete with two shared electrons. The Lewis structure is drawn with the carbon atom as the central atom, with the hydrogen atoms on each of the sides. The bonding pairs of electrons, which are single covalent bonds, may also be shown as single lines between the carbon atom and each of the hydrogen atoms.

CH H

H

H

H

H H HC

In addition to Lewis structures, three-dimensional models may be used to represent molecular compounds. A ball-and-stick model illustrates the shape of a molecule using spheres for atoms and sticks as bonding electron pairs. In a space-filling model, the atoms are shown as partial spheres that are pro- portional to the real size of the atoms. Because the bonds are within the filled spaces, they are not visible.

TABLE 10.4 gives examples of Lewis structures and molecular models for some molecules.

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10.1 Lewis Structures for Molecules and Polyatomic Ions 273

When we draw a Lewis structure for a molecule or polyatomic ion, we show the sequence of atoms, the bonding pairs of electrons shared between atoms, and the nonbonding (lone pairs) of electrons. From the formula, we identify the central atom, which is the ele- ment that has the fewer atoms. Then, the central atom is bonded to the other atoms, as shown in Sample Problem 10.2.

SAMPLE PROBLEM 10.2 Drawing Lewis Structures for Molecules

TRY IT FIRST

Draw the Lewis structures for PCl3, phosphorus trichloride, used to prepare insecticides and flame retardants.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

PCl3 Lewis structure total valence electrons

STEP 1 Determine the arrangement of atoms. In PCl3, the central atom is P because there is only one P atom.

Cl P Cl Cl

STEP 2 Determine the total number of valence electrons. We use the group number to determine the number of valence electrons for each of the atoms in the molecule.

Element Group Atoms Valence Electrons Total

P 5A (15) 1 P * 5 e- = 5 e-

Cl 7A (17) 3 Cl * 7 e- = 21 e-

Total valence electrons for PCl3 = 26 e -

STEP 3 Attach each bonded atom to the central atom with a pair of electrons. Each bonding pair can also be represented by a bond line.

PClor Cl

Cl Cl

Cl ClP

Six electrons (3 * 2 e-) are used to bond the central P atom to three Cl atoms. Twenty valence electrons are left.

26 valence e- - 6 bonding e- = 20 e- remaining

STEP 4 Use the remaining electrons to complete octets. We use the remaining 20 electrons as lone pairs, which are placed around the outer Cl atoms and on the P atom, such that all the atoms have octets.

PorP

Cl Cl

Cl ClCl Cl

SELF TEST 10.2

Draw the Lewis structures for Cl2O, one with electron dots and the other with lines for bonding pairs.

ANSWER

O OorCl Cl Cl Cl

ENGAGE 10.3 How are the total number of valence electrons determined for a molecule?

PCl3 Ball-and-stick model

PCl3 Space-filling model

Cl2O Ball-and-stick model

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274 CHAPTER 10 Bonding and Properties of Solids and Liquids

SAMPLE PROBLEM 10.3 Drawing Lewis Structures for Polyatomic Ions

TRY IT FIRST

Sodium chlorite, NaClO2, is a component of mouthwashes, toothpastes, and contact lens cleaning solutions. Draw the Lewis structures for the chlorite ion, ClO2

-.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connent

ClO2 - Lewis structure total valence electrons

STEP 1 Determine the arrangement of atoms. For the polyatomic ion ClO2 -,

the central atom is Cl because there is only one Cl atom. For a polyatomic ion, the atoms and electrons are placed in brackets, and the charge is written outside at the upper right.

[O Cl O]-

STEP 2 Determine the total number of valence electrons. We use the group numbers to determine the number of valence electrons for each of the atoms in the ion. Because the ion has a negative charge, one more electron is added to the valence electrons.

Element Group Atoms Valence Electrons Total

Cl 7A (17) 1 Cl * 7 e- = 7 e-

O 6A (16) 2 O * 6 e- = 12 e-

Ionic charge (negative) add 1 e- = 1 e-

Total valence electrons for ClO2 - = 20 e-

STEP 3 Attach each bonded atom to the central atom with a pair of electrons. Each bonding pair can also be represented by a line, which indicates a single bond.

[O [OO]- O]-orCl Cl

Four electrons (2 * 2 e-) are used to bond the O atoms to the central Cl atom, which leaves 16 valence electrons.

STEP 4 Use the remaining electrons to complete octets. Of the 16 remaining valence electrons, 12 are drawn as lone pairs to complete the octets of the O atoms.

O OOO Cl -

Cl -

or

The remaining four electrons are placed as two lone pairs on the central Cl atom.

O O orCl O OCl - -

SELF TEST 10.3

Draw the Lewis structures for the polyatomic ion NH2 -, one with electron dots and the

other with lines for bonding pairs.

ANSWER

or H HH HN NN --

ClO2 − Space-filling model

ClO2 − Ball-and-stick model

NH2 − Ball-and-stick model

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10.1 Lewis Structures for Molecules and Polyatomic Ions 275

Double and Triple Bonds Up to now, we have looked at bonding in molecules having only single bonds. In many molecular compounds, atoms share two or three pairs of electrons to complete their octets.

Double and triple bonds form when the number of valence electrons is not enough to complete the octets of all the atoms in the molecule. Then one or more lone pairs of electrons from the atoms attached to the central atom are shared with the central atom. A double bond occurs when two pairs of electrons are shared; in a triple bond, three pairs of electrons are shared. Atoms of carbon, oxygen, nitrogen, and sulfur are most likely to form multiple bonds.

Atoms of hydrogen and the halogens do not form double or triple bonds. The process of drawing a Lewis structure with multiple bonds is shown in Sample Problem 10.4.

SAMPLE PROBLEM 10.4 Drawing Lewis Structures with Multiple Bonds

TRY IT FIRST

Draw the Lewis structures for carbon dioxide, CO2.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

CO2 Lewis structure total valence electrons

STEP 1 Determine the arrangement of atoms. In CO2, the central atom is C because there is only one C atom.

O C O

STEP 2 Determine the total number of valence electrons. We use the group num- ber to determine the number of valence electrons for each of the atoms in the molecule.

Element Group Atoms Valence Electrons Total

C 4A (14) 1 C * 4 e- = 4 e-

O 6A (16) 2 O * 6 e- = 12 e-

Total valence electrons for CO2 = 16 e -

STEP 3 Attach each bonded atom to the central atom with a pair of electrons.

or O C OO OC

We use four valence electrons to attach the central C atom to two O atoms, which leaves 12 valence electrons.

STEP 4 Use the remaining electrons to complete octets, drawing multiple bonds if needed.

The 12 remaining electrons are placed as six lone pairs of electrons on the outside O atoms. However, this does not complete the octet of the C atom.

orC CO OO O

To obtain an octet, the C atom must share pairs of electrons from each of the O atoms. When two bonding pairs are drawn between two atoms, it is a double bond.

Lone pairs converted to bonding pairs

Double bonds Double bonds

orC O C

orC C

O O O

OO OO

CO2 Ball-and-stick model CO2 Space-filling model

INTERACTIVE VIDEO

Drawing Lewis Structures with Multiple Bonds

ENGAGE 10.4 How do you know when you need to draw one or more multiple bonds to complete a Lewis structure?

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276 CHAPTER 10 Bonding and Properties of Solids and Liquids

SELF TEST 10.4

Draw the Lewis structures for HCN, which has a triple bond. Draw one Lewis structure with electron dots and the other with lines for bonding pairs.

ANSWER

or CHH NC N

HCN Ball-and-stick model HCN Space-filling model B

ClCl

Cl

In BCl3, the central B atom is bonded to three Cl atoms.

S

F

F F

F F

F

In SF6, the central S atom is bonded to six F atoms.

Exceptions to the Octet Rule Although the octet rule is useful for bonding in many compounds, there are exceptions. We have already seen that a hydrogen (H2) molecule requires just two electrons or a single bond. Usually the nonmetals form octets. However, in BCl3, the B atom has only three valence electrons to share. Boron compounds typically have six valence electrons on the central B atoms and form just three bonds. Although we will generally see compounds of P, S, Cl, Br, and I with octets, they can form molecules in which they share more of their valence electrons. This expands their valence electrons to 10, 12, or even 14 electrons. For example, we have seen that the P atom in PCl3 has an octet, but in PCl5, the P atom has five bonds with 10 valence electrons. In H2S, the S atom has an octet, but in SF6, there are six bonds to sulfur with 12 valence electrons.

PRACTICE PROBLEMS

10.1 Lewis Structures for Molecules and Polyatomic Ions

10.1 Determine the total number of valence electrons for each of the following:

a. H2S b. I2 c. CCl4 d. OH -

10.2 Determine the total number of valence electrons for each of the following:

a. SBr2 b. NBr3 c. CH3OH d. NH4 +

10.3 When is it necessary to draw a multiple bond in a Lewis structure?

10.4 If the available number of valence electrons for a molecule or polyatomic ion does not complete all of the octets in a Lewis structure, what should you do?

10.5 Draw the Lewis structures for each of the following molecules or ions:

a. HF b. SF2 c. NBr3 d. BH4 -

10.6 Draw the Lewis structures for each of the following molecules or ions:

a. H2O b. CCl4 c. H3O + d. SiF4

10.7 Draw the Lewis structures for each of the following molecules or ions:

a. CO b. CN- c. H2CO (C is the central atom)

10.8 Draw the Lewis structures for each of the following molecules or ions:

a. HCCH (acetylene) b. CS2 c. NO +

10.2 Resonance Structures LEARNING GOAL Draw Lewis structures for molecules or polyatomic ions that have two or more resonance structures.

When a molecule or polyatomic ion contains multiple bonds, it may be possible to draw more than one Lewis structure. We can see how this happens when we draw the Lewis structure for ozone, O3, a component in the stratosphere that protects us from the ultraviolet rays of the Sun.

To draw the Lewis structure for O3, we determine the number of valence electrons for an O atom, and then the total number of valence electrons for O3. Because O is in Group 6A (16), it has six valence electrons. Therefore, the compound O3 would have a total of 18 valence electrons.

Element Group Atoms Valence Electrons Total

O 6A (16) 3 O * 6 e- = 18 e-

CORE CHEMISTRY SKILL Drawing Resonance Structures

PRACTICE PROBLEMS Try Practice Problems 10.1 to 10.8

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10.2 Resonance Structures 277

For the Lewis structure for O3, we place three O atoms in a row. Using four of the valence electrons, we draw a bonding pair between each of the O atoms on the ends and the central O atom. Two bonding pairs require four valence electrons.

O ¬ O ¬ O

The remaining valence electrons (14) are placed as lone pairs of electrons around the O atoms on both ends of the Lewis structure, and one lone pair goes on the central O atom.

OOO

To complete an octet for the central O atom, one lone pair of electrons from an end O atom needs to be shared. But which lone pair should be used? One possibility is to form a double bond between the central O atom and the O on the left, and the other possibility is to form a double bond between the central O atom and the O on the right.

O Oor OOOO

Thus it is possible to draw two or more Lewis structures for a molecule such as O3 or for a polyatomic ion. When this happens, all the Lewis structures are called resonance structures, and their relationship is shown by drawing a double-headed arrow between them.

O O Resonance structures

O O O O

Experiments show that the actual bonds in ozone are equivalent to a molecule with “one-and-a-half” bonds between the central O atom and each outside O atom. In an actual ozone molecule, the electrons are spread equally over all the O atoms. When we draw reso- nance structures for molecules or polyatomic ions, the true structure is really an average of those structures.

ENGAGE 10.5 Why are there two possible ways to form a double bond in ozone?

SAMPLE PROBLEM 10.5 Drawing Resonance Structures

TRY IT FIRST

Sulfur dioxide is produced by volcanic activity and the burning of sulfur-containing coal. Once in the atmosphere, the SO2 is converted to SO3, which combines with water, forming sulfuric acid, H2SO4, a component of acid rain. Draw two resonance structures for sulfur dioxide.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

SO2 resonance structures total valence electrons

STEP 1 Determine the arrangement of atoms. In SO2, the S atom is the central atom because there is only one S atom.

O S O

STEP 2 Determine the total number of valence electrons. We use the group number to determine the number of valence electrons for each of the atoms in the molecule.

Element Group Atoms Valence Electrons Total

S 6A (16) 1 S * 6 e- = 6 e-

O 6A (16) 2 O * 6 e- = 12 e-

Total valence electrons for SO2 = 18 e -

Representations of O3 molecule

Stratosphere

O3 molecule

Ozone, O3, is a component in the stratosphere that protects us from the ultraviolet rays of the Sun.

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278 CHAPTER 10 Bonding and Properties of Solids and Liquids

SO2 Ball-and-stick model

SO2 Space-filling model

TABLE 10.5 summarizes drawing Lewis structures for several molecules and ions.

TABLE 10.5 Using Valence Electrons to Draw Lewis Structures Molecule or Polyatomic Ion Total Valence Electrons

Form Single Bonds to Attach Atoms (electrons used) Electrons Remaining Completed Octets (or H:)

Cl2 2(7) = 14 Cl ¬ Cl (2 e-) 14 - 2 = 12 ClCl

HCl 1 + 7 = 8 H ¬ Cl (2 e-) 8 - 2 = 6 H Cl

H2O 2(1) + 6 = 8 H ¬ O ¬ H (4 e-) 8 - 4 = 4 H HO

PCl3 5 + 3(7) = 26 ClCl P

Cl

(6 e-) 26 - 6 = 20 P ClCl

Cl

ClO3 - 7 + 3(6) + 1 = 26 O Cl O

O -

(6 e-) 26 - 6 = 20

-

Cl OO

O

NO2 - 5 + 2(6) + 1 = 18 N O]-[O (4 e-) 18 - 4 = 14

-

-

OO

OO

N

N

STEP 3 Attach each bonded atom to the central atom with a pair of electrons.

O ¬ S ¬ O We use four electrons to form single bonds between the central S atom and the O atoms.

STEP 4 Place the remaining electrons using single or multiple bonds to complete octets. The remaining 14 electrons are drawn as lone pairs, which complete the octets for the O atoms but not the S atom.

S OO

To complete the octet for S, one lone pair of electrons from one of the O atoms is shared to form a double bond. One possibility is to form a double bond between the central S atom and the O on the left, and the other possibility is to form a double bond between the central S atom and the O on the right.

SSO O OO

SELF TEST 10.5

Draw three resonance structures for SO3.

ANSWER

S S S

O

OOOO

O O

O O

ENGAGE 10.6 Explain why SO2 has resonance structures but SCl2 does not.

PRACTICE PROBLEMS Try Practice Problems 10.9 to 10.12

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10.3 Shapes of Molecules and Polyatomic Ions (VSEPR Theory) 279

10.12 Draw two resonance structures for each of the following molecules or ions:

a. H2CNO2 -

b. N2O (N N O)

PRACTICE PROBLEMS

10.2 Resonance Structures

10.9 What is resonance?

10.10 When does a molecular compound have resonance?

10.11 Draw two resonance structures for each of the following molecules or ions:

a. ClNO2 (N is the central atom) b. OCN- (C is the central atom)

10.3 Shapes of Molecules and Polyatomic Ions (VSEPR Theory) LEARNING GOAL Predict the three-dimensional structure of a molecule or a polyatomic ion.

Using the Lewis structures, we can predict the three-dimensional shapes of many molecules and polyatomic ions. The shape is important in our understanding of how molecules interact with enzymes or certain antibiotics or produce our sense of taste and smell.

The three-dimensional shape of a molecule or polyatomic ion is determined by drawing its Lewis structure and identifying the number of electron groups (one or more electron pairs) around the central atom. We count lone pairs of electrons, single, double, and triple bonds as one electron group. In the valence shell electron-pair repulsion (VSEPR) theory, the electron groups are arranged as far apart as possible around the central atom to minimize the repulsion between the groups. Once we have counted the number of electron groups surrounding the central atom, we can determine its specific shape from the number of atoms bonded to the central atom.

Central Atoms with Two Electron Groups In the Lewis structure for CO2, there are two electron groups (two double bonds) attached to the central atom. According to VSEPR theory, minimal repulsion occurs when two electron groups are on opposite sides of the central C atom. This gives the CO2 molecule a linear electron-group geometry and a shape that is linear with a bond angle of 180°.

Linear electron-group

geometry

C OO

180°

Linear shape

Central Atoms with Three Electron Groups In the Lewis structure for formaldehyde, H2CO, the central atom C is attached to two H atoms by single bonds and to the O atom by a double bond. Minimal repulsion occurs when three electron groups are as far apart as possible around the central C atom, which gives 120° bond angles. This type of electron-group geometry is trigonal planar and gives a shape for H2CO also called trigonal planar. When the central atom has the same number of electron groups as bonded atoms, the shape and the electron-group geometry are the same.

H

120°

C

H

C HH

Trigonal planar electron-group

geometry

Lewis structure

Trigonal planar shape

O O

CORE CHEMISTRY SKILL Predicting Shape

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280 CHAPTER 10 Bonding and Properties of Solids and Liquids

In the Lewis structure for SO2, there are also three electron groups around the central S atom: a single bond to an O atom, a double bond to another O atom, and a lone pair of electrons. As in H2CO, three electron groups have minimal repulsion when they form a trigonal planar electron-group geometry. However, in SO2, one of the electron groups is a lone pair of electrons. Therefore, the shape of the SO2 molecule is determined by the two O atoms bonded to the central S atom, which gives the SO2 molecule a shape that is bent with a bond angle of 120°. When the central atom has more electron groups than bonded atoms, the shape and the electron-group geometry are different.

120°

Bent shapeLewis structure

Trigonal planar electron-group

geometry

S

S O OOO

Central Atom with Four Electron Groups In a molecule of methane, CH4, the central C atom is bonded to four H atoms. From the Lewis structure, you may think that CH4 is planar with 90° bond angles. However, the best geometry for minimal repulsion is tetrahedral, giving bond angles of 109°. When there are four atoms attached to four electron groups, the shape of the molecule is also tetrahedral.

C

H

H

H H

Tetrahedral electron-group

geometry

Lewis structure

Tetrahedral shape

Tetrahedral wedge–dash notation

H

H

H

H

C

109°

H

C HH

H

A way to represent the three-dimensional structure of methane is to use the wedge–dash notation. In this representation, the two bonds connecting carbon to hydrogen by solid lines are in the plane of the paper. The wedge represents a carbon-to-hydrogen bond coming out of the page toward us, whereas the dash represents a carbon-to-hydrogen bond going into the page away from us.

Now we can look at molecules that have four electron groups, of which one or more are lone pairs of electrons. Then the central atom is attached to only two or three atoms. For example, in the Lewis structure for ammonia, NH3, four electron groups have a tetrahedral electron-group geometry. However, in NH3, one of the electron groups is a lone pair of electrons. Therefore, the shape of NH3 is determined by the three H atoms bonded to the central N atom, which gives the NH3 molecule a shape that is trigonal pyramidal, with a bond angle of 109°. The wedge–dash notation can also represent this three-dimensional structure of ammonia with one N ¬ H bond in the plane, one N ¬ H bond coming toward us, and one N ¬ H bond going away from us.

N

H

H H

Lewis structure

Lone pair of electrons

Trigonal pyramidal shape

Trigonal pyramidal wedge–dash

notation

Tetrahedral electron-group

geometry

NH H

H H

N HH

In the Lewis structure for water, H2O, there are also four electron groups, which have minimal repulsion when the electron-group geometry is tetrahedral. However, in H2O, two

ENGAGE 10.7 If the four electron groups in a PH3 molecule have a tetrahedral geometry, why does a PH3 molecule have a trigonal pyramidal shape and not a tetrahedral shape?

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10.3 Shapes of Molecules and Polyatomic Ions (VSEPR Theory) 281

of the electron groups are lone pairs of electrons. Because the shape of H2O is determined by the two H atoms bonded to the central O atom, the H2O molecule has a bent shape with a bond angle of 109°. TABLE 10.6 gives the shapes of molecules with two, three, or four bonded atoms.

H H

OO

H

H

Two lone pairs of electrons

Lewis structure

Tetrahedral electron-group

geometry

Bent shape

Bent wedge–dash

notation

H

O H

PRACTICE PROBLEMS Try Practice Problems 10.13 and 10.14

TABLE 10.6 Molecular Shapes for a Central Atom with Two, Three, and Four Bonded Atoms

Electron Groups

Electron-Group Geometry

Bonded Atoms

Lone Pairs

Bond Angle* Molecular Shape Example

Three-Dimensional Models

Ball-and-Stick Space-Filling

2 Linear 2 0 180° Linear CO2

3 Trigonal planar 3 0 120° Trigonal planar H2CO

3 Trigonal planar 2 1 120° Bent SO2

4 Tetrahedral 4 0 109° Tetrahedral CH4

4 Tetrahedral 3 1 109° Trigonal pyramidal NH3

4 Tetrahedral 2 2 109° Bent H2O

*The bond angles in actual molecules may vary slightly.

SAMPLE PROBLEM 10.6 Shapes of Molecules

TRY IT FIRST

Use VSEPR theory to predict the shape of the molecule SiCl4.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

SiCl4 shape Lewis structure, electron groups, bonded atoms

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282 CHAPTER 10 Bonding and Properties of Solids and Liquids

SAMPLE PROBLEM 10.7 Predicting Shape of an Ion

TRY IT FIRST

Use VSEPR theory to predict the shape of the polyatomic ion NO3 -.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

NO3 - shape Lewis structure, electron groups,

bonded atoms

STEP 1 Draw the Lewis structure.

Element Group Atoms Valence Electrons Total

N 5A (15) 1 N * 5 e- = 5 e-

O 6A (16) 3 O * 6 e- = 18 e-

Ionic charge (negative) add 1 e- = 1 e-

Total valence electrons for NO3 - = 24 e-

The polyatomic ion NO3 - contains three electron groups (two single bonds between

the central N atom and O atoms, and one double bond between N and O). Note that the double bond can be drawn to any of the O atoms, which results in three resonance structures. However, we need just one of the structures to predict its shape.

N -

O

O

O

STEP 1 Draw the Lewis structure.

Element Group Atoms Valence Electrons Total

Si 4A (14) 1 Si * 4 e- = 4 e-

Cl 7A (17) 4 Cl * 7 e- = 28 e-

Total valence electrons for SiCl4 = 32 e -

Using 32 e-, we draw the bonds and lone pairs for the Lewis structure of SiCl4.

Si Cl

Cl ClCl

STEP 2 Arrange the electron groups around the central atom to minimize repulsion. In the Lewis structure of SiCl4, there are four electron groups around the central Si atom. To minimize repulsion, the electron-group geometry would be tetrahedral.

STEP 3 Use the atoms bonded to the central atom to determine the shape. Because the central Si atom has four bonded atoms and no lone pairs of electrons, the SiCl4 molecule has a tetrahedral shape.

SELF TEST 10.6

Use VSEPR theory to predict the shape of SCl2.

ANSWER

The central atom S has four electron groups: two bonded atoms and two lone pairs of electrons. The shape of SCl2 is bent, with a bond angle of 109°.

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10.4 Electronegativity and Bond Polarity 283

STEP 2 Arrange the electron groups around the central atom to minimize repulsion. In the Lewis structure of NO3

-, there are three electron groups around the central N atom. To minimize repulsion, the three electron groups have a trigonal planar geometry.

STEP 3 Use the atoms bonded to the central atom to determine the shape. Because NO3

- has three bonded atoms and no lone pairs, it has a trigonal planar shape.

SELF TEST 10.7

Use VSEPR theory to predict the shape of each of the following: a. ClO2

- b. PO2 +

ANSWER

a. With two bonded atoms and two lone pairs of electrons on the central Cl atom, the shape of ClO2

- is bent, with a bond angle of 109°. b. With two bonded atoms and no lone pairs on the central P atom, the shape of PO2

+ is linear.

PRACTICE PROBLEMS Try Practice Problems 10.15 to 10.22

10.16 Complete each of the following statements for a molecule of H2S:

a. There are _______ electron groups around the central S atom.

b. The electron-group geometry is _______. c. The number of atoms attached to the central S atom is

_______. d. The shape of the molecule is _______.

10.17 Compare the Lewis structures of CF4 and NF3. Why do these molecules have different shapes?

10.18 Compare the Lewis structures of CH4 and H2O. Why do these molecules have similar bond angles but different molecular shapes?

10.19 Use VSEPR theory to predict the shape of each of the following:

a. GaH3 b. OF2 c. HCN d. CCl4 10.20 Use VSEPR theory to predict the shape of each of the

following: a. CF4 b. NCl3 c. SeBr2 d. CS2

10.21 Draw the Lewis structure and predict the shape for each of the following:

a. AlH4 - b. SO4

2- c. NH4 + d. NO2

+

10.22 Draw the Lewis structure and predict the shape for each of the following:

a. NO2 - b. PO4

3- c. ClO4 - d. SF3

+

PRACTICE PROBLEMS

10.3 Shapes of Molecules and Polyatomic Ions (VSEPR Theory)

10.13 Choose the shape (1 to 6) that matches each of the following descriptions (a to c):

1. linear 2. bent (109°) 3. trigonal planar 4. bent (120°) 5. trigonal pyramidal 6. tetrahedral a. a molecule with a central atom that has four electron groups

and four bonded atoms b. a molecule with a central atom that has four electron groups

and three bonded atoms c. a molecule with a central atom that has three electron

groups and three bonded atoms

10.14 Choose the shape (1 to 6) that matches each of the following descriptions (a to c):

1. linear 2. bent (109°) 3. trigonal planar 4. bent (120°) 5. trigonal pyramidal 6. tetrahedral a. a molecule with a central atom that has four electron groups

and two bonded atoms b. a molecule with a central atom that has two electron groups

and two bonded atoms c. a molecule with a central atom that has three electron

groups and two bonded atoms

10.15 Complete each of the following statements for a molecule of SeO3:

a. There are _______ electron groups around the central Se atom.

b. The electron-group geometry is _______. c. The number of atoms attached to the central Se atom is

_______. d. The shape of the molecule is _______.

10.4 Electronegativity and Bond Polarity LEARNING GOAL Use electronegativity to determine the polarity of a bond.

We can learn more about the chemistry of compounds by looking at how bonding electrons are shared between atoms. The bonding electrons are shared equally in a bond between identical nonmetal atoms. However, when a bond is between atoms of different elements, the electron pairs are usually shared unequally. Then the shared pairs of electrons are attracted to one atom in the bond more than the other.

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284 CHAPTER 10 Bonding and Properties of Solids and Liquids

The electronegativity of an atom is its ability to attract the shared electrons in a chemical bond. Nonmetals have higher electronegativities than do metals, because non- metals have a greater attraction for electrons than metals. On the electronegativity scale, the nonmetal f luorine, located in the upper right corner of the periodic table, was assigned a value of 4.0, and the electronegativities for all other elements were determined relative to the attraction of f luorine for shared electrons. The metal cesium, which has the lowest electronegativity (0.7), is located in the lower left corner of the periodic table. The electronegativities for the representative elements are shown in FIGURE 10.2. Note that there are no electronegativity values for the noble gases because they do not typi- cally form bonds. The electronegativity values for transition elements are also low, but we have not included them in our discussion.

PRACTICE PROBLEMS Try Practice Problems 10.23 and 10.24

Polarity of Bonds The difference in the electronegativity values of two atoms can be used to predict the type of chemical bond, ionic or covalent, that forms. For the H ¬ H bond, the electro- negativity difference is zero (2.1 - 2.1 = 0), which means the bonding electrons are shared equally. We illustrate this by drawing a symmetrical electron cloud around the H atoms. A bond between atoms with identical or very similar electronegativity val- ues is a nonpolar covalent bond. However, when covalent bonds are between atoms with different electronegativity values, the electrons are shared unequally; the bond is a polar covalent bond. The electron cloud for a polar covalent bond is unsymmetrical. For the H ¬ Cl bond, there is an electronegativity difference of 0.9 (3.0 - 2.1 = 0.9), which means that the H ¬ Cl bond is polar covalent (see FIGURE 10.3). When finding the elec- tronegativity difference, the smaller electronegativity is always subtracted from the larger; thus, the difference is always a positive number.

The polarity of a bond depends on the difference in the electronegativity values of its atoms. In a polar covalent bond, the shared electrons are attracted to the more electronega- tive atom, which makes it partially negative, because of the negatively charged electrons around that atom. At the other end of the bond, the atom with the lower electronegativity becomes partially positive because of the lack of electrons at that atom.

A bond becomes more polar as the electronegativity difference increases. A polar covalent bond that has a separation of charges is called a dipole. The positive and nega- tive ends of the dipole are indicated by the lowercase Greek letter delta with a positive or negative sign, d+ and d-. Sometimes we use an arrow that points from the positive charge to the negative charge

Cl FN OC O d

+ d

- d

+ d

-d+ d-

to indicate the dipole.

Cl FN OC O d

+ d

- d

+ d

-d+ d-

Dipoles occur in polar covalent bonds containing N, O, or F.

FIGURE 10.2 The electronegativity values of representative elements in Group 1A (1) to Group 7A (17), which indicate the ability of atoms to attract shared electrons, increase going across a period from left to right and decrease going down a group.

E le

ct ro

ne ga

ti vi

ty D

ec re

as es

Electronegativity Increases

1 Group

1A

2 Group

2A

H 2.1

Li 1.0

Be 1.5

Na 0.9

Mg 1.2

K 0.8

Ca 1.0

B 2.0

C 2.5

N 3.0

O 3.5

F 4.0

Al 1.5

Si 1.8

P 2.1

S 2.5

Cl 3.0

Ga 1.6

Ge 1.8

As 2.0

Se 2.4

Br 2.8

Rb 0.8

Sr 1.0

In 1.7

Sn 1.8

Sb 1.9

Te 2.1

I 2.5

Cs 0.7

Ba 0.9

Tl 1.8

Pb 1.9

Bi 1.9

Po 2.0

At 2.1

13 Group

3A

14 Group

4A

15 Group

5A

16 Group

6A

17 Group

7A

18 Group

8A

ENGAGE 10.8 From their position on the periodic table, how you can determine that chlorine has a higher electronegativity than iodine?

CORE CHEMISTRY SKILL Using Electronegativity

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10.4 Electronegativity and Bond Polarity 285

FIGURE 10.3 In the nonpolar covalent bond of H2, electrons are shared equally. In the polar covalent bond of HCl, electrons are shared unequally.

H H

HH

Cl

Cl

H

H

Equal sharing of electrons in a nonpolar covalent bond

Unequal sharing of electrons in a polar covalent bond

H H Hd +

Cld -

Variations in Bonding The variations in bonding are continuous; there is no definite point at which one type of bond stops and the next starts. When the electronegativity difference is between 0.0 and 0.4, the electrons are considered to be shared equally in a nonpolar covalent bond. For example, the C ¬ C bond (2.5 - 2.5 = 0.0) and the C ¬ H bond (2.5 - 2.1 = 0.4) are classified as nonpolar covalent bonds.

As the electronegativity difference increases, the shared electrons are attracted more strongly to the more electronegative atom, which increases the polarity of the bond. When the electronegativity difference is from 0.5 to 1.8, the bond is a polar covalent bond. For example, the O ¬ H bond (3.5 - 2.1 = 1.4) is classified as a polar covalent bond (see TABLE 10.7).

ENGAGE 10.9 Use electronegativity differences to explain why a Si ¬ S bond is polar covalent and a Si ¬ P bond is non- polar covalent.

Formula Bond Electronegativity Difference* Type of Bond

H2 H ¬ H 2.1 - 2.1 = 0.0 Nonpolar covalent BrCl Br ¬ Cl 3.0 - 2.8 = 0.2 Nonpolar covalent

HBr Hd + ¬ Brd

- 2.8 - 2.1 = 0.7 Polar covalent

HCl Hd + ¬ Cld

- 3.0 - 2.1 = 0.9 Polar covalent

NaCl Na+Cl- 3.0 - 0.9 = 2.1 Ionic

MgO Mg2+O2- 3.5 - 1.2 = 2.3 Ionic *Values are taken from Figure 10.2.

TABLE 10.8 Predicting Bond Type from Electronegativity Differences

Electronegativity Difference 0.0 to 0.4 0.5 to 1.8 1.9 to 3.3

Bond Type Nonpolar covalent Polar covalent Ionic

Electron Bonding Electrons shared equally

Electrons shared unequally

Electrons transferred

+ -d+ d-

TABLE 10.7 Electronegativity Differences and Types of Bonds

When the electronegativity difference is greater than 1.8, electrons are transferred from one atom to another, which results in an ionic bond. For example, the electronegativity difference for the ionic compound NaCl is 3.0 - 0.9 = 2.1. Thus, for large differences in electronegativity, we would predict an ionic bond (see TABLE 10.8).

PRACTICE PROBLEMS Try Practice Problems 10.25 and 10.26

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286 CHAPTER 10 Bonding and Properties of Solids and Liquids

PRACTICE PROBLEMS

10.4 Electronegativity and Bond Polarity

10.26 Which electronegativity difference (a, b, or c) would you expect for a polar covalent bond?

a. from 0.0 to 0.4 b. from 0.5 to 1.8 c. from 1.9 to 3.3

10.27 Using the periodic table, arrange the atoms in each of the following sets in order of increasing electronegativity:

a. Li, Na, K b. Na, Cl, P c. Se, Ca, O

10.28 Using the periodic table, arrange the atoms in each of the following sets in order of increasing electronegativity:

a. Cl, F, Br b. B, O, N c. Mg, F, S

10.23 Describe the trend in electronegativity as increases or decreases for each of the following:

a. from B to F b. from Mg to Ba c. from F to I

10.24 Describe the trend in electronegativity as increases or decreases for each of the following:

a. from Al to Cl b. from Br to K c. from Li to Cs

10.25 Which electronegativity difference (a, b, or c) would you expect for a nonpolar covalent bond?

a. from 0.0 to 0.4 b. from 0.5 to 1.8 c. from 1.9 to 3.3

SAMPLE PROBLEM 10.8 Bond Polarity

TRY IT FIRST

Using electronegativity values, classify the bond between each of the following pairs of atoms as nonpolar covalent, polar covalent, or ionic and label any polar covalent bond with d+ and d- and show the direction of the dipole:

a. K and O b. As and Cl c. N and N d. P and Br

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

pairs of atoms type of bond electronegativity values

For each pair of atoms, we obtain the electronegativity values and calculate the difference in electronegativity.

Atoms Electronegativity Difference Type of Bond Dipole

a. K and O 3.5 - 0.8 = 2.7 Ionic

b. As and Cl 3.0 - 2.0 = 1.0 Polar covalent As d+

¬ Cl d-

c. N and N 3.0 - 3.0 = 0.0 Nonpolar covalent

d. P and Br 2.8 - 2.1 = 0.7 Polar covalent P d+

¬ Br d-

SELF TEST 10.8

Using electronegativity values, classify the bond between each of the following pairs of atoms as nonpolar covalent, polar covalent, or ionic and label any polar covalent bond with d+ and d- and show the direction of the dipole:

a. P and Cl b. Br and Br c. Na and O d. O and I

ANSWER

a. polar covalent (0.9) P d+

¬ Cl d-

b. nonpolar covalent (0.0)

c. ionic (2.6) d. polar covalent (1.0) O d-

¬ I d+

H Cl

A single dipole does not cancel; HCl is polar.

H Cl

A single dipole does not cancel; HCl is polar.

H Cl

A single dipole does not cancel; HCl is polar.

HCl

A single dipole does not cancel; HCl is polar.

PRACTICE PROBLEMS Try Practice Problems 10.27 to 10.32

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10.5 Polarity of Molecules 287

10.29 Predict whether the bond between each of the following pairs of atoms is nonpolar covalent, polar covalent, or ionic:

a. Si and Br b. Li and F c. Br and F d. I and I e. N and P f. C and P

10.30 Predict whether the bond between each of the following pairs of atoms is nonpolar covalent, polar covalent, or ionic:

a. Si and O b. K and Cl c. S and F d. P and Br e. Li and O f. N and S

10.31 For the bond between each of the following pairs of atoms, indicate the positive end with d+ and the negative end with d-. Draw an arrow to show the dipole for each.

a. N and F b. Si and Br c. C and O d. P and Br e. N and P

10.32 For the bond between each of the following pairs of atoms, indicate the positive end with d+ and the negative end with d-. Draw an arrow to show the dipole for each.

a. P and Cl b. Se and F c. Br and F d. N and H e. B and Cl

10.5 Polarity of Molecules LEARNING GOAL Use the three-dimensional structure of a molecule to classify it as polar or nonpolar.

We have seen that covalent bonds in molecules can be polar or nonpolar. Now we will look at how the bonds in a molecule and its shape determine whether that molecule is classified as polar or nonpolar.

Nonpolar Molecules In a nonpolar molecule, all the bonds are nonpolar or the polar bonds cancel each other out. Molecules such as H2, Cl2, and CH4 are nonpolar because they contain only nonpolar covalent bonds. A nonpolar molecule also occurs when polar bonds (dipoles) cancel each other because they are in a symmetrical arrangement. For example, CO2, a linear molecule, contains two equal polar covalent bonds whose dipoles point in opposite directions. As a result, the dipoles cancel, which makes a CO2 molecule nonpolar.

Another example of a nonpolar molecule is the CCl4 molecule, which has four polar bonds symmetrically arranged around the central C atom. Each of the C ¬ Cl bonds has the same polarity, but because they have a tetrahedral arrangement, their opposing dipoles cancel. As a result, a molecule of CCl4 is nonpolar.

Polar Molecules In a polar molecule, one end of the molecule is more negatively charged than the other end. Polarity in a molecule occurs when the dipoles from the individual polar bonds do not cancel each other. For example, HCl is a polar molecule because it has one covalent bond that is polar.

In molecules with two or more electron groups, the shape, such as bent or trigonal pyramidal, determines whether the dipoles cancel. For example, we have seen that H2O has a bent shape. Thus, a water molecule is polar because the individual dipoles do not cancel.

H H

More positive end of molecule

More negative end of molecule

The dipoles do not cancel; H2O is polar.

O

The NH3 molecule has a tetrahedral electron-group geometry with three bonded atoms, which gives it a trigonal pyramidal shape. Thus, the NH3 molecule is polar because the individual N ¬ H dipoles do not cancel.

N

H H

H

More positive end of molecule

More negative end of molecule

The dipoles do not cancel; NH3 is polar.

CORE CHEMISTRY SKILL Identifying Polarity of Molecules

O C O

The two C — O dipoles cancel; CO2 is nonpolar.

Cl

C Cl

The four C — Cl dipoles cancel; CCl4 is nonpolar.

Cl

Cl

H Cl

A single dipole does not cancel; HCl is polar.

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288 CHAPTER 10 Bonding and Properties of Solids and Liquids

SAMPLE PROBLEM 10.9 Polarity of Molecules

TRY IT FIRST

Determine whether a molecule of OF2 is polar or nonpolar.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

OF2 polarity Lewis structure, bond polarity

STEP 1 Determine if the bonds are polar covalent or nonpolar covalent. From Figure 10.2, F and O have an electronegativity difference of 0.5 (4.0 - 3.5 = 0.5), which makes each of the O ¬ F bonds polar covalent.

STEP 2 If the bonds are polar covalent, draw the Lewis structure, determine the molecular shape, and determine if the dipoles cancel. The Lewis structure for OF2 has four electron groups and two bonded atoms. The molecule has a bent shape in which the dipoles of the O ¬ F bonds do not cancel. The OF2 molecule would be polar.

F F

OO FF

The dipoles do not cancel; OF2 is polar.

SELF TEST 10.9

Determine if each of the following would be polar or nonpolar:

a. PCl3 b. Cl2

ANSWER

a. polar b. nonpolar PRACTICE PROBLEMS

Try Practice Problems 10.33 to 10.38

10.33 Why is F2 a nonpolar molecule, but HF is a polar molecule?

10.34 Why is CCl4 a nonpolar molecule, but PCl3 is a polar molecule?

10.35 Identify each of the following molecules as polar or nonpolar: a. CS2 b. NF3 c. CHF3 d. SO3

PRACTICE PROBLEMS

10.5 Polarity of Molecules

10.36 Identify each of the following molecules as polar or nonpolar: a. SeF2 b. PBr3 c. SiF4 d. SO2

10.37 The molecule CO2 is nonpolar, but CO is a polar molecule. Explain.

10.38 The molecules CH4 and CH3Cl both have tetrahedral shapes. Why is CH4 nonpolar whereas CH3Cl is polar?

C

F

H H H

The dipole does not cancel; CH3F is polar.

In the molecule CH3F, the C ¬ F bond is polar covalent, but the three C ¬ H bonds are nonpolar covalent. Because there is only one dipole in CH3F, which does not cancel, CH3F is a polar molecule.

10.6 Intermolecular Forces Between Atoms or Molecules LEARNING GOAL Describe the intermolecular forces between ions, polar covalent molecules, and nonpolar covalent molecules.

The forces that keep the atoms or ions together in a compound are called intramolecular forces. In ionic compounds, we have seen how large differences in electronegativity produce positive and negative ions that are attracted to each other. Ionic bonds are the strongest of the forces in

CORE CHEMISTRY SKILL Identifying Intermolecular Forces

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10.6 Intermolecular Forces Between Atoms or Molecules 289

H

Dipole–dipole attraction

HCl Cl

d +

d - d+ d-compounds. In molecules, we have seen that covalent bonds form by atoms sharing electrons.

Ionic bonds and covalent bonds are involved in the shapes and polarity of compounds. In solids and liquids, there are also intermolecular forces such as dipole–dipole

attractions, hydrogen bonding, and dispersion forces between covalent molecules, which hold them close together. They are much weaker than the intramolecular forces, but impor- tant in the physical properties such as melting and boiling point.

Dipole–Dipole Attractions For polar molecules, intermolecular forces called dipole–dipole attractions occur between the positive end of one molecule and the negative end of another. For a polar molecule with a dipole such as HCl, the partially positive H atom of one HCl molecule attracts the partially negative Cl atom in another HCl molecule.

Hydrogen Bonds Polar molecules containing hydrogen atoms bonded to highly electronegative atoms of nitrogen, oxygen, or fluorine form especially strong dipole–dipole attractions. This type of intermolecular force, called a hydrogen bond, occurs between the partially positive hydrogen atom in one molecule and the partially negative nitrogen, oxygen, or fluorine atom in another molecule. Hydrogen bonds are the strongest type of intermolecular force between polar covalent molecules. They are a major factor in the formation and structure of biological molecules such as proteins and DNA.

Dispersion Forces Very weak attractions called dispersion forces are the only intermolecular forces that occur between nonpolar molecules. Usually, the electrons in a nonpolar covalent molecule are distributed symmetrically. However, the movement of the electrons may place more of them in one part of the molecule than another, which forms a temporary dipole. These momen- tary dipoles align the molecules so that the positive end of one molecule is attracted to the negative end of another molecule. Although dispersion forces are very weak, they make it possible for nonpolar molecules to form liquids and solids.

H H

H H

H H

N N

Hydrogen bond

d +

d -

d +

d -

H

H

H H

H

O O

F H F

Hydrogen bond

Hydrogen bond

d +

d -

d +

d -

d +

d -

d +

d -

Nucleus

Electrons Symmetrical distribution

Unsymmetrical distribution

Temporary dipoles

d +

d -

d +

d -

d +

d -

Dispersion forces

Nonpolar covalent molecules have weak attractions when they form temporary dipoles.

TABLE 10.9 compares the intramolecular and intermolecular types of bonding.

Intermolecular Forces and Melting Points The melting point of a substance is related to the strength of the intermolecular forces between its particles. A compound with weak intermolecular forces, such as dispersion forces, has a low melting point because only a small amount of energy is needed to sepa- rate the molecules and form a liquid. A compound with dipole–dipole attractions requires more energy to break the intermolecular forces between the molecules. A compound with hydrogen bonds requires even more energy to overcome the intermolecular forces that exist between its molecules. TABLE 10.10 compares the melting points of some substances.

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290 CHAPTER 10 Bonding and Properties of Solids and Liquids

Type of Force Particle Arrangement Example Strength

Between Atoms or Ions (Intramolecular)

H

Y X

+ +-

+- -

Y X

XX

X

XX XX

H X

(temporary dipoles)

d + d

- d

+ d

-

d +

d -

d +

d -

d +

d -

d +

d -

H

Y X

+ +-

+- -

Y X

XX

X

XX XX

H X

(temporary dipoles)

d + d

- d

+ d

-

d +

d -

d +

d -

d +

d -

d +

d -

Strong

Ionic bonds Na+ Cl-

Weak

Covalent bonds (X = nonmetal) Cl ¬ Cl

Between Molecules (Intermolecular)

Hydrogen bonds (X = N, O, or F) Hd

+ ¬ Fd

- g Hd

+ ¬ Fd

-

Dipole–dipole attractions (X and Y = nonmetals) Hd

+ ¬ Cld

- g Hd

+ ¬ Cld

-

Dispersion forces (temporary shift of electrons in nonpolar bonds) Fd

+ ¬ Fd

- g Fd

+ ¬ Fd

-

TABLE 10.10 Melting Points of Selected Substances

Substance Melting Point (°C)

Hydrogen Bonds

H2O 0

NH3 - 78 Dipole–Dipole Attractions

HI - 51 HBr - 89 HCl - 115 Dispersion Forces

Br2 - 7 Cl2 - 101 F2 - 220

Size, Mass, and Melting and Boiling Points As the size and mass of similar types of molecular compounds increase, there are more electrons that produce stronger temporary dipoles. As the molar mass of similar compounds increases, the dispersion forces also increase due to the increase in the number of electrons. In general, larger nonpolar molecules with increased molar masses also have higher melting and boiling points.

SAMPLE PROBLEM 10.10 Intermolecular Forces Between Particles

TRY IT FIRST

Indicate the major type of intermolecular forces—dipole–dipole attractions, hydrogen bonds, or dispersion forces—expected of each of the following:

a. HF b. Br2 c. PCl3

SOLUTION

a. HF is a polar molecule that interacts with other HF molecules by hydrogen bonding. b. Br2 is nonpolar; only dispersion forces provide temporary intermolecular forces. c. The polarity of PCl3 molecules provides dipole–dipole attractions.

SELF TEST 10.10

Indicate the major type of intermolecular forces in each of the following:

a. H2S b. H2O

ANSWER

a. dipole–dipole attractions b. hydrogen bonds PRACTICE PROBLEMS

Try Practice Problems 10.39 to 10.44

ENGAGE 10.10 Why does GeH4 have a higher boiling point than CH4?

TABLE 10.9 Comparison of Bonding Intermolecular Forces

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10.7 Changes of State 291

PRACTICE PROBLEMS

10.6 Intermolecular Forces Between Atoms or Molecules 10.39 Identify the major type of intermolecular forces between the

particles of each of the following: a. BrF b. H2O2 c. NF3 d. Cl2

10.40 Identify the major type of intermolecular forces between the particles of each of the following:

a. HCl b. Br2 c. PBr3 d. NH3 10.41 Identify the strongest intermolecular forces between the

particles of each of the following: a. CH3OH b. CO c. CF4 d. CH3CH3

10.42 Identify the strongest intermolecular forces between the particles of each of the following:

a. O2 b. SiH4 c. CH3Cl d. (CH3)2NH

10.43 Identify the substance in each of the following pairs that would have the higher boiling point, and explain your choice:

a. HF or HBr b. HF or SF2 c. Br2 or PBr3 d. CH4 or CH3OH

10.44 Identify the substance in each of the following pairs that would have the higher boiling point, and explain your choice:

a. H2 or HCl b. H2O or H2Se c. NH3 or PH3 d. F2 or HF

10.7 Changes of State LEARNING GOAL Describe the changes of state between solids, liquids, and gases; calculate the energy released or absorbed.

The states and properties of gases, liquids, and solids depend on the types of forces between their particles. Matter undergoes a change of state when it is converted from one state to another (see FIGURE 10.4).

REVIEW Interpreting Graphs (1.4)

Using the Heat Equation (3.5)

ENGAGE 10.11 Is heat absorbed or released when liquid water freezes?

FIGURE 10.4 Changes of state include melting and freezing, vaporization and condensation, sublimation and deposition.

Melting

Freezing

Heat absorbed

Heat released

Condensation

Vaporization

D ep

os iti

on

Su bl

im

ati on

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292 CHAPTER 10 Bonding and Properties of Solids and Liquids

Melting and Freezing When heat is added to a solid, the particles move faster. At a temperature called the melting point (mp), the particles of a solid gain sufficient energy to overcome the inter- molecular forces that hold them together. The particles in the solid separate and move about in random patterns. The substance is melting, changing from a solid to a liquid.

If the temperature of a liquid is lowered, the reverse process takes place. Kinetic energy is lost, the particles slow down, and intermolecular forces pull the particles close together. The substance is freezing. A liquid changes to a solid at the freezing point (fp), which is the same temperature as its melting point. Every substance has its own freezing (melting) point: Solid water (ice) melts at 0 °C when heat is added, and liquid water freezes at 0 °C when heat is removed. Gold melts at 1064 °C when heat is added, and freezes at 1064 °C when heat is removed.

During a change of state, the temperature of a substance remains constant. Suppose we have a glass containing ice and water. The ice melts when heat is added at 0 °C, forming liquid. When heat is removed at 0 °C, the liquid water freezes, forming solid. The process of melting requires heat; the process of freezing releases heat. Melting and freezing are reversible at 0 °C.

Heat of Fusion During melting, the heat of fusion is the energy that must be added to convert exactly 1 g of solid to liquid at the melting point. For example, 334 J of heat is needed to melt exactly 1 g of ice at its melting point (0 °C).

H2O(s) + 334 J/g h H2O(l) Endothermic

Heat of Fusion for Water 334 J

1 g H2O and

1 g H2O

334 J

The heat of fusion is also the quantity of heat that must be removed to freeze exactly 1 g of water at its freezing point (0 °C).

Water is sometimes sprayed in fruit orchards during subfreezing weather. If the air temperature drops to 0 °C, the water begins to freeze. Because heat is released as the water molecules form solid ice, the air warms and protects the fruit.

H2O(l) h H2O(s) + 334 J/g Exothermic

The heat of fusion can be used as a conversion factor in calculations. For example, to determine the heat needed to melt a sample of ice, the mass of ice, in grams, is multiplied by the heat of fusion.

Calculating Heat to Melt (or Freeze) Water

Heat = mass * heat of fusion

J = g * 334 J

g

Sublimation and Deposition In a process called sublimation, the particles on the surface of a solid change directly to a gas with no temperature change and without going through the liquid state. In the reverse process called deposition, gas particles change directly to a solid. For example, dry ice, which is solid carbon dioxide, sublimes at - 78 °C. It is called “dry” because it does not form a liquid as it warms. In extremely cold areas, snow does not melt but sublimes directly to water vapor.

When frozen foods are left in the freezer for a long time, so much water sublimes that foods, especially meats, become dry and shrunken, a condition called freezer burn. Deposi- tion occurs in a freezer when water vapor forms ice crystals on the surface of freezer bags and frozen food.

Freeze-dried foods prepared by sublimation are convenient for long-term storage and for camping and hiking. A food that has been frozen is placed in a vacuum chamber where it dries as the ice sublimes. The dried food retains all of its nutritional value and needs only

CORE CHEMISTRY SKILL Calculating Heat for Change

of State

LiquidSolid

Melting

Freezing

+ Heat

- Heat

Melting and freezing are reversible processes.

Sublimation

Deposition

- Heat

Gas+ HeatSolid

Sublimation and deposition are reversible processes.

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10.7 Changes of State 293

water to be edible. A food that is freeze-dried does not need refrigeration because bacteria cannot grow without moisture.

Evaporation, Boiling, and Condensation Water in a mud puddle disappears, unwrapped food dries out, and clothes hung on a line dry. Evaporation is taking place as water molecules with sufficient energy escape from the liquid surface and enter the gas phase (see FIGURE 10.5). The loss of the “hot” water molecules removes heat, which cools the remaining liquid water. As heat is added, more and more water molecules gain sufficient energy to evaporate. At the boiling point (bp), the molecules within a liquid have enough energy to overcome their intermolecular forces and become a gas. We observe the boiling of a liquid such as water as gas bubbles form throughout the liquid, rise to the surface, and escape.

Water vapor will change to solid on contact with a cold surface.

When heat is removed, a reverse process takes place. In condensation, water vapor is converted to liquid as the water molecules lose kinetic energy and slow down. Condensation occurs at the same temperature as boiling but differs because heat is removed. You may have noticed that condensation occurs when you take a hot shower and the water vapor forms water droplets on a mirror. Because a substance loses heat as it condenses, its surroundings become warmer. That is why, when a rainstorm is approaching, you may notice a warming of the air as gaseous water molecules condense to rain.

Heat of Vaporization and Condensation The heat of vaporization is the energy that must be added to convert exactly 1 g of liquid to gas at its boiling point. For water, 2260 J is needed to convert 1 g of water to vapor at 100 °C.

H2O(l) + 2260 J/g h H2O(g) Endothermic

This same amount of heat is released when 1 g of water vapor (gas) changes to liquid at 100 °C.

H2O(g) h H2O(l) + 2260 J/g Exothermic

Therefore, 2260 J/g is also the heat of condensation of water.

FIGURE 10.5 Evaporation and boiling

Evaporation occurs at the surface of a liquid.

GasLiquid + Heat

- Heat

Vaporization

Condensation

Vaporization and condensation are reversible processes.

50 ºC

100 ºC: the boiling

point of water

Boiling occurs as bubbles of gas form throughout the liquid.

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294 CHAPTER 10 Bonding and Properties of Solids and Liquids

To determine the heat needed to boil a sample of water, the mass, in grams, is multiplied by the heat of vaporization. Because the temperature remains constant during a change of state (melting, freezing, boiling, or condensing), there is no temperature change used in the calculation, as shown in Sample Problem 10.11.

Heat of Vaporization for Water

2260 J 1 g H2O

and 1 g H2O

2260 J

Just as substances have different melting and boiling points, they also have different heats of fusion and vaporization. As seen in FIGURE 10.6, the heats of vaporization are larger than the heats of fusion.

FIGURE 10.6 For any substance, the heat of vaporization is greater than the heat of fusion.

Propane C3H8

336

18

Benzene C6H6

395

128

Acetic acid C2H4O2

390

192

Ethanol C2H6O

841

109

Ammonia NH3

1380

351

Water H2O

2260

334

Sodium chloride NaCl

518

13 000

Nonpolar covalent Polar covalent Ionic

Heat of vaporization (J/g)

Heat of fusion (J/g)

SAMPLE PROBLEM 10.11 Calculating Heat for Changes of State

TRY IT FIRST

In a sauna, 122 g of water is converted to steam at 100 °C. How many kilojoules are needed?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

122 g of water at 100 °C

kilojoules to convert to steam at 100 °C

heat of vaporization, 2260 J/g

STEP 2 Write a plan to convert the given quantity to the needed quantity.

grams of water joules kilojoules Heat of vaporization

Metric factor

STEP 3 Write the heat conversion factor and any metric factor.

2260 J and

1 g of H2O(l g) = 2260 J

1 g H2O

1 g H2O

2260 J

1000 J and

1 kJ = 1000 J

1 kJ

1 kJ

1000 J

ENGAGE 10.12 Why is temperature not used in calculations involving changes of state?

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10.7 Changes of State 295

STEP 4 Set up the problem and calculate the needed quantity.

122 g H2OHeat 276 kJ* == 2260 J

1 g H2O

1 kJ

1000 J *

Three SFs Exact Three SFsExact

Three SFs Exact

SELF TEST 10.11

a. When steam from a pan of boiling water reaches a cool window, it condenses. How much heat, in kilojoules, is released when 25.0 g of steam condenses at 100 °C?

b. Ice bags are used by sports trainers to treat muscle injuries. If 260. g of ice is placed in an ice bag, how much heat, in kilojoules, will be released when all the ice melts at 0 °C?

ANSWER

a. 56.5 kJ b. 86.8 kJ An ice bag is used to treat a sports injury.

Calculating Heat to Condense (or Boil) Water

Heat = mass * heat of vaporization

J = g * 2260 J

g

Heating and Cooling Curves All the changes of state during the heating or cooling of a substance can be illustrated visu- ally. On a heating curve or cooling curve, the temperature is shown on the vertical axis, and the loss or gain of heat is shown on the horizontal axis.

Steps on a Heating Curve The first diagonal line indicates that the temperature of the solid increases as heat is added. When the melting temperature is reached, a horizontal line, or plateau, indicates that the solid is melting. As melting takes place, the solid is changing to liquid without any change in temperature (see FIGURE 10.7).

PRACTICE PROBLEMS Try Practice Problems 10.45 to 10.48

Melting

Boiling

140

100

120

20

40

60

80

0

-20

-40

Gas

Liquid

Solid

Heat Added

T em

pe ra

tu re

( °C

)

Boiling point

Melting point

FIGURE 10.7 A heating curve diagrams the temperature increases and changes of state as heat is added.

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296 CHAPTER 10 Bonding and Properties of Solids and Liquids

When all the particles are in the liquid state, adding more heat will increase the tem- perature of the liquid. This increase is shown as a diagonal line from the melting point temperature to the boiling point temperature. Once the liquid reaches its boiling point, a horizontal line indicates that the liquid changes to gas at constant temperature. Because the heat of vaporization is greater than the heat of fusion, the horizontal line at the boiling point is longer than the line at the melting point. When all the liquid becomes gas, adding more heat increases the temperature of the gas.

Steps on a Cooling Curve A cooling curve is a diagram in which the temperature decreases as heat is removed. Ini- tially, a diagonal line is drawn to show that heat is removed from a gas until it begins to condense. At the condensation point, a horizontal line indicates a change of state as gas con- denses to form a liquid. When all the gas has changed into liquid, further cooling lowers the temperature. The decrease in temperature is shown as a diagonal line from the condensation point to the freezing point, where another horizontal line indicates that liquid is changing to solid. When all the substance is frozen, the removal of more heat decreases the temperature of the solid below its freezing point, which is shown as a diagonal line.

Combining Energy Calculations Up to now, we have calculated one step in a heating or cooling curve. However, a problem may require a combination of steps that include a temperature change as well as a change of state. The heat is calculated for each step separately, and the results are added together to give the total energy, as seen in Sample Problem 10.12.

140

120 100

80 60

40

20 0

-20

T em

pe ra

tu re

( °C

)

Water

ice Freezing

Water

Water + steam Condensation

Steam

Ice

+

Heat Removed

A cooling curve for water illustrates the temperature decreases and changes of state as heat is removed.

SAMPLE PROBLEM 10.12 Combining Heat Calculations

TRY IT FIRST

Charles has increased his activity by doing more exercise. After a session of using weights, he has a sore arm. An ice bag is filled with 125 g of ice at 0 °C. How much heat, in kilojoules, is absorbed to melt the ice and raise the temperature of the water to body temperature, 37.0 °C?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

125 g of ice at 0 °C

total kilojoules to melt ice at 0 °C and to raise temperature of water to 37.0 °C

combine heat from change of state (heat of fusion) and temperature change (specific heat of water)

STEP 2 Write a plan to convert the given quantity to the needed quantity.

Total heat = kilojoules needed to melt the ice at 0 °C and heat the water from 0 °C to 37.0 °C

STEP 3 Write the heat conversion factors and any metric factor.

=

334 J and

1 g of H2O (s l) = 334 J

1 g H2O

1 g H2O

334 J and

g °C

g °C

4.184 J

g °C 4.184 J

SHwater

4.184 J 1000 J and

1 kJ = 1000 J

1 kJ

1 kJ

1000 J

=

334 J and

1 g of H2O (s l) = 334 J

1 g H2O

1 g H2O

334 J and

g °C

g °C

4.184 J

g °C 4.184 J

SHwater

4.184 J 1000 J and

1 kJ = 1000 J

1 kJ

1 kJ

1000 J

=

334 J and

1 g of H2O (s l) = 334 J

1 g H2O

1 g H2O

334 J and

g °C

g °C

4.184 J

g °C 4.184 J

SHwater

4.184 J 1000 J and

1 kJ = 1000 J

1 kJ

1 kJ

1000 J

STEP 4 Set up the problem and calculate the needed quantity.

∆T = 37.0 °C - 0 °C = 37.0 °C

M10_TIMB8119_06_SE_C10.indd 296 11/27/18 11:59 AM

10.7 Changes of State 297

Heat needed to change ice (solid) to water (liquid) at 0 °C:

125 g ice * = 41.8 kJ

ExactThree SFs Three SFs

Three SFs

1 g ice 334 J

Exact

Exact

* 1 kJ

1000 J Heat =

Heat needed to warm water (liquid) from 0 °C to water (liquid) at 37.0 °C:

37.0 °C 125 g * *= = 19.4 kJ g °C

4.184 J

Three SFs Three SFs Three SFs

Exact

Exact

Exact

Exact

* 1 kJ

1000 J Heat

Calculate the total heat:

Melting ice at 0 °C 41.8 kJ

Heating water (0 °C to 37.0 °C) 19.4 kJ

Total heat needed 61.2 kJ

SELF TEST 10.12

a. How many kilojoules are released when 75.0 g of steam at 100 °C condenses, cools to 0 °C, and freezes at 0 °C? (Hint: The solution will require three energy calculations.)

b. How many kilocalories are needed to convert 18 g of ice at 0 °C to steam at 100 °C? (Hint: The solution will require three energy calculations.)

ANSWER

a. 226 kJ b. 13 kcal PRACTICE PROBLEMS

Try Practice Problems 10.49 to 10.52

Chemistry Link to Health Steam Burns

Hot water at 100 °C will cause burns and damage to the skin. However, getting steam on the skin is even more dangerous. If 25 g of hot water at 100 °C falls on a person’s skin, the temperature of the water will drop to body temperature, 37 °C. The heat released during cooling can cause severe burns. The amount of heat can be calculated from the mass, the temperature change, 100 °C - 37 °C = 63 °C, and the specific heat of water, 4.184 J/g °C.

25 g * 63 °C * 4.184 J

g °C = 6600 J of heat released

when water cools

The condensation of the same quantity of steam to liquid at 100 °C releases much more heat. The heat released when steam condenses can be calculated using the heat of vaporization, 2260 J/g.

25 g * 2260 J

1 g = 57 000 J released when water (gas)

condenses to water liquid) at 100 °C

The total heat released is calculated by combining the heat from the condensation at 100 °C and the heat from cooling of the steam from 100 °C to 37 °C (body temperature). We can see that most of the heat is from the condensation of steam.

Condensation (100 °C) = 57 000 J Cooling (100 °C to 37 °C) = 6 600 J Heat released = 64 000 J ( rounded off to

thousands place)

The amount of heat released from steam is almost ten times greater than the heat from the same amount of hot water. This large amount of heat released on the skin is what causes the extensive damage from steam burns.

When 1 g of steam condenses, 2260 J or 540 cal is released.

HEAT

H2O liquid

H2O steam

M10_TIMB8119_06_SE_C10.indd 297 11/27/18 11:59 AM

298 CHAPTER 10 Bonding and Properties of Solids and Liquids

PRACTICE PROBLEMS 10.7 Changes of State 10.45 Using Figure 10.6, calculate the heat change needed for each of

the following at the melting/freezing point: a. joules to melt 65.0 g of ice b. joules to melt 17.0 g of benzene c. kilojoules to freeze 225 g of benzene d. kilojoules to freeze 0.0500 kg of water

10.46 Using Figure 10.6, calculate the heat change needed for each of the following at the melting/freezing point:

a. joules to freeze 35.2 g of acetic acid b. joules to freeze 275 g of water c. kilojoules to melt 145 g of ammonia d. kilojoules to melt 5.00 kg of ice

10.47 Using Figure 10.6, calculate the heat change needed for each of the following at the boiling/condensation point:

a. joules to vaporize 10.0 g of water b. kilojoules to vaporize 50.0 g of ethanol c. joules to condense 8.00 g of acetic acid d. kilojoules to condense 0.175 kg of ammonia

10.48 Using Figure 10.6, calculate the heat change needed for each of the following at the boiling/condensation point:

a. joules to condense 10.0 g of steam b. kilojoules to condense 76.0 g of acetic acid c. joules to vaporize 44.0 g of ammonia d. kilojoules to vaporize 5.0 kg of water

10.49 Using Figure 10.6 and the specific heat of water, 4.184 J/g °C, calculate the total amount of heat for each of the following:

a. joules released when 2.00 g of water cools from 22.0 °C to 0 °C and freezes

b. joules needed to melt 50.0 g of ice at 0 °C and to warm the liquid to 65.0 °C

c. kilojoules released when 15.0 g of steam condenses at 100 °C and the liquid cools to 0 °C

d. kilojoules needed to melt 24.0 g of ice at 0 °C, warm the liquid to 100 °C, and change it to steam at 100 °C

10.50 Using Figure 10.6 and the specific heat of water, 4.184 J/g °C, calculate the total amount of heat for each of the following:

a. joules to condense 125 g of steam at 100 °C and to cool the liquid to 150 °C

b. joules needed to melt a 525-g ice sculpture at 0 °C and to warm the liquid to 15.0 °C

c. kilojoules released when 85.0 g of steam condenses at 100 °C, the liquid cools to 0 °C, and freezes

d. joules to warm 55.0 mL of water (density = 1.00 g/mL) from 10.0 °C to 100 °C and vaporize it at 100 °C

Applications 10.51 An ice bag containing 275 g of ice at 0 °C was used to treat

sore muscles. When the bag was removed, the ice had melted and the liquid water had a temperature of 24.0 °C. How many kilojoules of heat were absorbed?

10.52 A patient arrives in the emergency room with a burn caused by steam. Calculate the heat, in kilojoules, that is released when 18.0 g of steam at 100 °C hits the skin, condenses, and cools to body temperature of 37.0 °C.

UPDATE Histologist Stains Tissue with Dye

When Lisa receives the sample of tissue from Bill's surgery, she places it in liquid nitrogen, which has a heat of vaporization of 198 J/g at the boiling point, which is - 196 °C. When the tissue is frozen, Lisa places the tissue on a microtome, cuts thin slices, and mounts each

on a glass slide. One of the stains that Lisa uses to allow the pathologist to see any abnormalities is Eosin, which has a formula, C20H6Br4Na2O5. In preparing to stain the tissues, Lisa uses solutions of lithium carbonate and sodium bicarbonate.

The pathologist examines the tissue sample and reports back to the surgeon that no abnormal cells are present in the margin tissue. No further tissue removal is necessary.

Applications

10.53 At the boiling point of nitrogen, N2, how many joules are needed to vaporize 15.8 g of N2?

10.54 In the preparation of liquid nitrogen, how many kilojoules must be removed from 112 g of N2 gas at its boiling point to form liquid N2?

10.55 Using the electronegativity values in Figure 10.2, for each of the following bonds found in Eosin calculate the elec- tronegativity difference and predict if the bond is nonpolar covalent, polar covalent, or ionic:

a. C and C b. C and H c. C and O

10.56 Using the electronegativity values in Figure 10.2, for each of the following bonds found in Eosin, calculate the elec- tronegativity difference and predict if the bond is nonpolar covalent, polar covalent, or ionic:

a. O and H b. Na and O c. C and Br

10.57 a. Draw three resonance structures for carbonate (CO3 2-).

b. Determine the shape of the carbonate ion.

10.58 a. Draw two resonance structures for bicarbonate (HCO3 -)

with a central C atom and an H atom attached to one of the O atoms.

b. Determine the shape of the bicarbonate ion.

M10_TIMB8119_06_SE_C10.indd 298 11/27/18 11:59 AM

CONCEPT MAP

contain can be are

undergo

add/lose heat are drawn as

have

between

that are

contain

between

can be

can be

have

use

to determine

BONDING AND PROPERTIES OF COMPOUNDS

Ionic Compounds

Intermolecular Forces

Changes of State

Covalent Bonds

Lewis Structures

Molecular Compounds

States of Matter

Dipole–Dipole Attractions,

Hydrogen Bonds, or Dispersion Forces

Ionic Bonds

Molecules Nonmetals

Nonpolar

Single or Multiple Bonds

Solids, Liquids, Gases

Polar

Electronegativity

VSEPR Theory

Heating/ Cooling Curves

Vaporization/ Condensation

Melting/ Freezing

Polarity

Shape

CHAPTER REVIEW

10.1 Lewis Structures for Molecules and Polyatomic Ions LEARNING GOAL Draw the Lewis structures for molecular compounds and polyatomic ions with single and multiple bonds. • The Lewis symbol gives the symbol of the element

with the valence electrons shown as dots around the symbol. • The total number of valence electrons is determined for all the

atoms in the molecule or polyatomic ion. • Any negative charge is added to the total valence electrons,

whereas any positive charge is subtracted. • In the Lewis structure, a bonding pair of electrons is placed

between the central atom and each of the attached atoms. • Any remaining valence electrons are used as lone pairs to

complete the octets of the surrounding atoms and then the central atom.

• If octets are not completed, one or more lone pairs of electrons are placed as bonding pairs forming double or triple bonds.

10.2 Resonance Structures LEARNING GOAL Draw Lewis structures for molecules or polyatomic ions that have two or more resonance structures. • Resonance structures can be drawn when there are two or more

equivalent Lewis structures for a molecule or ion with multiple bonds.

10.3 Shapes of Molecules and Polyatomic Ions (VSEPR Theory) LEARNING GOAL Predict the three-dimensional structure of a molecule or a polyatomic ion. • The shape of a molecule is determined from the

Lewis structure, the electron-group geometry, and the number of bonded atoms.

• The electron-group geometry around a central atom with two electron groups is linear; with three electron groups, the geometry is trigonal planar; and with four electron groups, the geometry is tetrahedral.

• When all the electron groups are bonded to atoms, the shape has the same name as the electron arrangement.

• A central atom with three electron groups and two bonded atoms has a bent shape, 120°.

• A central atom with four electron groups and three bonded atoms has a trigonal pyramidal shape.

• A central atom with four electron groups and two bonded atoms has a bent shape, 109°.

10.4 Electronegativity and Bond Polarity LEARNING GOAL Use electronegativity to determine the polarity of a bond. • Electronegativity is the ability of an

atom to attract the electrons it shares with another atom. In general, the

Cl Cl

ClP

Tetrahedral shape

H Cl

Cl

H

Chapter Review 299

M10_TIMB8119_06_SE_C10.indd 299 11/27/18 11:59 AM

300 CHAPTER 10 Bonding and Properties of Solids and Liquids

H

Dipole–dipole attraction

HCl Cl

d +

d -

d +

d -

LiquidSolid

Melting

Freezing

+ Heat

- Heat

electronegativities of metals are low, whereas nonmetals have high electronegativities.

• In a nonpolar covalent bond, atoms share electrons equally. • In a polar covalent bond, the electrons are unequally shared

because they are attracted to the more electronegative atom. • In ionic bonds, atoms have large differences in electronegativites. • The atom in a polar bond with the lower electronegativity is par-

tially positive (d+), and the atom with the higher electronegativity is partially negative (d-).

10.5 Polarity of Molecules LEARNING GOAL Use the three-dimensional structure of a molecule to classify it as polar or nonpolar. • Nonpolar molecules contain nonpolar covalent

bonds or have an arrangement of bonded atoms that causes the dipoles to cancel.

• In polar molecules, the dipoles do not cancel.

10.6 Intermolecular Forces Between Atoms or Molecules LEARNING GOAL Describe the intermolecular forces between ions, polar covalent molecules, and nonpolar covalent molecules. • In ionic solids, oppositely charged ions are

held in a rigid structure by ionic bonds. • Intermolecular forces called dipole–dipole attractions and

hydrogen bonds hold the solid and liquid states of polar molecular compounds together.

• Nonpolar compounds form solids and liquids by weak intermolecular forces between temporary dipoles called dispersion forces.

10.7 Changes of State LEARNING GOAL Describe the changes of state between solids, liquids, and gases; calculate the energy released or absorbed. • Melting occurs when the

particles in a solid absorb enough energy to break apart and form a liquid.

• The amount of energy required to convert exactly 1 g of solid to liquid is called the heat of fusion.

• For water, 334 J is needed to melt 1 g of ice or must be removed to freeze 1 g of water.

• Sublimation is a process whereby a solid changes directly to a gas.

• Evaporation occurs when particles in a liquid state absorb enough energy to break apart and form gaseous particles.

• Boiling is the vaporization of liquid at its boiling point. The heat of vaporization is the amount of heat needed to convert exactly 1 g of liquid to vapor.

• For water, 2260 J is needed to vaporize 1 g of water or must be removed to condense 1 g of steam.

• A heating or cooling curve illustrates the changes in temperature and state as heat is added to or removed from a substance. Pla- teaus on the graph indicate changes of state.

• The total heat absorbed or removed from a substance undergoing temperature changes and changes of state is the sum of energy calculations for changes of state and changes in temperature.

bent The shape of a molecule with two bonded atoms and one lone pair or two lone pairs.

boiling The formation of bubbles of gas throughout a liquid. boiling point (bp) The temperature at which a liquid changes to gas

(boils) and gas changes to liquid (condenses). change of state The transformation of one state of matter to another,

for example, solid to liquid, liquid to solid, liquid to gas. condensation The change of state from a gas to a liquid. cooling curve A diagram that illustrates temperature changes and

changes of state for a substance as heat is removed. deposition The change of a gas directly to a solid; the reverse of

sublimation. dipole The separation of positive and negative charge in a polar bond

indicated by an arrow that is drawn from the more positive atom to the more negative atom.

dipole–dipole attractions Intermolecular forces between oppositely charged ends of polar molecules.

dispersion forces Weak dipole bonding that results from a momentary polarization of nonpolar molecules.

double bond A sharing of two pairs of electrons by two atoms. electronegativity The relative ability of an element to attract

electrons in a bond. evaporation The formation of a gas (vapor) by the escape of

high-energy molecules from the surface of a liquid. freezing A change of state from liquid to solid. freezing point (fp) The temperature at which a liquid changes to a

solid (freezes) and a solid changes to a liquid (melts).

heat of fusion The energy required to melt exactly 1 g of a substance at its melting point. For water, 334 J is needed to melt 1 g of ice; 334 J is released when 1 g of water freezes.

heat of vaporization The energy required to vaporize 1 g of a substance at its boiling point. For water, 2260 J is needed to vaporize exactly 1 g of water; 1 g of steam gives off 2260 J when it condenses.

heating curve A diagram that illustrates the temperature changes and changes of state of a substance as it is heated.

hydrogen bond The attraction between a partially positive H atom and a strongly electronegative atom of N, O, or F.

Lewis structure A structure drawn in which the valence electrons of all the atoms are arranged to give octets except two electrons for hydrogen.

Lewis symbol The representation of an atom that shows the valence electrons as dots placed around the symbol of an element.

linear The shape of a molecule that has two bonded atoms and no lone pairs.

melting The change of state from a solid to a liquid. melting point (mp) The temperature at which a solid becomes a

liquid (melts). It is the same temperature as the freezing point. nonpolar covalent bond A covalent bond in which the electrons are

shared equally between atoms. nonpolar molecule A molecule that has only nonpolar bonds or in

which the bond dipoles cancel. polar covalent bond A covalent bond in which the electrons are

shared unequally between atoms.

KEY TERMS

M10_TIMB8119_06_SE_C10.indd 300 11/27/18 11:59 AM

Core Chemistry Skills 301

polar molecule A molecule containing bond dipoles that do not cancel. polarity A measure of the unequal sharing of electrons, indicated by

the difference in electronegativities. resonance structures Two or more Lewis structures that can be

drawn for a molecule or polyatomic ion by placing a multiple bond between different atoms.

sublimation The change of state in which a solid is transformed directly to a gas without forming a liquid first.

tetrahedral The shape of a molecule with four bonded atoms.

trigonal planar The shape of a molecule with three bonded atoms and no lone pairs.

trigonal pyramidal The shape of a molecule that has three bonded atoms and one lone pair.

triple bond A sharing of three pairs of electrons by two atoms. valence shell electron-pair repulsion (VSEPR) theory A theory that

predicts the shape of a molecule by moving the electron groups on a central atom as far apart as possible to minimize the mutual repulsion of the electrons.

The chapter section containing each Core Chemistry Skill is shown in parentheses at the end of each heading.

Drawing Lewis Symbols (10.1) • The valence electrons are the electrons in the outermost energy level. • The number of valence electrons is the same as the group number

for the representative elements. • A Lewis symbol represents the number of valence electrons shown

as dots placed around the symbol for the element.

Example: Give the group number and number of valence electrons, and draw the Lewis symbol for each of the following:

a. Rb b. Se c. Xe

Answer: a. Group 1A (1), one valence electron,

b. Group 6A (16), six valence electrons,

c. Group 8A (18), eight valence electrons,

Drawing Lewis Structures (10.1) • The Lewis structure for a molecule or polyatomic ion shows the

sequence of atoms, the bonding pairs of electrons shared between atoms, and the nonbonding or lone pairs of electrons.

• Double or triple bonds result when a second or third electron pair is shared between the same atoms to complete octets.

Example: Draw the Lewis structures for CS2.

Answer: The central atom in the atom arrangement is C.

S C S

Determine the total number of valence electrons. 2 S * 6 e- = 12 e-

1 C * 4 e- = 4 e-

Total = 16 e-

Attach each bonded atom to the central atom using a pair of electrons. Two bonding pairs use four electrons.

S SC

Place the 12 remaining electrons as lone pairs around the S atoms.

CS S

To complete the octet for C, a lone pair of electrons from each of the S atoms is shared with C, which forms two double bonds.

S SSC CS or

Drawing Resonance Structures (10.2) • When a molecule or polyatomic ion contains multiple bonds, it

may be possible to draw multiple resonance structures.

Rb

Se

Xe

CORE CHEMISTRY SKILLS Example: Draw two resonance structures for NO2

-. Answer: The central atom in the atom arrangement is N.

O N O

Determine the total number of valence electrons.

1 N * 5 e- = 5 e-

2 O * 6 e- = 12 e-

negative charge = 1 e-

Total = 18 e-

Use electron pairs to attach each bonded atom to the central atom. Two bonding pairs use four electrons.

N

O

O O

ON

- Place the 14 remaining electrons as lone pairs around the O and N atoms.

N

O

O O

ON

-

To complete the octet for N, a lone pair from one O atom is shared with N, which forms one double bond. Because there are two O atoms, there are two resonance structures.

O ONO ON --

Predicting Shape (10.3) • The three-dimensional shape of a molecule or polyatomic ion

is determined by drawing a Lewis structure and identifying the number of electron groups (one or more electron pairs) around the central atom and the number of bonded atoms.

• In the valence shell electron-pair repulsion (VSEPR) theory, the electron groups are arranged as far apart as possible around a central atom to minimize the repulsion.

• A central atom with two electron groups bonded to two atoms is linear. A central atom with three electron groups bonded to three atoms is trigonal planar, and to two atoms is bent (120°). A central atom with four electron groups bonded to four atoms is tetrahedral, to three atoms is trigonal pyramidal, and to two atoms is bent (109°).

Example: Predict the shape of NO2 -.

Answer: Using the resonance structures we drew for NO2 -, we count

three electron groups around the central N atom: one double bond, one single bond, and a lone pair of electrons.

or O ONO ON --

The electron-group geometry is trigonal planar, but with the central N atom bonded to two O atoms, the shape is bent, with a bond angle of 120°.

O O

N -

M10_TIMB8119_06_SE_C10.indd 301 11/27/18 11:59 AM

302 CHAPTER 10 Bonding and Properties of Solids and Liquids

Using Electronegativity (10.4) • The electronegativity values indicate the ability of atoms to attract

shared electrons. • Electronegativity values increase going across a period from left to

right, and decrease going down a group. • A nonpolar covalent bond occurs between atoms with identical or

very similar electronegativity values such that the electronegativity difference is 0.0 to 0.4.

• A polar covalent bond occurs between atoms with differing electronegativity values such that the electronegativity difference is 0.5 to 1.8.

• An ionic bond occurs when the difference in electronegativity for two atoms is greater than 1.8.

Example: Use electronegativity values to classify each of the following bonds as nonpolar covalent, polar covalent, or ionic:

a. Sr and Cl b. C and S c. O and Br

Answer: a. An electronegativity difference of 2.0 (Cl 3.0 - Sr 1.0) makes this an ionic bond.

b. An electronegativity difference of 0.0 (C 2.5 - S 2.5) makes this a nonpolar covalent bond.

c. An electronegativity difference of 0.7 (O 3.5 - Br 2.8) makes this a polar covalent bond.

Identifying Polarity of Molecules (10.5) • A molecule is nonpolar if all of its bonds are nonpolar or it has

polar bonds that cancel. CCl4 is a nonpolar molecule that has four polar bonds that cancel.

Cl

C ClCl

Cl

• A molecule is polar if it contains polar bonds that do not cancel. H2O is a polar molecule that has polar bonds that do not cancel.

O

H H

Example: Predict whether AsCl3 is polar or nonpolar.

Answer: From its Lewis structure, we see that AsCl3 has four electron groups with three bonded atoms.

AsCl Cl Cl

The shape of a molecule of AsCl3 would be trigonal pyrami- dal with three polar bonds (As and Cl = 3.0 - 2.0 = 1.0) that do not cancel. Thus, it is a polar molecule.

As

ClCl Cl

Identifying Intermolecular Forces (10.6) • Dipole–dipole attractions occur between the dipoles in polar

compounds because the positively charged end of one molecule is attracted to the negatively charged end of another molecule.

H

Dipole–dipole attraction

HCl Cl

d +

d -

d +

d -

• Strong dipole–dipole attractions called hydrogen bonds occur in compounds in which H is bonded to N, O, or F. The partially positive H atom in one molecule has a strong attraction to the partially negative N, O, or F in another molecule.

• Dispersion forces are very weak intermolecular forces between nonpolar molecules that occur when temporary dipoles form as electrons are unsymmetrically distributed.

Example: Identify the strongest type of intermolecular forces in each of the following:

a. HF b. F2 c. NF3

Answer: a. HF molecules, which are polar with H bonded to F, have hydrogen bonding.

b. Nonpolar F2 molecules have only dispersion forces. c. NF3 molecules, which are polar, have dipole–dipole

attractions.

• Substances with hydrogen bonds have higher melting and boil- ing points than compounds with only dipole–dipole attractions. Substances with only dispersion forces would typically have the lowest melting and boiling points, which increase as molar mass increases.

Example: Identify the compound with the highest boiling point in each of the following:

a. HI, HBr, HF b. F2, Cl2, I2

Answer: a. HBr and HI have dipole–dipole attractions, but HF has hydrogen bonding, which gives HF the highest boiling point.

b. Because F2, Cl2, and I2 have only dispersion forces, I2 with the greatest molar mass would have the highest boiling point.

Calculating Heat for Change of State (10.7)

• At the melting/freezing point, the heat of fusion is absorbed/ released to convert 1 g of a solid to a liquid or 1 g of liquid to a solid.

• For example, 334 J of heat is needed to melt (freeze) exactly 1 g of ice at its melting (freezing) point (0 °C).

• At the boiling/condensation point, the heat of vaporization is absorbed/released to convert exactly 1 g of liquid to gas or 1 g of gas to liquid.

• For example, 2260 J of heat is needed to boil (condense) exactly 1 g of water/steam at its boiling (condensation) point, 100 °C.

Example: How many kilojoules are released when 45.8 g of steam (water) condenses at its boiling (condensation) point?

Answer: 45.8 g steam * 2260 J

1 g steam *

1 kJ 1000 J

= 104 kJ

Three SFs Exact Exact Three SFs

Three SFs Exact

M10_TIMB8119_06_SE_C10.indd 302 11/27/18 11:59 AM

Understanding the Concepts 303

UNDERSTANDING THE CONCEPTS

The chapter sections to review are shown in parentheses at the end of each problem.

10.59 State the number of valence electrons, bonding pairs, and lone pairs in each of the following Lewis structures: (10.1)

a. H H H Br Br Br

NOTE: 2nd pass

No changes were made. Base alignment issue must have been as a result of a PDF artifact

b. H H H Br Br Br

NOTE: 2nd pass

No changes were made. Base alignment issue must have been as a result of a PDF artifact

c. H H H Br Br Br

NOTE: 2nd pass

No changes were made. Base alignment issue must have been as a result of a PDF artifact

10.60 State the number of valence electrons, bonding pairs, and lone pairs in each of the following Lewis structures: (10.1)

a. Br Br H

H H H

H OO N b. Br Br H

H H H

H OO N c. Br Br H

H H H

H OO N

10.61 Match each of the Lewis structures (a to c) with the correct diagram (1 to 3) of its shape, and name the shape; indicate if each molecule is polar or nonpolar. Assume X and Y are non- metals and all bonds are polar covalent. (10.1, 10.3, 10.5)

1 2 3

a. X X

X

Y b.

X

XY c. X X

X

X

Y

10.62 Match each of the formulas (a to c) with the correct diagram (1 to 3) of its shape, and name the shape; indicate if each molecule is polar or nonpolar. (10.1, 10.3, 10.4, 10.5)

1 2 3

a. PBr3 c. OF2b. SiCl4

10.63 Consider the following bonds: Ca and O, C and O, K and O, O and O, N and O. (10.4)

a. Which bonds are polar covalent? b. Which bonds are nonpolar covalent? c. Which bonds are ionic? d. Arrange the covalent bonds in order of decreasing polarity.

10.64 Consider the following bonds: F and Cl, Cl and Cl, Cs and Cl, O and Cl, Ca and Cl. (10.4)

a. Which bonds are polar covalent? b. Which bonds are nonpolar covalent? c. Which bonds are ionic? d. Arrange the covalent bonds in order of decreasing polarity.

10.65 Identify the major intermolecular forces between each of the following atoms or molecules: (10.6)

a. PH3 b. NO2 c. CH3NH2 d. Ar

A spray is used to numb a sports injury.

a. How does perspiration during heavy exercise cool the body?

b. Why do towels dry more quickly on a hot summer day than on a cold winter day?

c. Why do wet clothes stay wet in a plastic bag?

10.68 Use your knowledge of changes of state to explain the following: (10.7)

a. Why is a spray that evaporates quickly, such as ethyl chloride, used to numb a sports injury during a game?

b. Why does water in a wide, flat, shallow dish evaporate more quickly than the same amount of water in a tall, narrow vase?

c. Why does a sandwich on a plate dry out faster than a sandwich in plastic wrap?

10.69 Draw a heating curve for a sample of ice that is heated from - 20 °C to 150 °C. Indicate the segment of the graph that corresponds to each of the following: (10.7)

a. solid b. melting c. liquid d. boiling e. gas

10.70 Draw a cooling curve for a sample of steam that cools from 110 °C to - 10 °C. Indicate the segment of the graph that cor- responds to each of the following: (10.7)

a. solid b. freezing c. liquid d. condensing e. gas

Perspiration forms on the skin during heavy exercise.

10.66 Identify the major intermolecular forces between each of the following atoms or molecules: (10.6)

a. He b. HBr c. SnH4 d. CH3CH2CH2OH

10.67 Use your knowledge of changes of state to explain the following: (10.7)

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304 CHAPTER 10 Bonding and Properties of Solids and Liquids

120

100

80

60

40

20

0

-20 A

B

C

1

D E

2 3 4 5

T °C

Heat Added

10.71 The following is a heating curve for chloroform, a solvent for fats, oils, and waxes: (10.5, 10.7)

ADDITIONAL PRACTICE PROBLEMS

10.73 Determine the total number of valence electrons in each of the following: (10.1)

a. HNO2 b. CH3CHO c. PH4 + d. SO3

2-

10.74 Determine the total number of valence electrons in each of the following: (10.1)

a. COCl2 b. N2O c. BrO2 - d. SeCl2

10.75 Draw the Lewis structures for each of the following: (10.1) a. BF4

- b. Cl2O c. H2NOH (N is the central atom) d. H2CCCl2

10.76 Draw the Lewis structures for each of the following: (10.1) a. H3COCH3 (the atoms are in the order C O C) b. HNO2 (the atoms are in the order H O N O)

c. IO3 - d. BrO -

10.77 Draw resonance structures for each of the following: (10.2) a. N3

- b. NO2 + c. HCO2

-

10.78 Draw resonance structures for each of the following: (10.2) a. NO3

- b. CO3 2- c. SCN-

10.79 Use the periodic table to arrange the following atoms in order of increasing electronegativity: (10.4)

a. I, F, Cl b. Li, K, S, Cl c. Mg, Sr, Ba, Be

10.80 Use the periodic table to arrange the following atoms in order of increasing electronegativity: (10.4)

a. Cl, Br, Se b. Na, Cs, O, S c. O, F, B, Li

10.81 Select the more polar bond in each of the following pairs: (10.4) a. C and N or C and O b. N and F or N and Br c. Br and Cl or S and Cl d. Br and Cl or Br and I e. N and F or N and O

10.82 Select the more polar bond in each of the following pairs: (10.4) a. C and C or C and O b. P and Cl or P and Br c. Si and S or Si and Cl d. F and Cl or F and Br e. P and O or P and S

10.83 Show the dipole arrow for each of the following bonds: (10.4) a. Si and Cl b. C and N c. F and Cl d. C and F e. N and O

10.84 Show the dipole arrow for each of the following bonds: (10.4) a. P and O b. N and F c. O and Cl d. S and Cl e. P and F

10.85 Calculate the electronegativity difference and classify each of the following bonds as nonpolar covalent, polar covalent, or ionic: (10.4)

a. Si and Cl b. C and C c. Na and Cl d. C and H e. F and F

10.86 Calculate the electronegativity difference and classify each of the following bonds as nonpolar covalent, polar covalent, or ionic: (10.4)

a. C and N b. Cl and Cl c. K and Br d. H and H e. N and F

10.87 For each of the following, draw the Lewis structure and determine the shape: (10.1, 10.3)

a. NF3 b. SiBr4 c. CSe2 d. SO2

10.88 For each of the following, draw the Lewis structure and determine the shape: (10.1, 10.3)

a. PCl4 + b. O2

2-

c. COCl2 (C is the central atom) d. HCCH

10.89 For each of the following, draw the Lewis structure and determine the shape: (10.1, 10.3)

a. BrO2 - b. H2O c. CBr4 d. PO3

3-

10.90 For each of the following, draw the Lewis structure and determine the shape: (10.1, 10.3)

a. PH3 b. NO3 -

c. HCN d. SO3 2-

10.91 Predict the shape and polarity of each of the following molecules, which have polar covalent bonds: (10.3, 10.5)

a. A central atom with three identical bonded atoms and one lone pair.

b. A central atom with two bonded atoms and two lone pairs.

100

60

20

-20

-60

-100 A

B

C

D E

T °C

Heat Added

a. What is the approximate melting point of chloroform? b. What is the approximate boiling point of chloroform?

c. On the heating curve, identify the segments A, B, C, D, and E as solid, liquid, gas, melting, or boiling.

d. At the following temperatures, is chloroform a solid, liquid, or gas? - 80 °C; - 40 °C; 25 °C; 80 °C

10.72 Associate the contents of the beakers (1 to 5) with segments (A to E) on the heating curve for water. (10.7)

M10_TIMB8119_06_SE_C10.indd 304 11/27/18 11:59 AM

Challenge Problems 305

10.92 Predict the shape and polarity of each of the following molecules, which have polar covalent bonds: (10.3, 10.5)

a. A central atom with four identical bonded atoms and no lone pairs.

b. A central atom with four bonded atoms that are not identical and no lone pairs.

10.93 Classify each of the following molecules as polar or nonpolar: (10.3, 10.4, 10.5)

a. HBr b. SiO2 c. NCl3 d. CH3Cl e. NI3 f. H2O

10.94 Classify each of the following molecules as polar or nonpolar: (10.3, 10.4, 10.5)

a. GeH4 b. I2 c. CF3Cl d. PCl3 e. BCl3 f. SCl2

10.95 Indicate the major type of intermolecular forces—(1) ionic bonds, (2) dipole–dipole attractions, (3) hydrogen bonds, (4) dispersion forces—that occurs between particles of the following: (10.6)

a. NF3 b. ClF c. Br2 d. Cs2O e. C4H10 f. CH3OH

10.96 Indicate the major type of intermolecular forces—(1) ionic bonds, (2) dipole–dipole attractions, (3) hydrogen bonds, (4) dispersion forces—that occurs between particles of the following: (10.6)

a. CHCl3 b. H2O c. LiCl d. OBr2 e. HBr f. IBr

10.97 When it rains or snows, the air temperature seems warmer. Explain. (10.7)

10.98 Water is sprayed on the ground of an orchard when temperatures are near freezing to keep the fruit from freezing. Explain. (10.7)

10.99 Using Figure 10.6, calculate the grams of ice that will melt at 0 °C if 1540 J is absorbed. (10.7)

10.100 Using Figure 10.6, calculate the grams of ethanol that will vaporize at its boiling point if 4620 J is absorbed. (10.7)

10.101 Using Figure 10.6, calculate the grams of acetic acid that will freeze at its freezing point if 5.25 kJ is removed. (10.7)

10.102 Using Figure 10.6, calculate the grams of benzene that will condense at its boiling point if 8.46 kJ is removed. (10.7)

The following problems are related to the topics in this chapter. However, they do not all follow the chapter order, and they require you to combine concepts and skills from several sections. These problems will help you increase your critical thinking skills and prepare for your next exam.

10.103 Complete the Lewis structure for each of the following: (10.1)

a. H N

H

C H

O

b.

Cl C

H

H

C N

c. H N N H

d.

Cl C

O

O C H

H

H

10.104 Identify the errors in each of the following Lewis structures, and draw the correct formula: (10.1)

a. O ClCl

b. H C H

O

c.

H

H

N HO

10.105 Predict the shape of each of the following molecules or ions: (10.3)

a. NH2Cl (N is the central atom) b. PH4 +

c. SCN-

10.106 Classify each of the following molecules as polar or nonpolar: (10.3, 10.4, 10.5)

a. N2 b. TeO2 c. NH2Cl (N is the central atom)

10.107 The melting point of dibromomethane is - 53 °C, and its boiling point is 97 °C. Sketch a heating curve for dibromomethane from - 100 °C to 120 °C. (10.7)

a. What is the state of dibromomethane at - 75 °C? b. What happens on the curve at - 53 °C? c. What is the state of dibromomethane at - 18 °C? d. What is the state of dibromomethane at 110 °C? e. At what temperature will both solid and liquid be

present?

10.108 The melting point of benzene is 5.5 °C, and its boiling point is 80.1 °C. Draw a heating curve for benzene from 0 °C to 100 °C. (10.7)

a. What is the state of benzene at 15 °C? b. What happens on the curve at 5.5 °C? c. What is the state of benzene at 63 °C? d. What is the state of benzene at 98 °C? e. At what temperature will both liquid and gas be present?

10.109 A 45.0-g piece of ice at 0.0 °C is added to a sample of water at 8.0 °C. All of the ice melts, and the temperature of the water decreases to 0.0 °C. How many grams of water were in the sample? (10.7)

10.110 An ice cube at 0 °C with a mass of 115 g is added to water in a beaker that has a temperature of 64.0 °C. If the final tem- perature of the mixture is 24.0 °C, what was the initial mass of the warm water? (10.7)

10.111 An ice cube tray holds 325 g of water. If the water initially has a temperature of 25 °C, how many kilojoules of heat must be removed to cool and freeze the water at 0 °C? (10.7)

10.112 A 3.0-kg block of lead is taken from a furnace at 300. °C and placed on a large block of ice at 0 °C. The specific heat of lead is 0.13 J/g °C. If all the heat given up by the lead is used to melt ice, how much ice is melted when the temperature of the lead drops to 0 °C? (10.7)

CHALLENGE PROBLEMS

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306 CHAPTER 10 Bonding and Properties of Solids and Liquids

ANSWERS TO ENGAGE QUESTIONS 10.8 Chlorine is above iodine in Group 7A (17). Thus, the valence

electrons are closer to the nucleus and have the greater attraction for shared electrons.

10.9 The difference in electronegativity between Si and S is 0.7, which makes the Si ¬ S bond polar covalent. The difference in electronegativity between Si and P is 0.3, which makes the Si ¬ P bond nonpolar covalent.

10.10 Both GeH4 and CH4 have nonpolar covalent bonds, but GeH4 has a larger molar mass and more electrons, which produce more dispersion forces.

10.11 When liquid water freezes, heat is released.

10.12 When a substance changes state, there is no change in temperature. Thus, the temperature at which vaporization occurs is not used in the calculation.

10.1 Nitrogen, phosphorus, and arsenic are all in Group 5A (15) and have five valence electrons.

10.2 The positive charge of the proton in the nucleus of each H atom is attracted to the negative electron in the other H atom.

10.3 The total number of valence electrons in a molecule is the sum of the valence electrons of all the atoms in the molecule.

10.4 When octets are not complete for all atoms, some electrons are shared, which form multiple bonds.

10.5 Two pairs of electrons may be shared between the central O atom and one of the other two O atoms.

10.6 SO2 can have a double bond from the S atom to one of the two O atoms; thus, it is possible to draw two Lewis structures for SO2. SCl2 has only single bonds between S and Cl; only one Lewis structure is possible.

10.7 There are four electron groups bonded to the central P in PH3. However, one group is a lone pair of electrons, and the three H atoms determine the shape, which is trigonal pyramidal.

10.1 a. 8 valence electrons b. 14 valence electrons c. 32 valence electrons d. 8 valence electrons

10.3 If complete octets cannot be formed by using all the valence electrons, it is necessary to draw multiple bonds.

10.5 a. HF (8 e -) H or HF F

b. SF2 (20 e -) orS SF FF F

c. NBr3 (26 e

-) N orBrBr Br

Br N

Br

Br

d. BH4 - (8 e-) H HB

H

H

-

orH H H

H

-

B

10.7 a. CO (10 e -) C C OorOO OC C

b. CN - (10 e-) or

- - C CN N

c. H2CO (12 e -) H H or H C H

O O

C

10.9 Resonance occurs when we can draw two or more Lewis structures for the same molecule or ion.

10.11 a. (24 e

-)ClNO2 N NCl

O O

O Cl O

b. (16 e -)

C

OCN- C

C

O N

O N O N --

-

(16 e-)

C

OCN- C

C

O N

O N O N --

-

10.13 a. 6, tetrahedral b. 5, trigonal pyramidal c. 3, trigonal planar

10.15 a. three b. trigonal planar c. three d. trigonal planar

10.17 In CF4, the central atom C has four bonded atoms and no lone pairs of electrons, which gives it a tetrahedral shape. In NF3, the central atom N has three bonded atoms and one lone pair of electrons, which gives NF3 a trigonal pyramidal shape.

10.19 a. trigonal planar b. bent (109°) c. linear d. tetrahedral

10.21 a.

H

H

H

-

AlH4 - (8 e-) H tetrahedralAl

b.

2-

SO4 2- (32 e-) tetrahedralSO O

O

O

c.

H

H

H

+

NH4 + (8 e-) H tetrahedralN

d. N

+ NO2

+ (16 e-) linearO O

10.23 a. increases b. decreases c. decreases

10.25 a. K, Na, Li b. Na, P, Cl c. Ca, Se, O

ANSWERS TO SELECTED PROBLEMS

M10_TIMB8119_06_SE_C10.indd 306 11/27/18 11:59 AM

Answers to Selected Problems 307

10.27 a. between 0.0 and 0.4

10.29 a. polar covalent b. ionic c. polar covalent d. nonpolar covalent e. polar covalent f. nonpolar covalent

10.31 a. N d+

¬ F d- b. Si

d+

¬ Br d-

c. C d+

¬ O d-

d. P d+

¬ Br d-

e. N d-

¬ P d+

10.33 Electrons are shared equally between two identical atoms and unequally between nonidentical atoms.

10.35 a. nonpolar b. polar c. polar d. nonpolar

10.37 In the molecule CO2, the two C ¬ O dipoles cancel; in CO, there is only one dipole.

10.39 a. dipole–dipole attractions b. hydrogen bonds c. dipole–dipole attractions d. dispersion forces

10.41 a. hydrogen bonds b. dipole–dipole attractions c. dispersion forces d. dispersion forces

10.43 a. HF; hydrogen bonds are stronger than the dipole–dipole attractions in HBr.

b. HF; hydrogen bonds are stronger than the dipole–dipole attractions in SF2.

c. PBr3, the dipole–dipole attractions in PBr3 are stronger than the dispersion forces in Br2.

d. CH3OH; hydrogen bonds are stronger than the dispersion forces in CH4.

10.45 a. 21 700 J b. 2180 J c. 28.8 kJ d. 16.7 kJ

10.47 a. 22 600 J b. 42.1 kJ c. 3100 J d. 242 kJ

10.49 a. 852 J b. 30 300 J c. 40.2 kJ d. 72.2 kJ

10.51 119.5 kJ

10.53 3130 J

10.55 a. 0.0, nonpolar covalent b. 0.4, nonpolar covalent c. 1.0, polar covalent

10.57 a.

C 2- 2- 2-

O O

O

CO O

O

CO O

O

b. Carbonate ion is trigonal planar.

10.59 a. two valence electrons, one bonding pair, no lone pairs b. eight valence electrons, one bonding pair, three lone pairs c. 14 valence electrons, one bonding pair, six lone pairs

10.61 a. 2, trigonal pyramidal, polar b. 1, bent (109°), polar c. 3, tetrahedral, nonpolar

10.63 a. C and O, N and O b. O and O c. Ca and O, K and O d. C and O, O and O, N and O

10.65 a. dispersion forces b. dipole–dipole attractions c. hydrogen bonds d. dispersion forces

H Cl

A single dipole does not cancel; HCl is polar.H Cl

A single dipole does not cancel; HCl is polar.

H Cl

A single dipole does not cancel; HCl is polar.

H Cl

A single dipole does not cancel; HCl is polar.HCl

A single dipole does not cancel; HCl is polar.

10.67 a. The heat from the skin is used to evaporate the water (perspiration). Therefore, the skin is cooled.

b. On a hot day, there are more molecules with sufficient energy to become water vapor.

c. In a closed bag, some molecules evaporate, but they cannot escape and will condense back to liquid; the clothes will not dry.

10.69 150

100

50

0

-20

T °C

Heat Added

Solid (a) Melting (b)

Liquid (c)

Boiling (d)

Gas (e)

10.71 a. about - 60 °C b. about 60 °C c. The diagonal line A represents the solid state as tempera-

ture increases. The horizontal line B represents the change from solid to liquid or melting of the substance. The diagonal line C represents the liquid state as temperature increases. The horizontal line D represents the change from liquid to gas or boiling of the liquid. The diagonal line E represents the gas state as temperature increases.

d. At - 80 °C, solid; at - 40 °C, liquid; at 25 °C, liquid; 80 °C, gas

10.73 a. 1 + 5 + 2(6) = 18 valence electrons b. 2(4) + 4(1) + 6 = 18 valence electrons c. 5 + 4(1) - 1 = 8 valence electrons d. 6 + 3(6) + 2 = 26 valence electrons

10.75 a.

-

BF4 - (32 e-) or B

-

F F F

F

F F

F FB

d.

b. Cl ClCl ClCl2O (20 e -) orO O

c. H

H HH2NOH (14 e -) or HN O

H

HON

10.77 a. (16 e-) orNN

N N N

-

N N - - -

NNN

N

N

b. orN +

N + +

N +

N

(16 e-)

OO

OO

OOOO

c.

-

H

or

C

-

(18 e- ) O

O

O

O

H C

-

O

O

H C

H2CCCl2 (24 e -) orH C

H H C C

H Cl

Cl

Cl

ClCl

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308 CHAPTER 10 Bonding and Properties of Solids and Liquids

HCl

A single dipole does not cancel; HCl is polar.

10.97 When water vapor condenses or liquid water freezes, heat is released, which warms the air.

10.99 4.61 g of water

10.101 27.3 g of acetic acid

10.103 a. (18 e-) H H

H

C

O

N

b. N

H

H

C C(22 e-) Cl

c. H H(12 e-) NN

d.

(30 e-) C

H

C

H

HCl

O

O

10.105 a. trigonal pyramidal b. tetrahedral c. linear

10.107 140

Heat Added

T °C

100

60

20

-20 -60

-100

10.79 a. I, Cl, F b. K, Li, S, Cl c. Ba, Sr, Mg, Be

10.81 a. C and O b. N and F c. S and Cl d. Br and I e. N and F

10.83 a. Si ¬ Cl b. C ¬ N

c. F ¬ Cl d. C ¬ F

e. N ¬ O

10.85 a. polar covalent b. nonpolar covalent c. ionic d. nonpolar covalent e. nonpolar covalent

10.87 a.

NF3 trigonal pyramidalN(26 e -)

F

F F

b.

SiBr4 tetrahedralSi(32 e -) Br Br

Br

Br

c. CSe2 (16 e

-) Se linearCSe

d.

SO2

bent (120°)

(18 e-) OO OO S S

10.89 a. bent (109°) b. bent (109°) c. tetrahedral d. trigonal pyramidal

10.91 a. trigonal pyramidal, polar b. bent (109°), polar

10.93 a. polar b. nonpolar c. nonpolar d. polar e. polar f. polar

10.95 a. (2) dipole–dipole attractions b. (2) dipole–dipole attractions c. (4) dispersion forces d. (1) ionic bonds e. (4) dispersion forces f. (3) hydrogen bonds

a. solid b. solid dibromomethane melts c. liquid d. gas e. - 53 °C 10.109 450 g of water

10.111 143 kJ of heat is removed.

H Cl

A single dipole does not cancel; HCl is polar.

H Cl

A single dipole does not cancel; HCl is polar. H Cl

A single dipole does not cancel; HCl is polar. H Cl

A single dipole does not cancel; HCl is polar.

M10_TIMB8119_06_SE_C10.indd 308 11/27/18 11:59 AM

CI.13 In an experiment, the mass of a piece of copper is determined to be 8.56 g. Then the copper is reacted with sufficient oxygen gas to produce solid copper(II) oxide. (7.1, 8.1, 8.2, 8.3, 9.1, 9.2, 9.3, 9.5)

a. How many copper atoms are in the sample? b. Write the balanced chemical equation for the reaction. c. Classify the type of reaction. d. How many grams of O2 are required to completely react

with the Cu? e. How many grams of CuO will result from the reaction of

8.56 g of Cu and 3.72 g of oxygen? f. How many grams of CuO will result in part e, if the yield

for the reaction is 85.0%?

CI.14 One of the components in gasoline is octane, C8H18, which has a density of 0.803 g/cm3 and ∆H = - 510 kJ/mol. Suppose a hybrid car has a fuel tank with a capacity of 11.9 gal and has a gas mileage of 45 mi/gal. (7.10, 6.7, 8.3, 8.5, 9.6)

a. Write a balanced chemical equation for the complete combus- tion of octane including the heat of reaction.

b. What is the energy, in kilojoules, produced from one tank of fuel assuming it is all octane?

c. How many molecules of C8H18 are present in one tank of fuel assuming it is all octane?

d. If this hybrid car is driven 24 500 miles in one year, how many kilograms of carbon dioxide will be produced from the combustion of the fuel assuming it is all octane?

CI.15 When clothes have stains, bleach may be added to the wash to react with the soil and make the stains colorless. The bleach solution is prepared by bubbling chlorine gas into a solution of sodium hydroxide to produce a solution of sodium hypochlorite, sodium chloride, and water. One brand of bleach contains 5.25% sodium hypochlorite by mass (active ingredient) with a density of 1.08 g/mL. (6.4, 7.1, 7.2, 8.2, 9.3, 9.4, 9.5, 10.1)

a. What is the formula and molar mass of sodium hypochlorite?

b. Draw the Lewis structure for the hypochlorite ion. c. How many hypochlorite ions are present in 1.00 gal of

bleach solution? d. Write the balanced chemical equation for the preparation

of bleach.

e. How many grams of NaOH are required to produce the sodium hypochlorite for 1.00 gal of bleach?

f. If 165 g of Cl2 is passed through a solution containing 275 g of NaOH and 162 g of sodium hypochlorite is produced, what is the percent yield of sodium hypochlorite for the reaction?

CI.16 Ethanol, C2H6O, is obtained from renewable crops such as corn, which use the Sun as their source of energy. In the United States, automobiles can now use a fuel known as E85 that contains 85.0% ethanol and 15.0% gasoline by volume. Ethanol has a melting point of - 115 °C, a boiling point of 78 °C, a heat of fusion of 109 J/g, and a heat of vaporization of 841 J/g. Liquid ethanol has a density of 0.796 g/mL and a specific heat of 2.46 J/g °C. (8.3, 8.5, 9.2, 9.3, 10.4, 10.5)

a. Draw a heating curve for ethanol from - 150 °C to 100 °C.

COMBINING IDEAS from Chapters 8 to 10

Cu

8.56 g

100

50

0

-50

-100

-150

Heat Added

T °C

Octane is one of the components of motor fuel.

Bleach is a solution of sodium hypochlorite.

b. When 20.0 g of ethanol at - 62 °C is heated and completely vaporized at 78 °C, how much energy, in kilojoules, is required?

c. If a 15.0-gal gas tank is filled with E85, how many liters of ethanol are in the gas tank?

d. Write the balanced chemical equation for the complete combustion of ethanol.

e. How many kilograms of CO2 are produced by the com- plete combustion of the ethanol in a full 15.0-gal gas tank?

f. What would be the strongest intermolecular force between liquid ethanol molecules?

Applications CI.17 Chloral hydrate, a sedative and hypnotic, was the first drug

used to treat insomnia. Chloral hydrate has a melting point of 57 °C. At its boiling point of 98 °C, it breaks down to chloral and water. (7.4, 7.5, 8.4, 10.1)

Cl OC C H

HOCl

Cl Chloral hydrate

H

Cl H +C C

OCl

Cl Chloral

H2O

a. Draw the Lewis structures for chloral hydrate and chloral. b. What are the empirical formulas of chloral hydrate and

chloral? c. What is the mass percent of Cl in chloral hydrate?

309

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310 CHAPTER 10 Bonding and Properties of Solids and Liquids

CI.18 Ethylene glycol, C2H6O2, used as a coolant and antifreeze, has a density of 1.11 g/mL. As a sweet-tasting liquid, it can be appealing to pets and small children, but it can be toxic, with an LD50 of 4700 mg/kg. Its accidental ingestion can cause kidney damage and difficulty with breathing. In the body, ethylene glycol is converted to another toxic substance, oxalic acid, H2C2O4. (2.8, 7.4, 7.5, 8.2, 8.4, 10.1, 10.3, 10.4)

Antifreeze often contains ethylene glycol.

Acetone has carbon atoms (black), hydrogen atoms (white), and an oxygen atom (red).

Dihydroxyacetone has carbon atoms (black), hydrogen atoms (white), and oxygen atoms (red).

CI.19 Acetone (propanone), a clear liquid solvent with an acrid odor, is used to remove nail polish, paints, and resins. It has a low boiling point and is highly flammable. Acetone has a density of 0.786 g/mL and a heat of combustion of - 1790 kJ/mol. (2.8, 7.2, 8.3, 8.4, 9.3, 9.6)

ANSWERS

O

C C H(36 e-)

Cl

Cl

Cl

b. chloral hydrate: C2H3O2Cl3 chloral: C2HOCl3 c. 64.33% Cl (by mass)

CI.19 a. C CH3CH3

O

b. C3H6O; 58.08 g/mol

c. C3H6O(l) + 4O2(g) h ∆

3CO2(g) + 3H2O(g) + 1790 kJ d. exothermic e. 79.5 kJ f. 26.0 g of O2

CI.13 a. 8.11 * 1022 atoms of copper b. 2Cu(s) + O2(g) h 2CuO(s) c. combination reaction d. 2.16 g of O2 e. 10.7 g of CuO f. 9.10 g of CuO

CI.15 a. NaOCl, 74.44 g/mol b. (14 e-) -

Cl O c. 1.74 * 1024 ClO- ions d. 2NaOH(aq) + Cl2(g) h

NaClO(aq) + NaCl(aq) + H2O(l) e. 231 g of NaOH f. 93.6%

CI.17 a. (44 e-) OC C H

HO

HCl

Cl

Cl

a. What are the empirical formulas of ethylene glycol and oxalic acid?

b. If ethylene glycol has a C ¬ C single bond with two H atoms attached to each C atom, what is its Lewis structure?

c. Which bonds in ethylene glycol are polar covalent and which are nonpolar covalent?

d. How many milliliters of ethylene glycol could be toxic for an 11.0-lb cat?

e. What would be the strongest intermolecular force in ethylene glycol?

f. If oxalic acid has two carbon atoms attached by a C ¬ C single bond with each carbon also attached to two oxygen atoms, what is its Lewis structure?

g. Write the balanced chemical equation for the reaction of ethylene glycol and oxygen (O2) to give oxalic acid and water.

a. Draw the Lewis structure for DHA. b. DHA has C ¬ C, C ¬ H, C ¬ O, and O ¬ H bonds.

Which of these bonds are polar covalent and which are nonpolar covalent?

c. What are the molecular formula and molar mass of DHA? d. A bottle of sunless tanning lotion contains 177 mL of

lotion. How many milligrams of DHA are in a bottle?

a. Draw the Lewis structure for acetone. b. What are the molecular formula and molar mass of

acetone? c. Write a balanced chemical equation for the complete

combustion of acetone, including the heat of reaction. d. Is the combustion of acetone an endothermic or

exothermic reaction? e. How much heat, in kilojoules, is released if 2.58 g of

acetone reacts completely with oxygen? f. How many grams of oxygen gas are needed to react with

15.0 mL of acetone? CI.20 The compound dihydroxyacetone (DHA) is used in “sunless”

tanning lotions, which darken the skin by reacting with the amino acids in the outer surface of the skin. A typical drug- store lotion contains 4.0% (mass/volume) DHA. (2.8, 7.2, 8.4)

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311

Gases After soccer practice, Louisa complains that she is having difficulty breathing. Her father takes her to the emergency room, where she is seen by Sam, a respiratory therapist, who listens to Louisa’s chest and tests her breathing capacity using a spirometer. Based on her limited breathing capacity and the wheezing noise in her chest, Sam diagnoses Louisa with asthma.

Sam gives Louisa a nebulizer containing a bronchodilator that opens the airways and allows more air to go into the lungs. During the breathing treatment, he measures the amount of oxygen (O2) in her blood and explains to Louisa and her father that air is a mixture of gases containing 78% nitrogen (N2) gas and 21% O2 gas. Because Louisa has difficulty obtaining sufficient oxygen, Sam gives her supplemental oxygen through a face mask. Within a short period of time, Louisa’s breathing returns to normal. Sam explains that the lungs work according to Boyle’s law: The volume of the lungs increases upon inhalation, and the pressure decreases to allow air to flow in. However, during an asthma attack, the airways become restricted, and it becomes more difficult to expand the volume of the lungs.

CAREER

Respiratory Therapist Respiratory therapists assess and treat a range of patients, including premature infants whose lungs have not developed, asthmatics, and patients with emphysema or cystic fibrosis. In assessing patients, they perform a variety of diagnostic tests including breathing capacity and concentrations of oxygen and carbon dioxide in a patient’s blood, as well as blood pH. In order to treat patients, therapists provide oxygen or aerosol medications to the patient, as well as chest physiotherapy to remove mucus from their lungs. Respiratory therapists also educate patients on how to use their inhalers correctly.

11

UPDATE Exercise-Induced Asthma

Louisa’s doctor prescribes an inhaled medication that opens up her airways before she starts exercise. In the UPDATE Exercise-Induced Asthma, page 341, you can view the impact of Louisa’s medication and learn about other treatments that help prevent exercise-induced asthma.

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312 CHAPTER 11 Gases

11.1 Properties of Gases LEARNING GOAL Describe the kinetic molecular theory of gases and the units of measurement used for gases.

We all live at the bottom of a sea of gases called the atmosphere. The most important of these gases is oxygen, which constitutes about 21% of the atmosphere. Without oxygen, life on this planet would be impossible: Oxygen is vital to all life processes of plants and animals. Ozone (O3), formed in the upper atmosphere by the interaction of oxygen with ultraviolet light, absorbs some of the harmful radiation before it can strike Earth’s surface. The other gases in the atmosphere include nitrogen (78%), argon, carbon dioxide (CO2), and water vapor. Carbon dioxide gas, a product of combustion and metabolism, is used by plants in photosynthesis, which produces the oxygen that is essential for humans and animals.

The behavior of gases is quite different from that of liquids and solids. Gas particles are far apart, whereas particles of both liquids and solids are held close together. Because there are great distances between gas particles, a gas is less dense than a solid or liquid, and easy to compress. A gas has no definite shape or volume and will completely fill any con- tainer. A model for the behavior of a gas, called the kinetic molecular theory of gases, helps us understand gas behavior.

Kinetic Molecular Theory of Gases 1. A gas consists of small particles (atoms or molecules) that move randomly with

high velocities. Gas molecules moving in random directions at high speeds cause a gas to fill the entire volume of a container.

2. The attractive forces between the particles of a gas are usually very small. Gas particles are far apart and fill a container of any size and shape.

3. The actual volume occupied by gas molecules is extremely small compared to the volume that the gas occupies. The volume of the gas is considered equal to the volume of the container. Most of the volume of a gas is empty space, which allows gases to be easily compressed.

4. Gas particles are in constant motion, moving rapidly in straight paths. When gas particles collide, they rebound and travel in new directions. Every time they hit the walls of the container, they exert pressure. An increase in the number or force of col- lisions against the walls of the container causes an increase in the pressure of the gas.

5. The average kinetic energy of gas molecules is proportional to the Kelvin temperature. Gas particles move faster as the temperature increases. At higher temperatures, gas particles hit the walls of the container more often and with more force, producing higher pressures.

The kinetic molecular theory helps explain some of the characteristics of gases. For example, you can smell perfume when a bottle is opened on the other side of a room because its particles move rapidly in all directions. At room temperature, the molecules in the air are moving at about 450 m/s, which is 1000 mi/h. They move faster at higher temperatures and more slowly at lower temperatures. Sometimes tires and gas-filled containers explode when temperatures are too high. From the kinetic molecular theory, you know that gas particles move faster when heated, hit the walls of a container with more force, and cause a buildup of pressure inside a container.

When we talk about a gas, we describe it in terms of four properties: pressure, volume, temperature, and the amount of gas.

Pressure (P) Gas particles are extremely small and move rapidly. When they hit the walls of a container, they exert a pressure (see FIGURE 11.1). If we heat the container, the molecules move faster and smash into the walls of the container more often and with increased force, thus increas- ing the pressure. The gas particles in the air, mostly oxygen and nitrogen, exert a pressure on us called atmospheric pressure (see FIGURE 11.2). As you go to higher altitudes, the

REVIEW Using Significant Figures in

Calculations (2.3)

Writing Conversion Factors from Equalities (2.5)

Using Conversion Factors (2.6)

INTERACTIVE VIDEO

Kinetic Molecular Theory

ENGAGE 11.1 Use the kinetic molecular theory to explain why a gas completely fills a container of any size and shape.

PRACTICE PROBLEMS Try Practice Problems 11.1 and 11.2

LOOKING AHEAD

11.1 Properties of Gases  312 11.2 Pressure and Volume

(Boyle’s Law)  317 11.3 Temperature and Volume

(Charles’s Law)  320 11.4 Temperature and Pressure

(Gay-Lussac’s Law)  322 11.5 The Combined Gas

Law  325 11.6 Volume and Moles

(Avogadro’s Law)  327 11.7 The Ideal Gas Law  329 11.8 Gas Laws and Chemical

Reactions  334 11.9 Partial Pressures (Dalton’s

Law)  335

FIGURE 11.1 Gas particles moving in straight lines within a container exert pressure when they collide with the walls of the container.

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11.1 Properties of Gases 313

atmospheric pressure is less because there are fewer gas particles. The most common units used to measure gas pressure are the atmosphere (atm) and millimeters of mercury (mmHg). On a TV weather report, the atmospheric pressure may be reported in inches of mercury, or in kilopascals in countries other than the United States. In a hospital, the unit torr or pounds per square inch (psi) may be used.

Volume (V ) The volume of gas equals the size of the container in which the gas is placed. When you inflate a balloon, you are adding more gas particles. The increase in the number of particles hitting the walls of the balloon increases the volume. The most common units for volume measurement are liters (L) and milliliters (mL).

Temperature (T ) The temperature of a gas is related to the kinetic energy of its particles. For example, if we have a gas at 200 K and heat it to a temperature of 400 K, the gas particles will have twice the kinetic energy that they did at 200 K. This also means that the gas at 400 K exerts twice the pressure of the gas at 200 K, if the volume and amount of gas do not change. Although we measure gas temperature using a Celsius thermometer, all comparisons of gas behavior and all calculations related to temperature must use the Kelvin temperature scale. No one has created the conditions for absolute zero (0 K), but scientists predict that the particles will have zero kinetic energy and exert zero pressure at absolute zero.

Amount of Gas (n ) When you add air to a bicycle tire, you increase the amount of gas, which results in a higher pressure in the tire. Usually, we measure the amount of gas by its mass, in grams. In gas law calculations, we need to change the grams of gas to moles.

A summary of the four properties of a gas is given in TABLE 11.1.

ENGAGE 11.2 Why is there less atmospheric pressure at a higher altitude?

FIGURE 11.2 A column of air extending from the top of the atmosphere to the surface of Earth produces pressure on each of us of about 1 atm.

Molecules in air Atmospheric

pressure

O2

21%

N2

78%

Other gases

1%

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314 CHAPTER 11 Gases

Property Description Units of Measurement

Pressure (P) The force exerted by a gas against the walls of the container

atmosphere (atm); millimeter of mercury (mmHg); torr (Torr); kilopascal (kPa)

Volume (V) The space occupied by a gas liter (L); milliliter (mL)

Temperature (T) The determining factor of the kinetic energy and rate of motion of gas particles

degree Celsius (°C); kelvin (K) is required in calculations

Amount (n) The quantity of gas present in a container

gram (g); mole (n) is required in calculations

TABLE 11.1 Properties That Describe a Gas

SAMPLE PROBLEM 11.1 Properties of Gases

TRY IT FIRST

Identify the property of a gas that is described by each of the following:

a. increases the kinetic energy of gas particles b. the force of the gas particles hitting the walls of the container c. the space that is occupied by a gas

SOLUTION

a. temperature b. pressure c. volume

SELF TEST 11.1

What property of a gas is described in each of the following?

a. Helium gas is added to a balloon. b. A balloon expands as it rises higher.

ANSWER

a. amount b. volume

Measurement of Gas Pressure When billions and billions of gas particles hit against the walls of a container, they exert pressure, which is a force acting on a certain area.

Pressure (P) = force area

The atmospheric pressure can be measured using a barometer (see FIGURE 11.3). At a pressure of exactly 1 atmosphere (atm), a mercury column in an inverted tube would be exactly 760 mm high. One atmosphere (atm) is defined as exactly 760 mmHg (millimeters of mercury). One atmosphere is also 760 Torr, a pressure unit named to honor Evangelista Torricelli, the inventor of the barometer. Because the units of Torr and mmHg are equal, they are used interchangeably. One atmosphere is also equivalent to 29.9 in. of mercury (inHg).

1 atm = 760 mmHg = 760 Torr (exact) 1 atm = 29.9 inHg 1 mmHg = 1 Torr (exact)

In SI units, pressure is measured in pascals (Pa); 1 atm is equal to 101 325 Pa. Because a pascal is a very small unit, pressures are usually reported in kilopascals.

1 atm = 101 325 Pa = 101.325 kPa

FIGURE 11.3 A barometer: The pressure exerted by the gases in the atmosphere is equal to the downward pressure of a mercury column in a closed glass tube. The height of the mercury column measured in mmHg is called atmospheric pressure.

Vacuum (no air particles)

Liquid mercury

760 mmHg

Gases of the atmosphere

at 1 atm

PRACTICE PROBLEMS Try Practice Problems 11.3 and 11.4

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11.1 Properties of Gases 315

The U.S. equivalent of 1 atm is 14.7 lb/in.2 When you use a pressure gauge to check the air pressure in the tires of a car, it may read 30 to 35 psi. This measurement is actually 30 to 35 psi above the pressure that the atmosphere exerts on the outside of the tire.

1 atm = 14.7 lb/in.2

TABLE 11.2 summarizes the various units used in the measurement of pressure.

TABLE 11.2 Units for Measuring Pressure Unit Abbreviation Unit Equivalent to 1 atm

Atmosphere atm 1 atm (exact)

Millimeters of Hg mmHg 760 mmHg (exact)

Torr Torr 760 Torr (exact)

Inches of Hg inHg 29.9 inHg

Pounds per square inch lb/in.2 (psi) 14.7 lb/in.2

Pascal Pa 101 325 Pa

Kilopascal kPa 101.325 kPa

Sea level

P = 1.0 atm (760 mmHg)

P = 0.70 atm (530 mmHg)

The atmospheric pressure decreases as the altitude increases.

A patient with severe COPD obtains oxygen from an oxygen tank.

Atmospheric pressure changes with variations in weather and altitude. On a hot, sunny day, the mercury column rises, indicating a higher atmospheric pressure. On a rainy day, the atmosphere exerts less pressure, which causes the mercury column to fall. In a weather report, this type of weather is called a low-pressure system. Above sea level, the density of the gases in the air decreases, which causes lower atmospheric pressures; the atmospheric pressure is greater than 760 mmHg at the Dead Sea because it is below sea level (see TABLE 11.3).

TABLE 11.3 Altitude and Atmospheric Pressure

Location Altitude (km) Atmospheric Pressure (mmHg)

Dead Sea - 0.40 800 Sea level 0.00 760

Los Angeles 0.09 752

Las Vegas 0.70 700

Denver 1.60 630

Mount Whitney 4.50 440

Mount Everest 8.90 253

Divers must be concerned about increasing pressures on their ears and lungs when they dive below the surface of the ocean. Because water is more dense than air, the pressure on a diver increases rapidly as the diver descends. At a depth of 33 ft below the surface of the ocean, an additional 1 atm of pressure is exerted by the water on a diver, which gives a total pressure of 2 atm. At 100 ft, there is a total pressure of 4 atm on a diver. The regulator that a diver uses continuously adjusts the pressure of the breathing mixture to match the increase in pressure.

SAMPLE PROBLEM 11.2 Units of Pressure

TRY IT FIRST

The oxygen in a tank in the hospital respiratory unit has a pressure of 4820 mmHg. Calculate the pressure, in atmospheres, of the oxygen gas.

SOLUTION

The equality 1 atm = 760 mmHg can be written as two conversion factors:

760 mmHg

1 atm and

1 atm 760 mmHg

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316 CHAPTER 11 Gases

Using the conversion factor that cancels mmHg and gives atm, we can set up the problem as

4820 mmHg * 1 atm

760 mmHg = 6.34 atm

SELF TEST 11.2

A tank of nitrous oxide (N2O) used as an anesthetic has a pressure of 48 psi.

a. What is that pressure in atmospheres? b. What is the pressure in torr?

ANSWER

a. 3.3 atm b. 2500 (2.5 * 103) Torr PRACTICE PROBLEMS

Try Practice Problems 11.5 to 11.8

Chemistry Link to Health Measuring Blood Pressure

Your blood pressure is one of the vital signs a doctor or nurse checks during a physical examination. It actually consists of two separate measurements. Acting like a pump, the heart contracts to create the pressure that pushes blood through the circulatory system. During contraction, the blood pressure is at its highest; this is your systolic pressure. When the heart muscles relax, the blood pressure falls; this is your diastolic pressure. The normal range for systolic pressure is 90  to 120 mmHg. For diastolic pressure, it is 60 to 80 mmHg. These two measurements are usually expressed as a ratio such as 100/80. These values are somewhat higher in older people. When blood pressures are elevated, such as 140/90, there is a greater risk of stroke, heart attack, or kidney damage. Low blood pressure prevents the brain from receiving adequate oxygen, causing dizziness and fainting.

The blood pressures are measured by a sphygmomanometer, an instrument consisting of a stethoscope and an inflatable cuff con- nected to a tube of mercury called a manometer. After the cuff is wrapped around the upper arm, it is pumped up with air until it cuts off the flow of blood. With the stethoscope over the artery, the air is

slowly released from the cuff, decreasing the pressure on the artery. When the blood flow first starts again in the artery, a noise can be heard through the stethoscope signifying the systolic blood pres- sure as the pressure shown on the manometer. As air continues to be released, the cuff deflates until no sound is heard in the artery. A second pressure reading is taken at the moment of silence and denotes the diastolic pressure, the pressure when the heart is not contracting.

The use of digital blood pressure monitors is becoming more com- mon. However, they have not been validated for use in all situations and can sometimes give inaccu- rate readings.

The measurement of blood pressure is part of a routine physical exam.

PRACTICE PROBLEMS

11.1 Properties of Gases

11.1 Use the kinetic molecular theory of gases to explain each of the following:

a. Gases move faster at higher temperatures. b. Gases can be compressed much more easily than

liquids or solids. c. Gases have low densities.

11.2 Use the kinetic molecular theory of gases to explain each of the following:

a. A container of nonstick cooking spray explodes when thrown into a fire.

b. The air in a hot-air balloon is heated to make the balloon rise.

c. You can smell the odor of cooking onions from far away.

11.3 Identify the property of a gas that is measured in each of the following:

a. 350 K b. 125 mL c. 2.00 g of O2 d. 755 mmHg

11.4 Identify the property of a gas that is measured in each of the following: a. 425 K b. 1.0 atm c. 10.0 L d. 0.50 mol of He

11.5 Which of the following statements describe the pressure of a gas? a. the force of the gas particles on the walls of the container b. the number of gas particles in a container c. 4.5 L of helium gas d. 750 Torr e. 28.8 lb/in.2

11.6 Which of the following statements describe the pressure of a gas? a. the temperature of a gas b. the volume of the container c. 3.00 atm d. 0.25 mol of O2 e. 101 kPa

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11.2 Pressure and Volume (Boyle’s Law) 317

FIGURE 11.4 Boyle’s law: As volume decreases, gas molecules become more crowded, which causes the pressure to increase. Pressure and volume are inversely related.

V1 = 4 L P1 = 1 atm

V2 = 2 L P2 = 2 atm

Piston

Oxygen therapy increases the oxygen available to the tissues of the body.

11.8 On a climb up Mount Whitney, the atmospheric pressure drops to 467 mmHg. What is the pressure in terms of the following units?

a. atm b. torr c. inHg d. Pa

11.7 A tank contains oxygen (O2) at a pressure of 2.00 atm. What is the pressure in the tank in terms of the following units?

a. torr b. lb/in.2 c. mmHg d. kPa

Applications

11.2 Pressure and Volume (Boyle’s Law) LEARNING GOAL Use the pressure–volume relationship (Boyle’s law) to calculate the unknown pressure or volume when the temperature and amount of gas do not change.

Imagine that you can see air particles hitting the walls inside a bicycle tire pump. What happens to the pressure inside the pump as you push down on the handle? As the volume decreases, there is a decrease in the surface area of the container. The air particles are crowded together, more collisions occur, and the pressure increases within the container.

When a change in one property (in this case, volume) causes a change in another property (in this case, pressure), the properties are related. If the changes occur in opposite directions, the properties have an inverse relationship. The inverse relationship between the pressure and volume of a gas is known as Boyle’s law. The law states that the volume (V) of a sample of gas changes inversely with the pressure (P) of the gas as long as there is no change in the temperature (T) and amount of gas (n), as illustrated in FIGURE 11.4.

If the volume or pressure of a gas changes without any change occurring in the temperature or in the amount of the gas, then the final pressure and volume will give the same PV product as the initial pressure and volume. Then we can set the initial and final PV products equal to each other. In the equation for Boyle’s law, the initial pressure and volume are written as P1 and V1 and the final pressure and volume are written as P2 and V2.

Boyle’s Law

P1V1 = P2V2 No change in temperature and amount of gas

REVIEW Solving Equations (1.4)

CORE CHEMISTRY SKILL Using the Gas Laws

ANALYZE THE

PROBLEM

Given Need Connect

P1 = 3800 mmHg P2 = 570 mmHg V1 = 12 L

Factors that do not change: T and n

V2 Boyle’s law, P1V1 = P2V2

Predict: V increases

SAMPLE PROBLEM 11.3 Calculating Volume When Pressure Changes

TRY IT FIRST

When Louisa had her asthma attack, she was given oxygen through a face mask. The gauge on a 12-L tank of compressed oxygen reads 3800 mmHg. How many liters would this same gas occupy at a final pressure of 570 mmHg when temperature and amount of gas do not change?

SOLUTION

STEP 1 State the given and needed quantities. We place the gas data in a table by writing the initial pressure and volume as P1 and V1 and the final pressure and volume as P2 and V2. We see that the pressure decreases from 3800 mmHg to 570 mmHg. Using Boyle’s law, we predict that the volume increases.

STEP 2 Rearrange the gas law equation to solve for the unknown quantity. For a PV relationship, we use Boyle’s law and solve for V2 by dividing both sides by P2.

P1V1 = P2V2

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318 CHAPTER 11 Gases

P1V1 P2

= P2V2 P2

V2 = V1 * P1 P2

STEP 3 Substitute values into the gas law equation and calculate. When we substitute in the values with pressures in units of mmHg, the ratio of pressures (pressure factor) is greater than 1, which increases the volume as predicted.

V2 = 12 L * 3800 mmHg

570 mmHg = 80. L

ENGAGE 11.3 If the volume of a gas increases, what will happen to its pressure, if its temperature and amount do not change?

Pressure factor increases volume

SELF TEST 11.3

In an underground gas reserve, a bubble of methane gas (CH4) has a volume of 45.0 mL at 1.60 atm pressure.

a. What volume, in milliliters, will the bubble occupy when it reaches the surface where the atmospheric pressure is 744 mmHg, if there is no change in the temperature and amount of gas?

b. What is the new pressure, in atmospheres, when the volume of the methane gas bubble expands to 125 mL, if there is no change in the temperature and amount of gas?

ANSWER

a. 73.5 mL b. 0.576 atm PRACTICE PROBLEMS

Try Practice Problems 11.9 to 11.24

Chemistry Link to Health Pressure–Volume Relationship in Breathing

The importance of Boyle’s law becomes apparent when you consider the mechanics of breathing. Our lungs are elastic, balloon-like structures contained within an airtight cham- ber called the thoracic cavity. The diaphragm, a muscle, forms the flexible floor of the cavity.

Inspiration The process of taking a breath of air begins when the diaphragm contracts and the rib cage expands, causing an increase in the volume of the thoracic cavity. The elasticity of the lungs allows them to expand when the thoracic cav- ity expands. According to Boyle’s law, the pressure inside the lungs decreases when their volume increases, causing the pressure inside the lungs to fall below the pressure of the atmo- sphere. This difference in pressures produces a pressure gradient between the lungs and the atmosphere. In a pressure gradient, molecules flow from an area of higher pressure to an area of lower pressure. During the inhalation phase of breathing, air flows into the lungs (inspiration), until the pressure within the lungs becomes equal to the pressure of the atmosphere.

Expiration Expiration, the exhalation phase of breathing, occurs when the dia- phragm relaxes and moves back up into the thoracic cavity to its

resting position. The volume of the thoracic cavity decreases, which squeezes the lungs and decreases their volume. Now the pressure in the lungs is higher than the pressure of the atmosphere, so air flows out of the lungs. Thus, breathing is a process in which pres- sure gradients are continuously created between the lungs and the environment because of the changes in the volume and pressure.

Air inhaled

Air exhaled

Inhalation Diaphragm contracts

(moves down)

Exhalation Diaphragm relaxes

(moves up)

Lung

Diaphragm

Rib cage expands as rib muscles

contract

Rib cage contracts as rib muscles

relax

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11.2 Pressure and Volume (Boyle’s Law) 319

PRACTICE PROBLEMS

11.2 Pressure and Volume (Boyle’s Law) 11.9 Why do scuba divers need to exhale air when they ascend to

the surface of the water?

11.10 Why does a sealed bag of chips expand when you take it to a higher altitude?

11.11 The air in a cylinder with a piston has a volume of 220 mL and a pressure of 650 mmHg.

a. To obtain a higher pressure inside the cylinder when the temperature and amount of gas do not change, would the cylinder change as shown in A or B? Explain your choice.

Initial A or B

b. If the pressure inside the cylinder increases to 1.2 atm, what is the final volume, in milliliters, of the cylinder?

11.12 A balloon is filled with helium gas. When each of the following changes are made with no change in temperature and amount of gas, which of these diagrams (A, B, or C) shows the final volume of the balloon?

Initial volume

A B C

a. The balloon floats to a higher altitude where the outside pressure is lower.

b. The balloon is taken inside the house, but the atmospheric pressure does not change.

c. The balloon is put in a hyperbaric chamber in which the pressure is increased.

11.13 A gas with a volume of 4.0 L is in a closed container. Indicate the changes (increases, decreases, does not change) in its pressure when the volume undergoes the following changes at the same temperature and amount of gas:

a. The volume is compressed to 2.0 L. b. The volume expands to 12 L. c. The volume is compressed to 0.40 L.

11.14 A gas at a pressure of 2.0 atm is in a closed container. Indicate the changes (increases, decreases, does not change) in its volume when the pressure undergoes the following changes at the same temperature and amount of gas:

a. The pressure increases to 6.0 atm. b. The pressure remains at 2.0 atm. c. The pressure drops to 0.40 atm.

11.15 A 10.0-L balloon contains helium gas at a pressure of 655 mmHg. What is the final pressure, in millimeters of mercury, when the helium is placed in tanks that have the following volumes, if there is no change in temperature and amount of gas?

a. 20.0 L b. 2.50 L c. 13 800 mL d. 1250 mL

11.16 The air in a 5.00-L tank has a pressure of 1.20 atm. What is the final pressure, in atmospheres, when the air is placed in tanks that have the following volumes, if there is no change in temperature and amount of gas?

a. 1.00 L b. 2500. mL c. 750. mL d. 8.00 L

11.17 A sample of nitrogen (N2) has a volume of 50.0 L at a pressure of 760. mmHg. What is the final volume, in liters, of the gas at each of the following pressures, if there is no change in temperature and amount of gas?

a. 725 mmHg b. 1520 mmHg c. 0.500 atm d. 850 Torr

11.18 A sample of methane (CH4) has a volume of 25 mL at a pressure of 0.80 atm. What is the final volume, in milliliters, of the gas at each of the following pressures, if there is no change in temperature and amount of gas?

a. 0.40 atm b. 2.00 atm c. 2500 mmHg d. 80.0 Torr

11.19 A sample of Ar gas has a volume of 5.40 L with an unknown pressure. The gas has a volume of 9.73 L when the pressure is 3.62 atm, with no change in temperature and amount of gas. What was the initial pressure, in atmospheres, of the gas?

11.20 A sample of Ne gas has a pressure of 654 mmHg with an unknown volume. The gas has a pressure of 345 mmHg when the volume is 495 mL, with no change in temperature and amount of gas. What was the initial volume, in milliliters, of the gas?

Applications

11.21 Cyclopropane, C3H6, is a general anesthetic. A 5.0-L sample has a pressure of 5.0 atm. What is the final volume, in liters, of this gas given to a patient at a pressure of 1.0 atm with no change in temperature and amount of gas?

11.22 A patient’s oxygen tank holds 20.0 L of oxygen (O2) at a pressure of 15.0 atm. What is the final volume, in liters, of this gas when it is released at a pressure of 1.00 atm with no change in temperature and amount of gas?

11.23 Use the words inspiration and expiration to describe the part of the breathing cycle that occurs as a result of each of the following:

a. The diaphragm contracts. b. The volume of the lungs decreases. c. The pressure within the lungs is less than that of the

atmosphere.

11.24 Use the words inspiration and expiration to describe the part of the breathing cycle that occurs as a result of each of the following:

a. The diaphragm relaxes, moving up into the thoracic cavity. b. The volume of the lungs expands. c. The pressure within the lungs is higher than that of the

atmosphere.

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320 CHAPTER 11 Gases

11.3 Temperature and Volume (Charles’s Law) LEARNING GOAL Use the temperature–volume relationship (Charles’s law) to calculate the unknown temperature or volume when the pressure and amount of gas do not change.

Suppose that you are going to take a ride in a hot-air balloon. The captain turns on a propane burner to heat the air inside the balloon. As the air is heated, it expands and becomes less dense than the air outside, causing the balloon and its passengers to lift off. In 1787, Jacques Charles, a balloonist as well as a physicist, proposed that the volume of a gas is related to the temperature. This proposal became Charles’s law, which states that the volume (V ) of a gas is directly related to the temperature (T ) when there is no change in the pressure (P) and amount (n) of gas. A direct relationship is one in which the related properties increase or decrease together. For two conditions, initial and final, we can write Charles’s law as follows:

Charles’s Law

V1 T1

= V2 T2

No change in pressure and amount of gas

All temperatures used in gas law calculations must be converted to their corresponding Kelvin (K) temperatures.

To determine the effect of changing temperature on the volume of a gas, the pressure and the amount of gas are not changed. If we increase the temperature of a gas sample, the volume of the container must increase (see FIGURE 11.5). If the temperature of the gas is decreased, the volume of the container must also decrease when pressure and the amount of gas do not change.

As the gas in a hot-air balloon is heated, it expands.

FIGURE 11.5 Charles’s law: The Kelvin temperature of a gas is directly related to the volume of the gas when there is no change in the pressure and amount of gas.

T1 = 200 K T2 = 400 K V1 = 1 L V2 = 2 L

SAMPLE PROBLEM 11.4 Calculating Volume When Temperature Changes

TRY IT FIRST

Helium gas is used to inf late the abdomen during laparoscopic surgery. A sample of helium gas has a volume of 5.40 L and a temperature of 15 °C. What is the final volume, in liters, of the gas after the temperature has been increased to 42 °C when the pressure and amount of gas do not change?

SOLUTION

STEP 1 State the given and needed quantities. We place the gas data in a table by writing the initial temperature and volume as T1 and V1 and the final temperature and volume as T2 and V2. We see that the temperature increases from 15 °C to 42 °C. Using Charles’s law, we predict that the volume increases.

T1 = 15 °C + 273 = 288 K

T2 = 42 °C + 273 = 315 K

ENGAGE 11.4 Why can we predict that the volume of the helium gas will increase when the temperature increases, if the pressure and amount of gas do not change?

ANALYZE THE PROBLEM

Given Need Connect

T1 = 288 K T2 = 315 K V1 = 5.40 L

Factors that do not change: P and n

V2 Charles’s law, V1 T1

= V2 T2

Predict: V increases

STEP 2 Rearrange the gas law equation to solve for the unknown quantity. In this problem, we want to know the final volume (V2) when the temperature increases. Using Charles’s law, we solve for V2 by multiplying both sides by T2.

V1 T1

= V2 T2

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11.3 Temperature and Volume (Charles’s Law) 321

V1 T1

* T2 = V2 T2

* T2

V2 = V1 * T2 T1

STEP 3 Substitute values into the gas law equation and calculate. From the table, we see that the temperature has increased. Because temperature is directly related to volume, the volume must increase. When we substitute in the values, we see that the ratio of the temperatures (temperature factor) is greater than 1, which increases the volume as predicted.

V2 = 5.40 L * 315 K 288 K

= 5.91 L Temperature factor increases volume

SELF TEST 11.4

a. A mountain climber inhales air that has a temperature of - 8 °C. If the final volume of air in the lungs is 569 mL at a body temperature of 37 °C, what was the initial volume of air, in milliliters, inhaled by the climber, if the pressure and amount of gas do not change?

b. A hiker inhales 598 mL of air. If the final volume of air in the lungs is 612 mL at a body temperature of 37 °C, what was the initial temperature of the air in degrees Celsius?

ANSWER

a. 486 mL b. 30. °C PRACTICE PROBLEMS

Try Practice Problems 11.25 to 11.32

PRACTICE PROBLEMS

11.3 Temperature and Volume (Charles’s Law)

11.25 Select the diagram that shows the final volume of a balloon when each of the following changes are made when the pressure and amount of gas do not change:

Initial volume

A B C

a. The temperature is changed from 100 K to 300 K. b. The balloon is placed in a freezer. c. The balloon is first warmed and then returned to its starting

temperature.

11.26 Indicate whether the final volume of gas in each of the following is the same, larger, or smaller than the initial volume, if pressure and amount of gas do not change:

a. A volume of 505 mL of air on a cold winter day at - 15 °C is breathed into the lungs, where body temperature is 37 °C.

b. The heater used to heat the air in a hot-air balloon is turned off.

c. A balloon filled with helium at the amusement park is left in a car on a hot day.

11.27 A sample of neon initially has a volume of 2.50 L at 15 °C. What final temperature, in degrees Celsius, is needed to change the volume of the gas to each of the following, if pressure and amount of gas do not change?

a. 5.00 L b. 1250 mL c. 7.50 L d. 3550 mL

11.28 A gas has a volume of 4.00 L at 0 °C. What final temperature, in degrees Celsius, is needed to change the volume of the gas to each of the following, if pressure and amount of gas do not change?

a. 1.50 L b. 1200 mL c. 10.0 L d. 50.0 mL

11.29 A balloon contains 2500 mL of helium gas at 75 °C. What is the final volume, in milliliters, of the gas when the temperature changes to each of the following, if pressure and amount of gas do not change?

a. 55 °C b. 680. K c. - 25 °C d. 240. K 11.30 An air bubble has a volume of 0.500 L at 18 °C. What is the

final volume, in liters, of the gas when the temperature changes to each of the following, if pressure and amount of gas do not change?

a. 0 °C b. 425 K c. - 12 °C d. 575 K 11.31 A gas sample has a volume of 0.256 L with an unknown tempera-

ture. The same gas has a volume of 0.198 L when the temperature is 32 °C, with no change in the pressure and amount of gas. What was the initial temperature, in degrees Celsius, of the gas?

11.32 A gas sample has a temperature of 22 °C with an unknown volume. The same gas has a volume of 456 mL when the temperature is 86 °C, with no change in the pressure and amount of gas. What was the initial volume, in milliliters, of the gas?

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322 CHAPTER 11 Gases

11.4 Temperature and Pressure (Gay-Lussac’s Law) LEARNING GOAL Use the temperature–pressure relationship (Gay-Lussac’s law) to calculate the unknown temperature or pressure when the volume and amount of gas do not change.

If we could observe the molecules of a gas as the temperature rises, we would notice that they move faster and hit the sides of the container more often and with greater force. If volume and amount of gas do not change, the pressure would increase. In the temperature– pressure relationship known as Gay-Lussac’s law, the pressure of a gas is directly related to its Kelvin temperature. This means that an increase in temperature increases the pressure of a gas, and a decrease in temperature decreases the pressure of the gas as long as the volume and amount of gas do not change (see FIGURE 11.6).

Gay-Lussac’s Law

P1 T1

= P2 T2

No change in volume and amount of gas

All temperatures used in gas law calculations must be converted to their corresponding Kelvin (K) temperatures.

FIGURE 11.6 Gay-Lussac’s law: When the Kelvin temperature of a gas is doubled and the volume and amount of gas do not change, the pressure also doubles.

T1 = 200 K T2 = 400 K P1 = 1 atm P2 = 2 atm

SAMPLE PROBLEM 11.5 Calculating Pressure When Temperature Changes

TRY IT FIRST

Home oxygen tanks can be dangerous if they are heated, because they can explode. Suppose an oxygen tank has a pressure of 120 atm at a room temperature of 25 °C. If a fire in the room causes the temperature of the gas inside the oxygen tank to reach 402 °C, what is its pressure, in atmospheres, if the volume and amount of gas do not change? The oxygen tank may rupture if the pressure inside exceeds 180 atm. Would you expect it to rupture?

SOLUTION

STEP 1 State the given and needed quantities. We place the gas data in a table by writing the initial temperature and pressure as T1 and P1 and the final temperature and pressure as T2 and P2. We see that the temperature increases from 25 °C to 402 °C. Using Gay-Lussac’s law, we predict that the pressure increases.

T1 = 25 °C + 273 = 298 K T2 = 402 °C + 273 = 675 K

ENGAGE 11.5 Why can we predict that the pressure in the oxygen tank will increase when the temperature increases, if the volume and amount of gas do not change?

ANALYZE THE PROBLEM

Given Need Connect

P1 = 120 atm T1 = 298 K T2 = 675 K

Factors that do not change: V and n

P2 Gay-Lussac’s law, P1 T1

= P2 T2

Predict: P increases

STEP 2 Rearrange the gas law equation to solve for the unknown quantity. Using Gay-Lussac’s law, we solve for P2 by multiplying both sides by T2.

P1 T1

= P2 T2

P1 T1

* T2 = P2 T2

* T2

P2 = P1 * T2 T1

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11.4 Temperature and Pressure (Gay-Lussac’s Law) 323

STEP 3 Substitute values into the gas law equation and calculate. When we substitute in the values, we see that the ratio of the temperatures (temperature factor) is greater than 1, which increases the pressure as predicted.

P2 = 120 atm * 675 K 298 K

= 270 atm

Cylinders of oxygen gas are placed in a hospital storage room.

Vapor Pressure and Boiling Point When liquid molecules with sufficient kinetic energy break away from the surface, they become gas particles or vapor. In an open container, all the liquid will eventually evaporate. In a closed container, the vapor accumulates and creates pressure called vapor pressure. Each liquid exerts its own vapor pressure at a given temperature. As temperature increases, more vapor forms, and vapor pressure increases. TABLE 11.4 lists the vapor pressure of water at various temperatures.

A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure. As boiling occurs, bubbles of the gas form within the liquid and quickly rise to the

760 mmHg

100 °C

Atmospheric pressure

760 mmHg

TABLE 11.4 Vapor Pressure of Water

Temperature (°C)

Vapor Pressure (mmHg)

0 5

10 9

20 18

30 32

37* 47

40 55

50 93

60 149

70 234

80 355

90 528

100 760

*Normal body temperature

Temperature factor

increases pressure

Because the calculated pressure of 270 atm exceeds the limit of 180 atm, we would expect the oxygen tank to rupture.

SELF TEST 11.5

In a storage area of a hospital where the temperature has reached 55 °C, the pressure of oxygen gas in a 15.0-L steel cylinder is 965 Torr.

a. To what temperature, in degrees Celsius, would the gas have to be cooled to reduce the pressure to 850. Torr, when the volume and the amount of the gas do not change?

b. What is the final pressure, in millimeters of mercury, when the temperature of the oxygen gas drops to 24 °C, and the volume and the amount of the gas do not change?

ANSWER

a. 16 °C b. 874 mmHg

Vapour pressure in the bubble equals the atmospheric pressure

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324 CHAPTER 11 Gases

surface. For example, at an atmospheric pressure of 760 mmHg, water will boil at 100 °C, the temperature at which its vapor pressure reaches 760 mmHg.

At high altitudes, where atmospheric pressures are lower than 760 mmHg, the boiling point of water is lower than 100 °C. For example, a typical atmospheric pressure in Denver is 630 mmHg. This means that water in Denver boils when the vapor pressure is 630 mmHg. From TABLE 11.5, we see that water has a vapor pressure of 630 mmHg at 95 °C, which means that water boils at 95 °C in Denver.

In a closed container such as a pressure cooker, a pressure greater than 1 atm can be obtained, which means that water boils at a temperature higher than 100 °C. Laboratories and hospitals use closed containers called autoclaves to sterilize laboratory and surgical equipment.

An autoclave used to sterilize equipment attains a temperature higher than 100 °C.

Water

Pressure (mmHg)

Boiling Point (°C)

270 70

467 87

630 95

752 99

760 100

900 105

1075 110

1520 (2 atm) 120

3800 (5 atm) 160

7600 (10 atm) 180

TABLE 11.5 Pressure and the Boiling Point of

PRACTICE PROBLEMS

11.4 Temperature and Pressure (Gay-Lussac’s Law)

11.33 Calculate the final pressure, in millimeters of mercury, for each of the following, if volume and amount of gas do not change:

a. A gas with an initial pressure of 1200 Torr at 155 °C is cooled to 0 °C.

b. A gas in an aerosol can at an initial pressure of 1.40 atm at 12 °C is heated to 35 °C.

11.34 Calculate the final pressure, in atmospheres, for each of the following, if volume and amount of gas do not change:

a. A gas with an initial pressure of 1.20 atm at 75 °C is cooled to - 32 °C.

b. A sample of N2 with an initial pressure of 780. mmHg at - 75 °C is heated to 28 °C.

11.35 Calculate the final temperature, in degrees Celsius, for each of the following, if volume and amount of gas do not change:

a. A sample of xenon gas at 25 °C and 740. mmHg is cooled to give a pressure of 620. mmHg.

b. A tank of argon gas with a pressure of 0.950 atm at - 18 °C is heated to give a pressure of 1250 Torr.

11.36 Calculate the final temperature, in degrees Celsius, for each of the following, if volume and amount of gas do not change:

a. A sample of helium gas with a pressure of 250 Torr at 0 °C is heated to give a pressure of 1500 Torr.

b. A sample of air at 40 °C and 740. mmHg is cooled to give a pressure of 680. mmHg.

11.37 A gas sample has a pressure of 744 mmHg when the tempera- ture is 22 °C. What is the final temperature, in degrees Celsius, when the pressure is 766 mmHg, with no change in the volume and amount of gas?

11.38 A gas sample has a pressure of 2.35 atm when the temperature is - 15 °C. What is the final pressure, in atmospheres, when the temperature is 46 °C, with no change in the volume and amount of gas?

11.39 Explain each of the following observations: a. Water boils at 87 °C on the top of Mount Whitney. b. Food cooks more quickly in a pressure cooker than in an

open pan.

11.40 Explain each of the following observations: a. Boiling water at sea level is hotter than boiling water in the

mountains. b. Water used to sterilize surgical equipment is heated to

120 °C at 2.0 atm in an autoclave.

Applications

11.41 A tank contains isoflurane, an inhaled anesthetic, at a pressure of 1.8 atm and 5 °C. What is the pressure, in atmospheres, if the gas is warmed to a temperature of 22 °C, if volume and amount of gas do not change?

11.42 Bacteria and viruses are inactivated by temperatures above 135 °C. An autoclave contains steam at 1.00 atm and 100 °C. What is the pressure, in atmospheres, when the temperature of the steam in the autoclave reaches 135 °C, if volume and amount of gas do not change?

PRACTICE PROBLEMS Try Practice Problems 11.33 to 11.42

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11.5 The Combined Gas Law 325

11.5 The Combined Gas Law LEARNING GOAL Use the combined gas law to calculate the unknown pressure, volume, or temperature of a gas when changes in two of these properties are given and the amount of gas does not change.

All of the pressure–volume–temperature relationships for gases that we have studied may be combined into a single relationship called the combined gas law. This expression is useful for studying the effect of changes in two of these variables on the third as long as the amount of gas (number of moles) does not change.

Combined Gas Law

= T1

P1V1 P2V2 T2

No change in number of moles of gas

By using the combined gas law, we can derive any of the gas laws by omitting those properties that do not change, as seen in TABLE 11.6.

ENGAGE 11.6 Why does the pressure of a gas decrease to one-fourth of its initial pressure when the volume of the gas doubles and the Kelvin temperature decreases by half, if the amount of gas does not change?

TABLE 11.6 Summary of Gas Laws

Combined Gas Law Properties That Do Not Change Relationship Name of Gas Law

P1V1 T1

= P2V2 T2

T, n P1V1 = P2V2 Boyle’s law

P1V1 T1

= P2V2 T2

P, n V1 T1

= V2 T2

Charles’s law

P1V1 T1

= P2V2 T2

V, n P1 T1

= P2 T2

Gay-Lussac’s law

SAMPLE PROBLEM 11.6 Using the Combined Gas Law

TRY IT FIRST

A 25.0-mL bubble is released from a diver’s air tank at a pressure of 4.00 atm and a tem- perature of 11 °C. What is the volume, in milliliters, of the bubble when it reaches the ocean surface where the pressure is 1.00 atm and the temperature is 18 °C? (Assume the amount of gas in the bubble does not change.)

SOLUTION

STEP 1 State the given and needed quantities. We list the properties that change, which are the pressure, volume, and temperature. The temperatures in degrees Celsius must be changed to kelvins.

T1 = 11 °C + 273 = 284 K T2 = 18 °C + 273 = 291 K

Under water, the pressure on a diver is greater than the atmospheric pressure.

ANALYZE THE PROBLEM

Given Need Connect

P1 = 4.00 atm P2 = 1.00 atm V1 = 25.0 mL T1 = 284 K T2 = 291 K

Factor that does not change: n

V2 combined gas law, P1V1 T1

= P2V2 T2

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326 CHAPTER 11 Gases

STEP 2 Rearrange the gas law equation to solve for the unknown quantity. Using the combined gas law, we solve for V2 by multiplying both sides by T2 and dividing both sides by P2.

P1V1 T1

= P2V2 T2

P1V1 T1

* T2 P2

= P2V2 T2

* T2 P2

V2 = V1 * P1 P2

* T2 T1

STEP 3 Substitute values into the gas law equation and calculate. From the data table, we determine that both the pressure decrease and the temperature increase will increase the volume.

V2 = 25.0 mL * 4.00 atm 1.00 atm

* 291 K 284 K

= 102 mL

ENGAGE 11.7 Rearrange the combined gas law to solve for T1.

Pressure factor

increases volume

Temperature factor

increases volume

However, in situations where the unknown value is decreased by one change but increased by the second change, it is difficult to predict the overall change for the unknown.

SELF TEST 11.6

A weather balloon is filled with 15.0 L of helium at a temperature of 25 °C and a pressure of 685 mmHg.

a. What is the pressure, in millimeters of mercury, of the helium in the balloon in the upper atmosphere when the final temperature is - 35 °C and the final volume becomes 34.0 L, if the amount of gas does not change?

b. What is the temperature, in degrees Celsius, if the helium in the balloon has a final vol- ume of 22.0 L and a final pressure of 426 mmHg, if the amount of gas does not change?

ANSWER

a. 241 mmHg b. - 1 °C PRACTICE PROBLEMS

Try Practice Problems 11.43 to 11.48

PRACTICE PROBLEMS

11.5 The Combined Gas Law

11.43 Rearrange the variables in the combined gas law to solve for T2.

11.44 Rearrange the variables in the combined gas law to solve for P2.

11.45 A sample of helium gas has a volume of 6.50 L at a pressure of 845 mmHg and a temperature of 25 °C. What is the final pressure of the gas, in atmospheres, when the volume and tem- perature of the gas sample are changed to the following, if the amount of gas does not change?

a. 1850 mL and 325 K b. 2.25 L and 12 °C c. 12.8 L and 47 °C

11.46 A sample of argon gas has a volume of 735 mL at a pressure of 1.20 atm and a temperature of 112 °C. What is the final volume of the gas, in milliliters, when the pressure and tem- perature of the gas sample are changed to the following, if the amount of gas does not change?

a. 658 mmHg and 281 K b. 0.55 atm and 75 °C c. 15.4 atm and - 15 °C

Applications

11.47 A 124-mL bubble of hot gas initially at 212 °C and 1.80 atm is emitted from an active volcano. What is the final temperature, in degrees Celsius, of the gas in the bubble outside the volcano when the final volume of the bubble is 138 mL and the pres- sure is 0.800 atm, if the amount of gas does not change?

11.48 A scuba diver 60 ft below the ocean surface inhales 50.0 mL of compressed air from a scuba tank at a pressure of 3.00 atm and a temperature of 8 °C. What is the final pressure of air, in atmospheres, in the lungs when the gas expands to 150.0 mL at a body temperature of 37 °C, if the amount of gas does not change?

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11.6 Volume and Moles (Avogadro’s Law) 327

FIGURE 11.7 Avogadro’s law: The volume of a gas is directly related to the number of moles of the gas. If the number of moles is doubled, the volume must double when temperature and pressure do not change.

n1 = 1 mol V1 = 1 L

n2 = 2 mol V2 = 2 L

11.6 Volume and Moles (Avogadro’s Law) LEARNING GOAL Use Avogadro’s law to calculate the unknown amount or volume of a gas when the pressure and temperature do not change.

In our study of the gas laws, we have looked at changes in properties for a specified amount (n) of gas. Now we consider how the properties of a gas change when there is a change in the number of moles or grams of the gas.

When you blow up a balloon, its volume increases because you add more air molecules. If the balloon has a small hole in it, air leaks out, causing its volume to decrease. In 1811, Amedeo Avogadro formulated Avogadro’s law, which states that the volume of a gas is directly related to the number of moles of a gas when temperature and pressure do not change. For example, if the number of moles of a gas is doubled, then the volume will also double as long as we do not change the pressure or the temperature (see FIGURE 11.7). When pressure and temperature do not change, we can write Avogadro’s law as follows:

Avogadro’s Law

V1 n1

= V2 n2

No change in pressure and temperature

REVIEW Using Molar Mass as a Conversion

Factor (7.3)

SAMPLE PROBLEM 11.7 Calculating Volume for a Change in Moles

TRY IT FIRST

A weather balloon with a volume of 44 L is filled with 2.0 mol of helium. What is the final volume, in liters, if helium is added to give a total of 5.0 mol of helium, if the pressure and temperature do not change?

SOLUTION

STEP 1 State the given and needed quantities. We list those properties that change, which are volume and amount (moles). Because there is an increase in the number of moles of gas, we can predict that the volume increases.

ANALYZE THE PROBLEM

Given Need Connect

V1 = 44 L n1 = 2.0 mol n2 = 5.0 mol

Factors that do not change: P and T

V2 Avogadro’s law, V1 n1

= V2 n2

Predict: V increases

STEP 2 Rearrange the gas law equation to solve for the unknown quantity. Using Avogadro’s law, we can solve for V2 by multiplying both sides of the equation by n2.

V1 n1

= V2 n2

V1 n1

* n2 = V2 n2

* n2

V2 = V1 * n2 n1

STEP 3 Substitute values into the gas law equation and calculate. When we substitute in the values, we see that the ratio of the moles (mole factor) is greater than 1, which increases the volume as predicted.

V2 = 44 L * 5.0 mol 2.0 mol

= 110 L

ENGAGE 11.8 Why can we predict that the volume of a balloon increases when the number of moles of gas increases, if the pressure and temperature do not change?

Mole factor increases volume

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328 CHAPTER 11 Gases

SAMPLE PROBLEM 11.8 Using Molar Volume

TRY IT FIRST

What is the volume, in liters, of 64.0 g of O2 gas at STP?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

64.0 g of O2(g) at STP

liters of O2 gas at STP

molar mass, molar volume (STP)

ENGAGE 11.9 Why would the molar volume of a gas be greater than 22.4 L if the pressure is 1 atm and the tempera- ture is 100 °C?

SELF TEST 11.7

A balloon containing 8.00 g of oxygen gas has a volume of 5.00 L.

a. What is the volume, in liters, after 4.00 g of oxygen gas is added to the 8.00 g of oxygen in the balloon, if the temperature and pressure do not change?

b. More oxygen gas is added to the balloon until the volume is 12.6 L. How many grams of oxygen gas are in the balloon, if the pressure and temperature do not change?

ANSWER

a. 7.50 L b. 20.2 g of oxygen PRACTICE PROBLEMS

Try Practice Problems 11.49 to 11.52

STP and Molar Volume Using Avogadro’s law, we can say that any two gases will have equal volumes if they contain the same number of moles of gas at the same temperature and pressure. To help us make comparisons between gases, arbitrary conditions called standard temperature (273 K) and standard pressure (1 atm), together abbreviated STP, were selected by scientists:

STP Conditions

Standard temperature is exactly 0 °C (273 K). Standard pressure is exactly 1 atm (760 mmHg).

At STP, 1 mol of any gas occupies a volume of 22.4 L, which is about the same as the volume of three basketballs. This volume, 22.4 L, of any gas is called the molar volume (see FIGURE 11.8). When a gas is at STP conditions (0 °C and 1 atm), its molar volume can be used to write conversion factors between the number of moles of gas and its volume, in liters.

The molar volume of a gas at STP is about the same as the volume of three basketballs.

Molar Volume Conversion Factors

1 mol of gas = 22.4 L (STP)

22.4 L (STP)

1 mol gas and

1 mol gas

22.4 L (STP)

FIGURE 11.8 Avogadro’s law indicates that 1 mol of any gas at STP has a volume of 22.4 L.

1 mol of He 4.003 g of He

273 K 1 atm

1 mol of O2 32.00 g of O2

273 K 1 atm

1 mol of N2 28.02 g of N2

273 K 1 atm

V = 22.4 L

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11.7 The Ideal Gas Law 329

a. A sample of 0.500 mol of O2 is added to the 4.80 g of O2 in the container.

b. A sample of 2.00 g of O2 is removed. c. A sample of 4.00 g of O2 is added to the 4.80 g of O2 gas in

the container.

11.53 Use the molar volume to calculate each of the following at STP:

a. the number of moles of O2 in 2.24 L of O2 gas b. the volume, in liters, occupied by 2.50 mol of N2 gas c. the volume, in liters, occupied by 50.0 g of Ar gas d. the number of grams of H2 in 1620 mL of H2 gas

11.54 Use the molar volume to calculate each of the following at STP:

a. the number of moles of CO2 in 4.00 L of CO2 gas b. the volume, in liters, occupied by 0.420 mol of He gas c. the volume, in liters, occupied by 6.40 g of O2 gas d. the number of grams of Ne contained in 11.2 L of Ne gas

PRACTICE PROBLEMS

11.6 Volume and Moles (Avogadro’s Law)

11.49 What happens to the volume of a bicycle tire or a basketball when you use an air pump to add air?

11.50 Sometimes when you blow up a balloon and release it, it flies around the room. What is happening to the air in the balloon and its volume?

11.51 A sample containing 1.50 mol of Ne gas has an initial volume of 11.00 L. What is the final volume, in liters, when each of the following occurs and pressure and temperature do not change?

a. A leak allows one-half of Ne atoms to escape. b. A sample of 3.50 mol of Ne is added to the 1.50 mol of Ne

gas in the container. c. A sample of 25.0 g of Ne is added to the 1.50 mol of Ne

gas in the container.

11.52 A sample containing 4.80 g of O2 gas has an initial volume of 15.0 L. What is the final volume, in liters, when each of the following occurs and pressure and temperature do not change?

STEP 2 Write a plan to calculate the needed quantity.

grams of O2 moles of O2Molar mass liters of O2Molar volume

STEP 3 Write the equalities and conversion factors including 22.4 L/mol at STP.

32.00 g O2 and

1 mol of O2 = 32.00 g of O2

1 mol O2

1 mol O2 32.00 g O2

22.4 L O2 (STP) and

1 mol of O2 = 22.4 L of O2 (STP)

1 mol O2

1 mol O2 22.4 L O2 (STP)

STEP 4 Set up the problem with factors to cancel units.

64.0 g O2 44.8 L of O2 (STP)* = 22.4 L O2 (STP)1 mol O2

32.00 g O2 *

1 mol O2

SELF TEST 11.8

a. How many grams of Cl2(g) are in 5.00 L of Cl2(g) at STP? b. What is the volume, in liters, of 42.4 g of O2(g) at STP?

ANSWER

a. 15.8 g of Cl2 b. 29.7 L of O2

PRACTICE PROBLEMS Try Practice Problems 11.53 and 11.54

11.7 The Ideal Gas Law LEARNING GOAL Use the ideal gas law equation to calculate the unknown P, V, T, or n of a gas when given three of the four values in the ideal gas law equation. Calculate the molar mass of a gas.

The ideal gas law is the relationship between the four properties used in the measurement of a gas—pressure (P), volume (V), temperature (T), and amount of a gas (n).

Ideal Gas Law

PV = nRT

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330 CHAPTER 11 Gases

Rearranging the ideal gas law equation shows that the four gas properties equal the ideal gas constant, R.

PV nT

= R

To calculate the value of R, we substitute the STP conditions for molar volume into the expression: 1.00 mol of any gas occupies 22.4 L at STP (273 K and 1.00 atm).

R = (1.00 atm)(22.4 L)

(1.00 mol)(273 K) =

0.0821 L # atm mol # K

The value for R is 0.0821 L # atm per mol # K. If we use 760. mmHg for the pressure, we obtain another useful value for R of 62.4 L # mmHg per mol # K.

R = (760. mmHg)(22.4 L)

(1.00 mol)(273 K) =

62.4 L # mmHg mol # K

The ideal gas law is a useful expression when you are given the quantities for any three of the four properties of a gas. Although real gases show some deviations in behavior, the ideal gas law closely approximates the behavior of real gases at typical conditions. In working problems using the ideal gas law, the units of each variable must match the units in the R you select.

ENGAGE 11.10 When you use the value of 0.0821 for R, what unit must you use for pressure?

Ideal Gas Constant (R) 0.0821 L # atm

mol # K 62.4 L # mmHg

mol # K Pressure (P) atm mmHg

Volume (V) L L

Amount (n) mol mol

Temperature (T) K K

CORE CHEMISTRY SKILL Using the Ideal Gas Law

SAMPLE PROBLEM 11.9 Using the Ideal Gas Law

TRY IT FIRST

Dinitrogen oxide, N2O, which is used in dentistry, is an anesthetic called laughing gas. What is the pressure, in atmospheres, of 0.350 mol of N2O gas at 22 °C in a 5.00-L container?

SOLUTION

STEP 1 State the given and needed quantities. When three of the four quantities (P, V, n, and T) are known, we use the ideal gas law equation to solve for the unknown quantity. It is helpful to organize the data in a table. The temperature is converted from degrees Celsius to kelvins so that the units of V, n, and T match the unit of the gas constant, R.

ANALYZE THE PROBLEM

Given Need Connect

V = 5.00 L n = 0.350 mol T = 22 °C + 273 = 295 K

P ideal gas law, PV = nRT

R = 0.0821 L # atm

mol # K STEP 2 Rearrange the ideal gas law equation to solve for the needed quantity.

By dividing both sides of the ideal gas law equation by V, we solve for pressure, P.

PV = nRT Ideal gas law equation

P V V

= nRT V

P = nRT V

Dinitrogen oxide is used as an anesthetic in dentistry.

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11.7 The Ideal Gas Law 331

STEP 3 Substitute the gas data into the equation, and calculate the needed quantity.

P = 0.350 mol *

0.0821 L # atm mol # K * 295 K

5.00 L = 1.70 atm

SELF TEST 11.9

a. Chlorine gas, Cl2, is used to purify water. How many moles of chlorine gas are in a 7.00-L tank if the gas has a pressure of 865 mmHg and a temperature of 24 °C?

b. Isobutane gas, C4H10, is used as a propellant in aerosol cans of whipping cream, shav- ing cream, and suntan spray. What is the pressure, in atmospheres, of 0.125 mol of isobutane in an aerosol can at 21 °C with a volume of 465 mL?

ANSWER

a. 0.327 mol of Cl2 b. 6.49 atm

SAMPLE PROBLEM 11.10 Calculating Mass Using the Ideal Gas Law

TRY IT FIRST

Butane, C4H10, is used as a fuel for camping stoves. If you have 108 mL of butane gas at 715 mmHg and 25 °C, what is the mass, in grams, of butane?

SOLUTION

STEP 1 State the given and needed quantities. When three of the quantities (P, V, and T) are known, we use the ideal gas law equation to solve for the unknown quantity moles (n). Because the pressure is given in mmHg, we will use R in mmHg. The volume given in milliliters (mL) is converted to a volume in liters (L). The tem- perature is converted from degrees Celsius to kelvins.

ANALYZE THE

PROBLEM

Given Need Connect

P = 715 mmHg V = 108 mL (0.108 L) T = 25 °C + 273 = 298 K

n ideal gas law, PV = nRT

R = 62.4 L # mmHg

mol # K molar mass

STEP 2 Rearrange the ideal gas law equation to solve for the needed quantity. By dividing both sides of the ideal gas law equation by RT, we solve for moles, n.

PV = n RT Ideal gas law equation

PV RT

= n RT RT

n = PV RT

Many times we need to know the amount of gas, in grams. Then the ideal gas law equation can be rearranged to solve for the amount (n) of gas, which is converted to mass in grams using its molar mass as shown in Sample Problem 11.10.

When camping, butane is used as a fuel for a portable stove.

An aerosol spray can contains a gas the propels the contents.

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332 CHAPTER 11 Gases

STEP 3 Substitute the gas data into the equation and calculate the needed quantity.

n = 715 mmHg * 0.108 L

62.4 L # mmHg mol # K * 298 K

= 0.004 15 mol (4.15 * 10-3 mol)

Now we can convert the moles of butane to grams using its molar mass of 58.12 g/mol:

0.004 15 mol C4H10 * 58.12 g C4H10 1 mol C4H10

= 0.241 g of C4H10

SELF TEST 11.10

a. What is the volume, in liters, of 1.20 g of carbon monoxide at 8 °C, if it has a pressure of 724 mmHg?

b. What is the pressure, in atmospheres, of 1.23 g of CH4 gas in a 3.22-L container at 37 °C?

ANSWER

a. 1.04 L b. 0.606 atm

PRACTICE PROBLEMS Try Practice Problems 11.55 to 11.60, 11.63 and 11.64

Molar Mass of a Gas Another use of the ideal gas law is to determine the molar mass of a gas. If the mass, in grams, of the gas is known, we calculate the number of moles of the gas using the ideal gas law equation. Then the molar mass (g/mol) can be determined.

STEP 2 Rearrange the ideal gas law equation to solve for the number of moles. To solve for moles, n, divide both sides of the ideal gas law equation by RT.

PV = n RT Ideal gas law equation

PV RT

= n RT RT

ANALYZE THE PROBLEM

Given Need Connect

P = 0.750 atm V = 2.05 L T = 45 °C + 273 = 318 K mass = 3.16 g

n molar mass

ideal gas law, PV = nRT

R = 0.0821 L # atm

mol # K

SAMPLE PROBLEM 11.11 Molar Mass of a Gas Using the Ideal Gas Law

TRY IT FIRST

What is the molar mass, in grams per mole, of a gas if a 3.16-g sample of gas at 0.750 atm and 45 °C occupies a volume of 2.05 L?

SOLUTION

STEP 1 State the given and needed quantities.

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11.7 The Ideal Gas Law 333

n = PV RT

n = 0.750 atm * 2.05 L

0.0821 L # atm mol # K * 318 K

= 0.0589 mol

STEP 3 Obtain the molar mass by dividing the given number of grams by the number of moles.

Molar mass = mass moles

= 3.16 g

0.0589 mol = 53.7 g/mol

SELF TEST 11.11

a. What is the molar mass, in grams per mole, of an unknown gas in a 1.50-L container, if 0.488 g of the gas has a pressure of 0.0750 atm at 19 °C?

b. What is the molar mass, in grams per mole, of an unknown gas with a volume of 0.583 L, when 0.213 g of the gas has a pressure of 46.5 mmHg at 25 °C?

ANSWER

a. 104 g/mol b. 146 g/mol

PRACTICE PROBLEMS Try Practice Problems 11.61 and 11.62

PRACTICE PROBLEMS

11.7 The Ideal Gas Law

11.55 Calculate the pressure, in atmospheres, of 2.00 mol of helium gas in a 10.0-L container at 27 °C.

11.56 What is the volume, in liters, of 4.00 mol of methane gas, CH4, at 18 °C and 1.40 atm?

11.57 An oxygen gas container has a volume of 20.0 L. How many grams of oxygen are in the container, if the gas has a pressure of 845 mmHg at 22 °C?

11.58 A 10.0-g sample of krypton has a temperature of 25 °C at 575 mmHg. What is the volume, in milliliters, of the krypton gas?

11.59 A 25.0-g sample of nitrogen, N2, has a volume of 50.0 L and a pressure of 630. mmHg. What is the temperature, in kelvins and degrees Celsius, of the gas?

11.60 A 0.226-g sample of carbon dioxide, CO2, has a volume of 525 mL and a pressure of 455 mmHg. What is the tempera- ture, in kelvins and degrees Celsius, of the gas?

11.61 Determine the molar mass of each of the following gases: a. 0.84 g of a gas that occupies 450 mL at 0 °C and 1.00 atm

(STP) b. 1.28 g of a gas that occupies 1.00 L at 0 °C and 760 mmHg

(STP)

c. 1.48 g of a gas that occupies 1.00 L at 685 mmHg and 22 °C d. 2.96 g of a gas that occupies 2.30 L at 0.95 atm and 24 °C

11.62 Determine the molar mass of each of the following gases: a. 2.90 g of a gas that occupies 0.500 L at 0 °C and 1.00 atm

(STP) b. 1.43 g of a gas that occupies 2.00 L at 0 °C and 760 mmHg

(STP) c. 0.726 g of a gas that occupies 855 mL at 1.20 atm and 18 °C d. 2.32 g of a gas that occupies 1.23 L at 685 mmHg and 25 °C

Applications 11.63 A single-patient hyperbaric chamber has a volume of 640 L.

At a temperature of 24 °C, how many grams of oxygen are needed to give a pressure of 1.6 atm?

11.64 A multipatient hyperbaric chamber has a volume of 3400 L. At a temperature of 22 °C, how many grams of oxygen are needed to give a pressure of 2.4 atm?

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334 CHAPTER 11 Gases

11.8 Gas Laws and Chemical Reactions LEARNING GOAL Calculate the mass or volume of a gas that reacts or forms in a chemical reaction.

Gases are involved as reactants and products in many chemical reactions. For example, we have seen that the combustion of organic fuels with oxygen gas produces carbon dioxide gas and water vapor. In combination reactions, we have seen that hydrogen gas and nitrogen gas react to form ammonia gas, and hydrogen gas and oxygen gas produce water. Typically, the information given for a gas in a reaction is its pressure (P), volume (V), and temperature (T). Then we can use the ideal gas law equation to determine the moles of a gas in a reaction. If we are given the number of moles for one of the gases in a reaction, we can use a mole–mole factor to determine the moles of any other substance.

CORE CHEMISTRY SKILL Calculating Mass or Volume of a

Gas in a Chemical Reaction

ANALYZE THE PROBLEM

Given Need Connect

25.0 g of CaCO3 P = 752 mmHg T = 24 °C + 273 = 297 K

V of CO2(g) ideal gas law, PV = nRT

R = 62.4 L # mmHg

mol # K molar mass

Equation

2HCl(aq) + CaCO3(s) h CO2(g) + H2O(I ) + CaCl2(aq)

STEP 2 Write a plan to convert the given quantity to the needed moles.

grams of CaCO3 moles of CaCO3Molar mass moles of CO2Mole–mole factor

STEP 3 Write the equalities and conversion factors for molar mass and mole–mole factors.

SAMPLE PROBLEM 11.12 Gases in Chemical Reactions

TRY IT FIRST

Limestone (CaCO3) reacts with HCl to produce carbon dioxide gas, water, and aqueous calcium chloride.

2HCl(aq) + CaCO3(s) h CO2(g) + H2O(l) + CaCl2(aq)

How many liters of CO2 are produced at 752 mmHg and 24 °C from a 25.0-g sample of limestone?

SOLUTION

STEP 1 State the given and needed quantities.

100.09 g CaCO3 and

1 mol CaCO3

1 mol CaCO3 100.09 g CaCO3

1 mol of CaCO3 = 100.09 g of CaCO3 1 mol CaCO3 and

1 mol CO2

1 mol CO2 1 mol CaCO3

1 mol of CaCO3 = 1 mol of CO2

STEP 4 Set up the problem to calculate moles of needed quantity.

25.0 g CaCO3 0.250 mol of CO2* = 1 mol CaCO3

100.09 g CaCO3

1 mol CO2* 1 mol CaCO3

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11.9 Partial Pressures (Dalton’s Law) 335

STEP 5 Convert the moles of needed quantity to volume using the ideal gas law equation.

V = nRT

P

V = 0.250 mol *

62.4 L # mmHg mol # K * 297 K

752 mmHg = 6.16 L of CO2

SELF TEST 11.12

a. If 12.8 g of aluminum reacts with HCl, how many liters of H2 would be formed at 715 mmHg and 19 °C?

6HCl(aq) + 2Al(s) h 3H2(g) + 2AlCl3(aq)

b. If 9.28 L of H2 is produced at 24 °C and 1.20 atm, how many grams of aluminum react?

ANSWER

a. 18.1 L of H2 b. 8.22 g of Al PRACTICE PROBLEMS

Try Practice Problems 11.65 to 11.70

When HCl reacts with aluminum, bubbles of H2 gas form.

PRACTICE PROBLEMS

11.8 Gas Laws and Chemical Reactions

11.65 HCl reacts with magnesium metal to produce hydrogen gas.

2HCl(aq) + Mg(s) h H2(g) + MgCl2(aq)

a. What volume, in liters, of hydrogen at 0 °C and 1.00 atm (STP) is released when 8.25 g of Mg reacts?

b. How many grams of magnesium are needed to prepare 5.00 L of H2 at 735 mmHg and 18 °C?

11.66 When heated to 350 °C at 0.950 atm, ammonium nitrate decomposes to produce nitrogen, water, and oxygen gases.

2NH4NO3(s) h ∆

2N2(g) + 4H2O(g) + O2(g)

a. How many liters of water vapor are produced when 25.8 g of NH4NO3 decomposes?

b. How many grams of NH4NO3 are needed to produce 10.0 L of oxygen?

11.67 Butane undergoes combustion when it reacts with oxygen to produce carbon dioxide and water. What volume, in liters, of oxygen is needed to react with 55.2 g of butane at 0.850 atm and 25 °C?

2C4H10(g) + 13O2(g) h ∆

8CO2(g) + 10H2O(g)

11.68 Potassium nitrate decomposes to potassium nitrite and oxygen. What volume, in liters, of O2 can be produced from the decom- position of 50.0 g of KNO3 at 35 °C and 1.19 atm?

2KNO3(s) h 2KNO2(s) + O2(g)

11.69 Aluminum and oxygen react to form aluminum oxide. How many liters of oxygen at 0 °C and 760 mmHg (STP) are required to completely react with 5.4 g of aluminum?

4Al(s) + 3O2(g) h 2Al2O3(s)

11.70 Nitrogen dioxide reacts with water to produce oxygen and ammonia. How many grams of NH3 can be produced when 4.00 L of NO2 reacts at 415 °C and 725 mmHg?

4NO2(g) + 6H2O(g) h 7O2(g) + 4NH3(g)

11.9 Partial Pressures (Dalton’s Law) LEARNING GOAL Use Dalton’s law of partial pressures to calculate the total pressure of a mixture of gases.

Many gas samples are a mixture of gases. For example, the air you breathe is a mixture of mostly oxygen and nitrogen gases. In gas mixtures, scientists observed that all gas particles behave in the same way. Therefore, the total pressure of the gases in a mixture is a result of the collisions of the gas particles regardless of what type of gas they are.

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336 CHAPTER 11 Gases

In a gas mixture, each gas exerts its partial pressure, which is the pressure it would exert if it were the only gas in the container. Dalton’s law states that the total pressure of a gas mixture is the sum of the partial pressures of the gases in the mixture.

Dalton’s Law

Ptotal = P1 + P2 + P3 + g

CORE CHEMISTRY SKILL Calculating Partial Pressure

Total pressure of

a gas mixture

= Sum of the partial pressures of the gases in the mixture

ENGAGE 11.11 Why does the combination of helium with a pressure of 2.0 atm and argon with a pressure of 4.0 atm produce a gas mixture with a total pressure of 6.0 atm?

Suppose we have two separate tanks, one filled with helium at a pressure of 2.0 atm and the other filled with argon at a pressure of 4.0 atm. When the gases are combined in a single tank with the same volume and temperature, the number of gas molecules, not the type of gas, determines the pressure in the container. The total pressure of the gas mixture would be 6.0 atm, which is the sum of their individual or partial pressures.

+

PHe = 2.0 atm PAr = 4.0 atm

Ptotal = PHe + PAr = 2.0 atm + 4.0 atm = 6.0 atm

The total pressure of two gases is the sum of their partial pressures.

Air Is a Gas Mixture The air you breathe is a mixture of gases. What we call the atmospheric pressure is actually the sum of the partial pressures of all the gases in the air. TABLE 11.7 lists partial pressures for the gases in air on a typical day.

Gas Partial Pressure (mmHg)

Percentage (%)

Nitrogen, N2 594 78.2

Oxygen, O2 160. 21.0

Carbon dioxide, CO2 Argon, Ar Water vapor, H2O

6 0.8

Total air 760. 100

TABLE 11.7 Typical Composition of Air

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11.9 Partial Pressures (Dalton’s Law) 337

Heart

Tissue cells Alveoli

Atmosphere PO2 = 160 PCO2 = 0.3

PCO2 = 40

PO2 = 100PCO2 7 50

PO2 6 30

PO2 = 40 PCO2 = 46

Oxygenated blood Deoxygenated blood

PO2 = 100 PCO2 = 40

Gas

Partial Pressure (mmHg)

Inspired Air Alveolar Air Expired Air

Nitrogen, N2 594 573 569

Oxygen, O2 160. 100. 116

Carbon dioxide, CO2 0.3 40. 28

Water vapor, H2O 5.7 47 47

Total 760. 760. 760.

TABLE 11.8 Partial Pressures of Gases During Breathing

Chemistry Link to Health Blood Gases

Our cells continuously use oxygen and produce carbon dioxide. Both gases move in and out of the lungs through the membranes of the alveoli, the tiny air sacs at the ends of the airways in the lungs. An exchange of gases occurs in which oxygen from the air diffuses into the lungs and into the blood, while carbon dioxide produced in the cells is carried to the lungs to be exhaled. In TABLE 11.8, partial pressures are given for the gases in air that we inhale (inspired air), air in the alveoli, and air that we exhale (expired air).

At sea level, oxygen normally has a partial pressure of 100 mmHg in the alveoli of the lungs. Because the partial pressure of oxygen in venous blood is 40 mmHg, oxygen diffuses from the alveoli into the

bloodstream. The oxygen combines with hemoglobin, which carries it to the tissues of the body where the partial pressure of oxygen can be very low, less than 30 mmHg. Oxygen diffuses from the blood, where the partial pressure of O2 is high, into the tissues where O2 pressure is low.

As oxygen is used in the cells of the body during metabolic pro- cesses, carbon dioxide is produced, so the partial pressure of CO2 may be as high as 50 mmHg or more. Carbon dioxide diffuses from the tissues into the bloodstream and is carried to the lungs. There it diffuses out of the blood, where CO2 has a partial pressure of 46 mmHg, into the alveoli, where the CO2 is at 40 mmHg and is exhaled. TABLE 11.9 gives the partial pressures of blood gases in the tissues and in oxygenated and deoxygenated blood.

Gas

Partial Pressure (mmHg)

Oxygenated Blood

Deoxygenated Blood Tissues

O2 100 40 30 or less

CO2 40 46 50 or greater

TABLE 11.9 Partial Pressures of Oxygen and Carbon Dioxide in Blood and Tissues

SAMPLE PROBLEM 11.13 Calculating the Partial Pressure of a Gas in a Mixture

TRY IT FIRST

A heliox breathing mixture of oxygen and helium is prepared for a patient with chronic obstructive pulmonary disease (COPD). The gas mixture has a total pressure of 7.00 atm. If the partial pressure of the oxygen in the tank is 1140 mmHg, what is the partial pressure, in atmospheres, of the helium in the breathing mixture?

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338 CHAPTER 11 Gases

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

Ptotal = 7.00 atm PO2 = 1140 mmHg

partial pressure of He

Dalton’s law

STEP 1 Write the equation for the sum of the partial pressures.

Ptotal = PO2 + PHe Dalton’s law

STEP 2 Rearrange the equation to solve for the unknown pressure. To solve for the partial pressure of helium (PHe), we rearrange the equation to give the following:

PHe = Ptotal - PO2 Convert units to match.

PO2 = 1140 mmHg * 1 atm

760 mmHg = 1.50 atm

STEP 3 Substitute known pressures into the equation, and calculate the unknown pressure.

PHe = Ptotal - PO2 PHe = 7.00 atm - 1.50 atm = 5.50 atm

SELF TEST 11.13

a. An anesthetic consists of a mixture of cyclopropane gas, C3H6, and oxygen gas, O2. If the mixture has a total pressure of 1.09 atm, and the partial pressure of the cyclopropane is 73 mmHg, what is the partial pressure, in millimeters of mercury, of the oxygen in the anesthetic?

b. Another anesthetic contains nitrous oxide gas, N2O, and oxygen gas, O2. If the mixture has a total pressure of 786 mmHg and the partial pressure of the nitrous oxide is 0.110 atm, what is the partial pressure, in millimeters of mercury, of the oxygen in the mixture?

ANSWER

a. 755 mmHg b. 702 mmHg

PRACTICE PROBLEMS Try Practice Problems 11.71 to 11.74, 11.77 to 11.80

Gases Collected Over Water In the laboratory, gases are often collected by bubbling them through water into a con- tainer (see FIGURE 11.9). In a reaction, magnesium (Mg) reacts with HCl to form H2 gas and MgCl2.

2HCl(aq) + Mg(s) h H2(g) + MgCl2(aq)

As hydrogen is produced during the reaction, it displaces some of the water in the container. Because of the vapor pressure of water, the gas that is collected is a mixture of hydrogen and water vapor. For our calculation, we need the pressure of the dry hydrogen gas. We use the vapor pressure of water (see Table 11.4) at the experimental temperature, and subtract it from the total gas pressure. Then we can use the ideal gas law to determine the moles or grams of the hydrogen gas that were collected.

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11.9 Partial Pressures (Dalton’s Law) 339

FIGURE 11.9 A gas from a reaction is collected by bubbling through water. Due to evaporation of water, the total pressure is equal to the partial pressure of the gas and the vapor pressure of water.

Patm =

Pgas

PH2O

Gas plus water vapor

+

Reacting metal

HCl

SAMPLE PROBLEM 11.14 Moles of Gas Collected Over Water

TRY IT FIRST

When magnesium reacts with HCl, a volume of 0.355 L of hydrogen gas is collected over water at 26 °C. The vapor pressure of water at 26 °C is 25 mmHg.

2HCl(aq) + Mg(s) h H2(g) + MgCl2(aq)

If the total pressure is 752 mmHg, how many moles of H2(g) were collected?

SOLUTION

ANALYZE THE

PROBLEM

Given Need Connect

V = 0.355 L of H2 P = 752 mmHg T = 26 °C + 273 = 299 K PH2O = 25 mmHg

moles of H2 n

Dalton’s law, ideal gas law, PV = nRT

R = 62.4 L # mmHg

mol # K

STEP 1 Obtain the vapor pressure of water. The vapor pressure of water at 26 °C is 25 mmHg.

STEP 2 Subtract the vapor pressure from the total gas pressure to give the partial pressure of the needed gas. Using Dalton’s law of partial pressures, deter- mine the partial pressure of H2.

Ptotal = PH2 + PH2O

Solving for the partial pressure of H2 gives

PH2 = Ptotal - PH2O PH2 = 752 mmHg - 25 mmHg

= 727 mmHg

ENGAGE 11.12 Why is the vapor pressure of water subtracted from the total pressure of a gas collected over water?

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340 CHAPTER 11 Gases

PRACTICE PROBLEMS Try Practice Problems 11.75 and 11.76

Chemistry Link to Health Hyperbaric Chambers

A burn patient may undergo treatment for burns and infections in a hyperbaric chamber, a device in which pressures can be obtained that are two to three times greater than atmospheric pressure. A greater oxygen pressure increases the level of dissolved oxygen in the blood and tissues, where it fights bacterial infections. High levels of oxygen are toxic to many strains of bacteria. The hyperbaric chamber may also be used during surgery, to help counteract carbon monoxide (CO) poisoning, and to treat some cancers.

The blood is normally capable of dissolving up to 95% of the oxygen. Thus, if the partial pressure of the oxygen in the hyperbaric chamber is 2280 mmHg (3 atm), about 2170 mmHg of oxygen can dissolve in the blood, saturating the tissues. In the treatment for carbon monoxide poisoning, oxygen at high pressure is used to displace the CO from the hemoglobin faster than breathing pure oxygen at 1 atm.

A patient undergoing treatment in a hyperbaric chamber must also undergo decompression (reduction of pressure) at a rate that slowly reduces the concentration of dissolved oxygen in the blood. If decompression is too rapid, the oxygen dissolved in the blood may form gas bubbles in the circulatory system.

Similarly, if a scuba diver does not decompress slowly, a condi- tion called the “bends” may occur. While below the surface of the ocean, a diver uses a breathing mixture with higher pressures. If there is nitrogen in the mixture, higher quantities of nitrogen gas

will dissolve in the blood. If the diver ascends to the surface too quickly, the dissolved nitrogen forms gas bubbles that can block a blood vessel and cut off the flow of blood in the joints and tissues of the body and be quite painful. A diver suffering from the bends is placed immediately into a hyperbaric chamber where pressure is first increased and then slowly decreased. The dissolved nitrogen can then diffuse through the lungs until atmospheric pressure is reached.

A hyperbaric chamber is used in the treatment of certain diseases.

STEP 3 Use the ideal gas law to convert P gas to moles of gas collected. By dividing both sides of the ideal gas law equation by RT, we solve for moles, n, of gas.

PV = nRT Ideal gas law equation

PV RT

= nRT RT

n = PV RT

Calculate the moles of H2 gas by placing the partial pressure of H2 (727 mmHg), volume of gas container (0.355 L), temperature (26 °C + 273 = 299 K), and R, using mmHg, into the ideal gas law equation.

n = 727 mmHg * 0.355 L

62.4 L # mmHg mol # K * 299 K

= 0.0138 mol of H2 (1.38 * 10-2 mol of H2)

SELF TEST 11.14

a. A 456-mL sample of oxygen gas (O2) is collected over water at a pressure of 744 mmHg and a temperature of 20. °C. How many grams of dry O2 are collected (see Table 11.4)?

b. A sample of nitrogen gas (N2) with a volume of 376 mL is collected over water at a pressure of 766 mmHg and a temperature of 30. °C. How many grams of dry N2 are collected (see Table 11.4)?

ANSWER

a. 0.579 g of O2 b. 0.409 g of N2

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Applications

11.77 An air sample in the lungs contains oxygen at 93 mmHg, nitrogen at 565 mmHg, carbon dioxide at 38 mmHg, and water vapor at 47 mmHg. What is the total pressure, in atmo- spheres, exerted by the gas mixture?

11.78 A nitrox II gas mixture for scuba diving contains oxygen gas at 53 atm and nitrogen gas at 94 atm. What is the total pres- sure, in torr, of the scuba gas mixture?

11.79 In certain lung ailments such as emphysema, there is a decrease in the ability of oxygen to diffuse into the blood.

a. How would the partial pressure of oxygen in the blood change?

b. Why does a person with severe emphysema sometimes use a portable oxygen tank?

11.80 Head trauma can affect the ability of a person to ventilate (breathe in and out).

a. What would happen to the partial pressures of oxygen and carbon dioxide in the blood if a person cannot properly ventilate?

b. When a person who cannot breathe properly is placed on a ventilator, an air mixture is delivered at pressures that are alternately above the air pressure in the person’s lung, and then below. How will this move oxygen gas into the lungs, and carbon dioxide out?

PRACTICE PROBLEMS

11.9 Partial Pressures (Dalton’s Law)

11.71 In a gas mixture, the partial pressures are nitrogen 425 Torr, oxygen 115 Torr, and helium 225 Torr. What is the total pres- sure, in torr, exerted by the gas mixture?

11.72 In a gas mixture, the partial pressures are argon 415 mmHg, neon 75 mmHg, and nitrogen 125 mmHg. What is the total pressure, in millimeters of mercury, exerted by the gas mixture?

11.73 A gas mixture containing oxygen, nitrogen, and helium exerts a total pressure of 925 Torr. If the partial pressures are oxygen 425 Torr and helium 75 Torr, what is the partial pressure, in torr, of the nitrogen in the mixture?

11.74 A gas mixture containing oxygen, nitrogen, and neon exerts a total pressure of 1.20 atm. If helium added to the mixture increases the pressure to 1.50 atm, what is the partial pressure, in atmospheres, of the helium?

11.75 When solid KClO3 is heated, it decomposes to give solid KCl and O2 gas. A volume of 256 mL of gas is collected over water at a total pressure of 765 mmHg and 24 °C. The vapor pressure of water at 24 °C is 22 mmHg.

2KClO3(s) h ∆

2KCl(s) + 3O2(g)

a. What was the partial pressure of the O2 gas? b. How many moles of O2 gas were produced in the reaction?

11.76 When Zn reacts with HCl solution, the products are H2 gas and ZnCl2. A volume of 425 mL of H2 gas is collected over water at a total pressure of 758 mmHg and 16 °C. The vapor pressure of water at 16 °C is 14 mmHg.

2HCl(aq) + Zn(s) h H2(g) + ZnCl2(aq)

a. What was the partial pressure of the H2 gas? b. How many moles of H2 gas were produced in the reaction?

UPDATE Exercise-Induced Asthma

Vigorous exercise can induce asthma, particularly in children. When Louisa had her asthma attack, her breathing became more rapid, the temperature within her airways increased, and the muscles around the bronchi

contracted, causing a narrowing of the airways. Louisa’s symptoms, which may occur within 5 to 20 min after the start of vigorous exercise, include shortness of breath, wheezing, and coughing.

Louisa now does several things to prevent exercise- induced asthma. She uses a pre-exercise inhaled

medication before she starts her activity. The medication relaxes the muscles that surround the airways and opens up the airways. Then she does a warm-up set of exercises. If pollen counts are high, she avoids exercising outdoors.

Applications

11.81 Louisa’s lung capacity was measured as 3.2 L at a body temperature of 37 °C and a pressure of 745 mmHg. What is her lung capacity, in liters, at STP?

11.82 Using the answer from problem 11.81, how many grams of nitrogen are in Louisa’s lungs at STP if air contains 78% nitrogen?

Update 341

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342 CHAPTER 11 Gases

CONCEPT MAP

Move Fast

Gas Mixture

Molar Mass of a Gas

Partial Pressures

Exert Pressure

Are Far Apart Molar Volume

Gas Laws

GASES

Kinetic Molecular Theory of Gases

P and V Boyle’s Law

V and T Charles’s Law Moles or Volume

of Gases P and T

Gay-Lussac’s Law

P, V, and T Combined Gas Law

PV = nRT Ideal Gas Law

states that gas particles

are described by

V and n Avogadro’s Law

volume and moles are related by

at STP gives

for a gas or

where gases exert

used to calculate

Reactions of Gases

used to find

in

behave according to

that relate

and

CHAPTER REVIEW

11.1 Properties of Gases LEARNING GOAL Describe the kinetic molecular theory of gases and the units of measurement used for gases. • In a gas, particles are so far apart and moving so

fast that their attractions are negligible. • A gas is described by the physical properties of

pressure (P), volume (V), temperature (T), and amount in moles (n).

• A gas exerts pressure, the force of the gas particles striking the surface of a container.

• Gas pressure is measured in units such as torr, mmHg, atm, and Pa.

11.2 Pressure and Volume (Boyle’s Law) LEARNING GOAL Use the pressure– volume relationship (Boyle’s law) to calculate the unknown pressure or volume when the temperature and amount of gas do not change. • The volume (V) of a gas changes inversely

with the pressure (P) of the gas, if there is no change in the temperature and the amount of gas.

P1V1 = P2V2

• The pressure increases if volume decreases; its pressure decreases if the volume increases.

11.3 Temperature and Volume (Charles’s Law) LEARNING GOAL Use the temperature– volume relationship (Charles’s law) to calculate the unknown temperature or volume when the pressure and amount of gas do not change. • The volume (V) of a gas is directly related to

its Kelvin temperature (T) when there is no change in the pressure and the amount of gas.

V1 T1

= V2 T2

• If the temperature of a gas increases, its volume increases; if its temperature decreases, the volume decreases.

11.4 Temperature and Pressure (Gay-Lussac’s Law) LEARNING GOAL Use the temperature– pressure relationship (Gay-Lussac’s law) to calculate the unknown temperature or pressure when the volume and amount of gas do not change. • The pressure (P) of a gas is directly related

to its Kelvin temperature (T) when there is no change in the volume and the amount of the gas.

P1 T1

= P2 T2

V1 = 4 L P1 = 1 atm

V2 = 2 L P2 = 2 atm

Piston

T1 = 200 K T2 = 400 K V1 = 1 L V2 = 2 L

T1 = 200 K T2 = 400 K P1 = 1 atm P2 = 2 atm

M11_TIMB8119_06_SE_C11.indd 342 11/30/18 8:06 AM

Key Terms 343

• As temperature of a gas increases, its pressure increases; if its temperature decreases, its pressure decreases.

• Vapor pressure is the pressure of the gas that forms when a liquid evaporates.

• At the boiling point of a liquid, the vapor pressure equals the external pressure.

11.5 The Combined Gas Law LEARNING GOAL Use the combined gas law to calculate the unknown pressure, volume, or temperature of a gas when changes in two of these properties are given and the amount of gas does not change. • The combined gas law is the relationship of

pressure (P), volume (V), and temperature (T) when the amount of gas does not change.

P1V1 T1

= P2V2 T2

• The combined gas law is used to determine the effect of changes in two of the variables on the third.

11.6 Volume and Moles (Avogadro’s Law) LEARNING GOAL Use Avogadro’s law to calculate the unknown amount or volume of a gas when the pressure and temperature do not change. • The volume (V) of a gas is directly related to the

number of moles (n) of the gas when the pres- sure and temperature of the gas do not change.

V1 n1

= V2 n2

• If the moles of gas increase, the volume must increase; if the moles of gas decrease, the volume must decrease.

• At standard temperature (273 K) and standard pressure (1 atm), abbreviated STP, 1 mol of any gas has a volume of 22.4 L.

11.7 The Ideal Gas Law LEARNING GOAL Use the ideal gas law equation to solve for P, V, T, or n of a gas when given three of the four values in the ideal gas law equation. Calculate the molar mass of a gas. • The ideal gas law gives the relationship of the quantities P, V, n,

and T that describe and measure a gas.

PV = nRT • Any of the four variables can be calculated if the values of the

other three are known. • The molar mass of a gas can be calculated using the ideal gas law

equation.

11.8 Gas Laws and Chemical Reactions LEARNING GOAL Calculate the mass or volume of a gas that reacts or forms in a chemical reaction. • The ideal gas law equation is used to convert the

quantities (P, V, and T) of gases to moles in a chemical reaction.

• The moles of gases can be used to determine the number of moles or grams of other substances in the reaction.

11.9 Partial Pressures (Dalton’s Law) LEARNING GOAL Use Dalton’s law of partial pressures to calculate the total pressure of a mixture of gases. • In a mixture of two or more gases, the total

pressure is the sum of the partial pressures of the individual gases.

Ptotal = P1 + P2 + P3 + g

• The partial pressure of a gas in a mixture is the pressure it would exert if it were the only gas in the container.

• For gases collected over water, the vapor pressure of water is subtracted from the total pressure of the gas mixture to obtain the partial pressure of the dry gas.

1 mol of O2 32.00 g of O2

273 K 1 atm

V = 22.4 L

Ptotal = PHe + PAr = 2.0 atm + 4.0 atm = 6.0 atm

atmosphere (atm) A unit equal to the pressure exerted by a column of mercury 760 mm high.

atmospheric pressure The pressure exerted by the atmosphere. Avogadro’s law A gas law stating that the volume of a gas is directly

related to the number of moles of gas when pressure and tem- perature do not change.

Boyle’s law A gas law stating that the pressure of a gas is inversely related to the volume when temperature and moles of the gas do not change.

Charles’s law A gas law stating that the volume of a gas is directly related to the Kelvin temperature when pressure and moles of the gas do not change.

combined gas law A relationship that combines several gas laws relating pressure, volume, and temperature when the amount of gas does not change. P1V1 T1

= P2V2 T2

Dalton’s law A gas law stating that the total pressure exerted by a mixture of gases in a container is the sum of the partial pressures that each gas would exert alone.

direct relationship A relationship in which two properties increase or decrease together.

Gay-Lussac’s law A gas law stating that the pressure of a gas is directly related to the Kelvin temperature when the number of moles of a gas and its volume do not change.

ideal gas constant, R A numerical value that relates the quantities P, V, n, and T in the ideal gas law, PV = nRT.

ideal gas law A law that combines the four measured properties of a gas. PV = nRT

inverse relationship A relationship in which two properties change in opposite directions.

KEY TERMS

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344 CHAPTER 11 Gases

kinetic molecular theory of gases A model used to explain the behavior of gases.

molar volume A volume of 22.4 L occupied by 1 mol of a gas at STP conditions of 0 °C (273 K) and 1 atm.

partial pressure The pressure exerted by a single gas in a gas mixture.

pressure The force exerted by gas particles that hit the walls of a container.

STP Standard conditions of exactly 0 °C (273 K) temperature and 1 atm pressure used for the comparison of gases.

vapor pressure The pressure exerted by the particles of vapor above a liquid.

The chapter section containing each Core Chemistry Skill is shown in parentheses at the end of each heading.

Using the Gas Laws (11.2) • Boyle’s, Charles’s, Gay-Lussac’s, and Avogadro’s laws show the

relationships between two properties of a gas.

P1V1 = P2V2 Boyle’s law

V1 T1

= V2 T2

Charles’s law

P1 T1

= P2 T2

Gay-Lussac’s law

V1 n1

= V2 n2

Avogadro’s law

• The combined gas law shows the relationship between P, V, and T for a gas.

P1V1 T1

= P2V2 T2

• When two properties of a gas vary and the other two do not change, we list the initial and final conditions of each property in a table.

Example: A sample of helium gas (He) has a volume of 6.8 L and a pressure of 2.5 atm. What is the final volume, in liters, if it has a final pressure of 1.2 atm with no change in tempera- ture and amount of gas?

Answer:

CORE CHEMISTRY SKILLS Example: What is the volume, in liters, of 0.750 mol of CO2

at a pressure of 1340 mmHg and a temperature of 295 K?

Answer: V = nRT

P

= 0.750 mol *

62.4 L # mmHg mol # K * 295 K

1340 mmHg = 10.3 L

Calculating Mass or Volume of a Gas in a Chemical Reaction (11.8) • The ideal gas law equation can be used to calculate the volume or

mass of a gas in a chemical reaction.

Example: What is the volume, in liters, of N2 required to react with 18.5 g of magnesium at a pressure of 1.20 atm and a tem- perature of 303 K?

3Mg(s) + N2(g) h Mg3N2(s)

Answer: Initially, we convert the grams of Mg to moles and use a mole–mole factor from the balanced equation to calculate the moles of N2 gas.

18.5 g Mg * 1 mol Mg

24.31 g Mg *

1 mol N2 3 mol Mg

= 0.254 mol of N2

Now, we use the moles of N2 in the ideal gas law equation and solve for liters, the needed quantity.

V = nRT

P

= 0.254 mol N2 *

0.0821 L # atm mol # K * 303 K

1.20 atm = 5.27 L

Calculating Partial Pressure (11.9) • In a gas mixture, each gas exerts its partial pressure, which

is the pressure it would exert if it were the only gas in the container.

• Dalton’s law states that the total pressure of a gas mixture is the sum of the partial pressures of the gases in the mixture.

Ptotal = P1 + P2 + P3 + g

Example: A gas mixture with a total pressure of 1.18 atm contains helium gas at a partial pressure of 465 mmHg and nitrogen gas. What is the partial pressure, in atmospheres, of the nitrogen gas?

ANALYZE THE PROBLEM

Given Need Connect

P1 = 2.5 atm P2 = 1.2 atm V1 = 6.8 L

V2 Boyle’s law, P1V1 = P2V2

Factors that do not change: T and n

Predict: V increases

Using Boyle’s law, we can write the relationship for V2,

which we predict will increase.

V2 = V1 * P1 P2

V2 = 6.8 L * 2.5 atm 1.2 atm

= 14 L

Using the Ideal Gas Law (11.7) • The ideal gas law equation combines the relationships of the four

properties of a gas into one equation.

PV = nRT

• When three of the four properties are given, we rearrange the ideal gas law equation for the needed quantity.

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Understanding the Concepts 345

UNDERSTANDING THE CONCEPTS

The chapter sections to review are shown in parentheses at the end of each problem.

11.83 Two flasks of equal volume and at the same temperature contain different gases. One flask contains 10.0 g of Ne, and the other flask contains 10.0 g of He. Is each of the following statements true or false? Explain. (11.1)

a. The flask that contains He has a higher pressure than the flask that contains Ne.

b. The densities of the gases are the same.

11.84 Two flasks of equal volume and at the same temperature contain different gases. One flask contains 5.0 g of O2, and the other flask contains 5.0 g of H2. Is each of the following statements true or false? Explain. (11.1)

a. Both flasks contain the same number of molecules. b. The pressures in the flasks are the same.

11.85 At 100 °C, which of the following diagrams (1, 2, or 3) represents a gas sample that exerts the: (11.1)

a. lowest pressure? b. highest pressure?

11.87 A balloon is filled with helium gas with a partial pressure of 1.00 atm and neon gas with a partial pressure of 0.50 atm. For each of the following changes (a to e) of the initial balloon, select the diagram (A, B, or C) that shows the final volume of the balloon: (11.2, 11.3, 11.6)

Answer: Initially, we convert the partial pressure of helium gas from mmHg to atm.

465 mmHg * 1 atm

760 mmHg = 0.612 atm of He gas

Using Dalton’s law, we solve for the needed quantity, PN2.

Ptotal = PN2 + PHe

PN2 = Ptotal - PHe

PN2 = 1.18 atm - 0.612 atm = 0.57 atm

2 31Initial gas

1 2 3

Initial volume

A B C

11.86 Indicate which diagram (1, 2, or 3) represents the volume of the gas sample in a flexible container when each of the following changes (a to d) takes place: (11.2, 11.3)

a. Temperature increases, if pressure does not change. b. Temperature decreases, if pressure does not change. c. Atmospheric pressure decreases, if temperature does not

change. d. Doubling the atmospheric pressure and doubling the

Kelvin temperature.

a. The balloon is put in a cold storage unit (pressure and amount of gas do not change).

b. The balloon floats to a higher altitude where the pressure is less (temperature and amount of gas do not change).

c. All of the neon gas is removed (temperature and pressure do not change).

d. The Kelvin temperature doubles and half of the gas atoms leak out (pressure does not change).

e. 2.0 mol of O2 gas is added (temperature and pressure do not change).

11.88 Indicate if pressure increases, decreases, or stays the same in each of the following: (11.2, 11.4, 11.6)

a.

b.

c.

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346 CHAPTER 11 Gases

ADDITIONAL PRACTICE PROBLEMS

11.89 A gas sample has a volume of 4250 mL at 15 °C and 745 mmHg. What is the final temperature, in degrees Celsius, after the sample is transferred to a different container with a volume of 2.50 L and a pressure of 1.20 atm, when the amount of gas does not change? (11.5)

11.90 In the fermentation of glucose (wine making), 780 mL of CO2 gas was produced at 37 °C and 1.00 atm. What is the final volume, in liters, of the gas when measured at 22 °C and 675 mmHg, when the amount of gas does not change? (11.5)

11.91 During laparoscopic surgery, carbon dioxide gas is used to expand the abdomen to help create a larger working space. If 4.80 L of CO2 gas at 18 °C at 785 mmHg is used, what is the final volume, in liters, of the gas at 37 °C and a pressure of 745 mmHg, if the amount of CO2 does not change? (11.5)

11.92 A weather balloon has a volume of 750 L when filled with helium at 8 °C at a pressure of 380 Torr. What is the final volume, in liters, of the balloon when the pressure is 0.20 atm, the temperature is - 45 °C, and the amount of gas does not change? (11.5)

11.93 In 1783, Jacques Charles launched his first balloon filled with hydrogen gas (H2), which he chose because it was lighter than air. The balloon had a volume of 31 000 L when it was filled at a pressure of 755 mmHg and a temperature of 22 °C. How many kilograms of hydrogen were needed to fill the balloon? (11.7)

11.97 A 2.00-L container is filled with methane gas, CH4, at a pressure of 2500. mmHg and a temperature of 18 °C. How many grams of methane are in the container? (11.7)

11.98 A steel cylinder with a volume of 15.0 L is filled with 50.0 g of nitrogen gas at 25 °C. What is the pressure, in atmospheres, of the N2 gas in the cylinder? (11.7)

11.99 A sample of gas with a mass of 1.62 g occupies a volume of 941 mL at a pressure of 748 Torr and a temperature of 20.0 °C. What is the molar mass of the gas? (11.7)

11.100 What is the molar mass of a gas if 1.15 g of the gas has a volume of 8 mL at 0 °C and 1.00 atm (STP)? (11.7)

11.101 How many grams of CO2 are in 35.0 L of CO2(g) at 1.20 atm and 5 °C? (11.7)

11.102 A container is filled with 0.644 g of O2 at 5 °C and 845 mmHg. What is the volume, in milliliters, of the container? (11.7)

11.103 How many liters of H2 gas can be produced at 0 °C and 1.00 atm (STP) from 25.0 g of Zn? (11.7, 11.8)

2HCl(aq) + Zn(s) h H2(g) + ZnCl2(aq)

11.104 In the formation of smog, nitrogen and oxygen gas react to form nitrogen dioxide. How many grams of NO2 will be produced when 2.0 L of nitrogen at 840 mmHg and 24 °C are completely reacted? (11.7, 11.8)

N2(g) + 2O2(g) h 2NO2(g)

11.105 Nitrogen dioxide reacts with water to produce oxygen and ammonia. A 5.00-L sample of H2O(g) reacts at a temperature of 375 °C and a pressure of 725 mmHg. How many grams of NH3 can be produced? (11.7, 11.8)

4NO2(g) + 6H2O(g) h 7O2(g) + 4NH3(g)

11.106 Hydrogen gas can be produced in the laboratory through the reaction of magnesium metal with hydrochloric acid. When 12.0 g of Mg reacts, what volume, in liters, of H2 gas is produced at 24 °C and 835 mmHg? (11.7, 11.8)

2HCl(aq) + Mg(s) h H2(g) + MgCl2(aq)

11.107 A gas mixture with a total pressure of 2400 Torr is used by a scuba diver. If the mixture contains 2.0 mol of helium and 6.0 mol of oxygen, what is the partial pressure, in torr, of each gas in the sample? (11.9)

11.108 A gas mixture with a total pressure of 4.6 atm is used in a hospital. If the mixture contains 5.4 mol of nitrogen and 1.4 mol of oxygen, what is the partial pressure, in atmospheres, of each gas in the sample? (11.9)

11.109 A gas mixture contains oxygen and argon at partial pressures of 0.60 atm and 425 mmHg. If nitrogen gas added to the sample increases the total pressure to 1250 Torr, what is the partial pressure, in torr, of the nitrogen added? (11.9)

11.110 What is the total pressure, in millimeters of mercury, of a gas mixture containing argon gas at 0.25 atm, helium gas at 350 mmHg, and nitrogen gas at 360 Torr? (11.9)

Jacques Charles used hydrogen to launch his balloon in 1783.

11.94 When the balloon in problem 11.93 reached an altitude of 1000 m, the pressure was 658 mmHg and the temperature was - 8 °C. What is the volume of the balloon at these conditions, if the amount of hydrogen does not change? (11.5)

11.95 You are doing research on planet X. The temperature inside the space station is a carefully controlled 24 °C, and the pressure is 755 mmHg. Suppose that a balloon, which has a volume of 850. mL inside the space station, is placed into the airlock and floats out to planet X. If planet X has an atmo- spheric pressure of 0.150 atm and the volume of the balloon changes to 3.22 L, what is the temperature, in degrees Celsius, on planet X (the amount of gas does not change)? (11.5)

11.96 Your spaceship has docked at a space station above Mars. The temperature inside the space station is a carefully controlled 24 °C at a pressure of 745 mmHg. A balloon with a volume of 425 mL drifts into the airlock where the temperature is - 95 °C and the pressure is 0.115 atm. What is the final vol- ume, in milliliters, of the balloon if the amount of gas does not change and the balloon is very elastic? (11.5)

M11_TIMB8119_06_SE_C11.indd 346 11/30/18 8:06 AM

Answers to Engage Questions 347

The following problems are related to the topics in this chapter. However, they do not all follow the chapter order, and they require you to combine concepts and skills from several sections. These problems will help you increase your critical thinking skills and prepare for your next exam.

11.111 Solid aluminum reacts with aqueous H2SO4 to form H2 gas and aluminum sulfate. When a sample of Al is allowed to react, 415 mL of gas is collected over water at 23 °C, at a pressure of 755 mmHg. At 23 °C, the vapor pressure of water is 21 mmHg. (11.7, 11.8, 11.9)

3H2SO4(aq) + 2Al(s) h 3H2(g) + Al2(SO4)3(aq) a. What is the pressure, in millimeters of mercury, of the

dry H2 gas? b. How many moles of H2 were produced? c. How many grams of Al were reacted?

11.112 When heated, KClO3 forms KCl and O2. When a sample of KClO3 is heated, 226 mL of gas with a pressure of 744 mmHg is collected over water at 26 °C. At 26 °C, the vapor pressure of water is 25 mmHg. (11.7, 11.8, 11.9)

2KClO3(s) h ∆

2KCl(s) + 3O2(g) a. What is the pressure, in millimeters of mercury, of the

dry O2 gas? b. How many moles of O2 were produced? c. How many grams of KClO3 were reacted?

11.113 A sample of gas with a mass of 1.020 g occupies a volume of 762 mL at 0 °C and 1.00 atm (STP). What is the molar mass of the gas? (11.7)

11.114 A sample of an unknown gas with a mass of 3.24 g occupies a volume of 1.88 L at a pressure of 748 mmHg and a tem- perature of 20. °C. What is the molar mass of the gas? (11.7)

Applications

11.115 The propane, C3H8, in a fuel cylinder, undergoes combustion with oxygen in the air. How many liters of CO2 are produced at STP if the cylinder contains 881 g of propane? (11.8)

C3H8(g) + 5O2(g) h ∆

3CO2(g) + 4H2O(g)

11.116 When sensors in a car detect a collision, they cause the reaction of sodium azide, NaN3, which generates nitro- gen gas to fill the airbags within 0.03 s. How many liters of N2 are produced at STP if the airbag contains 132 g of NaN3? (11.8)

2NaN3(s) h 2Na(s) + 3N2(g)

11.117 Glucose, C6H12O6, is metabolized in living systems. How many grams of water can be produced from the reaction of 18.0 g of glucose and 7.50 L of O2 at 1.00 atm and 37 °C? (11.7, 11.8)

C6H12O6(s) + 6O2(g) h 6CO2(g) + 6H2O(l)

11.118 2.00 L of N2, at 25 °C and 1.08 atm, is mixed with 4.00 L of O2, at 25 °C and 0.118 atm, and the mixture is allowed to react. How much NO, in grams, is produced? (11.7, 11.8)

N2(g) + O2(g) h 2NO(g)

11.119 Hyperbaric therapy uses 100% oxygen at pressure to help heal wounds and infections, and to treat carbon monoxide poisoning. If the pressure inside a hyperbaric chamber is 3.0 atm, what is the volume, in liters, of the chamber containing 2400 g of O2 at 28 °C? (11.7)

11.120 A hyperbaric chamber has a volume of 1510 L. How many kilograms of O2 gas are needed to give an oxygen pressure of 2.04 atm at 25 °C? (11.7)

11.121 Laparoscopic surgery involves inflating the abdomen with carbon dioxide gas to separate the internal organs and the abdominal wall. If the CO2 injected into the abdomen pro- duces a pressure of 20. mmHg and a volume of 4.00 L at 32 °C, how many grams of CO2 were used? (11.7)

11.122 In another laparoscopic surgery, helium is used to inflate the abdomen to separate the internal organs and the abdominal wall. If the He injected into the abdomen produces a pres- sure of 15 mmHg and a volume of 3.1 L at 28 °C, how many grams of He were used? (11.7)

11.123 At a restaurant, a customer chokes on a piece of food. You put your arms around the person’s waist and use your fists to push up on the person’s abdomen, an action called the Heimlich maneuver. (11.2)

a. How would this action change the volume of the chest and lungs?

b. Why does it cause the person to expel the food item from the airway?

11.124 An airplane is pressurized with air to 650. mmHg. (11.9) a. If air is 21% oxygen, what is the partial pressure of

oxygen on the plane? b. If the partial pressure of oxygen drops below

100. mmHg, passengers become drowsy. If this happens, oxygen masks are released. What is the total cabin pressure at which oxygen masks are dropped?

CHALLENGE PROBLEMS

ANSWERS TO ENGAGE QUESTIONS 11.4 According to Charles’s law, the volume of a gas increases when

the temperature increases, if the pressure and amount of gas do not change.

11.5 According to Gay-Lussac’s law, the pressure of a gas increases when the temperature increases, if the volume and amount of gas do not change.

11.6 The pressure of a gas decreases to one-half when its volume doubles, if the amount of gas does not change. When the

11.1 According to the kinetic molecular theory, the molecules of a gas are moving rapidly in all directions to completely fill the size and shape of its container.

11.2 At a higher altitude there are less gas molecules, which means there is a lower pressure.

11.3 According to Boyle’s law, the pressure of a gas decreases when the volume increases, if the temperature and amount of gas do not change.

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348 CHAPTER 11 Gases

11.9 Molar volume increases when the temperature increases from 0 °C to 100 °C.

11.10 If you use the value of 0.0821 for R, the pressure units must be atmospheres.

11.11 According to Dalton’s law, partial pressures are additive. The total pressure in a container with 2.0 atm of helium and 4.0 atm of argon would be 6.0 atm.

11.12 According to Dalton’s law, the total pressure of the gases col- lected over water includes the partial pressure of water vapor. To obtain the partial pressure of a gas, the partial pressure of water is subtracted from the total pressure.

Kelvin temperature of that gas also decreases by one-half, its pressure decreases to half. The total decrease in pressure is one-half times one-half or one-fourth of the initial pressure of the gas.

11.7 T1 = T2 * P1 P2

* V1 V2

11.8 According to Avogadro’s law, the volume of a balloon increases when the number of moles (or molecules) of gas in the balloon increases, if the pressure and temperature of the gas do not change.

ANSWERS TO SELECTED PROBLEMS 11.37 31 °C

11.39 a. On top of a mountain, water boils below 100 °C because the atmospheric (external) pressure is less than 1 atm.

b. Because the pressure inside a pressure cooker is greater than 1 atm, water boils above 100 °C. At a higher tempera- ture, food cooks faster.

11.41 1.9 atm

11.43 T2 = T1 * P2 P1

* V2 V1

11.45 a. 4.26 atm b. 3.07 atm c. 0.606 atm

11.47 - 33 °C 11.49 The volume increases because the number of gas particles is

increased.

11.51 a. 4.00 L b. 26.7 L c. 14.6 L

11.53 a. 0.100 mol of O2 b. 56.0 L c. 28.0 L d. 0.146 g of H2 11.55 4.93 atm

11.57 29.4 g of O2 11.59 566 K (293 °C)

11.61 a. 42 g/mol b. 28.7 g/mol c. 39.8 g/mol d. 33 g/mol

11.63 1300 g of O2 11.65 a. 7.60 L of H2 b. 4.92 g of Mg

11.67 178 L of O2 11.69 3.4 L of O2 11.71 765 Torr

11.73 425 Torr

11.75 a. 743 mmHg b. 0.0103 mol of O2 11.77 0.978 atm

11.79 a. The partial pressure of oxygen will be lower than normal. b. Breathing a higher concentration of oxygen will help to

increase the supply of oxygen in the lungs and blood and raise the partial pressure of oxygen in the blood.

11.81 2.8 L

11.1 a. At a higher temperature, gas particles have greater kinetic energy, which makes them move faster.

b. Because there are great distances between the particles of a gas, they can be pushed closer together and still remain a gas.

c. Gas particles are very far apart, which means that the mass of a gas in a certain volume is very small, resulting in a low density.

11.3 a. temperature b. volume c. amount d. pressure

11.5 Statements a, d, and e describe the pressure of a gas.

11.7 a. 1520 Torr b. 29.4 lb/in.2

c. 1520 mmHg d. 203 kPa

11.9 As a diver ascends to the surface, external pressure decreases. If the air in the lungs were not exhaled, its volume would expand and severely damage the lungs. The pressure in the lungs must adjust to changes in the external pressure.

11.11 a. The pressure is greater in cylinder A. According to Boyle’s law, a decrease in volume pushes the gas particles closer together, which will cause an increase in the pressure.

b. 160 mL

11.13 a. increases b. decreases c. increases

11.15 a. 328 mmHg b. 2620 mmHg c. 475 mmHg d. 5240 mmHg

11.17 a. 52.4 L b. 25.0 L c. 100. L d. 45 L

11.19 6.52 atm

11.21 25 L of cyclopropane

11.23 a. inspiration b. expiration c. inspiration

11.25 a. C b. A c. B

11.27 a. 303 °C b. - 129 °C c. 591 °C d. 136 °C

11.29 a. 2400 mL b. 4900 mL c. 1800 mL d. 1700 mL

11.31 121 °C

11.33 a. 770 mmHg b. 1150 mmHg

11.35 a. - 23 °C b. 168 °C

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Answers to Selected Problems 349

11.105 1.02 g of NH3 11.107 He 600 Torr, O2 1800 Torr

11.109 370 Torr

11.111 a. 734 mmHg b. 0.0165 mol of H2 c. 0.297 g of Al

11.113 30.0 g/mol

11.115 1340 L of CO2 11.117 5.31 g of water

11.119 620 L

11.121 0.18 g of CO2 11.123 a. By pushing in on the abdomen, the volume of the chest

and lungs is decreased. b. Because the volume is decreased, Boyle’s law states that

the pressure is increased, and this will expel the food blocking the airway.

11.83 a. True. The flask containing helium has more moles of helium and thus more helium atoms.

b. True. The mass and volume of each are the same, which means the mass/volume ratio or density is the same in both flasks.

11.85 a. 2 b. 1

11.87 a. A b. C c. A

d. B e. C

11.89 - 66 °C 11.91 5.39 L of CO2 11.93 2.6 kg of H2 11.95 - 103 °C 11.97 4.41 g of CH4 11.99 42.1 g/mol

11.101 81.0 g of CO2 11.103 8.56 L of H2

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350

Our kidneys produce urine, which carries waste products and excess fluid from the body. They also reabsorb electrolytes, such as potassium, and produce hormones that regulate blood pressure and the levels of calcium in the blood. Diseases such as diabetes and high blood pressure can cause a decrease in kidney function. Symptoms of kidney malfunction include protein in the urine, an abnormal level of urea in the blood, frequent urination, and swollen feet. If kidney failure occurs, it may be treated with dialysis or transplantation.

Michelle suffers from kidney disease because of severe strep throat she contracted as a child. When her kidneys stopped functioning, Michelle was placed on dialysis three times a week. As she enters the dialysis unit, her dialysis nurse, Amanda, asks Michelle how she is feeling. Michelle indicates that she feels tired today and has considerable swelling around her ankles. Amanda informs her that these side effects occur because of her body’s inability to regulate the amount of water in her cells. Amanda explains that the retention of water is regulated by the concentration of electrolytes in her body fluids and the rate at which waste products are removed from her body. Amanda explains that although water is essential for the many chemical reactions that occur in the body, the amount of water can become too high or too low because of various diseases and conditions. Because Michelle’s kidneys no longer function properly, she cannot regulate the amount of electrolytes or waste in her body fluids. As a result, she has an electrolyte imbalance and a buildup of waste products, so her body retains water. Amanda then explains that the dialysis machine does the work of her kidneys to reduce the high levels of electrolytes and waste products.

CAREER

Dialysis Nurse A dialysis nurse specializes in assisting patients with kidney disease undergoing dialysis. This requires monitoring the patient before, during, and after dialysis for any complications such as a drop in blood pressure or cramping. The dialysis nurse connects the patient to the dialysis unit via a dialysis catheter that is inserted into the neck or chest. A dialysis nurse must have considerable knowledge about how the dialysis machine functions to ensure that it is operating correctly at all times.

Solutions

Michelle continues to have dialysis treatment three times a week. You can see more details of Michelle’s dialysis in the UPDATE Using Dialysis for Renal Failure, page 388, and discover that dialysis requires 120 L of fluid containing electrolytes to adjust Michelle’s blood level to that of normal serum.

UPDATE Using Dialysis for Renal Failure

12

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12.1 Solutions 351

12.1 Solutions LEARNING GOAL Identify the solute and solvent in a solution; describe the formation of a solution.

Solutions are everywhere around us. Most of the gases, liquids, and solids we see are mix- tures of at least one substance dissolved in another. There are different types of solutions. The air we breathe is a solution that is primarily oxygen and nitrogen gases. Carbon dioxide gas dissolved in water makes carbonated drinks. When we make solutions of coffee or tea, we use hot water to dissolve substances from coffee beans or tea leaves. The ocean is also a solution, consisting of many ionic compounds such as sodium chloride dissolved in water. In your medicine cabinet, the antiseptic tincture of iodine is a solution of iodine dissolved in ethanol.

A solution is a homogeneous mixture in which one substance, called the solute, is uni- formly dispersed in another substance called the solvent. Because the solute and the solvent do not react with each other, they can be mixed in varying proportions. A solution with a small amount of salt dissolved in water tastes slightly salty. When a large amount of salt is dissolved in water, the solution tastes very salty. Usually, the solute (in this case, salt) is the substance present in the lesser amount, whereas the solvent (in this case, water) is present in the greater amount. For example, in a solution composed of 5.0 g of salt and 50. g of water, salt is the solute and water is the solvent. In a solution, the particles of the solute are evenly dispersed among the molecules within the solvent (see FIGURE 12.1).

REVIEW Identifying Polarity of Molecules

(10.5)

Identifying Intermolecular Forces (10.6)

ENGAGE 12.1 What does the uniform blue color in the graduated cylinder on the right indicate?

FIGURE 12.1 A solution of copper(II) sulfate (CuSO4) forms as particles of solute dissolve, move away from the crystal, and become evenly dispersed among the solvent (water) molecules.

CuSO4

H2O

LOOKING AHEAD

12.1 Solutions  351 12.2 Electrolytes and

Nonelectrolytes  355 12.3 Solubility  357 12.4 Solution

Concentrations  363 12.5 Dilution of Solutions  371 12.6 Chemical Reactions in

Solution  374 12.7 Molality and Freezing

Point Lowering/Boiling Point Elevation  378

12.8 Properties of Solutions: Osmosis  385

A solution has at least one solute dispersed in a solvent.

Solute: The substance present in lesser amount

Solvent: The substance present in greater amount

Salt

Water

Types of Solutes and Solvents Solutes and solvents may be solids, liquids, or gases. The solution that forms has the same physical state as the solvent. When sugar crystals are dissolved in water, the resulting sugar solution is liquid. Sugar is the solute, and water is the solvent. Soda water and soft drinks are prepared by dissolving carbon dioxide gas in water. The carbon dioxide gas is the solute, and water is the solvent. TABLE 12.1 lists some solutions and their solutes and solvents.

Water as a Solvent Water is one of the most common solvents in nature. In the H2O molecule, an oxygen atom shares electrons with two hydrogen atoms. Because oxygen is much more electronegative than hydrogen, the O ¬ H bonds are polar. In each polar bond, the oxygen atom has a partial

M12_TIMB8119_06_SE_C12.indd 351 11/27/18 12:07 PM

352 CHAPTER 12 Solutions

Type Example Solute Solvent

Gas Solutions

Gas in a gas Air O2(g) N2(g)

Liquid Solutions

Gas in a liquid Soda water CO2(g) H2O(l)

Household ammonia NH3(g) H2O(l)

Liquid in a liquid Vinegar HC2H3O2(l) H2O(l)

Solid in a liquid Seawater NaCl(s) H2O(l)

Tincture of iodine I2(s) C2H6O(l)

Solid Solutions

Solid in a solid Brass Zn(s) Cu(s)

Steel C(s) Fe(s)

TABLE 12.1 Some Examples of Solutions

negative (d-) charge, and the hydrogen atom has a partial positive (d+) charge. Because the shape of a water molecule is bent, its dipoles do not cancel out. Thus, water is polar and is a polar solvent.

Attractive forces known as hydrogen bonds occur between molecules where partially positive hydrogen atoms are attracted to the partially negative atoms N, O, or F. As seen in the diagram, the hydrogen bonds are shown as a series of dots. Although hydrogen bonds are much weaker than covalent or ionic bonds, there are many of them linking water molecules together. Hydrogen bonds are important in the properties of biological compounds such as proteins, carbohydrates, and DNA.

PRACTICE PROBLEMS Try Practice Problems 12.1 and 12.2

O

O O

H

Hydrogen bonds

Partial positive charge on H

Partial negative charge on O

H

H

HH

H

d -

d -

d + d

+

In water, hydrogen bonds form between an oxygen atom in one water molecule and the hydrogen atom in another.

Chemistry Link to Health Water in the Body

The average adult is about 60% water by mass, and the average infant about 75%. About 60% of the body’s water is contained within the cells as intracellular f luids; the other 40% makes up extracellular fluids, which include the interstitial fluid in tissue and the plasma in the blood. These external fluids carry nutrients and waste materials between the cells and the circulatory system.

Every day you lose between 1500 and 3000 mL of water from the kidneys as urine, from the skin as perspiration, from the lungs as you exhale, and from the gastrointestinal tract. Serious dehydration

Water Gain

Liquid Food Metabolism

Total

1000 mL 1200 mL 300 mL

2500 mL

Water Loss Urine Perspiration Breath Feces

Total

1500 mL 300 mL 600 mL 100 mL

2500 mL

Typical water gain and loss during 24 hours

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12.1 Solutions 353

can occur in an adult if there is a 10% net loss in total body fluid; a 20% loss of fluid can be fatal. An infant suffers severe dehydration with only a 5 to 10% loss in body fluid.

Water loss is continually replaced by the liquids and foods in the diet and from metabolic processes that produce water in the cells of the body. TABLE 12.2 lists the percentage by mass of water con- tained in some foods.

The water lost from the body is replaced by the intake of fluids.

TABLE 12.2 Percentage of Water in Some Foods

Food Water (% by mass) Food

Water (% by mass)

Vegetables Meats/Fish

Carrot 88 Chicken, cooked 71

Celery 94 Hamburger, broiled 60

Cucumber 96 Salmon 71

Tomato 94

Fruits Milk Products

Apple 85 Cottage cheese 78

Cantaloupe 91 Milk, whole 87

Orange 86 Yogurt 88

Strawberry 90

Watermelon 93

Formation of Solutions The interactions between solute and solvent will determine whether a solution will form. Ini- tially, energy is needed to separate the particles in the solute and the solvent particles. Then energy is released as solute particles move between the solvent particles to form a solution. However, there must be attractions between the solute and the solvent particles to provide the energy for the initial separation. These attractions occur when the solute and the solvent have similar polarities. The expression “like dissolves like” is a way of saying that the polarities of a solute and a solvent must be similar for a solution to form (see FIGURE 12.2). In the absence of attractions between a solute and a solvent, there is insufficient energy to form a solution (see TABLE 12.3).

Solutions Will Form Solutions Will Not Form

Solute Solvent Solute Solvent

Polar Polar Polar Nonpolar

Nonpolar Nonpolar Nonpolar Polar

TABLE 12.3 Possible Combinations of Solutes and Solvents

FIGURE 12.2 Like dissolves like. In each test tube, the lower layer is CH2Cl2 (more dense), and the upper layer is water (less dense).

CH2Cl2 is nonpolar and water is polar; the two layers do not mix.

The nonpolar solute I2 (purple) is soluble in the nonpolar solvent CH2Cl2.

The ionic solute Ni(NO3)2 (green) is soluble in the polar solvent water.

Water (less dense)

CH2Cl2 (more dense)

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354 CHAPTER 12 Solutions

Solutions with Ionic and Polar Solutes In ionic solutes such as sodium chloride, NaCl, there are strong ionic bonds between posi- tively charged Na+ ions and negatively charged Cl- ions. In water, a polar solvent, the hydrogen bonds provide strong solvent–solvent attractions. When NaCl crystals are placed in water, partially negative oxygen atoms in water molecules attract positive Na+ ions, and the partially positive hydrogen atoms in other water molecules attract negative Cl- ions (see FIGURE 12.3). As soon as the Na+ ions and the Cl- ions form a solution, they undergo hydration as water molecules surround each ion. Hydration of the ions diminishes their attraction to other ions and keeps them in solution.

In the equation for the formation of the NaCl solution, the solid and aqueous NaCl are shown with the formula H2O over the arrow, which indicates that water is needed for the dissociation process but is not a reactant.

NaCl(s) Na+(aq) H2O + Cl-(aq)

Dissociation

In another example, we find that a polar molecular compound such as methanol, CH4O, is soluble in water because methanol has a polar ¬ OH group that forms hydrogen bonds with water (see FIGURE 12.4). Polar solutes require polar solvents for a solution to form.

ENGAGE 12.2 Why does KCl(s) form a solution with water, but nonpolar hexane (C6H14) does not form a solution with water?

PRACTICE PROBLEMS Try Practice Problems 12.3 to 12.6

FIGURE 12.3 Ions on the surface of a crystal of NaCl dissolve in water as they are attracted to the polar water molecules that pull the ions into solution and surround them.

O

O

O

O

OH

H H

H

H

H

H H

H

H

H H

O

H2O

Na+ Cl-

NaCl

Hydrated ions

+ +

+ +

+

+

+

+

-

- -

- -

-

-

-

FIGURE 12.4 Polar molecules of methanol, CH4O, form hydrogen bonds with polar water molecules to form a methanol–water solution.

Methanol (CH4O) solute Methanol–water solution with hydrogen bonding

Water solvent

Solutions with Nonpolar Solutes Compounds containing nonpolar molecules, such as iodine (I2), oil, or grease, do not dissolve in water because there are no attractions between the particles of a nonpolar solute and the polar solvent. Nonpolar solutes require nonpolar solvents for a solution to form.

PRACTICE PROBLEMS

12.1 Solutions

12.1 Identify the solute and the solvent in each solution composed of the following:

a. 10.0 g of NaCl and 100.0 g of H2O b. 50.0 mL of ethanol, C2H6O, and 10.0 mL of H2O c. 0.20 L of O2 and 0.80 L of N2

12.2 Identify the solute and the solvent in each solution composed of the following:

a. 10.0 mL of acetic acid, HC2H3O2, and 200. mL of H2O b. 100.0 mL of H2O and 5.0 g of sugar, C12H22O11 c. 1.0 g of Br2 and 50.0 mL of methylene chloride, CH2Cl2 12.3 Describe the formation of an aqueous KI solution, when solid

KI dissolves in water.

12.4 Describe the formation of an aqueous LiBr solution, when solid LiBr dissolves in water.

Applications 12.5 Water is a polar solvent and carbon tetrachloride (CCl4) is a

nonpolar solvent. In which solvent is each of the following, which is found or used in the body, more likely to be soluble?

a. CaCO3 (calcium supplement), ionic b. retinol (vitamin A), nonpolar c. sucrose (table sugar), polar d. cholesterol (lipid), nonpolar 12.6 Water is a polar solvent and hexane (C6H14) is a nonpolar

solvent. In which solvent is each of the following, which is found or used in the body, more likely to be soluble?

a. vegetable oil, nonpolar b. oleic acid (lipid), nonpolar c. niacin (vitamin B3), polar d. FeSO4 (iron supplement), ionic

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12.2 Electrolytes and Nonelectrolytes 355

Cl-

Cl-

Cl-Na+

Na+ Na+

- +

Strong electrolyte

HF

HF

F- H+

HF

HF

- +

Weak electrolyte

+-

CH4O

Nonelectrolyte

12.2 Electrolytes and Nonelectrolytes LEARNING GOAL Identify solutes as electrolytes or nonelectrolytes.

Solutes can be classified by their ability to conduct an electrical current. When electrolytes dissolve in water, the process of dissociation separates them into ions forming solutions that conduct electricity. When nonelectrolytes dissolve in water, they do not separate into ions, and their solutions do not conduct electricity.

To test solutions for the presence of ions, we can use an apparatus that consists of a battery and a pair of electrodes connected by wires to a light bulb. The light bulb glows when electricity can flow, which can only happen when electrolytes provide ions that move between the electrodes to complete the circuit.

Types of Electrolytes Electrolytes can be further classified as strong electrolytes or weak electrolytes. For a strong electrolyte, such as sodium chloride (NaCl), there is 100% dissociation of the solute into ions. When the electrodes from the light bulb apparatus are placed in the NaCl solution, the light bulb glows very brightly.

In an equation for dissociation of a compound in water, the charges must balance. For example, magnesium nitrate dissociates to give one magnesium ion for every two nitrate ions. However, only the ionic bonds between Mg2+ and NO3

- are broken, not the covalent bonds within the polyatomic ion. The equation for the dissociation of Mg(NO3)2 is written as follows:

Mg(NO3)2(s) Mg 2+(aq) 2NO3

-(aq)+ H2O

Dissociation

A weak electrolyte is a compound that dissolves in water mostly as molecules. Only a few of the dissolved solute molecules undergo dissociation, producing a small number of ions in solu- tion. Thus, solutions of weak electrolytes do not conduct electrical current as well as solutions of strong electrolytes. When the electrodes are placed in a solution of a weak electrolyte, the glow of the light bulb is very dim. In an aqueous solution of the weak electrolyte HF, a few HF molecules dissociate to produce H+ and F - ions. As more H+ and F - ions form, some recombine to give HF molecules. These forward and reverse reactions of molecules to ions and back again are indicated by two arrows between reactant and products that point in opposite directions:

HF(aq) H+(aq) Dissociation

Recombination F-(aq)+

A nonelectrolyte such as methanol (CH4O) dissolves in water only as molecules, which do not dissociate. When electrodes of the light bulb apparatus are placed in a solution of a nonelectrolyte, the light bulb does not glow because the solution does not contain ions and cannot conduct electricity.

CH4O(l) CH4O(aq) H2O

TABLE 12.4 summarizes the classification of solutes in aqueous solutions.

REVIEW Writing Conversion Factors from

Equalities (2.5)

Using Conversion Factors (2.6)

Writing Positive and Negative Ions (6.1)

Type of Solute In Solution Type(s) of Particles in Solution

Conducts Electricity? Examples

Strong electrolyte

Dissociates completely

Only ions Yes Ionic compounds such as NaCl, KBr, MgCl2, NaNO3; bases such as NaOH, KOH; acids such as HCl, HBr, HI, HNO3, HClO4, H2SO4

Weak electrolyte

Dissociates partially

Mostly molecules and a few ions

Weakly HF, H2O, NH3, HC2H3O2 (acetic acid)

Nonelectrolyte No dissocia- tion

Only molecules No Carbon compounds such as CH4O (methanol), C2H6O (ethanol), C12H22O11 (sucrose), CH4N2O (urea)

TABLE 12.4 Classification of Solutes in Aqueous Solutions

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356 CHAPTER 12 Solutions

An intravenous solution is used to replace electrolytes in the body.

Chemistry Link to Health Electrolytes in Body Fluids

Electrolytes in the body play an important role in maintaining the proper function of the cells and organs in the body. Typically, the electrolytes sodium, potassium, chloride, and bicarbonate are mea- sured in a blood test. Sodium ions regulate the water content in the body and are important in carrying electrical impulses through the nervous system. Potassium ions are also involved in the transmission of electrical impulses and play a role in the maintenance of a regular heartbeat. Chloride ions balance the charges of the positive ions and also control the balance of fluids in the body. Bicarbonate is important in maintaining the proper pH of the blood. Sometimes when vomit- ing, diarrhea, or sweating is excessive, the concentrations of certain electrolytes may decrease. Then fluids such as Pedialyte may be given to return electrolyte levels to normal.

SAMPLE PROBLEM 12.1 Solutions of Electrolytes and Nonelectrolytes

TRY IT FIRST

Indicate whether solutions of each of the following contain only ions, only molecules, or mostly molecules and a few ions. Write the equation for the formation of a solution for each of the following:

a. Na2SO4(s), a strong electrolyte b. sucrose, C12H22O11(s), a nonelectrolyte c. acetic acid, HC2H3O2(l ), a weak electrolyte

SOLUTION

a. An aqueous solution of Na2SO4(s) contains only the ions Na + and SO4

2-.

Na2SO4(s) 2Na +(aq) SO4

2-(aq)+ H2O

b. A nonelectrolyte such as sucrose, C12H22O11(s), produces only molecules when it dis- solves in water.

C12H22O11(s) C12H22O11(aq) H2O

c. A weak electrolyte such as HC2H3O2(l) produces mostly molecules and a few ions when it dissolves in water.

HC2H3O2(l) H +(aq) C2H3O2

-(aq)+ H2O

SELF TEST 12.1

a. Boric acid, H3BO3(s), is a weak electrolyte. Would you expect a boric acid solution to contain only ions, only molecules, or mostly molecules and a few ions?

b. Hydrochloric acid, HCl, is a strong electrolyte. Would you expect a solution of hydro- chloric acid to contain only ions, only molecules, or mostly molecules and a few ions?

ANSWER

a. A solution of a weak electrolyte would contain mostly molecules and a few ions. b. A solution of a strong electrolyte would contain only ions.

ENGAGE 12.3 Why does a solution of LiNO3, a strong electrolyte, contain only ions, whereas a solution of urea, CH4N2O, a nonelectrolyte, contains only molecules?

PRACTICE PROBLEMS Try Practice Problems 12.7 to 12.14

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12.3 Solubility 357

PRACTICE PROBLEMS

12.2 Electrolytes and Nonelectrolytes

12.7 KF is a strong electrolyte, and HF is a weak electrolyte. How is the solution of KF different from that of HF?

12.8 NaOH is a strong electrolyte, and CH4O is a nonelectrolyte. How is the solution of NaOH different from that of CH4O?

12.9 Write a balanced equation for the dissociation of each of the following strong electrolytes in water:

a. KCl b. CaCl2 c. K3PO4 d. Fe(NO3)3 12.10 Write a balanced equation for the dissociation of each of the

following strong electrolytes in water: a. LiBr b. NaNO3 c. CuCl2 d. K2CO3 12.11 Indicate whether aqueous solutions of each of the following

solutes contain only ions, only molecules, or mostly molecules and a few ions:

a. acetic acid, HC2H3O2, a weak electrolyte b. NaBr, a strong electrolyte c. fructose, C6H12O6, a nonelectrolyte

12.12 Indicate whether aqueous solutions of each of the following solutes contain only ions, only molecules, or mostly molecules and a few ions:

a. NH4Cl, a strong electrolyte b. ethanol, C2H6O, a nonelectrolyte c. hydrocyanic acid, HCN, a weak electrolyte

12.13 Classify the solute represented in each of the following equations as a strong, weak, or nonelectrolyte:

a.

K2SO4(s) 2K

+(aq) + SO42-(aq) H2O

b. NH3(g) H2O(l) NH4

+(aq) OH-(aq)++

c.

C6H12O6(s) C6H12O6(aq) H2O

12.14 Classify the solute represented in each of the following equations as a strong, weak, or nonelectrolyte:

a.

C4H10O(l) C4H10O(aq) H2O

b. MgCl2(s) Mg

2+(aq) + 2Cl-(aq) H2O

c. HClO(aq) H+(aq) + ClO-(aq)

H2O

Saturated solution

Unsaturated solution

Dissolved solute

Undissolved solute

D is

so lv

in g

C ry

st al

liz at

io n

Additional solute can dissolve in an unsaturated solution, but not in a saturated solution.

12.3 Solubility LEARNING GOAL Define solubility; distinguish between an unsaturated and a saturated solution. Identify an ionic compound as soluble or insoluble.

The term solubility is used to describe the amount of a solute that can dissolve in a given amount of solvent. Many factors, such as the type of solute, the type of solvent, and the temperature, affect the solubility of a solute. Solubility, usually expressed in grams of solute in 100. g of solvent, is the maximum amount of solute that can dissolve at a certain temperature. If a solute readily dissolves when added to the solvent, the solution does not contain the maximum amount of solute. We call this solution an unsaturated solution.

A solution that contains all the solute that can dissolve is a saturated solution. When a solution is saturated, the rate at which the solute dissolves becomes equal to the rate at which solid forms, a process known as crystallization. Then there is no further change in the amount of dissolved solute in solution.

Solute solvent+ saturated solution Solute dissolves

Solute crystallizes

We can prepare a saturated solution by adding an amount of solute greater than that needed to reach solubility. Stirring the solution will dissolve the maximum amount of solute and leave the excess on the bottom of the container. The addition of more solute to the saturated solution will only increase the amount of undissolved solute.

REVIEW Interpreting Graphs (1.4)

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358 CHAPTER 12 Solutions

Chemistry Link to Health Gout and Kidney Stones: Saturation in Body Fluids

The conditions of gout and kidney stones involve compounds in the body that exceed their solubility levels and form solid products. Gout affects adults, primarily men, over the age of 40. Attacks of gout may occur when the concentration of uric acid in blood plasma exceeds its solubility, which is 7 mg/100 mL of plasma at 37 °C. Insoluble deposits of needle-like crystals of uric acid can form in the cartilage, tendons, and soft tissues, where they cause painful gout attacks. They may also form in the tissues of the kidneys, where they can cause renal damage. High levels of uric acid in the body can be caused by an increase in uric acid production, failure of the kidneys to remove uric acid, or a diet with an overabundance of foods containing purines, which are metabolized to uric acid in the body. Foods in the diet that contribute to high levels of uric acid include certain meats, sardines, mushrooms, asparagus, and beans. Drinking alcoholic beverages may also significantly increase uric acid levels and bring about gout attacks.

Treatment for gout involves diet changes and drugs. Medications, such as probenecid, which helps the kidneys eliminate uric acid, or allopurinol, which blocks the production of uric acid by the body, may be useful.

Kidney stones are solid materials that form in the urinary tract. Most kidney stones are composed of calcium phosphate and calcium oxalate, although they can be solid uric acid. Insufficient water intake and high levels of calcium, oxalate, and phosphate in the urine can lead to the formation of kidney stones. When a kidney stone passes through the urinary tract, it causes considerable pain and discomfort, necessitat- ing the use of painkillers and surgery. Sometimes ultrasound is used to break up kidney stones. Persons prone to kidney stones are advised to drink six to eight glasses of water every day to prevent saturation levels of minerals in the urine.

SAMPLE PROBLEM 12.2 Saturated Solutions

TRY IT FIRST

At 20 °C, the solubility of KCl is 34 g/100. g of H2O. In the laboratory, a student mixes 75 g of KCl with 200. g of H2O at a temperature of 20 °C.

a. How much of the KCl will dissolve? b. Is the solution saturated or unsaturated? c. What is the mass, in grams, of any solid KCl left undissolved on the bottom of the

container?

SOLUTION

a. At 20 °C, KCl has a solubility of 34 g of KCl in 100. g of water. Using the solubility as a conversion factor, we can calculate the maximum amount of KCl that can dissolve in 200. g of water as follows:

200. g H2O * 34 g KCl

100. g H2O = 68 g of KCl

b. Because 75 g of KCl exceeds the maximum amount (68 g) that can dissolve in 200. g of water, the KCl solution is saturated.

c. If we add 75 g of KCl to 200. g of water and only 68 g of KCl can dissolve, there is 7 g (75 g - 68 g) of solid (undissolved) KCl on the bottom of the container.

SELF TEST 12.2

At 40 °C, the solubility of KNO3 is 65 g/100. g of H2O. a. How many grams of KNO3 will dissolve in 120 g of H2O at 40 °C? b. How many grams of H2O are needed to completely dissolve 110 g of KNO3 at

20 °C?

ANSWER

a. 78 g of KNO3 b. 170 g of H2O

Gout occurs when uric acid exceeds its solubility in blood plasma.

Kidney stones form when calcium phosphate exceeds its solubility.

PRACTICE PROBLEMS Try Practice Problems 12.15 to 12.20

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12.3 Solubility 359

Effect of Temperature on Solubility The solubility of most solids is greater as temperature increases, which means that solutions usually contain more dissolved solute at higher temperature. A few substances show little change in solubility at higher temperatures, and a few are less soluble (see FIGURE 12.5). For example, when you add sugar to iced tea, some undissolved sugar may form on the bottom of the glass. But if you add sugar to hot tea, many teaspoons of sugar are needed before solid sugar appears. Hot tea dissolves more sugar than does cold tea because the solubility of sugar is much greater at a higher temperature.

When a saturated solution is carefully cooled, it becomes a supersaturated solution because it contains more solute than the solubility allows. Such a solution is unstable, and if the solution is agitated or if a solute crystal is added, the excess solute will crystallize to give a saturated solution again.

Conversely, the solubility of a gas in water decreases as the temperature increases. At higher temperatures, more gas molecules have the energy to escape from the solution. Perhaps you have observed the bubbles escaping from a carbonated soft drink as it warms. At high temperatures, bottles containing carbonated solutions may burst as more gas molecules leave the solution and increase the gas pressure inside the bottle. Biologists have found that increased temperatures in rivers and lakes cause the amount of dissolved oxygen to decrease until the warm water can no longer support a biological community. Electricity-generating plants are required to have their own ponds to use with their cooling towers to lessen the threat of thermal pollution in surrounding waterways.

Henry’s Law Henry’s law states that the solubility of gas in a liquid is directly related to the pressure of that gas above the liquid. At higher pressures, there are more gas molecules available to enter and dissolve in the liquid. A can of soda is carbonated by using CO2 gas at high pressure to increase the solubility of the CO2 in the beverage. When you open the can at atmospheric pressure, the pressure on the CO2 drops, which decreases the solubility of CO2. As a result, bubbles of CO2 rapidly escape from the solution. The burst of bubbles is even more noticeable when you open a warm can of soda.

ENGAGE 12.4 Compare the solubility of NaNO3 at 20 °C and 60 °C.

Soluble and Insoluble Ionic Compounds In our discussion up to now, we have considered ionic compounds that dissolve in water. However, some ionic compounds do not dissociate into ions and remain as solids even in contact with water. The solubility rules give some guidelines about the solubility of ionic compounds in water.

CO2 dissolved in soda

CO2 bubbles out of solution

Cola

Gas molecule

Cola

When the pressure of a gas above a solution decreases, the solubility of that gas in the solution also decreases.

PRACTICE PROBLEMS Try Practice Problems 12.21 and 12.22

FIGURE 12.5 In water, most common solids are more soluble as the temperature increases.

S ol

ub il

it y

(g s

ol ut

e/ 10

0. g

H 2O

)

0 20 40

60 80

100 120 140 160

200

0 20 40 60 80 100

180

G lu

co se

K N

O 3

NaCl

Na NO

3

KI

Temperature (°C)

Na 3 PO

4

8

6

10 20 30 40

4

2

0 0

Temperature (°C)

S ol

ub il

it y

(m g

so lu

te /1

00 . g

H 2O

)

O2

N2

CO

In water, gases are less soluble as the temperature increases.

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360 CHAPTER 12 Solutions

ENGAGE 12.6 Why are each of the ionic compounds in Figure 12.6 insoluble in water?

TABLE 12.5 Solubility Rules for Ionic Compounds in Water An ionic compound is soluble in water if it contains one of the following:

Positive Ions Li+, Na+, K+, Rb+, Cs+, NH4 +

Negative Ions NO3 -, C2H3O2

-

Cl-, Br-, I- except when combined with Ag +, Pb2+, or Hg2 2+

SO4 2- except when combined with Ba2+, Pb2+, Ca2+, Sr2+, or Hg2

2+

Ionic compounds that do not contain at least one of these ions are usually insoluble.

FIGURE 12.6 If an ionic compound contains a combination of a cation and an anion that are not soluble, that ionic compound is insoluble. For example, combinations of cadmium and sulfide, iron and sulfide, lead and iodide, and nickel and hydroxide do not contain any soluble ions. Thus, they form insoluble ionic compounds.

CdS FeS PbI2 Ni(OH)2

Ionic compounds that are soluble in water typically contain at least one of the ions in TABLE 12.5. Only an ionic compound containing a soluble cation or anion will dissolve in water. Most ionic compounds containing Cl- are soluble, but AgCl, PbCl2, and Hg2Cl2 are insoluble. Similarly, most ionic compounds containing SO4

2- are soluble, but a few are insoluble. Most other ionic compounds are insoluble (see FIGURE 12.6). In an insoluble ionic compound, the ionic bonds between its positive and negative ions are too strong for the polar water molecules to break. We can use the solubility rules to predict whether a solid ionic compound would be soluble or not. TABLE 12.6 illustrates the use of these rules.

CORE CHEMISTRY SKILL Using Solubility Rules

ENGAGE 12.5 Why is K2S soluble in water whereas PbCl2 is not soluble?

FIGURE 12.7 A barium sulfate-enhanced X-ray of the abdomen shows the lower gastrointestinal (GI) tract.

Ionic Compound Solubility in Water Reasoning

K2S Soluble Contains K +

Ca(NO3)2 Soluble Contains NO3 -

PbCl2 Insoluble Is an insoluble chloride

NaOH Soluble Contains Na+

AlPO4 Insoluble Contains no soluble ions

TABLE 12.6 Using Solubility Rules

In medicine, insoluble BaSO4 is used as an opaque substance to enhance X-rays of the gastrointestinal tract (see FIGURE 12.7). BaSO4 is so insoluble that it does not dissolve in gastric fluids. Other ionic barium compounds cannot be used because they would dissolve in water, releasing Ba2+, which is poisonous.

SAMPLE PROBLEM 12.3 Soluble and Insoluble Ionic Compounds

TRY IT FIRST

Predict whether each of the following ionic compounds is soluble in water or not and explain your answer:

a. Na3PO4 b. CaCO3

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12.3 Solubility 361

SOLUTION

a. The ionic compound Na3PO4 is soluble in water because any compound that contains Na+ is soluble.

b. The ionic compound CaCO3 is not soluble because it does not contain a soluble positive or negative ion.

SELF TEST 12.3

a. In some electrolyte drinks, MgCl2 is added to provide magnesium. Why would you expect MgCl2 to be soluble in water?

b. Even though barium ions are toxic if ingested, why can barium sulfate be given to a patient for an X-ray of the lower gastrointestinal tract?

ANSWER

a. MgCl2 is soluble in water because ionic compounds that contain chloride are soluble unless they contain Ag +, Pb2+, or Hg2

2+. b. Barium sulfate is insoluble in water; it does not produce barium ions as it passes

through the GI tract during an X-ray.

PRACTICE PROBLEMS Try Practice Problems 12.23 and 12.24

Formation of a Solid We can use solubility rules to predict whether a solid, called a precipitate, forms when two solutions containing soluble reactants are mixed, as shown in Sample Problem 12.4.

SAMPLE PROBLEM 12.4 Writing Equations for the Formation of an Insoluble Ionic Compound

TRY IT FIRST

When solutions of NaCl and AgNO3 are mixed, a white solid forms. Write the ionic and net ionic equations for the reaction.

SOLUTION

STEP 1 Write the ions of the reactants.

Reactants (initial combinations)

+

+

Cl-(aq)

NO3 -(aq)

Na+(aq)

Ag+(aq)

STEP 2 Write the combinations of ions, and determine if any are insoluble. When we look at the ions of each solution, we see that the combination of Ag + and Cl- forms an insoluble ionic compound.

Mixture (new combinations) Product Soluble

Ag +(aq) + Cl-(aq) AgCl No

Na+(aq) + NO3 -(aq) NaNO3 Yes

STEP 3 Write the ionic equation including any solid. In the ionic equation, we show all the ions of the reactants. The products include the solid AgCl that forms along with the remaining ions Na+ and NO3

-.

Ag +(aq) + NO3 -(aq) + Na+(aq) + Cl-(aq) h AgCl(s) + Na+(aq) + NO3 -(aq)

ENGAGE 12.7 When mixing solutions of Pb(NO3)2 and NaBr, how do we know that PbBr2(s) is the solid that forms?

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362 CHAPTER 12 Solutions

STEP 4 Write the net ionic equation. We remove the Na+ and NO3 - ions, known

as spectator ions, which are unchanged. This gives the net ionic equation, which only shows the ions that form a solid precipitate.

Insoluble ionic compound

Ag+

NO3 -Na+

Cl-

AgCl(s)Ag+(aq) + Cl-(aq) Net ionic equation

AgCl(s)Ag+(aq) Na+(aq) Na+(aq)+ + + + +NO3-(aq) NO3 -(aq)Cl-(aq)

Spectator ionsSpectator ions

Type of Equation

Chemical AgNO3(aq) + NaCl(aq) iiiiih AgCl(s) + NaNO3(aq) Ionic Ag +(aq) + NO3 -(aq) + Na+(aq) + Cl-(aq) S AgCl(s) + Na+(aq) + NO3 -(aq)

Net Ionic Ag +(aq) + Cl-(aq) iiiiiih AgCl(s)

PRACTICE PROBLEMS

12.3 Solubility

12.15 State whether each of the following refers to a saturated or an unsaturated solution:

a. A crystal added to a solution does not change in size. b. A sugar cube completely dissolves when added to a cup of

coffee. c. A uric acid concentration of 4.6 mg/100 mL in the kidney

does not cause gout.

12.16 State whether each of the following refers to a saturated or an unsaturated solution:

a. A spoonful of salt added to boiling water dissolves. b. A layer of sugar forms on the bottom of a glass of tea as ice

is added. c. A kidney stone of calcium phosphate forms in the kidneys

when urine becomes concentrated.

SELF TEST 12.4

Predict whether a solid might form in each of the following mixtures of solutions. If so, write the net ionic equation for the reaction.

a. NH4Cl(aq) + Ca(NO3)2(aq) b. Pb(NO3)2(aq) + KCl(aq)

ANSWER

a. No solid forms because the products, NH4NO3(aq) and CaCl2(aq), are soluble. b. Pb2+(aq) + 2Cl-(aq) h PbCl2(s)

PRACTICE PROBLEMS Try Practice Problems 12.25 and 12.26

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12.4 Solution Concentrations 363

12.4 Solution Concentrations LEARNING GOAL Calculate the concentration of a solute in a solution; use concentration units to calculate the amount of solute or solution.

Our body f luids contain water and dissolved substances including glucose, urea, and electrolytes such as K+, Na+, Cl-, Mg2+, HCO3

-, and HPO 2-4 . Proper amounts of each of these dissolved substances and water must be maintained in the body fluids. Small changes in elec- trolyte levels can seriously disrupt cellular processes and endanger our health. Solutions can be described by their concentration, which is the amount of solute in a specific amount of that solution.

We will look at different ways to express a concentration as a ratio of a certain amount of solute in a given amount of solution as shown in TABLE 12.7. The amount of a solute may be expressed in units of grams, milliliters, or moles. The amount of a solution may be expressed in units of grams, milliliters, or liters.

Concentration of a solution = amount of solute

amount of solution

REVIEW Calculating Percentages (1.4)

Using Molar Mass as a Conversion Factor (7.3)

Using Mole–Mole Factors (9.2)

CORE CHEMISTRY SKILL Calculating Concentration

Mass Percent (m/m)

Volume Percent (v/v)

Mass/Volume Percent (m/v)

Molarity (M)

Solute Unit g mL g mole

Solvent Unit g mL mL L

TABLE 12.7 Summary of Types of Concentration Expressions and Their Units

Use the following table for problems 12.17 to 12.20:

Solubility (g/100. g H2O)

Substance 20 °C 50 °C

KCl 34 43

NaNO3 88 110

C12H22O11 (sugar) 204 260

12.17 Determine whether each of the following solutions will be saturated or unsaturated at 20 °C:

a. adding 25 g of KCl to 100. g of H2O b. adding 11 g of NaNO3 to 25 g of H2O c. adding 400. g of sugar to 125 g of H2O

12.18 Determine whether each of the following solutions will be saturated or unsaturated at 50 °C:

a. adding 25 g of KCl to 50. g of H2O b. adding 150. g of NaNO3 to 75 g of H2O c. adding 80. g of sugar to 25 g of H2O

12.19 A solution containing 80. g of KCl in 200. g of H2O at 50 °C is cooled to 20 °C.

a. How many grams of KCl remain in solution at 20 °C? b. How many grams of solid KCl crystallized after cooling?

12.20 A solution containing 80. g of NaNO3 in 75 g of H2O at 50 °C is cooled to 20 °C.

a. How many grams of NaNO3 remain in solution at 20 °C? b. How many grams of solid NaNO3 crystallized after cooling?

12.21 Explain the following observations: a. More sugar dissolves in hot tea than in iced tea. b. Champagne in a warm room goes flat. c. A warm can of soda has more spray when opened than a

cold one.

12.22 Explain the following observations: a. An open can of soda loses its “fizz” faster at room

temperature than in the refrigerator. b. Chlorine gas in tap water escapes as the sample warms to

room temperature. c. Less sugar dissolves in iced coffee than in hot coffee.

12.23 Predict whether each of the following ionic compounds is soluble in water:

a. LiCl b. PbS c. BaCO3 d. K2O e. Fe(NO3)3

12.24 Predict whether each of the following ionic compounds is soluble in water:

a. AgCl b. KI c. Na2S d. Ag2O e. CaSO4

12.25 Determine whether a solid forms when solutions containing the following ionic compounds are mixed. If so, write the ionic equation and the net ionic equation.

a. KCl(aq) and Na2S(aq) b. AgNO3(aq) and K2S(aq) c. CaCl2(aq) and Na2SO4(aq) d. CuCl2(aq) and Li3PO4(aq)

12.26 Determine whether a solid forms when solutions containing the following ionic compounds are mixed. If so, write the ionic equation and the net ionic equation.

a. Na3PO4(aq) and AgNO3(aq) b. K2SO4(aq) and Na2CO3(aq) c. Pb(NO3)2(aq) and Na2CO3(aq) d. BaCl2(aq) and KOH(aq)

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364 CHAPTER 12 Solutions

Mass Percent (m/m) Concentration Mass percent (m/m) describes the mass of the solute in grams for 100. g of solution. The mass percent is calculated by dividing the mass of a solute by the mass of the solution multiplied by 100% to give the percentage. In the calculation of mass percent (m/m), the units of mass of the solute and solution must be the same. If the mass of the solute is given as grams, then the mass of the solution must also be grams. The mass of the solution is the sum of the mass of the solute and the mass of the solvent.

Mass percent (m/m) = mass of solute (g)

mass of solute (g) + mass of solvent (g) * 100%

= mass of solute (g)

mass of solution (g) * 100%

Suppose we prepared a solution by mixing 8.00 g of KCl (solute) with 42.00 g of water (solvent). Together, the mass of the solute and mass of solvent give the mass of the solution (8.00 g + 42.00 g = 50.00 g). Mass percent is calculated by substituting the mass of the solute and the mass of the solution into the mass percent expression.

8.00 g KCl

50.00 g solution * 100% = 16.0% (m/m) KCl solution

$+++++%+++++& 8.00 g KCl + 42.00 g H2O

When water is added to 8.00 g of KCl to form 50.00 g of a KCl solution, the mass percent concentration is 16.0% (m/m).

SAMPLE PROBLEM 12.5 Calculating Mass Percent (m/m) Concentration

TRY IT FIRST

What is the mass percent of NaOH in a solution prepared by dissolving 30.0 g of NaOH in 120.0 g of H2O?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

30.0 g of NaOH, 120.0 g of H2O

mass percent (m/m)

mass of solute mass of solution

* 100%

STEP 2 Write the concentration expression.

Mass percent (m/m) = grams of solute

grams of solution * 100%

STEP 3 Substitute solute and solution quantities into the expression and calculate. The mass of the solution is obtained by adding the mass of the solute and the mass of the solution.

mass of solution = 30.0 g NaOH + 120.0 g H2O = 150.0 g of NaOH solution

Mass percent (m/m) = 30.0 g NaOH

150.0 g solution * 100%

= 20.0% (m/m) NaOH solution

SELF TEST 12.5

a. What is the mass percent (m/m) of NaCl in a solution made by dissolving 2.0 g of NaCl in 56.0 g of H2O?

b. What is the mass percent (m/m) of MgCl2 in a solution prepared by dissolving 1.8 g of MgCl2 in 18.5 g of H2O?

ANSWER

a. 3.4% (m/m) NaCl solution b. 8.9% (m/m) MgCl2 solution

Three SFs

Four SFs

Three SFs

PRACTICE PROBLEMS Try Practice Problems 12.29 and 12.30

Solute + Solvent

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12.4 Solution Concentrations 365

Using Mass Percent Concentration as a Conversion Factor In the preparation of solutions, we often need to calculate the amount of solute or solution. Then the concentration of a solution is useful as a conversion factor as shown in Sample Problem 12.6.

CORE CHEMISTRY SKILL Using Concentration as a

Conversion Factor

SAMPLE PROBLEM 12.6 Using Mass Percent to Calculate Mass of Solute

TRY IT FIRST

The topical antibiotic ointment Neosporin is 3.5% (m/m) neomycin solution. How many grams of neomycin are in a tube containing 64 g of ointment?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

64 g of 3.5% (m/m) neomycin solution

grams of neomycin

mass percent factor g of solute

100. g of solution

STEP 2 Write a plan to calculate the mass.

% (m/m) factorgrams of ointment grams of neomycin

STEP 3 Write equalities and conversion factors. The mass percent (m/m) indicates the grams of a solute in every 100. g of a solution. The mass percent (3.5% m/m) can be written as two conversion factors.

3.5 g neomycin 100. g ointment

and 100. g ointment 3.5 g neomycin

3.5 g of neomycin = 100. g of ointment

STEP 4 Set up the problem to calculate the mass.

64 g ointment * 3.5 g neomycin

100. g ointment = 2.2 g of neomycin

Two SFs

Two SFsTwo SFs Exact

SELF TEST 12.6

a. Calculate the grams of KCl in 225 g of an 8.00% (m/m) KCl solution. b. Calculate the grams of KNO3 in 45.0 g of a 1.85% (m/m) KNO3 solution.

ANSWER

a. 18.0 g of KCl b. 0.833 g of KNO3

ENGAGE 12.8 How is the mass percent (m/m) of a solution used to convert the mass of the solution to the grams of solute?

Volume Percent (v/v) Concentration Because the volumes of liquids or gases are easily measured, the concentrations of their solutions are often expressed as volume percent (v/v). The units of volume used in the ratio must be the same, for example, both in milliliters or both in liters.

Volume percent (v/v) = volume of solute

volume of solution * 100%

We interpret a volume percent as the volume of solute in 100. mL of solution. On a bottle of extract of vanilla, a label that reads alcohol 35% (v/v) means 35 mL of alcohol solute in 100. mL of vanilla solution.

The label indicates that vanilla extract contains 35% (v/v) alcohol.

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366 CHAPTER 12 Solutions

Lemon extract is a solution of lemon flavor and alcohol.

SAMPLE PROBLEM 12.7 Calculating Volume Percent (v/v) Concentration

TRY IT FIRST

A bottle contains 59 mL of lemon extract solution. If the extract contains 49 mL of alcohol, what is the volume percent (v/v) of the alcohol in the solution?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

49 mL of alcohol, 59 mL of solution

volume percent (v/v)

volume of solute volume of solution

* 100%

STEP 2 Write the concentration expression.

Volume percent (v/v) = volume of solute

volume of solution * 100%

STEP 3 Substitute solute and solution quantities into the expression and calculate.

Volume percent (v/v) = 49 mL alcohol 59 mL solution

* 100% = 83% (v/v) alcohol solution

SELF TEST 12.7

a. What is the volume percent (v/v) of Br2 in a solution prepared by dissolving 12 mL of liquid bromine (Br2) in the solvent carbon tetrachloride (CCl4) to make 250 mL of solution?

b. What is the volume percent (v/v) of a “rubbing alcohol” solution that contains 14 mL of isopropyl alcohol in 20. mL of solution?

ANSWER

a. 4.8% (v/v) Br2 solution b. 70.% (v/v) isopropyl alcohol solution

Two SFs Two SFs

Two SFs

PRACTICE PROBLEMS Try Practice Problems 12.35 and 12.36

Mass/Volume Percent (m/v) Concentration Mass/volume percent (m/v) describes the mass of the solute in grams for 100. mL of solu- tion. In the calculation of mass/volume percent, the unit of mass of the solute is grams, and the unit of the solution volume is milliliters.

Mass/volume percent (m/v) = grams of solute

milliliters of solution * 100%

The mass/volume percent is widely used in hospitals and pharmacies for the preparation of intravenous solutions and medicines. For example, a 5% (m/v) glucose solution contains 5 g of glucose in 100. mL of solution. The volume of solution represents the combined volumes of the glucose and H2O.

ENGAGE 12.9 What units should be used for a 3.0% (m/v) solution?

SAMPLE PROBLEM 12.8 Calculating Mass/Volume Percent (m/v) Concentration

TRY IT FIRST

A potassium iodide solution may be used in a diet that is low in iodine. A KI solution is prepared by dissolving 5.0 g of KI in enough water to give a final volume of 250 mL. What is the mass/volume percent (m/v) of KI in the solution?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

5.0 g of KI solute, 250 mL of KI solution

mass/volume percent (m/v)

mass of solute volume of solution

* 100%

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12.4 Solution Concentrations 367

STEP 2 Write the concentration expression.

Mass/volume percent (m/v) = mass of solute

volume of solution * 100%

STEP 3 Substitute solute and solution quantities into the expression and calculate.

Mass/volume percent (m/v) = 5.0 g KI

250 mL solution * 100% = 2.0% (m/v) KI solution

SELF TEST 12.8

a. What is the mass/volume percent (m/v) of NaOH in a solution prepared by dissolving 12 g of NaOH in enough water to make 220 mL of solution?

b. What is the mass/volume percent (m/v) of MgCl2 in a solution prepared by dissolving 3.25 g of MgCl2 in enough water to make 125 mL of solution?

ANSWER

a. 5.5% (m/v) NaOH solution b. 2.60% (m/v) MgCl2 solution

Two SFsTwo SFs

Two SFs

SAMPLE PROBLEM 12.9 Using Mass/Volume Percent to Calculate Mass of Solute

TRY IT FIRST

A topical antibiotic is 1.0% (m/v) clindamycin. How many grams of clindamycin are in 60. mL of the 1.0% (m/v) clindamycin solution?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

60. mL of 1.0% (m/v) clindamycin solution

grams of clindamycin

% (m/v) factor

STEP 2 Write a plan to calculate the mass.

milliliters of solution grams of clindamycin% (m/v) factor

STEP 3 Write equalities and conversion factors. The percent (m/v) indicates the grams of a solute in every 100. mL of a solution. The 1.0% (m/v) can be written as two conversion factors.

1.0 g clindamycin

100. mL solution and

100. mL solution 1.0 g clindamycin

1.0 g of clindamycin = 100. mL of solution

STEP 4 Set up the problem to calculate the mass. The volume of the solution is converted to mass of solute using the conversion factor that cancels mL.

60. mL solution * 1.0 g clindamycin

100. mL solution = 0.60 g of clindamycin

Two SFs

Two SFsTwo SFs Exact

SELF TEST 12.9

a. In 2010, the FDA approved a 2.0% (m/v) morphine oral solution to treat severe or chronic pain. How many grams of morphine does a patient receive if 0.60 mL of 2.0% (m/v) morphine solution was ordered?

250 mL of KI solution

5.0 g of KI

250 mL Water added to make a solution

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368 CHAPTER 12 Solutions

Molarity (M) Concentration When chemists work with solutions, they often use molarity (M), a concentration that states the number of moles of solute in exactly 1 L of solution.

Molarity (M) = moles of solute liters of solution

The molarity of a solution can be calculated when we know the moles of solute and the volume of solution in liters. For example, if 1.0 mol of NaCl were dissolved in enough water to prepare 1.0 L of solution, the resulting NaCl solution has a molarity of 1.0 M. The abbreviation M indicates the units of mole per liter (mol/L).

M = moles of solute liters of solution

= 1.0 mol NaCl 1 L solution

= 1.0 M NaCl solution

b. For a skin infection, the doctor orders 300 mg of ampicillin. How many milliliters of 5.0% (m/v) ampicillin should be given?

ANSWER

a. 0.012 g of morphine b. 6 mL

PRACTICE PROBLEMS Try Practice Problems 12.27 and 12.28, 12.31 to 12.34, 12.37 and 12.38, 12.47 to 12.50

SAMPLE PROBLEM 12.10 Calculating Molarity

TRY IT FIRST

What is the molarity (M) of 60.0 g of NaOH in 0.250 L of NaOH solution?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

60.0 g of NaOH, 0.250 L of NaOH solution

molarity (mol/L)

molar mass of NaOH, moles of solute liters of solution

To calculate the moles of NaOH, we need to write the equality and conversion factors for the molar mass of NaOH. Then the moles in 60.0 g of NaOH can be determined.

1 mol NaOH

40.00 g NaOH and

40.00 g NaOH 1 mol NaOH

40.00 g of NaOH=1 mol of NaOH

60.0 g NaOH * 1 mol NaOH

40.00 g NaOH =

1.50 mol of NaOH=

moles of NaOH

0.250 L of NaOH solution=volume of solution

STEP 2 Write the concentration expression.

Molarity 1M2 = moles of solute liters of solution

STEP 3 Substitute solute and solution quantities into the expression and calculate.

M = 1.50 mol NaOH 0.250 L solution

= 6.00 mol NaOH

1 L solution = 6.00 M NaOH solution

1.0 mol of NaCl

Volumetric flask

Mix

Add water until 1-liter mark is reached.

A 1.0 molar (M) NaCl solution A 1.0 molar (M) NaCl solution

ENGAGE 12.10 Why does a solution containing 2.0 mol of KCl in 4.0 L of KCl solution have a molarity of 0.50 M?

M12_TIMB8119_06_SE_C12.indd 368 11/27/18 12:07 PM

12.4 Solution Concentrations 369

A summary of percent concentrations and molarity, their meanings, and conversion factors are given in TABLE 12.8.

SAMPLE PROBLEM 12.11 Using Molarity to Calculate Volume of Solution

TRY IT FIRST

How many liters of a 2.00 M NaCl solution are needed to provide 67.3 g of NaCl?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

67.3 g of NaCl, 2.00 M NaCl solution

liters of NaCl solution

molar mass of NaCl, molarity

STEP 2 Write a plan to calculate the volume.

grams of NaCl moles of NaCl liters of NaCl solutionMolar mass Molarity

STEP 3 Write equalities and conversion factors.

1 mol NaCl 58.44 g NaCl

and 58.44 g NaCl

1 mol NaCl

1 mol of NaCl = 58.44 g of NaCl 1 L NaCl solution

2.00 mol NaCl and

2.00 mol NaCl 1 L NaCl solution

1 L of NaCl solution = 2.00 mol of NaCl

STEP 4 Set up the problem to calculate the volume.

67.3 g NaCl = 0.576 L of NaCl solution* * 1 mol NaCl

58.44 g NaCl 1 L NaCl solution

2.00 mol NaCl

SELF TEST 12.11

a. How many milliliters of a 6.0 M HCl solution will provide 164 g of HCl? b. What volume, in milliliters, of a 2.50 M NaOH solution contains 12.5 g of NaOH?

ANSWER

a. 750 mL of HCl solution b. 125 mL of NaOH solution PRACTICE PROBLEMS

Try Practice Problems 12.39 to 12.46

SELF TEST 12.10

a. What is the molarity of a solution that contains 75.0 g of KNO3 dissolved in 0.350 L of solution?

b. What is the molarity of a mouthwash solution that contains 6.40 g of the disinfectant thymol, C10H14O, in 2.00 L of solution?

ANSWER

a. 2.12 M KNO3 solution b. 0.0213 M thymol solution

M12_TIMB8119_06_SE_C12.indd 369 11/27/18 12:07 PM

370 CHAPTER 12 Solutions

TABLE 12.8 Conversion Factors from Concentrations Percent Concentration Meaning Conversion Factors

10% (m/m) KCl solution

10 g of KCl in 100. g of KCl solution

10 g KCl

100. g solution and 100. g solution

10 g KCl

12% (v/v) ethanol solution

12 mL of ethanol in 100. mL of ethanol solution

12 mL ethanol 100. mL solution

and 100. mL solution 12 mL ethanol

5% (m/v) glucose solution

5 g of glucose in 100. mL of glucose solution

5 g glucose

100. mL solution and 100. mL solution

5 g glucose

Molarity Meaning Conversion Factors

6.0 M HCl solution 6.0 mol of HCl in 1 L of HCl solution

6.0 mol HCl 1 L solution

and 1 L solution 6.0 mol HCl

PRACTICE PROBLEMS

12.4 Solution Concentrations

12.27 What is the difference between a 5.00% (m/m) glucose solution and a 5.00% (m/v) glucose solution?

12.28 What is the difference between a 10.0% (v/v) methanol solution and a 10.0% (m/m) methanol solution?

12.29 Calculate the mass percent (m/m) for the solute in each of the following:

a. 25 g of KCl and 125 g of H2O b. 12 g of sucrose in 225 g of tea solution c. 8.0 g of CaCl2 in 80.0 g of CaCl2 solution

12.30 Calculate the mass percent (m/m) for the solute in each of the following:

a. 75 g of NaOH in 325 g of NaOH solution b. 2.0 g of KOH and 20.0 g of H2O c. 48.5 g of Na2CO3 in 250.0 g of Na2CO3 solution

12.31 Calculate the mass/volume percent (m/v) for the solute in each of the following:

a. 75 g of Na2SO4 in 250 mL of Na2SO4 solution b. 39 g of sucrose in 355 mL of a carbonated drink

12.32 Calculate the mass/volume percent (m/v) for the solute in each of the following:

a. 2.50 g of LiCl in 40.0 mL of LiCl solution b. 7.5 g of casein in 120 mL of low-fat milk

12.33 Calculate the grams or milliliters of solute needed to prepare the following:

a. 50. g of a 5.0% (m/m) KCl solution b. 1250 mL of a 4.0% (m/v) NH4Cl solution c. 250. mL of a 10.0% (v/v) acetic acid solution

12.34 Calculate the grams or milliliters of solute needed to prepare the following:

a. 150. g of a 40.0% (m/m) LiNO3 solution b. 450 mL of a 2.0% (m/v) KOH solution c. 225 mL of a 15% (v/v) isopropyl alcohol solution

12.35 A mouthwash contains 22.5% (v/v) alcohol. If the bottle of mouthwash contains 355 mL, what is the volume, in milliliters, of alcohol?

12.36 A bottle of champagne is 11% (v/v) alcohol. If there are 750 mL of champagne in the bottle, what is the volume, in milliliters, of alcohol?

12.37 For each of the following solutions, calculate the: a. grams of a 25% (m/m) LiNO3 solution that contains 5.0 g

of LiNO3 b. milliliters of a 10.0% (m/v) KOH solution that contains

40.0 g of KOH c. milliliters of a 10.0% (v/v) formic acid solution that

contains 2.0 mL of formic acid

12.38 For each of the following solutions, calculate the: a. grams of a 2.0% (m/m) NaCl solution that contains 7.50 g

of NaCl b. milliliters of a 25% (m/v) NaF solution that contains 4.0 g

of NaF c. milliliters of an 8.0% (v/v) ethanol solution that contains

20.0 mL of ethanol

12.39 Calculate the molarity of each of the following: a. 2.00 mol of glucose in 4.00 L of a glucose solution b. 4.00 g of KOH in 2.00 L of a KOH solution c. 5.85 g of NaCl in 400. mL of a NaCl solution

12.40 Calculate the molarity of each of the following: a. 0.500 mol of glucose in 0.200 L of a glucose solution b. 73.0 g of HCl in 2.00 L of a HCl solution c. 30.0 g of NaOH in 350. mL of a NaOH solution

12.41 Calculate the grams of solute needed to prepare each of the following:

a. 2.00 L of a 1.50 M NaOH solution b. 4.00 L of a 0.200 M KCl solution c. 25.0 mL of a 6.00 M HCl solution

12.42 Calculate the grams of solute needed to prepare each of the following:

a. 2.00 L of a 6.00 M NaOH solution b. 5.00 L of a 0.100 M CaCl2 solution c. 175 mL of a 3.00 M NaNO3 solution

12.43 For each of the following solutions, calculate the: a. liters of a 2.00 M KBr solution to obtain 3.00 mol of KBr b. liters of a 1.50 M NaCl solution to obtain 15.0 mol of NaCl c. milliliters of a 0.800 M Ca(NO3)2 solution to obtain

0.0500 mol of Ca(NO3)2

M12_TIMB8119_06_SE_C12.indd 370 11/27/18 12:07 PM

12.5 Dilution of Solutions 371

FIGURE 12.8 When water is added to a concentrated solution, there is no change in the number of particles. However, the solute particles spread out as the volume of the diluted solution increases.

12.5 Dilution of Solutions LEARNING GOAL Describe the dilution of a solution; calculate the unknown concentration or volume when a solution is diluted.

In chemistry and biology, we often prepare diluted solutions from more concentrated solu- tions. In a process called dilution, a solvent, usually water, is added to a solution, which increases the volume. As a result, the concentration of the solution decreases. In an everyday example, you are making a dilution when you add three cans of water to a can of concen- trated orange juice.

REVIEW Solving Equations (1.4)

Although the addition of solvent increases the volume, the amount of solute does not change; it is the same in the concentrated solution and the diluted solution (see FIGURE 12.8).

Grams or moles of solute = grams or moles of solute Concentrated solution Diluted solution

12.44 For each of the following solutions, calculate the: a. liters of a 4.00 M KCl solution to obtain 0.100 mol of KCl b. liters of a 6.00 M HCl solution to obtain 5.00 mol of HCl c. milliliters of a 2.50 M K2SO4 solution to obtain 1.20 mol of

K2SO4

12.45 Calculate the volume, in milliliters, for each of the following that provides the given amount of solute:

a. 12.5 g of Na2CO3 from a 0.120 M Na2CO3 solution b. 0.850 mol of NaNO3 from a 0.500 M NaNO3 solution c. 30.0 g of LiOH from a 2.70 M LiOH solution

12.46 Calculate the volume, in liters, for each of the following that provides the given amount of solute:

a. 5.00 mol of NaOH from a 12.0 M NaOH solution b. 15.0 g of Na2SO4 from a 4.00 M Na2SO4 solution c. 28.0 g of NaHCO3 from a 1.50 M NaHCO3 solution

Applications

12.47 A patient receives 100. mL of 20.% (m/v) mannitol solution every hour.

a. How many grams of mannitol are given in 1 h? b. How many grams of mannitol does the patient receive in 12 h?

12.48 A patient receives 250 mL of a 4.0% (m/v) amino acid solution twice a day.

a. How many grams of amino acids are in 250 mL of solution?

b. How many grams of amino acids does the patient receive in 1 day?

12.49 A patient needs 100. g of glucose in the next 12 h. How many liters of a 5% (m/v) glucose solution must be given?

12.50 A patient received 2.0 g of NaCl in 8 h. How many milliliters of a 0.90% (m/v) NaCl (saline) solution were delivered?

1 can of orange juice concentrate

3 cans of water 4 cans of orange juice+ =

Mix

In the dilution of a solution, solvent is added to increase the volume and decrease the concentration.

M12_TIMB8119_06_SE_C12.indd 371 11/27/18 12:07 PM

372 CHAPTER 12 Solutions

We can write this equality in terms of the concentration, C, and the volume, V. The concentration, C, may be percent concentration or molarity.

C1V1 = C2V2 Concentrated

solution Diluted solution

If we are given any three of the four variables (C1, C2, V1, or V2) we can rearrange the dilution expression to solve for the unknown quantity as seen in Sample Problems 12.12 and 12.13.

PRACTICE PROBLEMS Try Practice Problems 12.51 and 12.52

SAMPLE PROBLEM 12.12 Calculations Involving Concentration of Solutions

TRY IT FIRST

What is the molarity of a solution when 75.0 mL of a 4.00 M KCl solution is diluted to a volume of 500. mL?

SOLUTION

STEP 1 Prepare a table of the concentrations and volumes of the solutions.

ANALYZE THE PROBLEM

Given Need Connect

C1 = 4.00 M V1 = 75.0 mL V2 = 500. mL

C2 C1V1 = C2V2 Predict: C2 decreases

STEP 2 Rearrange the dilution expression to solve for the unknown quantity.

C1V1 = C2V2

C1V1

V2 =

C2V2 V2

Divide both sides by V2

C2 = C1 * V1 V2

STEP 3 Substitute the known quantities into the dilution expression and calculate.

ENGAGE 12.11 How do we know that the dilution of this KCl solution has a lower molarity?

When the initial molarity (C1) is multiplied by a ratio of the volumes (volume factor) that is less than 1, the molarity of the diluted solution (C2) decreases as predicted in Step 1.

SELF TEST 12.12

a. What is the molarity of a solution when 50.0 mL of a 4.00 M KOH solution is diluted to 200. mL?

b. What is the final concentration (m/v) when 18 mL of a 12% (m/v) NaCl solution is diluted to a volume of 85 mL?

ANSWER

a. 1.00 M KOH solution b. 2.5% (m/v) NaCl solution PRACTICE PROBLEMS

Try Practice Problems 12.53 to 12.56

SAMPLE PROBLEM 12.13 Calculations Involving Volume of Solutions

TRY IT FIRST

A doctor orders 1000. mL of a 35.0% (m/v) dextrose solution. If you have a 50.0% (m/v) dextrose solution, how many milliliters would you use to prepare 1000. mL of 35.0% (m/v) dextrose solution?

Three SFs

Three SFs

Three SFsThree SFs Volume factor

decreases concentration

C2 = 4.00 M * 75.0 mL 500. mL

= 0.600 M (diluted KCl solution)

M12_TIMB8119_06_SE_C12.indd 372 11/27/18 12:07 PM

12.5 Dilution of Solutions 373

PRACTICE PROBLEMS

12.5 Dilution of Solutions

12.51 To make tomato soup, you add one can of water to the con- densed soup. Why is this a dilution?

12.52 A can of frozen lemonade calls for the addition of three cans of water to make a pitcher of the beverage. Why is this a dilution?

12.53 Calculate the final concentration of each of the following: a. 2.0 L of a 6.0 M HCl solution is added to water so that the

final volume is 6.0 L. b. Water is added to 0.50 L of a 12 M NaOH solution to make

3.0 L of a diluted NaOH solution. c. A 10.0-mL sample of a 25% (m/v) KOH solution is diluted

with water so that the final volume is 100.0 mL. d. A 50.0-mL sample of a 15% (m/v) H2SO4 solution is added

to water to give a final volume of 250 mL.

12.54 Calculate the final concentration of each of the following: a. 1.0 L of a 4.0 M HNO3 solution is added to water so that

the final volume is 8.0 L. b. Water is added to 0.25 L of a 6.0 M NaF solution to make

2.0 L of a diluted NaF solution. c. A 50.0-mL sample of an 8.0% (m/v) KBr solution is

diluted with water so that the final volume is 200.0 mL. d. A 5.0-mL sample of a 50.0% (m/v) acetic acid (HC2H3O2)

solution is added to water to give a final volume of 25 mL.

STEP 3 Substitute the known quantities into the dilution expression and calculate.

When the final volume (V2) is multiplied by a ratio of the percent concentrations (concentration factor) that is less than 1, the initial volume (V1) is less than the final volume (V2) as predicted in Step 1.

SELF TEST 12.13

a. What initial volume, in milliliters, of a 15% (m/v) mannose solution is needed to prepare 125 mL of a 3.0% (m/v) mannose solution?

b. What initial volume, in milliliters, of a 2.40 M NaCl solution is needed to prepare 80.0 mL of a 0.600 M NaCl solution?

ANSWER

a. 25 mL of a 15% (m/v) mannose solution b. 20.0 mL of a 2.40 M NaCl solution

SOLUTION

STEP 1 Prepare a table of the concentrations and volumes of the solutions.

ANALYZE THE PROBLEM

Given Need Connect

C1 = 50.0% (m/v) C2 = 35.0% (m/v) V2 = 1000. mL

V1 C1V1 = C2V2 Predict: V1 decreases

STEP 2 Rearrange the dilution expression to solve for the unknown quantity.

C1V1 = C2V2

C1V1 C1

= C2V2 C1

Divide both sides by C1

V1 = V2 * C2 C1

PRACTICE PROBLEMS Try Practice Problems 12.57 to 12.60

V1 = 1000. mL * 35.0% 50.0%

= 700. mL of dextrose solution

Three SFs

Three SFs

Four SFs Three SFs Concentration factor

decreases volume

M12_TIMB8119_06_SE_C12.indd 373 11/27/18 12:08 PM

374 CHAPTER 12 Solutions

12.55 Determine the final volume, in milliliters, of each of the following:

a. a 1.5 M HCl solution prepared from 20.0 mL of a 6.0 M HCl solution

b. a 2.0% (m/v) LiCl solution prepared from 50.0 mL of a 10.0% (m/v) LiCl solution

c. a 0.500 M H3PO4 solution prepared from 50.0 mL of a 6.00 M H3PO4 solution

d. a 5.0% (m/v) glucose solution prepared from 75 mL of a 12% (m/v) glucose solution

12.56 Determine the final volume, in milliliters, of each of the following:

a. a 1.00% (m/v) H2SO4 solution prepared from 10.0 mL of a 20.0% H2SO4 solution

b. a 0.10 M HCl solution prepared from 25 mL of a 6.0 M HCl solution

c. a 1.0 M NaOH solution prepared from 50.0 mL of a 12 M NaOH solution

d. a 1.0% (m/v) CaCl2 solution prepared from 18 mL of a 4.0% (m/v) CaCl2 solution

12.57 Determine the initial volume, in milliliters, required to prepare each of the following:

a. 255 mL of a 0.200 M HNO3 solution using a 4.00 M HNO3 solution

b. 715 mL of a 0.100 M MgCl2 solution using a 6.00 M MgCl2 solution

c. 0.100 L of a 0.150 M KCl solution using an 8.00 M KCl solution

12.58 Determine the initial volume, in milliliters, required to prepare each of the following:

a. 20.0 mL of a 0.250 M KNO3 solution using a 6.00 M KNO3 solution

b. 25.0 mL of a 2.50 M H2SO4 solution using a 12.0 M H2SO4 solution

c. 0.500 L of a 1.50 M NH4Cl solution using a 10.0 M NH4Cl solution

Applications

12.59 You need 500. mL of a 5.0% (m/v) glucose solution. If you have a 25% (m/v) glucose solution on hand, how many milliliters do you need?

12.60 A doctor orders 100. mL of 2.0% (m/v) ibuprofen. If you have 8.0% (m/v) ibuprofen on hand, how many milliliters do you need?

12.6 Chemical Reactions in Solution LEARNING GOAL Given the volume and concentration of a solution in a chemical reaction, calculate the amount of a reactant or product in the reaction.

When chemical reactions involve aqueous solutions, we use the balanced chemical equation, the molarity, and the volume to determine the moles or grams of a reactant or product. For example, we can determine the volume of a solution from the molarity and the grams of reactant as seen in Sample Problem 12.14.

SAMPLE PROBLEM 12.14 Calculations Involving Solutions in Reactions

TRY IT FIRST

HCl reacts with zinc to produce hydrogen gas, H2, and ZnCl2.

2HCl(aq) + Zn(s) h H2(g) + ZnCl2(aq) How many liters of a 1.50 M HCl solution completely react with 5.32 g of zinc?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

5.32 g of Zn, 1.50 M HCl solution

liters of HCl solution

molar mass of Zn, molarity, mole–mole factor

Equation

2HCl(aq) + Zn(s) h H2(g) + ZnCl2(aq)

STEP 2 Write a plan to calculate the needed quantity.

Zinc reacts when placed in a solution of HCl.

INTERACTIVE VIDEO

Solutions

grams of Zn moles of Zn moles of HCl liters of HCl solutionMolar mass Mole–mole factor Molarity

M12_TIMB8119_06_SE_C12.indd 374 11/27/18 12:08 PM

12.6 Chemical Reactions in Solution 375

When a BaCl2 solution is added to a Na2SO4 solution, BaSO4, a white solid, forms.

SAMPLE PROBLEM 12.15 Calculations Involving Solutions in Reactions

TRY IT FIRST

How many milliliters of a 0.250 M BaCl2 solution are needed to react with 0.0325 L of a 0.160 M Na2SO4 solution?

Na2SO4(aq) + BaCl2(aq) h BaSO4(s) + 2NaCl(aq)

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

0.0325 L of 0.160 M Na2SO4 solution, 0.250 M BaCl2 solution

milliliters of BaCl2 solution

mole–mole factor, metric factor

Equation

Na2SO4(aq) + BaCl2(aq) h BaSO4(s) + 2NaCl(aq)

STEP 2 Write a plan to calculate the needed quantity.

STEP 3 Write equalities and conversion factors including mole–mole and concentration factors.

1 mol Zn 65.41 g Zn

and 65.41 g Zn 1 mol Zn

1 mol of Zn = 65.41 g of Zn 1 mol Zn 2 mol HCl

and 2 mol HCl

1 mol Zn

1 mol of Zn = 2 mol of HCl

1 L solution 1.50 mol HCl

and 1.50 mol HCl

1 L solution

1 L of solution = 1.50 mol of HCl

STEP 4 Set up the problem to calculate the needed quantity.

5.32 g Zn = 0.108 L of HCl solution* * * 1 mol Zn

65.41 g Zn 2 mol HCl 1 mol Zn

1 L solution 1.50 mol HCl

SELF TEST 12.14

a. Using the reaction in Sample Problem 12.14, how many grams of zinc can react with 225 mL of a 0.200 M HCl solution?

b. Using the reaction in Sample Problem 12.14, how many milliliters of a 0.235 M HCl solution will react with 0.426 g of zinc ?

ANSWER

a. 1.47 g of Zn b. 28.6 mL of HCl solution

liters of Na2SO4 solution moles of Na2SO4 moles of BaCl2

liters of BaCl2 solution

Molarity Mole–mole factor

milliliters of BaCl2 solution Metric factor

Molarity

CORE CHEMISTRY SKILL Calculating the Quantity of a

Reactant or Product for a Chemical Reaction in Solution

INTERACTIVE VIDEO

Calculations Involving Solutions in Reaction

M12_TIMB8119_06_SE_C12.indd 375 11/27/18 12:08 PM

376 CHAPTER 12 Solutions

STEP 4 Set up the problem to calculate the needed quantity.

1 L solution 0.160 mol Na2SO41 L solution

0.160 mol Na2SO4 and

1 L of solution = 0.160 mol of Na2SO4 1 mol Na2SO4 1 mol BaCl21 mol Na2SO4

1 mol BaCl2 and

1 mol of Na2SO4 = 1 mol of BaCl2

1 L solution

0.250 mol BaCl21 L solution

0.250 mol BaCl2 and

1 L of solution = 0.250 mol of BaCl2 1 L

1000 mL1 L

1000 mL and

1 L = 1000 mL

1 L solution 0.160 mol Na2SO41 L solution

0.160 mol Na2SO4 and

1 L of solution = 0.160 mol of Na2SO4 1 mol Na2SO4 1 mol BaCl21 mol Na2SO4

1 mol BaCl2 and

1 mol of Na2SO4 = 1 mol of BaCl2

1 L solution

0.250 mol BaCl21 L solution

0.250 mol BaCl2 and

1 L of solution = 0.250 mol of BaCl2 1 L

1000 mL1 L

1000 mL and

1 L = 1000 mL

STEP 3 Write equalities and conversion factors including mole–mole and concentration factors.

SAMPLE PROBLEM 12.16 Volume of a Gas from a Reaction in Solution

TRY IT FIRST

Acid rain results from the reaction of nitrogen dioxide with water in the air.

3NO2(g) + H2O(l) h 2HNO3(aq) + NO(g)

At STP, how many liters of NO2 gas are required to produce 0.275 L of a 0.400 M HNO3 solution?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

0.275 L of 0.400 M HNO3 solution

liters of NO2(g) at STP

mole–mole factor, molar volume

Equation

3NO2(g) + H2O(l ) h 2HNO3(aq) + NO(g)

STEP 2 Write a plan to calculate the needed quantity. We start the problem with the volume and molarity of the HNO3 solution to calculate moles. Then we can use the mole–mole factor and the molar volume to calculate the liters of NO2 gas.

22.4 L NO2 (STP) 1 mol NO2 22.4 L NO2 (STP)

1 mol NO2and

1 mol of NO2 = 22.4 L of NO2 (STP)

liters of solution moles of HNO3 moles of NO2 Mole–mole factor

liters of NO2 (at STP) Molar volume

Molarity

1 L solution 0.400 mol HNO31 L solution

0.400 mol HNO3 and

1 L of solution = 0.400 mol of HNO3 2 mol HNO3 3 mol NO22 mol HNO3

3 mol NO2 and

3 mol of NO2 = 2 mol of HNO3

0.275 L solution 3.70 L of NO2 (STP)* * = 0.400 mol HNO3

1 L solution 22.4 L NO2 (STP)

1 mol NO2 *

3 mol NO2 2 mol HNO3

0.0325 L solution * * * 0.160 mol Na2SO4

1 L solution 1 L solution

0.250 mol BaCl2 *

1 mol BaCl2 1 mol Na2SO4

1000 mL BaCl2 solution 1 L solution

= 20.8 mL of BaCl2 solution

SELF TEST 12.15

a. For the reaction in Sample Problem 12.15, how many milliliters of a 0.330 M Na2SO4 solution are needed to react with 26.8 mL of a 0.216 M BaCl2 solution?

b. Using the reaction in Sample Problem 12.15, what is the molarity of a BaCl2 solution if 19.4 mL of the BaCl2 solution is needed to react with 15.0 mL of a 0.225 M Na2SO4 solution?

ANSWER

a. 17.5 mL of Na2SO4 solution b. 0.174 M BaCl2 solution

M12_TIMB8119_06_SE_C12.indd 376 11/27/18 12:08 PM

12.6 Chemical Reactions in Solution 377

FIGURE 12.9 In calculations involving chemical reactions, substance A is converted to moles of A using molar mass (if solid), gas laws (if gas), or molarity (if solution). Then moles of A are converted to moles of substance B, which are converted to grams of solid, liters of gas, or liters of solution, as needed.

Gas volume (L) of B

Grams of B

Solution volume (L) of B

Molar mass of B

Molarity of B

Moles of B Gas law Gas volume

(L) of A Gas

Grams of ASolid

Solution volume (L) of A

Solution

Molar mass of A

Molarity of A

Moles of AGas law Mole–mole factor

STEP 3 Write equalities and conversion factors including mole–mole and concentration factors.

22.4 L NO2 (STP) 1 mol NO2 22.4 L NO2 (STP)

1 mol NO2and

1 mol of NO2 = 22.4 L of NO2 (STP)

liters of solution moles of HNO3 moles of NO2 Mole–mole factor

liters of NO2 (at STP) Molar volume

Molarity

1 L solution 0.400 mol HNO31 L solution

0.400 mol HNO3 and

1 L of solution = 0.400 mol of HNO3 2 mol HNO3 3 mol NO22 mol HNO3

3 mol NO2 and

3 mol of NO2 = 2 mol of HNO3

0.275 L solution 3.70 L of NO2 (STP)* * = 0.400 mol HNO3

1 L solution 22.4 L NO2 (STP)

1 mol NO2 *

3 mol NO2 2 mol HNO3

STEP 4 Set up the problem to calculate the needed quantity.

22.4 L NO2 (STP) 1 mol NO2 22.4 L NO2 (STP)

1 mol NO2and

1 mol of NO2 = 22.4 L of NO2 (STP)

liters of solution moles of HNO3 moles of NO2 Mole–mole factor

liters of NO2 (at STP) Molar volume

Molarity

1 L solution 0.400 mol HNO31 L solution

0.400 mol HNO3 and

1 L of solution = 0.400 mol of HNO3 2 mol HNO3 3 mol NO22 mol HNO3

3 mol NO2 and

3 mol of NO2 = 2 mol of HNO3

0.275 L solution 3.70 L of NO2 (STP)* * = 0.400 mol HNO3

1 L solution 22.4 L NO2 (STP)

1 mol NO2 *

3 mol NO2 2 mol HNO3

22.4 L NO2 (STP) 1 mol NO2 22.4 L NO2 (STP)

1 mol NO2and

1 mol of NO2 = 22.4 L of NO2 (STP)

liters of solution moles of HNO3 moles of NO2 Mole–mole factor

liters of NO2 (at STP) Molar volume

Molarity

1 L solution 0.400 mol HNO31 L solution

0.400 mol HNO3 and

1 L of solution = 0.400 mol of HNO3 2 mol HNO3 3 mol NO22 mol HNO3

3 mol NO2 and

3 mol of NO2 = 2 mol of HNO3

0.275 L solution 3.70 L of NO2 (STP)* * = 0.400 mol HNO3

1 L solution 22.4 L NO2 (STP)

1 mol NO2 *

3 mol NO2 2 mol HNO3

SELF TEST 12.16

a. Using the equation in Sample Problem 12.16, determine the volume of NO produced at 100 °C and 1.20 atm, when 2.20 L of a 1.50 M HNO3 solution is produced.

b. Using the reaction in Sample Problem 12.16, how many milliliters of 0.344 M HNO3 solution are produced when 1.40 L of NO2 gas at 0.800 atm and 20. °C reacts?

ANSWER

a. 42.1 L of NO b. 90.3 mL of HNO3 solution

PRACTICE PROBLEMS Try Practice Problems 12.61 to 12.66

ENGAGE 12.12 What sequence of conversion factors would you use to calculate the number of grams of CaCO3 needed to react with 1.50 L of a 2.00 M HCl solution in the reaction 2HCl(aq) + CaCO3(s) h CaCl2(aq) + CO2(g) + H2O(l )?

FIGURE 12.9 gives a summary of the pathways and conversion factors needed for substances including solutions involved in chemical reactions.

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378 CHAPTER 12 Solutions

12.7 Molality and Freezing Point Lowering/ Boiling Point Elevation LEARNING GOAL Identify a mixture as a solution, a colloid, or a suspension. Using the molality, calculate the freezing point and boiling point for a solution.

The size and number of solute particles in different types of mixtures play an important role in determining the properties of that mixture.

Solutions In the solutions discussed up to now, the solute was dissolved as small particles that are uniformly dispersed throughout the solvent to give a homogeneous solution. When you observe a solution, such as salt water, you cannot visually distinguish the solute from the solvent. The solution appears transparent, although it may have a color. The particles are so small that they go through filters and through semipermeable membranes. A semipermeable membrane allows solvent molecules such as water and very small solute particles to pass through, but does not allow the passage of large solute molecules.

Colloids The particles in a colloid are much larger than solute particles in a solution. Colloidal par- ticles are large molecules, such as proteins, or groups of molecules or ions. Colloids, similar to solutions, are homogeneous mixtures that do not separate or settle out. Colloidal particles are small enough to pass through filters, but too large to pass through semipermeable mem- branes. TABLE 12.9 lists several examples of colloids.

12.64 Answer the following for the reaction:

2HCl(aq) + CaCO3(s) h CO2(g) + H2O(l) + CaCl2(aq) a. How many milliliters of a 0.200 M HCl solution can react

with 8.25 g of CaCO3? b. How many liters of CO2 gas can form at STP when 15.5 mL

of a 3.00 M HCl solution reacts with excess CaCO3? c. What is the molarity of a HCl solution if the reaction of

200. mL of the HCl solution with excess CaCO3 produces 12.0 L of CO2 gas at 725 mmHg and 18 °C?

12.65 Answer the following for the reaction:

2HBr(aq) + Zn(s) h H2(g) + ZnBr2(aq) a. How many milliliters of a 3.50 M HBr solution are required

to react with 8.56 g of zinc? b. How many liters of hydrogen gas can form at STP when

0.750 L of a 6.00 M HBr solution reacts with excess zinc? c. What is the molarity of a HBr solution if the reaction of

28.7 mL of the HBr solution with excess zinc produces 0.620 L of H2 gas at 725 mmHg and 24 °C?

12.66 Answer the following for the reaction: 3AgNO3(aq) + Na3PO4(aq) h Ag3PO4(s) + 3NaNO3(aq)

a. How many milliliters of a 0.265 M AgNO3 solution are required to react with 31.2 mL of 0.154 M Na3PO4 solution?

b. How many grams of silver phosphate are produced from the reaction of 25.0 mL of a 0.165 M AgNO3 solution and excess Na3PO4?

c. What is the molarity of 20.0 mL of a Na3PO4 solution that reacts completely with 15.0 mL of a 0.576 M AgNO3 solution?

PRACTICE PROBLEMS

12.6 Chemical Reactions in Solution

12.61 Answer the following for the reaction:

Pb(NO3)2(aq) + 2KCl(aq) h PbCl2(s) + 2KNO3(aq) a. How many grams of PbCl2 will be formed from 50.0 mL of

a 1.50 M KCl solution? b. How many milliliters of a 2.00 M Pb(NO3)2 solution will

react with 50.0 mL of a 1.50 M KCl solution? c. What is the molarity of 20.0 mL of a KCl solution that reacts

completely with 30.0 mL of a 0.400 M Pb(NO3)2 solution?

12.62 Answer the following for the reaction:

NiCl2(aq) + 2NaOH(aq) h Ni(OH)2(s) + 2NaCl(aq) a. How many milliliters of a 0.200 M NaOH solution are

needed to react with 18.0 mL of a 0.500 M NiCl2 solution? b. How many grams of Ni(OH)2 are produced from the reac-

tion of 35.0 mL of a 1.75 M NaOH solution and excess NiCl2?

c. What is the molarity of 30.0 mL of a NiCl2 solution that reacts completely with 10.0 mL of a 0.250 M NaOH solution?

12.63 Answer the following for the reaction:

2HCl(aq) + Mg(s) h H2(g) + MgCl2(aq) a. How many milliliters of a 6.00 M HCl solution are required

to react with 15.0 g of magnesium? b. How many liters of hydrogen gas can form at STP when

0.500 L of a 2.00 M HCl solution reacts with excess magnesium?

c. What is the molarity of a HCl solution if the reaction of 45.2 mL of the HCl solution with excess magnesium produces 5.20 L of H2 gas at 735 mmHg and 25 °C?

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12.7 Molality and Freezing Point Lowering/Boiling Point Elevation 379

Suspensions separate into solids and water that can be removed by filtration.

TABLE 12.9 Examples of Colloids Colloid Substance Dispersed Dispersing Medium

Fog, clouds, hair sprays Liquid Gas

Dust, smoke Solid Gas

Shaving cream, whipped cream, soapsuds Gas Liquid

Styrofoam, marshmallows Gas Solid

Mayonnaise, homogenized milk Liquid Liquid

Cheese, butter Liquid Solid

Blood plasma, paints (latex), gelatin Solid Liquid

TABLE 12.10 Comparison of Solutions, Colloids, and Suspensions Type of Mixture Type of Particle Settling Separation

Solution Small particles such as atoms, ions, or small molecules

Particles do not settle

Particles cannot be separated by filters or semipermeable membranes

Colloid Larger molecules or groups of molecules or ions

Particles do not settle

Particles can be separated by semipermeable membranes but not by filters

Suspension Very large particles that may be visible

Particles settle rapidly

Particles can be separated by filters

Suspensions Suspensions are heterogeneous, nonuniform mixtures that are very different from solutions or colloids. The particles of a suspension are so large that they can often be seen with the naked eye. They are trapped by filters and semipermeable membranes.

The weight of the suspended solute particles causes them to settle out soon after mix- ing. If you stir muddy water, it mixes but then quickly separates as the suspended particles settle to the bottom and leave clear liquid at the top. You can find suspensions among the medications in a hospital or in your medicine cabinet. These include Kaopectate, calamine lotion, antacid mixtures, and liquid penicillin. It is important to follow the instructions on the label that state “shake well before using” so that the particles form a suspension.

Water-treatment plants make use of the properties of suspensions to purify water. When chemicals such as aluminum sulfate or iron(III) sulfate are added to untreated water, they react with impurities to form large suspended particles called floc. In the water-treatment plant, a system of filters traps the suspended particles, but clean water passes through.

TABLE 12.10 compares the different types of mixtures, and FIGURE 12.10 illustrates some properties of solutions, colloids, and suspensions.

FIGURE 12.10 Properties of different types of mixtures.

Solution

Suspension Colloid

Settling

Filter Semipermeable membrane

Suspensions settle out.

Suspensions are separated by a filter.

Solution particles go through a semipermeable membrane, but colloids and suspensions do not.

PRACTICE PROBLEMS Try Practice Problems 12.67 and 12.68

ENGAGE 12.13 A filter can be used to separate suspension particles from a solution, but a semipermeable membrane is needed to separate colloids from a solution. Explain.

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380 CHAPTER 12 Solutions

Solute and solventPure solvent

Vapor pressure Lower vapor pressure Freezing Point Lowering and Boiling Point Elevation When we add a solute to water, it changes the vapor pressure, freezing point, and boiling point of the solvent pure water. These types of changes in physical properties, known as colligative properties, depend only on the concentration of solute particles in the solution.

We can illustrate how these changes in physical properties occur by comparing the number of evaporating solvent molecules in pure solvent with those in a solution with a nonvolatile solute in the solvent. In the solution, there are fewer solvent molecules at the surface because the solute that has been added takes up some of the space at the surface. As a result, fewer solvent molecules can evaporate compared to the pure

solvent. Vapor pressure is lowered for the solution. If we add more solute molecules, the vapor pressure will be lowered even more.

The freezing point of a solvent is lowered when a nonvolatile solute is added. In this case, the solute particles prevent the organization of solvent molecules needed to form the solid state. Thus, a lower temperature is required before the molecules of the solvent can become organized enough to freeze. When you spread salt on an icy sidewalk when the temperature is below freezing, the particles from the salt combine with water to lower the freezing point, which causes the ice to melt.

Insects and fish in climates with subfreezing temperatures control ice formation by pro- ducing biological antifreezes made of glycerol, proteins, and sugars, such as glucose, within their bodies. Some insects can survive temperatures below - 60 °C. These forms of biologi- cal antifreezes may one day be applied to the long-term preservation of human organs.

The boiling point of a solvent is raised when a nonvolatile solute is added. The vapor pressure of a solvent must reach atmospheric pressure before it begins boiling. However, because the solute lowers the vapor pressure of the solvent, a temperature higher than the normal boiling point of the pure solvent is needed to cause the solution to boil. Antifreeze, which is a compound such as ethylene glycol, C2H6O2, is added to the water in a car radiator. If an ethylene glycol and water mixture is about 50–50% by mass, it does not freeze until the temperature drops to about - 30 °F, and does not boil unless the temperature reaches about 225 °F. The solution in the radiator prevents the water in the radiator from forming ice in cold weather or boiling over on a hot desert highway.

Particles in Solution A solute that is a nonelectrolyte dissolves as molecules, whereas a solute that is a strong elec- trolyte dissolves entirely as ions. The antifreeze ethylene glycol, C2H6O2, a nonelectrolyte, dissolves as molecules.

Nonelectrolyte: 1 mol of C2H6O2(l) = 1 mol of C2H6O2(aq2 However, when 1 mol of a strong electrolyte, such as NaCl or CaCl2, dissolves in water,

the NaCl solution will contain 2 mol of particles, and the CaCl2 solution will contain 3 mol of particles.

Strong electrolytes:

1 mol of NaCl(s) = 1 mol of Na+(aq) + 1 mol of Cl-(aq) (++++111++)++11+++111*

2 mol of particles (aq)

1 mol of CaCl2(s) = 1 mol of Ca 2+(aq) + 2 mol of Cl-(aq)

(++++111++)++11+++111* 3 mol of particles (aq)

Molality (m) The calculation for freezing point lowering or boiling point elevation uses a concentration unit of molality. The molality, abbreviation m, of a solution is the number of moles of solute particles per kilogram of solvent. This may seem similar to molarity, but the denominator for molality refers to the mass of the solvent, not the volume of the solution.

Molality (m) = moles of solute particles

kilograms of solvent

The Alaska Upis beetle produces biological antifreeze to survive subfreezing temperatures.

Ethylene glycol is added to a radiator to form an aqueous solution that has a lower freezing point than water and a higher boiling point.

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12.7 Molality and Freezing Point Lowering/Boiling Point Elevation 381

Freezing Point Lowering The change in the freezing point temperature of a solvent (∆Tf) is determined from the molality (m) of the particles in the solution and the freezing point constant, Kf, for the solvent.

∆Tf = m * Kf The freezing point constant (Kf) for water, which is determined experimentally, is

1.86 °C m

.

For a 1 m solution, we can calculate the change in the freezing point temperature as

∆Tf = m * Kf = 1 m * 1.86 °C

m = 1.86 °C

Then we can calculate the new, lower freezing point of the solution:

Tsolution = Twater - ∆Tf = 0.00 °C - 1.86 °C = - 1.86 °C

CORE CHEMISTRY SKILL Calculating the Freezing Point/

Boiling Point of a Solution

SAMPLE PROBLEM 12.17 Calculating Molality

TRY IT FIRST

Calculate the molality of a solution containing 35.5 g of the nonelectrolyte glucose (C6H12O6) in 0.400 kg of water.

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

35.5 g of glucose (C6H12O6), 0.400 kg of water

molality (m) molar mass of glucose

STEP 2 Write the molality expression.

Molality (m) of glucose solution = moles of glucose

kilograms of water

STEP 3 Substitute solute and solvent quantities into the expression and calculate. To match the unit of moles in the molality expression, we convert the grams of glucose to moles of glucose, using its molar mass.

1 mol glucose 180.16 g glucose

and 180.16 g glucose

1 mol glucose

1 mol of glucose = 180.16 g of glucose

moles of glucose 35.5 g glucose = 0.197 mol of glucose* 1 mol glucose

180.16 g glucose =

Molality (m) = 0.197 mol glucose

0.400 kg water = 0.493 m

SELF TEST 12.17

Calculate the molality of each of the following solutions: a. 15.8 g of urea, CH4N2O, a nonelectrolyte, in 250. g of water b. 44.6 g of ethanol, C2H6O, a nonelectrolyte, in 340. g of water

ANSWER

a. 1.05 m b. 2.84 m

PRACTICE PROBLEMS Try Practice Problems 12.69 and 12.70

ENGAGE 12.14 Why does a 1 m solution of KBr, a strong electrolyte, lower the freez- ing point of water more than a 1 m solution of urea, a nonelectrolyte?

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382 CHAPTER 12 Solutions

A truck spreads calcium chloride on the road to melt ice and snow.

SAMPLE PROBLEM 12.18 Calculating the Freezing Point of a Solution

TRY IT FIRST

In the northeastern United States during freezing weather, CaCl2 is spread on highways to melt the ice. Calculate the freezing point lowering and freezing point of a solution containing 225 g of CaCl2 in 500. g of water.

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

225 g of CaCl2, 500. g of water = 0.500 kg of water

∆Tf , freezing point

molar mass, ∆Tf = m * Kf

1 mol CaCl2 110.98 g NaCl

and 110.98 g CaCl2

1 mol CaCl2

1 mol of CaCl2 = 110.98 g of CaCl2 1 mol CaCl2

3 mol solute particles and

3 mol solute particles 1 mol CaCl2

1 mol of CaCl2 = 3 mol of solute particles

3 mol of solute particles =CaCl2(s) Ca 2+(aq) 2Cl-(aq)+

STEP 2 Determine the number of moles of solute particles and calculate the molality.

We use molar mass to calculate the moles of CaCl2. Then we multiply by three to obtain the number of moles of ions (particles) produced by 1 mol of CaCl2 in solution.

3 mol particles 225 g CaCl2

=

=

6.08 mol of particles

moles of particles * * 1 mol CaCl2

1 mol CaCl2110.98 g CaCl2

The molality (m) of the particles in solution is obtained by dividing the moles of particles by the number of kilograms of water in the solution.

m = moles of particles

kilograms of water

Molality (m) = 6.08 mol particles

0.500 kg water = 12.2 m

STEP 3 Calculate the temperature change and subtract from the freezing point. The freezing point lowering is calculated using the molality and the freezing point constant. The freezing point lowering is subtracted from 0.00 °C to obtain the new freezing point of the CaCl2 solution.

∆Tf = m * Kf

∆Tf = 12.2 m * 1.86 °C

m = 22.7 °C

Tsolution = Twater - ∆Tf = 0.00 °C - 22.7 °C = - 22.7 °C

SELF TEST 12.18

Calculate the freezing point of each of the following solutions: a. ethylene glycol, C2H6O2, a nonelectrolyte, is added to the water in a radiator to give a

solution containing 315 g of ethylene glycol in 565 g of water b. sodium nitrate, NaNO3, a strong electrolyte, is added to water to give a solution containing

315 g of sodium nitrate in 565 g of water

ANSWER

a. - 16.7 °C b. - 24.4 °C

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12.7 Molality and Freezing Point Lowering/Boiling Point Elevation 383

Boiling Point Elevation A similar change occurs with the boiling point of water. The boiling point elevation (∆Tb) is determined from the molality (m) of the particles in the solution and the boiling point

constant, Kb, 0.51 °C

m for water, which is determined experimentally.

∆Tb = m * Kb The effect of some solutions on freezing and boiling point is summarized in TABLE 12.11.

SAMPLE PROBLEM 12.19 Calculating the Boiling Point of a Solution

TRY IT FIRST

Propylene glycol, C3H8O2, is a nonelectrolyte that is added to the water in a radiator to increase the boiling point. If 4.6 mol of propylene glycol is added to 1.55 kg of water (solvent) in a radiator, what is the boiling point of the solution?

TABLE 12.11 Effect of Solute Concentration on Freezing and Boiling Points of 1 kg of Water

Substance/kg water Type of Solute Molality of Particles Freezing Point Boiling Point

Pure water None 0 0.00 °C 100.00 °C

1 mol of C2H6O2 Nonelectrolyte 1 m - 1.86 °C 100.51 °C 1 mol of NaCl Strong electrolyte 2 m - 3.72 °C 101.02 °C 1 mol of CaCl2 Strong electrolyte 3 m - 5.58 °C 101.53 °C

PRACTICE PROBLEMS Try Practice Problems 12.71 to 12.74

Pure water 1 mol particles 1 kg water

2 mol particles 1 kg water

3 mol particles 1 kg water

Bp 100.00 °C

Fp 0.00 °C

Fp -1.86 °C

Fp -3.72 °C

Fp -5.58 °C T decreases

Bp 100.51 °C Bp 101.02 °C

Bp 101.53 °C T increases

As the concentration of a solution increases, the freezing point is lowered and the boiling point is raised.

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384 CHAPTER 12 Solutions

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

4.6 mol of propylene glycol, 1.55 kg of water

boiling point ∆Tb = m * Kb

STEP 2 Determine the number of moles of solute particles and calculate the molality. Since propylene glycol is a nonelectrolyte, the moles of propylene glycol is equal to the moles of particles.

m = moles of particles

kilograms of water =

4.6 mol 1.55 kg

= 3.0 m

STEP 3 Calculate the temperature change and add to the boiling point.

∆Tb = m * Kb = 3.0 m * 0.51 °C

m = 1.5 °C

Tsolution = Twater + ∆Tb = 100.0 °C + 1.5 °C = 101.5 °C

SELF TEST 12.19

Calculate the boiling point of each of the following solutions: a. 1.2-mol of potassium phosphate, K3PO4, a strong electrolyte, added to 0.26 kg of water b. 36 g of ethylene glycol, C2H6O2, a nonelectrolyte, added to 130 g of water

ANSWER

a. 109.6 °C b. 102.3 °C

12.71 Calculate the freezing point and boiling point of each solution in problem 12.69.

12.72 Calculate the freezing point and boiling point of each solution in problem 12.70.

12.73 In each pair, identify the solution that will have a lower freezing point. Explain.

a. 1.0 mol of glycerol (nonelectrolyte) and 2.0 mol of ethylene glycol (nonelectrolyte) each in 1.0 kg of water

b. 0.50 mol of KCl (strong electrolyte) and 0.50 mol of MgCl2 (strong electrolyte) each in 1.0 kg of water

12.74 In each pair, identify the solution that will have a higher boiling point. Explain.

a. 1.50 mol of LiOH (strong electrolyte) and 3.00 mol of KOH (strong electrolyte) each in 1.0 kg of water

b. 0.40 mol of Al(NO3)3 (strong electrolyte) and 0.40 mol of CsCl (strong electrolyte) each in 1.0 kg of water

PRACTICE PROBLEMS

12.7 Molality and Freezing Point Lowering/Boiling Point Elevation

12.67 Identify each of the following as characteristic of a solution, colloid, or suspension:

a. a mixture that cannot be separated by a semipermeable membrane

b. a mixture that settles out upon standing

12.68 Identify each of the following as characteristic of a solution, colloid, or suspension:

a. particles of this mixture remain inside a semipermeable membrane but pass through filters

b. particles of solute in this mixture are very large and visible

12.69 Calculate the molality (m) of the following solutions: a. 325 g of methanol, CH4O, a nonelectrolyte, added to 455 g

of water b. 640. g of the antifreeze propylene glycol, C3H8O2, a

nonelectrolyte, dissolved in 1.22 kg of water

12.70 Calculate the molality (m) of the following solutions: a. 65.0 g of glucose, C6H12O6, a nonelectrolyte, dissolved in

112 g of water b. 110. g of sucrose, C12H22O11, a nonelectrolyte, dissolved in

1.50 kg of water

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12.8 Properties of Solutions: Osmosis 385

12.8 Properties of Solutions: Osmosis LEARNING GOAL Describe how the number of particles in solution affects osmosis.

The movement of water into and out of the cells of plants as well as the cells of our bod- ies is an important biological process that also depends on the solute concentration. In a process called osmosis, water molecules move through a semipermeable membrane from the solution with the lower concentration of solute into a solution with the higher solute concentration. In an osmosis apparatus, water is placed on one side of a semipermeable membrane and a sucrose (sugar) solution on the other side. The semipermeable membrane allows water molecules to flow back and forth but blocks the sucrose molecules because they cannot pass through the membrane. Because the sucrose solution has a higher solute concentration, more water molecules flow into the sucrose solution than out of the sucrose solution. The volume level of the sucrose solution rises as the volume level on the water side falls. The increase of water dilutes the sucrose solution to equalize (or attempt to equalize) the concentrations on both sides of the membrane.

Eventually the height of the sucrose solution creates sufficient pressure to equalize the flow of water between the two compartments. This pressure, called osmotic pressure, prevents the flow of additional water into the more concentrated solution. Then there is no further change in the volumes of the two solutions. The osmotic pressure depends on the

Semipermeable membrane

Water (solvent)

Sucrose (solute)

Time

Semipermeable membrane

H2O

H2O

H2O

H2O Water flows into the solution with a higher solute concentration until the flow of water becomes equal in both directions.

concentration of solute particles in the solution. The greater the number of particles dis- solved, the higher its osmotic pressure. In this example, the sucrose solution has a higher osmotic pressure than pure water, which has an osmotic pressure of zero.

In a process called reverse osmosis, a pressure greater than the osmotic pressure is applied to a solution so that it is forced through a purification membrane. The flow of water is reversed because water flows from an area of lower water concentration to an area of higher water concentration. The molecules and ions in solution stay behind, trapped by the membrane, while water passes through the membrane. This process of reverse osmosis is used in a few desalination plants to obtain pure water from sea (salt) water. However, the pressure that must be applied requires so much energy that reverse osmosis is not yet an economical method for obtaining pure water in most parts of the world.

Isotonic Solutions Because the cell membranes in biological systems are semipermeable, osmosis is an ongoing process. The solutes in body solutions such as blood, tissue fluids, lymph, and plasma all exert osmotic pressure. Most intravenous (IV) solutions used in a hospital are isotonic solutions, which exert the same osmotic pressure as body fluids such as blood. The percent concentration typically used in IV solutions is mass/volume percent (m/v), which is a type of percent con- centration we have already discussed. The most typical isotonic solutions are 0.9% (m/v) NaCl solution, or 0.9 g of NaCl/100. mL of solution, and 5% (m/v) glucose, or 5 g of glucose/100. mL of solution. Although they do not contain the same kinds of particles, a 0.9% (m/v) NaCl

A 0.9% NaCl solution is isotonic with the solute concentration of the blood cells of the body.

PRACTICE PROBLEMS Try Practice Problems 12.75 to 12.78

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386 CHAPTER 12 Solutions

solution as well as a 5% (m/v) glucose solution both have the same osmotic pressure. A red blood cell placed in an isotonic solution retains its volume because there is an equal flow of water into and out of the cell (see FIGURE 12.11a).

Hypotonic and Hypertonic Solutions If a red blood cell is placed in a solution that is not isotonic, the differences in osmotic pressure inside and outside the cell can drastically alter the volume of the cell. When a red blood cell is placed in a hypotonic solution, which has a lower solute concentration (hypo means “lower than”), water flows into the cell by osmosis. The increase in fluid causes the cell to swell, and possibly burst—a process called hemolysis (see FIGURE 12.11b). A similar process occurs when you place dehydrated food, such as raisins or dried fruit, in water. The water enters the cells, and the food becomes plump and smooth.

FIGURE 12.11 Red blood cells in isotonic, hypotonic, and hypertonic solutions

Isotonic solution

Hypotonic solution

Hypertonic solution

(a) Normal (b) Hemolysis (c) Crenation

In an isotonic solution, a red blood cell retains its normal volume.

In a hypotonic solution, water flows into a red blood cell, causing it to swell and burst.

In a hypertonic solution, water leaves the red blood cell, causing it to shrink.

If a red blood cell is placed in a hypertonic solution, which has a higher solute concen- tration (hyper means “greater than”), water flows out of the cell into the hypertonic solution by osmosis. Suppose a red blood cell is placed in a 10% (m/v) NaCl solution. Because the osmotic pressure in the red blood cell is the same as a 0.9% (m/v) NaCl solution, the 10% (m/v) NaCl solution has a much greater osmotic pressure. As water leaves the cell, it shrinks, a process called crenation (see FIGURE 12.11c). A similar process occurs when making pickles, which uses a hypertonic salt solution that causes the cucumbers to shrivel as they lose water.

ENGAGE 12.15 What happens to a red blood cell placed in a 4% NaCl solution?

SAMPLE PROBLEM 12.20 Isotonic, Hypotonic, and Hypertonic Solutions

TRY IT FIRST

Describe each of the following solutions as isotonic, hypotonic, or hypertonic. Indicate whether a red blood cell placed in each solution will undergo hemolysis, crenation, or no change.

a. a 5% (m/v) glucose solution b. a 0.2% (m/v) NaCl solution

SOLUTION

a. A 5% (m/v) glucose solution is isotonic. A red blood cell will not undergo any change. b. A 0.2% (m/v) NaCl solution is hypotonic. A red blood cell will undergo hemolysis.

PRACTICE PROBLEMS Try Practice Problems 12.79 and 12.80

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12.8 Properties of Solutions: Osmosis 387

Initial Final

Solution particles such as Na+, Cl-, glucose

Colloidal particles such as protein, starch

Solution particles pass through a dialyzing membrane, but colloidal particles are retained.

Dialysis Dialysis is a process that is similar to osmosis. In dialysis, a semipermeable membrane, called a dialyzing membrane, permits small solute molecules and ions as well as solvent water molecules to pass through, but it retains large particles, such as colloids.

Suppose we fill a cellophane bag with a solution containing NaCl, glucose, starch, and protein and place it in pure water. Cellophane is a dialyzing membrane, and the sodium ions, chloride ions, and glucose molecules will pass through it into the surrounding water. However, large colloidal particles, like starch and protein, remain inside. Water molecules will flow into the cellophane bag. Eventually the concentrations of sodium ions, chloride ions, and glucose molecules inside and outside the dialysis bag become equal. To remove more NaCl or glucose, the cellophane bag must be placed in a fresh sample of pure water.

SELF TEST 12.20

a. What will happen to a red blood cell placed in a 10% (m/v) glucose solution? b. What will happen to a red blood cell placed in a 1.0% (m/v) glucose solution?

ANSWER

a. The red blood cell will shrink (crenate). b. The red blood cell will undergo hemolysis.

PRACTICE PROBLEMS Try Practice Problems 12.81 and 12.82

Chemistry Link to Health Hemodialysis and the Artificial Kidney

In an adult, each kidney contains about 2 million nephrons. At the top of each nephron, there is a network of arterial capillaries called the glomerulus. As blood flows into the glomerulus, small particles, such as amino acids, glucose, urea, water, and certain ions dialyze through the capillary membranes into the nephron. As this solution moves through the nephron, substances still of value to the body (such as amino acids, glucose, certain ions, and 99% of the water) are reabsorbed. The major waste product, urea, is excreted in the urine. If the kidneys fail to remove waste products, increased levels of urea can become life-threatening in a relatively short time.

A person with kidney failure must use an artificial kidney, which cleanses the blood by hemodialysis. A typical artificial kidney machine contains a large tank filled with water containing selected electrolytes. In the dialyzing bath (dialysate), there is a dialyzing membrane made of cellulose tubing. As the patient’s blood f lows through the dialyzing membrane, the concentrated waste products dialyze out of the blood. No blood is lost because the membrane is not permeable to red blood cells. Dialysis patients do not produce much urine. As a result, they retain large amounts of water between dialysis treatments, which produces a strain on the heart. For some

dialysis patients, 2 to 10 L of water may be removed during one treatment. At a treatment center, dialysis patients typically have three treatments a week for about 4 h each. For many patients, dialysis is done at home using a home dialysis unit for 1.5 to 2 h every day.

Blood out (filtered)

Urine to ureter

Collecting duct

Blood in

Glomerulus

In the kidneys, each nephron contains a glomerulus where urea and waste products are removed to form urine.

Dialyzed blood

Blood

Urea and other waste products

Pump

Dialyzing coil

Dialysate

=

During dialysis, waste products and excess water are removed from the blood.

M12_TIMB8119_06_SE_C12.indd 387 11/27/18 12:08 PM

388 CHAPTER 12 Solutions

PRACTICE PROBLEMS

12.8 Properties of Solutions: Osmosis

12.75 A 10% (m/v) starch solution is separated from a 1% (m/v) starch solution by a semipermeable membrane. (Starch is a colloid.)

a. Which compartment has the higher osmotic pressure? b. In which direction will water flow initially? c. In which compartment will the volume level rise?

12.76 A 0.1% (m/v) albumin solution is separated from a 2% (m/v) albumin solution by a semipermeable membrane. (Albumin is a colloid.)

a. Which compartment has the higher osmotic pressure? b. In which direction will water flow initially? c. In which compartment will the volume level rise?

12.77 Indicate the compartment (A or B) that will increase in volume for each of the following pairs of solutions separated by a semipermeable membrane:

A B

a. 5% (m/v) sucrose 10% (m/v) sucrose

b. 8% (m/v) albumin 4% (m/v) albumin

c. 0.1% (m/v) starch 10% (m/v) starch

12.78 Indicate the compartment (A or B) that will increase in volume for each of the following pairs of solutions separated by a semipermeable membrane:

A B

a. 20% (m/v) starch 10% (m/v) starch

b. 10% (m/v) albumin 2% (m/v) albumin

c. 0.5% (m/v) sucrose 5% (m/v) sucrose

Applications

12.79 Are the following solutions isotonic, hypotonic, or hypertonic compared with a red blood cell?

a. distilled H2O b. 1% (m/v) glucose c. 0.9% (m/v) NaCl d. 15% (m/v) glucose

12.80 Will a red blood cell undergo crenation, hemolysis, or no change in each of the following solutions?

a. 1% (m/v) glucose b. 2% (m/v) NaCl c. 5% (m/v) glucose d. 0.1% (m/v) NaCl

12.81 Each of the following mixtures is placed in a dialyzing bag and immersed in distilled water. Which substances will be found outside the bag in the distilled water?

a. NaCl solution b. starch solution (colloid) and alanine (an amino acid)

solution c. NaCl solution and starch solution (colloid) d. urea solution

12.82 Each of the following mixtures is placed in a dialyzing bag and immersed in distilled water. Which substances will be found outside the bag in the distilled water?

a. KCl solution and glucose solution b. albumin solution (colloid) c. an albumin solution (colloid), KCl solution, and glucose

solution d. urea solution and NaCl solution

UPDATE Using Dialysis for Renal Failure

As a dialysis patient, Michelle has a 4-h dialysis treatment three times a week. When she arrives at the dialysis clinic, her weight, temperature, and blood pressure are taken and blood tests are done to determine the level of elec- trolytes and urea in her blood. In the dialysis center, tubes to the dialyzer are connected to the catheter she has had implanted. Blood is then pumped out of her body, through the dialyzer where it is filtered, and returned to her body. As Michelle’s blood flows through the dialyzer, electrolytes from the dialysate move into her blood, and waste products in her blood move into the dialysate, which is continually renewed.

To achieve normal serum electrolyte levels, dialysate fluid contains sodium, chloride, and magnesium levels that are equal to serum concentrations. These electrolytes are removed from the blood only if their concentrations are higher than normal. Typically, in dialysis patients, the potassium ion level is higher than normal. Therefore, initial dialysis may start with a low concentration of potassium ion in the dialysate. During dialysis excess fluid is removed

by osmosis. A 4-h dialysis session requires at least 120 L of dialysis fluid. During dialysis, the electrolytes in the dialy- sate are adjusted until the electrolytes have the same levels as normal serum. Initially the dialysate solution prepared for Michelle contains the following: HCO3

-, K+, Na+, Ca2+, Mg2+, Cl-, and glucose.

Applications 12.83 A doctor orders 0.075 g of chlorpromazine, which is used

to treat schizophrenia. If the stock solution is 2.5% (m/v), how many milliliters are administered to the patient?

12.84 A doctor orders 5.0 mg of compazine, which is used to treat nausea, vertigo, and migraine headaches. If the stock solution is 2.5% (m/v), how many milliliters are administered to the patient?

12.85 A CaCl2 solution is given to increase blood levels of cal- cium. If a patient receives 5.0 mL of a 10.% (m/v) CaCl2 solution, how many grams of CaCl2 were given?

12.86 An intravenous solution of mannitol is used as a diuretic to increase the loss of sodium and chloride by a patient. If a patient receives 30.0 mL of a 25% (m/v) mannitol solution, how many grams of mannitol were given?

M12_TIMB8119_06_SE_C12.indd 388 11/27/18 12:08 PM

Chapter Review 389

CONCEPT MAP

Strong Weak Moles

Volume

Dilutions

Electrolytes

Saturated

Solute ConcentrationSolvent

SOLUTIONS

Solubility Nonelectrolytes Percent Molarity

Solute Particles

Vapor Pressure, Freezing Point, Boiling Point,

Osmotic Pressure

m>m, v>v, m>v mol>L

do not dissociate

of

consist of a

is used to calculate

Freezing and

Boiling Point

Change

Molality

mol>kg

is used to calculate

asamount dissolved is

the maximum

amount that dissolves is

dissociate

100% slightly

change the

add water for

CHAPTER REVIEW

12.1 Solutions LEARNING GOAL Identify the solute and solvent in a solution; describe the formation of a solution. • A solution forms when a solute dissolves

in a solvent. • In a solution, the particles of solute are

evenly dispersed in the solvent. • The solute and solvent may be solid,

liquid, or gas. • The polar O ¬ H bond leads to hydrogen

bonding between water molecules. • An ionic solute dissolves in water, a polar solvent, because the

polar water molecules attract and pull the ions into solution, where they become hydrated.

• The expression like dissolves like means that a polar or an ionic solute dissolves in a polar solvent while a nonpolar solute dissolves in a nonpolar solvent.

12.2 Electrolytes and Nonelectrolytes LEARNING GOAL Identify solutes as electrolytes or nonelectrolytes. • Substances that produce ions in water

are called electrolytes because their solu- tions will conduct an electrical current.

• Strong electrolytes are completely dis- sociated, whereas weak electrolytes are only partially dissociated.

• Nonelectrolytes are substances that dissolve in water to produce only molecules and cannot conduct electrical currents.

12.3 Solubility LEARNING GOAL Define solubility; distinguish between an unsaturated and a saturated solution. Identify an ionic compound as soluble or insoluble. • The solubility of a solute is the

maximum amount of a solute that can dissolve in 100. g of solvent.

• A solution that contains the maxi- mum amount of dissolved solute is a saturated solution.

• A solution containing less than the maximum amount of dissolved solute is unsaturated.

• An increase in temperature increases the solubility of most solids in water, but decreases the solubility of gases in water.

• Ionic compounds that are soluble in water usually contain Li+, Na+, K+, NH4

+, NO3 -, or acetate, C2H3O2

-. • When two solutions are mixed, solubility rules can be used to

predict whether a precipitate will form.

12.4 Solution Concentrations LEARNING GOAL Calculate the concentration of a solute in a solution; use concentration units to calculate the amount of solute or solution. • Mass percent expresses the mass/mass

(m/m) ratio of the mass of solute to the mass of solution multiplied by 100%.

• Percent concentration can also be expressed as volume/volume (v/v) and mass/volume (m/v) ratios.

• Molarity is the moles of solute per liter of solution. • In calculations of grams or milliliters of solute or solution, the

concentration is used as a conversion factor. • Molarity (or mol/L) is written as conversion factors to solve for

moles of solute or volume of solution.

Solute: The substance present in lesser amount

Solvent: The substance present in greater amount

Salt

Water

Cl-

Cl-

Cl-Na+

Na+ Na+

Strong electrolyte

+-

Saturated solution

Dissolved solute

Undissolved solute D

is so

lv in

g C

ry st

al liz

at io

n

250 mL Water added to make a solution

250 mL of KI solution

5.0 g of KI

M12_TIMB8119_06_SE_C12.indd 389 11/27/18 12:08 PM

390 CHAPTER 12 Solutions

Semipermeable membrane

12.5 Dilution of Solutions LEARNING GOAL Describe the dilution of a solution; calculate the unknown concentration or volume when a solution is diluted. • In dilution, a solvent such as

water is added to a solution, which increases its volume and decreases its concentration.

12.6 Chemical Reactions in Solution LEARNING GOAL Given the volume and concentration of a solution in a chemical reaction, calculate the amount of a reactant or product in the reaction. • When solutions are involved in chemical

reactions, the moles of a substance in solution can be determined from the vol- ume and molarity of the solution.

• When mass, volume, and molarities of substances in a reaction are given, the balanced equation is used to determine the quantities or concentrations of other substances in the reaction.

12.7 Molality and Freezing Point Lowering/Boiling Point Elevation LEARNING GOAL Identify a mixture as a solution, a colloid, or a suspension. Using the molality, calculate the freezing point and boiling point for a solution.

• Colloids contain particles that pass through most filters but do not settle out or pass through semipermeable membranes.

• Suspensions have very large particles that settle out.

• The particles in a solution lower the vapor pressure, lower the freezing point, and raise the boiling point.

• Molality is the moles of solute per kilogram of solvent, usually water.

12.8 Properties of Solutions: Osmosis LEARNING GOAL Describe how the number of particles in solution affects osmosis. • The particles in a solution increase

the osmotic pressure. • In osmosis, solvent (water) passes

through a semipermeable mem- brane from a solution with a lower osmotic pressure (lower solute concentration) to a solution with a higher osmotic pressure (higher solute concentration).

• Isotonic solutions have osmotic pressures equal to that of body fluids.

• A red blood cell maintains its volume in an isotonic solution but swells in a hypotonic solution, and shrinks in a hypertonic solution.

• In dialysis, water and small solute particles pass through a dialyzing membrane, while larger particles are retained.

colloid A mixture having particles that are moderately large. Col- loids pass through filters but cannot pass through semipermeable membranes.

concentration A measure of the amount of solute that is dissolved in a specified amount of solution.

dialysis A process in which water and small solute particles pass through a semipermeable membrane.

dilution A process by which water (solvent) is added to a solu- tion to increase the volume and decrease (dilute) the solute concentration.

electrolyte A substance that produces ions when dissolved in water; its solution conducts electricity.

Henry’s law The solubility of a gas in a liquid is directly related to the pressure of that gas above the liquid.

hydration The process of surrounding dissolved ions by water molecules.

mass percent (m/m) The grams of solute in 100. g of solution. mass/volume percent (m/v) The grams of solute in 100. mL of

solution. molality (m) The number of moles of solute particles in exactly 1 kg

of solvent. molarity (M) The number of moles of solute in exactly 1 L of solution. nonelectrolyte A substance that dissolves in water as molecules; its

solution does not conduct an electrical current. osmosis The flow of a solvent, usually water, through a semiperme-

able membrane into a solution of higher solute concentration.

osmotic pressure The pressure that prevents the flow of water into the more concentrated solution.

saturated solution A solution containing the maximum amount of solute that can dissolve at a given temperature. Any additional solute will remain undissolved in the container.

solubility The maximum amount of solute that can dissolve in 100. g of solvent, usually water, at a given temperature.

solubility rules A set of guidelines that states whether an ionic compound is soluble or insoluble in water.

solute The component in a solution that is present in the lesser amount. solution A homogeneous mixture in which the solute is made up of

small particles (ions or molecules) that can pass through filters and semipermeable membranes.

solvent The substance in which the solute dissolves; usually the component present in greater amount.

strong electrolyte A compound that ionizes completely when it dis- solves in water; its solution is a good conductor of electricity.

suspension A mixture in which the solute particles are large enough and heavy enough to settle out and be retained by both filters and semipermeable membranes.

unsaturated solution A solution that contains less solute than can be dissolved.

volume percent (v/v) The milliliters of solute in 100. mL of solution. weak electrolyte A substance that produces only a few ions along

with many molecules when it dissolves in water; its solution is a weak conductor of electricity.

KEY TERMS

M12_TIMB8119_06_SE_C12.indd 390 11/27/18 12:08 PM

Core Chemistry Skills 391

The chapter section containing each Core Chemistry Skill is shown in parentheses at the end of each heading.

Using Solubility Rules (12.3) • Ionic compounds that are soluble in water contain Li+, Na+, K+,

NH4 +, NO3

-, or C2H3O2 - (acetate).

• Ionic compounds containing Cl-, Br-, or I- are soluble, but if they contain Ag +, Pb2+, or Hg2

2+, they are insoluble. • Most ionic compounds containing SO4

2- are soluble, but if they contain Ba2+, Pb2+, Ca2+, Sr2+, or Hg2

2+, they are insoluble. • Most other ionic compounds, including those containing the anions

CO3 2-, S2-, PO4

3-, or OH-, are insoluble. • To write an equation or ionic equation for the formation of an insol-

uble ionic compound, we write the cations and anions to identify any combination that would form an insoluble ionic compound.

Example: Determine if an ionic compound forms when solutions of CaCl2 and K2CO3 are mixed. If so, write the net ionic equation for the reaction.

Answer: In the ionic equation, we show all the ions of the reactants and show the ionic compound CaCO3 as a solid.

Ca2+(aq) + 2Cl-(aq) + 2K+(aq) + CO3 2-(aq) h CaCO3(s) + 2K+(aq) + 2Cl-(aq) For the net ionic equation, spectator ions that appear on both sides of the equation are removed.

Ca2+(aq) + CO3 2-(aq) h CaCO3(s)

Calculating Concentration (12.4) The amount of solute dissolved in a certain amount of solution is called the concentration of the solution.

• Mass percent (m/m) = mass of solute

mass of solution * 100%

• Volume percent (v/v) = volume of solute

volume of solution * 100%

• Mass/volume percent (m/v) = grams of solute

milliliters of solution * 100%

• Molarity (M) = moles of solute liters of solution

Example: What is the mass/volume percent (m/v) and the molarity (M) of 225 mL (0.225 L) of a LiCl solution that contains 17.1 g of LiCl?

Answer: Mass/volume % (m/v) = grams of solute

milliliters of solution * 100%

= 17.1 g LiCl

225 mL solution * 100%

= 7.60% (m/v) LiCl solution

moles of LiCl = 17.1 g LiCl * 1 mol LiCl

42.39 g LiCl

= 0.403 mol of LiCl

Molarity (M) = moles of solute liters of solution

= 0.403 mol LiCl 0.225 L solution

= 1.79 M LiCl solution

CORE CHEMISTRY SKILLS

Using Concentration as a Conversion Factor (12.4) • When we need to calculate the amount of solute or solution, we use

the concentration as a conversion factor. • For example, the concentration of a 4.50 M HCl solution means

there are 4.50 mol of HCl in 1 L of HCl solution, which gives two conversion factors written as

4.50 mol HCl 1 L solution

and 1 L solution

4.50 mol HCl

Example: How many milliliters of a 4.50 M HCl solution will provide 41.2 g of HCl?

Answer:

41.2 g HCl * 1 mol HCl

36.46 g HCl *

1 L solution 4.50 mol HCl

* 1000 mL solution

1 L solution

= 251 mL of HCl solution

Calculating the Quantity of a Reactant or Product for a Chemical Reaction in Solution (12.6) • When chemical reactions involve aqueous solutions of reactants

or products, we use the balanced chemical equation, the molarity, and the volume to determine the moles or grams of the reactants or products.

Example: How many grams of zinc metal will react with 0.315 L of a 1.20 M HCl solution?

2HCl(aq) + Zn(s) h H2(g) + ZnCl2(aq) Answer:

0.315 L solution * 1.20 mol HCl 1 L solution

* 1 mol Zn

2 mol HCl *

65.41 g Zn

1 mol Zn

= 12.4 g of Zn

Calculating the Freezing Point/Boiling Point of a Solution (12.7)

• The particles in a solution lower the freezing point, raise the boiling point, and increase the osmotic pressure.

• The freezing point lowering (∆Tf) is determined from the molality (m) of the particles in the solution and the freezing point constant, Kf.

∆Tf = m * Kf • The boiling point elevation (∆Tb) is determined from the molality (m)

of the particles in the solution and the boiling point constant, Kb.

∆Tb = m * Kb

Example: What is the boiling point of a solution that contains 1.5 mol of the strong electrolyte KCl in 1 kg of water?

Answer: A solution of 1.5 mol of KCl in 1 kg of water, which contains 3.0 mol of particles (1.5 mol of K+ and 1.5 mol of Cl-), is a 3.0-m solution.

∆Tb = m * Kb = 3.0 m * 0.51 °C

m = 1.5 °C

Tsolution = Twater + ∆Tb = 100.0 °C + 1.5 °C = 101.5 °C

M12_TIMB8119_06_SE_C12.indd 391 11/27/18 12:08 PM

392 CHAPTER 12 Solutions

UNDERSTANDING THE CONCEPTS

The chapter sections to review are shown in parentheses at the end of each problem.

12.87 Match the diagrams with the following: (12.1) a. a polar solute and a polar solvent b. a nonpolar solute and a polar solvent c. a nonpolar solute and a nonpolar solvent

12.89 Select the diagram that represents the solution formed by a solute that is a (12.2)

a. nonelectrolyte b. weak electrolyte c. strong electrolyte

1 2

Solid

1 2 3

1 2 3

12.88 If all the solute is dissolved in diagram 1, how would heating or cooling the solution cause each of the following changes? (12.3)

a. 2 to 3 b. 2 to 1

12.91 Use the following ions: (12.3)

21 3 4

A B

12.90 Select the container that represents the dilution of a 4% (m/v) KCl solution to give each of the following: (12.5)

a. a 2% (m/v) KCl solution b. a 1% (m/v) KCl solution

1 2 34% (m/v) KCl

Use the following illustration of beakers and solutions for problems 12.91 and 12.92:

Na+ Cl- Ag+ NO3 -

a. Select the beaker (1, 2, 3, or 4) that contains the products after the solutions in beakers A and B are mixed.

b. If an insoluble ionic compound forms, write the ionic equation.

c. If a reaction occurs, write the net ionic equation.

K+ NO3 - NH4

+ Br-

12.92 Use the following ions: (12.3)

a. Select the beaker (1, 2, 3, or 4) that contains the products after the solutions in beakers A and B are mixed.

b. If an insoluble ionic compound forms, write the ionic equation.

c. If a reaction occurs, write the net ionic equation.

12.93 A pickle is made by soaking a cucumber in brine, a salt- water solution. What makes the smooth cucumber become wrinkled like a prune? (12.8)

12.94 Why do lettuce leaves in a salad wilt after a vinaigrette dressing containing salt is added? (12.8)

M12_TIMB8119_06_SE_C12.indd 392 11/27/18 12:08 PM

Additional Practice Problems 393

Solution in A Solution in B

a. 2% (m/v) starch 8% (m/v) starch

b. 1% (m/v) starch 1% (m/v) starch

c. 5% (m/v) sucrose 1% (m/v) sucrose

d. 0.1% (m/v) sucrose 1% (m/v) sucrose

1

A B

2

A B

3

A B

1 Normal red blood cell

2 3

12.96 Select the diagram that represents the shape of a red blood cell when placed in each of the following a to e: (12.8)

a. 0.9% (m/v) NaCl solution b. 10% (m/v) glucose solution c. 0.01% (m/v) NaCl solution d. 5% (m/v) glucose solution e. 1% (m/v) glucose solution

ADDITIONAL PRACTICE PROBLEMS

12.97 Why does iodine dissolve in hexane, but not in water? (12.1)

12.98 How do temperature and pressure affect the solubility of solids and gases in water? (12.3)

12.99 Potassium nitrate has a solubility of 32 g of KNO3 in 100. g of H2O at 20 °C. Determine if each of the following forms an unsaturated or saturated solution at 20 °C: (12.3)

a. adding 32 g of KNO3 to 200. g of H2O b. adding 19 g of KNO3 to 50. g of H2O c. adding 68 g of KNO3 to 150. g of H2O

12.100 Potassium chloride has a solubility of 43 g of KCl in 100. g of H2O at 50 °C. Determine if each of the following forms an unsaturated or saturated solution at 50 °C: (12.3)

a. adding 25 g of KCl to 100. g of H2O b. adding 15 g of KCl to 25 g of H2O c. adding 86 g of KCl to 150. g of H2O

12.101 Indicate whether each of the following ionic compounds is soluble or insoluble in water: (12.3)

a. KCl b. MgSO4 c. CuS d. AgNO3 e. Ca(OH)2 12.102 Indicate whether each of the following ionic compounds is

soluble or insoluble in water: (12.3) a. CuCO3 b. FeO c. Mg3(PO4)2 d. (NH4)2SO4 e. NaHCO3 12.103 Write the net ionic equation to show the formation of a solid

(insoluble ionic compound) when the following solutions are mixed. Write none if no solid forms. (12.3)

a. AgNO3(aq) and LiCl(aq) b. NaCl(aq) and KNO3(aq) c. Na2SO4(aq) and BaCl2(aq)

12.104 Write the net ionic equation to show the formation of a solid (insoluble ionic compound) when the following solutions are mixed. Write none if no solid forms. (12.3)

a. Ca(NO3)2(aq) and Na2S(aq) b. Na3PO4(aq) and Pb(NO3)2(aq) c. FeCl3(aq) and NH4NO3(aq)

12.105 If NaCl has a solubility of 36.0 g in 100. g of H2O at 20 °C, how many grams of water are needed to prepare a saturated solution containing 80.0 g of NaCl? (12.3)

12.106 If the solid NaCl in a saturated solution of NaCl continues to dissolve, why is there no change in the concentration of the NaCl solution? (12.3)

12.107 Calculate the mass percent (m/m) of a solution containing 15.5 g of Na2SO4 and 75.5 g of H2O. (12.4)

12.108 Calculate the mass percent (m/m) of a solution containing 26 g of K2CO3 and 724 g of H2O. (12.4)

12.109 How many milliliters of a 12% (v/v) propyl alcohol solution would you need to obtain 4.5 mL of propyl alcohol? (12.4)

12.110 An 80-proof brandy is a 40.% (v/v) ethanol solution. The “proof” is twice the percent concentration of alcohol in the beverage. How many milliliters of alcohol are present in 750 mL of brandy? (12.4)

12.111 How many liters of a 12% (m/v) KOH solution would you need to obtain 86.0 g of KOH? (12.4)

12.112 How many liters of a 5.0% (m/v) glucose solution would you need to obtain 75 g of glucose? (12.4)

12.113 What is the molarity of a solution containing 8.0 g of NaOH in 400. mL of NaOH solution? (12.4)

12.114 What is the molarity of a solution containing 15.6 g of KCl in 274 mL of KCl solution? (12.4)

12.115 How many milliliters of a 1.75 M LiCl solution contain 15.2 g of LiCl? (12.4)

12.116 How many milliliters of a 1.50 M NaBr solution contain 75.0 g of NaBr? (12.4)

12.117 How many liters of a 2.50 M KNO3 solution can be prepared from 60.0 g of KNO3? (12.4)

12.118 How many liters of a 4.00 M NaCl solution will provide 25.0 g of NaCl? (12.4)

12.95 A semipermeable membrane separates two compartments, A and B. If the levels in A and B are equal initially, select the diagram that illustrates the final levels in a to d: (12.8)

M12_TIMB8119_06_SE_C12.indd 393 11/27/18 12:08 PM

394 CHAPTER 12 Solutions

12.119 How many grams of solute are in each of the following solutions? (12.4)

a. 2.5 L of a 3.0 M Al(NO3)3 solution b. 75 mL of a 0.50 M C6H12O6 solution c. 235 mL of a 1.80 M LiCl solution

12.120 How many grams of solute are in each of the following solutions? (12.4)

a. 0.428 L of a 0.450 M K2CO3 solution b. 10.5 mL of a 2.50 M AgNO3 solution c. 28.4 mL of a 6.00 M H3PO4 solution

12.121 Calculate the final concentration of the solution when water is added to prepare each of the following: (12.5)

a. 25.0 mL of a 0.200 M NaBr solution is diluted to 50.0 mL b. 15.0 mL of a 12.0% (m/v) K2SO4 solution is diluted to

40.0 mL c. 75.0 mL of a 6.00 M NaOH solution is diluted to 255 mL

12.122 Calculate the final concentration of the solution when water is added to prepare each of the following: (12.5)

a. 25.0 mL of an 18.0 M HCl solution is diluted to 500. mL b. 50.0 mL of a 15.0% (m/v) NH4Cl solution is diluted to

125 mL c. 4.50 mL of an 8.50 M KOH solution is diluted to 75.0 mL

12.123 What is the initial volume, in milliliters, needed to prepare each of the following diluted solutions? (12.5)

a. 250 mL of a 3.0% (m/v) HCl solution using a 10.0% (m/v) HCl solution

b. 500. mL of a 0.90% (m/v) NaCl solution using a 5.0% (m/v) NaCl solution

c. 350. mL of a 2.00 M NaOH solution using a 6.00 M NaOH solution

12.124 What is the initial volume, in milliliters, needed to prepare each of the following diluted solutions? (12.5)

a. 250 mL of a 5.0% (m/v) glucose solution using a 20.% (m/v) glucose solution

b. 45.0 mL of a 1.0% (m/v) CaCl2 solution using a 5.0% (m/v) CaCl2 solution

c. 100. mL of a 6.00 M H2SO4 solution using an 18.0 M H2SO4 solution

12.125 What is the final volume, in milliliters, when 25.0 mL of each of the following solutions is diluted to provide the given concentration? (12.5)

a. 10.0% (m/v) HCl solution to give a 2.50% (m/v) HCl solution

b. 5.00 M HCl solution to give a 1.00 M HCl solution c. 6.00 M HCl solution to give a 0.500 M HCl solution

12.126 What is the final volume, in milliliters, when 5.00 mL of each of the following solutions is diluted to provide the given concentration? (12.5)

a. 20.0% (m/v) NaOH solution to give a 4.00% (m/v) NaOH solution

b. 0.600 M NaOH solution to give a 0.100 M NaOH solution

c. 16.0% (m/v) NaOH solution to give a 2.00% (m/v) NaOH solution

12.127 The antacid Amphojel contains aluminum hydroxide Al(OH)3. How many milliliters of a 6.00 M HCl solution are required to react with 60.0 mL of a 2.00 M Al(OH)3 solution? (12.6)

3HCl(aq) + Al(OH)3(s) h 3H2O(l) + AlCl3(aq) 12.128 Cadmium reacts with HCl to produce hydrogen gas and cad-

mium chloride. What is the molarity (M) of the HCl solution if 250. mL of the HCl solution reacts with excess cadmium to produce 4.20 L of H2 gas measured at STP? (12.6)

2HCl(aq) + Cd(s) h H2(g) + CdCl2(aq) 12.129 Calculate the freezing point of each of the following

solutions: (12.7) a. 0.580 mol of lactose, a nonelectrolyte, added to 1.00 kg

of water b. 45.0 g of KCl, a strong electrolyte, dissolved in 1.00 kg

of water c. 1.5 mol of K3PO4, a strong electrolyte, dissolved in 1.00 kg

of water

12.130 Calculate the boiling point of each of the following solutions: (12.7)

a. 175 g of glucose, C6H12O6, a nonelectrolyte, added to 1.00 kg of water

b. 1.8 mol of CaCl2, a strong electrolyte, dissolved in 1.00 kg of water

c. 50.0 g of LiNO3, a strong electrolyte, dissolved in 1.00 kg of water

Applications 12.131 An antacid tablet, such as Amphojel, may be taken to reduce

excess stomach acid, which is a 0.20 M HCl solution. If one dose of Amphojel contains 450 mg of Al(OH)3, what volume, in milliliters, of stomach acid will be neutralized? (12.6)

3HCl(aq) + Al(OH)3(s) h 3H2O(l) + AlCl3(aq) 12.132 Calcium carbonate, CaCO3, reacts with stomach acid, (HCl,

hydrochloric acid) according to the following equation: (12.6)

2HCl(aq) + CaCO3(s) h CO2(g) + H2O(l) + CaCl2(aq)

Tums, an antacid, contains CaCO3. If Tums is added to 20.0 mL of a 0.400 M HCl solution, how many grams of CO2 gas are produced?

M12_TIMB8119_06_SE_C12.indd 394 11/27/18 12:08 PM

Challenge Problems 395

The following problems are related to the topics in this chapter. However, they do not all follow the chapter order, and they require you to combine concepts and skills from several sections. These problems will help you increase your critical thinking skills and prepare for your next exam.

12.133 Write the net ionic equation to show the formation of a solid when the following solutions are mixed. Write none if no solid forms. (12.3)

a. AgNO3(aq) + Na2SO4(aq) b. KCl(aq) + Pb(NO3)2(aq) c. CaCl2(aq) + (NH4)3PO4(aq) d. K2SO4(aq) + BaCl2(aq) 12.134 Write the net ionic equation to show the formation of a solid

when the following solutions are mixed. Write none if no solid forms. (12.3)

a. Pb(NO3)2(aq) + NaBr(aq) b. AgNO3(aq) + (NH4)2CO3(aq) c. Na3PO4(aq) + Al(NO3)3(aq) d. NaOH(aq) + CuCl2(aq) 12.135 In a laboratory experiment, a 10.0-mL sample of NaCl

solution is poured into an evaporating dish with a mass of 24.10 g. The combined mass of the evaporating dish and NaCl solution is 36.15 g. After heating, the evaporating dish and dry NaCl have a combined mass of 25.50 g. (12.4)

a. What is the mass percent (m/m) of the NaCl solution? b. What is the molarity (M) of the NaCl solution? c. If water is added to 10.0 mL of the initial NaCl solution

to give a final volume of 60.0 mL, what is the molarity of the diluted NaCl solution?

12.136 In a laboratory experiment, a 15.0-mL sample of KCl solution is poured into an evaporating dish with a mass of 24.10 g. The combined mass of the evaporating dish and KCl solution is 41.50 g. After heating, the evaporating dish and dry KCl have a combined mass of 28.28 g. (12.4)

a. What is the mass percent (m/m) of the KCl solution? b. What is the molarity (M) of the KCl solution? c. If water is added to 10.0 mL of the initial KCl solution to

give a final volume of 60.0 mL, what is the molarity of the diluted KCl solution?

12.137 Potassium fluoride has a solubility of 92 g of KF in 100. g of H2O at 18 °C. Determine if each of the following mixtures forms an unsaturated or saturated solution at 18 °C: (12.3)

a. adding 35 g of KF to 25 g of H2O b. adding 42 g of KF to 50. g of H2O c. adding 145 g of KF to 150. g of H2O

12.138 Lithium chloride has a solubility of 55 g of LiCl in 100. g of H2O at 25 °C. Determine if each of the following mixtures forms an unsaturated or saturated solution at 25 °C: (12.3)

a. adding 10 g of LiCl to 15 g of H2O b. adding 25 g of LiCl to 50. g of H2O c. adding 75 g of LiCl to 150. g of H2O

12.139 A solution is prepared with 70.0 g of HNO3 and 130.0 g of H2O. The HNO3 solution has a density of 1.21 g/mL. (12.4)

a. What is the mass percent (m/m) of the HNO3 solution? b. What is the total volume, in milliliters, of the solution? c. What is the mass/volume percent (m/v) of the solution? d. What is the molarity (M) of the solution?

12.140 A solution is prepared by dissolving 22.0 g of NaOH in 118.0 g of water. The NaOH solution has a density of 1.15 g/mL. (12.4)

a. What is the mass percent (m/m) of the NaOH solution? b. What is the total volume, in milliliters, of the solution? c. What is the mass/volume percent (m/v) of the solution? d. What is the molarity (M) of the solution?

12.141 A 355-mL sample of a HCl solution reacts with excess Mg to produce 4.20 L of H2 gas measured at 745 mmHg and 35 °C. What is the molarity (M) of the HCl solution? (12.6)

2HCl(aq) + Mg(s) h H2(g) + MgCl2(aq)

12.142 A 255-mL sample of a HCl solution reacts with excess Ca to produce 14.0 L of H2 gas measured at 732 mmHg and 22 °C. What is the molarity (M) of the HCl solution? (12.6)

2HCl(aq) + Ca(s) h H2(g) + CaCl2(aq)

12.143 How many moles of each of the following strong electrolytes are needed to give the same freezing point lowering as 1.2 mol of the nonelectrolyte ethylene glycol in 1 kg of water? (12.7)

a. NaCl b. K3PO4 12.144 How many moles of each of the following are needed to give

the same freezing point lowering as 3.0 mol of the nonelec- trolyte ethylene glycol in 1 kg of water? (12.7)

a. CH4O, a nonelectrolyte b. KNO3, a strong electrolyte

12.145 The boiling point of a NaCl solution is 101.04 °C. (12.7) a. What is the molality (m) of the solute particles in the

NaCl solution? b. What is the freezing point of the NaCl solution?

12.146 The freezing point of a CaCl2 solution is - 25 °C. (12.7) a. What is the molality (m) of the solute particles in the

CaCl2 solution? b. What is the boiling point of the CaCl2 solution?

12.147 Calculate the freezing point of each of the following solutions:

a. 1.36 mol of methanol, CH4O, a nonelectrolyte, added to 1.00 kg of water

b. 640. g of the antifreeze propylene glycol, C3H8O2, a nonelectrolyte, dissolved in 1.00 kg of water

c. 111 g of KCl, a strong electrolyte, dissolved in 1.00 kg of water

12.148 Calculate the boiling point of each of the following solutions:

a. 2.12 mol of glucose, C6H12O6, a nonelectrolyte, added to 1.00 kg of water

b. 110. g of sucrose, C12H22O11, a nonelectrolyte, dissolved in 1.00 kg of water

c. 146 g of NaNO3, a strong electrolyte, dissolved in 1.00 kg of water

CHALLENGE PROBLEMS

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396 CHAPTER 12 Solutions

12.1 The uniform blue color indicates that the solute particles are evenly dispersed among the solvent particles.

12.2 KCl is a polar solute, and water is a polar solvent.

12.3 In an aqueous solution, a strong electrolyte such as LiNO3 dissociates into ions Li+ and NO3

-, whereas a polar nonelec- trolyte such as CH4N2O dissolves as molecules only.

12.4 At 20 °C, the solubility of NaNO3 is 85 g/100 g H2O, and at 60 °C, it is 120 g/100 g H2O.

12.5 K2S is soluble in water because the K + is a soluble cation,

which makes any compound soluble.

12.6 Each of the test tubes contains a cation or an anion or both that form a compound that is not soluble.

12.7 The combination of Pb2+ and Br- forms an insoluble compound.

12.8 The mass percent (m/m) of a solution is an equality of mass of solute in exactly 100 g of solution. By multiplying the mass of a solution and the percent factor (g solute/100. g solution), the mass in grams of the solute is obtained.

12.9 The units for a mass/volume percent solution are g of solute in 100. mL of solution. A 3.0% solution contains 3.0 g of solute in 100. mL of solution.

12.10 2.0 mol of KCl in 4.00 L of solution is 2.0 mol/4.0 L = 0.50 mol/L = 0.50 M.

12.11 Increasing the volume during dilution decreases the mol/L concentration of KCl.

12.12 2.00 mol HCl 1 L solution

, 1 mol CaCO3

2 mol HCl

, 100.09 g CaCO3

1 mol CaCO3 12.13 Suspension particles are very large and can be trapped by

filters. Because colloids are smaller, a semipermeable mem- brane is needed to separate colloids from the solute.

12.14 A 1 m solution of KBr contains 2 mol of particles (K+ and Br-) in 1 L of water, whereas a 1 m solution of the nonelectrolyte urea contains 1 mol of molecules in 1 L of water. Thus, 2 mol of particles (KBr) lowers the freezing point of water more than 1 mol of particles (urea).

12.15 A 4% NaCl solution is hypertonic; a red blood cell will crenate (shrivel up) in this solution.

ANSWERS TO ENGAGE QUESTIONS

ANSWERS TO SELECTED PROBLEMS

c. Ca2+(aq) + 2Cl-(aq) + 2Na+(aq) + SO4 2-(aq) h CaSO4(s) + 2Na+(aq) + 2Cl-(aq)

Ca2+(aq) + SO4 2-(aq) h CaSO4(s)

12.1 a. NaCl, solute; water, solvent b. water, solute; ethanol, solvent c. oxygen, solute; nitrogen, solvent

12.3 The polar water molecules pull the K+ and I- ions away from the solid and into solution, where they are hydrated.

12.5 a. water b. CCl4 c. water d. CCl4 12.7 In a solution of KF, only the ions of K+ and F - are present in

the solvent. In a HF solution, there are a few ions of H+ and F - present but mostly dissolved HF molecules.

12.9 a. KCl(s) K +(aq) Cl-(aq)+

H2O

b. CaCl2(s) Ca 2+(aq) 2Cl-(aq)+

H2O

c. K3PO4(s) 3K +(aq) PO4

3-(aq)+ H2O

d. Fe(NO3)3(s) Fe 3+(aq) 3NO3

-(aq)+ H2O

12.11 a. mostly molecules and a few ions b. only ions c. only molecules

12.13 a. strong electrolyte b. weak electrolyte c. nonelectrolyte

12.15 a. saturated b. unsaturated

12.17 a. unsaturated b. unsaturated c. saturated

12.19 a. 68 g of KCl b. 12 g of KCl

12.21 a. The solubility of solid solutes typically increases as temperature increases.

b. The solubility of a gas is less at a higher temperature. c. Gas solubility is less at a higher temperature and the CO2

pressure in the can is increased.

12.23 a. soluble b. insoluble c. insoluble d. soluble e. soluble

12.25 a. No solid forms. b. 2Ag +(aq) + 2NO3-(aq) + 2K+(aq) + S2-(aq) h

Ag2S(s) + 2K+(aq) + 2NO3 -(aq) 2Ag +(aq) + S2-(aq) h Ag2S(s)

d. 3Cu2+(aq) + 6Cl-(aq) + 6Li+(aq) + 2PO4 3-(aq) h Cu3(PO4)2(s) + 6Li+(aq) + 6Cl-(aq)

3Cu2+(aq) + 2PO4 3-(aq) h Cu3(PO4)2(s) 12.27 A 5.00% (m/m) glucose solution can be made by adding

5.00 g of glucose to 95.00 g of water, while a 5.00% (m/v) glucose solution can be made by adding 5.00 g of glucose to enough water to make 100.0 mL of solution.

12.29 a. 17% (m/m) KCl solution b. 5.3% (m/m) sucrose solution c. 10.% (m/m) CaCl2 solution

12.31 a. 30.% (m/v) Na2SO4 solution b. 11% (m/v) sucrose solution

12.33 a. 2.5 g of KCl b. 50. g of NH4Cl c. 25.0 mL of acetic acid

12.35 79.9 mL of alcohol

12.37 a. 20. g of LiNO3 solution b. 400. mL of KOH solution c. 20. mL of formic acid solution

12.39 a. 0.500 M glucose solution b. 0.0356 M KOH solution c. 0.250 M NaCl solution

12.41 a. 120. g of NaOH b. 59.6 g of KCl c. 5.47 g of HCl

12.43 a. 1.50 L of KBr solution b. 10.0 L of NaCl solution c. 62.5 mL of Ca(NO3)2 solution

12.45 a. 983 mL b. 1.70 * 103 mL c. 464 mL 12.47 a. 20. g of mannitol b. 240 g of mannitol

12.49 2 L of glucose solution

12.51 Adding water (solvent) to the soup increases the volume and dilutes the tomato soup concentration.

M12_TIMB8119_06_SE_C12.indd 396 11/27/18 12:08 PM

Answers to Selected Problems 397

12.53 a. 2.0 M HCl solution b. 2.0 M NaOH solution c. 2.5% (m/v) KOH solution d. 3.0% (m/v) H2SO4 solution

12.55 a. 80. mL of HCl solution b. 250 mL of LiCl solution c. 600. mL of H3PO4 solution d. 180 mL of glucose solution

12.57 a. 12.8 mL of 4.00 M HNO3 solution b. 11.9 mL of 6.00 M MgCl2 solution c. 1.88 mL of 8.00 M KCl solution

12.59 1.0 * 102 mL 12.61 a. 10.4 g of PbCl2

b. 18.8 mL of Pb(NO3)2 solution c. 1.20 M KCl solution

12.63 a. 206 mL of HCl solution b. 11.2 L of H2 gas c. 9.12 M HCl solution

12.65 a. 74.8 mL of 3.50 M HBr solution b. 50.4 L of H2 c. 1.69 M HBr

12.67 a. solution b. suspension

12.69 a. 22.3 m b. 6.89 m

12.71 a. freezing point: - 41.5 °C; boiling point: 111.6 °C b. freezing point: - 12.8 °C; boiling point: 103.6 °C

12.73 a. 2.0 mol of ethylene glycol in 1.0 kg of water will have a lower freezing point because it has more particles in solution.

b. 0.50 mol of MgCl2 in 1.0 kg of water has a lower freezing point because each formula unit of MgCl2 dissociates in water to give three particles, whereas each formula unit of KCl dissociates to give only two particles.

12.75 a. 10% (m/v) starch solution b. from the 1% (m/v) starch solution into the 10% (m/v)

starch solution c. 10% (m/v) starch solution

12.77 a. B 10% (m/v) sucrose solution b. A 8% (m/v) albumin solution c. B 10% (m/v) starch solution

12.79 a. hypotonic b. hypotonic c. isotonic d. hypertonic

12.81 a. Na+, Cl- b. alanine c. Na+, Cl- d. urea

12.83 3.0 mL of chlorpromazine solution

12.85 0.50 g of CaCl2 12.87 a. 1 b. 2

c. 1

12.89 a. 3 b. 1 c. 2

12.91 a. beaker 3 b. Na+(aq) + Cl-(aq) + Ag +(aq) + NO3 -(aq) h

AgCl(s) + Na+(aq) + NO3 -(aq) c. Ag+(aq) + Cl-(aq) h AgCl(s)

12.93 The skin of the cucumber acts like a semipermeable membrane, and the more dilute solution inside flows into the brine solution.

12.95 a. 2 b. 1 c. 3 d. 2

12.97 Because iodine is a nonpolar molecule, it will dissolve in hexane, a nonpolar solvent. Iodine does not dissolve in water because water is a polar solvent.

12.99 a. unsaturated solution b. saturated solution c. saturated solution

12.101 a. soluble b. soluble c. insoluble d. soluble e. insoluble

12.103 a. Ag +(aq) + Cl-(aq) S AgCl(s) b. none c. Ba2+(aq) + SO4 2-(aq) h BaSO4(s)

12.105 222 g of water

12.107 17.0% (m/m) Na2SO4 solution

12.109 38 mL of propyl alcohol solution

12.111 0.72 L of KOH solution

12.113 0.500 M NaOH solution

12.115 205 mL of LiCl solution

12.117 0.237 L of KNO3 solution

12.119 a. 1600 g of Al(NO3)3 b. 6.8 g of C6H12O6 c. 17.9 g of LiCl

12.121 a. 0.100 M NaBr solution b. 4.50% (m/v) K2SO4 solution c. 1.76 M NaOH solution

12.123 a. 75 mL of 10.0% (m/v) HCl solution b. 90. mL of 5.0% (m/v) NaCl solution c. 117 mL of 6.00 M NaOH solution

12.125 a. 100. mL b. 125 mL c. 300. mL

12.127 60.0 mL of HCl solution

12.129 a. - 1.08 °C b. - 2.25 °C c. - 11 °C

12.131 87 mL of HCl solution

12.133 a. 2Ag +(aq) + SO4 2-(aq) h Ag2SO4(s) b. Pb2+(aq) + 2Cl-(aq) h PbCl2(s) c. 3Ca2+(aq) + 2PO4 3-(aq) h Ca3(PO4)2(s) d. Ba2+(aq) + SO4 2-(aq) h BaSO4(s)

12.135 a. 11.6% (m/m) NaCl solution b. 2.40 M NaCl solution c. 0.400 M NaCl solution

12.137 a. saturated b. unsaturated c. saturated

12.139 a. 35.0% (m/m) HNO3 solution b. 165 mL c. 42.4% (m/v) HNO3 solution d. 6.73 M HNO3 solution

12.141 0.917 M HCl solution

12.143 a. 0.60 mol of NaCl b. 0.30 mol of K3PO4 12.145 a. 2.0 m b. - 3.7 °C 12.147 a. - 2.53 °C b. - 15.6 °C c. - 5.54 °C

M12_TIMB8119_06_SE_C12.indd 397 11/27/18 12:08 PM

398

13 Peter, a chemical oceanographer, is collecting data concerning the amount of dissolved gases, specifically carbon dioxide (CO2) in the Atlantic Ocean. Studies indicate that CO2 in the atmosphere has increased as much as 45% since the eighteenth century. Peter’s research involves measuring the amount of dissolved CO2 in the oceans and trying to determine its impact. The oceans are a complex mixture of many different chemicals including gases, elements and minerals, and organic and particulate matter. Peter understands that CO2 is absorbed in the ocean through a series of equilibrium reactions that convert CO2 to HCO3

-. CO2(g) + H2O(l ) vh HCO3

-(aq) + H+(aq) An equilibrium reaction is a reversible reaction in which both the products and the reactants

are present. Because the equilibrium reaction shifts according to Le Châtelier’s principle, an increase in the CO2 concentration could eventually increase the amount of dissolved calcium carbonate, CaCO3, which makes up coral reefs and shells.

CAREER

Chemical Oceanographer A chemical oceanographer, also called a marine chemist, studies the chemistry of the ocean. One area of study includes how chemicals or pollutants enter into and affect the ocean. These include sewage, oil or fuels, chemical fertilizers, plastics, and storm drain overflows. Oceanographers analyze how these chemicals interact with seawater, marine life, and sediments, as they can behave differently due to the ocean’s varied environmental conditions. Chemical oceanographers also study how the various elements are cycled within the ocean. For instance, oceanographers quantify the amount and rate at which carbon dioxide is absorbed at the ocean’s surface and eventually transferred to deep waters. Chemical oceanographers also aid ocean engineers in the development of instruments and vessels that enable researchers to collect data and discover previously unknown marine life.

Reaction Rates and Chemical Equilibrium

Peter studies the effect of changes in the acidity of ocean water on coral. Read how CO2 dissolved in oceans changes the pH in the UPDATE Equilibrium of CO2 in the Ocean, page 424.

UPDATE Equilibrium of CO2 in the Ocean

M13_TIMB8119_06_SE_C13.indd 398 11/27/18 12:06 PM

13.1 Rates of Reactions 399

13.1 Rates of Reactions LEARNING GOAL Describe how temperature, concentration, and catalysts affect the rate of a reaction.

Earlier we looked at chemical reactions and determined the amounts of substances that react and the products that form. Now we are interested in how fast a reaction goes. If we know how fast a medication acts on the body, we can adjust the time over which the medication is taken. Some reactions such as explosions or the formation of precipitates in a solution are very fast. When we roast a turkey or bake a cake, the reaction is slower. Some reactions such as the tarnishing of silver and the aging of the body are much slower (see FIGURE 13.1). We have seen that some reactions need energy while other reactions produce energy. In this chapter, we will also look at the effect of changing the concentrations of reactants or products on the rate of reaction.

FIGURE 13.1 Reaction rates vary greatly for everyday processes. A banana ripens in a few days, silver tarnishes in a few months, while the aging process of humans takes many years.

5 days

R ea

ct io

n R

at e

In cr

ea se

s

5 months

50 years

LOOKING AHEAD

13.1 Rates of Reactions 399 13.2 Chemical Equilibrium 403 13.3 Equilibrium

Constants 406 13.4 Using Equilibrium

Constants 410 13.5 Changing Equilibrium

Conditions: Le Châtelier’s Principle 414

13.6 Equilibrium in Saturated Solutions 420

For a chemical reaction to take place, the molecules of the reactants must come in con- tact with each other. The collision theory indicates that a reaction takes place only when molecules collide with the proper orientation and sufficient energy. Many collisions can occur, but only a few actually lead to the formation of product. For example, consider the reaction of nitrogen and oxygen molecules (see FIGURE 13.2). To form nitrogen oxide (NO), the collisions between N2 and O2 molecules must place the atoms in the proper alignment. If the molecules are not aligned properly, no reaction takes place.

Activation Energy Even when a collision has the proper orientation, there still must be sufficient energy to break the bonds between the atoms of the reactants. The activation energy is the minimum amount of energy required to break the bonds between atoms of the reactants. The concept of activa- tion energy is analogous to climbing a hill. To reach a destination on the other side, we must have the energy needed to climb to the top of the hill. Once we are at the top, we can run down the other side. The energy needed to get us from our starting point to the top of the hill would be our activation energy.

In the same way, a collision must provide enough energy to push the reactants to the top of the energy hill. Then the reactants may be converted to products. If the energy

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400 CHAPTER 13 Reaction Rates and Chemical Equilibrium

provided by the collision is less than the activation energy, the molecules simply bounce apart, and no reaction occurs. The features that lead to a successful reaction are summarized next (see FIGURE 13.3).

ENGAGE 13. 1 What happens in a collision of reacting molecules that have the proper orientation, but not the minimum energy of activation?

FIGURE 13.2 Reacting molecules must collide, have a minimum amount of energy, and have the proper orientation to form product.

Collisions that do not form products

Insufficient energy

Wrong orientation

N2 +

+

+

+

+

+

+

O2 NO +

NO

Collision that forms products

FIGURE 13.3 The activation energy is the minimum energy needed to convert the colliding molecules into product.

E ne

rg y

In cr

ea se

s

Reactants

Products

Activation energy

N2

NO NO

O2

+

+

Progress of Reaction

Three Conditions Required for a Reaction to Occur

1. Collision The reactants must collide. 2. Orientation The reactants must align properly to break and form bonds. 3. Energy The collision must provide the energy of activation.

Rate of Reaction The rate (or speed) of reaction is determined by measuring the amount of a reactant used up, or the amount of a product formed, in a certain period of time.

Rate of reaction = change in concentration of reactant or product

change in time

M13_TIMB8119_06_SE_C13.indd 400 11/27/18 12:06 PM

13.1 Rates of Reactions 401

We can describe the rate of reaction with the analogy of eating a pizza. When we start to eat, we have a whole pizza. As time goes by, there are fewer slices of pizza left. If we know how long it took to eat the pizza, we could determine the rate at which the pizza was consumed. Let’s assume 4 slices are eaten every 8 minutes. That gives a rate of 12 slice per minute. After 16 minutes, all 8 slices are gone.

Rate at Which Pizza Slices Are Eaten

Slices Eaten 0 4 slices 6 slices 8 slices

Time (min) 0 8 min 12 min 16 min

Factors That Affect the Rate of a Reaction Reactions with low activation energies go faster than reactions with high activation ener- gies. Some reactions go very fast, while others are very slow. For any reaction, the rate is affected by changes in temperature, changes in the concentration of the reactants, and the addition of catalysts.

Temperature At higher temperatures, the increase in kinetic energy of the reactants makes them move faster and collide more often, and it provides more collisions with the required energy of acti- vation. Reactions almost always go faster at higher temperatures. For every 10 °C increase in temperature, most reaction rates approximately double. If we want food to cook faster, we increase the temperature. When body temperature rises, there is an increase in the pulse rate, rate of breathing, and metabolic rate. If we are exposed to extreme heat, we may experience heat stroke, which is a condition that occurs when body temperature goes above 40.5 °C (105 °F). If the body loses its ability to regulate temperature, body temperature continues to rise and may cause damage to the brain and internal organs. On the other hand, we slow down a reaction by decreasing the temperature. For example, we refrigerate perishable foods to make them last longer. In some cardiac surgeries, body temperature is decreased to 28 °C so the heart can be stopped and less oxygen is required by the brain. This is also the reason why some people have survived submersion in icy lakes for long periods of time. Cool water or an ice blanket may also be used to decrease the body temperature of a person with hyperthermia or heat stroke.

Concentrations of Reactants For virtually all reactions, the rate of a reaction increases when the concentration of the reactants increases. When there are more reacting molecules, more collisions that form products can occur, and the reaction goes faster (see FIGURE 13.4). For example, a patient having difficulty breathing may be given a breathing mixture with a higher oxygen content than the atmosphere. The increase in the number of oxygen molecules in the lungs increases the rate at which oxygen combines with hemoglobin. The increased rate of oxygenation of the blood means that the patient can breathe more easily. FIGURE 13.4 Increasing the

concentration of a reactant increases the number of collisions that are possible.

Reactants in Reaction Flask

Reaction:

+

Possible Collisions

1

2

4

Hemoglobin Oxygen Oxyhemoglobin

Hb(aq) + O2(g) h HbO2(aq)

Catalysts Another way to speed up a reaction is to lower the energy of activation. The energy of activation is the minimum energy needed to break apart the bonds of the reacting molecules. If a collision provides less than the activation energy, the bonds do not break, and the reactant molecules bounce apart. A catalyst speeds up a reaction by providing an alternative

ENGAGE 13.2 Why would an increase in temperature increase the rate of a reaction?

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402 CHAPTER 13 Reaction Rates and Chemical Equilibrium

pathway that has a lower energy of activation. When activation energy is lowered, more col- lisions provide sufficient energy for reactants to form product. During a reaction, a catalyst is not changed or consumed.

E ne

rg y

In cr

ea se

s

Progress of Reaction

Activation energy for uncatalyzed reaction

Activation energy for catalyzed reaction

Energy released by reaction

Reactants

Products When a catalyst lowers the activation energy, the reaction occurs at a faster rate.

Catalysts have many uses in industry. In the manufacturing of margarine, hydrogen (H2) is added to vegetable oils. Normally, the reaction is very slow because it has a high activa- tion energy. However, when platinum (Pt) is used as a catalyst, the reaction occurs rapidly. In the body, biocatalysts called enzymes make metabolic reactions proceed at rates necessary for proper cellular activity. Enzymes are added to laundry detergents to break down proteins (proteases), starches (amylases), or greases (lipases) that have stained clothes. Such enzymes function at the low temperatures that are used in home washing machines, and they are bio- degradable as well.

The factors affecting reaction rates are summarized in TABLE 13.1.

TABLE 13.1 Factors That Increase Reaction Rate

Factor Reason

Increasing temperature

More collisions, more collisions with energy of activation

Increasing reactant concentration

More collisions

Adding a catalyst Lowers energy of activation

SAMPLE PROBLEM 13.1 Factors That Affect the Rate of Reaction

TRY IT FIRST

Explain why each of the following changes will increase, decrease, or have no effect on the rate of reaction:

a. increasing the temperature b. decreasing the number of reacting molecules c. adding a catalyst

SOLUTION

a. A higher temperature increases the kinetic energy of the particles, which increases the number of collisions and makes more collisions effective, causing an increase in the rate of reaction.

b. Decreasing the number of reacting molecules decreases the number of collisions and the rate of the reaction.

c. Adding a catalyst increases the rate of reaction by lowering the activation energy, which increases the number of collisions that form product.

SELF TEST 13.1

a. How does using an ice blanket on a patient affect the rate of metabolism in the body? b. How does adding more reacting molecules affect the rate of the reaction?

ANSWER

a. Lowering the temperature will decrease the rate of metabolism. b. Adding more reacting molecules will increase the rate of the reaction.

PRACTICE PROBLEMS Try Practice Problems 13.1 to 13.6

ENGAGE 13.3 Why does decreasing the concen- tration of a reactant decrease the rate of reaction?

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13.2 Chemical Equilibrium 403

PRACTICE PROBLEMS

13.1 Rates of Reactions

13.1 a. What is meant by the rate of a reaction? b. Why does bread grow mold more quickly at room

temperature than in the refrigerator?

13.2 a. How does a catalyst affect the activation energy? b. Why is pure oxygen used in respiratory distress?

13.3 In the following reaction, what happens to the number of collisions when more Br2(g) molecules are added?

H2(g) + Br2(g) h 2HBr(g) 13.4 In the following reaction, what happens to the number of

collisions when the temperature of the reaction is decreased?

2H2(g) + CO(g) h CH4O(g)

13.5 How would each of the following change the rate of the reaction shown here?

2SO2(g) + O2(g) h 2SO3(g)

a. adding more SO2(g) b. increasing the temperature c. adding a catalyst d. removing some O2(g)

13.6 How would each of the following change the rate of the reaction shown here?

2NO(g) + 2H2(g) h N2(g) + 2H2O(g)

a. adding more NO(g) b. decreasing the temperature c. removing some H21g) d. adding a catalyst

13.2 Chemical Equilibrium LEARNING GOAL Use the concept of reversible reactions to explain chemical equilibrium.

Previously we assumed that all of the reactants in a chemical reaction were converted to products. However, most of the time reactants are not completely converted to products because a reverse reaction takes place in which products collide to form the reactants. When a reaction proceeds in both a forward and reverse direction, it is said to be reversible. We have looked at other reversible processes. For example, the melt- ing of solids to form liquids and the freezing of liquids to solids is a reversible physical change. Even in our daily life we have reversible events. We go from home to school and we return from school to home. We go up an escalator and we come back down. We put money in our bank account and we take money out.

An analogy for a forward and reverse reaction can be found in the phrase “We are going to the grocery store.” Although we mention our trip in one direction, we know that we will also return home from the store. Because our trip has both a forward and reverse direction, we can say the trip is reversible. It is not very likely that we would stay at the store forever.

A trip to the grocery store can be used to illustrate another aspect of reversible reactions. Perhaps the grocery store is nearby and we usually walk. However, we can change our rate. Suppose that one day we drive to the store, which increases our rate and gets us to the store faster. Correspondingly, a car also increases the rate at which we return home.

Reversible Chemical Reactions A reversible reaction proceeds in both the forward and reverse direction. That means there are two reaction rates: one is the rate of the forward reaction, and the other is the rate of the reverse reaction. When molecules begin to react, the rate of the forward reaction is faster than the rate of the reverse reaction. As reactants are consumed and products accumulate, the rate of the forward reaction decreases, and the rate of the reverse reaction increases.

Equilibrium Eventually, the rates of the forward and reverse reactions become equal; the reactants form products at the same rate that the products form reactants. A reaction reaches chemical equilibrium when no further change takes place in the concentrations of the reactants and products, even though the two reactions continue at equal but opposite rates.

As a reaction progresses, the rate of the forward reaction decreases and that of the reverse reaction increases. At equilibrium, the rates of the forward and reverse reactions are equal.

R ea

ct io

n R

at e

Rate of forward reaction

Equilibrium reached

Rate of reverse reaction

Progress of Reaction

[Reactants] decrease

Equilibrium reached

[Products] increase

C on

ce nt

ra ti

on (

m ol

/L )

Progress of Reaction

Equilibrium is reached when there are no further changes in the concentrations of reactants and products.

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404 CHAPTER 13 Reaction Rates and Chemical Equilibrium

Let us look at the process as the reaction of H2 and I2 proceeds to equilibrium. Initially, only the reactants H2 and I2 are present. Soon, a few molecules of HI are produced by the forward reaction. With more time, additional HI molecules are produced. As the concentra- tion of HI increases, more HI molecules collide and react in the reverse direction. As HI product builds up, the rate of the reverse reaction increases, while the rate of the forward reaction decreases. Eventually, the rates become equal, which means the reaction has reached equilibrium. Even though the concentrations remain constant at equilibrium, the forward and reverse reactions continue to occur. The forward and reverse reactions are usually shown together in a single equation by using a double arrow. A reversible reaction is two opposing reactions that occur at the same time (see FIGURE 13.5).

H2(g) + I2(g) 2HI(g) Forward reaction

Reverse reaction

FIGURE 13.5 Changes in reactant and product concentrations

Time (h)

Concentration of Reactants

Concentration of Products

Rates of Forward and Reverse Reactions

8 (4 + 4)

0

6 (3 + 3)

2

4 (2+ 2)

4

2 (1+ 1)

6

2 (1+ 1)

6

The forward reaction between H2 and I2 begins to produce HI.

As the reaction proceeds, there are fewer molecules of H2 and I2 and more molecules of HI, which increase the rate of the reverse reaction.

At equilibrium, the concentra- tions of reactants H2 and I2 and product HI are constant.

Initially, the reaction flask contains the reactants H2 (white) and I2 (purple).

The reaction continues with the rate of the forward reaction equal to the rate of the reverse reaction.

At Equilibrium: • The rate of the forward reaction is equal to the rate of the reverse reaction.

• No further changes occur in the concentrations of reactants and products, even though the two reactions continue at equal but opposite rates.

ENGAGE 13.5 How do the rates of the forward and reverse reactions compare once a chemical reaction reaches equilibrium?

ENGAGE 13.4 Why do the concentrations of the reactants decrease before equilibrium is reached?

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13.2 Chemical Equilibrium 405

SAMPLE PROBLEM 13.2 Reaction Rates and Equilibrium

TRY IT FIRST

Complete each of the following with change or do not change, faster or slower, equal or not equal:

a. Before equilibrium is reached, the concentrations of the reactants and products _______________.

b. Initially, reactants placed in a container have a _______________ rate of reaction than the rate of reaction of the products.

c. At equilibrium, the rate of the forward reaction is _______________ to the rate of the reverse reaction.

SOLUTION

a. Before equilibrium is reached, the concentrations of the reactants and products change. b. Initially, reactants placed in a container have a faster rate of reaction than the rate of

reaction of the products. c. At equilibrium, the rate of the forward reaction is equal to the rate of the reverse

reaction.

SELF TEST 13.2

Complete the following statements with changes or does not change: a. At equilibrium, the concentration of the reactant _______________. b. The rate of the forward reaction _______________ as the reaction proceeds toward

equilibrium.

ANSWER

a. At equilibrium, the concentration of the reactant does not change. b. The rate of the forward reaction changes as the reaction proceeds toward equilibrium.

PRACTICE PROBLEMS Try Practice Problems 13.7 to 13.12

We can set up a reaction starting with only reactants or with only products. Let’s look at the initial reactions in each, the forward and reverse reactions, and the equilibrium mixture that forms (see FIGURE 13.6).

2SO2(g) + O2(g) H 2SO3(g) If we start with only the reactants SO2 and O2 in the container, the forward reaction to form SO3 takes place until equilibrium is reached. However, if we start with only the product SO3 in the container, the reverse reaction to form SO2 and O2 takes place until equilibrium is reached. In both containers, the equilibrium mixture contains the same concentrations of SO2, O2, and SO3.

FIGURE 13.6 Reactions of SO2(g) and O2(g) or SO3(g) reach the same equilibrium mixture

Initially, the reaction flask contains SO2(g) and O2(g).

At equilibrium, the reaction flask contains mostly SO3(g) and small amounts of SO2(g) and O2(g).

Initially, the reaction flask contains SO3(g).

Forward Reaction

Reverse Reaction

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406 CHAPTER 13 Reaction Rates and Chemical Equilibrium

PRACTICE PROBLEMS

13.2 Chemical Equilibrium

13.7 What is meant by the term reversible reaction?

13.8 When does a reversible reaction reach equilibrium?

13.9 Which of the following are at equilibrium? a. The rate of the forward reaction is twice as fast as the rate

of the reverse reaction. b. The concentrations of the reactants and the products do

not change. c. The rate of the reverse reaction does not change.

13.10 Which of the following are not at equilibrium? a. The rates of the forward and reverse reactions are equal. b. The rate of the forward reaction does not change. c. The concentrations of reactants and the products are not

constant.

13.11 The following diagrams show the chemical reaction with time:

A vh B

If A is blue and B is orange, state whether or not the reaction has reached equilibrium in this time period and explain why.

13.12 The following diagrams show the chemical reaction with time:

C vh D

If C is blue and D is yellow, state whether or not the reaction has reached equilibrium in this time period and explain why.

1 h 2 h 3 h 4 h

1 h 2 h 3 h 4 h

13.3 Equilibrium Constants LEARNING GOAL Calculate the equilibrium constant for a reversible reaction given the concentrations of reactants and products at equilibrium.

At equilibrium, the concentrations of the reactants and products are constant. We can use a ski lift as an analogy. Early in the morning, skiers at the bottom of the mountain begin to ride the ski lift up to the slopes. After the skiers reach the top of the mountain, they ski down. Eventually, the number of people riding up the ski lift becomes equal to the number of people skiing down the mountain. There is no further change in the number of skiers on the slopes; the system is at equilibrium.

Equilibrium Expression At equilibrium, the concentrations can be used to set up a relationship between the products and the reactants. Suppose we write a general equation for reactants A and B that form products C and D. The small italic letters are the coefficients in the balanced equation.

aA + bB vh cC + dD

An equilibrium expression for a reversible chemical reaction multiplies the concen- trations of the products together and divides by the concentrations of the reactants. Each concentration is raised to a power that is equal to its coefficient in the balanced chemical equation. The square bracket around each substance indicates that the concentration is expressed in moles per liter (M). For our general reaction, this is written as:

Equilibrium expression

CoefficientsKc = [Products]

[Reactants]

[C]c [D]d

[A]a [B]b =

We can now describe how to write the equilibrium expression for the reaction of H2 and I2 that forms HI. The balanced chemical equation is written with a double arrow between the reactants and the products.

H2(g) + I2(g) vh 2HI(g)

REVIEW Using Significant Figures

in Calculations (2.3)

Balancing a Chemical Equation (8.2)

Calculating Concentration (12.4)

CORE CHEMISTRY SKILL Writing the Equilibrium Expression

At equilibrium, the number of people riding up the lift and the number of people skiing down the slope are constant.

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13.3 Equilibrium Constants 407

We show the concentration of the products using brackets in the numerator and the concentrations of the reactants in brackets in the denominator and write any coefficient as an exponent of its concentration (a coefficient 1 is understood).

Coefficient of HI

Kc = [HI]2

[H2][I2]

ENGAGE 13.6 How is a coefficient in a chemical equation indicated in the equilibrium expression?

Calculating Equilibrium Constants The equilibrium constant, Kc, is the numerical value obtained by substituting experimen- tally measured molar concentrations at equilibrium into the equilibrium expression. For example, the equilibrium expression for the reaction of H2 and I2 is written

H2(g) + I2(g) vh 2HI(g) Kc = [HI]2

[H2][I2]

In the first experiment, the molar concentrations for the reactants and products at equilibrium are found to be [H2] = 0.10 M, [I2] = 0.20 M, and [HI] = 1.04 M. When we substitute these values into the equilibrium expression, we obtain the numerical value of Kc.

CORE CHEMISTRY SKILL Calculating an Equilibrium Constant

SAMPLE PROBLEM 13.3 Writing Equilibrium Expressions

TRY IT FIRST

Write the equilibrium expression for the following reaction:

2SO2(g) + O2(g) vh 2SO3(g)

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

equation equilibrium expression [products]

[reactants]

STEP 1 Write the balanced chemical equation.

2SO2(g) + O2(g) vh 2SO3(g)

STEP 2 Write the concentrations of the products as the numerator and the reactants as the denominator.

[Products] [Reactants]

h h

[SO3]

[SO2][O2]

STEP 3 Write any coefficient in the equation as an exponent.

Kc = [SO3]

2

[SO2] 2[O2]

SELF TEST 13.3

Write the equilibrium expression for each of the following reactions:

a. 2NO(g) + O2(g) vh 2NO2(g) b. 2N2O5(g) vh 4NO2(g) + O2(g)

ANSWER

a. Kc = [NO2]

2

[NO]2[O2] b. Kc =

[NO2] 4[O2]

[N2O5] 2

PRACTICE PROBLEMS Try Practice Problems 13.13 and 13.14

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408 CHAPTER 13 Reaction Rates and Chemical Equilibrium

In additional experiments 2 and 3, the mixtures have different equilibrium concentrations for the system at equilibrium at the same temperature. However, when these concentrations are used to calculate the equilibrium constant, we obtain the same value of Kc for each (see TABLE 13.2). Thus, a reaction at a specific temperature can have only one value for the equilibrium constant.

TABLE 13.2 Equilibrium Constant for H2(g) + I2(g) H 2HI(g) at 427 °C Experiment Concentrations at Equilibrium Equilibrium Constant

[H2] [I2] [HI] Kc = [HI]2

[H2][I2]

1 0.10 M 0.20 M 1.04 M Kc = [1.04]2

[0.10][0.20] = 54

2 0.20 M 0.20 M 1.47 M Kc = [1.47]2

[0.20][0.20] = 54

3 0.30 M 0.17 M 1.66 M Kc = [1.66]2

[0.30][0.17] = 54

The units of Kc depend on the specific equation. In this example, the units of [M] 2/[M]2

cancel to give a value of 54. In other equations, the concentration units do not cancel. How- ever, in this text, the numerical value will be given without any units as shown in Sample Problem 13.4.

SAMPLE PROBLEM 13.4 Calculating an Equilibrium Constant

TRY IT FIRST

The decomposition of dinitrogen tetroxide forms nitrogen dioxide.

N2O4(g) vh 2NO2(g)

What is the numerical value of Kc at 100 °C if a reaction mixture at equilibrium contains 0.45 M N2O4 and 0.31 M NO2?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

0.45 M N2O4, 0.31 M NO2

Kc equilibrium expression

Equation

N2O4(g) vh 2NO2(g)

STEP 2 Write the equilibrium expression.

Kc = [NO2]

2

[N2O4]

STEP 3 Substitute equilibrium (molar) concentrations and calculate Kc.

Kc = [0.31]2

[0.45] = 0.21

SELF TEST 13.4

a. Calculate the numerical value of Kc if an equilibrium mixture contains 0.040 M NH3, 0.60 M H2, and 0.20 M N2.

2NH3(g) vh 3H2(g) + N2(g)

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13.3 Equilibrium Constants 409

FIGURE 13.7 At equilibrium at constant temperature, the concentration of CO2 is the same regardless of the amounts of CaCO3(s) and CaO(s) in the container.

CaO(s) + CO2(g)CaCO3(s)

CO2

T = 800 °C

T = 800 °C

CaO CaCO3

b. Calculate the numerical value of Kc if an equilibrium mixture contains 0.20 M ICl, 0.010 M I2, and 0.048 M Cl2.

2ICl(g) vh I2(g) + Cl2(g)

ANSWER

a. Kc = 27 b. Kc = 0.012

Heterogeneous Equilibrium Up to now, our examples have been reactions that involve only gases. A reaction in which all the reactants and products are in the same state reaches homogeneous equilibrium. When the reactants and products are in two or more states, the equilibrium is termed a heterogeneous equilibrium. In the following reaction, solid calcium carbonate reaches het- erogeneous equilibrium with solid calcium oxide and carbon dioxide gas (see FIGURE 13.7).

CaCO3(s) vh CaO(s) + CO2(g)

In contrast to gases, the concentrations of pure solids and pure liquids are constant; they do not change. Therefore, pure solids and liquids are not included in the equilibrium expression. For this heterogeneous equilibrium, the Kc expression does not include the concentration of CaCO3(s) or CaO(s). It is written as Kc = [CO2].

ENGAGE 13.7 Why are the concentrations of CaO(s) and CaCO3(s) not included in the equilibrium expression for Kc for the decomposition of CaCO3(s)?

SAMPLE PROBLEM 13.5 Heterogeneous Equilibrium Expression

TRY IT FIRST

Write the equilibrium expression for the following reaction at equilibrium:

4HCl(g) + O2(g) vh 2H2O(l) + 2Cl2(g)

SOLUTION

STEP 1 Write the balanced chemical equation.

4HCl(g) + O2(g) vh 2H2O(l) + 2Cl2(g)

STEP 2 Write the concentrations of the products as the numerator and the reactants as the denominator. In this heterogeneous reaction, the concentration of the H2O, which is a pure liquid, is not included in the equilibrium expression.

[Products] [Reactants]

h h

[Cl2]

[HCl][O2]

STEP 3 Write any coefficient in the equation as an exponent.

Kc = [Cl2]

2

[HCl]4[O2]

SELF TEST 13.5

a. Solid iron(II) oxide and carbon monoxide gas react to produce solid iron and carbon dioxide gas. Write the balanced chemical equation and the equilibrium expression for this reaction at equilibrium.

b. When heated, solid sodium hydrogen carbonate decomposes to form solid sodium carbonate, gaseous water, and carbon dioxide gas. Write the balanced chemical equa- tion and the equilibrium expression for this reaction at equilibrium.

ANSWER

a. FeO(s) + CO(g) vh Fe(s) + CO2(g) Kc = [CO2]

[CO]

b. 2NaHCO3(s) vh Na2CO3(s) + H2O(g) + CO2(g) Kc = [H2O][CO2] PRACTICE PROBLEMS

Try Practice Problems 13.21 to 13.26

PRACTICE PROBLEMS Try Practice Problems 13.15 to 13.20

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410 CHAPTER 13 Reaction Rates and Chemical Equilibrium

PRACTICE PROBLEMS

13.3 Equilibrium Constants

13.13 Write the equilibrium expression for each of the following reactions:

a. CH4(g) + 2H2S(g) vh CS2(g) + 4H2(g) b. 2NO(g) vh N2(g) + O2(g) c. 2SO3(g) + CO2(g) vh CS2(g) + 4O2(g) d. CH4(g) + H2O(g) vh 3H2(g) + CO(g)

13.14 Write the equilibrium expression for each of the following reactions:

a. 2HBr(g) vh H2(g) + Br2(g) b. 2BrNO(g) vh Br2(g) + 2NO(g) c. CH4(g) + Cl2(g) vh CH3Cl(g) + HCl(g) d. Br2(g) + Cl2(g) vh 2BrCl(g)

13.15 Write the equilibrium expression for the reaction in the diagram, and calculate the numerical value of Kc. In the diagram, X atoms are orange and Y atoms are blue; the volume of the container is 1.0 L.

X2(g) + Y2(g) vh 2XY(g)

13.18 What is the numerical value of Kc for the following reaction if the equilibrium mixture contains 0.30 M CO2, 0.033 M H2, 0.20 M CO, and 0.30 M H2O?

CO2(g) + H2(g) vh CO(g) + H2O(g)

13.19 What is the numerical value of Kc for the following reaction if the equilibrium mixture contains 0.51 M CO, 0.30 M H2, 1.8 M CH4, and 2.0 M H2O?

CO(g) + 3H2(g) vh CH4(g) + H2O(g)

13.20 What is the numerical value of Kc for the following reaction if the equilibrium mixture contains 0.44 M N2, 0.40 M H2, and 2.2 M NH3?

N2(g) + 3H2(g) vh 2NH3(g)

13.21 Identify each of the following as a homogeneous or heterogeneous equilibrium:

a. 2O3(g) vh 3O2(g) b. 2NaHCO3(s) vh Na2CO3(s) + CO2(g) + H2O(g) c. C6H6(g) + 3H2(g) vh C6H12(g) d. 4HCl(g) + Si(s) vh SiCl4(l) + 2H2(g)

13.22 Identify each of the following as a homogeneous or heterogeneous equilibrium:

a. CO(g) + H2(g) vh C(s) + H2O(g) b. NH4Cl(s) vh NH3(g) + HCl(g) c. CS2(g) + 4H2(g) vh CH4(g) + 2H2S(g) d. Ti(s) + 2Cl2(g) vh TiCl4(g)

13.23 Write the equilibrium expression for each of the reactions in problem 13.21.

13.24 Write the equilibrium expression for each of the reactions in problem 13.22.

13.25 What is the numerical value of Kc for the following reaction if the equilibrium mixture at 750 °C contains 0.20 M CO and 0.052 M CO2?

FeO(s) + CO(g) vh Fe(s) + CO2(g)

13.26 What is the numerical value of Kc for the following reaction if the equilibrium mixture at 800 °C contains 0.030 M CO2?

CaCO3(s) vh CaO(s) + CO2(g)

Equilibrium mixture

Equilibrium mixture

13.16 Write the equilibrium expression for the reaction in the diagram, and calculate the numerical value of Kc. In the diagram, A atoms are red and B atoms are green; the volume of the container is 1.0 L.

2AB(g) vh A2(g) + B2(g)

13.17 What is the numerical value of Kc for the following reaction if the equilibrium mixture contains 0.030 M N2O4 and 0.21 M NO2?

N2O4(g) vh 2NO2(g)

13.4 Using Equilibrium Constants LEARNING GOAL Use an equilibrium constant to predict the extent of reaction and to calculate equilibrium concentrations.

The values of Kc can be large or small. The size of the equilibrium constant depends on whether equilibrium is reached with more products than reactants, or more reactants than products. However, the size of an equilibrium constant does not affect how fast equilibrium is reached.

REVIEW Solving Equations (1.4)

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13.4 Using Equilibrium Constants 411

Equilibrium with a Large Kc When a reaction has a large equilibrium constant, it means that the forward reaction produced a large amount of products when equilibrium was reached. Then the equilib- rium mixture contains mostly products, which makes the concentrations of the products in the numerator higher than the concentrations of the reactants in the denominator. Thus at equilibrium, this reaction has a large Kc. Consider the reaction of SO2 and O2, which has a large Kc. At equilibrium, the reaction mixture contains mostly product and few reactants (see FIGURE 13.8).

2SO2(g) + O2(g) H 2SO3(g)

Kc = [SO3]

2

[SO2] 2[O2]

Mostly product

Few reactants = 3.4 * 102

ENGAGE 13.8 Why does an equilibrium mixture containing mostly products have a large Kc?

FIGURE 13.8 In the reaction of SO2(g) and O2(g), the equilibrium mixture contains mostly product SO3(g), which results in a large Kc.

O2

O2

SO2

Initially At equilibrium

SO2

SO3

C on

ce nt

ra ti

on (

m ol

/L )

2SO2(g) 2SO3(g)+ O2(g)

Equilibrium with a Small Kc When a reaction has a small equilibrium constant, the equilibrium mixture contains a high concentration of reactants and a low concentration of products. Then the equilibrium expression has a small number in the numerator and a large number in the denominator. Thus at equilibrium, this reaction has a small Kc. Consider the reaction for the formation of NO(g) from N2(g) and O2(g), which has a small Kc (see FIGURE 13.9).

N2(g) + O2(g) H 2NO(g)

Kc = [NO]2

[N2][O2]

Few products

Mostly reactants = 2 * 10-9

FIGURE 13.9 The equilibrium mixture contains a very small amount of the product NO and a large amount of the reactants N2 and O2, which results in a small Kc.

Initially At equilibrium

NO

N2 O2 N2 O2

C on

ce nt

ra ti

on (

m ol

/L )

N2(g) 2NO(g)O2(g)+

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412 CHAPTER 13 Reaction Rates and Chemical Equilibrium

A few reactions have equilibrium constants close to 1, which means they have about equal concentrations of reactants and products. Moderate amounts of reactants have been converted to products upon reaching equilibrium (see FIGURE 13.10).

FIGURE 13.10 At equilibrium, a reaction with a large Kc contains mostly products, whereas a reaction with a small Kc contains mostly reactants.

Mostly productsMostly reactants

Small Kc Large KcKc L 1

Products < < Reactants Little reaction takes place

Products > > Reactants Reaction essentially

complete

Reactants L Products Moderate reaction

Reactants Products K c Equilibrium Mixture Contains

2CO(g) + O2(g) vh 2CO2(g) 2 * 1011 Mostly product

2H2(g) + S2(g) vh 2H2S(g) 1.1 * 107 Mostly product

N2(g) + 3H2(g) vh 2NH3(g) 1.6 * 102 Mostly product

PCl5(g) vh PCl3(g) + Cl2(g) 1.2 * 10-2 Mostly reactant

N2(g) + O2(g) vh 2NO(g) 2 * 10 -9 Mostly reactants

TABLE 13.3 Examples of Reactions with Large and Small Kc Values

TABLE 13.3 lists some equilibrium constants and the extent of their reaction.PRACTICE PROBLEMS Try Practice Problems 13.27 to 13.30

ENGAGE 13.10 Why does the equilibrium mixture for a reaction with a Kc = 2 * 10-9 contain mostly reactants and only a few products?

Calculating Concentrations at Equilibrium When we know the numerical value of the equilibrium constant and all the equilibrium concentrations except one, we can calculate the unknown concentration as shown in Sample Problem 13.6.

CORE CHEMISTRY SKILL Calculating Equilibrium

Concentrations

SAMPLE PROBLEM 13.6 Calculating Concentration Using an Equilibrium Constant

TRY IT FIRST

For the reaction of carbon dioxide and hydrogen, the equilibrium concentrations are 0.25 M CO2, 0.80 M H2, and 0.50 M H2O. What is the equilibrium concentration of CO(g)?

CO2(g) + H2(g) vh CO(g) + H2O(g) Kc = 0.11

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

0.25 M CO2, 0.80 M H2, 0.50 M H2O

[CO] equilibrium expression, value of Kc

Equation

CO2(g) + H2(g) vh CO(g) + H2O(g) Kc = 0.11

STEP 2 Write the equilibrium expression and solve for the needed concentration.

Kc = [CO][H2O]

[CO2][H2]

ENGAGE 13.9 Does a reaction with a Kc = 1.2 contain mostly reactants, mostly products, or about equal amounts of both reactants and products at equilibrium?

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13.4 Using Equilibrium Constants 413

We rearrange the equilibrium expression to solve for the unknown [CO] as follows:

Kc = [CO][H2O]

[CO2][H2] * [CO2][H2] *

= [CO][H2O]

[CO2][H2]

[H2O] =

[CO][H2O]

[H2O]

[CO] Kc [CO2][H2]

[H2O] =

Multiply both sides by [CO2][H2].

Divide both sides by [H2O].

Kc * [CO2][H2]

Kc *

*

[CO2][H2]

STEP 3 Substitute the equilibrium (molar) concentrations and calculate the needed concentration.

[CO] = Kc * [CO2][H2]

[H2O] = 0.11 *

[0.25][0.80] [0.50]

= 0.044 M

SELF TEST 13.6

a. When ethene (C2H4) reacts with water vapor, ethanol (C2H6O) is produced. If an equilibrium mixture contains 0.020 M C2H4 and 0.015 M H2O, what is the equilibrium concentration of C2H6O? At 327 °C, the Kc is 9.0 * 103.

C2H4(g) + H2O(g) vh C2H6O(g)

b. NO(g) and Br2(g) react to form NOBr(g).

2NO(g) + Br2(g) vh 2NOBr(g)

If an equilibrium mixture contains 0.70 M NO and 0.060 M NOBr, what is the equi- librium concentration of Br2? At 1000 K, the Kc is 1.3 * 10-2.

ANSWER

a. [C2H6O] = 2.7 M b. [Br2] = 0.57 M PRACTICE PROBLEMS

Try Practice Problems 13.31 to 13.34

PRACTICE PROBLEMS

13.4 Using Equilibrium Constants

13.27 If the Kc for this reaction is 4, which of the following diagrams represents the molecules in an equilibrium mixture? In the diagrams, X atoms are orange and Y atoms are blue.

X2(g) + Y2(g) H 2XY(g)

13.28 If the Kc for this reaction is 2, which of the following diagrams represents the molecules in an equilibrium mixture? In the diagrams, A atoms are orange and B atoms are green.

2AB(g) H A2(g) + B2(g)

A B C A B C

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414 CHAPTER 13 Reaction Rates and Chemical Equilibrium

13.29 Indicate whether each of the following equilibrium mixtures contains mostly products or mostly reactants:

a. Cl2(g) + NO(g) vh 2NOCl(g) Kc = 3.7 * 108

b. 2H2(g) + S2(g) vh 2H2S(g) Kc = 1.1 * 107

c. 3O2(g) vh 2O3(g) Kc = 1.7 * 10-56

13.30 Indicate whether each of the following equilibrium mixtures contains mostly products or mostly reactants:

a. CO(g) + Cl2(g) vh COCl2(g) Kc = 5.0 * 10-9

b. 2HF(g) vh H2(g) + F2(g) Kc = 1.0 * 10-95

c. 2NO(g) + O2(g) vh 2NO2(g) Kc = 6.0 * 1013

13.31 The numerical value of the equilibrium constant, Kc, for this reaction is 54.

H2(g) + I2(g) vh 2HI(g) If the equilibrium mixture contains 0.015 M I2 and 0.030 M HI, what is the molar concentration of H2?

13.32 The numerical value of the equilibrium constant, Kc, for the fol- lowing reaction is 4.6 * 10-3. If the equilibrium mixture contains 0.050 M NO2, what is the molar concentration of N2O4?

N2O4(g) vh 2NO2(g)

13.33 The numerical value of the equilibrium constant, Kc, for the fol- lowing reaction is 2.0. If the equilibrium mixture contains 2.0 M NO and 1.0 M Br2, what is the molar concentration of NOBr?

2NOBr(g) vh 2NO(g) + Br2(g)

13.34 The numerical value of the equilibrium constant, Kc, for the fol- lowing reaction is 1.7 * 102. If the equilibrium mixture contains 0.18 M H2 and 0.020 M N2, what is the molar concentration of NH3?

3H2(g) + N2(g) vh 2NH3(g)

13.5 Changing Equilibrium Conditions: Le Châtelier’s Principle LEARNING GOAL Use Le Châtelier’s principle to describe the changes made in equilibrium concentrations when reaction conditions change.

We have seen that when a reaction reaches equilibrium, the rates of the forward and reverse reactions are equal and the concentrations remain constant. Now we will look at what hap- pens to a system at equilibrium when changes occur in reaction conditions, such as changes in concentration, volume, or temperature.

Le Châtelier’s Principle When we alter any of the conditions of a system at equilibrium, the rates of the forward and reverse reactions may no longer be equal. We say that a stress is placed on the equilibrium. Then the system responds by changing the rate of the forward or reverse reaction in the direction that relieves that stress to reestablish equilibrium. We can use Le Châtelier’s principle, which states that when a system at equilibrium is disturbed, the system will shift in the direction that will reduce that stress.

Le Châtelier’s Principle When a stress (change in conditions) is placed on a reaction at equilibrium, the equilibrium will shift in the direction that relieves the stress.

Suppose we have two water tanks connected by a pipe. When the water levels in the tanks are equal, water flows in the forward direction from Tank A to Tank B at the same rate as it flows in the reverse direction from Tank B to Tank A. Suppose we add more water

At equilibrium, the water levels are equal.

Tank A Tank B Tank A Tank B Tank A Tank B

Water added to Tank A increases the rate of flow in the forward direction.

Equilibrium is reached again when the water levels are equal.

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13.5 Changing Equilibrium Conditions: Le Châtelier’s Principle 415

to Tank A. With a higher level of water in Tank A, more water flows in the forward direc- tion from Tank A to Tank B than in the reverse direction from Tank B to Tank A, which is shown with a longer arrow. Eventually, equilibrium is reached as the levels in both tanks become equal, but higher than before. Then the rate of water flows equally between Tank A and Tank B.

Effect of Concentration Changes on Equilibrium We will now use the reaction of H2 and I2 to illustrate how a change in concentration disturbs the equilibrium and how the system responds to that stress.

H2(g) + I2(g) vh 2HI(g)

Suppose that more of the reactant H2 is added to the equilibrium mixture, which increases the concentration of H2. Because Kc cannot change for a reaction at a given tem- perature, adding more H2 places a stress on the system (see FIGURE 13.11). Then the system

CORE CHEMISTRY SKILL Using Le Châtelier’s Principle

relieves this stress by increasing the rate of the forward reaction, which is indicated by the direction of the large arrow. Thus, more product is formed until the system is again at equi- librium. According to Le Châtelier’s principle, adding more reactant causes the system to shift in the direction of the product until equilibrium is reestablished.

H2(g) + I2(g) 2HI(g) Add H2

Suppose now that some H2 is removed from the reaction mixture at equilibrium, which lowers the concentration of H2 and slows the rate of the forward reaction. Using Le Châte- lier’s principle, we know that when some of the reactants are removed, the system will shift in the direction of the reactants until equilibrium is reestablished.

H2(g) + I2(g) 2HI(g) Remove H2

The concentrations of the products of an equilibrium mixture can also increase or decrease. For example, if more HI is added, there is an increase in the rate of the reaction in the reverse direction, which converts some of the product to reactants. The concentration

H2(g) + I2(g) 2HI(g)

The addition of H2 places stress on the equilibrium system.

To relieve stress, the forward reaction converts some reactants H2 and I2 to form more product HI.

A new equilibrium is established when the rates of the forward and the reverse reactions become equal.

FIGURE 13.11 Effect of adding reactant to a system at equilibrium

ENGAGE 13.11 If more product HI is added, will the equilibrium shift in the direction of the product or reactants?

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416 CHAPTER 13 Reaction Rates and Chemical Equilibrium

TABLE 13.4 Effect of Concentration Changes on Equilibrium H2(g) + I2(g) H 2HI(g)

Stress Shift in the Direction of

Increasing [H2] Product

Decreasing [H2] Reactants

Increasing [I2] Product

Decreasing [I2] Reactants

Increasing [HI] Reactants

Decreasing [HI] Product

of the products decreases and the concentration of the reactants increases until equilibrium is reestablished. Using Le Châtelier’s principle, we see that the addition of a product causes the system to shift in the direction of the reactants.

H2(g) + I2(g) 2HI(g) Add HI

In another example, some HI is removed from an equilibrium mixture, which decreases the concentration of the product. Then there is a shift in the direction of the product to reestablish equilibrium.

H2(g) + I2(g) 2HI(g) Remove HI

In summary, Le Châtelier’s principle indicates that a stress caused by adding a substance at equilibrium is relieved when the equilibrium system shifts the reaction away from that substance. Adding more reactant causes an increase in the forward reaction to products. Adding more product causes an increase in the reverse reaction to reactants. When some of a substance is removed, the equilibrium system shifts in the direction of that substance. These features of Le Châtelier’s principle are summarized in TABLE 13.4.

Effect of a Catalyst on Equilibrium Sometimes a catalyst is added to a reaction to speed up a reaction by low- ering the activation energy. As a result, the rates of both the forward and reverse reactions increase. The time required to reach equilibrium is shorter, but the same ratios of products and reactants are attained. Therefore, a cata- lyst speeds up the forward and reverse reactions, but it has no effect on the equilibrium mixture.

Effect of Volume Change on Equilibrium If there is a change in the volume of a gas mixture at equilibrium, there will also be a change in the concentrations of those gases. Decreasing the vol- ume will increase the concentration of gases, whereas increasing the volume will decrease their concentration. Then the system responds to reestablish equilibrium.

Let’s look at the effect of decreasing the volume of the equilibrium mix- ture of the following reaction:

2CO(g) + O2(g) vh 2CO2(g)

If we decrease the volume, all the concentrations increase. According to Le Châtelier’s principle, the increase in concentration is relieved when the system shifts in the direction of the smaller number of moles.

O2(g)2CO(g) + 2CO2(g) 3 mol of gas 2 mol of gas

Decrease V

On the other hand, when the volume of the equilibrium gas mixture increases, the concentrations of all the gases decrease. Then the system shifts in the direction of the greater number of moles to reestablish equilibrium.

Volume increase

Volume decrease

2CO(g) + O2(g) 2CO2(g)

Decreasing the volume of the container shifts the equilibrium in the direction of fewer moles of gas.

Increasing the volume of the container shifts the equilibrium in the direction of more moles of gas.

O2(g)+2CO(g) 2CO2(g) 3 mol of gas 2 mol of gas

Increase V

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13.5 Changing Equilibrium Conditions: Le Châtelier’s Principle 417

Chemistry Link to Health Oxygen–Hemoglobin Equilibrium and Hypoxia

The transport of oxygen in the blood involves an equilibrium between hemoglobin (Hb), oxygen, and oxyhemoglobin (HbO2). Hemoglobin, present in our red blood cells, is an iron-containing protein that carries oxygen from our lungs to the muscles and tissues of our body. Each hemoglobin molecule contains four iron-containing heme groups. Because each heme group can attach to one O2 in the lungs, one hemoglobin can carry as many as four O2 molecules. The transport of oxygen involves an equilibrium reaction between the reactants Hb and O2 and the product HbO2.

Hb(aq) + O2(g) vh HbO2(aq) The equilibrium expression can be written

Kc = [HbO2]

[Hb][O2]

When a reaction has the same number of moles of reactants as products, a volume change does not affect the equilibrium mixture because the concentrations of the reactants and products change in the same way.

H2(g) + I2(g) vh 2HI(g) 2 mol of gas 2 mol of gas

Hb

In the lungs, Hb binds O2.

Red cells carry O2 as HbO2 from the lungs to the tissues and muscles.

In the tissues and muscles, O2 is released from HbO2.

Hb Hb

HbO2 HbO2

HbO2

O2 O2

In the alveoli of the lungs where the O2 concentration is high, the forward reaction is faster, which shifts the equilibrium in the direction of the product HbO2. As a result, O2 binds to hemoglobin in the lungs.

Hb(aq) + O2(g) h HbO2(aq)

In the tissues and the muscles where the O2 concentration is low, the reverse reaction is faster, which shifts the equilibrium in the direction of the reactant Hb and releases the oxygen from the hemoglobin.

Hb(aq) + O2(g) v HbO2(aq)

At normal atmospheric pressure, oxygen diffuses into the blood because the partial pressure of oxygen in the alveoli is higher than that in the blood. At an altitude above 8000 ft, a decrease in the atmo- spheric pressure results in a significant reduction in the partial pressure of oxygen, which means that less oxygen is available for the blood and body tissues. At an altitude of 18 000 ft, a person will obtain 29% less oxygen. When oxygen levels are lowered, a person may experi- ence hypoxia, characterized by increased respiratory rate, headache, decreased mental acuteness, fatigue, decreased physical coordination, nausea, vomiting, and cyanosis. A similar problem occurs in persons with a history of lung disease that impairs gas diffusion in the alveoli or in persons with a reduced number of red blood cells, such as in smokers.

Hypoxia may occur at high altitudes where the oxygen concentration is lower.

someone at sea level. This increase in hemoglobin causes a shift in the equilibrium back in the direction of HbO2 product. Eventually, the higher concentration of HbO2 will provide more oxygen to the tissues and the symptoms of hypoxia will lessen.

O2(g)+Hb(aq) HbO2(aq)

Add Hb

For some who climb high mountains, it is important to stop and acclimatize for several days at increasing altitudes. At very high alti- tudes, it may be necessary to use an oxygen tank.

According to Le Châtelier’s principle, we see that a decrease in oxygen will shift the equilibrium in the direction of the reactants. Such a shift depletes the concentration of HbO2 and causes the hypoxia.

O2(g)+Hb(aq) HbO2(aq)

Remove O2

Immediate treatment of altitude sickness includes hydration, rest, and if necessary, descending to a lower altitude. The body’s adaptation to lowered oxygen levels requires about 10 days. Dur- ing this time, the bone marrow increases red blood cell production, providing more red blood cells and more hemoglobin. A person living at a high altitude can have 50% more red blood cells than

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418 CHAPTER 13 Reaction Rates and Chemical Equilibrium

TABLE 13.5 Effects of Condition Changes on Equilibrium Condition Change (Stress) Shift in the Direction of

Concentration Adding a reactant Removing a reactant Adding a product Removing a product

Products (forward reaction) Reactants (reverse reaction) Reactants (reverse reaction) Products (forward reaction)

Volume (container) Decreasing the volume Increasing the volume

Fewer moles of gas More moles of gas

Temperature Endothermic Reaction

Increasing the temperature Decreasing the temperature

Products (forward reaction) Reactants (reverse reaction)

Exothermic Reaction

Increasing the temperature Decreasing the temperature

Reactants (reverse reaction) Products (forward reaction)

Catalyst Increasing the rates equally No effect

SAMPLE PROBLEM 13.7 Using Le Châtelier’s Principle

TRY IT FIRST

Methanol, CH4O, is finding use as a fuel additive. Describe the effect of each of the following changes on the equilibrium mixture for the combustion of methanol:

2CH4O(g) + 3O2(g) vh 2CO2(g) + 4H2O(g) + 1450 kJ

Effect of a Change in Temperature on Equilibrium We can think of heat as a reactant or a product in a reaction. For example, in the equation for an endothermic reaction, heat is written on the reactant side. When the temperature of an endothermic reaction increases, the system responds by shifting in the direction of the products to remove heat.

heatN2(g) ++ O2(g) 2NO(g) Increase T

If the temperature is decreased for an endothermic reaction, there is a decrease in heat. Then the system shifts in the direction of the reactants to add heat.

heat+O2(g)+N2(g) 2NO(g) Decrease T

In the equation for an exothermic reaction, heat is written on the product side. When the temperature of an exothermic reaction increases, the system responds by shifting in the direction of the reactants to remove heat.

Increase T

O2(g)+2SO2(g) 2SO3(g) + heat

If the temperature is decreased for an exothermic reaction, there is a decrease in heat. Then the system shifts in the direction of the products to add heat.

O2(g)+2SO2(g) 2SO3(g) + heat Decrease T

TABLE 13.5 summarizes the ways we can use Le Châtelier’s principle to determine the shift in equilibrium that relieves a stress caused by the change in a condition.

ENGAGE 13.12 If you want to decrease the amount of product in an exothermic reaction, would you increase or decrease the temperature?

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13.5 Changing Equilibrium Conditions: Le Châtelier’s Principle 419

a. adding more CO2(g) b. adding more O2(g) c. increasing the volume of the container d. increasing the temperature e. adding a catalyst

SOLUTION

a. When the concentration of the product CO2 increases, the equilibrium shifts in the direction of the reactants.

b. When the concentration of the reactant O2 increases, the equilibrium shifts in the direction of the products.

c. When the volume increases, the equilibrium shifts in the direction of the greater number of moles of gas, which is the products.

d. When the temperature is increased for an exothermic reaction, the equilibrium shifts in the direction of the reactants.

e. When a catalyst is added, there is no change in the equilibrium mixture.

SELF TEST 13.7

Describe the effect of each of the following changes on the equilibrium mixture for the following reaction:

2HF(g) + Cl2(g) + 357 kJ vh 2HCl(g) + F2(g)

a. adding more Cl2(g) b. decreasing the volume of the container c. decreasing the temperature

ANSWER

a. When the concentration of the reactant Cl2 increases, the equilibrium shifts in the direction of the products.

b. There is no change in equilibrium mixture because the moles of reactants are equal to the moles of products.

c. When the temperature for an endothermic reaction decreases, the equilibrium shifts in the direction of the reactants.

Chemistry Link to Health Homeostasis: Regulation of Body Temperature

In a physiological system of equilibrium called homeostasis, changes in our environment are balanced by changes in our bodies. It is cru- cial to our survival that we balance heat gain with heat loss. If we do not lose enough heat, our body temperature rises. At high tempera- tures, the body can no longer regulate our metabolic reactions. If we lose too much heat, body temperature drops. At low temperatures, essential functions proceed too slowly.

The skin plays an important role in the maintenance of body tem- perature. When the outside temperature rises, receptors in the skin send signals to the brain. The temperature-regulating part of the brain stimulates the sweat glands to produce perspiration. As perspiration evaporates from the skin, heat is removed and the body temperature is decreased.

In cold temperatures, epinephrine is released, causing an increase in metabolic rate, which increases the production of heat. Receptors on the skin signal the brain to constrict the blood vessels. Less blood f lows through the skin, and heat is conserved. The production of perspiration stops, thereby lessening the heat lost by evaporation.

H ea

t G

ai n

In cr

ea se

s

H ea

t L

os s

Blood vessels dilate • sweat production increases • sweat evaporates • skin cools

Blood vessels constrict and epinephrine is released • metabolic activity increases • muscular activity increases • shivering occurs • sweat production stops

D ec

re as

es

B od

y H

ea t

B od

y T

em p

er at

u re

PRACTICE PROBLEMS Try Practice Problems 13.35 to 13.42

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420 CHAPTER 13 Reaction Rates and Chemical Equilibrium

13.6 Equilibrium in Saturated Solutions LEARNING GOAL Write the solubility product expression for a slightly soluble ionic compound and calculate the Ksp; use Ksp to determine the solubility.

Until now, we have looked primarily at equilibrium systems that involve gases. However, there are also equilibrium systems that involve aqueous saturated solutions between solid solutes of slightly soluble ionic compounds and their ions. Everyday examples of solubility equilibrium in solution are found in the slightly soluble ionic compounds that are found in bone and kidney stones. Bone is composed of calcium phosphate, Ca3(PO4)2, which produces ions during bone loss. Kidney stones are composed of compounds such as calcium oxalate, CaC2O4.

CORE CHEMISTRY SKILL Writing the Solubility Product

Expression

PRACTICE PROBLEMS

13.5 Changing Equilibrium Conditions: Le Châtelier’s Principle

13.35 In the lower atmosphere, oxygen is converted to ozone (O3) by the energy provided from lightning.

3O2(g) + heat vh 2O3(g) For each of the following changes at equilibrium, indicate whether the equilibrium shifts in the direction of product, r eactants, or does not change:

a. adding more O2(g) b. adding more O3(g) c. increasing the temperature d. increasing the volume of the container e. adding a catalyst

13.36 Ammonia is produced by reacting nitrogen gas and hydrogen gas.

N2(g) + 3H2(g) vh 2NH3(g) + 92 kJ For each of the following changes at equilibrium, indicate whether the equilibrium shifts in the direction of product, reactants, or does not change:

a. removing some N2(g) b. decreasing the temperature c. adding more NH3(g) d. adding more H2(g) e. increasing the volume of the container

13.37 Hydrogen chloride can be made by reacting hydrogen gas and chlorine gas.

H2(g) + Cl2(g) + heat vh 2HCl(g) For each of the following changes at equilibrium, indicate whether the equilibrium shifts in the direction of product, reactants, or does not change:

a. adding more H2(g) b. increasing the temperature c. removing some HCl(g) d. adding a catalyst e. removing some Cl2(g)

13.38 When heated, carbon monoxide reacts with water to produce carbon dioxide and hydrogen.

CO(g) + H2O(g) vh CO2(g) + H2(g) + heat For each of the following changes at equilibrium, indicate whether the equilibrium shifts in the direction of products, reactants, or does not change:

a. decreasing the temperature b. adding more H2(g) c. removing CO2(g) as its forms d. adding more H2O(g) e. decreasing the volume of the container

Applications Use the following equation for the equilibrium of hemoglobin in the blood to answer problems 13.39 to 13.42:

Hb(aq) + O2(g) vh HbO2(aq) 13.39 For the hemoglobin equilibrium, indicate if each of the

following changes shifts the equilibrium in the direction of the product, the reactants, or does not change:

a. increasing [O2] b. increasing [HbO2] c. decreasing [Hb]

13.40 For the hemoglobin equilibrium, indicate if each of the following changes shifts the equilibrium in the direction of the product, the reactants, or does not change:

a. decreasing [O2] b. increasing [Hb] c. decreasing [HbO2]

13.41 Athletes who train at high altitudes initially experience hypoxia, which causes their body to produce more hemoglobin. When an athlete first arrives at high altitude,

a. What is the stress on the hemoglobin equilibrium? b. In what direction does the hemoglobin equilibrium shift?

13.42 A person who has been a smoker and has a low oxygen blood saturation uses an oxygen tank for supplemental oxygen. When oxygen is first supplied,

a. What is the stress on the hemoglobin equilibrium? b. In what direction does the hemoglobin equilibrium shift?

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13.6 Equilibrium in Saturated Solutions 421

Solubility Product Expression In a saturated solution, a solid slightly soluble ionic compound is in equilibrium with its ions. As long as the temperature remains constant, the concentration of the ions in the satu- rated solution is constant. Let us look at the solubility equilibrium equation for CaC2O4, which is written with the solid solute on the left and the ions in solution on the right.

CaC2O4(s) vh Ca 2+(aq) + C2O4 2-(aq)

The solubility of a substance is the quantity that dissolves to form a saturated solu- tion. We represent the solubility in a saturated aqueous solution of solid CaC2O4 by the solubility product expression, which is the product of the ion concentrations. As in other heterogeneous equilibria, the concentration of the solid CaC2O4 is constant and is not included in the solubility product expression.

Ksp = [Ca 2+][C2O4

2-]

In another example, we look at the equilibrium of solid calcium phosphate and its ions Ca2+ and PO4

3-.

Ca3(PO4)2(s) vh 3Ca 2+(aq) + 2PO4 3-(aq)

As with other equilibrium expressions, the molar concentration of each product ion is raised to a power that is equal to its coefficient in the balanced equilibrium equation. For this equilibrium equation, the solubility product expression consists of [Ca2+] raised to the power of 3 and [PO4

3-] raised to the power of 2.

Ksp = [Ca 2+]3[PO4

3-]2

Solubility Product Constant The numerical value of the solubility product expression is the solubility product constant, Ksp. In this text, solubility will be expressed as the molar solubility, which is the moles of solute that dissolve in 1 liter of saturated solution. The calculation of a solubility product constant is shown in Sample Problem 13.8.

PRACTICE PROBLEMS Try Practice Problems 13.43 and 13.44

CORE CHEMISTRY SKILL Calculating a Solubility Product

Constant

Calcium oxalate produces small amounts of Ca2+ and C2O4

2- in aqueous solution.

Terraces in Pamukkale, Turkey, are composed of calcium carbonate, which is slightly soluble.

SAMPLE PROBLEM 13.8 Calculating the Solubility Product Constant

TRY IT FIRST

We can make a saturated solution of BaCO3 by adding solid BaCO3 to water and stirring until equilibrium is reached. What is the numerical value of Ksp for BaCO3 if the equilib- rium mixture contains 5.1 * 10-5 M Ba2+ and 5.1 * 10-5 M CO3 2-?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

[Ba2+] = 5.1 * 10-5 M, [CO3

2-] = 5.1 * 10-5 M numerical

value of Ksp solubility product

expression

STEP 2 Write the equilibrium equation for the dissociation of the slightly soluble ionic compound.

BaCO3(s) vh Ba 2+(aq) + CO3 2-(aq)

STEP 3 Write the solubility product expression.

Ksp = [Ba 2+][CO3

2-]

ENGAGE 13.13 Why is the solubility product expression for Cu2CO3 written Ksp = [Cu

+]2[CO3 2-]?

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422 CHAPTER 13 Reaction Rates and Chemical Equilibrium

SAMPLE PROBLEM 13.9 Calculating the Solubility Product Constant Using Coefficients

TRY IT FIRST

A saturated solution of strontium f luoride, SrF2, contains 8.7 * 10-4 M Sr2+ and 1.7 * 10-3 M F -. What is the numerical value of Ksp for SrF2?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

[Sr2+] = 8.7 * 10-4 M, [F-] = 1.7 * 10-3 M

numerical value of Ksp

solubility product expression

STEP 2 Write the equilibrium equation for the dissociation of the slightly soluble ionic compound.

SrF2(s) vh Sr 2+(aq) + 2F -(aq)

STEP 3 Write the solubility product expression.

Ksp = [Sr 2+][F -]2

STEP 4 Substitute the molar concentration of each ion into the solubility product expression and calculate.

Ksp = [8.7 * 10-4][1.7 * 10-3]2 = 2.5 * 10-9

SELF TEST 13.9

a. What is the numerical value of Ksp for silver oxalate, Ag2C2O4, if a saturated solution contains 2.2 * 10-4 M Ag + and 1.1 * 10-4 M C2O4 2-?

b. A saturated solution of PbI2 contains 1.2 * 10-3 M Pb2+ and 2.4 * 10-3 M I-. What is the numerical value of Ksp for PbI2?

ANSWER

a. Ksp = 5.3 * 10-12 b. Ksp = 6.9 * 10-9

PRACTICE PROBLEMS Try Practice Problems 13.45 to 13.48

TABLE 13.6 gives values of Ksp for a selected group of slightly soluble ionic compounds at 25 °C.

TABLE 13.6 Examples of Solubility Product Constants (Ksp)

Formula Ksp

AgCl 1.8 * 10-10

Ag2SO4 1.2 * 10-5

BaSO4 1.1 * 10-10

CaCO3 5.0 * 10-9

CaF2 3.2 * 10-11

Ca(OH)2 6.5 * 10-6

CaSO4 2.4 * 10-5

PbCl2 1.5 * 10-6

PbCO3 7.4 * 10-14

STEP 4 Substitute the molar concentration of each ion into the solubility product expression and calculate.

Ksp = [5.1 * 10-5][5.1 * 10-5] = 2.6 * 10-9

SELF TEST 13.8

a. A saturated solution of AgBr contains 7.3 * 10-7 M Ag + and 7.3 * 10-7 M Br-. What is the numerical value of Ksp for AgBr?

b. A saturated solution of CoS contains 6.3 * 10-11 M Co2+ and 6.3 * 10-11 M S2-. What is the numerical value of Ksp for CoS?

ANSWER

a. Ksp = 5.3 * 10-13 b. Ksp = 4.0 * 10-21

M13_TIMB8119_06_SE_C13.indd 422 11/27/18 12:07 PM

13.6 Equilibrium in Saturated Solutions 423

Cadmium sulfide is slightly soluble.

Molar Solubility, S The molar solubility, S, of a slightly soluble ionic compound is the number of moles of solute that dissolves in 1 liter of solution. For example, the molar solubility of cadmium sulfide, CdS, is found experimentally to be 1 * 10-12 M.

CdS(s) vh Cd 2+(aq) + S2-(aq)

Because CdS dissociates into Cd2+ and S2- ions, they each have a concentration equal to the solubility S.

S = [Cd2+] = [S2-] = 1 * 10-12 M

If we know the Ksp of a slightly soluble ionic compound, we can determine its molar solubility, as shown in Sample Problem 13.10.

PRACTICE PROBLEMS

13.6 Equilibrium in Saturated Solutions

13.43 For each of the following slightly soluble ionic compounds, write the equilibrium equation for dissociation and the solubility product expression:

a. MgCO3 b. CaF2 c. Ag3PO4

13.44 For each of the following slightly soluble ionic compounds, write the equilibrium equation for dissociation and the solubility product expression:

a. Ag2S b. Al(OH)3 c. BaF2

SAMPLE PROBLEM 13.10 Calculating the Molar Solubility from K sp

TRY IT FIRST

Calculate the molar solubility, S, of PbSO4 if it has a Ksp = 1.6 * 10-8.

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

Ksp = 1.6 * 10-8 molar solubility (S) of PbSO4(s)

solubility product expression

STEP 2 Write the equilibrium equation for the dissociation of the slightly soluble ionic compound.

PbSO4(s) vh Pb 2+(aq) + SO4 2-(aq)

STEP 3 Write the solubility product expression using S.

Ksp = [Pb 2+][SO4

2-] = S * S = S2 = 1.6 * 10-8

STEP 4 Calculate the molar solubility, S.

S2 = 1.6 * 10-8

S = 21.6 * 10-8 = 1.3 * 10-4 M Thus, 1.3 * 10-4 mol of PbSO4 will dissolve in 1 L of solution.

SELF TEST 13.10

a. Calculate the molar solubility, S, of MnS if it has a Ksp = 2.5 * 10-10. b. Calculate the molar solubility, S, of AgOH if it has a Ksp = 2.0 * 10-8.

ANSWER

a. S = 1.6 * 10-5 M b. S = 1.4 * 10-4 M

PRACTICE PROBLEMS Try Practice Problems 13.49 and 13.50

CORE CHEMISTRY SKILL Calculating the Molar Solubility

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424 CHAPTER 13 Reaction Rates and Chemical Equilibrium

13.45 A saturated solution of barium sulfate, BaSO4, has [Ba2+] = 1 * 10-5 M and [SO4 2-] = 1 * 10-5 M. What is the numerical value of Ksp for BaSO4?

13.46 A saturated solution of copper(II) sulfide, CuS, has [Cu2+] = 1.1 * 10-18 M and [S2-] = 1.1 * 10-18 M. What is the numerical value of Ksp for CuS?

13.47 A saturated solution of silver carbonate, Ag2CO3, has [Ag +] = 2.6 * 10-4 M and [CO3 2-] = 1.3 * 10-4 M. What is the numerical value of Ksp for Ag2CO3?

13.48 A saturated solution of barium fluoride, BaF2, has [Ba2+] = 3.6 * 10-3 M and [F -] = 7.2 * 10-3 M. What is the numerical value of Ksp for BaF2?

13.49 Calculate the molar solubility, S, of CuI if it has a Ksp of 1 * 10-12.

13.50 Calculate the molar solubility, S, of SnS if it has a Ksp of 1 * 10-26.

UPDATE Equilibrium of CO2 in the Ocean

The CO2 level in the atmosphere has increased over the last few decades due to burning fossil fuels and deforestation. About 35% of the CO2 released into the atmosphere dissolves in the oceans, where it forms H2CO3(aq), which breaks up into H+(aq) and HCO3

-(aq). The increase in H+ makes the ocean water more acidic. One of Peter’s research projects is to study the influence of a change in ocean water acidity on coral species, which have calcium carbonate skeletons.

Applications

13.51 a. Write the equation for the dissociation of the slightly soluble ionic compound calcium carbonate, CaCO3.

b. Write the solubility product expression for CaCO3.

13.52 a. What is the numerical value of Ksp for CaCO3 if the equilibrium mixture has [Ca2+] and [CO3

2-] equal to 7.1 * 10-5 M?

b. If increasing the acidity of the ocean has the effect of decreasing the concentration of CO3

2-, will calcium carbonate be more or less soluble if the acidity increases?

Coral reefs are affected by the changing acidity of the ocean.

CONCEPT MAP

Reaction Rates

Temperature

Catalyst Kc =

indicates that equilibrium shifts

for changes in

are affected by

Concentrations of Reactants

Solubility Product Expression

Equilibrium Expression

Concentration, Temperature, and Volume

Le Châtelier’s PrincipleReversible Reactions Saturated Solutions

is written as

[Products] [Reactants]

Ksp = [cation][anion]

Large Kc has Mostly Products

Small Kc has Mostly Reactants

involves in

when equal in rate give

is written as

is called

REACTION RATES AND CHEMICAL EQUILIBRIUM

M13_TIMB8119_06_SE_C13.indd 424 11/27/18 12:07 PM

Key Terms 425

CHAPTER REVIEW

13.1 Rates of Reactions LEARNING GOAL Describe how temperature, concentration, and catalysts affect the rate of a reaction. • The rate of a reaction is

the speed at which the reactants are converted to products.

• Increasing the concentrations of reactants, increasing the tempera- ture, or adding a catalyst can increase the rate of a reaction.

13.2 Chemical Equilibrium LEARNING GOAL Use the concept of reversible reactions to explain chemical equilibrium. • Chemical equilibrium occurs

in a reversible reaction when the rate of the forward reaction becomes equal to the rate of the reverse reaction.

• At equilibrium, no further change occurs in the concentrations of the reactants and products as the forward and reverse reactions continue.

13.3 Equilibrium Constants LEARNING GOAL Calculate the equilibrium constant for a reversible reaction given the concentrations of reactants and products at equilibrium. • An equilibrium constant, Kc, is the

ratio of the concentrations of the products to the concentrations of the reactants, with each concentration raised to a power equal to its coefficient in the balanced chemical equation.

• For heterogeneous reactions, only the molar concentrations of gases are placed in the equilibrium expression.

13.4 Using Equilibrium Constants LEARNING GOAL Use an equilibrium constant to predict the extent of reaction and to calculate equilibrium concentrations. • A large value of Kc

indicates that an

equilibrium mixture contains mostly products, whereas a small value of Kc indicates that the equilibrium mixture contains mostly reactants.

• Equilibrium constants can be used to calculate the concentration of a component in the equilibrium mixture.

13.5 Changing Equilibrium Conditions: Le Châtelier’s Principle LEARNING GOAL Use Le Châtelier’s principle to describe the changes made in equilibrium concentrations when reaction conditions change. • When reactants are removed or products are added to an

equilibrium mixture, the system shifts in the direction of the reactants.

• When reactants are added or products are removed from an equilibrium mixture, the system shifts in the direction of the products.

• A decrease in the volume of a reaction container causes a shift in the direction of the smaller number of moles of gas.

• An increase in the volume of a reaction container causes a shift in the direction of the greater number of moles of gas.

• Increasing the temperature of an endothermic reaction or decreasing the temperature of an exothermic reaction will cause the system to shift in the direction of the products.

• Decreasing the temperature of an endothermic reaction or increasing the temperature of an exothermic reaction will cause the system to shift in the direction of the reactants.

13.6 Equilibrium in Saturated Solutions LEARNING GOAL Write the solubility product expression for a slightly soluble ionic compound and calculate Ksp; use Ksp to determine the solubility. • In a saturated solution, a slightly soluble

ionic compound is in equilibrium with its ions.

• In a saturated solution, the concentrations of the ions from the slightly soluble ionic compound are constant and can be used to calculate the solubility product constant, Ksp.

• If Ksp for a slightly soluble ionic compound is known, its solubility can be calculated.

E ne

rg y

In cr

ea se

s

Activation energy for uncatalyzed reaction

Activation energy for catalyzed reaction

Reactants

Products

Progress of Reaction

Energy released by reaction

[Reactants] decrease

Equilibrium reached

[Products] increase

C on

ce nt

ra ti

on (

m ol

/L )

Progress of Reaction

O2

O2

SO2

Initially At equilibrium

2SO2(g) + O2(g) 2SO3(g)

SO2

SO3

C on

ce nt

ra ti

on (

m ol

/L )

activation energy The energy that must be provided by a collision to break apart the bonds of the reacting molecules.

catalyst A substance that increases the rate of reaction by lowering the activation energy.

chemical equilibrium The point at which the rate of forward and reverse reactions are equal so that no further change in concen- trations of reactants and products takes place.

collision theory A model for a chemical reaction stating that molecules must collide with sufficient energy and proper orientation to form products.

equilibrium constant, Kc The numerical value obtained by substituting the equilibrium concentrations of the components into the equilibrium expression.

KEY TERMS

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426 CHAPTER 13 Reaction Rates and Chemical Equilibrium

equilibrium expression The ratio of the concentrations of products to the concentrations of reactants, with each component raised to an exponent equal to the coefficient of that compound in the balanced chemical equation.

heterogeneous equilibrium An equilibrium system in which the components are in different states.

homogeneous equilibrium An equilibrium system in which all components are in the same state.

Le Châtelier’s principle When a stress is placed on a system at equilibrium, the equilibrium shifts to relieve that stress.

rate of reaction The speed at which reactants form products.

reversible reaction A reaction in which a forward reaction occurs from reactants to products, and a reverse reaction occurs from products back to reactants.

solubility product constant, Ksp The product of the concentrations of the ions in a saturated solution of a slightly soluble ionic compound, with each concentration raised to a power equal to its coefficient in the balanced equilibrium equation.

solubility product expression The product of the ion concentrations, with each concentration raised to an exponent equal to the coefficient in the balanced chemical equation.

The chapter section containing each Core Chemistry Skill is shown in parentheses at the end of each heading.

Writing the Equilibrium Expression (13.3) • An equilibrium expression for a reversible reaction is written by

multiplying the concentrations of the products in the numerator and dividing by the product of the concentrations of the reactants in the denominator.

• Each concentration is raised to a power equal to its coefficient in the balanced chemical equation:

CoefficientsKc = [Products]

[Reactants]

[C]c [D]d

[A]a [B]b =

Example: Write the equilibrium expression for the following chemical reaction:

2NO2(g) vh N2O4(g)

Answer: Kc = [N2O4]

[NO2] 2

Calculating an Equilibrium Constant (13.3) • The equilibrium constant, Kc, is the numerical value obtained by

substituting experimentally measured molar concentrations at equilibrium into the equilibrium expression.

Example: Calculate the numerical value of Kc for the following reac- tion when the equilibrium mixture contains 0.025 M NO2 and 0.087 M N2O4:

2NO2(g) vh N2O4(g)

Answer: Write the equilibrium expression, substitute the molar concentrations, and calculate.

Kc = [N2O4]

[NO2] 2

= [0.087]

[0.025]2 = 140

Calculating Equilibrium Concentrations (13.4) • To determine the concentration of a product or a reactant at

equilibrium, we use the equilibrium expression to solve for the unknown concentration.

Example: Calculate the equilibrium concentration for CF4 if Kc = 2.0, and the equilibrium mixture contains 0.10 M COF2 and 0.050 M CO2.

2COF2(g) vh CO2(g) + CF4(g)

Answer: Write the equilibrium expression.

Kc = [CO2][CF4]

[COF2] 2

CORE CHEMISTRY SKILLS Solve the equation for the unknown concentration, substitute the molar concentrations, and calculate.

[CF4] = Kc * [COF2]

2

[CO2] = 2.0 *

[0.10]2

[0.050] = 0.40 M

Using Le Châtelier’s Principle (13.5) • Le Châtelier’s principle states that when a system at equilibrium is

disturbed by changes in concentration, volume, or temperature, the system will shift in the direction that will reduce that stress.

Example: Nitrogen and oxygen form dinitrogen pentoxide in an exothermic reaction.

2N2(g) + 5O2(g) vh 2N2O5(g) + heat

For each of the following changes at equilibrium, indicate whether the equilibrium shifts in the direction of the product, the reactants, or does not change:

a. removing some N2(g) b. decreasing the temperature c. increasing the volume of the container

Answer: a. Removing a reactant shifts the equilibrium in the direction of the reactants.

b. Decreasing the temperature shifts the equilibrium of an exothermic reaction in the direction of the product.

c. Increasing the volume of the container shifts the equilibrium in the direction of the greater number of moles of gas, which is in the direction of the reactants.

Writing the Solubility Product Expression (13.6) • In a saturated solution, a solid slightly soluble ionic compound is

in equilibrium with its ions.

FeF2(s) vh Fe 2+(aq) + 2F -(aq)

• We represent the solubility of solid FeF2 in a saturated aqueous solution by the solubility product expression, which is the prod- uct of the ion concentrations raised to the powers equal to the coefficients.

Ksp = [Fe 2+][F -]2

Example: Write the equilibrium equation for the dissociation of the slightly soluble ionic compound PbI2, and its solubility product expression.

Answer: PbI2(s) vh Pb 2+(aq) + 2I-(aq) Ksp = [Pb2+][I-]2

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Understanding the Concepts 427

Calculating a Solubility Product Constant (13.6) • The solubility product constant, Ksp, is calculated by substituting

the molar concentrations of the ions into the solubility product expression.

Example: What is the numerical value of Ksp for a saturated solution of PbI2 that contains 1.3 * 10-3 M Pb2+ and 2.6 * 10-3 M I-?

Answer: The Ksp is calculated by substituting the molar concentrations into the solubility product expression.

Ksp = [Pb 2+][I-]2 = [1.3 * 10-3][2.6 * 10-3]2

= 8.8 * 10-9

Calculating the Molar Solubility (13.6) • The molar solubility, S, of a slightly soluble ionic compound is the

number of moles of solute that dissolves in 1 liter of solution. • If we know the Ksp of a slightly soluble ionic compound, we

can calculate the molar solubility, S, of the slightly soluble ionic compound.

Example: What is the solubility product expression and the molar solubility, S, of NiS, which has a Ksp of 4.0 * 10-20?

Answer: Ksp = [Ni 2+][S2-] = S * S = S2 = 4.0 * 10-20

S = 24.0 * 10-20 = 2.0 * 10-10 M Therefore, 2.0 * 10-10 mol of NiS will dissolve in 1 L of solution.

UNDERSTANDING THE CONCEPTS

The chapter sections to review are shown in parentheses at the end of each problem.

13.53 Write the equilibrium expression for each of the following reactions: (13.3)

a. CH4(g) + 2O2(g) vh CO2(g) + 2H2O(g) b. 4NH3(g) + 3O2(g) vh 2N2(g) + 6H2O(g) c. C(s) + 2H2(g) vh CH4(g)

13.54 Write the equilibrium expression for each of the following reactions: (13.3)

a. 2C2H6(g) + 7O2(g) vh 4CO2(g) + 6H2O(g) b. 2KHCO3(s) vh K2CO3(s) + CO2(g) + H2O(g) c. 4NH3(g) + 5O2(g) vh 4NO(g) + 6H2O(g)

13.55 Would the equilibrium constant, Kc, for the reaction in the diagrams have a large or small value? (13.4)

13.57 Would T2 be higher or lower than T1 for the reaction shown in the diagrams? (13.5)

+ +

Initially At equilibrium

+ +

Initially At equilibrium

T1 = 300 °C T2 = ?

heat++ +

T1 = 100 °C T2 = 200 °C

+ +

13.56 Would the equilibrium constant, Kc, for the reaction in the diagrams have a large or small value? (13.4)

13.58 Would the reaction shown in the diagrams be exothermic or endothermic? (13.5)

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428 CHAPTER 13 Reaction Rates and Chemical Equilibrium

ADDITIONAL PRACTICE PROBLEMS

13.59 For each of the following changes at equilibrium, indicate whether the equilibrium shifts in the direction of products, reactants, or does not change: (13.5)

C2H4(g) + Cl2(g) vh C2H4Cl2(g) + heat a. increasing the temperature b. decreasing the volume of the container c. adding a catalyst d. adding more Cl2(g)

13.60 For each of the following changes at equilibrium, indicate whether the equilibrium shifts in the direction of products, reactants, or does not change: (13.5)

N2(g) + O2(g) + heat vh 2NO(g) a. increasing the temperature b. decreasing the volume of the container c. adding a catalyst d. adding more N2(g)

13.61 For each of the following reactions, indicate if the equilibrium mixture contains mostly products, mostly reactants, or both reactants and products: (13.4)

a. H2(g) + Cl2(g) vh 2HCl(g) Kc = 1.3 * 10 34

b. 2NOBr(g) vh 2NO(g) + Br2(g) Kc = 2.0 c. 2H2S(g) + CH4(g) vh CS2(g) + 4H2(g)

Kc = 5.3 * 10-8

d. C(s) + H2O(g) vh CO(g) + H2(g) Kc = 6.3 * 10 -1

13.62 For each of the following reactions, indicate if the equilibrium mixture contains mostly products, mostly reactants, or both reactants and products: (13.4)

a. 2H2O(g) vh 2H2(g) + O2(g) Kc = 4 * 10 -48

b. N2(g) + 3H2(g) vh 2NH3(g) Kc = 0.30 c. 2SO2(g) + O2(g) vh 2SO3(g) Kc = 1.2 * 10

9

d. H2(g) + S(s) vh 2H2S(g) Kc = 7.8 * 10 5

13.63 Consider the reaction: (13.3) 2NH3(g) vh N2(g) + 3H2(g)

a. Write the equilibrium expression. b. What is the numerical value of Kc for the reaction if the

concentrations at equilibrium are 0.20 M NH3, 3.0 M N2, and 0.50 M H2?

13.64 Consider the reaction: (13.3) 2SO2(g) + O2(g) vh 2SO3(g)

a. Write the equilibrium expression. b. What is the numerical value of Kc for the reaction if the

concentrations at equilibrium are 0.10 M SO2, 0.12 M O2, and 0.60 M SO3?

13.65 The Kc for the following reaction is 5.0 at 100 °C. If an equilibrium mixture contains 0.50 M NO2, what is the molar concentration of N2O4? (13.3, 13.4)

2NO2(g) vh N2O4(g) 13.66 The Kc for the following reaction is 15 at 220 °C. If an

equilibrium mixture contains 0.40 M CO and 0.20 M H2, what is the molar concentration of CH4O? (13.3, 13.4)

CO(g) + 2H2(g) vh CH4O(g) 13.67 According to Le Châtelier’s principle, does the equilibrium shift

in the direction of products or reactants when O2 is added to the equilibrium mixture of each of the following reactions? (13.5)

a. 3O2(g) vh 2O3(g) b. 2CO2(g) vh 2CO(g) + O2(g) c. 2SO2(g) + O2(g) vh 2SO3(g) d. 2SO2(g) + 2H2O(g) vh 2H2S(g) + 3O2(g)

13.68 According to Le Châtelier’s principle, does the equilibrium shift in the direction of products or reactants when N2 is added to the equilibrium mixture of each of the following reactions? (13.5)

a. 2NH3(g) vh 3H2(g) + N2(g) b. N2(g) + O2(g) vh 2NO(g) c. 2NO2(g) vh N2(g) + 2O2(g) d. 4NH3(g) + 3O2(g) vh 2N2(g) + 6H2O(g)

13.69 Would decreasing the volume of the container for each of the following reactions cause the equilibrium to shift in the direction of products, reactants, or not change? (13.5)

a. 3O2(g) vh 2O3(g) b. 2CO2(g) vh 2CO(g) + O2(g) c. P4(g) + 5O2(g) vh P4O10(s) d. 2SO2(g) + 2H2O(g) vh 2H2S(g) + 3O2(g)

13.70 Would increasing the volume of the container for each of the following reactions cause the equilibrium to shift in the direction of products, reactants, or not change? (13.5)

a. 2NH3(g) vh 3H2(g) + N2(g) b. N2(g) + O2(g) vh 2NO(g) c. N2(g) + 2O2(g) vh 2NO2(g) d. 4NH3(g) + 3O2(g) vh 2N2(g) + 6H2O(g)

13.71 The numerical value of the equilibrium constant, Kc, for the decomposition of COCl2 to CO and Cl2 is 0.68. If an equilib- rium mixture contains 0.40 M CO and 0.74 M Cl2, what is the molar concentration of COCl2? (13.3, 13.4)

COCl2(g) vh CO(g) + Cl2(g) 13.72 The numerical value of the equilibrium constant, Kc, for the

reaction of carbon and water to form carbon monoxide and hydrogen is 0.20 at 1000 °C. If an equilibrium mixture con- tains solid carbon, 0.40 M H2O, and 0.40 M CO, what is the molar concentration of H2? (13.3, 13.4)

C(s) + H2O(g) vh CO(g) + H2(g) 13.73 For each of the following slightly soluble ionic compounds,

write the equilibrium equation for dissociation and the solubility product expression: (13.6)

a. CuCO3 b. PbF2 c. Fe(OH)3 13.74 For each of the following slightly soluble ionic compounds,

write the equilibrium equation for dissociation and the solubility product expression: (13.6)

a. CuS b. Ag2SO4 c. Zn(OH)2 13.75 A saturated solution of iron(II) sulfide, FeS, has

[Fe2+] = 7.7 * 10-10 M and [S2-] = 7.7 * 10-10 M. What is the numerical value of Ksp for FeS? (13.6)

13.76 A saturated solution of copper(I) chloride, CuCl, has [Cu+] = 1.1 * 10-3 M and [Cl-] = 1.1 * 10-3 M. What is the numerical value of Ksp for CuCl? (13.6)

13.77 A saturated solution of manganese(II) hydroxide, Mn(OH)2, has [Mn2+] = 3.7 * 10-5 M and [OH-] = 7.4 * 10-5 M. What is the numerical value of Ksp for Mn(OH)2? (13.6)

13.78 A saturated solution of silver chromate, Ag2CrO4, has [Ag +] = 1.3 * 10-4 M and [CrO4 2-] = 6.5 * 10-5 M. What is the numerical value of Ksp for Ag2CrO4? (13.6)

13.79 What is the molar solubility, S, of CdS if it has a Ksp of 1.0 * 10-24? (13.6)

13.80 What is the molar solubility, S, of CuCO3 if it has Ksp of 1 * 10-26? (13.6)

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Answers to Engage Questions 429

ANSWERS TO ENGAGE QUESTIONS mixture containing mostly products will have a large numeri- cal value for Kc.

13.9 An equilibrium reaction with a Kc = 1.2, which is close to 1, contains almost equal amounts of both reactants and products.

13.10 The equilibrium expression has the concentrations of prod- ucts over the concentrations of reactants. An equilibrium mixture with a Kc of 2 * 10-9 means the numerator, which is reactants, is much smaller than the denominator.

13.11 Because HI is a product, adding more HI will shift the equilib- rium in the direction of the reactants.

13.12 In an exothermic reaction, heat is a product. To decrease the amount of product, the temperature should be increased.

13.13 In the equation for the reaction when the slightly soluble ionic compound Cu2CO3 dissolves, Cu

+ has a coefficient of 2 and CO3

2- has a coefficient of 1. Thus, the Ksp expression is written as Ksp = [Cu

+]2[CO3 2-].

13.1 If molecules collide without the minimum activation energy, they do not react.

13.2 An increase in temperature will increase the rate of reaction because the molecules move faster with more energy, which means there are more collisions with the necessary activation energy.

13.3 A decrease in the concentration of a reactant decreases the rate of reaction because there are fewer molecules to collide.

13.4 Before equilibrium is reached, the forward reaction is faster, decreasing the concentrations of reactants.

13.5 Once a chemical reaction reaches equilibrium, the rates of the forward and reverse reactions are equal.

13.6 In the equilibrium expression, any coefficient in the chemical equation is indicated as an exponent.

13.7 The concentrations of solids or liquids are not included in the equilibrium expression for Kc because they are constant.

13.8 The equilibrium expression has the concentrations of prod- ucts over the concentrations of reactants, thus an equilibrium

The following problems are related to the topics in this chapter. However, they do not all follow the chapter order, and they require you to combine concepts and skills from several sections. These problems will help you increase your critical thinking skills and prepare for your next exam.

13.81 The Kc at 100 °C is 2.0 for the decomposition reaction of NOBr. (13.3, 13.4, 13.5)

2NOBr(g) vh 2NO(g) + Br2(g) In an experiment, 1.0 mol of NOBr, 1.0 mol of NO, and 1.0 mol of Br2 were placed in a 1.0-L container.

a. Write the equilibrium expression for the reaction. b. Is the system at equilibrium? c. If not, will the rate of the forward or reverse reaction

initially speed up? d. At equilibrium, which concentration(s) will be greater than

1.0 mol/L, and which will be less than 1.0 mol/L?

13.82 Consider the following reaction: (13.3, 13.4, 13.5) PCl5(g) vh PCl3(g) + Cl2(g)

a. Write the equilibrium expression for the reaction. b. Initially, 0.60 mol of PCl5 is placed in a 1.0-L flask. At

equilibrium, there is 0.16 mol of PCl3 in the flask. What are the equilibrium concentrations of PCl5 and Cl2?

c. What is the numerical value of the equilibrium constant, Kc, for the reaction?

d. If 0.20 mol of Cl2 is added to the equilibrium mixture, will the concentration of PCl5 increase or decrease?

13.83 Indicate how each of the following will affect the equilibrium concentration of CO in the following reaction: (13.3, 13.5)

C(s) + H2O(g) + 31 kcal vh CO(g) + H2(g) a. adding more H2(g) b. increasing the temperature c. increasing the volume of the container d. decreasing the volume of the container e. adding a catalyst

13.84 Indicate how each of the following will affect the equilibrium concentration of NH3 in the following reaction: (13.3, 13.5)

4NH3(g) + 5O2(g) vh 4NO(g) + 6H2O(g) + 906 kJ a. adding more O2(g) b. increasing the temperature c. increasing the volume of the container d. adding more NO(g) e. removing some H2O(g)

13.85 Indicate if you would increase or decrease the volume of the container to increase the yield of the products in each of the following: (13.5)

a. 2C(s) + O2(g) vh 2CO(g) b. 2CH4(g) vh C2H2(g) + 3H2(g) c. 2H2(g) + O2(g) vh 2H2O(g)

13.86 Indicate if you would increase or decrease the volume of the container to increase the yield of the products in each of the following: (13.5)

a. Cl2(g) + 2NO(g) vh 2NOCl(g) b. N2(g) + 2H2(g) vh N2H4(g) c. N2O4(g) vh 2NO2(g)

13.87 The antacid milk of magnesia, which contains Mg(OH)2, is used to neutralize excess stomach acid. If the solubility of Mg(OH)2 in water is 9.7 * 10-3 g/L, what is the numerical value of Ksp for Mg(OH)2? (13.6)

13.88 The slightly soluble ionic compound PbF2 has a solubility in water of 0.74 g/L. What is the numerical value of Ksp for PbF2? (13.6)

CHALLENGE PROBLEMS

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430 CHAPTER 13 Reaction Rates and Chemical Equilibrium

ANSWERS TO SELECTED PROBLEMS 13.45 Ksp = 1 * 10-10

13.47 Ksp = 8.8 * 10-12

13.49 S = 1 * 10-6 M 13.51 a. CaCO3(s) vh Ca

2+(aq) + CO3 2-(aq) b. Ksp = [Ca

2+][CO3 2-]

13.53 a. Kc = [CO2][H2O]

2

[CH4][O2] 2

b. Kc = [N2]

2[H2O] 6

[NH3] 4[O2]

3

c. Kc = [CH4]

[H2] 2

13.55 The equilibrium constant for the reaction would have a small value.

13.57 T2 is lower than T1.

13.59 a. Equilibrium shifts in the direction of the reactants. b. Equilibrium shifts in the direction of the product. c. There is no change in equilibrium. d. Equilibrium shifts in the direction of the product.

13.61 a. mostly products b. both reactants and products c. mostly reactants d. both reactants and products

13.63 a. Kc = [N2][H2]

3

[NH3] 2

b. Kc = 9.4

13.65 [N2O4] = 1.3 M

13.67 a. Equilibrium shifts in the direction of the product. b. Equilibrium shifts in the direction of the reactant. c. Equilibrium shifts in the direction of the product. d. Equilibrium shifts in the direction of the reactants.

13.69 a. Equilibrium shifts in the direction of the product. b. Equilibrium shifts in the direction of the reactant. c. Equilibrium shifts in the direction of the product. d. Equilibrium shifts in the direction of the reactants.

13.71 [COCl2] = 0.44 M 13.73 a. CuCO3(s) vh Cu

2+(aq) + CO3 2-(aq); Ksp = [Cu

2+][CO3 2-]

b. PbF2(s) vh Pb 2+(aq) + 2F -(aq);

Ksp = [Pb 2+][F -]2

c. Fe(OH)3(s) vh Fe 3+(aq) + 3OH-(aq);

Ksp = [Fe 3+][OH-]3

13.75 Ksp = 5.9 * 10-19

13.77 Ksp = 2.0 * 10-13

13.79 S = 1.0 * 10-12 M

13.81 a. Kc = [NO]2[Br2]

[NOBr]2

b. When the concentrations are placed in the equilibrium expression, the result is 1.0, which is not equal to Kc. The system is not at equilibrium.

c. The rate of the forward reaction will increase. d. The [Br2] and [NO] will increase and [NOBr] will

decrease.

13.83 a. decrease b. increase c. increase d. decrease e. no change

13.85 a. increase b. increase c. decrease

13.87 Ksp = 2.0 * 10-11

13.1 a. The rate of the reaction indicates how fast the products form or how fast the reactants are used up.

b. At room temperature, the reactions involved in the growth of bread mold will proceed at a faster rate than at the lower temperature of the refrigerator.

13.3 The number of collisions will increase when the number of Br2 molecules is increased.

13.5 a. increase b. increase c. increase d. decrease

13.7 A reversible reaction is one in which a forward reaction converts reactants to products, whereas a reverse reaction converts products to reactants.

13.9 a. not at equilibrium b. at equilibrium c. at equilibrium

13.11 The reaction has reached equilibrium because the number of reactants and products does not change.

13.13 a. Kc = [CS2][H2]

4

[CH4][H2S] 2 b. Kc =

[N2][O2]

[NO]2

c. Kc = [CS2][O2]

4

[SO3] 2[CO2]

d. Kc = [H2]

3[CO]

[CH4][H2O]

13.15 Kc = [XY]2

[X2][Y2] = 36

13.17 Kc = 1.5 13.19 Kc = 260 13.21 a. homogeneous equilibrium b. heterogeneous equilibrium c. homogeneous equilibrium d. heterogeneous equilibrium

13.23 a. Kc = [O2]

3

[O3] 2 b. Kc = [CO2][H2O]

c. Kc = [C6H12]

[C6H6][H2] 3 d. Kc =

[H2] 2

[HCl]4

13.25 Kc = 0.26 13.27 Diagram B represents the equilibrium mixture.

13.29 a. mostly products b. mostly products c. mostly reactants

13.31 [H2] = 1.1 * 10-3 M 13.33 [NOBr] = 1.4 M 13.35 a. Equilibrium shifts in the direction of the product.

b. Equilibrium shifts in the direction of the reactants. c. Equilibrium shifts in the direction of the product. d. Equilibrium shifts in the direction of the reactants. e. No shift in equilibrium occurs.

13.37 a. Equilibrium shifts in the direction of the product. b. Equilibrium shifts in the direction of the product. c. Equilibrium shifts in the direction of the product. d. No shift in equilibrium occurs. e. Equilibrium shifts in the direction of the reactants.

13.39 a. Equilibrium shifts in the direction of the product. b. Equilibrium shifts in the direction of the reactants. c. Equilibrium shifts in the direction of the reactants.

13.41 a. The oxygen concentration is lowered. b. Equilibrium shifts in the direction of the reactants.

13.43 a. MgCO3(s) H Mg2+(aq) + CO3 2-(aq); Ksp = [Mg

2+][CO3 2-]

b. CaF2(s) H Ca2+(aq) + 2F -(aq); Ksp = [Ca2+][F -]2 c. Ag3PO4(s) H 3Ag +(aq) + PO4 3-(aq);

Ksp = [Ag +]3[PO4

3-]

M13_TIMB8119_06_SE_C13.indd 430 11/27/18 12:07 PM

431

After Larry was discharged from the hospital, he complained of a sore throat and dry cough, which his doctor diagnosed as acid reflux. You can view the symptoms of acid reflux disease (GERD) in the UPDATE Acid Reflux Disease,  pages 463–464, and learn about the pH changes in the stomach and how the condition is treated.

After an automobile accident, Larry, a 30-year-old man, is brought to the emergency room, where he is unresponsive. One of the emergency room nurses takes a blood sample, which is then sent to Brianna, a medical laboratory technologist, who begins the process of analyzing the pH, the partial pressures of O2 and CO2, and the concentrations of glucose and electrolytes.

Brianna determines that Larry’s blood pH is 7.30, and the partial pressure of CO2 gas is above the desired level. Blood pH is typically in the range of 7.35 to 7.45, and a value less than 7.35 indicates a state of acidosis. Respiratory acidosis occurs because an increase in the partial pressure of CO2 gas in the bloodstream increases the concentration of H2CO3, which leads to an increase in the concentration of H3O

+ . Brianna recognizes these signs and immediately contacts the emergency room to inform

them that Larry’s airway may be blocked. In the emergency room, they provide Larry with an IV containing bicarbonate to increase the blood pH and begin the process of unblocking his airway. Shortly afterward, Larry’s airway is cleared, and his blood pH and partial pressure of CO2 gas return to normal.

CAREER

Medical Laboratory Technologist Medical laboratory technologists perform a wide variety of tests on body fluids and cells that help in the diagnosis and treatment of patients. These tests range from determining blood concentrations of glucose and cholesterol to determining drug levels in the blood for transplant patients or a patient undergoing treatment. Medical laboratory technologists also prepare specimens in the detection of cancerous tumors and type blood samples for transfusions. Medical laboratory technologists must also interpret and analyze the test results, which are then passed on to the physician.

Acids and Bases

UPDATE Acid Reflux Disease

14

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432 CHAPTER 14 Acids and Bases

14.1 Acids and Bases LEARNING GOAL Describe and name acids and bases.

Acids and bases are important substances in health, industry, and the environment. One of the most common characteristics of acids is their sour taste. Lemons and grapefruits taste sour because they contain acids such as citric and ascorbic acid (vitamin C). Vinegar tastes sour because it contains acetic acid. We produce lactic acid in our muscles when we exercise. Acid from bacteria turns milk sour in the production of yogurt and cottage cheese. We have hydrochloric acid in our stomachs that helps us digest food. Sometimes we take antacids, which are bases such as sodium bicarbonate or milk of magnesia, to neutralize the effects of too much stomach acid.

The term acid comes from the Latin word acidus, which means “sour.” You are prob- ably already familiar with the sour tastes of vinegar and lemons.

In 1887, Swedish chemist Svante Arrhenius was the first to describe acids as substances that produce hydrogen ions (H+) when they dissolve in water. Because acids produce ions in water, they are electrolytes. For example, hydrogen chloride dissociates in water to give hydrogen ions, H+, and chloride ions, Cl-. The hydrogen ions give acids a sour taste, change the blue litmus indicator to red, and corrode some metals.

HCl(g) H+(aq) Dissociation

H2O

Hydrogen ion

Polar molecular compound

+ Cl-(aq)

Naming Acids Acids dissolve in water to produce hydrogen ions, along with a negative ion that may be a simple nonmetal anion or a polyatomic ion. When an acid dissolves in water to produce a hydrogen ion and a simple nonmetal anion, the prefix hydro is used before the name of the nonmetal, and its ide ending is changed to ic acid. For example, hydrogen chloride (HCl) dissolves in water to form HCl(aq), which is named hydrochloric acid. An exception is hydrogen cyanide (HCN), which as an acid is named hydrocyanic acid.

When an acid contains oxygen, it dissolves in water to produce a hydrogen ion and an oxygen-containing polyatomic anion. The most common form of an oxygen-containing acid has a name that ends with ic acid. The name of its polyatomic anion ends in ate. If the acid contains a polyatomic ion with an ite ending, its name ends in ous acid. When the acid has one oxygen atom less than the common form, the suffix ous is used and the polyatomic ion is named with an ite ending. The names of some common acids and their anions are listed in TABLE 14.1.

LOOKING AHEAD

14.1 Acids and Bases 432 14.2 Brønsted–Lowry Acids

and Bases 434 14.3 Strengths of Acids and

Bases 437 14.4 Dissociation of Weak

Acids and Bases 442 14.5 Dissociation of Water 444 14.6 The pH Scale 447 14.7 Reactions of Acids and

Bases 454 14.8 Acid–Base Titration 457 14.9 Buffers 459

REVIEW Writing Ionic Formulas (6.2)

TABLE 14.1 Names of Common Acids and Their Anions Acid Name of Acid Anion Name of Anion

HF Hydrofluoric acid F - Fluoride

HCl Hydrochloric acid Cl- Chloride

HBr Hydrobromic acid Br- Bromide

HI Hydroiodic acid I- Iodide

HCN Hydrocyanic acid CN- Cyanide

HNO3 Nitric acid NO3 - Nitrate

HNO2 Nitrous acid NO2 - Nitrite

H2SO4 Sulfuric acid SO4 2- Sulfate

H2SO3 Sulfurous acid SO3 2- Sulfite

H2CO3 Carbonic acid CO3 2- Carbonate

HC2H3O2 Acetic acid C2H3O2 - Acetate

H3PO4 Phosphoric acid PO4 3- Phosphate

H3PO3 Phosphorous acid PO3 3- Phosphite

Citrus fruits are sour because of the presence of acids.

Sulfuric acid dissolves in water to produce one or two H+ and an anion.

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14.1 Acids and Bases 433

The halogens in Group 7A (17) can form more than two oxygen-containing acids. For chlorine, the common form is chloric acid, HClO3, which contains the polyatomic ion chlorate, ClO3

-. For the acid that contains one more oxygen atom than the common form, the prefix per is used; HClO4 is perchloric acid. When the polyatomic ion in the acid has one oxygen atom less than the common form, the suffix ous is used. Thus, HClO2 is chlorous acid; it contains the chlorite ion, ClO2

-. The prefix hypo is used for the acid that has two oxygen atoms less than the common form; HClO is hypochlorous acid.

ENGAGE 14.1 Why is HBr named hydrobromic acid?

Names of Common Acids and Their Anions

Acid Name of Acid Anion Name of Anion

HClO4 Perchloric acid ClO4 - Perchlorate

HClO3 Chloric acid ClO3 - Chlorate

HClO2 Chlorous acid ClO2 - Chlorite

HClO Hypochlorous acid ClO- Hypochlorite

Bases You may be familiar with household bases such as antacids, drain openers, and oven clean- ers. According to the Arrhenius theory, bases are ionic compounds that dissociate into cations and hydroxide ions (OH-) when they dissolve in water. They are strong electrolytes. For example, sodium hydroxide is an Arrhenius base that dissociates completely in water to give sodium ions Na+, and hydroxide ions OH-.

Most Arrhenius bases are formed from Groups 1A (1) and 2A (2) metals, such as NaOH, KOH, LiOH, and Ca(OH)2. The hydroxide ions (OH

-) give Arrhenius bases common char- acteristics, such as a bitter taste and a slippery feel. A base turns litmus indicator blue and phenolphthalein indicator pink. TABLE 14.2 compares some characteristics of acids and bases.

An Arrhenius base produces cations and OH- anions in an aqueous solution.

Ionic compound

Hydroxide ion

Dissociation

OH -

Na+

NaOH(s) Na+(aq) OH -(aq)+ H2O

NaOH(s)

Water

+ +

+

+ ++

+ +

+

-

- -

-

- -

- -

-

Calcium hydroxide, Ca(OH)2, is used in the food industry to produce beverages, and in dentistry as a filler for root canals.

A soft drink contains H3PO4 and H2CO3.

Characteristic Acids Bases

Arrhenius Produce H+ Produce OH-

Electrolytes Yes Yes

Taste Sour Bitter, chalky

Feel May sting Soapy, slippery

Litmus Red Blue

Phenolphthalein Colorless Pink

Neutralization Neutralize bases Neutralize acids

TABLE 14.2 Some Characteristics of Acids and Bases

Naming Bases Typical Arrhenius bases are named as hydroxides. Base Name

LiOH Lithium hydroxide

NaOH Sodium hydroxide

KOH Potassium hydroxide

Ca(OH)2 Calcium hydroxide

Al(OH)3 Aluminum hydroxide

SAMPLE PROBLEM 14.1 Names and Formulas of Acids and Bases

TRY IT FIRST

a. Identify each of the following as an acid or a base, and give its name: 1. H3PO4, ingredient in soft drinks 2. NaOH, ingredient in oven cleaner

b. Write the formula for each of the following: 1. magnesium hydroxide, ingredient in antacids 2. hydrobromic acid, used industrially to prepare bromide compounds

M14_TIMB8119_06_SE_C14.indd 433 11/27/18 12:18 PM

434 CHAPTER 14 Acids and Bases

SOLUTION

a. 1. acid, phosphoric acid 2. base, sodium hydroxide b. 1. Mg(OH)2 2. HBr

SELF TEST 14.1

a. Determine whether H2CO3 is an acid or base, and give its name. b. Determine whether iron(III) hydroxide is an acid or base, and write its formula.

ANSWER

a. acid, carbonic acid b. base, Fe(OH)3

PRACTICE PROBLEMS Try Practice Problems 14.1 to 14.6

PRACTICE PROBLEMS

14.1 Acids and Bases

14.1 Indicate whether each of the following statements is characteristic of an acid, a base, or both:

a. has a sour taste b. neutralizes bases c. produces H+ ions in water d. is named barium hydroxide e. is an electrolyte

14.2 Indicate whether each of the following statements is characteristic of an acid, a base, or both:

a. neutralizes acids b. produces OH- ions in water c. has a slippery feel d. conducts an electrical current in solution e. turns litmus red

14.3 Name each of the following acids or bases: a. HCl b. Ca(OH)2 c. HClO4 d. Sr(OH)2 e. H2SO3 f. HBrO2 14.4 Name each of the following acids or bases: a. Al(OH)3 b. HBr c. H2SO4 d. KOH e. HNO2 f. HClO2 14.5 Write formulas for each of the following acids and bases: a. rubidium hydroxide b. hydrofluoric acid c. phosphoric acid d. lithium hydroxide e. ammonium hydroxide f. periodic acid

14.6 Write formulas for each of the following acids and bases: a. barium hydroxide b. hydroiodic acid c. nitric acid d. strontium hydroxide e. acetic acid f. hypochlorous acid

14.2 Brønsted–Lowry Acids and Bases LEARNING GOAL Identify conjugate acid–base pairs for Brønsted–Lowry acids and bases.

In 1923, J. N. Brønsted in Denmark and T. M. Lowry in Great Britain expanded the definition of acids and bases to include bases that do not contain OH- ions. A Brønsted–Lowry acid can donate a hydrogen ion, H+, and a Brønsted–Lowry base can accept a hydrogen ion.

A Brønsted–Lowry acid is a substance that donates H+.

A Brønsted–Lowry base is a substance that accepts H+.

A free hydrogen ion, H+, does not actually exist in water. Its attraction to polar water  molecules is so strong that the H+ bonds to a water molecule and forms a hydronium ion, H3O

+.

H O + H + H O H

HH

Water Hydrogen ion

Hydronium ion

+

We can write the formation of a hydrochloric acid solution as a transfer of H+ from hydrogen chloride to water. By accepting an H+ in the reaction, water is acting as a base according to the Brønsted–Lowry concept.

M14_TIMB8119_06_SE_C14.indd 434 11/27/18 12:18 PM

14.2 Brønsted–Lowry Acids and Bases 435

HCl + H2O H3O+ + Cl- Hydrogen chloride

Water Hydronium ion

Chloride ion

Acid (H+ donor)

Base (H+ acceptor) Acidic solution

-+

In another reaction, ammonia, NH3, acts as a base by accepting H + when it reacts with

water. Because the nitrogen atom of NH3 has a stronger attraction for H + than oxygen, water

acts as an acid by donating H+.

Base (H+ acceptor)

Acid (H+ donor) Basic solution

NH3 + H2O NH4+ + OH- Ammonia Water Ammonium

ion Hydroxide

ion

- +

SAMPLE PROBLEM 14.2 Acids and Bases

TRY IT FIRST

In each of the following equations, identify the reactant that is a Brønsted–Lowry acid and the reactant that is a Brønsted–Lowry base:

a. HBr(aq) + H2O(l) h H3O+(aq) + Br-(aq) b. CN-(aq) + H2O(l) vh  HCN(aq) + OH-(aq)

SOLUTION

a. HBr, Brønsted–Lowry acid; H2O, Brønsted–Lowry base b. H2O, Brønsted–Lowry acid; CN

-, Brønsted–Lowry base

SELF TEST 14.2

a. When HNO3 reacts with water, water acts as a Brønsted–Lowry base. Write the equation for the reaction.

b. When hypochlorite ion, ClO-, reacts with water, water acts as a Brønsted–Lowry acid. Write the equation for the reaction.

ANSWER

a. HNO3(aq) + H2O(l) h H3O+(aq) + NO3 -(aq) b. ClO-(aq) + H2O(l) vh  HClO(aq) + OH-(aq)

Conjugate Acid–Base Pairs According to the Brønsted–Lowry theory, a conjugate acid–base pair consists of molecules or ions related by the loss of one H+ by an acid, and the gain of one H+ by a base. Every acid–base reaction contains two conjugate acid–base pairs because an H+ is transferred in both the forward and reverse directions. When an acid such as HF loses one H+, the conjugate base F - is formed. When the base H2O gains an H

+, its conjugate acid, H3O +,

is formed. Because the overall reaction of HF is reversible, the conjugate acid H3O

+ can donate H+ to the conjugate base F - and re-form the acid HF and the base H2O. Using the relationship of loss and gain of one H+, we can now identify the conjugate acid–base pairs as HF/F - along with H3O

+/H2O.

PRACTICE PROBLEMS Try Practice Problems 14.7 and 14.8

F-

Conjugate acid–base pair

H2O H3O +

Donates H+

Accepts H+

HF

Conjugate acid–base pair

+

-

M14_TIMB8119_06_SE_C14.indd 435 11/27/18 12:18 PM

436 CHAPTER 14 Acids and Bases

HF(aq) H2O(l)

Acid Conjugate base

Conjugate acid

+ F-(aq) H3O+(aq)+

H+ loss

Base

H+ gain

H+ loss

H+ gain

HF, an acid, loses one H+ to form its conjugate base F-. Water acts as a base by gaining one H+ to form its conjugate acid H3O

+.

In another reaction, ammonia (NH3) accepts H + from H2O to form the conjugate acid

NH4 + and conjugate base OH-. Each of these conjugate acid–base pairs, NH4

+/NH3 and H2O/OH

-, is related by the loss and gain of one H+.

ENGAGE 14.2 Why is HBrO2 the conjugate acid of BrO2

-?

Conjugate acid–base pair

Conjugate acid–base pair

NH3(g) + H2O(l) NH4+(aq) + OH-(aq)

Ammonia, NH3, acts as a base when it gains one H+ to form its conjugate acid NH4

+. Water acts as an acid by losing one H+ to form its conjugate base OH-.

In these two examples, we see that water can act as an acid when it donates H+ or as a base when it accepts H+. Substances that can act as both acids and bases are amphoteric or amphiprotic. For water, the most common amphoteric substance, the acidic or basic behavior depends on the other reactant. Water donates H+ when it reacts with a stronger base, and it accepts H+ when it reacts with a stronger acid. Another example of an ampho- teric substance is bicarbonate (HCO3

-). With a base, HCO3 - acts as an acid and donates

one H+ to give CO3 2-. However, when HCO3

- reacts with an acid, it acts as a base and accepts one H+ to form H2CO3.

ENGAGE 14.3 Why can H2O be both the conjugate base of H3O

+ and the conjugate acid of OH-?

Acts as a base Acts as an acid H2O

HCO3 -

H3O +

H2CO3

OH-

CO3 2-

Amphoteric substances act as both acids and bases.

SAMPLE PROBLEM 14.3 Identifying Conjugate Acid–Base Pairs

TRY IT FIRST

Identify the conjugate acid–base pairs in the following reaction:

HBr(aq) + NH3(aq) h Br-(aq) + NH4 +(aq)

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

HBr NH3

Br- NH4

+ conjugate

acid–base pairs lose/gain

one H+

STEP 1 Identify the reactant that loses H+ as the acid. In the reaction, HBr loses H+ to form the product Br-. Thus HBr is the acid and Br- is its conjugate base.

STEP 2 Identify the reactant that gains H+ as the base. In the reaction, NH3 gains H+ to form the product NH4

+. Thus, NH3 is the base and NH4 + is its conjugate

acid.

STEP 3 Write the conjugate acid–base pairs.

HBr/Br- and NH4 +/NH3

CORE CHEMISTRY SKILL Identifying Conjugate Acid–Base

Pairs

M14_TIMB8119_06_SE_C14.indd 436 11/27/18 12:18 PM

14.3 Strengths of Acids and Bases 437

PRACTICE PROBLEMS

14.2 Brønsted–Lowry Acids and Bases

14.7 Identify the reactant that is a Brønsted–Lowry acid and the reac- tant that is a Brønsted–Lowry base in each of the following:

a. HI(aq) + H2O(l) h I-(aq) + H3O+(aq) b. F -(aq) + H2O(l) vh HF(aq) + OH

-(aq) c. H2S(aq) + C2H5 ¬ NH2(aq) vh

HS-(aq) + C2H5 ¬ NH3 +(aq) 14.8 Identify the reactant that is a Brønsted–Lowry acid and the reac-

tant that is a Brønsted–Lowry base in each of the following: a. CO3

2-(aq) + H2O(l) vh HCO3 -(aq) + OH-(aq)

b. H2SO4(aq) + H2O(l) h HSO4 -(aq) + H3O+(aq) c. C2H3O2

-(aq) + H3O+(aq) vh HC2H3O2(aq) + H2O(l) 14.9 Write the formula for the conjugate base for each of the

following acids: a. HF b. H2O c. H2PO3

-

d. HSO4 - e. HClO2

14.10 Write the formula for the conjugate base for each of the following acids:

a. HCO3 - b. CH3 ¬ NH3 + c. HPO4 2-

d. HNO2 e. HBrO

14.11 Write the formula for the conjugate acid for each of the following bases:

a. CO3 2- b. H2O c. H2PO4

-

d. Br- e. ClO4 -

14.12 Write the formula for the conjugate acid for each of the following bases:

a. SO4 2- b. CN- c. NH3

d. ClO2 - e. HS-

14.13 Identify the Brønsted–Lowry acid–base pairs in each of the following equations:

a. H2CO3(aq) + H2O(l) vh HCO3 -(aq) + H3O+(aq)

b. HCN(aq) + NO2 -(aq) vh CN -(aq) + HNO2(aq)

c. CHO2 -(aq) + HF(aq) vh HCHO2(aq) + F

-(aq)

14.14 Identify the Brønsted–Lowry acid–base pairs in each of the following equations:

a. H3PO4(aq) + H2O(l) vh H2PO4 -(aq) + H3O+(aq)

b. H3PO4(aq) + NH3(aq) vh H2PO4 -(aq) + NH4 +(aq)

c. HNO2(aq) + C2H5 ¬ NH2(aq) vh C2H5 ¬ NH3 +(aq) + NO2 -(aq)

14.15 When ammonium chloride dissolves in water, the ammonium ion, NH4

+, acts as an acid. Write a balanced equation for the reaction of the ammonium ion with water.

14.16 When sodium carbonate dissolves in water, the carbonate ion, CO3

2-, acts as a base. Write a balanced equation for the reaction of the carbonate ion with water.

SELF TEST 14.3

Identify the conjugate acid–base pairs in each of the following reactions:

a. HCN(aq) + SO4 2-(aq) vh CN-(aq) + HSO4 -(aq) b. H2O(l) + S2-(aq) vh OH-(aq) + HS-(aq)

ANSWER

a. The conjugate acid–base pairs are HCN/CN- and HSO4 -/SO4

2-. b. The conjugate acid–base pairs are H2O/OH

- and HS-/S2-.

14.3 Strengths of Acids and Bases LEARNING GOAL Write equations for the dissociation of strong and weak acids; identify the direction of reaction.

In the process called dissociation, an acid or a base separates into ions in water. The strength of an acid is determined by the moles of H3O

+ that are produced for each mole of acid that dissolves. The strength of a base is determined by the moles of OH- that are produced for each mole of base that dissolves. Strong acids and strong bases dissociate completely in water, whereas weak acids and weak bases dissociate only slightly, leaving most of the initial acid or base undissociated.

Strong and Weak Acids Strong acids are examples of strong electrolytes because they donate H+ so easily that their dissociation in water is essentially complete. For example, when HCl, a strong acid, dissociates in water, H+ is transferred to H2O; the resulting solution contains essentially only the ions H3O

+ and Cl-. We consider the reaction of HCl in H2O as going 100% to

PRACTICE PROBLEMS Try Practice Problems 14.9 to 14.16

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438 CHAPTER 14 Acids and Bases

products. Thus, one mole of a strong acid dissociates in water to yield one mole of H3O +

and one mole of its conjugate base. We write the equation for a strong acid such as HCl with a single arrow.

HCl(g) + H2O(l) h H3O+(aq) + Cl-(aq)

There are only six common strong acids, which are stronger acids than H3O +. All other

acids are weak. TABLE 14.3 lists the relative strengths of acids and bases. Weak acids are weak electrolytes because they dissociate slightly in water, forming only a small amount of H3O

+ ions. A weak acid has a strong conjugate base, which is why the reverse reaction is more prevalent. Even at high concentrations, weak acids produce low concentrations of H3O

+ ions (see FIGURE 14.1). Many of the products you use at home contain weak acids. Citric acid is a weak

acid found in fruits and fruit juices such as lemons, oranges, and grapefruit. The vin- egar used in salad dressings is typically a 5% (m/v) acetic acid, HC2H3O2, solution. In water, a few HC2H3O2 molecules donate H

+ to H2O to form H3O + ions and acetate

ions C2H3O2 -. The reverse reaction also takes place, which converts the H3O

+ ions and acetate ions C2H3O2

- back to reactants. The formation of hydronium ions from vinegar is the reason we notice the sour taste of vinegar. We write the equation for a weak acid in an aqueous solution with a double arrow to indicate that the forward and reverse reactions are at equilibrium.

HC2H3O2(aq) + H2O(l) vh C2H3O2 -(aq) + H3O+(aq)

ENGAGE 14.4 What is the difference between a strong acid and a weak acid?

ENGAGE 14.5 Which is the weaker acid: H2SO4 or H2S?

PRACTICE PROBLEMS Try Practice Problems 14.17 to 14.22

Acetic acid Acetate ion

Weak acids are found in foods and household products.

TABLE 14.3 Relative Strengths of Acids and Bases

A ci

d S

tr en

gt h

In cr

ea se

s

Acid Conjugate Base

B as

e S

tr en

gt h

In cr

ea se

s

Strong Acids Weak Bases

Hydroiodic acid HI I- Iodide ion

Hydrobromic acid HBr Br- Bromide ion

Perchloric acid HClO4 ClO4 - Perchlorate ion

Hydrochloric acid HCl Cl- Chloride ion

Sulfuric acid H2SO4 HSO4 - Hydrogen sulfate ion

Nitric acid HNO3 NO3 - Nitrate ion

Hydronium ion H3O + H2O Water

Weak Acids Strong Bases

Hydrogen sulfate ion HSO4 - SO4

2- Sulfate ion

Phosphoric acid H3PO4 H2PO4 - Dihydrogen phosphate ion

Nitrous acid HNO2 NO2 - Nitrite ion

Hydrofluoric acid HF F - Fluoride ion

Acetic acid HC2H3O2 C2H3O2 - Acetate ion

Carbonic acid H2CO3 HCO3 - Bicarbonate ion

Hydrosulfuric acid H2S HS - Hydrogen sulfide ion

Dihydrogen phosphate ion H2PO4 - HPO4

2- Hydrogen phosphate ion

Ammonium ion NH4 + NH3 Ammonia

Hydrocyanic acid HCN CN- Cyanide ion

Bicarbonate ion HCO3 - CO3

2- Carbonate ion

Methylammonium ion CH3 ¬ NH3 + CH3 ¬ NH2 Methylamine Hydrogen phosphate ion HPO4

2- PO4 3- Phosphate ion

Hydrogen sulfide ion HS- S2- Sulfide ion

Water H2O OH - Hydroxide ion

M14_TIMB8119_06_SE_C14.indd 438 11/27/18 12:18 PM

14.3 Strengths of Acids and Bases 439

Diprotic Acids Some weak acids, such as carbonic acid, are diprotic acids that have two H+, which dis- sociate one at a time. For example, carbonated soft drinks are prepared by dissolving CO2 in water to form carbonic acid, H2CO3. A weak acid such as H2CO3 reaches equilibrium between the mostly undissociated H2CO3 molecules and the ions H3O

+ and HCO3 -.

H2CO3(aq) + H2O(l) vh H3O +(aq) + HCO3 -(aq)

Because HCO3 - is also a weak acid, a second dissociation can take place to produce

another hydronium ion and the carbonate ion, CO3 2-.

FIGURE 14.1 A strong acid such as HCl is completely dissociated (≈100%), whereas a weak acid such as HC2H3O2 contains mostly molecules and a few ions.

+

+

+

+ +

+

+

+

+

+

-

-

-

- -

-

- -

-

-

HCl 1 M

HC2H3O2 1 M

HCl: Completely dissociated

HC2H3O2: Mostly molecules and a

few ions

Carbonic acid

Sulfuric acid

Bisulfate ion (hydrogen sulfate)

Bicarbonate ion (hydrogen carbonate)

H2CO3 HCO3 - CO3

2-

Bicarbonate ion ( hydrogen carbonate)

Carbonate ion

HCO3 -(aq) H2O(l) H3O

+(aq) CO3 2-(aq)++

Carbonic acid, a weak acid, loses one H+ to form hydrogen carbonate ion, which loses a second H+ to form carbonate ion.

Sulfuric acid, H2SO4, is also a diprotic acid. However, its first dissociation is complete (100%), which means H2SO4 is a strong acid. The product, hydrogen sulfate HSO4

-, can dissociate again but only slightly, which means that the hydrogen sulfate ion is a weak acid.

H2SO4(aq) + H2O(l) h H3O+(aq) + HSO4 -(aq)

HSO4 -(aq) + H2O(l) vh H3O

+(aq) + SO4 2-(aq)

Bisulfate ion (hydrogen sulfate)

Sulfate ion

M14_TIMB8119_06_SE_C14.indd 439 11/27/18 12:18 PM

440 CHAPTER 14 Acids and Bases

In summary, a strong acid such as HI in water dissociates completely to form an aque- ous solution of the ions H3O

+ and I-. A weak acid such as HF dissociates only slightly in water to form an aqueous solution that consists mostly of HF molecules and a few H3O

+ and F - ions (see FIGURE 14.2).

Strong acid: HI(aq) + H2O(l) h H3O+(aq) + I-(aq) Completely dissociated

Weak acid: HF(aq) + H2O(l) vh H3O+(aq) + F -(aq) Slightly dissociated

Bases in household products are used to remove grease.

Bases in Household Products

Weak Bases

Window cleaner, ammonia, NH3 Bleach, NaOCl

Laundry detergent, Na2CO3, Na3PO4 Toothpaste and baking soda, NaHCO3 Baking powder, scouring powder, Na2CO3 Lime for lawns and agriculture, CaCO3 Laxatives, antacids, Mg(OH)2, Al(OH)3

Strong Bases Drain cleaner, oven cleaner, NaOH

Strong and Weak Bases As strong electrolytes, strong bases dissociate completely in water. Because these strong bases are ionic compounds, they dissociate in water to give an aqueous solution of metal ions and hydroxide ions. The Group 1A (1) hydroxides are very soluble in water, which can give high concentrations of OH- ions. A few strong bases are less soluble in water, but what does dissolve dissociates completely as ions. For example, when KOH forms a KOH solution, it contains only the ions K+ and OH-.

H2O KOH(s) K+(aq) + OH-(aq)

Strong Bases

Lithium hydroxide LiOH

Sodium hydroxide NaOH

Potassium hydroxide KOH

Rubidium hydroxide RbOH

Cesium hydroxide CsOH

Calcium hydroxide Ca(OH)2*

Strontium hydroxide Sr(OH)2*

Barium hydroxide Ba(OH)2*

*Low solubility, but dissociates completely

Weak bases are weak electrolytes that are poor acceptors of hydrogen ions and pro- duce very few ions in solution. A typical weak base, ammonia, NH3, is found in window cleaners. In an aqueous solution, only a few ammonia molecules accept hydrogen ions to form NH4

+ and OH-.

NH3(g) + H2O(l) vh NH4 +(aq) + OH-(aq) Ammonium hydroxideAmmonia

FIGURE 14.2 Strong and weak acids in water

HI H3O + I -

H2O

Strong acid

HF HF

H3O +

I -

I -

I -

I - H3O +

H3O +

H3O +

H3O +

H3O +

HF

HF

HF

HF

HF

HF

H2O

Weak acid

F -

F -

C on

ce nt

ra ti

on (

m ol

/L )

At equilibrium, the strong acid HI has high concentrations of H3O

+ and I-.

At equilibrium, the weak acid HF has a high concentration of HF and low concentrations of H3O

+ and F-.

M14_TIMB8119_06_SE_C14.indd 440 11/27/18 12:18 PM

14.3 Strengths of Acids and Bases 441

PRACTICE PROBLEMS

14.3 Strengths of Acids and Bases

14.17 What is meant by the phrase “A strong acid has a weak conju- gate base”?

14.18 What is meant by the phrase “A weak acid has a strong conju- gate base”?

14.19 Identify the stronger acid in each of the following pairs: a. HBr or HNO2 b. H3PO4 or HSO4

- c. HCN or H2CO3

14.20 Identify the stronger acid in each of the following pairs: a. NH4

+ or H3O + b. H2SO4 or HCl c. H2O or H2CO3

14.21 Identify the weaker acid in each of the following pairs: a. HCl or HSO4

- b. HNO2 or HF c. HCO3 - or NH4

+

14.22 Identify the weaker acid in each of the following pairs: a. HNO3 or HCO3

- b. HSO4 - or H2O c. H2SO4 or H2CO3

Direction of Reaction There is a relationship between the components in each conjugate acid–base pair. Strong acids have weak conjugate bases that do not readily accept H+. As the strength of the acid decreases, the strength of its conjugate base increases.

In any acid–base reaction, there are two acids and two bases. However, one acid is stronger than the other acid, and one base is stronger than the other base. By comparing their relative strengths, we can determine the direction of the reaction. For example, the strong acid H2SO4 readily gives up H

+ to water. The hydronium ion H3O + produced is a weaker

acid than H2SO4, and the conjugate base HSO4 - is a weaker base than water.

H2SO4(aq) + H2O(l) h H3O+(aq) + HSO4 -(aq) Mostly products

Let’s look at another reaction in which water donates one H+ to carbonate, CO3 2-, to

form HCO3 - and OH-. From TABLE 14.3, we see that HCO3

- is a stronger acid than H2O. We also see that OH- is a stronger base than CO3

2-. To reach equilibrium, the stronger acid and stronger base react in the direction of the weaker acid and weaker base.

CO3 2-(aq) + H2O(l) v HCO3 -(aq) + OH-(aq) Mostly reactants

Stronger acid

Stronger base

Weaker acid

Weaker base

Weaker base

Weaker acid

Stronger acid

Stronger base

H3O +

HF

F -

Hydrofluoric acid is the only halogen acid that is a weak acid.

SAMPLE PROBLEM 14.4 Direction of Reaction

TRY IT FIRST

Does the mixture of the following reaction contain mostly reactants or products?

HF(aq) + H2O(l) vh H3O+(aq) + F -(aq)

SOLUTION

From TABLE 14.3, we see that HF is a weaker acid than H3O + and that H2O is a weaker

base than F -. Thus, the mixture contains mostly reactants.

HF(aq) + H2O(l) vh H3O +(aq) + F -(aq)

SELF TEST 14.4

Does the mixture from the reaction of nitric acid and water contain mostly reactants or products?

ANSWER

HNO3(aq) + H2O(l) h H3O+(aq) + NO3 -(aq)

The mixture contains mostly products because HNO3 is a stronger acid than H3O +, and

H2O is a stronger base than NO3 -.

Weaker acid

Weaker base

Stronger acid

Stronger base

PRACTICE PROBLEMS Try Practice Problems 14.23 to 14.26

M14_TIMB8119_06_SE_C14.indd 441 11/27/18 12:18 PM

442 CHAPTER 14 Acids and Bases

14.4 Dissociation of Weak Acids and Bases LEARNING GOAL Write the expression for the dissociation of a weak acid or weak base.

As we have seen, acids have different strengths depending on how much they dissociate in water. Because the dissociation of strong acids in water is essentially complete, the reaction is not considered to be an equilibrium situation. However, because weak acids in water dissociate only slightly, the ion products reach equilibrium with the undissociated weak acid molecules. For example, formic acid HCHO2, the acid found in bee and ant stings, is a weak acid. Formic acid is a weak acid that dissociates in water to form hydronium ion, H3O

+, and formate ion, CHO2 -.

REVIEW Writing the Equilibrium

Expression (13.3)

Using Le Châtelier’s Principle (13.5)

14.23 Predict whether each of the following reactions contains mostly reactants or products at equilibrium:

a. H2CO3(aq) + H2O(l) vh HCO3 -(aq) + H3O+(aq)

b. NH4 +(aq) + H2O(l) vh NH3(aq) + H3O+(aq)

c. HNO2(aq) + NH3(aq) vh NO2 -(aq) + NH4 +(aq)

14.24 Predict whether each of the following reactions contains mostly reactants or products at equilibrium:

a. H3PO4(aq) + H2O(l) vh H3O +(aq) + H2PO4 -(aq)

b. CO3 2-(aq) + H2O(l) vh OH

-(aq) + HCO3 -(aq) c. HS-(aq) + F -(aq) vh HF(aq) + S

2-(aq)

14.25 Write an equation for the acid–base reaction between ammo- nium ion and sulfate ion. Why does the equilibrium mixture contain mostly reactants?

14.26 Write an equation for the acid–base reaction between nitrous acid and hydroxide ion. Why does the equilibrium mixture con- tain mostly products?

Formic acid, a weak acid, loses one H+ to form formate ion.HCHO2 CHO2

-

HCHO2(aq) H2O(l) H3O +(aq) CHO2

-(aq)+ +

Writing Dissociation Expressions An acid dissociation expression can be written for weak acids that gives the ratio of the concentrations of products to the reactants. As with other equilibrium expressions, the molar concentration of the products is divided by the molar concentration of the reactants. (Recall that the brackets in an equilibrium expression represent the molar concentrations of the reac- tants and products.) Because water is a pure liquid with a constant concentration, it is omitted.

Ka = [H3O

+][CHO2 -]

[HCHO2] Acid dissociation expression

The numerical value of the acid dissociation expression is the acid dissociation constant, Ka. The value of the Ka for formic acid at 25 °C is determined by experiment to be 1.8 * 10-4. Thus, for the weak acid HCHO2, the Ka is written

Ka = [H3O

+][CHO2 -]

[HCHO2] = 1.8 * 10-4 Acid dissociation constant

The Ka for formic acid is small, which means that the equilibrium mixture of formic acid in water contains mostly reactants and only small amounts of the products. Weak acids have small Ka values. However, strong acids, which are essentially 100% dissociated, have very large Ka values, but these values are not usually given. TABLE 14.4 gives Ka and Kb values for selected weak acids and bases.

Similar to what we saw for weak acids, weak bases have small Kb values. Strong bases, which are essentially 100% dissociated, have very large Kb values, but these values are not usually given.

Let us now consider the dissociation of the weak base methylamine:

CH3 ¬ NH2(aq) + H2O(l) vh CH3 ¬ NH3 +(aq) + OH-(aq)

M14_TIMB8119_06_SE_C14.indd 442 11/27/18 12:18 PM

14.4 Dissociation of Weak Acids and Bases 443

As we did with the acid dissociation expression, the concentration of water is omitted from the base dissociation expression. The base dissociation constant, Kb, for methylamine is written

Kb = [CH3 ¬ NH3 +][OH-]

[CH3 ¬ NH2] = 4.4 * 10-4

TABLE 14.4 gives Ka and Kb values for selected weak acids and bases.

Acids Ka

Phosphoric acid H3PO4 7.5 * 10-3

Nitrous acid HNO2 4.5 * 10-4

Hydrofluoric acid HF 3.5 * 10-4

Formic acid HCHO2 1.8 * 10-4

Acetic acid HC2H3O2 1.8 * 10-5

Carbonic acid H2CO3 4.3 * 10-7

Hydrosulfuric acid H2S 9.1 * 10-8

Dihydrogen phosphate H2PO4 - 6.2 * 10-8

Hypochlorous acid HClO 3.0 * 10-8

Hydrocyanic acid HCN 4.9 * 10-10

Hydrogen carbonate HCO3 - 5.6 * 10-11

Hydrogen phosphate HPO4 2- 2.2 * 10-13

Bases Kb

Methylamine CH3 ¬ NH2 4.4 * 10-4

Carbonate CO3 2- 2.2 * 10-4

Ammonia NH3 1.8 * 10-5

TABLE 14.4 Ka and Kb Values for Selected Weak Acids and Bases

TABLE 14.5 summarizes the characteristics of acids and bases in terms of strength and equilibrium position.

Characteristic Strong Acids Weak Acids

Equilibrium Position Toward products Toward reactants

Ka Large Small

[H3O +] and [A-] 100% of [HA] dissociates Small percent of [HA] dissociates

Conjugate Base Weak Strong

Characteristic Strong Bases Weak Bases

Equilibrium Position Toward products Toward reactants

Kb Large Small

[BH+] and [OH-] 100% of [B] reacts Small percent of [B] reacts

Conjugate Acid Weak Strong

TABLE 14.5 Characteristics of Acids and Bases

SAMPLE PROBLEM 14.5 Writing an Acid Dissociation Expression

TRY IT FIRST

Write the acid dissociation expression for the weak acid, nitrous acid.

SOLUTION

The equation for the dissociation of nitrous acid is written

HNO2(aq) + H2O(l) vh H3O +(aq) + NO2 -(aq)

M14_TIMB8119_06_SE_C14.indd 443 11/27/18 12:18 PM

444 CHAPTER 14 Acids and Bases

The acid dissociation expression is written as the concentration of the products divided by the concentration of the undissociated weak acid.

Ka = [H3O

+][NO2 -]

[HNO2]

SELF TEST 14.5

Write the acid dissociation expression for each of the following:

a. hydrogen phosphate b. hypochlorous acid

ANSWER

a. Ka = [H3O

+][PO4 3-]

[HPO4 2-]

b. Ka = [H3O

+][ClO-]

[HClO] PRACTICE PROBLEMS

Try Practice Problems 14.27 to 14.32

PRACTICE PROBLEMS

14.4 Dissociation of Weak Acids and Bases

14.27 Answer True or False for each of the following: A strong acid a. is completely dissociated in aqueous solution b. has a small value of Ka c. has a strong conjugate base d. has a weak conjugate base e. is slightly dissociated in aqueous solution

14.28 Answer True or False for each of the following: A weak acid a. is completely dissociated in aqueous solution b. has a small value of Ka c. has a strong conjugate base d. has a weak conjugate base e. is slightly dissociated in aqueous solution

14.29 Consider the following acids and their dissociation constants:

H2SO3(aq) + H2O(l) vh H3O +(aq) + HSO3 -(aq)

Ka = 1.2 * 10-2

HS-(aq) + H2O(l) vh H3O +(aq) + S2-(aq)

Ka = 1.3 * 10-19

a. Which is the stronger acid, H2SO3 or HS -?

b. What is the conjugate base of H2SO3? c. Which acid has the weaker conjugate base?

d. Which acid has the stronger conjugate base? e. Which acid produces more ions?

14.30 Consider the following acids and their dissociation constants:

HPO4 2-(aq) + H2O(l) vh H3O+(aq) + PO4 3-(aq)

Ka = 2.2 * 10-13

HCHO2(aq) + H2O(l) vh H3O +(aq) + CHO2 -(aq)

Ka = 1.8 * 10-4

a. Which is the weaker acid, HPO4 2- or HCHO2?

b. What is the conjugate base of HPO4 2-?

c. Which acid has the weaker conjugate base? d. Which acid has the stronger conjugate base? e. Which acid produces more ions?

14.31 Phosphoric acid dissociates to form hydronium ion and dihydro- gen phosphate. Phosphoric acid has a Ka of 7.5 * 10-3. Write the equation for the reaction and the acid dissociation expres- sion for phosphoric acid.

14.32 Aniline, C6H5 ¬ NH2, a weak base with a Kb of 4.0 * 10-10, reacts with water to form C6H5 ¬ NH3 + and hydroxide ion. Write the equation for the reaction and the base dissociation expression for aniline.

14.5 Dissociation of Water LEARNING GOAL Use the water dissociation expression to calculate the [H3O

+] and [OH-] in an aqueous solution.

In many acid–base reactions, water is amphoteric, which means that it can act either as an acid or as a base. In pure water, there is a forward reaction between two water molecules that transfers H+ from one water molecule to the other. One molecule acts as an acid by losing H+, and the water molecule that gains H+ acts as a base. Every time H+ is transferred between two water molecules, the products are one H3O

+ and one OH-, which react in the reverse direction to re-form two water molecules. Thus, equilibrium is reached between the conjugate acid–base pairs of water molecules.

ENGAGE 14.6 Why is the [H3O

+] equal to the [OH-] in pure water?

M14_TIMB8119_06_SE_C14.indd 444 11/27/18 12:18 PM

14.5 Dissociation of Water 445

+ O

Conjugate acid–base pair

Conjugate acid–base pair

Acid (H+ donor)

Base (H+ acceptor)

Base (H+ acceptor)

Acid (H+ donor)

+O H H

H

O H H+

H

O

H

H -

Writing the Water Dissociation Expression Using the equation for the dissociation of water, we can write the equilibrium expres- sion that shows the concentrations of the products divided by the concentrations of the reactants.

H2O(l) + H2O(l) vh H3O +(aq) + OH-(aq)

K = [H3O

+][OH-]

[H2O][H2O]

By omitting the constant concentration of pure water, we can write the water dissociation expression.

Kw = [H3O +][OH-]

Experiments have determined that, in pure water, the concentration of H3O + and OH- at

25 °C are each 1.0 * 10-7 M.

Pure water [H3O +] = [OH-] = 1.0 * 10-7 M

When we place the [H3O +] and [OH-] into the water dissociation expression, we obtain the

numerical value of the water dissociation constant, Kw, which is 1.0 * 10-14 at 25 °C. As before, the concentration units are omitted in the Kw value.

Kw = [H3O +][OH-]

= [1.0 * 10-7][1.0 * 10-7] = 1.0 * 10-14

Neutral, Acidic, and Basic Solutions The Kw value (1.0 * 10-14) applies to any aqueous solution at 25 °C because all aqueous solutions contain both H3O

+ and OH- (see FIGURE 14.3). When the [H3O +] and [OH-] in

a solution are equal, the solution is neutral. However, most solutions are not neutral; they have different concentrations of H3O

+ and OH-. If acid is added to water, there is an increase

PRACTICE PROBLEMS Try Practice Problems 14.33 and 14.34

FIGURE 14.3 In a neutral solution, [H3O

+] and [OH-] are equal. In acidic solutions, the [H3O

+] is greater than the [OH-]. In basic solutions, the [OH-] is greater than the [H3O

+].Acidic solution Neutral solution Basic solution

C on

ce nt

ra ti

on (

m ol

/L )

10-14

10-7

100

[H3O +] > [OH-] [H3O

+] = [OH-] [H3O +] < [OH-]

H3O +

OH-

H3O + OH-

OH-

H3O +

M14_TIMB8119_06_SE_C14.indd 445 11/27/18 12:18 PM

446 CHAPTER 14 Acids and Bases

in [H3O +] and a decrease in [OH-], which makes an acidic solution. If base is added, [OH-]

increases and [H3O +] decreases, which gives a basic solution. However, for any aqueous

solution, whether it is neutral, acidic, or basic, the product [H3O +][OH-] is equal to Kw

(1.0 * 10-14 at 25 °C) (see TABLE 14.6).

ENGAGE 14.7 Is a solution that has a [H3O

+] of 1.0 * 10-3 M acidic, basic, or neutral?

PRACTICE PROBLEMS Try Practice Problems 14.35 and 14.36

Type of Solution [H3O +] [OH−] Kw (25 °C)

Neutral 1.0 * 10-7 M 1.0 * 10-7 M 1.0 * 10-14

Acidic 1.0 * 10-2 M 1.0 * 10-12 M 1.0 * 10-14

Acidic 2.5 * 10-5 M 4.0 * 10-10 M 1.0 * 10-14

Basic 1.0 * 10-8 M 1.0 * 10-6 M 1.0 * 10-14

Basic 5.0 * 10-11 M 2.0 * 10-4 M 1.0 * 10-14

TABLE 14.6 Examples of [H3O +] and [OH−] in Neutral, Acidic, and Basic

Solutions

Using the Kw to Calculate [H3O +] and [OH−] in a Solution

If we know the [H3O +] of a solution, we can use the Kw to calculate [OH

-]. If we know the [OH-] of a solution, we can calculate [H3O

+] from their relationship in the Kw, as shown in Sample Problem 14.6.

Kw = [H3O +][OH-]

[OH-] = Kw

[H3O +] [H3O

+] = Kw

[OH-]

ENGAGE 14.9 Why does the [H3O

+] of an aqueous solution increase if the [OH-] decreases?

SAMPLE PROBLEM 14.6 Calculating the [H3O +] of a Solution

TRY IT FIRST

A vinegar solution has a [OH-] = 5.0 * 10-12 M at 25 °C. What is the [H3O+] of the vinegar solution? Is the solution acidic, basic, or neutral?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

[OH-] = 5.0 * 10-12 M [H3O+] Kw = [H3O+][OH-]

STEP 2 Write the Kw for water and solve for the unknown [H3O +].

Kw = [H3O +][OH-] = 1.0 * 10-14

Solve for [H3O +] by dividing both sides by [OH-].

Kw

[OH-] =

[H3O +][OH-]

[OH-]

[H3O +] =

1.0 * 10-14

[OH-]

STEP 3 Substitute the known [OH−] into the equation and calculate.

[H3O +] =

1.0 * 10-14

[5.0 * 10-12] = 2.0 * 10-3 M

Because the [H3O +] of 2.0 * 10-3 M is larger than the [OH-] of 5.0 * 10-12 M,

the solution is acidic.

ENGAGE 14.8 If you know the [H3O

+] of a solution, how do you use the Kw to calculate the [OH-]?

CORE CHEMISTRY SKILL Calculating [H3O

+] and [OH-] in Solutions

M14_TIMB8119_06_SE_C14.indd 446 11/27/18 12:18 PM

14.6 The pH Scale 447

SELF TEST 14.6

a. What is the [H3O +] of an ammonia cleaning solution with [OH-] = 4.0 * 10-4 M? Is

the solution acidic, basic, or neutral? b. The [H3O

+] of tomato juice is 6.3 * 10-5 M. What is the [OH-] of the juice? Is the tomato juice acidic, basic, or neutral?

ANSWER

a. [H3O +] = 2.5 * 10-11 M, basic b. [OH-] = 1.6 * 10-10 M, acidic

PRACTICE PROBLEMS

14.5 Dissociation of Water

14.33 Why are the concentrations of H3O + and OH- equal in pure water?

14.34 What is the meaning and value of Kw at 25 °C?

14.35 In an acidic solution, how does the concentration of H3O +

compare to the concentration of OH-?

14.36 If a base is added to pure water, why does the [H3O +] decrease?

14.37 Indicate whether each of the following solutions is acidic, basic, or neutral:

a. [H3O +] = 2.0 * 10-5 M

b. [H3O +] = 1.4 * 10-9 M

c. [OH-] = 8.0 * 10-3 M d. [OH-] = 3.5 * 10-10 M 14.38 Indicate whether each of the following solutions is acidic, basic,

or neutral: a. [H3O

+] = 6.0 * 10-12 M b. [H3O

+] = 1.4 * 10-4 M c. [OH-] = 5.0 * 10-12 M d. [OH-] = 4.5 * 10-2 M 14.39 Calculate the [H3O

+] of each aqueous solution with the following [OH-]:

a. coffee, 1.0 * 10-9 M b. soap, 1.0 * 10-6 M c. cleanser, 2.0 * 10-5 M d. lemon juice, 4.0 * 10-13 M

14.40 Calculate the [H3O +] of each aqueous solution with the

following [OH-]: a. NaOH solution, 1.0 * 10-2 M b. milk of magnesia, 1.0 * 10-5 M c. aspirin, 1.8 * 10-11 M d. seawater, 2.5 * 10-6 M

Applications

14.41 Calculate the [OH-] of each aqueous solution with the following [H3O

+]: a. stomach acid, 4.0 * 10-2 M b. urine, 5.0 * 10-6 M c. orange juice, 2.0 * 10-4 M d. bile, 7.9 * 10-9 M 14.42 Calculate the [OH-] of each aqueous solution with the

following [H3O +]:

a. baking soda, 1.0 * 10-8 M b. blood, 4.2 * 10-8 M c. milk, 5.0 * 10-7 M d. pancreatic juice, 4.0 * 10-9 M

14.6 The pH Scale LEARNING GOAL Calculate pH from [H3O

+]; given the pH, calculate the [H3O +] and

[OH-] of a solution.

In the environment, the acidity, or pH, of rain can have significant effects. When rain becomes too acidic, it can dissolve marble statues and accelerate the corrosion of metals. In lakes and ponds, the acidity of water can affect the ability of plants and fish to survive. The acidity of soil around plants affects their growth. If the soil pH is too acidic or too basic, the roots of the plant cannot take up some nutrients. Most plants thrive in soil with a nearly neutral pH, although certain plants, such as orchids, camellias, and blueberries, require a more acidic soil.

Although we have expressed H3O + and OH- as molar concentrations, it is more conve-

nient to describe the acidity of solutions using the pH scale. On this scale, a number between 0 and 14 represents the H3O

+ concentration for common solutions. A neutral solution has a pH of 7.0 at 25 °C. An acidic solution has a pH less than 7.0; a basic solution has a pH greater than 7.0 (see FIGURE 14.4).

When we relate acidity and pH, we are using an inverse relationship, which is when one component increases while the other component decreases. When an acid is added to pure water, the [H3O

+] (acidity) of the solution increases but its pH decreases. When a base is added to pure water, it becomes more basic, which means its acidity decreases and the pH increases.

PRACTICE PROBLEMS Try Practice Problems 14.37 to 14.42

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448 CHAPTER 14 Acids and Bases

FIGURE 14.4 On the pH scale, values below 7.0 are acidic, a value of 7.0 is neutral, and values above 7.0 are basic.

Potato 5.8

1 M HCl solution 0.0

Gastric juice 1.6

Lemon juice 2.2

Vinegar 2.8

Orange 3.5

Tomato 4.2

Coffee 5.0

Milk 6.4

Water (pure) 7.0

Seawater 8.5

Bleach 12.0

Urine 6.0

Milk of magnesia 10.5

1 M NaOH solution (lye) 14.0

Blood plasma 7.4

Carbonated beverage 3.0

Apple juice 3.8

Bread 5.5

Drinking water 7.2

Ammonia 11.0

Acidic

pH Value

Neutral

Basic

Detergents 8.0–9.0

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

Bile 8.1

In the laboratory, a pH meter is commonly used to determine the pH of a solution. There are also various indicators and pH papers that turn specific colors when placed in solutions of different pH values. The pH is found by comparing the color on the test paper or the color of the solution to a color chart (see FIGURE 14.5).

FIGURE 14.5 The pH of a solution can be determined using different methods. pH meter pH paper pH indicator

ENGAGE 14.10 Is apple juice an acidic, a basic, or a neutral solution?

M14_TIMB8119_06_SE_C14.indd 448 11/27/18 12:18 PM

14.6 The pH Scale 449

SAMPLE PROBLEM 14.7 pH of Solutions

TRY IT FIRST

Consider the pH of the following body fluids:

Body Fluid pH

Stomach acid 1.4

Pancreatic juice 8.4

Sweat 4.8

Urine 5.3

Cerebrospinal fluid 7.3

a. Place the pH values of the body fluids on the list in order of most acidic to most basic. b. Which body fluid has the highest [H3O

+]?

SOLUTION

a. The most acidic body fluid is the one with the lowest pH, and the most basic is the body fluid with the highest pH: stomach acid (1.4), sweat (4.8), urine (5.3), cerebrospinal fluid (7.3), pancreatic juice (8.4).

b. The body f luid with the highest [H3O +] would have the lowest pH value, which is

stomach acid.

SELF TEST 14.7

a. Which body fluid has the highest [OH-]? b. Which body fluid is more acidic, urine or cerebrospinal fluid?

ANSWER

a. pancreatic juice b. urine

Calculating the pH of Solutions The pH scale is a logarithmic scale that corresponds to the [H3O

+] of aqueous solutions. Mathematically, pH is the negative logarithm (base 10) of the [H3O

+].

pH = - log[H3O+]

Essentially, the negative powers of 10 in the molar concentrations are converted to positive numbers. For example, a lemon juice solution with [H3O

+] = 1.0 * 10-2 M has a pH of 2.00. This can be calculated using the pH equation:

pH = - log[1.0 * 10-2] pH = - ( - 2.00)

= 2.00

The number of decimal places in the pH value is the same as the number of significant figures in the [H3O

+]. The number to the left of the decimal point in the pH value is the power of 10.

1.0 00

Two SFs Two SFs

[H3O +] = * 10-2 pH = 2.

Because pH is a log scale, a change of one pH unit corresponds to a tenfold change in [H3O

+]. It is important to recall that the pH decreases as the [H3O +] increases. For example,

a solution with a pH of 2.00 has a [H3O +] that is ten times greater than a solution with a pH

of 3.00 and 100 times greater than a solution with a pH of 4.00. The pH of a solution is calculated from the [H3O

+] by using the log key and changing the sign as shown in Sample Problem 14.8.

ENGAGE 14.11 If a pH meter reads 4.00, why is the solution acidic?

PRACTICE PROBLEMS Try Practice Problems 14.43 to 14.46

KEY MATH SKILL Calculating pH from [H3O

+]

ENGAGE 14.12 Explain why 6.00, but not 6.0, is the correct pH for [H3O

+] = 1.0 * 10-6 M.

A dipstick is used to measure the pH of a urine sample.

If soil is too acidic, nutrients are not absorbed by crops. Then lime (CaCO3), which acts as a base, may be added to increase the soil pH.

M14_TIMB8119_06_SE_C14.indd 449 11/27/18 12:18 PM

450 CHAPTER 14 Acids and Bases

SAMPLE PROBLEM 14.8 Calculating pH from [H3O +]

TRY IT FIRST

Aspirin, which is acetylsalicylic acid, was the first nonsteroidal anti-inf lammatory drug (NSAID) used to alleviate pain and fever. If a solution of aspirin has a [H3O

+] = 1.7 * 10-3 M, what is the pH of the solution?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

[H3O +] = 1.7 * 10-3 M pH pH equation

STEP 2 Enter the [H3O + ] into the pH equation and calculate.

pH = -log[H3O+] = -log[1.7 * 10-3]

EE or EXPlog 1.7 3EE or EXP log1.7 or3 == +/-+/-+/- +/-

Calculator DisplayCalculator Procedure

Be sure to check the instructions for your calculator.

STEP 3 Adjust the number of SFs on the right of the decimal point. In a pH value, the number to the left of the decimal point is an exact number derived from the power of 10. Thus, the two SFs in the coefficient determine that there are two SFs after the decimal point in the pH value.

Two SFs Two SFsExact Exact

pH = -log[1.7 * 10-3] = 2.771.7 Coefficient

10-3 M* Power of ten

SELF TEST 14.8

a. What is the pH of bleach with [H3O +] = 4.2 * 10-12 M?

b. What is the pH of a borax solution with [H3O +] = 6.1 * 10-10 M?

ANSWER

a. pH = 11.38 b. pH = 9.21

PRACTICE PROBLEMS Try Practice Problems 14.47 and 14.48

Acidic H that dissociates in aqueous solution

Aspirin, acetylsalicylic acid, is a weak acid.

M14_TIMB8119_06_SE_C14.indd 450 11/27/18 12:18 PM

14.6 The pH Scale 451

pOH The pOH scale is similar to the pH scale except that pOH is associated with the [OH-] of an aqueous solution.

pOH = - log[OH-]

Solutions with high [OH-] have low pOH values; solutions with low [OH-] have high pOH values. In any aqueous solution, the sum of the pH and pOH is equal to 14.00, which is the negative logarithm of the Kw.

pH + pOH = 14.00

For example, if the pH of a solution is 3.50, the pOH can be calculated as follows:

pH + pOH = 14.00 pOH = 14.00 - pH = 14.00 - 3.50 = 10.50

A comparison of [H3O +], [OH-], and their corresponding pH and pOH values is given

in TABLE 14.7.

pH [H3O +] [OH−] pOH

0 100 10-14 14

Acidic

Basic

Neutral

1 10-1 10-13 13

2 10-2 10-12 12

3 10-3 10-11 11

4 10-4 10-10 10

5 10-5 10-9 9

6 10-6 10-8 8

7 10-7 10-7 7

8 10-8 10-6 6

9 10-9 10-5 5

10 10-10 10-4 4

11 10-11 10-3 3

12 10-12 10-2 2

13 10-13 10-1 1

14 10-14 100 0

TABLE 14.7 A Comparison of pH and pOH Values at 25 °C, [H3O

+], and [OH−]

Acids produce the sour taste of the fruits we eat.

pH = 3.4

pH = 2.4 pH = 2.0

pH = 3.9

When we need to calculate the pH from [OH-], we use the Kw to calculate [H3O +], place

it in the pH equation, and calculate the pH of the solution as shown in Sample Problem 14.9.

SAMPLE PROBLEM 14.9 Calculating pH from [OH−]

TRY IT FIRST

What is the pH of an ammonia solution with [OH-] = 3.7 * 10-3 M?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

[OH-] = 3.7 * 10-3 M [H3O+], pH Kw = [H3O+][OH-], pH equation

M14_TIMB8119_06_SE_C14.indd 451 11/27/18 12:18 PM

452 CHAPTER 14 Acids and Bases

Calculating [H3O +] from pH

If we are given the pH of the solution and asked to determine the [H3O +], we need to reverse

the calculation of pH.

[H3O +] = 10-pH

For example, if the pH of a solution is 3.0, we can substitute it into this equation. The num- ber of significant figures in [H3O

+] is equal to the number of decimal places in the pH value.

[H3O +] = 10-pH = 10-3.0 = 1 * 10-3 M

For pH values that are not whole numbers, the calculation requires the use of the 10x key, which is usually a 2nd function key. On some calculators, this operation is done using the inverse log equation as shown in Sample Problem 14.10.

PRACTICE PROBLEMS Try Practice Problems 14.49 and 14.50

KEY MATH SKILL Calculating [H3O

+] from pH

SAMPLE PROBLEM 14.10 Calculating [H3O +] from pH

TRY IT FIRST

Calculate [H3O +] for a urine sample, which has a pH of 7.5.

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

pH = 7.5 [H3O +] [H3O

+] = 10-pH

STEP 2 Enter the pH value into the inverse log equation and calculate.

[H3O +] = 10-pH = 10-7.5

log 7.5 = 7.5+/-2nd log2ndor =+/-

Calculator DisplayCalculator Procedure

STEP 2 Enter the [H3O +] into the pH equation and calculate. Because [OH-] is

given for the ammonia solution, we have to calculate [H3O +]. Using the water disso-

ciation expression, we divide both sides by [OH-] to obtain [H3O +].

Kw = [H3O +][OH-] = 1.0 * 10-14

Kw

[OH-] =

[H3O +][OH-]

[OH-]

[H3O +] =

1.0 * 10-14

[3.7 * 10-3] = 2.7 * 10-12 M

Now, we enter the [H3O +] into the pH equation.

pH = -log[H3O+] = -log[2.7 * 10-12]

EE or EXPlog 2.7 12EE or EXP log2.7 or12 == +/-+/-+/- +/-

Calculator DisplayCalculator Procedure

STEP 3 Adjust the number of SFs on the right of the decimal point.

2.7 * 10-12 M pH = 11.57 Two SFs Two SFs to the right of the decimal point

SELF TEST 14.9

a. Calculate the pH of a sample of bile that has [OH-] = 1.3 * 10-6 M. b. What is the pH of egg whites with [OH-] = 1.2 * 10-5 M?

ANSWER

a. pH = 8.11 b. pH = 9.08

M14_TIMB8119_06_SE_C14.indd 452 11/27/18 12:18 PM

14.6 The pH Scale 453

STEP 3 Adjust the SFs for the coefficient. Because the pH value 7.5 has one digit to the right of the decimal point, the coefficient for [H3O

+] is written with one SF.

[H3O +] = 3 * 10-8 M

SELF TEST 14.10

a. What are the [H3O +] and [OH-] of cola that has a pH of 3.17?

b. What are the [H3O +] and [OH-] of pancreatic juice that has a pH of 8.3?

ANSWER

a. [H3O +] = 6.8 * 10-4 M, [OH-] = 1.5 * 10-11 M

b. [H3O +] = 5 * 10-9 M, [OH] = 2 * 10-6 M

One SF

PRACTICE PROBLEMS Try Practice Problems 14.51 to 14.54

The pH of a liquid is measured using a pH meter.

Chemistry Link to Health Stomach Acid, HCl

Gastric acid, which contains HCl, is produced by parietal cells that line the stomach. When the stomach expands with the intake of food, the gastric glands begin to secrete a strongly acidic solution of HCl. In a single day, a person may secrete 2000 mL of gastric juice, which contains hydrochloric acid, mucins, and the enzymes pep- sin and lipase.

The HCl in the gastric juice activates a diges- tive enzyme from the chief cells called pepsinogen to form pepsin, which breaks down proteins in food entering the stomach. The secretion of HCl continues until the stomach has a pH of about 2, which is the optimum for activating the digestive enzymes without ulcerating the stomach lining. In addition, the low pH destroys bacteria that reach the stomach. Normally, large quantities of viscous mucus are secreted within the stomach to protect its lining from acid and enzyme damage. Gastric acid may also form under conditions of stress when the nervous system activates the production of HCl. As the contents of the stomach move into the small intestine, cells produce bicarbonate that neutralizes the gastric acid until the pH is about 5.

Duodenum

Esophagus

Gastric gland

Parietal cell

Stomach lining

Mucous cell

Chief cell

Parietal cells in the lining of the stomach secrete gastric acid HCl.

PRACTICE PROBLEMS

14.6 The pH Scale

14.43 Why does a neutral solution have a pH of 7.0?

14.44 If you know the [OH-], how can you determine the pH of a solution?

14.45 State whether each of the following solutions is acidic, basic, or neutral:

a. blood plasma, pH 7.38 b. vinegar, pH 2.8 c. drain cleaner, pOH 2.8 d. coffee, pH 5.52 e. tomatoes, pH 4.2 f. chocolate cake, pH 7.6

14.46 State whether each of the following solutions is acidic, basic, or neutral:

a. soda, pH 3.22 b. shampoo, pOH 8.3 c. laundry detergent, pOH 4.56 d. rain, pH 5.8 e. honey, pH 3.9 f. cheese, pH 4.9

M14_TIMB8119_06_SE_C14.indd 453 11/27/18 12:18 PM

454 CHAPTER 14 Acids and Bases

14.47 A solution with a pH of 3 is 10 times more acidic than a solu- tion with pH 4. Explain.

14.48 A solution with a pH of 10 is 100 times more basic than a solu- tion with pH 8. Explain.

14.49 Calculate the pH of each solution given the following: a. [H3O

+] = 1 * 10-4 M b. [H3O+] = 3 * 10-9 M c. [OH-] = 1 * 10-5 M d. [OH-] = 2.5 * 10-11 M e. [H3O

+] = 6.7 * 10-8 M f. [OH-] = 8.2 * 10-4 M 14.50 Calculate the pOH of each solution given the following: a. [H3O

+] = 1 * 10-8 M b. [H3O+] = 5 * 10-6 M c. [OH-] = 1 * 10-2 M d. [OH-] = 8.0 * 10-3 M e. [H3O

+] = 4.7 * 10-2 M f. [OH-] = 3.9 * 10-6 M

Applications 14.51 Complete the following table:

[H3O +] [OH−] pH pOH

Acidic, Basic, or Neutral?

1.0 * 10-6 M 3.49

2.8 * 10-5 M 2.00

14.52 Complete the following table:

[H3O +] [OH−] pH pOH

Acidic, Basic, or Neutral?

10.00

Neutral

5.66

6.4 * 10-12 M

14.53 A patient with severe metabolic acidosis has a blood plasma pH of 6.92. What is the [H3O

+] of the blood plasma?

14.54 A patient with respiratory alkalosis has a blood plasma pH of 7.58. What is the [H3O

+] of the blood plasma?

Magnesium reacts rapidly with acid and forms H2 gas and a salt of magnesium.

14.7 Reactions of Acids and Bases LEARNING GOAL Write balanced equations for reactions of acids with metals, carbonates or bicarbonates, and bases.

Typical reactions of acids and bases include the reactions of acids with metals, carbonates or bicarbonates, and bases. For example, when you drop an antacid tablet in water, the bicarbonate ion and citric acid in the tablet react to produce carbon dioxide bubbles, water, and a salt. A salt is an ionic compound that does not have H+ as the cation or OH- as the anion.

Acids and Metals Acids react with certain metals to produce hydrogen gas (H2) and a salt. Active metals include potassium, sodium, calcium, magnesium, aluminum, zinc, iron, and tin. In these single replacement reactions, the metal ion replaces the hydrogen in the acid.

2HCl(aq) + Mg(s) h H2(g) + MgCl2(aq)

2HNO3(aq) + Zn(s) h H2(g) + Zn(NO3)2(aq)

Acids React with Carbonates or Bicarbonates When an acid is added to a carbonate or bicarbonate, the products are carbon dioxide gas, water, and a salt. The acid reacts with CO3

2- or HCO3 - to produce carbonic acid, H2CO3,

which breaks down rapidly to CO2 and H2O.

2HCl(aq) + Na2CO3(aq) h CO2(g) + H2O(l) + 2NaCl(aq)

HBr(aq) + NaHCO3(aq) h CO2(g) + H2O(l) + NaBr(aq)

REVIEW Balancing a Chemical Equation (8.2)

Acid Metal Hydrogen Salt

Acid Metal Hydrogen Salt

Acid Carbonate Carbon dioxide

Water Salt

Acid Bicarbonate Carbon dioxide

Water Salt

Sodium bicarbonate (baking soda) reacts with an acid (vinegar) to form carbon dioxide gas, water, and a salt.

M14_TIMB8119_06_SE_C14.indd 454 11/27/18 12:18 PM

14.7 Reactions of Acids and Bases 455

Acids and Hydroxides: Neutralization Neutralization is a reaction between a strong or weak acid with a strong base to produce a water and a salt. The H+ of the acid and the OH- of the base combine to form water. The salt is the combination of the cation from the base and the anion from the acid. We can write the following equation for the neutralization reaction between HCl and NaOH:

HCl(aq) + NaOH(aq) h H2O(l) + NaCl(aq)

If we write the strong acid HCl and the strong base NaOH as ions, we see that H+ combines with OH- to form water, leaving the ions Na+ and Cl- in solution.

H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) h H2O(l) + Na+(aq) + Cl-(aq)

When we omit the ions that do not change during the reaction (spectator ions), we obtain the net ionic equation.

H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) h H2O(l) + Na+(aq) + Cl-(aq)

The net ionic equation for the neutralization of H+ and OH- to form H2O is

H+(aq) + OH-(aq) h H2O(l) Net ionic equation

Balancing Neutralization Equations In a neutralization reaction, one H+ always reacts with one OH-. Therefore, a neutralization equation may need coefficients to balance the H+ from the acid with the OH- from the base as shown in Sample Problem 14.11.

Acid Base Water Salt

PRACTICE PROBLEMS Try Practice Problems 14.55 to 14.60

SAMPLE PROBLEM 14.11 Balancing Equations for Acids

TRY IT FIRST

Write the balanced equation for the neutralization of HCl(aq) and Ba(OH)2(s).

SOLUTION

STEP 1 Write the reactants and products.

HCl(aq) + Ba(OH)2(s) h H2O(l) + salt

STEP 2 Balance the H+ in the acid with the OH− in the base. Placing a coefficient of 2 in front of the HCl provides 2H+ for the 2OH- from Ba(OH)2.

2HCl(aq) + Ba(OH)2(s) h H2O(l) + salt

STEP 3 Balance the H2O with the H + and the OH−. Use a coefficient of 2 in

front of H2O to balance 2H + and 2OH-.

2HCl(aq) + Ba(OH)2(s) h 2H2O(l) + salt

STEP 4 Write the salt from the remaining ions. Use the ions Ba2+ and 2Cl- and write the formula for the salt as BaCl2.

2HCl(aq) + Ba(OH)2(s) h 2H2O(l) + BaCl2(aq)

SELF TEST 14.11

Write the balanced equation for the reaction between each of the following:

a. H2SO4(aq) and K2CO3(s) b. H3PO4(aq) and LiOH(aq)

ANSWER

a. H2SO4(aq) + K2CO3(s) h CO2(g) + H2O(l) + K2SO4(aq) b. H3PO4(aq) + 3LiOH(aq) h Li3PO4(aq) + 3H2O(l)

CORE CHEMISTRY SKILL Writing Equations for Reactions

of Acids and Bases

ENGAGE 14.13 What salt is produced when KOH completely neutralizes H2SO4?

M14_TIMB8119_06_SE_C14.indd 455 11/27/18 12:18 PM

456 CHAPTER 14 Acids and Bases

Chemistry Link to Health Antacids

Antacids are substances used to neutralize excess stomach acid (HCl). Some antacids are mixtures of aluminum hydroxide and mag- nesium hydroxide. These hydroxides are not very soluble in water, so the levels of available OH- are not damaging to the intestinal tract. However, aluminum hydroxide has the side effects of produc- ing constipation and binding phosphate in the intestinal tract, which may cause weakness and loss of appetite. Magnesium hydroxide has a laxative effect. These side effects are less likely when a combina- tion of the antacids is used.

3HCl(aq) + Al(OH)3(s) h 3H2O(l) + AlCl3(aq)

2HCl(aq) + Mg(OH)2(s) h 2H2O(l) + MgCl2(aq)

Some antacids use calcium carbonate to neutralize excess stom- ach acid. About 10% of the calcium is absorbed into the bloodstream, where it elevates the level of serum calcium. Calcium carbonate is not recommended for patients who have peptic ulcers or a tendency to form kidney stones, which typically consist of an insoluble cal- cium salt.

2HCl(aq) + CaCO3(s) h CO2(g) + H2O(l) + CaCl2(aq)

Still other antacids contain sodium bicarbonate. This type of ant- acid neutralizes excess gastric acid, increases blood pH, but also elevates sodium levels in the body fluids. It also is not recommended in the treatment of peptic ulcers.

HCl(aq) + NaHCO3(s) h CO2(g) + H2O(l) + NaCl(aq)

The neutralizing substances in some antacid preparations are given in TABLE 14.8.

Antacid Base(s)

Amphojel Al(OH)3 Milk of magnesia Mg(OH)2 Mylanta, Maalox, Di-Gel, Gelusil, Riopan

Mg(OH)2, Al(OH)3

Bisodol, Rolaids CaCO3, Mg(OH)2 Titralac, Tums, Pepto-Bismol CaCO3 Alka-Seltzer NaHCO3, KHCO3

TABLE 14.8 Basic Compounds in Some Antacids

Antacids neutralize excess stomach acid.

PRACTICE PROBLEMS

14.7 Reactions of Acids and Bases

14.55 Complete and balance the equation for each of the following reactions:

a. ZnCO3(s) + HBr(aq) h b. Zn(s) + HCl(aq) h c. HCl(aq) + NaHCO3(s) h d. H2SO4(aq) + Mg(OH)2(s) h 14.56 Complete and balance the equation for each of the following

reactions: a. KHCO3(s) + HBr(aq) h b. Ca(s) + H2SO4(aq) h c. H2SO4(aq) + Ca(OH)2(s) h d. Na2CO3(s) + H2SO4(aq) h 14.57 Balance each of the following neutralization reactions: a. HCl(aq) + Mg(OH)2(s) h H2O(l ) + MgCl2(aq) b. H3PO4(aq) + LiOH(aq) h H2O(l ) + Li3PO4(aq)

14.58 Balance each of the following neutralization reactions: a. HNO3(aq) + Ba(OH)2(s) h H2O(l ) + Ba(NO3)2(aq) b. H2SO4(aq) + Al(OH)3(s) h H2O(l ) + Al2(SO4)3(aq) 14.59 Write a balanced equation for the neutralization of each of the

following: a. H2SO4(aq) and NaOH(aq) b. HCl(aq) and Fe(OH)3(s) c. H2CO3(aq) and Mg(OH)2(s)

14.60 Write a balanced equation for the neutralization of each of the following:

a. H3PO4(aq) and NaOH(aq) b. HI(aq) and LiOH(aq) c. HNO3(aq) and Ca(OH)2(s)

M14_TIMB8119_06_SE_C14.indd 456 11/27/18 12:18 PM

14.8 Acid–Base Titration 457

14.8 Acid–Base Titration LEARNING GOAL Calculate the molarity or volume of an acid or base solution from titration information.

Suppose we need to find the molarity of a solution of HCl, which has an unknown concen- tration. We can do this by a laboratory procedure called titration in which we neutralize an acid sample with a known amount of base. In a titration, we place a measured volume of the acid in a flask and add a few drops of an indicator, such as phenolphthalein. An indica- tor is a compound that dramatically changes color when pH of the solution changes. In an acidic solution, phenolphthalein is colorless. Then we fill a buret with a NaOH solution of known molarity and carefully add NaOH solution to neutralize the acid in the flask (see FIGURE 14.6). We know that neutralization has taken place when the phenolphthalein in the solution changes from colorless to pink. This is called the neutralization endpoint. From the measured volume of the NaOH solution and its molarity, we calculate the number of moles of NaOH, the moles of acid, and use the measured volume of acid to calculate its concentration.

REVIEW Using Concentration as a

Conversion Factor (12.4)

ENGAGE 14.14 What data is needed to determine the molarity of the acid in the flask?

FIGURE 14.6 The titration of an acid. A known volume of an acid is placed in a flask with an indicator and titrated with a measured volume of a base solution, such as NaOH, to the neutralization endpoint.

SAMPLE PROBLEM 14.12 Titration of an Acid

TRY IT FIRST

If 16.3 mL of a 0.185 M Sr(OH)2 solution is used to titrate the HCl in 0.0250 L of gastric juice, what is the molarity of the HCl solution?

2HCl(aq) + Sr(OH)2(aq) h 2H2O(l ) + SrCl2(aq)

SOLUTION

STEP 1 State the given and needed quantities and concentrations.

ANALYZE THE PROBLEM

Given Need Connect

16.3 mL of 0.185 M Sr(OH)2 solution,

0.0250 L of HCl solution

molarity of the HCl solution

molarity, mole–mole factor

Neutralization Equation

2HCl(aq) + Sr(OH)2(aq) h 2H2O(l ) + SrCl2(aq)

STEP 2 Write a plan to calculate the molarity.

mL of Sr(OH)2 solution

L of Sr(OH)2 solution

moles of Sr(OH)2

moles of HCl

molarity of HCl solution

Molarity Metric factor

Mole–mole factor

Divide by liters

CORE CHEMISTRY SKILL Calculating Molarity or Volume of an Acid or Base in a Titration

M14_TIMB8119_06_SE_C14.indd 457 11/27/18 12:18 PM

458 CHAPTER 14 Acids and Bases

STEP 3 State equalities and conversion factors, including concentrations.

2 mol HCl 1 mol Sr(OH)2

1 mol Sr(OH)2 2 mol HCl

2 mol of HCl = 1 mol of Sr(OH)2

1 L Sr(OH)2 solution 0.185 mol Sr(OH)2

0.185 mol Sr(OH)2 1 L Sr(OH)2 solution

1 L Sr(OH)2 solution 1000 mL Sr(OH)2 solution

1000 mL Sr(OH)2 solution 1 L Sr(OH)2 solution

1 L of Sr(OH)2 solution = 0.185 mol of Sr(OH)21 L of Sr(OH)2 solution = 1000 mL of Sr(OH)2 solution

and

and

and

STEP 4 Set up the problem to calculate the needed quantity.

INTERACTIVE VIDEO

Acid–Base Titration

PRACTICE PROBLEMS Try Practice Problems 14.61 to 14.68

SELF TEST 14.12

a. What is the molarity of an HCl solution if 28.6 mL of a 0.175 M NaOH solution is needed to titrate a 25.0-mL sample of the HCl solution?

b. What volume, in milliliters, of a 0.220 M KOH solution is needed to titrate 8.00 mL of 0.425 M H2SO4 solution?

ANSWER

a. 0.200 M HCl solution b. 30.9 mL of KOH solution

PRACTICE PROBLEMS

14.8 Acid–Base Titration

14.61 If you need to determine the molarity of a formic acid solution, HCHO2, how would you proceed?

14.62 If you need to determine the molarity of an acetic acid solution, HC2H3O2, how would you proceed?

14.63 What is the molarity of a solution of HCl if 5.00 mL of the HCl solution is titrated with 28.6 mL of a 0.145 M NaOH solution?

HCl(aq) + NaOH(aq) h H2O(l) + NaCl(aq)

14.64 What is the molarity of an acetic acid solution if 25.0 mL of the HC2H3O2 solution is titrated with 29.7 mL of a 0.205 M KOH solution?

HC2H3O2(aq) + KOH(aq) h H2O(l ) + KC2H3O2(aq)

14.65 If 38.2 mL of a 0.163 M KOH solution is required to neutralize completely 25.0 mL of a solution of H2SO4, what is the molar- ity of the H2SO4 solution?

H2SO4(aq) + 2KOH(aq) h 2H2O(l) + K2SO4(aq)

14.66 A solution of 0.162 M NaOH is used to titrate 25.0 mL of a solu- tion of H2SO4. If 32.8 mL of the NaOH solution is required to reach the endpoint, what is the molarity of the H2SO4 solution?

H2SO4(aq) + 2NaOH(aq) h 2H2O(l) + Na2SO4(aq)

14.67 A solution of 0.204 M NaOH is used to titrate 50.0 mL of a 0.0224 M H3PO4 solution. What volume, in milliliters, of the NaOH solution is required?

H3PO4(aq) + 3NaOH(aq) h 3H2O(l) + Na3PO4(aq)

14.68 A solution of 0.312 M KOH is used to titrate 15.0 mL of a 0.186 M H3PO4 solution. What volume, in milliliters, of the KOH solution is required?

H3PO4(aq) + 3KOH(aq) h 3H2O(l) + K3PO4(aq)

2 mol HCl 1 mol Sr(OH)2

16.3 mL Sr(OH)2 solution

0.006 03 mol of HCl

* 1 L Sr(OH)2 solution

1000 mL Sr(OH)2 solution *

0.185 mol Sr(OH)2 *

=

1 L Sr(OH)2 solution

0.241 M HCl solution=molarity of HCl solution 0.006 03 mol HCl

0.0250 L HCl solution =

M14_TIMB8119_06_SE_C14.indd 458 11/27/18 12:18 PM

14.9 Buffers 459

Plasma

Red blood cells

White blood cells and platelets

Whole blood consists of plasma, white blood cells and platelets, and red blood cells.

FIGURE 14.7 Adding an acid or a base to water changes the pH drastically, but a buffer resists pH change when small amounts of acid or base are added.

Add

H2O

Add OH -

H 3 O

+

OH -

H 3 O

+

Buffer

pH

pH

pH

pH

pH

pH

14.9 Buffers LEARNING GOAL Describe the role of buffers in maintaining the pH of a solution; calculate the pH of a buffer.

The lungs and the kidneys are the primary organs that regulate the pH of body f luids, including blood and urine. Major changes in the pH of the body fluids can severely affect biological activities within the cells. Buffers are present to prevent large fluctuations in pH.

The pH of water and most solutions changes drastically when a small amount of acid or base is added. However, when an acid or a base is added to a buffer solution, there is little change in pH. A buffer solution maintains the pH of a solution by neutralizing small amounts of added acid or base. In the human body, whole blood contains plasma, white blood cells and platelets, and red blood cells. Blood plasma contains buffers that maintain a consistent pH of about 7.4. If the pH of the blood plasma goes slightly above or below 7.4, changes in our oxygen levels and our metabolic processes can be drastic enough to cause death. Even though we obtain acids and bases from foods and cellular reactions, the buf- fers in the body neutralize those compounds so effectively that the pH of our blood plasma remains essentially unchanged (see FIGURE 14.7).

Large amount Large amount

In a buffer, an acid must be present to react with any OH- that is added, and a base must be available to react with any added H3O

+. However, that acid and base must not neutralize each other. Therefore, a combination of an acid–base conjugate pair is used in buffers. Most buffer solutions consist of nearly equal concentrations of a weak acid and a salt containing its conjugate base. Buffers may also contain a weak base and the salt of the weak base, which contains its conjugate acid.

For example, a typical buffer can be made from the weak acid acetic acid (HC2H3O2) and its salt, sodium acetate (NaC2H3O2). As a weak acid, acetic acid dissociates slightly in water to form H3O

+ and a very small amount of C2H3O2 -. The addition of its salt, sodium

acetate, provides a much larger concentration of acetate ion (C2H3O2 -), which is necessary

for its buffering capability.

HC2H3O2(aq) + H2O(l) vh H3O +(aq) + C2H3O2 -(aq)

ENGAGE 14.15 Why does a buffer require the pres- ence of a weak acid or weak base and the salt of that weak acid or weak base?

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460 CHAPTER 14 Acids and Bases

FIGURE 14.8 The buffer described here consists of about equal concentrations of acetic acid (HC2H3O2) and its conjugate base acetate ion (C2H3O2

-). The pH of the solution is maintained as long as the added amount of acid or base is small compared to the concentrations of the buffer components.

HC2H3O2

C2H3O2 -

C2H3O2 -HC2H3O2

C2H3O2 -

HC2H3O2 H3O

+

Weak acid Conjugate base Weak acid Conjugate base Weak acid Conjugate base

OH -

Adding H3O +

to a buffer neutralizes

some C2H3O2 -.

Adding OH -

to a buffer neutralizes

some HC2H3O2.

INTERACTIVE VIDEO

Calculating the pH of a Buffer

We can now describe how this buffer solution maintains the [H3O +]. When a small

amount of acid is added, the additional H3O + combines with the acetate ion, C2H3O2

-, causing the equilibrium to shift in the direction of the reactants, acetic acid and water. There will be a slight decrease in the [C2H3O2

-] and a slight increase in the [HC2H3O2], but both the [H3O

+] and pH are maintained.

HC2H3O2(aq) + H2O(l) v H3O+(aq) + C2H3O2 -(aq)

If a small amount of base is added to this same buffer solution, it is neutralized by the acetic acid, HC2H3O2, which shifts the equilibrium in the direction of the products, acetate ion and water. The [HC2H3O2] decreases slightly and the [C2H3O2

-] increases slightly, but again the [H3O

+] and thus the pH of the solution are maintained (see FIGURE 14.8).

HC2H3O2(aq) + OH-(aq) h H2O(l) + C2H3O2 -(aq)

ENGAGE 14.16 Which part of a buffer neutralizes any H3O

+ that is added?

Equilibrium shifts in the direction of the reactants

PRACTICE PROBLEMS Try Practice Problems 14.69 to 14.72 Equilibrium shifts in the

direction of the products

Weak acid

Conjugate base

SAMPLE PROBLEM 14.13 Calculating the pH of a Buffer

TRY IT FIRST

The Ka for acetic acid, HC2H3O2, is 1.8 * 10-5. What is the pH of a buffer prepared with 1.0 M HC2H3O2 and 1.0 M C2H3O2

-?

HC2H3O2(aq) + H2O(l) vh H3O +(aq) + C2H3O2 -(aq)

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

1.0 M HC2H3O2, 1.0 M C2H3O2

- pH Ka expression

Equation

HC2H3O2(aq) + H2O(l ) vh H3O +(aq) + C2H3O2 -(aq)

Calculating the pH of a Buffer By rearranging the Ka expression to give [H3O

+], we can obtain the ratio of the acetic acid/ acetate buffer.

Ka = [H3O

+][C2H3O2 -]

[HC2H3O2]

Solving for [H3O +] gives:

[H3O +] = Ka *

[HC2H3O2]

[C2H3O2 -]

In this rearrangement of Ka, the weak acid is in the numerator and the conjugate base in the denominator. We can now calculate the [H3O

+] and pH for an acetic acid buffer as shown in Sample Problem 14.13.

v v

CORE CHEMISTRY SKILL Calculating the pH of a Buffer

M14_TIMB8119_06_SE_C14.indd 460 11/27/18 12:18 PM

14.9 Buffers 461

Because Ka is a constant at a given temperature, the [H3O +] is determined by the

[HC2H3O2]/[C2H3O2 -] ratio. As long as the addition of small amounts of either acid or

base changes the ratio of [HC2H3O2]/[C2H3O2 -] only slightly, the changes in [H3O

+] will be small, and the pH will be maintained. If a large amount of acid or base is added, the buffering capacity of the system may be exceeded. Buffers can be prepared from conjugate acid–base pairs such as H2PO4

-/HPO4 2-, HPO4

2-/PO4 3-, HCO3

-/CO3 2-, or

NH4 +/NH3. The pH of the buffer solution will depend on the conjugate acid–base pair

chosen. Using a common phosphate buffer for biological specimens, we can look at the effect

of using different ratios of [H2PO4 -]/[HPO4

2-] on the [H3O +] and pH. The Ka of H2PO4

- is 6.2 * 10-8. The equation and the [H3O+] are written as follows:

H2PO4 -(aq) + H2O(l) vh H3O

+(aq) + HPO4 2-(aq)

[H3O +] = Ka *

[H2PO4 -]

[HPO4 2-]

Ka

[H2PO4 −]

[HPO4 2−] Ratio [H3O

+] pH

6.2 * 10-8 1.0 M 0.10 M

10 1

6.2 * 10-7 6.21

6.2 * 10-8 1.0 M 1.0 M

1 1

6.2 * 10-8 7.21

6.2 * 10-8 0.10 M 1.0 M

1 10

6.2 * 10-9 8.21

To prepare a phosphate buffer with a pH close to the pH of a biological sample, 7.4, we would choose concentrations that are about equal, such as 1.0 M H2PO4

- and 1.0 M HPO4

2-.

ENGAGE 14.17 How would a solution composed of H2PO4

- and HPO4 2- act as a buffer?

PRACTICE PROBLEMS Try Practice Problems 14.73 to 14.78

STEP 2 Write the Ka expression and rearrange for [H3O +].

Ka = [H3O

+][C2H3O2 -]

[HC2H3O2]

[H3O +] = Ka *

[HC2H3O2]

[C2H3O2 -]

STEP 3 Substitute [HA] and [A−] into the Ka expression.

[H3O +] = 1.8 * 10-5 *

[1.0] [1.0]

[H3O +] = 1.8 * 10-5 M

STEP 4 Use [H3O +] to calculate pH. Placing the [H3O

+] into the pH equation gives the pH of the buffer.

pH = - log[1.8 * 10-5] = 4.74

SELF TEST 14.13

a. One of the conjugate acid–base pairs that buffers the blood plasma is H2PO4 -/HPO4

2-. The Ka for H2PO4

- is 6.2 * 10-8. What is the pH of a buffer that is prepared from 0.10 M H2PO4

- and 0.50 M HPO4 2-?

b. What is the pH of a buffer made from 0.10 M formic acid (HCHO2) and 0.010 M formate (CHO2

-)? The Ka for formic acid is 1.8 * 10-4.

ANSWER

a. pH = 7.91 b. pH = 2.74

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462 CHAPTER 14 Acids and Bases

Chemistry Link to Health Buffers in the Blood Plasma

The arterial blood plasma has a normal pH of 7.35 to 7.45. If changes in H3O

+ lower the pH below 6.8 or raise it above 8.0, cells cannot function properly, and death may result. In our cells, CO2 is continu- ally produced as an end product of cellular metabolism. Some CO2 is carried to the lungs for elimination, and the rest dissolves in body fluids such as plasma and saliva, forming carbonic acid, H2CO3. As a weak acid, carbonic acid dissociates to give bicarbonate, HCO3

-, and H3O

+. More of the anion HCO3 - is supplied by the kidneys to give an

important buffer system in the body fluid—the H2CO3/HCO3 - buffer.

CO2(g) H2CO3(aq)+ H2O(l )

H3O +(aq) HCO3

-(aq)+

Excess H3O + entering the body fluids reacts with the HCO3

-, and excess OH- reacts with the carbonic acid.

H2CO3(aq) H3O +(aq) HCO3

-(aq)+ +H2O(l) Equilibrium shifts in the direction of the reactants

H2CO3(aq) H2O (l) HCO3 -(aq)+ +OH-(aq)

Equilibrium shifts in the direction of the products

For carbonic acid, we can write the equilibrium expression as

Ka = [H3O

+][HCO3 -]

[H2CO3]

To maintain the normal blood plasma pH (7.35 to 7.45), the ratio of [H2CO3]/[HCO3

-] needs to be about 1 to 10, which is obtained by the concentrations in the blood plasma of 0.0024 M H2CO3 and 0.024 M HCO3

-.

[H3O +] = Ka *

[H2CO3]

[HCO3 -]

[H3O +] = 4.3 * 10-7 *

[0.0024]

[0.024] = 4.3 * 10-7 * 0.10 = 4.3 * 10-8 M

pH = - log[4.3 * 10-8] = 7.37 In the body, the concentration of carbonic acid is closely associated

with the partial pressure of CO2, PCO2. TABLE 14.9 lists the normal values for arterial blood. If the CO2 level rises, increasing [H2CO3], the equilibrium shifts to produce more H3O

+, which lowers the pH. This condition is called acidosis. Difficulty with ventilation or gas diffusion can lead to respiratory acidosis, which can happen in emphysema or when an accident or depressive drugs affect the medulla of the brain.

A lowering of the CO2 level leads to a high blood pH, a condition called alkalosis. Excitement, trauma, or a high temperature may cause a person to hyperventilate, which expels large amounts of CO2. As the partial pressure of CO2 in the blood falls below normal, the equilibrium shifts from H2CO3 to CO2 and H2O. This shift decreases the [H3O

+] and raises the pH. The kidneys also regulate H3O

+ and HCO3 -, but they do

so more slowly than the adjustment made by the lungs during ventilation. TABLE 14.10 lists some of the conditions that lead to changes

in the blood pH and some possible treatments.

PCO2 40 mmHg

H2CO3 2.4 mmol/L of plasma

HCO3 - 24 mmol/L of plasma

pH 7.35 to 7.45

TABLE 14.9 Normal Values for Blood Buffer in Arterial Blood

Respiratory Acidosis: CO2 c pH T Symptoms Failure to ventilate, suppression of breathing, disorientation, weakness, coma

Causes Lung disease blocking gas diffusion (e.g., emphysema, pneumonia, bronchitis, asthma); depression of respiratory center by drugs, cardiopulmonary arrest, stroke, poliomyelitis, or nervous system disorders

Treatment Correction of disorder, infusion of bicarbonate

Metabolic Acidosis: H+c pH T Symptoms Increased ventilation, fatigue, confusion

Causes Renal disease, including hepatitis and cirrhosis; increased acid production in diabetes mellitus, hyperthyroidism, alcoholism, and starvation; loss of alkali in diarrhea; acid retention in renal failure

Treatment Sodium bicarbonate given orally, dialysis for renal failure, insulin treatment for diabetic ketosis

Respiratory Alkalosis: CO2 T pH c Symptoms Increased rate and depth of breathing, numbness, light-headedness, tetany

Causes Hyperventilation because of anxiety, hysteria, fever, exercise; reaction to drugs such as salicylate, quinine, and antihistamines; conditions causing hypoxia (e.g., pneumonia, pulmonary edema, heart disease)

Treatment Elimination of anxiety-producing state, rebreathing into a paper bag

Metabolic Alkalosis: H+T pH c Symptoms Depressed breathing, apathy, confusion

Causes Vomiting, diseases of the adrenal glands, ingestion of excess alkali

Treatment Infusion of saline solution, treatment of underlying diseases

TABLE 14.10 Acidosis and Alkalosis: Symptoms, Causes, and Treatments

M14_TIMB8119_06_SE_C14.indd 462 11/27/18 12:18 PM

PRACTICE PROBLEMS

14.9 Buffers

14.69 Which of the following make a buffer system when added to water? Explain.

a. NaOH and NaCl b. H2CO3 and NaHCO3 c. HF and KF d. KCl and NaCl

14.70 Which of the following make a buffer system when added to water? Explain.

a. HClO2 b. NaNO3 c. HC2H3O2 and NaC2H3O2 d. HCl and NaOH

14.71 Consider the buffer system of hydrofluoric acid, HF, and its salt, NaF.

HF(aq) + H2O(l) vh H3O +(aq) + F -(aq)

a. The purpose of this buffer system is to: 1. maintain [HF] 2. maintain [F -] 3. maintain pH b. The salt of the weak acid is needed to: 1. provide the conjugate base 2. neutralize added H3O

+

3. provide the conjugate acid c. If OH- is added, it is neutralized by: 1. the salt 2. H2O 3. H3O

+

d. When H3O + is added, the equilibrium shifts in the direction

of the: 1. reactants 2. products 3. does not change

14.72 Consider the buffer system of nitrous acid, HNO2, and its salt, NaNO2.

HNO2(aq) + H2O(l) vh H3O +(aq) + NO2 -(aq)

a. The purpose of this buffer system is to: 1. maintain [HNO2] 2. maintain [NO2

-] 3. maintain pH

b. The weak acid is needed to: 1. provide the conjugate base 2. neutralize added OH-

3. provide the conjugate acid c. If H3O

+ is added, it is neutralized by: 1. the salt 2. H2O 3. OH

-

d. When OH- is added, the equilibrium shifts in the direction of the:

1. reactants 2. products 3. does not change

14.73 Nitrous acid has a Ka of 4.5 * 10-4. What is the pH of a buffer solution containing 0.10 M HNO2 and 0.10 M NO2

-?

14.74 Acetic acid has a Ka of 1.8 * 10-5. What is the pH of a buffer solution containing 0.15 M HC2H3O2 and 0.15 M C2H3O2

-?

14.75 Using Table 14.4 for Ka values, compare the pH of a HF buffer that contains 0.10 M HF and 0.10 M NaF with another HF buf- fer that contains 0.060 M HF and 0.120 M NaF.

14.76 Using Table 14.4 for Ka values, compare the pH of a H2CO3 buffer that contains 0.10 M H2CO3 and 0.10 M NaHCO3 with another H2CO3 buffer that contains 0.15 M H2CO3 and 0.050 M NaHCO3.

Applications

14.77 Someone with kidney failure excretes urine with large amounts of HCO3

-. How would this loss of HCO3 - affect the pH of the

blood plasma?

14.78 Someone with severe diabetes obtains energy by the breakdown of fats, which produce large amounts of acidic substances. How would this affect the pH of the blood plasma?

UPDATE Acid Reflux Disease

Larry has not been feeling well lately. He tells his doctor that he has discomfort and a burning feel- ing in his chest, and a sour taste in his throat and mouth. At times, Larry says he feels bloated after a big meal, has a dry cough, is hoarse, and sometimes has a sore throat. He has tried antacids, but they do not bring any relief.

The doctor tells Larry that he thinks he has acid reflux. At the

top of the stomach there is a valve, the lower esophageal sphincter, that normally closes after food passes through it. However, if the valve does not close completely, acid pro- duced in the stomach to digest food can move up into the esophagus, a condition called acid reflux. The acid, which is hydrochloric acid, HCl, is produced in the stomach to kill microorganisms, and activate the enzymes we need to break down food.

If acid reflux occurs, the strong acid HCl comes in contact with the lining of the esophagus, where it causes irritation and produces a burning feeling in the chest. Sometimes the pain in the chest is called heartburn. If the HCl reflux goes high enough to reach the throat, a sour taste may be noticed in the mouth. If Larry’s symptoms occur three or more times a week, he may have a chronic condition known as acid reflux disease or gastroesophageal reflux disease (GERD).

Larry’s doctor orders an esophageal pH test in which the amount of acid entering the esophagus from the stomach is measured over 24 h. A probe that measures the pH is inserted into the lower esophagus above the esophageal sphincter. The pH measurements indicate a reflux episode each time the pH drops to 4 or less.

In the 24-h period, Larry has several reflux episodes, and his doctor determines that he has chronic GERD. He and Larry discuss treatment for GERD, which includes eating smaller meals, not lying down for 3 h after eating, making dietary changes, and losing weight. Antacids may be used

Update 463

M14_TIMB8119_06_SE_C14.indd 463 11/27/18 12:18 PM

464 CHAPTER 14 Acids and Bases

to neutralize the acid coming up from the stomach. Other medications known as proton pump inhibitors (PPIs), such as Prilosec and Nexium, may be used to suppress the pro- duction of HCl in the stomach (gastric parietal cells), which raises the pH in the stomach to between 4 and 5, and gives the esophagus time to heal. Nexium may be given in oral doses of 40 mg once a day for 4 weeks. In severe GERD cases, an artificial valve may be created at the top of the stomach to strengthen the lower esophageal sphincter.

Applications

14.79 At rest, the [H3O +] of the stomach fluid is 2.0 * 10-4 M.

What is the pH of the stomach fluid?

14.80 When food enters the stomach, HCl is released and the [H3O

+] of the stomach fluid rises to 4.2 * 10-2 M. What is the pH of the stomach fluid while eating?

14.81 In Larry’s esophageal pH test, a pH value of 3.60 was recorded in the esophagus. What is the [H3O

+] in his esophagus?

14.82 After Larry had taken Nexium for 4 weeks, the pH in his stom- ach was raised to 4.52. What is the [H3O

+] in his stomach?

14.83 Write the balanced chemical equation for the neutralization reaction of stomach acid HCl with CaCO3, an ingredient in some antacids.

14.84 Write the balanced chemical equation for the neutralization reaction of stomach acid HCl with Al(OH)3, an ingredient in some antacids.

14.85 How many grams of CaCO3 are required to neutralize 100. mL of stomach acid HCl, which is 0.0400 M HCl?

14.86 How many grams of Al(OH)3 are required to neutralize 150. mL of stomach acid HCl with a pH of 1.50?

Lower esophageal

sphincter open, allowing

reflux

Liquid Stomach

Pylorus

Lower esophageal sphincter

closed

Esophagus

In acid reflux disease, the lower esophageal sphincter opens, allowing acidic fluid from the stomach to enter the esophagus.

CONCEPT MAP

[H3O +] [A–]

[HA] Ka =

Kw = [H3O +] [OH–]

–log[OH–]

Acid

H+ AcceptorH+ Donor

Strong Acid

Weak Acid

Strong Base

Weak Base

Base Neutralization

Salt

Titration

Concentration of an Acid Solution

Buffer pH

pH

pOH

Weak Acid or

Base

Water

Dissociation of H2O

H3O +

OH–

–log[H3O +]

ACIDS AND BASES

gives

gives

is

is

product

gives

to form

and

of a

determines

in ais ais a

100% small %

has a

with its conjugate forms a

100% small %

undergo

[BH+] [OH–]

[B] Kb =

has a

to maintain

M14_TIMB8119_06_SE_C14.indd 464 11/27/18 12:18 PM

Chapter Review 465

CHAPTER REVIEW

14.1 Acids and Bases LEARNING GOAL Describe and name acids and bases. • An Arrhenius acid produces H+ and

an Arrhenius base produces OH- in aqueous solutions.

• Acids taste sour, may sting, and neutralize bases.

• Bases taste bitter, feel slippery, and neutralize acids.

• Acids containing a simple anion use a hydro prefix, whereas acids with oxygen-containing polyatomic anions are named as ic or ous acids.

14.2 Brønsted–Lowry Acids and Bases LEARNING GOAL Identify conjugate acid–base pairs for Brønsted–Lowry acids and bases. • According to the

Brønsted–Lowry theory, acids are H+ donors and bases are H+ acceptors.

• A conjugate acid–base pair is related by the loss or gain of one H+. • For example, when the acid HF donates H+, the F - is its conjugate

base. The other acid–base pair would be H3O +/H2O.

HF(aq) + H2O(l) vh H3O+(aq) + F -(aq)

14.3 Strengths of Acids and Bases LEARNING GOAL Write equations for the dissociation of strong and weak acids; identify the direction of reaction. • Strong acids dissociate completely in

water, and the H+ is accepted by H2O acting as a base.

• A weak acid dissociates slightly in water, producing only a small percentage of H3O

+. • Strong bases are hydroxides of Groups 1A

(1) and 2A (2) that dissociate completely in water.

• An important weak base is ammonia, NH3.

14.4 Dissociation of Weak Acids and Bases LEARNING GOAL Write the expression for the dissociation of a weak acid or weak base. • In water, weak acids

and weak bases produce only a few ions when equilibrium is reached.

• Weak acids have small Ka values whereas strong acids, which are essentially 100% dissociated, have very large Ka values.

• The reaction for a weak acid can be written:

HA + H2O vh H3O + + A-

• The acid dissociation expression is written:

Ka = [H3O

+][A-]

[HA]

• For a weak base, B + H2O vh BH+ + OH-, the base dissociation expression is written:

Kb = [BH+][OH-]

[B]

14.5 Dissociation of Water LEARNING GOAL Use the water dissociation expression to calculate the [H3O

+] and [OH-] in an aqueous solution. • In pure water, a few water molecules trans-

fer H+ to other water molecules, producing small, but equal, [H3O

+] and [OH-]. • In pure water, the molar concentrations of

H3O + and OH- are each 1.0 * 10-7 mol/L.

• The water dissociation expression, is written as Kw = [H3O +][OH-].

• At 25 °C, Kw = 1.0 * 10-14. • In acidic solutions, the [H3O

+] is greater than the [OH-]. • In basic solutions, the [OH-] is greater than the [H3O

+].

14.6 The pH Scale LEARNING GOAL Calculate pH from [H3O

+]; given the pH, calculate the [H3O

+] and [OH-] of a solution. • The pH scale is a range of numbers

typically from 0 to 14, which represents the [H3O

+] of the solution. • A neutral solution has a pH of 7.0. In

acidic solutions, the pH is below 7.0; in basic solutions, the pH is above 7.0.

• Mathematically, pH is the negative logarithm of the hydronium ion concentration, pH = - log[H3O+].

• The pOH is the negative log of the hydroxide ion concentration, pOH = - log[OH-].

• The sum of the pH + pOH is 14.00.

14.7 Reactions of Acids and Bases LEARNING GOAL Write balanced equations for reactions of acids with metals, carbonates or bicarbonates, and bases. • An acid reacts with a metal to produce

hydrogen gas and a salt. • The reaction of an acid with a carbonate

or bicarbonate produces carbon dioxide, water, and a salt.

• In neutralization, an acid reacts with a base to produce water and a salt.

14.8 Acid–Base Titration LEARNING GOAL Calculate the molarity or volume of an acid or base solution from titration information. • In a titration, an acid sample is

neutralized with a known amount of a base.

• From the volume and molarity of the base, the concentration of the acid is calculated.

Ionic compound

Hydroxide ion

Dissociation

OH -

Na+

NaOH(s) Na+(aq) OH -(aq)+ H2O

NaOH(s)

Water

+ +

+

+ ++

+ +

+

-

- -

-

- -

- -

-

HCl + H2O H3O+ + Cl- Hydrogen chloride

Water Hydronium ion

Chloride ion

Acid (H+ donor)

Base (H+ acceptor) Acidic solution

-+

+

-

HC2H3O2 1 M

HCHO2 CHO2 -

[H3O +] = [OH-]

H3O + OH-

M14_TIMB8119_06_SE_C14.indd 465 11/27/18 12:18 PM

466 CHAPTER 14 Acids and Bases

Calculating [H3O +] from pH (14.6)

• The calculation of [H3O +] from the pH is done by reversing the pH

calculation using the negative pH.

[H3O +] = 10-pH

Example: What is the [H3O +] of a solution with a pH of 4.80?

Answer: [H3O +] = 10-pH

= 10-4.80

2nd log +>- 4.80 = = 1.6 * 10-5 M

Two SFs in the [H3O +] equal the two decimal places in the pH.

KEY MATH SKILLS

Calculator Display

10.61978876

Calculator Display

1.584893192 05

The chapter section containing each Key Math Skill is shown in paren- theses at the end of each heading.

Calculating pH from [H3O +] (14.6)

• The pH of a solution is calculated from the negative log of the [H3O +].

pH = - log[H3O+]

Example: What is the pH of a solution that has

[H3O +] = 2.4 * 10-11 M?

Answer: We substitute the given [H3O +] into the pH equation and

calculate the pH. pH = - log[H3O+]

= - log[2.4 * 10-11] +>- log 2.4 EE or EXP +>- 11 =

= 10.62 Two decimal places in the pH equal the two SFs in the [H3O

+] coefficient.

Add OH -

H 3 O

+

Buffer

pH

pH

pH

14.9 Buffers LEARNING GOAL Describe the role of buffers in maintaining the pH of a solution; calculate the pH of a buffer. • A buffer solution resists changes in pH when small amounts of an

acid or a base are added.

• A buffer contains either a weak acid and its salt or a weak base and its salt.

• In a buffer, the weak acid reacts with added OH-, and the anion of the salt reacts with added H3O

+. • Most buffer solutions consist of nearly equal concentrations of a

weak acid and a salt containing its conjugate base. • The pH of a buffer is calculated by solving the Ka expression for

[H3O +].

acid A substance that dissolves in water and produces hydrogen ions (H+), according to the Arrhenius theory. All acids are hydrogen ion donors, according to the Brønsted–Lowry theory.

acid dissociation constant, Ka The numerical value of the product of the ions from the dissociation of a weak acid divided by the concentration of the weak acid.

amphoteric Substances that can act as either an acid or a base in water. base A substance that dissolves in water and produces hydroxide ions

(OH-), according to the Arrhenius theory. All bases are hydrogen ion acceptors, according to the Brønsted–Lowry theory.

base dissociation constant, Kb The numerical value of the product of the ions from the dissociation of a weak base divided by the concentration of the weak base.

Brønsted–Lowry acids and bases An acid is a hydrogen ion donor; a base is a hydrogen ion acceptor.

buffer solution A solution of a weak acid and its conjugate base or a weak base and its conjugate acid that maintains the pH by neutralizing added acid or base.

conjugate acid–base pair An acid and a base that differ by one H+. When an acid donates a hydrogen ion, the product is its conju- gate base, which is capable of accepting a hydrogen ion in the reverse reaction.

dissociation The separation of an acid or a base into ions in water. endpoint The point at which an indicator changes color. For the indicator

phenolphthalein, the color change occurs when the number of moles of OH- is equal to the number of moles of H3O

+ in the sample.

KEY TERMS

hydronium ion, H3O + The ion formed by the attraction of a

hydrogen ion, H+, to a water molecule. indicator A substance added to a titration sample that changes color

when the pH of the solution changes. neutral The term that describes a solution with equal concentrations

of H3O + and OH-.

neutralization A reaction between an acid and a base to form water and a salt.

pH A measure of the [H3O +] in a solution; pH = - log[H3O+].

pOH A measure of the [OH-] in a solution; pOH = - log[OH-]. salt An ionic compound that contains a metal ion or NH4

+ and a non- metal or polyatomic ion other than OH-.

strong acid An acid that completely dissociates in water. strong base A base that completely dissociates in water. titration The addition of base to an acid sample to determine the

concentration of the acid. water dissociation constant, Kw The numerical value of the product

of [H3O +] and [OH-] in solution; Kw = 1.0 * 10-14.

water dissociation expression The product of [H3O +] and [OH-] in

solution; Kw = [H3O +][OH-].

weak acid An acid that is a poor donor of H+ and dissociates only slightly in water.

weak base A base that is a poor acceptor of H+ and produces only a small number of ions in water.

log 7.5 = 7.5+/-2nd log2ndor =+/-

Calculator DisplayCalculator Procedure

M14_TIMB8119_06_SE_C14.indd 466 11/27/18 12:18 PM

Core Chemistry Skills 467

The chapter section containing each Core Chemistry Skill is shown in parentheses at the end of each heading.

Identifying Conjugate Acid–Base Pairs (14.2) • According to the Brønsted–Lowry theory, a conjugate acid–base

pair consists of molecules or ions related by the loss of one H+ by an acid, and the gain of one H+ by a base.

• Every acid–base reaction contains two conjugate acid–base pairs because an H+ is transferred in both the forward and reverse directions.

• When an acid such as HF loses one H+, the conjugate base F - is formed. When H2O acts as a base, it gains one H

+, which forms its conjugate acid, H3O

+.

Example: Identify the conjugate acid–base pairs in the following reaction: H2SO4(aq) + H2O(l) vh HSO4

-(aq) + H3O+(aq)

Answer: H2SO4(aq) + H2O(l) vh HSO4 -(aq) + H3O+(aq)

Acid Base Conjugate base Conjugate acid

Conjugate acid–base pairs: H2SO4/HSO4 - and H3O

+/H2O

Calculating [H3O +] and [OH−] in Solutions (14.5)

• For all aqueous solutions, the product of [H3O +] and [OH-] is

equal to the water dissociation constant, Kw.

Kw = [H3O +][OH-]

• Because pure water contains equal numbers of OH- ions and H3O +

ions, each with molar concentrations of 1.0 * 10-7 M, the numeri- cal value of Kw is 1.0 * 10-14 at 25 °C.

Kw = [H3O +][OH-] = [1.0 * 10-7][1.0 * 10-7]

= 1.0 * 10-14

• If we know the [H3O +] of a solution, we can use the Kw expression

to calculate the [OH-]. If we know the [OH-] of a solution, we can calculate the [H3O

+] using the Kw expression.

[OH-] = Kw

[H3O +] [H3O

+] = Kw

[OH-]

Example: What is the [OH-] in a solution that has [H3O

+] = 2.4 * 10-11 M? Is the solution acidic or basic? Answer: We solve the Kw expression for [OH

-] and substitute in the known values of Kw and [H3O

+].

[OH-] = Kw

[H3O +]

= 1.0 * 10-14

[2.4 * 10-11] = 4.2 * 10-4 M

Because the [OH-] is greater than the [H3O +], this is a

basic solution.

Writing Equations for Reactions of Acids and Bases (14.7) • Acids react with certain metals to produce hydrogen gas (H2) and

a salt.

2HCl(aq) + Mg(s) h H2(g) + MgCl2(aq) Acid Metal Hydrogen Salt

• When an acid is added to a carbonate or bicarbonate, the products are carbon dioxide gas, water, and a salt.

2HCl(aq) + Na2CO3(aq) h CO2(g) + H2O(l) + 2NaCl(aq) Acid Carbonate Carbon Water Salt dioxide

CORE CHEMISTRY SKILLS • Neutralization is a reaction between a strong or weak acid with a

strong base to produce water and a salt.

HCl(aq) + NaOH(aq) h H2O(l) + NaCl(aq) Acid Base Water Salt

Example: Write the balanced chemical equation for the reaction of hydrobromic acid HBr(aq) and ZnCO3(s).

Answer:

2HBr(aq) + ZnCO3(s) h CO2(g) + H2O(l) + ZnBr2(aq)

Calculating Molarity or Volume of an Acid or Base in a Titration (14.7) • In a titration, a measured volume of acid is neutralized by a strong

base solution of known molarity. • From the measured volume of the strong base solution required for

titration and its molarity, the number of moles of the strong base, the moles of acid, and the concentration of the acid are calculated.

Example: A 15.0-mL sample of a H2SO4 solution is titrated with 24.0 mL of a 0.245 M NaOH solution. What is the molarity of the H2SO4 solution?

H2SO4(aq) + 2NaOH(aq) h 2H2O(l) + Na2SO4(aq)

Answer:

24.0 mL NaOH solution * 1 L NaOH solution

1000 mL NaOH solution

* 0.245 mol NaOH

1 L NaOH solution *

1 mol H2SO4 2 mol NaOH

= 0.002 94 mol of H2SO4

Molarity (M) = 0.002 94 mol H2SO4

0.0150 L H2SO4 solution = 0.196 M H2SO4 solution

Calculating the pH of a Buffer (14.9) • A buffer solution maintains pH by neutralizing small amounts of

added acid or base. • Most buffer solutions consist of nearly equal concentrations of a

weak acid and a salt containing its conjugate base such as acetic acid, HC2H3O2, and its salt, NaC2H3O2.

• The [H3O +] is calculated by solving the Ka expression for [H3O

+], then substituting the values of [H3O

+], [HA], and Ka into the equation.

Ka = [H3O

+][C2H3O2 -]

[HC2H3O2]

Solving for [H3O +] gives:

[H3O +] = Ka *

[HC2H3O2]

[C2H3O2 -]

• The pH of the buffer is calculated from the [H3O +].

pH = - log[H3O+]

Example: What is the pH of a buffer prepared with 0.40 M HC2H3O2 and 0.20 M C2H3O2

-, if the Ka of acetic acid is 1.8 * 10-5?

Answer: [H3O +] = Ka *

[HC2H3O2]

[C2H3O2 -]

= 1.8 * 10-5 * [0.40]

[0.20]

= 3.6 * 10-5 M pH = - log[3.6 * 10-5] = 4.44

Weak acidv Conjugate basev

M14_TIMB8119_06_SE_C14.indd 467 11/27/18 12:18 PM

468 CHAPTER 14 Acids and Bases

UNDERSTANDING THE CONCEPTS

The chapter sections to review are shown in parentheses at the end of each problem.

14.87 Determine if each of the following diagrams represents a strong acid or a weak acid. The acid has the formula HX. (14.3)

A

HX HX HX

HXH3O +

HXHX

B

H3O +

H3O +

H3O +

H3O +

H3O +

X-

X-

X- X-

X-

X-

HX HX

14.88 Adding a few drops of a strong acid to water will lower the pH appreciably. However, adding the same number of drops to a buffer does not appreciably alter the pH. Why? (14.9)

Water

pH = 7.0

pH = 3.0

Buffer

pH = 7.0

pH = 6.9

14.89 Identify each of the following as an acid or a base: (14.1) a. H2SO4 b. RbOH c. Ca(OH)2 d. HI

14.90 Identify each of the following as an acid or a base: (14.1) a. Sr(OH)2 b. H2SO3 c. HC2H3O2 d. CsOH

14.91 Complete the following table: (14.2)

Acid Conjugate Base

H2O

CN-

HNO2 H2PO4

-

14.92 Complete the following table: (14.2)

Base Conjugate Acid

HS-

H3O +

NH3 HCO3

-

Applications

14.93 State whether each of the following solutions is acidic, basic, or neutral: (14.6)

a. sweat, pH 5.2 b. tears, pH 7.5 c. bile, pH 8.1 d. stomach acid, pH 2.5

14.94 State whether each of the following solutions is acidic, basic, or neutral: (14.6)

a. saliva, pH 6.8 b. urine, pH 5.9 c. pancreatic juice, pH 8.0 d. blood, pH 7.45

14.95 Sometimes, during stress or trauma, a person can start to hyper- ventilate. Then the person might breathe into a paper bag to avoid fainting. (14.9)

a. What changes occur in the blood pH during hyperventilation?

b. How does breathing into a paper bag help return blood pH to normal?

Breathing into a paper bag can help a person who is hyperventilating.

ADDITIONAL PRACTICE PROBLEMS

14.97 Identify each of the following as an acid, base, or salt, and give its name: (14.1)

a. HBrO2 b. CsOH c. Mg(NO3)2 d. HClO4 14.98 Identify each of the following as an acid, base, or salt, and give

its name: (14.1) a. HNO2 b. MgBr2 c. NH3 d. Li2SO3

14.99 Complete the following table: (14.2)

Acid Conjugate Base

HI

Cl-

NH4 +

HS-

14.96 In the blood plasma, pH is maintained by the carbonic acid– bicarbonate buffer system. (14.9)

a. How is pH maintained when acid is added to the buffer system?

b. How is pH maintained when base is added to the buffer system?

M14_TIMB8119_06_SE_C14.indd 468 11/27/18 12:18 PM

Challenge Problems 469

14.100 Complete the following table: (14.2)

Base Conjugate Acid

F -

HC2H3O2 HSO3

-

ClO-

14.101 Using Table 14.3, identify the stronger acid in each of the following pairs: (14.3)

a. HF or HCN b. H3O + or H2S

c. HNO2 or HC2H3O2 d. H2O or HCO3 -

14.102 Using Table 14.3, identify the stronger base in each of the following pairs: (14.3)

a. H2O or Cl - b. OH- or NH3

c. SO4 2- or NO2

- d. CO3 2- or H2O

14.103 Determine the pH and pOH for each of the following solutions: (14.6)

a. [H3O +] = 2.0 * 10-8 M

b. [H3O +] = 5.0 * 10-2 M

c. [OH-] = 3.5 * 10-4 M d. [OH-] = 0.0054 M 14.104 Determine the pH and pOH for each of the following

solutions: (14.6) a. [OH-] = 1.0 * 10-7 M b. [H3O

+] = 4.2 * 10-3 M c. [H3O

+] = 0.0001 M d. [OH-] = 8.5 * 10-9 M 14.105 Are the solutions in problem 14.103 acidic, basic, or neutral?

(14.6)

The following problems are related to the topics in this chapter. However, they do not all follow the chapter order, and they require you to combine concepts and skills from several sections. These problems will help you increase your critical thinking skills and prepare for your next exam.

14.115 For each of the following: (14.2, 14.3) 1. H2S 2. H3PO4 a. Write the formula for the conjugate base. b. Write the Ka expression. c. Which is the weaker acid?

14.116 For each of the following: (14.2, 14.3) 1. HCO3

- 2. HC2H3O2 a. Write the formula for the conjugate base. b. Write the Ka expression. c. Which is the stronger acid?

14.117 Using Table 14.3, identify the conjugate acid–base pairs in each of the following equations and whether the equilibrium mixture contains mostly products or mostly reactants: (14.2, 14.3)

a. NH3(aq) + HNO3(aq) vh NH4 +(aq) + NO3 -(aq)

b. HBr(aq) + H2O(l) vh H3O +(aq) + Br-(aq)

14.118 Using Table 14.3, identify the conjugate acid–base pairs in each of the following equations and whether the equilibrium mixture contains mostly products or mostly reactants: (14.2, 14.3)

a. HNO2(aq) + HS-(aq) vh H2S(g) + NO2 -(aq)

b. Cl-(aq) + H2O(l) vh OH -(aq) + HCl(aq)

14.119 Complete and balance each of the following: (14.7) a. H2SO4(aq) + ZnCO3(s) h b. HNO3(aq) + Al(s) h

14.120 Complete and balance each of the following: (14.7) a. H3PO4(aq) + Ca(OH)2(s) h b. HNO3(aq) + KHCO3(s) h

14.121 Determine each of the following for a 0.050 M KOH solution: (14.6, 14.7)

a. [H3O +]

b. pH c. pOH d. the balanced equation for the reaction with H2SO4 e. milliliters of KOH solution required to neutralize 40.0 mL

of a 0.035 M H2SO4 solution

14.122 Determine each of the following for a 0.10 M HBr solution: (14.6, 14.7)

a. [H3O +]

b. pH c. pOH d. the balanced equation for the reaction with LiOH e. milliliters of HBr solution required to neutralize 36.0 mL

of a 0.25 M LiOH solution

14.123 Calculate the volume, in milliliters, of a 0.150 M NaOH solu- tion that will completely neutralize each of the following: (14.8)

a. 25.0 mL of a 0.288 M HCl solution b. 10.0 mL of a 0.560 M H2SO4 solution

14.124 Calculate the volume, in milliliters, of a 0.215 M NaOH solu- tion that will completely neutralize each of the following: (14.8)

a. 3.80 mL of a 1.25 M HNO3 solution b. 8.50 mL of a 0.825 M H3PO4 solution

CHALLENGE PROBLEMS

14.106 Are the solutions in problem 14.104 acidic, basic, or neutral? (14.6)

14.107 Calculate the [H3O +] and [OH-] for a solution with each of

the following pH values: (14.6) a. 3.00 b. 6.2 c. 8.85 d. 11.00

14.108 Calculate the [H3O +] and [OH-] for a solution with each of

the following pH values: (14.6) a. 10.00 b. 5.0 c. 6.54 d. 1.82

14.109 Solution A has a pH of 4.5, and solution B has a pH of 6.7. (14.6) a. Which solution is more acidic? b. What is the [H3O

+] in each? c. What is the [OH-] in each?

14.110 Solution X has a pH of 9.5, and solution Y has a pH of 7.5. (14.6) a. Which solution is more acidic? b. What is the [H3O

+] in each? c. What is the [OH-] in each?

14.111 What is the [OH-] in a solution that contains 0.225 g of NaOH in 0.250 L of solution? (14.7)

14.112 What is the [H3O +] in a solution that contains 1.54 g of HNO3

in 0.500 L of solution? (14.7)

14.113 What are the pH and pOH values of a solution prepared by dissolving 2.5 g of HCl in water to make 425 mL of solution? (14.6)

14.114 What are the pH and pOH values of a solution prepared by dissolving 1.0 g of Ca(OH)2 in water to make 875 mL of solution? (14.6)

M14_TIMB8119_06_SE_C14.indd 469 11/27/18 12:19 PM

470 CHAPTER 14 Acids and Bases

14.125 A solution of 0.205 M NaOH is used to titrate 20.0 mL of a H2SO4 solution. If 45.6 mL of the NaOH solution is required to reach the endpoint, what is the molarity of the H2SO4 solution? (14.8)

H2SO4(aq) + 2NaOH(aq) h 2H2O(l) + Na2SO4(aq)

14.126 A 10.0-mL sample of vinegar, which is an aqueous solution of acetic acid, HC2H3O2, requires 16.5 mL of a 0.500 M NaOH solution to reach the endpoint in a titration. What is the molarity of the acetic acid solution? (14.8)

HC2H3O2(aq) + NaOH(aq) h H2O(l) + NaC2H3O2(aq)

14.127 A buffer solution is made by dissolving H3PO4 and NaH2PO4 in water. (14.9)

a. Write an equation that shows how this buffer neutralizes added acid.

b. Write an equation that shows how this buffer neutralizes added base.

c. Calculate the pH of this buffer if it contains 0.50 M H3PO4 and 0.20 M H2PO4

-. The Ka for H3PO4 is 7.5 * 10-3. 14.128 A buffer solution is made by dissolving HC2H3O2 and

NaC2H3O2 in water. (14.9) a. Write an equation that shows how this buffer neutralizes

added acid. b. Write an equation that shows how this buffer neutralizes

added base. c. Calculate the pH of this buffer if it contains 0.20 M

HC2H3O2 and 0.40 M C2H3O2 -. The Ka for HC2H3O2 is

1.8 * 10-5.

Applications 14.129 One of the most acidic lakes in the United States is Little Echo

Pond in the Adirondacks in New York. Recently, this lake had a pH of 4.2, well below the recommended pH of 6.5. (14.6, 14.8)

a. What are the [H3O +] and [OH-] of Little Echo Pond?

b. What are the [H3O +] and [OH-] of a lake that has a pH

of 6.5?

c. One way to raise the pH of an acidic lake (and restore aquatic life) is to add limestone (CaCO3). How many grams of CaCO3 are needed to neutralize 1.0 kL of the acidic water from the lake if the acid is sulfuric acid?

H2SO4(aq) + CaCO3(s) h CO2(g) + H2O(l) + CaSO4(aq)

14.130 The daily output of stomach acid (gastric juice) is 1000 mL to 2000 mL. Prior to a meal, stomach acid (HCl) typically has a pH of 1.42. (14.6, 14.7, 14.8)

a. What is the [H3O +] of the stomach acid?

b. One chewable tablet of the antacid Maalox contains 600. mg of CaCO3. Write the neutralization equation, and calculate the milliliters of stomach acid neutralized by two tablets of Maalox.

c. The antacid milk of magnesia contains 400. mg of Mg(OH)2 per teaspoon. Write the neutralization equation, and calculate the number of milliliters of stomach acid that are neutralized by 1 tablespoon of milk of magnesia (1 tablespoon = 3 teaspoons).

A helicopter drops calcium carbonate on an acidic lake to increase its pH.

ANSWERS TO ENGAGE QUESTIONS 14.10 According to Figure 14.4, apple juice is acidic.

14.11 A pH of 4.00, which is less than 7.0, is acidic.

14.12 A solution with [H3O +] of 1.0 * 10-6 M has a pH of 6.00.

There are two zeros after the decimal point because there are two significant figures in the coefficient 1.0.

14.13 When KOH completely neutralizes H2SO4, the salt produced is K2SO4.

14.14 To determine the molarity of an acid solution, the volume of the acid and the volume and molarity of the base used to neutralize the acid are needed.

14.15 A weak acid or base and the salt of that weak acid or base are necessary in a buffer to react with added acid or base.

14.16 When H3O + is added to a buffer, it is neutralized by the weak

base or the salt of the weak acid.

14.17 A solution of H2PO - 4 and HPO

2- 4 acts as a buffer; H2PO

- 4

will neutralize added OH-, and HPO 2-4 will neutralize added H3O

+.

14.1 An acid composed of a hydrogen ion and a simple nonmetal anion is named with the prefix hydro before the name of the nonmetal, and the ide ending of the most common form is changed to ic acid.

14.2 The base BrO2 - accepts H+ to form its conjugate acid HBrO2.

14.3 H2O loses one H + to form OH- but gains one H+ to form H3O

+.

14.4 A strong acid dissociates nearly 100% to form H+ and a negative ion, whereas a weak acid dissociates slightly in water, forming a small amount of H+ and a negative ion.

14.5 According to Table 14.3, H2SO4 is a stronger acid than H2S.

14.6 In pure water, when one H2O molecule donates H + to another

water molecule, forming equal amounts of H3O + and OH-.

14.7 A solution that has a [H3O +] of 1.0 * 10-3 M would have

a [OH-] of 1.0 * 10-11 M. Because [H3O+] 7 [OH-], the solution is acidic.

14.8 You obtain the [OH-] of a solution by dividing the Kw by the [H3O

+] of the solution.

14.9 Because there is an inverse relationship between [H3O +] and

[OH-], if [OH-] decreases, [H3O +] must increase.

M14_TIMB8119_06_SE_C14.indd 470 11/27/18 12:19 PM

Answers to Selected Problems 471

ANSWERS TO SELECTED PROBLEMS

14.47 An increase or decrease of one pH unit changes the [H3O +] by

a factor of 10. Thus a pH of 3 is 10 times more acidic than a pH of 4.

14.49 a. 4.0 b. 8.5 c. 9.0 d. 3.40 e. 7.17 f. 10.92

14.51

[H3O +] [OH−] pH pOH

Acidic, Basic, or Neutral?

1.0 * 10-8 M 1.0 * 10-6 M 8.00 6.00 Basic

3.2 * 10-4 M 3.1 * 10-11 M 3.49 10.51 Acidic

2.8 * 10-5 M 3.6 * 10-10 M 4.55 9.45 Acidic

1.0 * 10-12 M 1.0 * 10-2 M 12.00 2.00 Basic

14.53 1.2 * 10-7 M 14.55 a. 2HBr(aq) + ZnCO3(s) h

CO2(g) + H2O(l) + ZnBr2(aq) b. 2HCl(aq) + Zn(s) h H2(g) + ZnCl2(aq) c. HCl(aq) + NaHCO3(s) h

CO2(g) + H2O(l) + NaCl(aq) d. H2SO4(aq) + Mg(OH)2(s) h 2H2O(l) + MgSO4(aq) 14.57 a. 2HCl(aq) + Mg(OH)2(s) h 2H2O(l) + MgCl2(aq) b. H3PO4(aq) + 3LiOH(aq) h 3H2O(l) + Li3PO4(aq) 14.59 a. H2SO4(aq) + 2NaOH(aq) h 2H2O(l) + Na2SO4(aq) b. 3HCl(aq) + Fe(OH)3(s) h 3H2O(l) + FeCl3(aq) c. H2CO3(aq) + Mg(OH)2(s) h 2H2O(l) + MgCO3(s) 14.61 To a known volume of a formic acid solution, add a few drops

of indicator. Place a solution of NaOH of known molarity in a buret. Add base to acid until one drop changes the color of the solution. Use the volume and molarity of NaOH and the volume of a formic acid solution to calculate the concentration of the formic acid in the sample.

14.63 0.829 M HCl solution

14.65 0.124 M H2SO4 solution

14.67 16.5 mL

14.69 b and c are buffer systems. b contains the weak acid H2CO3 and its salt NaHCO3. c contains HF, a weak acid, and its salt, KF.

14.71 a. 3 b. 1 and 2 c. 3 d. 1

14.73 pH = 3.35 14.75 The pH of the 0.10 M HF/0.10 M NaF buffer is 3.46.

The pH of the 0.060 M HF/0.120 M NaF buffer is 3.76.

14.77 If large amounts of HCO3 - are lost, equilibrium shifts to higher

H3O +, which lowers the pH.

14.79 pH = 3.70 14.81 2.5 * 10-4 M 14.83 2HCl(aq) + CaCO3(s) h CO2(g) + H2O(l) + CaCl2(aq) 14.85 0.200 g of CaCO3 14.87 a. This diagram represents a weak acid; only a few HX mol-

ecules separate into H3O + and X- ions.

b. This diagram represents a strong acid; all the HX molecules separate into H3O

+ and X- ions.

14.89 a. acid b. base c. base d. acid

14.1 a. acid b. acid c. acid d. base e. both

14.3 a. hydrochloric acid b. calcium hydroxide c. perchloric acid d. strontium hydroxide e. sulfurous acid f. bromous acid

14.5 a. RbOH b. HF c. H3PO4 d. LiOH e. NH4OH f. HIO4 14.7 a. HI is the acid (hydrogen ion donor), and H2O is the base

(hydrogen ion acceptor). b. H2O is the acid (hydrogen ion donor), and F

- is the base (hydrogen ion acceptor).

c. H2S is the acid (hydrogen ion donor), and C2H5 ¬ NH2 is the base (hydrogen ion acceptor).

14.9 a. F - b. OH- c. HPO3 2-

d. SO4 2- e. ClO2

-

14.11 a. HCO3 - b. H3O

+ c. H3PO4 d. HBr e. HClO4 14.13 a. The conjugate acid–base pairs are H2CO3/HCO3

- and H3O

+/H2O. b. The conjugate acid–base pairs are HCN/CN- and

HNO2/NO2 -.

c. The conjugate acid–base pairs are HF/F - and HCHO2/CHO2

-.

14.15 NH4 +(aq) + H2O(l) vh NH3(aq) + H3O

+(aq)

14.17 A strong acid is a good hydrogen ion donor, whereas its conju- gate base is a poor hydrogen ion acceptor.

14.19 a. HBr b. HSO4 - c. H2CO3

14.21 a. HSO4 - b. HF c. HCO3

-

14.23 a. reactants b. reactants c. products

14.25 NH4 +(aq) + SO4 2-(aq) vh NH3(aq) + HSO4

-(aq)

The equilibrium mixture contains mostly reactants because NH4

+ is a weaker acid than HSO4 -, and SO4

2- is a weaker base than NH3.

14.27 a. True b. False c. False d. True e. False

14.29 a. H2SO3 b. HSO3 - c. H2SO3

d. HS- e. H2SO3 14.31 H3PO4(aq) + H2O(l) vh H3O

+(aq) + H2PO4 -(aq)

Ka = [H3O

+][H2PO4 -]

[H3PO4]

14.33 In pure water, [H3O +] = [OH-] because one of each is pro-

duced every time a hydrogen ion is transferred from one water molecule to another.

14.35 In an acidic solution, the [H3O +] is greater than the [OH-].

14.37 a. acidic b. basic c. basic d. acidic

14.39 a. 1.0 * 10-5 M b. 1.0 * 10-8 M c. 5.0 * 10-10 M d. 2.5 * 10-2 M 14.41 a. 2.5 * 10-13 M b. 2.0 * 10-9 M c. 5.0 * 10-11 M d. 1.3 * 10-6 M 14.43 In a neutral solution, the [H3O

+] is 1.0 * 10-7 M and the pH is 7.00, which is the negative value of the power of 10.

14.45 a. basic b. acidic c. acidic d. acidic e. basic

M14_TIMB8119_06_SE_C14.indd 471 11/27/18 12:19 PM

472 CHAPTER 14 Acids and Bases

14.91 Acid Conjugate Base

H2O OH -

HCN CN-

HNO2 NO2 -

H3PO4 H2PO4 -

14.93 a. acidic b. basic c. basic d. acidic

14.95 a. During hyperventilation, a person will lose CO2 and the blood pH will rise.

b. Breathing into a paper bag will increase the CO2 concen- tration and lower the blood pH.

14.97 a. acid, bromous acid b. base, cesium hydroxide c. salt, magnesium nitrate d. acid, perchloric acid

14.111 [OH-] = 0.0225 M 14.113 pH = 0.80; pOH = 13.20 14.115 a. 1. HS-

2. H2PO4 -

b. 1. [H3O

+][HS-]

[H2S]

2. [H3O

+][H2PO4 -]

[H3PO4]

c. H2S

14.117 a. HNO3 /NO3 - and NH4

+/NH3; equilibrium mixture contains mostly products

b. HBr/Br- and H3O +/H2O; equilibrium mixture contains

mostly products

14.119 a. H2SO4(aq) + ZnCO3(s) h CO2(g) + H2O(l) + ZnSO4(aq)

b. 6HNO3(aq) + 2Al(s) h 3H2(g) + 2Al(NO3)3(aq) 14.121 a. [H3O

+] = 2.0 * 10-13 M b. pH = 12.70 c. pOH = 1.30 d. H2SO4(aq) + 2KOH(aq) h 2H2O(l) + K2SO4(aq) e. 56 mL of the KOH solution

14.123 a. 48.0 mL of NaOH solution b. 74.7 mL of NaOH solution

14.125 0.234 M H2SO4 14.127 a. acid: H2PO4

-(aq) + H3O+(aq) h H3PO4(aq) + H2O(l) b. base: H3PO4(aq) + OH-(aq) h H2PO4 -(aq) + H2O(l) c. pH = 1.72 14.129 a. [H3O

+] = 6 * 10-5 M; [OH-] = 2 * 10-10 M b. [H3O

+] = 3 * 10-7 M; [OH-] = 3 * 10-8 M c. 3 g of CaCO3

14.99 Acid Conjugate Base

HI I-

HCl Cl-

NH4 + NH3

H2S HS -

14.101 a. HF b. H3O +

c. HNO2 d. HCO3 -

14.103 a. pH = 7.70; pOH = 6.30 b. pH = 1.30; pOH = 12.70 c. pH = 10.54; pOH = 3.46 d. pH = 11.73; pOH = 2.27 14.105 a. basic b. acidic c. basic d. basic

14.107 a. [H3O +] = 1.0 * 10-3 M; [OH-] = 1.0 * 10-11 M

b. [H3O +] = 6 * 10-7 M; [OH-] = 2 * 10-8 M

c. [H3O +] = 1.4 * 10-9 M; [OH-] = 7.1 * 10-6 M

d. [H3O +] = 1.0 * 10-11 M; [OH-] = 1.0 * 10-3 M

14.109 a. Solution A b. Solution A [H3O

+] = 3 * 10-5 M; Solution B [H3O

+] = 2 * 10-7 M c. Solution A [OH-] = 3 * 10-10 M; Solution B [OH-] = 5 * 10-8 M

M14_TIMB8119_06_SE_C14.indd 472 11/27/18 12:19 PM

473

typical automobile uses 550 gal of gasoline and produces 41 lb of nitrogen oxide. (2.7, 7.2, 7.3, 8.2, 8.3, 9.2, 11.6)

a. Write balanced chemical equations for the production of nitrogen oxide and nitrogen dioxide.

b. If all the nitrogen oxide emitted by one automobile is converted to nitrogen dioxide in the atmosphere, how many kilograms of nitrogen dioxide are produced in one year by a single automobile?

c. Write a balanced chemical equation for the combustion of octane.

d. How many moles of C8H18 are present in 15.2 gal of octane? e. How many liters of CO2 at STP are produced in one year

from the gasoline used by the typical automobile?

CI.23 A mixture of 25.0 g of CS2 gas and 30.0 g of O2 gas is placed in a 10.0-L container and heated to 125 °C. The products of the reaction are carbon dioxide gas and sulfur dioxide gas. (7.2, 7.3, 8.2, 9.2, 9.3, 11.1, 11.7)

a. Write a balanced chemical equation for the reaction. b. How many grams of CO2 are produced? c. What is the partial pressure, in millimeters of mercury, of

the remaining reactant? d. What is the final pressure, in millimeters of mercury, in

the container?

CI.24 In wine-making, glucose (C6H12O6) from grapes undergoes fermentation in the absence of oxygen to produce ethanol and carbon dioxide. A bottle of vintage port wine has a volume of 750 mL and contains 135 mL of ethanol (C2H6O). Ethanol has a density of 0.789 g/mL. In 1.5 lb of grapes, there are 26 g of glucose. (2.7, 7.2, 7.3, 8.2, 9.2, 12.4)

a. Calculate the volume percent (v/v) of ethanol in the port wine. b. What is the molarity (M) of ethanol in the port wine? c. Write the balanced chemical equation for the fermentation

reaction of sugar in grapes. d. How many grams of sugar from grapes are required to

produce one bottle of port wine? e. How many bottles of port wine can be produced from

1.0 ton of grapes (1 ton = 2000 lb)? CI.25 Consider the following reaction at equilibrium:

2H2(g) + S2(g) vh 2H2S(g) + heat In a 10.0-L container, an equilibrium mixture contains 2.02 g of H2, 10.3 g of S2, and 68.2 g of H2S. (7.2, 7.3, 13.2, 13.3, 13.4, 13.5)

a. What is the numerical value of Kc for this equilibrium mixture? b. If more H2 is added to the equilibrium mixture, how will

the equilibrium shift?

CI.21 Methane is a major component of purified natural gas used for heating and cooking. When 1.0 mol of methane gas burns with oxygen to produce carbon dioxide and water vapor, 883 kJ of heat is produced. At STP, methane gas has a density of 0.715 g/L. For transport, the natural gas is cooled to - 163 °C to form liquefied natural gas (LNG) with a density of 0.45 g/mL. A tank on a ship can hold 7.0 million gallons of LNG. (2.7, 7.2, 7.3, 8.2, 8.3, 9.5, 10.1, 11.6)

a. Draw the Lewis structures for methane, which has the formula CH4.

b. What is the mass, in kilograms, of LNG (assume that LNG is all methane) transported in one tank on a ship?

c. What is the volume, in liters, of LNG (methane) from one tank when the LNG (methane) from one tank is converted to methane gas at STP?

d. Write the balanced chemical equation for the combustion of methane and oxygen in a gas burner, including the heat of reaction.

e. How many kilograms of oxygen are needed to react with all of the methane in one tank of LNG?

f. How much heat, in kilojoules, is released after burning all of the methane from one tank of LNG?

CI.22 Automobile exhaust is a major cause of air pollution. One pol- lutant is nitrogen oxide, which forms from nitrogen and oxy- gen gases in the air at the high temperatures in an automobile engine. Once emitted into the air, nitrogen oxide reacts with oxygen to produce nitrogen dioxide, a reddish brown gas with a sharp, pungent odor that makes up smog. One com- ponent of gasoline is octane, C8H18, which has a density of 0.803 g/mL. In one year, a

An LNG carrier transports liquefied natural gas.

Methane is the fuel burned in a gas cooktop.

Two gases found in automobile exhaust are carbon dioxide and nitrogen oxide.

Port is a type of fortified wine that is produced in Portugal.

When the glucose in grapes is fermented, ethanol is produced.

COMBINING IDEAS from Chapters 11 to 14

M14_TIMB8119_06_SE_C14.indd 473 11/27/18 12:19 PM

474 CHAPTER 14 Acids and Bases

c. How will the equilibrium shift if the mixture is placed in a 5.00-L container with no change in temperature?

d. If a 5.00-L container has an equilibrium mixture of 0.300 mol of H2 and 2.50 mol of H2S, what is the [S2] if temperature does not change?

CI.26 A saturated solution of silver hydroxide has a pH of 10.15. (7.2, 7.3, 13.2, 13.6)

a. Write the solubility product expression for silver hydroxide.

b. Calculate the numerical value of Ksp for silver hydroxide. c. How many grams of silver hydroxide will dissolve in

2.0 L of water?

CI.27 A metal M with a mass of 0.420 g completely reacts with 34.8 mL of a 0.520 M HCl solution to form H2 gas and aqueous MCl3. (7.2, 7.3, 8.2, 9.2, 11.7)

a. Write a balanced chemical equation for the reaction of the metal M(s) and HCl(aq).

b. What volume, in milliliters, of H2 at 720. mmHg and 24 °C is produced?

c. How many moles of metal M reacted? d. Using your results from part c, determine the molar mass

and name of metal M. e. Write the balanced chemical equation for the reaction.

CI.28 In a teaspoon (5.0 mL) of a liquid antacid, there are 400. mg of Mg(OH)2 and 400. mg of Al(OH)3. A 0.080 M HCl solution, which is similar to stomach acid, is used to neutralize 5.0 mL of the liquid antacid. (12.4, 14.6, 14.7, 14.8)

CI.29 A KOH solution is prepared by dissolving 8.57 g of KOH in enough water to make 850. mL of KOH solution. (12.4, 14.6, 14.7, 14.8)

a. What is the molarity of the KOH solution? b. What is the [H3O

+] and pH of the KOH solution? c. Write the balanced chemical equation for the

neutralization of KOH by H2SO4. d. How many milliliters of a 0.250 M H2SO4 solution is

required to neutralize 10.0 mL of the KOH solution?

CI.30 A solution of HCl is prepared by diluting 15.0 mL of a 12.0 M HCl solution with enough water to make 750. mL of HCl solution. (12.4, 14.6, 14.7, 14.8)

a. What is the molarity of the HCl solution? b. What is the [H3O

+] and pH of the HCl solution? c. Write the balanced chemical equation for the reaction of

HCl and MgCO3. d. How many milliliters of the diluted HCl solution is

required to completely react with 350. mg of MgCO3?

CI.31 A volume of 200.0 mL of a carbonic acid buffer for blood plasma is prepared that contains 0.403 g of NaHCO3 and 0.0149 g of H2CO3. At body temperature (37 °C), the Ka of carbonic acid is 7.9 * 10-7. (7.2, 7.3, 8.4, 14.9)

H2CO3(aq) + H2O(l) vh HCO3 -(aq) + H3O+(aq)

a. What is the [H2CO3]? b. What is the [HCO3

- ]? c. What is the [H3O

+ ]? d. What is the pH of the buffer? e. Write a balanced chemical equation that shows how this

buffer neutralizes added acid. f. Write a balanced chemical equation that shows how this

buffer neutralizes added base.

CI.32 In the kidneys, the ammonia buffer system buffers high H3O +.

Ammonia, which is produced in renal tubules from amino acids, combines with H+ to be excreted as NH4Cl. At body temperature (37 °C), the Ka = 5.6 * 10-10. A buffer solution with a volume of 125 mL contains 3.34 g of NH4Cl and 0.0151 g of NH3. (7.2, 7.3, 8.4, 14.9)

NH4 +(aq) + H2O(l) vh NH3(aq) + H3O+(aq)

a. What is the [NH4 +]?

b. What is the [NH3]? c. What is the [H3O

+]? d. What is the pH of the buffer? e. Write a balanced chemical equation that shows how this

buffer neutralizes added acid. f. Write a balanced chemical equation that shows how this

buffer neutralizes added base.

When a metal reacts with a strong acid, bubbles of hydrogen gas form.

An antacid neutralizes stomach acid and raises the pH.

a. Write the equation for the neutralization of HCl and Mg(OH)2.

b. Write the equation for the neutralization of HCl and Al(OH)3.

c. What is the pH of the HCl solution? d. How many milliliters of the HCl solution are needed to

neutralize the Mg(OH)2? e. How many milliliters of the HCl solution are needed to

neutralize the Al(OH)3?

M14_TIMB8119_06_SE_C14.indd 474 11/27/18 12:19 PM

Combining Ideas from Chapters 11 to 14 475

ANSWERS

CI.27 a. 6HCl(aq) + 2M(s) h 3H2(g) + 2MCl3(aq) b. 233 mL of H2 c. 6.03 * 10-3 mol of M d. 69.7 g/mol; gallium e. 6HCl(aq) + 2Ga(s) h 3H2(g) + 2GaCl3(aq)

CI.29 a. 0.180 M b. [H3O

+] = 5.56 * 10-14 M; pH = 13.255 c. H2SO4(aq) + 2KOH(aq) h 2H2O(l) + K2SO4(aq) d. 3.60 mL

CI.31 a. 1.20 * 10-3 M b. 0.0240 M c. 4.0 * 10-8 M d. 7.40 e. HCO3

-(aq) + H3O+(aq) h H2CO3(aq) + H2O(l) f. H2CO3(aq) + OH-(aq) h HCO -3 (aq) + H2O(l)

CI.21 a.

H H orC H

H H HC

H

H

b. 1.2 * 107 kg of LNG (methane) c. 1.7 * 1010 L of LNG (methane) d. CH4(g) + 2O2(g) h

∆ CO2(g) + 2H2O(g) + 883 kJ

e. 4.8 * 107 kg of O2 f. 6.6 * 1011 kJ

CI.23 a. CS2(g) + 3O2(g) h ∆

CO2(g) + 2SO2(g) b. 13.8 g of CO2 c. 37 mmHg d. 2370 mmHg

CI.25 a. Kc = 248 b. If H2 is added, the equilibrium will shift in the direction of

the products. c. If the volume decreases, the equilibrium will shift in the

direction of the products. d. [S2] = 0.280 M

M14_TIMB8119_06_SE_C14.indd 475 11/27/18 12:19 PM

476

After Kimberly’s teeth are cleaned and whitened, her dentist

recommends she use a toothpaste to strengthen her teeth. You can see more details about the whitening process in the UPDATE Whitening Kimberly’s Teeth, page 499, and learn about compounds in teeth and toothpaste.

UPDATE Whitening Kimberly’s Teeth

Kimberly’s teeth are badly stained by food, coffee, and drinks that form a layer on top of the enamel of her teeth. Kimberly’s dentist, Jane, removes some of this layer by scraping her teeth, then brushes Kimberly’s teeth with an abrasive toothpaste. After that, Jane begins the process of whitening Kimberly’s teeth. She explains to Kimberly that she uses a gel containing 35% hydrogen peroxide (H2O2) that penetrates into the enamel of the tooth, where it undergoes a chemical reaction that whitens the teeth. The chemical reaction is an oxidation–reduction reaction where one chemical (hydrogen peroxide) is reduced and the other chemical (the coffee stain) is oxidized. During the oxidation, the coffee stains on the teeth become lighter or colorless, and therefore, the teeth are whiter.

CAREER

Dentist A dentist assesses and maintains the health of the teeth, gums, mouth, and bones of the jaw. A dentist examines X-rays to determine if there is disease or decay of the teeth and gums. Dentists remove decay and fill cavities, or they may extract teeth if necessary. Local anesthetics are often used for dental procedures. Dentists also educate patients on proper care of teeth and gums. Many dentists operate their own businesses with a staff that includes dental hygienists and a dental technician. A dentist may specialize in certain areas of dentistry. An orthodontist fits a patient’s teeth with wires to straighten teeth or correct a bite. A periodontist treats problems with soft tissues of the gums. Pediatric dentists treat children's teeth.

Oxidation and Reduction 15

M15_TIMB8119_06_SE_C15.indd 476 11/26/18 6:43 PM

15.1 Oxidation and Reduction 477

Glasses with photochromic lenses darken when exposed to sunlight.

LOOKING AHEAD

15.1 Oxidation and Reduction 477

15.2 Balancing Oxidation– Reduction Equations Using Half-Reactions 483

15.3 Electrical Energy from Oxidation–Reduction Reactions 488

15.4 Oxidation–Reduction Reactions That Require Electrical Energy 497

REVIEW Using Positive and Negative

Numbers in Calculations (1.4)

Writing Positive and Negative Ions (6.1)

Writing Ionic Formulas (6.2)

Balancing a Chemical Equation (8.2)

Identifying Oxidized and Reduced Substances (8.4)

15.1 Oxidation and Reduction LEARNING GOAL Identify a reaction as an oxidation or a reduction. Assign and use oxidation numbers to identify elements that are oxidized or reduced.

Oxidation and reduction are types of chemical reactions that are prevalent in our everyday lives. When we use natural gas to heat our home, turn on a computer, start a car, or eat food, we are using energy from the processes of oxidation and reduction.

In all of these processes, electrons are transferred during oxidation and reduction. In an oxidation reaction, a reactant loses one or more electrons. We say the substance is oxidized. In a reduction reaction, a reactant gains one or more electrons. We say the substance is reduced. When a substance undergoes oxidation, the electrons it loses must be gained by another substance, which is reduced. In an oxidation–reduction reaction, processes of oxidation and reduction must occur together. Sometimes we shorten this to redox reactions.

Writing Oxidation and Reduction Reactions Perhaps you have seen glasses with photochromic lenses that darken when exposed to UV light from the Sun. Once the wearer returns indoors, the lenses become clear again. The change from clear to dark and back to clear is the result of oxidation and reduction reac- tions. We can use the compounds in the photochromic lenses to illustrate some oxidation and reduction reactions.

To make the lenses photosensitive, silver chloride (AgCl) and copper(I) chloride (CuCl) are embedded in the lens material. Away from sunlight, visible light goes through the clear lens. However, when exposed to sunlight, there is an oxidation–reduction reaction when the UV light is absorbed by AgCl, and Cl atoms and Ag atoms are produced. It is the silver atoms that are black and darken the lenses. This redox reaction is written:

AgCl h Ag + Cl or

Ag+ + Cl- h Ag + Cl Oxidation–reduction reaction

The chloride (Cl-) ions lose electrons; they are oxidized. The silver (Ag+) ions gain electrons; they are reduced.

Half-Reactions To identify the oxidation and the reduction reactions, we write each of the reactants with its product as a half-reaction. The half-reaction with Cl- is an oxidation because Cl- loses an electron when it forms Cl. The half-reaction with Ag+ is a reduction because the Ag+ gains an electron when it forms Ag.

Cl- h Cl + e- Oxidation Loss of electron

Ag+ + e- h Ag Reduction Gain of electron Lenses darken

To prevent the lenses from clearing immediately, some Cu+ ions are added to the lenses. These Cu+ ions convert the Cl atoms produced from the darkening process to Cl- ions.

Cu+ + Cl h Cu2+ + Cl-

Cu+ h Cu2+ + e- Oxidation

Cl + e- h Cl- Reduction

When the wearer of the lenses moves out of sunlight, the reactions take place in the opposite direction. The Cu2+ ions combine with the silver atoms to re-form Cu+ and Ag+, which causes the lens to become clear again.

Cu2+ + Ag h Cu+ + Ag+

Ag h Ag+ + e- Oxidation Lenses clear

Cu2+ + e- h Cu+ Reduction

ENGAGE 15.1 How can you identify a half-reaction that is a reduction?

M15_TIMB8119_06_SE_C15.indd 477 11/26/18 6:43 PM

478 CHAPTER 15 Oxidation and Reduction

PRACTICE PROBLEMS Try Practice Problems 15.1 and 15.2

ENGAGE 15.2 Why is the oxidation number 0 assigned to Mg and to Ba, whereas the oxidation number + 2 is assigned to Mg2+ and to Ba2+?

We can now look at how these rules are used to assign oxidation numbers. For each formula, the oxidation numbers are written below the symbols of the elements (see TABLE 15.2).

TABLE 15.1 Rules for Assigning Oxidation Numbers 1. The sum of the oxidation numbers in a molecule is zero (0), or for a polyatomic ion is

equal to its charge.

2. The oxidation number of an element (monatomic or diatomic) is zero (0).

3. The oxidation number of a monatomic ion is equal to its charge.

4. In compounds, the oxidation number of Group 1A (1) metals is + 1, and that of Group 2A (2) metals is + 2.

5. In compounds, the oxidation number of fluorine is - 1. Other nonmetals in Group 7A (17) are - 1 except when combined with oxygen or fluorine.

6. In compounds, the oxidation number of oxygen is - 2 except in OF2, where O is + 1; in H2O2 and other peroxides, O is - 1.

7. In compounds with nonmetals, the oxidation number of hydrogen is + 1; in compounds with metals, the oxidation number of hydrogen is - 1.

SAMPLE PROBLEM 15.1 Identifying Oxidation and Reduction Reactions

TRY IT FIRST

For each of the following, identify the reaction as an oxidation or a reduction:

a. Cu2+(aq) + 2 e- h Cu(s) b. Fe(s) h Fe3+(aq) + 3 e-

c. Cr2+(aq) h Cr3+(aq) + e-

d. Li+(aq) + e- h Li(s)

SOLUTION

a. Since Cu2+ gained electrons, this is a reduction. b. Since Fe lost electrons, this is an oxidation. c. Since Cr2+ lost an electron, this is an oxidation. d. Since Li+ gained an electron, this is a reduction.

SELF TEST 15.1

Identify each of the following reactions as an oxidation or a reduction:

a. Na(s) h Na+(aq) + e- b. Zn2+(aq) + 2 e- h Zn(s)

ANSWER

a. oxidation b. reduction

Oxidation Numbers In more complex oxidation–reduction reactions, the identification of the substances oxidized and reduced is not always obvious. To help identify the atoms or ions that are oxidized or reduced, we assign values called oxidation numbers (sometimes called oxidation states) to the elements of the reactants and products. It is important to recognize that oxidation numbers do not always represent actual charges, but they help us identify loss or gain of electrons.

Rules for Assigning Oxidation Numbers The rules for assigning oxidation numbers to the atoms or ions in the reactants and products are given in TABLE 15.1.

CORE CHEMISTRY SKILL Assigning Oxidation Numbers

M15_TIMB8119_06_SE_C15.indd 478 11/26/18 6:43 PM

15.1 Oxidation and Reduction 479

TABLE 15.2 Examples of Using Rules to Assign Oxidation Numbers Formula Oxidation Numbers Explanation

Br2 Br2 0

Each Br atom in diatomic bromine has an oxidation number of 0 (Rule 2).

Ba2+ Ba2+ + 2

The oxidation number of a monatomic ion is equal to its charge (Rule 3).

CO2 CO2 + 4 - 2

In compounds, O has an oxidation number of - 2 (Rule 6). Because CO2 is neutral, the oxidation number of C is calculated as + 4 (Rule 1).

C + 2O = 0 C + 2( - 2) = 0 C = + 4

Al2O3 Al2O3 + 3 - 2

In compounds, the oxidation number of O is - 2 (Rule 6). For Al2O3 (neutral), the oxidation number of Al is calculated as + 3 (Rule 1).

2Al + 3O = 0 2Al + 3( - 2) = 0 2Al = + 6 Al = + 3

HClO3 HClO3 + 1 + 5 - 2

In compounds, the oxidation number of H is + 1 (Rule 7), and O is - 2 (Rule 6). For HClO3 (neutral), the oxidation number of Cl is calculated as + 5 (Rule 1).

H + Cl + 3O = 0 ( + 1) + Cl + 3( - 2) = 0 Cl - 5 = 0 Cl = + 5

SO4 2- SO4

2-

+ 6 - 2 The oxidation number of O is - 2 (Rule 6). For SO4 2- ( - 2

charge), the oxidation number of S is calculated as + 6 (Rule 1).

S + 4O = - 2 S + 4( - 2) = - 2 S = + 6

CH2O CH2O 0 + 1 - 2

In compounds, the oxidation number of H is + 1 (Rule 7), and O is - 2 (Rule 6). For CH2O (neutral), the oxidation number of C is calculated as 0 (Rule 1).

C + 2H + O = 0 C + 2( + 1) + ( - 2) = 0 C = 0

ENGAGE 15.3 Why is the S in the formula H2SO3 assigned an oxidation number of + 4?

SAMPLE PROBLEM 15.2 Assigning Oxidation Numbers

TRY IT FIRST

Assign oxidation numbers to the elements in each of the following:

a. NCl3 b. CO3 2-

SOLUTION

a. NCl3: The oxidation number of Cl is - 1 (Rule 5). For NCl3 (neutral), the sum of the oxidation numbers of N and 3Cl must be equal to zero (Rule 1). Thus, the oxidation number of N is calculated as + 3.

N + 3Cl = 0 N + 3( - 1) = 0 N = + 3

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480 CHAPTER 15 Oxidation and Reduction

The oxidation numbers are written as

NCl3 + 3 - 1

b. CO3 2-: The oxidation number of O is - 2 (Rule 6). For CO3 2-, the sum of the oxidation

numbers is equal to - 2 (Rule 1). The oxidation number of C is calculated as + 4. C + 3O = - 2 C + 3( - 2) = - 2 C = + 4 The oxidation numbers are written as CO3

2-

+ 4 - 2

SELF TEST 15.2

Assign oxidation numbers to the elements in each of the following:

a. H3PO4 b. MnO4 -

ANSWER

a. Because H has an oxidation number of + 1, and O is - 2, P has an oxidation number of + 5 to maintain a neutral charge.

H3PO4 + 1 + 5 - 2

b. Because O has an oxidation number of - 2, Mn has an oxidation number of + 7 to give an overall charge of - 1. MnO4

-

+ 7 - 2

PRACTICE PROBLEMS Try Practice Problems 15.3 to 15.14

Using Oxidation Numbers to Identify Oxidation–Reduction Oxidation numbers can be used to identify the elements that are oxidized and the elements that are reduced in a reaction. In oxidation, the loss of electrons increases the oxidation number so that it is higher (more positive) in the product than in the reactant. In reduction, the gain of electrons decreases the oxidation number so that it is lower (more negative) in the product than in the reactant.

CORE CHEMISTRY SKILL Using Oxidation Numbers

An oxidation reaction occurs when the charge becomes more positive. A reduction reaction occurs when the charge becomes more negative.

Reduction: oxidation number decreases

-7 -6 -5 -4 -3 -2 -1 0 +1 +2 +3 +4 +5 +6 +7

Oxidation: oxidation number increases

ENGAGE 15.4 Why is the reaction of the hydrogen atoms in H2 to hydrogen ions (2H

+) called an oxidation reaction?

SAMPLE PROBLEM 15.3 Using Oxidation Numbers to Determine Oxidation and Reduction

TRY IT FIRST

Identify the element that is oxidized and the element that is reduced in the following equation:

CO2(g) + H2(g) h CO(g) + H2O(g)

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15.1 Oxidation and Reduction 481

Oxidation–Reduction Terminology

Oxidation Loss of electrons Oxidation number increases Reducing agent

Reduction Gain of electrons Oxidation number decreases Oxidizing agent

SOLUTION

STEP 1 Assign oxidation numbers to each element. In H2, the oxidation number of H is 0. In H2O, the oxidation number of H is + 1. In CO2, CO, and H2O, the oxida- tion number of O is - 2. Using the oxidation number - 2 for O, the oxidation number of C would be + 4 in CO2, and + 2 in CO.

H2(g)+CO2(g) CO(g) H2O(g)+ Oxidation numbers-2 0+4 -2 -2+1+2

Oxidation number of H increases; oxidation

Oxidation number of C decreases; reduction

SELF TEST 15.3

a. In the following equation, which reactant is oxidized?

2Al(s) + 3Sn2+(aq) h 2Al3+(aq) + 3Sn(s)

b. In the following equation, which reactant is reduced?

Al(s) + 3Ag +(aq) h Al3+(aq) + 3Ag(s)

ANSWER

a. The oxidation number of Al increases from 0 to + 3; Al is oxidized. b. The oxidation number of Ag + decreases from + 1 to 0; Ag + is reduced.

Oxidizing and Reducing Agents We have seen that an oxidation reaction must always be accompanied by a reduction reaction. The substance that loses electrons is oxidized, and the substance that gains electrons is reduced. For example, Zn is oxidized to Zn2+ by losing 2 electrons and Cl2 reduced to 2Cl

- by gaining 2 electrons.

Zn(s) + Cl2(g) h ZnCl2(s) In an oxidation–reduction reaction, the substance that is oxidized is called the

reducing agent because it provides electrons for reduction. The substance that is reduced is called the oxidizing agent because it accepts electrons from oxidation. Because Zn is oxidized in this reaction, it is the reducing agent. In the same reaction Cl2 is reduced, which makes it the oxidizing agent.

Zn(s) h Zn2+(aq) + 2 e- Zn is oxidized; Zn is the reducing agent. Cl2(g) + 2 e- h 2Cl-(aq) Cl (in Cl2) is reduced; Cl2 is the oxidizing agent. The terms we use to describe oxidation and reduction are listed in the margin.

ENGAGE 15.5 In the following oxidation– reduction reaction, why is Zn the reducing agent, and Co2+ the oxidizing agent?

Zn(s) + CoBr2(aq) h ZnBr2(aq) + Co(s)

CORE CHEMISTRY SKILL Identifying Oxidizing and Reducing

Agents

SAMPLE PROBLEM 15.4 Identifying Oxidizing and Reducing Agents

TRY IT FIRST

Identify the oxidizing agent and the reducing agent for the reaction of lead(II) oxide and carbon monoxide.

PbO(s) + CO(g) h Pb(s) + CO2(g)

H2(g)+CO2(g) CO(g) H2O(g)+ Oxidation numbers-2 0+4 -2 -2+1+2

STEP 2 Identify the increase in oxidation number as oxidation; the decrease as reduction. H is oxidized because its oxidation number increases from 0 to + 1. C is reduced because its oxidation number decreases from + 4 to + 2.

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482 CHAPTER 15 Oxidation and Reduction

Lead oxides, once used in paint, are now banned due to the toxicity of lead.

PRACTICE PROBLEMS

15.1 Oxidation and Reduction

15.1 Identify each of the following as an oxidation or a reduction reaction:

a. Al3+(aq) + 3 e- h Al(s) b. Ca(s) h Ca2+(aq) + 2 e- c. Fe3+(aq) + e- h Fe2+(aq)

15.2 Identify each of the following as an oxidation or a reduction reaction:

a. Ni2+(aq) h Ni3+(aq) + e-

b. K+(aq) + e- h K(s) c. Br2(l) + 2 e- h 2Br-(aq)

15.3 In each of the following reactions, identify the reactant that is oxidized and the reactant that is reduced:

a. 4Al(s) + 3O2(g) h 2Al2O3(s) b. Zn(s) + 2H+(aq) h Zn2+(aq) + H2(g) c. F2(g) + 2Br-(aq) h 2F -(aq) + Br2(l)

15.4 In each of the following reactions, identify the reactant that is oxidized and the reactant that is reduced:

a. 2Ag+(aq) + Zn(s) h 2Ag(s) + Zn2+(aq) b. Ca(s) + S(s) h CaS(s) c. Sn2+(aq) + 2Cr3+(aq) h Sn4+(aq) + 2Cr2+(aq)

PRACTICE PROBLEMS Try Practice Problems 15.15 to 15.20

CO(g)+PbO(s) Pb(s) CO2(g) Oxidation numbers-2 0-2 -2+4+2 +2

+

STEP 2 Identify the increase in oxidation number as oxidation; the decrease as reduction. The C in CO is oxidized because the oxidation number of C increases from + 2 to + 4. The Pb in PbO is reduced because the oxidation number of Pb decreases from + 2 to 0.

Oxidation

Reduction

CO(g)+PbO(s) Pb(s) CO2(g) Oxidation numbers-2 0-2 -2+4+2 +2

+

STEP 3 Identify the oxidized substance as the reducing agent and the reduced substance as the oxidizing agent. The compound CO is oxidized; it is the reducing agent. The compound PbO is reduced; it is the oxidizing agent.

SELF TEST 15.4

Identify the oxidizing agent and the reducing agent in each of the following reactions:

a. 2Al(s) + 3CuO(s) h Al2O3(s) + 3Cu(s) b. 2HCl(aq) + Mg(s) h MgCl2(aq) + H2(g)

ANSWER

a. CuO is the oxidizing agent; Al is the reducing agent. b. HCl is the oxidizing agent; Mg is the reducing agent.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

equation oxidizing agent, reducing agent

oxidation numbers

STEP 1 Assign oxidation numbers to each element. Using the oxidation number - 2 for O, the oxidation number of Pb in PbO is + 2, the oxidation number of C in CO is + 2, and the C in CO2 would have an oxidation number of + 4. The element Pb has an oxidation number of 0.

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15.2 Balancing Oxidation–Reduction Equations Using Half-Reactions 483

15.5 Assign oxidation numbers to each of the following: a. Cu b. F2 c. Fe

2+ d. Cl-

15.6 Assign oxidation numbers to each of the following: a. Al b. Al3+ c. F - d. N2 15.7 Assign oxidation numbers to all the elements in each of the

following: a. KCl b. MnO2 c. NO d. Mn2O3 15.8 Assign oxidation numbers to all the elements in each of the

following: a. H2S b. NO2 c. CCl4 d. PCl3 15.9 Assign oxidation numbers to all the elements in each of the

following: a. Li3PO4 b. SO3

2- c. Cr2S3 d. NO3 -

15.10 Assign oxidation numbers to all the elements in each of the following:

a. C2H3O2 - b. AlCl3 c. NH4

+ d. HBrO4 15.11 Assign oxidation numbers to all the elements in each of the

following: a. HSO4

- b. H3PO3 c. Cr2O7 2- d. Na2CO3

15.12 Assign oxidation numbers to all the elements in each of the following:

a. N2O b. LiOH c. SbO2 - d. IO4

-

15.13 What is the oxidation number of the specified element in each of the following?

a. N in HNO3 b. C in C3H6 c. P in K3PO4 d. Cr in CrO4

2-

15.14 What is the oxidation number of the specified element in each of the following?

a. C in BaCO3 b. Fe in FeBr3 c. Cl in ClF4

- d. S in S2O3 2-

15.15 Indicate whether each of the following describes the oxidizing agent or the reducing agent in an oxidation–reduction reaction:

a. the substance that is oxidized b. the substance that gains electrons

15.16 Indicate whether each of the following describes the oxidizing agent or the reducing agent in an oxidation–reduction reaction:

a. the substance that is reduced b. the substance that loses electrons

15.17 For each of the following reactions, identify the substance that is oxidized, the substance that is reduced, the oxidizing agent, and the reducing agent:

a. 2Li(s) + Cl2(g) h 2LiCl(s) b. Cl2(g) + 2NaBr(aq) h 2NaCl(aq) + Br2(l) c. 2Pb(s) + O2(g) h 2PbO(s)

d. Al(s) + 3Ag +(aq) h Al3+(aq) + 3Ag(s) 15.18 For each of the following reactions, identify the substance that

is oxidized, the substance that is reduced, the oxidizing agent, and the reducing agent:

a. 2Li(s) + F2(g) h 2LiF(s) b. Cl2(g) + 2KI(aq) h 2KCl(aq) + I2(s)

c. Zn(s) + Ni2+(aq) h Zn2+(aq) + Ni(s) d. Fe(s) + CuSO4(aq) h FeSO4(aq) + Cu(s) 15.19 For each of the following reactions, identify the substance that

is oxidized, the substance that is reduced, the oxidizing agent, and the reducing agent:

a. 2NiS(s) + 3O2(g) h 2NiO(s) + 2SO2(g) b. Sn2+(aq) + 2Fe3+(aq) h Sn4+(aq) + 2Fe2+(aq)

c. CH4(g) + 2O2(g) h CO2(g) + 2H2O(g) d. 2Cr2O3(s) + 3Si(s) h 4Cr(s) + 3SiO2(s) 15.20 For each of the following reactions, identify the substance that

is oxidized, the substance that is reduced, the oxidizing agent, and the reducing agent:

a. 2HgO(s) h 2Hg(l) + O2(g) b. Zn(s) + 2HCl(aq) h ZnCl2(aq) + H2(g)

c. 2Na(s) + 2H2O(l) h 2Na+ (aq) + 2OH-(aq) + H2(g) d. 14H+(aq) + 6Fe2+(aq) + Cr2O7 2-(aq) h 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l)

CORE CHEMISTRY SKILL Using Half-Reactions to Balance

Redox Equations

15.2 Balancing Oxidation–Reduction Equations Using Half-Reactions LEARNING GOAL Balance oxidation–reduction equations using the half-reaction method.

In the half-reaction method for balancing equations, an oxidation–reduction reaction is writ- ten as two half-reactions. The half-reaction method uses ionic charges and electrons to balance each half-reaction. Oxidation numbers are not used. The loss of electrons by one reactant is used to identify the oxidized substance, and the gain of electrons by another reactant is used to identify the reduced substance. Once the loss and gain of electrons are equalized for the half- reactions, they are combined to obtain the overall balanced equation. The half-reaction method is typically used to balance equations that are written as ionic equations. Let us consider the reaction between aluminum metal and a solution of Cu2+ as shown in Sample Problem 15.5.

SAMPLE PROBLEM 15.5 Using Half-Reactions to Balance Equations

TRY IT FIRST

Use half-reactions to balance the following equation:

Al(s) + Cu2+(aq) h Al3+(aq) + Cu(s)

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484 CHAPTER 15 Oxidation and Reduction

ENGAGE 15.6 How do we know that the final oxidation–reduction equation is balanced?

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

Equation: Al(s) + Cu2+(aq) h Al3+(aq) + Cu(s)

balanced equation

half-reactions, electron balance

STEP 1 Write two half-reactions for the equation.

Al(s) h Al3+(aq) Cu2+(aq) h Cu(s)

STEP 2 For each half-reaction, balance the elements other than H and O. In these half-reactions, the Al and Cu are already balanced.

Al(s) h Al3+(aq) Cu2+(aq) h Cu(s)

STEP 3 Balance each half-reaction for charge by adding electrons. For the aluminum half-reaction, we need to add three electrons on the product side to balance the charge. With a loss of electrons, this is an oxidation.

Al(s) h Al3+(aq) + 3 e− Oxidation 0 charge = 0 charge For the Cu2+ half-reaction, we need to add two electrons on the reactant side to balance the charge. With a gain of electrons, this is a reduction. Cu2+(aq) + 2 e− h Cu(s) Reduction

0 charge = 0 charge

STEP 4 Multiply each half-reaction by factors that equalize the loss and gain of electrons. To obtain the same number of electrons in each half-reaction, we need to multiply the oxidation half-reaction by 2 and the reduction half-reaction by 3.

2 * [Al(s) h Al3+(aq) + 3 e−] 2Al(s) h 2Al3+(aq) + 6 e− 6 e- lost

3 * [Cu2+(aq) + 2 e− h Cu(s)] 3Cu2+(aq) + 6 e− h 3Cu(s) 6 e- gained

STEP 5 Add half-reactions and cancel electrons, and any identical ions or molecules. Check the balance of atoms and charge.

2Al(s) h 2Al3+(aq) + 6 e−

3Cu2+(aq) + 6 e− h 3Cu(s)

2Al(s) + 3Cu2+(aq) + 6 e − h 2Al3+(aq) + 6 e − + 3Cu(s)

Final balanced equation:

2Al(s) + 3Cu2+(aq) h 2Al3+(aq) + 3Cu(s) Check the balance of atoms and charge.

Reactants Products 2Al = 2Al 3Cu = 3Cu

Charge: 6+ = 6+

SELF TEST 15.5

Use the half-reaction method to balance each of the following equations:

a. Zn(s) + Fe3+(aq) h Zn2+(aq) + Fe2+(aq) b. Ag +(aq) + Sn(s) h Ag(s) + Sn4+(aq)

ANSWER

a. Zn(s) + 2Fe3+(aq) h Zn2+(aq) + 2Fe2+(aq) b. 4Ag +(aq) + Sn(s) h 4Ag(s) + Sn4+(aq)

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15.2 Balancing Oxidation–Reduction Equations Using Half-Reactions 485

Balancing Oxidation–Reduction Equations in Acidic Solution When we use the half-reaction method for balancing equations for reactions in acidic solution, we include balancing O by adding H2O and balancing H by adding H

+ as shown in Sample Problem 15.6.

6I-(aq) h 3I2(s) + 6 e−

6 e− + 14H+(aq) + Cr2O7 2-(aq) h 2Cr3+(aq) + 7H2O(l)

6 e− + 14H+(aq) + Cr2O7 2-(aq) + 6I-(aq) h 2Cr3+(aq) + 3I2(s) + 7H2O(l) + 6 e−

SAMPLE PROBLEM 15.6 Using Half-Reactions to Balance Equations in Acidic Solution

TRY IT FIRST

Use half-reactions to balance the following equation for a reaction that takes place in acidic solution:

I-(aq) + Cr2O7 2-(aq) h I2(s) + Cr3+(aq)

SOLUTION

STEP 1 Write two half-reactions for the equation.

I-(aq) h I2(s)

Cr2O7 2-(aq) h Cr3+(aq)

STEP 2 For each half-reaction, balance the elements other than H and O. Balance O by adding H2O, and H by adding H

+. The two I atoms in I2 are balanced with a coefficient of 2 for I-.

2I-(aq) h I2(s)

The two Cr atoms are balanced with a coefficient of 2 for Cr3+.

Cr2O7 2-(aq) h 2Cr3+(aq)

Now balance O by adding H2O to the product side.

Cr2O7 2-(aq) h 2Cr3+(aq) + 7H2O(l) H2O balances O

Balance H by adding H+ to the reactant side.

14H+(aq) + Cr2O7 2-(aq) h 2Cr3+(aq) + 7H2O(l) H+ balances H

STEP 3 Balance each half-reaction for charge by adding electrons. A charge of - 2 is balanced with two electrons on the product side. 2I-(aq) h I2(s) + 2 e− Oxidation

- 2 = - 2

A charge of + 6 is obtained on the reactant side by adding six electrons to the reactant side.

6 e− + 14H+(aq) + Cr2O7 2-(aq) h 2Cr3+(aq) + 7H2O(l) Reduction + 6 = + 6

STEP 4 Multiply each half-reaction by factors that equalize the loss and gain of electrons. The half-reaction with I is multiplied by 3 to equal the gain of 6 e- by the Cr half-reaction.

3 * [2I-(aq) h I2(s) + 2 e−] 6I-(aq) h 3I2(s) + 6 e− 6 e- lost

6 e− + 14H+(aq) + Cr2O7 2-(aq) h 2Cr3+(aq) + 7H2O(l) 6 e- gained

STEP 5 Add half-reactions and cancel electrons, and any identical ions or molecules. Check the balance of atoms and charge.

A dichromate solution (yellow) and an iodide solution (colorless) form a brown solution of Cr3+ and iodine (I2).

ENGAGE 15.7 Why are 7H2O added to the product side of this half-reaction?

ENGAGE 15.8 Why are 6 e- added to the reactant side of this half-reaction?

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486 CHAPTER 15 Oxidation and Reduction

Final balanced equation:

14H+(aq) + Cr2O7 2-(aq) + 6I-(aq) h 2Cr3+(aq) + 3I2(s) + 7H2O(l)

Check the balance of atoms and charge.

Reactants Products 6I = 6I 2Cr = 2Cr 14H = 14H 7O = 7O

Charge: 6+ = 6+

SELF TEST 15.6

Use half-reactions to balance each of the following equations in acidic solution:

a. Fe2+(aq) + IO3 -(aq) h Fe3+(aq) + I2(s) b. Mn2+(aq) + BiO3 -(aq) h MnO4 -(aq) + Bi2+(aq)

ANSWER

a. 12H+(aq) + 10Fe2+(aq) + 2IO3 -(aq) h 10Fe3+(aq) + I2(s) + 6H2O(l) b. 6H+(aq) + 3Mn2+(aq) + 5BiO3 -(aq) h 3MnO4 -(aq) + 5Bi2+(aq) + 3H2O(l)

PRACTICE PROBLEMS Try Practice Problems 15.21 to 15.24

Balancing Oxidation–Reduction Equations in Basic Solution An oxidation–reduction reaction can also take place in basic solution. In that case, we use the same half-reaction method, but once we have the balanced equation, we will neutralize the H+ with OH- to form water. The H+ is neutralized by adding OH- to both sides of the equation to form H2O as shown in Sample Problem 15.7.

SAMPLE PROBLEM 15.7 Using Half-Reactions to Balance Equations in Basic Solution

TRY IT FIRST

Use half-reactions to balance the following equation that takes place in basic solution:

Fe2+(aq) + MnO4 -(aq) h Fe3+(aq) + MnO2(s)

SOLUTION

STEP 1 Write two half-reactions for the equation.

Fe2+(aq) h Fe3+(aq) MnO4

-(aq) h MnO2(s)

STEP 2 For each half-reaction, balance the elements other than H and O. Balance O by adding H2O, and H by adding H

+.

Fe2+(aq) h Fe3+(aq) MnO4

-(aq) h MnO2(s) + 2H2O(l) H2O balances O 4H+(aq) + MnO4 -(aq) h MnO2(s) + 2H2O(l) H+ balances H

STEP 3 Balance each half-reaction for charge by adding electrons.

Fe2+(aq) h Fe3+(aq) + e− Oxidation

+ 2 = + 2

3 e− + 4H+(aq) + MnO4 -(aq) h MnO2(s) + 2H2O(l) Reduction 0 charge = 0 charge

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15.2 Balancing Oxidation–Reduction Equations Using Half-Reactions 487

STEP 4 Multiply each half-reaction by factors that equalize the loss and gain of electrons. The half-reaction with Fe is multiplied by 3 to equal the gain of 3 e- by Mn.

3 * [Fe2+(aq) h Fe3+(aq) + e−]

3Fe2+(aq) h 3Fe3+(aq) + 3 e− 3 e- lost

3 e− + 4H+(aq) + MnO4 -(aq) h MnO2(s) + 2H2O(l) 3 e- gained

STEP 5 Add half-reactions and cancel electrons, and any identical ions or molecules. In base, add OH− to neutralize H+. Check the balance of atoms and charge.

3Fe2+(aq) h 3Fe3+(aq) + 3 e−

3 e− + 4H+(aq) + MnO4 -(aq) h MnO2(s) + 2H2O(l)

3 e− + 4H+(aq) + 3Fe2+(aq) + MnO4 -(aq) h 3Fe3+(aq) + MnO2(s) + 2H2O(l) + 3 e−

Final balanced equation:

4H+(aq) + 3Fe2+(aq) + MnO4 -(aq) h 3Fe3+(aq) + MnO2(s) + 2H2O(l)

To convert the equation to an oxidation–reduction reaction in basic solution, we neutralize H+ with OH- to form H2O. For this equation, we add 4OH

-(aq) to both sides of the equation.

4OH−(aq) + 4H+(aq) + 3Fe2+(aq) + MnO4 -(aq) h 3Fe3+(aq) + MnO2(s) + 2H2O(l) + 4OH−(aq)

Combining 4H+ and 4OH- gives 4H2O on the reactant side.

4H2O(l) + 3Fe2+(aq) + MnO4 -(aq) h 3Fe3+(aq) + MnO2(s) + 2H2O(l) + 4OH-(aq)

Canceling 2H2O on both the reactant and the product side gives the balanced equation in basic solution.

Final balanced equation in basic solution:

2H2O(l) + 3Fe2+(aq) + MnO4 -(aq) h 3Fe3+(aq) + MnO2(s) + 4OH-(aq)

Check the balance of atoms and charge.

Reactants Products 3Fe = 3Fe 1Mn = 1Mn 4H = 4H 6O = 6O

Charge: 5+ = 5+

SELF TEST 15.7

For the following equation in basic solution:

N2O(g) + ClO-(aq) h NO2 -(aq) + Cl-(aq) a. write the balanced half-reactions b. write the balanced equation

ANSWER

a. 3H2O(l) + N2O(g) h 2NO2 -(aq) + 6H+(aq) + 4 e-

2 e- + 2H+(aq) + ClO-(aq) h Cl-(aq) + H2O(l) b. N2O(g) + 2ClO-(aq) + 2OH-(aq) h 2Cl-(aq) + 2NO2 -(aq) + H2O(l)

PRACTICE PROBLEMS Try Practice Problems 15.25 and 15.26

ENGAGE 15.9 Why is 4OH- added to the product side of this equation?

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488 CHAPTER 15 Oxidation and Reduction

PRACTICE PROBLEMS

15.2 Balancing Oxidation–Reduction Equations Using Half-Reactions

15.21 Balance each of the following half-reactions in acidic solution: a. Sn2+(aq) h Sn4+(aq) b. Mn2+(aq) h MnO4

-(aq) c. NO2

-(aq) h NO3 -(aq)

d. ClO3 -(aq) h ClO2(aq)

15.22 Balance each of the following half-reactions in acidic solution: a. Cu(s) h Cu2+(aq) b. SO4

2-(aq) h SO3 2-(aq)

c. BrO3 -(aq) h Br-(aq)

d. IO3 -(aq) h I2(s)

15.23 Use the half-reaction method to balance each of the following in acidic solution:

a. Ag(s) + NO3 -(aq) h Ag+(aq) + NO2(g) b. NO3

-(aq) + S(s) h NO(g) + SO2(g) c. S2O3

2-(aq) + Cu2+(aq) h S4O6 2-(aq) + Cu(s)

15.24 Use the half-reaction method to balance each of the following in acidic solution:

a. Mn(s) + NO3 -(aq) h Mn2+(aq) + NO2(g) b. C2O4

2-(aq) + MnO4 -(aq) h CO2(g) + Mn2+(aq) c. ClO3

-(aq) + SO3 2-(aq) h Cl-(aq) + SO4 2-(aq) 15.25 Use the half-reaction method to balance each of the following

in basic solution: a. Fe(s) + CrO4 2-(aq) h Fe2O3(s) + Cr2O3(s) b. CN-(aq) + MnO4 -(aq) h CNO-(aq) + MnO2(s)

15.26 Use the half-reaction method to balance each of the following in basic solution:

a. Al(s) + ClO-(aq) h AlO2 -(aq) + Cl-(aq) b. Sn2+(aq) + IO4 -(aq) h Sn4+(aq) + I-(aq)

Zinc strip

Cu2+ solution

Cu(s)

Because Zn is above Cu on the activity series, a spontaneous oxidation–reduction reaction occurs.

15.3 Electrical Energy from Oxidation– Reduction Reactions LEARNING GOAL Use the activity series to determine if an oxidation–reduction reaction is spontaneous. Write the half-reactions that occur in a voltaic cell and the cell notation.

When we place a zinc metal strip in a solution of Cu2+, reddish-brown Cu metal accumulates on the Zn strip according to the following spontaneous reaction:

Zn(s) + Cu2+(aq) h Zn2+(aq) + Cu(s) Spontaneous

However, if we place a Cu metal strip in a Zn2+ solution, nothing will happen. The reaction does not occur spontaneously in the reverse direction because Cu does not lose electrons as easily as Zn.

We can determine the direction of a spontaneous reaction from the activity series, which ranks the metals and H2 in terms of how easily they lose electrons.

In the activity series, the metals that lose electrons most easily are placed at the top, and the metals that do not lose electrons easily are at the bottom. Thus the metals that are more easily oxidized are above the metals whose ions are more easily reduced (see TABLE 15.3). Active metals include K, Na, Ca, Mg, Al, Zn, Fe, and Sn. In single replacement reactions, the metal ion replaces the H in the acid. Metals listed below H2(g) will not react with H

+ from acids. According to the activity series, a metal will oxidize spontaneously when it is combined

with the reverse of the half-reaction for any metal below it on the list. We use the activity series to predict the direction of the spontaneous reaction. Suppose we have two beakers. In one, we place a Mg strip in a solution containing Ni2+ ions. In the other, we place a Ni strip in a solution containing Mg2+ ions. Looking at the activity series we see that the half- reaction for the oxidation of Mg is listed above that for Ni, which means that Mg is the more active metal and loses electrons more easily than Ni. Using the activity series table, we write these two half-reactions as follows:

Mg(s) h Mg2+(aq) + 2 e-

Ni(s) h Ni2+(aq) + 2 e-

The reaction that will be spontaneous is the oxidation of Mg combined with the reverse (reduction) of Ni2+.

Mg(s) h Mg2+(aq) + 2 e-

Ni2+(aq) + 2 e- h Ni(s)

CORE CHEMISTRY SKILL Identifying Spontaneous Reactions

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15.3 Electrical Energy from Oxidation– Reduction Reactions 489

ENGAGE 15.10 Why would the following oxidation– reduction reaction be spontaneous?

Fe(s) + Pb2+(aq) h Fe2+(aq) + Pb(s)

Therefore, we combine the half-reactions, which gives the following overall reaction that occurs spontaneously:

Mg(s) + Ni2+(aq) + 2 e- h Mg2+(aq) + Ni(s) + 2 e-

Mg(s) + Ni2+(aq) h Mg2+(aq) + Ni(s) Spontaneous However, a reaction between a Mg strip and a solution containing K+ ions will not

occur spontaneously. We determine this by looking at the two half-reactions needed.

Mg(s) h Mg2+(aq) + 2 e-

K+(aq) + e- h K(s) Because the half-reaction for the oxidation of K is above that for Mg, there will not be

a spontaneous reaction between Mg and K+.

Mg(s) + 2K+(aq) h Mg2+(aq) + 2K(s) Not spontaneous

SAMPLE PROBLEM 15.8 Predicting Spontaneous Reactions

TRY IT FIRST

Determine if the reaction for each of the following metals with an HCl (H+) solution is spontaneous:

a. Zn(s) + 2H+(aq) h Zn2+(aq) + H2(g) b. Cu(s) + 2H+(aq) h Cu2+(aq) + H2(g)

SOLUTION

a. Using the activity series (see Table 15.3), we see that Zn oxidizes more easily than H2. Thus the oxidation half-reaction for Zn, combined with the reverse half-reaction for H2, is spontaneous.

Zn(s) h Zn2+(aq) + 2 e-

2H+(aq) + 2 e- h H2(g) Zn(s) + 2H+(aq) h Zn2+(aq) + H2(g) Spontaneous

TABLE 15.3 Activity Series for Some Metals and Hydrogen Metal Ion

Li(s) h Li+(aq) + e-

K(s) h K+(aq) + e-

Ca(s) h Ca2+(aq) + 2 e-

Na(s) h Na+(aq) + e-

Mg(s) h Mg2+(aq) + 2 e-

Al(s) h Al3+(aq) + 3 e-

Zn(s) h Zn2+(aq) + 2 e-

Cr(s) h Cr3+(aq) + 3 e-

Fe(s) h Fe2+(aq) + 2 e-

Ni(s) h Ni2+(aq) + 2 e-

Sn(s) h Sn2+(aq) + 2 e-

Pb(s) h Pb2+(aq) + 2 e-

H2(g) h 2H +(aq) + 2 e-

Cu(s) h Cu2+(aq) + 2 e-

Ag(s) h Ag+(aq) + e-

Au(s) h Au3+(aq) + 3 e-

Most Reactive E

as e

of O

xi d

at io

n D

ec re

as es

Least Reactive

M15_TIMB8119_06_SE_C15.indd 489 11/26/18 6:43 PM

490 CHAPTER 15 Oxidation and Reduction

PRACTICE PROBLEMS Try Practice Problems 15.27 and 15.28

b. Using the activity series (see Table 15.3), we see that H2 oxidizes more easily than Cu. In this reaction, we would be combining an oxidation half-reaction with one that is above it in the activity series.

2H+(aq) + 2 e- h H2(g) Cu(s) h Cu2+(aq) + 2 e-

Cu(s) + 2H+(aq) h Cu2+(aq) + H2(g) Not spontaneous

SELF TEST 15.8

Determine if each of the following reactions is spontaneous:

a. 2Al(s) + 3Cu2+(aq) h 2Al3+(aq) + 3Cu(s) b. Fe(s) + Mg2+(aq) h Fe2+(aq) + Mg(s)

ANSWER

a. Yes, this reaction is spontaneous. b. No, this reaction is not spontaneous.

Voltaic Cells We can generate electrical energy from a spontaneous oxidation–reduction reaction by using a voltaic cell. A piece of zinc metal placed in a Cu2+ solution becomes coated with a rusty-brown coating of Cu, while the blue color (Cu2+) of the solution fades. The oxidation of the zinc metal provides electrons for the reduction of the Cu2+ ions. We can write the two half-reactions as

Zn(s) h Zn2+(aq) + 2 e- Oxidation Cu2+(aq) + 2 e- h Cu(s) Reduction

The overall reaction is

Zn(s) + Cu2+(aq) h Zn2+(aq) + Cu(s)

As long as the Zn metal and Cu2+ ions are in the same container, the electrons are trans- ferred directly from Zn to Cu2+. In a voltaic cell, the components of the two half-reactions are separated into half-cells. The electrons can only flow from one half-cell to the other through an external circuit, producing an electrical current. In each half-cell, there is a strip of metal, called an electrode, in contact with the ionic solution. The electrode where oxida- tion takes place is called the anode; the electrode where reduction takes place is called the cathode. In this example, the anode is a zinc metal strip placed in a Zn2+(ZnSO4) solution. The cathode is a copper metal strip placed in a Cu2+(CuSO4) solution. In this voltaic cell, the Zn anode and Cu cathode are connected by a wire that allows electrons to move from the oxidation half-cell to the reduction half-cell.

Anode is where

oxidation takes place electrons are produced

f Zn(s) h Zn2+(aq) + 2 e-

Cathode is where reduction takes place electrons are used up

f Cu2+(aq) + 2 e- h Cu(s)

The circuit is completed by a salt bridge containing positive and negative ions that are placed in the half-cell solutions. The purpose of the salt bridge is to provide ions, such as Na+ and SO4

2- ions, to maintain an electrical balance in each half-cell solution. As oxidation occurs, there is an increase in Zn2+ ions, which is balanced by SO4

2- anions from the salt bridge. At the cathode, there is a loss of Cu2+, which is balanced by SO4

2- moving into the salt bridge. The complete circuit involves the flow of electrons from the anode to the cathode and the flow of anions from the cathode solution to the anode solution (see FIGURE 15.1).

ENGAGE 15.11 What is the purpose of a salt bridge in a voltaic cell?

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15.3 Electrical Energy from Oxidation– Reduction Reactions 491

FIGURE 15.1 In this voltaic cell, the Zn anode is in a Zn2+ solution, and the Cu cathode is in a Cu2+ solution. Electrons produced by the oxidation of Zn flow from the anode through the wire to the cathode where they reduce Cu2+ to Cu. The circuit is completed by the flow of SO4

2- through the salt bridge.

Salt bridge Cu cathode (reduction)

Zn anode (oxidation)

+ +

Zn2+

Zn2+

Cu2+ Cu

Zn

+ +

SO4 2-

SO4 2-

2Na+

Electrons

SO4 2-

SO4 2-

Cu2+ SO4

2-

2 e-

e- e-

2 e-

Zn(s)

Oxidation Half-Reaction Reduction Half-Reaction

Overall Oxidation–Reduction Reaction

Zn2+(aq) 2 e- Cu2+(aq) 2 e- Cu(s)

Cu2+(aq)Zn(s) Cu(s)Zn2+(aq)

Zn atoms in the anode form Zn2+ ions in solution.

Cu2+ ions form Cu metal on the cathode.

ENGAGE 15.12 Which electrode will be heavier when the reaction in Figure 15.1 ends?

We can diagram the oxidation and reduction reactions that take place in the cell using a shorthand notation as follows:

Zn(s)� Zn2+(aq) ∣∣ Cu2+(aq) �Cu(s)

The components of the oxidation half-cell (anode) are written on the left side, and the components of the reduction half-cell (cathode) are written on the right. A single vertical line separates the solid Zn anode from the Zn2+ solution, and another vertical line separates the Cu2+ solution from the Cu cathode. A double vertical line separates the two half-cells.

Oxidation half-cell

Anode

Electrons

Salt bridge

Zn(s) Zn2+(aq) Cu2+(aq) Cu(s)

Reduction half-cell

Cathode

In some voltaic cells, there is no component in the half-reactions that can be used as an electrode. When this is the case, inert electrodes made of graphite or platinum are used for the transfer of electrons. If there are two ionic components in a cell, their symbols are separated by a comma. For example, suppose a voltaic cell consists of a platinum anode placed in a Sn2+ solution, and a silver cathode placed in a Ag+ solution. The notation for the cell would be written as

Pt(s) � Sn2+(aq), Sn4+(aq) ∣∣ Ag+(aq) � Ag(s)

The oxidation reaction at the anode is

Sn2+(aq) h Sn4+(aq) + 2 e-

The reduction reaction at the cathode is

Ag+(aq) + e- h Ag(s)

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492 CHAPTER 15 Oxidation and Reduction

To balance the overall cell reaction, we multiply the cathode reduction by 2 and combine the two half-reactions.

2Ag+(aq) + 2 e− h 2Ag(s) Reduction

Sn2+(aq) h Sn4+(aq) + 2 e− Oxidation

Sn2+(aq) + 2Ag +(aq) h Sn4+(aq) + 2Ag(s) Oxidation–reduction reaction

PRACTICE PROBLEMS Try Practice Problems 15.29 to 15.32

SAMPLE PROBLEM 15.9 Diagramming a Voltaic Cell

TRY IT FIRST

A voltaic cell consists of an iron (Fe) anode in a Fe2+ solution and a tin (Sn) cathode placed in a Sn2+ solution. Write the cell notation, the oxidation and reduction half- reactions, and the overall cell reaction.

SOLUTION

The notation for the cell would be written as

Fe(s) � Fe2+(aq) ∣∣ Sn2+(aq) � Sn(s)

The oxidation reaction at the anode is

Fe(s) h Fe2+(aq) + 2 e-

The reduction reaction at the cathode is

Sn2+(aq) + 2 e- h Sn(s)

To write the overall cell reaction, we combine the two half-reactions.

Fe(s) h Fe2+(aq) + 2 e- Oxidation Sn2+(aq) + 2 e- h Sn(s) Reduction

Fe(s) + Sn2+(aq) h Fe2+(aq) + Sn(s) Oxidation–reduction reaction

SELF TEST 15.9

Using the following notation for a voltaic cell, write the following:

Co(s) � Co2+(aq) ∣∣ Cu2+(aq) � Cu(s)

a. the half-reactions at the anode and cathode b. the overall cell reaction

ANSWER

a. Anode reaction: Co(s) h Co2+(aq) + 2 e-

Cathode reaction: Cu2+(aq) + 2 e- h Cu(s) b. Overall cell reaction: Co(s) + Cu2+(aq) h Co2+(aq) + Cu(s)

Batteries Batteries are needed to power your cell phone, watch, and calculator. Batteries also are needed to make cars start, and flashlights produce light. Within each of these batteries are voltaic cells that produce electrical energy. Let’s look at some examples of commonly used batteries.

Lead Storage Battery A lead storage battery is used to operate the electrical system in a car. We need a car battery to start the engine, turn on the lights, or operate the radio. If the battery runs down, the car won’t start and the lights won’t turn on. A car battery or a lead storage battery is a type of voltaic cell. In a typical 12-V battery, there are six voltaic cells

Batteries come in many shapes and sizes.

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15.3 Electrical Energy from Oxidation– Reduction Reactions 493

Cathode: PbO2 grids

H2SO4(aq) solution

Anode: Pb metal grids

e- Flow

A 12-volt car battery is also known as a lead storage battery.

Graphite core

MnO2 paste around graphite

Zinc metal can

NH4Cl and ZnCl2 paste

Acidic dry-cell battery

In an acidic dry-cell battery, the cathode is graphite, and the anode is a zinc case.

An alkaline battery has similar components except that NaOH or KOH replaces the NH4Cl electrolyte. Under basic conditions, the product of oxidation is zinc oxide (ZnO). Alkaline batteries tend to be more expensive, but they last longer and produce more power than acidic dry-cell batteries.

Anode (oxidation): Zn(s) + 2OH-(aq) h ZnO(s) + H2O(l) + 2 e−

Cathode (reduction): 2 e− + 2MnO2(s) + H2O(l) h Mn2O3(s) + 2OH-(aq)

Overall cell reaction: Zn(s) + 2MnO2(s) h ZnO(s) + Mn2O3(s)

PRACTICE PROBLEMS Try Practice Problems 15.33 to 15.36

Anode (oxidation): Zn(s) h Zn2+(aq) + 2 e−

Cathode (reduction): 2 e− + 2MnO2(s) + 2NH4 +(aq) h Mn2O3(s) + 2NH3(aq) + H2O(l)

Overall cell reaction: Zn(s) + 2MnO2(s) + 2NH4 +(aq) h Zn2+(aq) + Mn2O3(s) + 2NH3(aq) + H2O(l)

linked together. Each of the cells consists of a lead (Pb) plate that acts as the anode and a lead(IV) oxide (PbO2) plate that acts as the cathode. Both half-cells contain a sulfuric acid (H2SO4) solution. When the car battery is producing electrical energy (discharging), the following half-reactions take place:

Anode (oxidation): Pb(s) + SO4 2-(aq) h PbSO4(s) + 2 e−

Cathode (reduction): 2 e− + 4H+(aq) + PbO2(s) + SO4 2-(aq) h PbSO4(s) + 2H2O(l)

Overall cell reaction: 4H+(aq) + Pb(s) + PbO2(s) + 2SO4 2-(aq) h 2PbSO4(s) + 2H2O(l)

In both half-reactions, Pb2+ is produced, which combines with SO4 2- to form

an insoluble ionic compound PbSO4. As a car battery is used, there is a buildup of PbSO4 on the electrodes. At the same time, there is a decrease in the concentrations of the sulfuric acid components, H+ and SO4

2-. As a car runs, the battery is continu- ously recharged by an alternator, which is powered by the engine. The recharging reactions restore the Pb and PbO2 electrodes as well as H2SO4. Without recharging, the car battery cannot continue to produce electrical energy.

Dry-Cell Batteries Dry-cell batteries are used in calculators, watches, flashlights, and battery-operated toys. The term dry cell describes a battery that uses a paste rather than an aqueous solution. Dry cells can be acidic or alkaline. In an acidic dry cell, the anode is a zinc metal case that contains a paste of MnO2, NH4Cl, ZnCl2, H2O, and starch. Within this MnO2 electrolyte mixture is a graphite cathode.

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494 CHAPTER 15 Oxidation and Reduction

Nickel–Cadmium (NiCad) Batteries Nickel–cadmium (NiCad) batteries, used in cordless power tools, can be recharged. They consist of a cadmium anode and a cathode of solid nickel oxide NiO(OH)(s).

Anode (oxidation): Cd(s) + 2OH-(aq) h Cd(OH)2(s) + 2 e−

Cathode (reduction): 2 e− + 2NiO(OH)(s) + 2H2O(l) h 2Ni(OH)2(s) + 2OH-(aq)

Overall cell reaction: Cd(s) + 2NiO(OH)(s) + 2H2O(l) h Cd(OH)2(s) + 2Ni(OH)2(s)

Chemistry Link to the Environment Corrosion: Oxidation of Metals

Metals used in building materials, such as iron, eventually oxidize, which causes deterioration of the metal. This oxidation process, known as corrosion, produces rust on cars, bridges, ships, and under- ground pipes.

4Fe(s) + 3O2(g) h 2Fe2O3(s) Rust

The formation of rust requires both oxygen and water. The process of rusting requires an anode and cathode in different places on the surface of a piece of iron. In one area of the iron surface, called the anode region, the oxidation half-reaction takes place (see FIGURE 15.2).

Anode (oxidation): Fe(s) h Fe2+(aq) + 2 e-

or 2Fe(s) h 2Fe2+(aq) + 4 e-

The electrons move through the iron metal from the anode to an area called the cathode region where oxygen dissolved in water is reduced to water.

Cathode (reduction): 4 e- + 4H+(aq) + O2(g) h 2H2O(l)

By combining the half-reactions that occur in the anode and cathode regions, we can write the overall oxidation–reduction equation.

2Fe(s) + 4H+(aq) + O2(g) h 2Fe2+(aq) + 2H2O(l)

The formation of rust occurs as Fe2+ ions move out of the anode region and come in contact with dissolved oxygen (O2). The Fe

2+ oxidizes to Fe3+, which reacts with oxygen to form rust.

4H2O(l) + 4Fe2+(aq) + O2(g) h 2Fe2O3(s) + 8H+(aq) Rust

We can write the formation of rust starting with solid Fe reacting with O2 as follows. There is no H

+ in the overall equation because H+ is used and produced in equal quantities.

Corrosion of iron

4Fe(s) + 3O2(g) h 2Fe2O3(s) Rust Other metals such as aluminum, copper, and silver also undergo

corrosion, but at a slower rate than iron. The oxidation of Al on the surface of an aluminum object produces Al3+, which reacts with oxygen in the air to form a protective coating of Al2O3. This Al2O3 coating prevents further oxidation of the aluminum underneath it.

Al(s) h Al3+(aq) + 3 e-

FIGURE 15.2 Rust forms when electrons from the oxidation of Fe flow from the anode region to the cathode region where oxygen is reduced. As Fe2+ ions come in contact with O2 and H2O, rust forms.

Rust (Fe2O3)

Water droplet

Air

Surface of iron metal

O2 O2O2

Anode region

2Fe(s) 2Fe2+(aq) + 4 e- Cathode region

4 e- + 4H+(aq) + O2(g) 2H2O(l)

2Fe2+ H+ H+H2O

H2O

A battery in a cell phone can be recharged many times.

NiCad batteries are expensive, but they can be recharged many times. A charger provides an electrical current that converts the solid Cd(OH)2 and Ni(OH)2 products in the NiCad battery back to the reactants. Because NiCad batteries contain cadmium, which is toxic, they need to be disposed of properly.

Lithium-Ion Batteries Lithium-ion batteries are now used in many of our electronic devices including cell phones, laptops, digital cameras, and tablets. Lithium-ion batteries charge quickly, last longer, are rechargeable, and are light due to the low density of lithium. When the battery is in use and discharging, lithium ions, which are part of a lithium cobalt oxide compound, LiCoO2(s), move from the anode to the cathode. The electrons move through a circuit that provides the energy for the device. When the lithium ion battery is charging, the lithium ions move in the opposite direction from cathode to anode to build up the charge. A simplified reaction is:

Li(s) + CoO2(s) vh LiCoO2(s) charging

discharging

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15.3 Electrical Energy from Oxidation– Reduction Reactions 495

When copper is used on a roof, dome, or a steeple, it oxidizes to Cu2+, which is converted to a green patina of Cu2(OH)2CO3.

Cu(s) h Cu2+(aq) + 2 e-

When we use silver dishes and utensils, the Ag+ ion from oxidation reacts with sulfides in food to form Ag2S, which we call “tarnish.”

Ag(s) h Ag+(aq) + e-

Prevention of Corrosion Billions of dollars are spent each year to prevent corrosion and repair building materials made of iron. One way to prevent corrosion is to paint the bridges, cars, and ships with paints containing materials that seal the iron surface from H2O and O2. But it is necessary to repaint often; a scratch in the paint exposes the iron, which then begins to rust.

A more effective way to prevent corrosion is to place the iron in contact with a metal that substitutes for the anode region of iron. Metals such as Zn, Mg, or Al lose electrons more easily than iron. When one of these metals is in contact with iron, the metal acts as the anode instead of iron. For example, in a process called galvanization, an iron object is coated with zinc. The zinc becomes the anode because zinc is more easily oxidized than Fe. As long as Fe does not act as an anode, rust cannot form.

In a method called cathodic protection, structures such as iron pipes and underground storage containers are placed in contact with a piece of metal such as Mg, Al, or Zn, which is called the sacrificial anode. Again, because these metals lose electrons more easily than Fe, they become the anode, thereby preventing the rusting of the iron. A magnesium plate that is welded or bolted to a ship’s hull loses electrons more easily than iron or steel and protects the hull from rusting. Occasionally, a new magnesium plate is added to replace the magnesium as it is used up. Magnesium stakes placed in the ground are connected to underground pipelines and storage containers to prevent corrosion damage.

Oxidation (corrosion) is prevented by painting a metal surface.

Ground level

Mg stake (sacrificial

anode)

Iron (Fe) pipe or storage tank

Fe

2Mg 2Mg2+ + 4 e-

4 e-+ 4H+ + O2(g) 2H2O(l)

Rusting of an iron pipe is prevented by attaching a piece of magnesium, which oxidizes more easily than iron.

PRACTICE PROBLEMS

15.3 Electrical Energy from Oxidation–Reduction Reactions

15.27 Use the activity series in Table 15.3 to predict whether each of the following reactions will occur spontaneously:

a. 2Au(s) + 6H+(aq) h 2Au3+(aq) + 3H2(g) b. Ni2+(aq) + Fe(s) h Ni(s) + Fe2+(aq)

c. 2Ag(s) + Cu2+(aq) h 2Ag+(aq) + Cu(s) 15.28 Use the activity series in Table 15.3 to predict whether each of

the following reactions will occur spontaneously: a. 2Ag(s) + 2H+(aq) h 2Ag+(aq) + H2(g) b. Mg(s) + Cu2+(aq) h Mg2+(aq) + Cu(s)

c. 2Al(s) + 3Cu2+(aq) h 2Al3+(aq) + 3Cu(s) 15.29 Write the half-reactions and the overall cell reaction for each

of the following voltaic cells: a. Pb(s) � Pb2+(aq) ∣∣ Cu2+(aq) � Cu(s) b. Cr(s) � Cr2+(aq) ∣∣ Ag+(aq) � Ag(s)

15.30 Write the half-reactions and the overall cell reaction for each of the following voltaic cells:

a. Al(s) � Al3+(aq) ∣∣ Cd2+(aq) � Cd(s) b. Sn(s) � Sn2+(aq) ∣∣ Fe3+(aq), Fe2+(aq) � C (electrode)

15.31 Describe the voltaic cell and half-cell components, and write the shorthand notation for the following oxidation–reduction reactions:

a. Cd(s) + Sn2+(aq) h Cd2+(aq) + Sn(s) b. Zn(s) + Cl2(g) h Zn2+(aq) + 2Cl-(aq) � C (electrode)

15.32 Describe the voltaic cell and half-cell components, and write the shorthand notation for the following oxidation–reduction reactions:

a. Mn(s) + Sn2+(aq) h Mn2+(aq) + Sn(s) b. Ni(s) + 2Ag+(aq) h Ni2+(aq) + 2Ag(s)

Applications

15.33 The following half-reaction takes place in a nickel–cadmium battery used in a cordless drill:

Cd(s) + 2OH-(aq) h Cd(OH)2(s) + 2 e-

a. Is the half-reaction an oxidation or a reduction? b. What substance is oxidized or reduced? c. At which electrode would this half-reaction occur?

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496 CHAPTER 15 Oxidation and Reduction

FIGURE 15.3 With a supply of hydrogen and oxygen, a fuel cell can generate electricity continuously.

Oxidation 2H2(g) 4 e

- + 4H+(aq) Reduction 4 e- + 4H+(aq) + O2(g) 2H2O(l)

Catalyst

Cathode Anode

H2O

O2 gasH2 gas

H+

H+

H+

H+

H+

Electric current

O2-

O2-

O2-

e-

e-

e- e-

e-

e-

H+

Plastic membrane

H+

15.34 The following half-reaction takes place in a mercury battery used in hearing aids:

HgO(s) + H2O(l) + 2 e- h Hg(l) + 2OH-(aq)

a. Is the half-reaction an oxidation or a reduction? b. What substance is oxidized or reduced? c. At which electrode would this half-reaction occur?

15.35 The following half-reaction takes place in a mercury battery used in pacemakers and watches:

Zn(s) + 2OH-(aq) h ZnO(s) + H2O(l) + 2 e-

a. Is the half-reaction an oxidation or a reduction? b. What substance is oxidized or reduced? c. At which electrode would this half-reaction occur?

15.36 The following half-reaction takes place in a lead storage bat- tery used in automobiles:

Pb(s) + SO4 2-(aq) h PbSO4(s) + 2 e-

a. Is the half-reaction an oxidation or a reduction? b. What substance is oxidized or reduced? c. At which electrode would this half-reaction occur?

Chemistry Link to the Environment Fuel Cells: Clean Energy for the Future

Fuel cells are of interest to scientists because they provide an alterna- tive source of electrical energy that is more efficient, does not use up oil reserves, and generates products that do not pollute the atmo- sphere. Fuel cells are considered to be a clean way to produce energy.

Like other electrochemical cells, a fuel cell consists of an anode and a cathode connected by a wire. But unlike other cells, the reactants must continuously enter the fuel cell to produce energy; electrical cur- rent is generated only as long as the fuels are supplied. One type of hydrogen–oxygen fuel cell has been used in automobile prototypes. In this hydrogen cell, gas enters the fuel cell and comes in contact with a platinum catalyst embedded in a plastic membrane. The catalyst

assists in the oxidation of hydrogen atoms to hydrogen ions and elec- trons (see FIGURE 15.3).

The electrons produce an electric current as they travel through the wire from the anode to the cathode. The hydrogen ions f low through the plastic membrane to the cathode. At the cathode, oxygen molecules are reduced to oxide ions that combine with the hydrogen ions to form water. The overall hydrogen–oxygen fuel cell reaction can be written as

2H2(g) + O2(g) h 2H2O(l) Fuel cells have already been used for power on the space station

and are now used to produce energy for cars and buses. An advantage of fuel cell cars is that the hydrogen fuel produces water and heat, which gives zero pollution.

In homes, fuel cells may one day replace the batteries currently used to provide electrical power for cell phones, DVD players, and laptop computers. Fuel cell design is still in the prototype phase, although there is much interest in their development. We already know they can work, but modifications must still be made before they become reasonably priced and part of our everyday lives.

Motor

Fuel Cell

Hydrogen Tanks

Hydrogen Station

Battery H2

O2

Fuel cells in cars use hydrogen as the fuel and produce water and heat.

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15.4 Oxidation–Reduction Reactions That Require Electrical Energy 497

15.4 Oxidation–Reduction Reactions That Require Electrical Energy LEARNING GOAL Describe the half-cell reactions and the overall reactions that occur in electrolysis.

When we look at the activity series in Table 15.3, we see that the oxidation of Cu is below Zn. This means that the following oxidation–reduction reaction is not spontaneous:

Zn2+(aq)Cu(s) Cu2+(aq) Zn(s) Not spontaneous Less active More active

+ +

To make this reaction take place, we need an electrolytic cell that uses an electrical current to drive a nonspontaneous oxidation–reduction reaction. This process is called electrolysis (see FIGURE 15.4).

ENGAGE 15.13 Why does the reaction in an electrolytic cell require an electrical current?

FIGURE 15.4 In this electrolytic cell, the Cu anode is in a Cu2+ solution, and the Zn cathode is in a Zn2+ solution. Electrons provided by a battery reduce Zn2+ to Zn and drive the oxidation of Cu to Cu2+ at the Cu anode.

Salt bridge

Cu(s) Zn2+(aq)

Cu2+ Zn2+

Cu(s) + Zn2+(aq) Zn(s)

SO4 2-

Electrons

Battery (electrical current)

SO4 2-2 e- 2 e-

SO4 2-2Na

+

ZnCu

e-e-

+ 2 e-Cu2+(aq) + 2 e- Zn(s)

Zn cathode (reduction)

Cu anode (oxidation)

Cu atoms in the anode form Cu2+ ions in solution.

Zn2+ ions form Zn metal on the cathode.

Cu2+(aq) +

Electrolysis of Sodium Chloride When molten sodium chloride is electrolyzed, the products are sodium metal and chlorine gas. In this electrolytic cell, electrodes are placed in the mixture of Na+ and Cl- and con- nected to a battery. The products are separated to prevent them from reacting spontaneously with each other. As electrons flow to the cathode, Na+ is reduced to sodium metal. At the same time, electrons leave the anode as Cl- is oxidized to Cl2. The half-reactions and the overall reactions are

Electrical energy

2Na+(l) + 2Cl-(l)

2Cl-(l)Anode (oxidation):

Overall cell reaction:

Cl2(g) + 2 e-

2Na+(l) + 2 e-Cathode (reduction): 2Na(l)

2Na(l) + Cl2(g) Not spontaneous

M15_TIMB8119_06_SE_C15.indd 497 11/26/18 6:43 PM

498 CHAPTER 15 Oxidation and Reduction

Electroplating In industry, the process of electroplating uses electrolysis to coat an object with a thin layer of a metal such as silver, platinum, or gold. Steel objects are electroplated with chromium to prevent rusting. Silver-plated utensils, bowls, and platters are made by electroplating objects with a layer of silver.

Ag(s) Anode

NO3 -

Ag+

Battery (electrical current)

e-e-

Cathode

Utensils are silver-plated by electrolysis.

SAMPLE PROBLEM 15.10 Electrolysis

TRY IT FIRST

Electrolysis is used to chrome plate an iron hubcap by placing the hubcap in a Cr3+ solution.

a. What half-reaction takes place to plate the hubcap with metallic chromium? b. Is the iron hubcap the anode or the cathode?

SOLUTION

a. The Cr3+ ions in solution would gain electrons (reduction).

Cr3+(aq) + 3 e- h Cr(s)

b. The iron hubcap is the cathode where reduction takes place.

SELF TEST 15.10

a. Why is energy needed to chrome plate the iron in Sample Problem 15.10? b. Why is a chromium bar used as the anode in the reaction?

ANSWER

a. Since Fe is below Cr in the activity series, the plating of Cr3+ onto Fe is not spontaneous. Energy is needed to make the reaction proceed.

b. A chromium bar is used as the anode in the reaction in order to supply Cr3+ for the plating reaction. At the anode, the reaction is Cr(s) h Cr3+(aq) + 3 e-.

Using electrolysis, a thin layer of chromium is plated onto a hubcap.

PRACTICE PROBLEMS Try Practice Problems 15.37 to 15.40

M15_TIMB8119_06_SE_C15.indd 498 11/26/18 6:43 PM

Update 499

PRACTICE PROBLEMS

15.4 Oxidation–Reduction Reactions That Require Electrical Energy

Applications

15.37 What we call “tin cans” are really iron cans coated with a thin layer of tin. The anode is a bar of tin, and the cathode is the iron can. An electrical current is used to oxidize the Sn to Sn2+ in solution, where the Sn2+ is reduced to produce a thin coating of Sn on the can.

a. What half-reaction takes place to tin plate an iron can?

b. Why is the iron can the cathode?

c. Why is the tin bar the anode?

15.38 Electrolysis is used to gold plate jewelry made of stainless steel.

a. What half-reaction takes place when a Au3+ solution is used to gold plate a stainless steel ring?

b. Is the ring the anode or the cathode? c. Why is energy needed to gold plate the ring?

15.39 When the tin coating on an iron can is scratched, rust will form. Use the activity series in Table 15.3 to explain why this happens.

15.40 When the zinc coating on an iron can is scratched, rust does not form. Use the activity series in Table 15.3 to explain why this happens.

An iron can is coated with tin to prevent rusting.

UPDATE Whitening Kimberly’s Teeth

15.41 The simplified reaction for the production of aqueous H2O2 involves the combination of H2 gas and O2 gas.

a. What reactant is oxidized and what reactant is reduced in this reaction?

b. What is the oxidizing agent? c. What is the reducing agent? d. Write a balanced chemical equation for the reaction.

15.42 The reaction of solid Sn and F2 gas forms solid SnF2. a. What reactant is oxidized and what reactant is reduced

in this reaction? b. What is the oxidizing agent? c. What is the reducing agent? d. Write a balanced chemical equation for the reaction.

The reaction of hydrogen peroxide, H2O2(aq), placed on Kimberly’s teeth produces water, H2O(l ), and oxygen gas. After Kimberly had her teeth cleaned and whitened, Jane recommended that Kimberly use a toothpaste containing tin(II) fluoride. Tooth enamel, which is

composed of hydroxyapatite, Ca5(PO4)3OH, is strengthened when it reacts with fluoride ions to form fluoroapatite, Ca5(PO4)3F.

M15_TIMB8119_06_SE_C15.indd 499 11/26/18 6:43 PM

500 CHAPTER 15 Oxidation and Reduction

CHAPTER REVIEW

15.1 Oxidation and Reduction LEARNING GOAL Identify a reaction as an oxidation or a reduction. Assign and use oxidation numbers to identify elements that are oxidized or reduced. • In an oxidation–reduction reaction,

electrons are transferred from one reactant to another.

• The reactant that loses electrons is oxidized, and the reactant that gains electrons is reduced.

• Oxidation numbers assigned to elements keep track of the changes in the loss and gain of electrons.

• Oxidation is an increase in oxidation number; reduction is a decrease in oxidation number.

• The reducing agent is the substance that provides electrons for reduction.

• The oxidizing agent is the substance that accepts the electrons from oxidation.

• In molecular compounds and polyatomic ions, oxidation numbers are assigned using a set of rules.

• The oxidation number of an element is zero, and the oxidation number of a monatomic ion is the same as the ionic charge of the ion.

• The sum of the oxidation numbers for a compound is equal to zero and for a polyatomic ion is equal to the overall charge.

• Balancing oxidation–reduction equations using oxidation numbers involves the following:

1. assigning oxidation numbers 2. determining the loss and gain of electrons 3. equalizing the loss and gain of electrons 4. balancing the remaining substances by inspection

15.2 Balancing Oxidation–Reduction Equations Using Half-Reactions LEARNING GOAL Balance oxidation–reduction equations using the half-reaction method. • Balancing oxidation–reduction

equations using half-reactions involves the following:

1. separate the equation into half-reactions

2. balance elements other than H and O 3. balance O with H2O and H with H

+

4. balance charge with electrons 5. equalize the loss and gain of electrons 6. combine half-reactions, cancele electrons, and combine

H2O and H +

7. use OH- to neutralize H+ to H2O for an oxidation–reduction reaction in basic solution

15.3 Electrical Energy from Oxidation– Reduction Reactions LEARNING GOAL Use the activity series to determine if an oxidation–reduction reaction is spontaneous. Write the half-reactions that occur in a voltaic cell and the cell notation. • The activity series, which lists metals with the most

easily oxidized metal at the top, is used to predict the direction of a spontaneous reaction.

• In a voltaic cell, the components of the two half- reactions of a spontaneous oxidation–reduction reaction are placed in separate containers called half-cells.

• With a wire connecting the half-cells, an electrical current is gener- ated as electrons move from the anode where oxidation takes place to the cathode where reduction takes place.

to make using a using ato make

are balanced using as determined by the

when that produces that requires

OXIDATION AND REDUCTION

Electrons Are Lost

The Oxidation Numbers Increase

Oxidation Numbers

and

or

involve involve

Electrons Are Gained

Half-Reactions

Electrical Energy

Spontaneous Reaction

Electrical Energy

Nonspontaneous Reaction

Activity Series

Voltaic Cell Electrolytic Cell

The Oxidation Numbers Decrease

ReductionOxidation

when

CONCEPT MAP

M15_TIMB8119_06_SE_C15.indd 500 11/26/18 6:43 PM

Core Chemistry Skills 501

The chapter section containing each Core Chemistry Skill is shown in parentheses at the end of each heading.

Assigning Oxidation Numbers (15.1) • Oxidation numbers are assigned to atoms or ions in a reaction

to determine the substance that is oxidized and the substance that is reduced.

1. The sum of the oxidation numbers for a molecule is 0; for a polyatomic ion the sum is its charge

The oxidation number for: 2. an element is 0 3. a monatomic ion is its charge 4. Group 1A (1) metals is + 1, Group 2A (2) metals is + 2 5. fluorine is - 1; for other Group 7A (17) metals is - 1,

except when combined with O or F 6. oxygen is - 2, except in OF2 or peroxides 7. H in nonmetal compounds is + 1; for H with a metal is - 1 Example: Assign oxidation numbers to each of the elements in the

following:

a. Sn4+ b. Mn c. MnO2 d. MgCO3

Answer: a. Sn4+ b. Mn c. MnO2 d. MgCO3 + 4 0 + 4 - 2 + 2 + 4 - 2

Using Oxidation Numbers (15.1) • An atom is oxidized when its oxidation number increases from

reactant to product. • An atom is reduced when its oxidation number decreases from

reactant to product.

Example: Assign oxidation numbers to each element and identify which is oxidized and which is reduced.

Fe2O3(s) + 2Al(s) h Al2O3(s) + 2Fe(l)

CORE CHEMISTRY SKILLS Answer: Fe2O3(s) + 2Al(s) h Al2O3(s) + 2Fe(l) + 3 - 2 0 + 3 - 2 0

Fe( + 3) h Fe(0) Fe is reduced. Al(0) h Al( + 3) Al is oxidized.

Identifying Oxidizing and Reducing Agents (15.1) • In an oxidation–reduction reaction, the substance that is reduced is

the oxidizing agent. • In an oxidation–reduction reaction, the substance that is oxidized is

the reducing agent.

Example: Identify the oxidizing agent and the reducing agent in the following:

SnO2(s) + 2H2(g) h Sn(s) + 2H2O(l)

Answer: Oxidation numbers are assigned to identify the substance that is oxidized as the reducing agent and the substance that is reduced as the oxidizing agent.

SnO2(s) + 2H2(g) h Sn(s) + 2H2O(l) + 4 - 2 0 0 + 1 - 2

Sn( + 4) h Sn(0) Sn is reduced; SnO2 is the oxidizing agent.

H(0) h H( + 1) H is oxidized; H2 is the reducing agent.

Using Half-Reactions to Balance Redox Equations (15.2) • The half-reaction method for balancing equations involves: 1. separate the oxidation–reduction reaction into two

half-reactions 2. balance the elements in each except H and O; then if the reaction

takes place in acid or base balancing O with H2O and H with H +

3. balance charge by adding electrons

15.4 Oxidation–Reduction Reactions That Require Electrical Energy LEARNING GOAL Describe the half- cell reactions and the overall reactions that occur in electrolysis. • In an electrolytic cell, electrical

energy from an external source is used to make reactions take place that are not spontaneous.

• A method called electrolysis is used to plate chrome on hubcaps, zinc on iron, or gold on stainless steel jewelry.

activity series A table of half-reactions with the metals that oxidize most easily at the top, and the metals that do not oxidize easily at the bottom.

anode The electrode where oxidation takes place. cathode The electrode where reduction takes place. electrolysis The use of electrical energy to run a nonspontaneous

oxidation–reduction reaction in an electrolytic cell. electrolytic cell A cell in which electrical energy is used to make a

nonspontaneous oxidation–reduction reaction happen. half-reaction method A method of balancing oxidation–reduction

reactions in which the half-reactions are balanced separately and then combined to give the complete reaction.

oxidation The loss of electrons by a substance.

oxidation number A number equal to zero in an element or the charge of a monatomic ion; in molecular compounds and polyatomic ions, oxidation numbers are assigned using a set of rules.

oxidation–reduction reaction A reaction in which electrons are transferred from one reactant to another.

oxidizing agent The reactant that gains electrons and is reduced. reducing agent The reactant that loses electrons and is oxidized. reduction The gain of electrons by a substance. voltaic cell A type of cell with two compartments that uses

spontaneous oxidation–reduction reactions to produce electrical energy.

KEY TERMS

M15_TIMB8119_06_SE_C15.indd 501 11/26/18 6:43 PM

502 CHAPTER 15 Oxidation and Reduction

Identifying Spontaneous Reactions (15.3)

• The activity series in Table 15.3 places the metals that are easily oxidized at the top, and the metals that do not oxidize easily at the bottom.

• A metal will oxidize spontaneously when it is combined with the reverse of the half-reaction of any metal below it on the activity series in Table 15.3.

Example: Write a balanced redox equation for the spontaneous reaction for the following half-reactions on the activity series:

Ca(s) h Ca2+(aq) + 2 e-

Ni(s) h Ni2+(aq) + 2 e-

Answer: The spontaneous reaction is a combination of the oxidation half-reaction higher on the activity series with the reverse of the half-reaction below it.

4. multiply each half-reaction by factors that equalize the loss and gain of electrons

5. combine half-reactions and cancel electrons, identical ions or molecules, and use OH- in base to neutralize H+

Example: Use the half-reaction method to balance the following oxidation–reduction equation in acidic solution:

NO3 -(aq) + Sn2+(aq) h NO(g) + Sn4+(aq)

Answer:

1. NO3 -(aq) h NO(g)

Sn2+(aq) h Sn4+(aq)

2. 4H+(aq) + NO3 -(aq) h NO(g) + 2H2O(l) Sn2+(aq) h Sn4+(aq)

3. 3 e− + 4H+(aq) + NO3 -(aq) h NO(g) + 2H2O(l) Sn2+(aq) h Sn4+(aq) + 2 e−

4. 2 * [3 e− + 4H+(aq) + NO3 -(aq) h NO(g) + 2H2O(l)] 6 e− + 8H+(aq) + 2NO3 -(aq) h 2NO(g) + 4H2O(l)

3 * [Sn2+(aq) h Sn4+(aq) + 2 e−] 3Sn2+(aq) h 3Sn4+(aq) + 6 e−

5. 8H+(aq) + 2NO3 -(aq) + 3Sn2+(aq) h 2NO(g) + 3Sn4+(aq) + 4H2O(l)

UNDERSTANDING THE CONCEPTS

The chapter sections to review are shown in parentheses at the end of each problem.

15.43 Classify each of the following as oxidation or reduction: (15.1)

a. Electrons are lost. b. Reaction of an oxidizing agent. c. O2(g) h OH

-(aq) d. Br2(l) h 2Br

-(aq) e. Sn2+(aq) h Sn4+(aq)

15.44 Classify each of the following as oxidation or reduction: (15.1)

a. Electrons are gained. b. Reaction of a reducing agent. c. Ni(s) h Ni2+(aq) d. MnO4

-(aq) h MnO2(s) e. Sn4+(aq) h Sn2+(aq)

15.45 Assign oxidation numbers to the elements in each of the following: (15.1)

a. VO2 b. Ag2CrO4 c. S2O8

2- d. FeSO4

15.46 Assign oxidation numbers to the elements in each of the following: (15.1)

a. NbCl3 b. NbO c. NbO2 d. Nb2O5

15.47 Which of the following are oxidation–reduction reactions? (15.1)

a. Ca(s) + 2H2O(l) h Ca(OH)2(aq) + H2(g) b. CaCO3(s) h CaO(s) + CO2(g) c. 4Al(s) + 3O2(g) h 2Al2O3(s)

15.48 Which of the following are oxidation–reduction reactions? (15.1)

a. BaCl2(aq) + Na2SO4(aq) h BaSO4(s) + 2NaCl(aq) b. Cl2(g) + 2NaBr(aq) h Br2(l) + 2NaCl(aq) c. 2KClO3(s) h 2KCl(s) + 3O2(g)

15.49 For this reaction, identify each of the following: (15.1)

2Cr2O3(s) + 3Si(s) h 4Cr(s) + 3SiO2(s) a. the substance reduced b. the substance oxidized c. the oxidizing agent d. the reducing agent

Chromium(III) oxide and silicon undergo an oxidation–reduction reaction.

15.50 For this reaction, identify each of the following: (15.1)

C2H4(g) + 3O2(g) h ∆

2CO2(g) + 2H2O(g) a. the substance reduced b. the substance oxidized c. the oxidizing agent d. the reducing agent

15.51 Balance each of the following half-reactions in acidic solution: (15.2)

a. Zn(s) h Zn2+(aq) b. SnO2

2-(aq) h SnO3 2-(aq)

c. SO3 2-(aq) h SO4

2-(aq)

d. NO3 -(aq) h NO(g)

15.52 Balance each of the following half-reactions in acidic solution: (15.2)

a. I2(s) h I -(aq)

b. MnO4 -(aq) h Mn2+(aq)

c. Br2(l) h BrO3 -(aq)

d. ClO3 -(aq) h ClO4

-(aq)

Ca(s) h Ca2+(aq) + 2 e- Oxidation Ni2+(aq) + 2 e- h Ni(s) Reduction

Ca(s) + Ni2+(aq) h Ca2+(aq) + Ni(s) Spontaneous reaction

M15_TIMB8119_06_SE_C15.indd 502 11/26/18 6:43 PM

Additional Practice Problems 503

Al Cu

Al(NO3)3(aq)

Salt bridge

Al3+ Cu2+

Cu(NO3)2(aq)

Fe Ni

Ni2+Fe2+

FeSO4(aq) NiSO4(aq)

Salt bridge

15.53 Consider the following voltaic cell: (15.3) 15.54 Consider the following voltaic cell: (15.3)

a. What is the oxidation half-reaction? b. What is the reduction half-reaction? c. What metal is the anode? d. What metal is the cathode? e. What is the direction of electron flow? f. What is the overall reaction that takes place? g. Write the shorthand cell notation.

a. What is the oxidation half-reaction? b. What is the reduction half-reaction? c. What metal is the anode? d. What metal is the cathode? e. What is the direction of electron flow? f. What is the overall reaction that takes place? g. Write the shorthand cell notation.

ADDITIONAL PRACTICE PROBLEMS

15.55 Which of the following are oxidation–reduction reactions? (15.1)

a. AgNO3(aq) + NaCl(aq) h AgCl(s) + NaNO3(aq) b. 6Li(s) + N2(g) h 2Li3N(s) c. Ni(s) + Pb(NO3)2(aq) h Ni(NO3)2(aq) + Pb(s) d. 2K(s) + 2H2O(l) h 2KOH(aq) + H2(g) 15.56 Which of the following are oxidation–reduction reactions?

(15.1)

a. Ca(s) + F2(g) h CaF2(s) b. Fe(s) + 2HCl(aq) h FeCl2(aq) + H2(g) c. 2NaCl(aq) + Pb(NO3)2(aq) h

PbCl2(s) + 2NaNO3(aq) d. 2CuCl(aq) h Cu(s) + CuCl2(aq) 15.57 In the mitochondria of human cells, energy is provided by the

oxidation and reduction of the iron ions in the cytochromes. Identify each of the following reactions as an oxidation or reduction: (15.1)

a. Fe3+(aq) + e- h Fe2+(aq) b. Fe2+(aq) h Fe3+(aq) + e-

15.58 Chlorine (Cl2) is used as a germicide to kill microbes in swimming pools. If the product is Cl-, was the elemental chlorine oxidized or reduced? (15.1)

15.59 Assign oxidation numbers to all the elements in each of the following: (15.1)

a. Co2O3 b. KMnO4 c. SbCl5 d. ClO3

- e. PO4 3-

15.60 Assign oxidation numbers to all the elements in each of the following: (15.1)

a. PO3 2- b. NH4

+ c. Fe(OH)2 d. HNO3 e. H2CO

15.61 Assign oxidation numbers to all the elements in each of the following reactions, and identify the reactant that is oxidized, the reactant that is reduced, the oxidizing agent, and the reducing agent: (15.1)

a. 2FeCl2(aq) + Cl2(g) h 2FeCl3(aq) b. 2H2S(g) + 3O2(g) h 2H2O(l) + 2SO2(g) c. P2O5(s) + 5C(s) h 2P(s) + 5CO(g)

15.62 Assign oxidation numbers to all the elements in each of the following reactions, and identify the reactant that is oxidized, the reactant that is reduced, the oxidizing agent, and the reducing agent: (15.1)

a. 4Al(s) + 3O2(g) h 2Al2O3(s) b. I2O5(s) + 5CO(g) h I2(g) + 5CO2(g) c. 2Cr2O3(s) + 3C(s) h 4Cr(s) + 3CO2(g)

15.63 Write the balanced half-reactions and a balanced redox equation for each of the following reactions in acidic solution: (15.2)

a. Zn(s) + NO3 -(aq) h Zn2+(aq) + NO2(g) b. MnO4

-(aq) + SO3 2-(aq) h Mn2+(aq) + SO4 2-(aq) c. ClO3

-(aq) + I-(aq) h Cl-(aq) + I2(s) d. Cr2O7

2-(aq) + C2O4 2-(aq) h Cr3+(aq) + CO2(g)

15.64 Write the balanced half-reactions and a balanced redox equation for each of the following reactions in acidic solution: (15.2)

a. Sn2+(aq) + IO4 -(aq) h Sn4+(aq) + I-(aq) b. S2O3

2-(aq) + I2(s) h S4O6 2-(aq) + I-(aq) c. Mg(s) + VO4 3-(aq) h Mg2+(aq) + V2+(aq) d. Al(s) + Cr2O7 2-(aq) h Al3+(aq) + Cr3+(aq)

15.65 Use the activity series in Table 15.3 to predict whether each of the following reactions will occur spontaneously: (15.3)

a. 2Cr(s) + 3Ni2+(aq) h 2Cr3+(aq) + 3Ni(s) b. Cu(s) + Zn2+(aq) h Cu2+(aq) + Zn(s) c. Zn(s) + Pb2+(aq) h Zn2+(aq) + Pb(s)

M15_TIMB8119_06_SE_C15.indd 503 11/26/18 6:43 PM

504 CHAPTER 15 Oxidation and Reduction

15.66 Use the activity series in Table 15.3 to predict whether each of the following reactions will occur spontaneously: (15.3)

a. Zn(s) + Mg2+(aq) h Zn2+(aq) + Mg(s) b. 3Na(s) + Al3+(aq) h 3Na+(aq) + Al(s) c. Mg(s) + Ni2+(aq) h Mg2+(aq) + Ni(s) 15.67 In a voltaic cell, one half-cell consists of nickel metal in a Ni2+

solution, and the other half-cell consists of magnesium metal in a Mg2+ solution. Give each of the following: (15.3)

a. the anode b. the cathode c. the half-reaction at the anode d. the half-reaction at the cathode e. the overall reaction f. the shorthand cell notation

15.68 In a voltaic cell, one half-cell consists of zinc metal in a Zn2+ solution, and the other half-cell consists of copper metal in a Cu2+ solution. Give each of the following: (15.3)

a. the anode b. the cathode c. the half-reaction at the anode d. the half-reaction at the cathode e. the overall reaction f. the shorthand cell notation

15.69 Use the activity series in Table 15.3 to determine which of the following ions will be reduced when an iron strip is placed in an aqueous solution of that ion: (15.3)

a. Ca2+(aq) b. Ag+(aq) c. Ni2+(aq) d. Al3+(aq) e. Pb2+(aq)

15.70 Use the activity series in Table 15.3 to determine which of the following ions will be reduced when an aluminum strip is placed in an aqueous solution of that ion: (15.3)

a. Fe2+(aq) b. Au3+(aq) c. Mg2+(aq) d. H+(aq) e. Pb2+(aq)

15.71 In a lead storage battery, the following unbalanced half- reaction takes place: (15.3)

Pb(s) + SO4 2-(aq) h PbSO4(s) a. Balance the half-reaction. b. Is Pb(s) oxidized or reduced? c. Indicate whether the half-reaction takes place at the anode

or cathode.

15.72 In an acidic dry-cell battery, the following unbalanced half- reaction takes place in acidic solution: (15.3)

MnO2(s) h Mn2O3(s)

a. Balance the half-reaction. b. Is MnO2(s) oxidized or reduced? c. Indicate whether the half-reaction takes place at the anode

or cathode.

15.73 Steel bolts made for sailboats are coated with zinc. Add the necessary components (electrodes, wires, batteries) to this diagram of an electrolytic cell with a zinc nitrate solution to show how it could be used to zinc plate a steel bolt. (15.3, 15.4)

Cu2+(aq) 2NO3

-(aq)

Zn2+(aq) 2NO3

-(aq)

a. What is the anode? b. What is the cathode? c. What is the half-reaction that takes place at the anode? d. What is the half-reaction that takes place at the cathode? e. If steel is mostly iron, what is the purpose of the zinc

coating?

15.74 Copper cooking pans are stainless steel pans plated with a layer of copper. Add the necessary components (electrodes,  wires, batteries) to this diagram of an electrolytic cell with a copper(II) nitrate solution to show how it could be used to copper plate a stainless steel (iron) pan. (15.3, 15.4)

a. What is the anode? b. What is the cathode? c. What is the half-reaction that takes place at the anode? d. What is the half-reaction that takes place at the cathode?

The following problems are related to the topics in this chapter. However, they do not all follow the chapter order, and they require you to combine concepts and skills from several sections. These problems will help you increase your critical thinking skills and prepare for your next exam.

15.75 Determine the oxidation number of Br in each of the following: (15.1)

a. Br2 b. HBrO2 c. BrO3

- d. NaBrO4

15.76 Determine the oxidation number of Cr in each of the following: (15.1)

a. CrO b. HCrO4 -

c. CrO3 - d. CrF4

15.77 Use half-reactions to balance the following equation in acidic solution: (15.2)

Cr2O7 2-(aq) + NO2 -(aq) h Cr3+(aq) + NO3 -(aq)

15.78 Use half-reactions to balance the following equation in acidic solution: (15.2)

Mn(s) + Cr3+(aq) h Mn2+(aq) + Cr(s)

15.79 The following unbalanced reaction takes place in acidic solu- tion: (11.6, 15.2)

Ag(s) + NO3 -(aq) h Ag+(aq) + NO(g) a. Write the balanced equation. b. How many liters of NO(g) are produced at STP when

15.0 g of silver reacts with excess nitric acid?

CHALLENGE PROBLEMS

M15_TIMB8119_06_SE_C15.indd 504 11/26/18 6:44 PM

Answers to Selected Problems 505

15.80 The following unbalanced reaction takes place in acidic solution: (12.6, 15.2)

MnO4 -(aq) + Fe2+(aq) h Mn2+(aq) + Fe3+(aq)

a. Write the balanced equation. b. How many milliliters of a 0.150 M KMnO4 solution

are needed to react with 25.0 mL of a 0.400 M FeSO4 solution?

15.81 The following unbalanced reaction takes place in acidic solution: (12.6, 15.2)

S2-(aq) + NO3-(aq) h SO42-(aq) + NO(g) a. Write the balanced equation. b. How many milliliters of a 16.0 M HNO3 solution are

needed to react with 24.8 g of CuS?

15.82 The following unbalanced reaction takes place in acidic solution: (12.6, 15.2)

Cr2O7 2-(aq) + Fe2+(aq) h Cr3+(aq) + Fe3+(aq)

a. Write the balanced equation. b. How many milliliters of a 0.211 M K2Cr2O7 solution are

needed to react with 5.00 g of FeSO4?

15.83 Use half-reactions to balance the following equation in basic solution: (15.2)

Br-(aq) + MnO4 -(aq) h BrO3 -(aq) + MnO2(s)

15.84 Use half-reactions to balance the following equation in basic solution: (15.2)

CN-(aq) + IO3 -(aq) h CNO-(aq) + I-(aq) 15.85 Using the activity series in Table 15.3, indicate whether each of

the following reactions is spontaneous: (15.3) a. Zn(s) + Ca2+(aq) h Zn2+(aq) + Ca(s)

b. 2Al(s) + 3Sn2+(aq) h 2Al3+(aq) + 3Sn(s) c. Mg(s) + 2H+(aq) h Mg2+(aq) + H2(g) 15.86 Using the activity series in Table 15.3, indicate whether each of

the following reactions is spontaneous: (15.3) a. Cu(s) + Ni2+(aq) h Cu2+(aq) + Ni(s)

b. 2Cr(s) + 3Fe2+(aq) h 2Cr3+(aq) + 3Fe(s) c. Fe(s) + Mg2+(aq) h Fe2+(aq) + Mg(s) 15.87 Draw a diagram of a voltaic cell for

Ni(s) � Ni2+(aq) ∣∣ Ag+(aq) � Ag(s). (15.3) a. What is the anode?

b. What is the cathode? c. What is the half-reaction that takes place at the anode? d. What is the half-reaction that takes place at the cathode?

e. What is the overall reaction for the cell?

15.88 Draw a diagram of a voltaic cell for Mg(s) � Mg2+(aq) ∣∣ Al3+(aq) � Al(s). (15.3)

a. What is the anode? b. What is the cathode? c. What is the half-reaction that takes place at the anode? d. What is the half-reaction that takes place at the cathode?

e. What is the overall reaction for the cell?

ANSWERS TO ENGAGE QUESTIONS 15.7 The 7 H2O are added to the product side of the half-reaction

to balance the 7 O in Cr2O7 2- on the reactant side.

15.8 On the product side of the half-reaction, the charge is + 6. On the reactant side, the charge is + 14 + ( - 2) = + 12. To bal- ance the two sides, - 6 (6 e-) must be added to the reactant side.

15.9 The 4OH- added to the product side of the equation to neutralize the 4H+.

15.10 Because Fe is above Pb in the activity series, the reaction Fe(s) + Pb2+(aq) h Fe2+(aq) + Pb(s) is spontaneous.

15.11 A salt bridge allows ions to move from one side of the cell to the other to complete the electrical circuit.

15.12 The copper electrode will be heavier because solid copper is deposited on it.

15.13 An electrolytic cell uses a nonspontaneous reaction, and it needs an outside source of energy, an electrical current, in order to proceed.

15.1 In a reduction half-reaction, there is a gain of electrons by one of the reactants.

15.2 Mg and Ba are elements, which have oxidation numbers of 0. The ions of Mg and Ba have the oxidation number of the monoatomic ions Mg2+, Ba2+.

15.3 The oxidation number of S in H2SO3 is calculated by adding + 2 from 2 H and - 6 from 3 O, which gives - 4. Because H2SO3 has an overall 0 charge, S must have an oxidation number of + 4.

15.4 The reaction of H2 to 2H + is an oxidation because there is an

increase in the oxidation number of H, from 0 to + 1. 15.5 Zn is oxidized when it loses electrons to form Zn2+; it is the

reducing agent. Co2+ is reduced when it gains electrons to form Co; it is the oxidizing agent.

15.6 To be sure the final oxidation–reduction reaction is balanced, check the balance of all the atoms and the overall charge.

ANSWERS TO SELECTED PROBLEMS 15.9 a. Li is + 1, P is + 5, and O is - 2. b. S is + 4, O is - 2.

c. Cr is + 3, S is - 2. d. N is + 5, O is - 2. 15.11 a. H is + 1, S is + 6, and O is - 2. b. H is + 1, P is + 3, and O is - 2. c. Cr is + 6, O is - 2. d. Na is + 1, C is + 4, and O is - 2. 15.13 a. + 5 b. - 2 c. + 5 d. + 6 15.15 a. The substance that is oxidized is the reducing agent. b. The substance that gains electrons is reduced and is the

oxidizing agent.

15.1 a. reduction b. oxidation c. reduction

15.3 a. Al is oxidized and O2 is reduced. b. Zn is oxidized and H+ is reduced. c. Br- is oxidized and F2 is reduced.

15.5 a. 0 b. 0 c. + 2 d. - 1 15.7 a. K is + 1, Cl is - 1. b. Mn is + 4, O is - 2.

c. N is + 2, O is - 2. d. Mn is + 3, O is - 2.

M15_TIMB8119_06_SE_C15.indd 505 11/26/18 6:44 PM

506 CHAPTER 15 Oxidation and Reduction

15.17 a. Li is oxidized; Li is the reducing agent. Cl (in Cl2) is reduced; Cl2 is the oxidizing agent.

b. Br- (in NaBr) is oxidized; NaBr is the reducing agent. Cl (in Cl2) is reduced; Cl2 is the oxidizing agent.

c. Pb is oxidized; Pb is the reducing agent. O (in O2) is reduced; O2 is the oxidizing agent.

d. Al is oxidized; Al is the reducing agent. Ag+ is reduced; Ag+ is the oxidizing agent.

15.19 a. S2- (in NiS) is oxidized; NiS is the reducing agent. O (in O2) is reduced; O2 is the oxidizing agent.

b. Sn2+ is oxidized; Sn2+ is the reducing agent. Fe3+ is reduced; Fe3+ is the oxidizing agent.

c. C (in CH4) is oxidized; CH4 is the reducing agent. O (in O2) is reduced; O2 is the oxidizing agent.

d. Si is oxidized; Si is the reducing agent. Cr3+ (in Cr2O3) is reduced; Cr2O3 is the oxidizing agent.

15.21 a. Sn2+(aq) h Sn4+(aq) + 2 e-

b. Mn2+(aq) + 4H2O(l) h MnO4 -(aq) + 8H+(aq) + 5 e-

c. NO2 -(aq) + H2O(l) h NO3 -(aq) + 2H+(aq) + 2 e-

d. e- + 2H+(aq) + ClO3 -(aq) h ClO2(aq) + H2O(l) 15.23 a. 2H+(aq) + Ag(s) + NO3 -(aq) h

Ag+(aq) + NO2(g) + H2O(l) b. 4H+(aq) + 4NO3 -(aq) + 3S(s) h

4NO(g) + 3SO2(g) + 2H2O(l) c. 2S2O3

2-(aq) + Cu2+(aq) h S4O6 2-(aq) + Cu(s)

15.25 a. 2Fe(s) + 2CrO4 2-(aq) + 2H2O(l) h Fe2O3(s) + Cr2O3(s) + 4OH-(aq)

b. 3CN-(aq) + 2MnO4 -(aq) + H2O(l) h 3CNO-(aq) + 2MnO2(s) + 2OH-(aq)

15.27 a. Since Au is below H2 in the activity series, the reaction will not be spontaneous.

b. Since Fe is above Ni in the activity series, the reaction will be spontaneous.

c. Since Ag is below Cu in the activity series, the reaction will not be spontaneous.

15.29 a. Anode reaction: Pb(s) h Pb2+(aq) + 2 e-

Cathode reaction: Cu2+(aq) + 2 e- h Cu(s) Overall cell reaction: Pb(s) + Cu2+(aq) h

Pb2+(aq) + Cu(s) b. Anode reaction: Cr(s) h Cr2+(aq) + 2 e-

Cathode reaction: Ag+(aq) + e- h Ag(s) Overall cell reaction: Cr(s) + 2Ag+(aq) h

Cr2+(aq) + 2Ag(s) 15.31 a. The anode is a Cd metal electrode in a Cd2+ solution. The

anode reaction is

Cd(s) h Cd2+(aq) + 2 e- The cathode is a Sn metal electrode in a Sn2+ solution. The cathode reaction is

Sn2+(aq) + 2 e- h Sn(s) The shorthand notation for this cell is

Cd(s) � Cd2+(aq) ∣∣ Sn2+(aq) � Sn(s) b. The anode is a Zn metal electrode in a Zn2+ solution. The

anode reaction is Zn(s) h Zn2+(aq) + 2 e-

The cathode is a C (graphite) electrode, where Cl2 gas is reduced to Cl-. The cathode reaction is

Cl2(g) + 2 e- h 2Cl-(aq) The shorthand notation for this cell is

Zn(s) � Zn2+(aq) ∣∣ Cl2(g) � Cl -(aq) � C

15.33 a. The half-reaction is an oxidation. b. Cd metal is oxidized. c. Oxidation takes place at the anode.

15.35 a. The half-reaction is an oxidation. b. Zn metal is oxidized. c. Oxidation takes place at the anode.

15.37 a. Sn2+(aq) + 2 e- h Sn(s) b. The reduction of Sn2+ to Sn occurs at the cathode, which is

the iron can. c. The oxidation of Sn to Sn2+ occurs at the anode, which is

the tin bar.

15.39 Since Fe is above Sn in the activity series, if the Fe is exposed to air and water, Fe will be oxidized and rust will form. To pro- tect iron, Sn would have to be more active than Fe and it is not.

15.41 a. H2 is oxidized and O2 is reduced. b. O2 is the oxidizing agent. c. H2 is the reducing agent. d. H2(g) + O2(g) h H2O2(aq) 15.43 a. oxidation b. reduction c. reduction d. reduction e. oxidation

15.45 a. V = + 4, O = - 2 b. Ag = + 1, Cr = + 6, O = - 2 c. S = + 7, O = - 2 d. Fe = + 2, S = + 6, O = - 2 15.47 Reactions a and c involve loss and gain of electrons; a and c

are oxidation–reduction reactions.

15.49 a. Cr in Cr2O3 is reduced. b. Si is oxidized. c. Cr2O3 is the oxidizing agent. d. Si is the reducing agent.

15.51 a. Zn(s) h Zn2+(aq) + 2 e-

b. SnO2 2-(aq) + H2O(l) h SnO3 2-(aq) + 2H+(aq) + 2 e-

c. SO3 2-(aq) + H2O(l) h SO4 2-(aq) + 2H+(aq) + 2 e-

d. 3 e- + 4H+(aq) + NO3 -(aq) h NO(g) + 2H2O(l) 15.53 a. Fe(s) h Fe2+(aq) + 2 e-

b. Ni2+(aq) + 2 e- h Ni(s) c. Fe is the anode. d. Ni is the cathode. e. The electrons flow from Fe to Ni. f. Fe(s) + Ni2+(aq) h Fe2+(aq) + Ni(s) g. Fe(s) � Fe2+(aq) ∣∣ Ni2+(aq) � Ni(s) 15.55 Reactions b, c, and d all involve loss and gain of electrons;

b, c, and d are oxidation–reduction reactions.

15.57 a. Fe3+ is gaining electrons; this is a reduction. b. Fe2+ is losing electrons; this is an oxidation.

15.59 a. Co = + 3, O = - 2 b. K = + 1, Mn = + 7, O = - 2 c. Sb = + 5, Cl = - 1 d. Cl = + 5, O = - 2 e. P = + 5, O = - 2

15.61 a. 2FeCl2(aq) + Cl2(g) h 2FeCl3(aq) + 2 - 1 0 + 3 - 1

Fe in FeCl2 is oxidized; FeCl2 is the reducing agent. Cl in Cl2 is reduced; Cl2 is the oxidizing agent.

b. 2H2S(g) + 3O2(g) h 2H2O(l) + 2SO2(g) + 1 - 2 0 + 1 - 2 + 4 - 2

S in H2S is oxidized; H2S is the reducing agent. O in O2 is reduced; O2 is the oxidizing agent.

c. P2O5(s) + 5C(s) h 2P(s) + 5CO(g) + 5 - 2 0 0 + 2 - 2

C is oxidized; C is the reducing agent. P in P2O5 is reduced; P2O5 is the oxidizing agent.

M15_TIMB8119_06_SE_C15.indd 506 11/26/18 6:44 PM

Answers to Selected Problems 507

15.63 a. Zn(s) h Zn2+(aq) + 2 e-; e- + 2H+(aq) + NO3 -(aq) h NO2(g) + H2O(l) Overall:

4H+(aq) + Zn(s) + 2NO3 -(aq) h Zn2+(aq) + 2NO2(g) + 2H2O(l)

b. 5 e- + 8H+(aq) + MnO4 -(aq) h Mn2+(aq) + 4H2O(l); SO3

2-(aq) + H2O(l) h SO4 2-(aq) + 2H+(aq) + 2 e-

Overall: 6H+(aq) + 2MnO4 -(aq) + 5SO3 2-(aq) h 2Mn2+(aq) + 5SO4 2-(aq) + 3H2O(l)

c. 2I-(aq) h I2(s) + 2 e-; 6 e- + 6H+(aq) + ClO3 -(aq) h Cl-(aq) + 3H2O(l) Overall: 6H+(aq) + ClO3 -(aq) + 6I-(aq) h Cl-(aq) + 3I2(s) + 3H2O(l)

d. C2O4 2-(aq) h 2CO2(g) + 2 e-;

6 e- + 14H+(aq) + Cr2O7 2-(aq) h 2Cr3+(aq) + 7H2O(l)

Overall:

14H+(aq) + Cr2O7 2-(aq) + 3C2O4 2-(aq) h 2Cr3+(aq) + 6CO2(g) + 7H2O(l)

15.65 a. Since Cr is above Ni in the activity series, the reaction will be spontaneous.

b. Since Cu is below Zn in the activity series, the reaction will not be spontaneous.

c. Since Zn is above Pb in the activity series, the reaction will be spontaneous.

15.67 a. The anode is Mg. b. The cathode is Ni. c. The half-reaction at the anode is

Mg(s) h Mg2+(aq) + 2 e-

d. The half-reaction at the cathode is Ni2+(aq) + 2 e- h Ni(s)

e. The overall reaction is

Mg(s) + Ni2+(aq) h Mg2+(aq) + Ni(s) f. The shorthand cell notation is

Mg(s) � Mg2+(aq) ∣∣ Ni2+(aq) � Ni(s)

15.69 a. Ca2+(aq) will not be reduced by an iron strip. b. Ag+(aq) will be reduced by an iron strip. c. Ni2+(aq) will be reduced by an iron strip. d. Al3+(aq) will not be reduced by an iron strip. e. Pb2+(aq) will be reduced by an iron strip.

15.71 a. Pb(s) + SO4 2-(aq) h PbSO4(s) + 2 e- b. Pb(s) is oxidized. c. The half-reaction takes place at the anode.

15.73

a. Ni(s) is the anode. b. Ag(s) is the cathode. c. The half-reaction at the anode is

Ni(s) h Ni2+(aq) + 2 e- d. The half-reaction at the cathode is

Ag+(aq) + e- h Ag(s) e. The overall cell reaction is

Ni(s) + 2Ag+(aq) h Ni2+(aq) + 2Ag(s) e-

e-

Zn2+(aq) 2NO3

-(aq)

Steel bolt cathode

Zn anode

Battery

a. The anode is a bar of zinc. b. The cathode is the steel bolt. c. The half-reaction at the anode is

Zn(s) h Zn2+(aq) + 2 e-

d. The half-reaction at the cathode is

Zn2+(aq) + 2 e- h Zn(s) e. The purpose of the zinc coating is to prevent rusting of

the bolt by H2O and O2.

15.75 a. 0 b. + 3 c. + 5 d. + 7 15.77 8H+(aq) + Cr2O7 2-(aq) + 3NO2 -(aq) h

2Cr3+(aq) + 3NO3 -(aq) + 4H2O(l) 15.79 a. 4H+(aq) + 3Ag(s) + NO3 -(aq) h

3Ag+(aq) + NO(g) + 2H2O(l) b. 1.04 L of NO(g) are produced at STP.

15.81 a. 8H+(aq) + 3S2-(aq) + 8NO3-(aq) h 3SO4

2-(aq) + 8NO(g) + 4H2O(l) b. 43.2 mL of HNO3 solution

15.83 H2O(l) + Br-(aq) + 2MnO4-(aq) h BrO3

-(aq) + 2MnO2(s) + 2OH-(aq)

15.85 a. Since Zn is below Ca in the activity series, the reaction will not be spontaneous.

b. Since Al is above Sn in the activity series, the reaction will be spontaneous.

c. Since Mg is above H2 in the activity series, the reaction will be spontaneous.

15.87

Ni2+(aq) Ag+(aq)

Salt bridge Ag(s)Ni(s)

M15_TIMB8119_06_SE_C15.indd 507 11/26/18 6:44 PM

508

Simone’s doctor is concerned about her elevated cholesterol, which could lead to coronary heart disease and a heart attack. He sends her to a nuclear medicine center to undergo a cardiac stress test. Pauline, the radiation technologist, explains to Simone that a stress test measures the blood flow to her heart muscle at rest and then during stress. The test is performed in a similar way as a routine exercise stress test, but images are produced that show areas of blood flow through the heart. It involves taking two sets of images of her heart, one when the heart is at rest, and one when she is walking on a treadmill.

Pauline tells Simone that she will inject thallium-201 into her bloodstream. She explains that Tl-201 is a radioactive isotope that has a half-life of 3.0 days. Simone is curious about the term “half-life.” Pauline explains that a half-life is the amount of time it takes for one-half of a radioactive sample to break down. She assures Simone that after four half-lives, the radiation emitted will be almost zero. Pauline tells Simone that Tl-201 decays to Hg-201 and emits energy similar to X-rays. When the Tl-201 reaches any areas within her heart with restricted blood supply, smaller amounts of the radioisotope will accumulate.

CAREER

Radiation Technologist A radiation technologist works in a hospital or imaging center where nuclear medicine is used to diagnose and treat a variety of medical conditions. In a diagnostic test, a radiation technologist uses a scanner, which converts radiation into images. The images are evaluated to determine any abnormalities in the body. A radiation technologist performs tests such as computed tomography (CT), magnetic resonance imaging (MRI), and positron emission tomography (PET) and operates the instrumentation and computers associated with these tests. In addition, they must physically and mentally prepare patients for imaging. A radiation technologist must know how to handle radioisotopes safely, use the necessary type of shielding, and give radioactive isotopes to patients. A patient may be given radioactive tracers such as technetium-99m, iodine-131, gallium-67, and thallium-201 that emit gamma radiation, which is detected and used to develop an image of the kidneys or thyroid or to follow the blood flow in the heart muscle.

Nuclear Chemistry

UPDATE Cardiac Imaging Using a Radioisotope

16

When Simone arrives at the clinic for her nuclear stress test, a

radioactive isotope is injected that helps the technologist make images

of her heart muscle. You can read about Simone’s scans and the results

of her nuclear stress test in the UPDATE Cardiac Imaging Using a

Radioisotope, page 532.

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16.1 Natural Radioactivity 509

2 4 He orAlpha particle a

16.1 Natural Radioactivity LEARNING GOAL Describe alpha, beta, positron, and gamma radiation.

Most naturally occurring isotopes of elements up to atomic number 19 have stable nuclei. Elements with atomic numbers 20 and higher usually have one or more isotopes that have unstable nuclei in which the nuclear forces cannot offset the repulsions between the protons. An unstable nucleus is radioactive, which means that it spontaneously emits small amounts of energy called radiation to become more stable. Radiation may take the form of alpha (a) and beta (b) particles, positrons (b+), or pure energy such as gamma (g) rays. An isotope of an element that emits radiation is called a radioisotope. For most types of radiation, there is a change in the number of protons in the nucleus, which means that an atom is converted into an atom of a different element. Elements with atomic num- bers of 93 and higher are produced artificially in nuclear laboratories and consist only of radioactive isotopes.

Symbols for Radioisotopes The atomic symbols for the different isotopes are written with the mass number in the upper left corner and the atomic number in the lower left corner. The mass number is the sum of the number of protons and neutrons in the nucleus, and the atomic number is equal to the number of protons. For example, a radioactive isotope of carbon used for archaeological dating has a mass number of 14 and an atomic number of 6.

REVIEW Writing Atomic Symbols for

Isotopes (4.5)

LOOKING AHEAD

16.1 Natural Radioactivity 509 16.2 Nuclear Reactions 512 16.3 Radiation

Measurement 519 16.4 Half-Life of a

Radioisotope 522 16.5 Medical Applications

Using Radioactivity 526 16.6 Nuclear Fission and

Fusion 529

TABLE 16.1 Stable and Radioactive Isotopes of Some Elements Magnesium Iodine Uranium

Stable Isotopes

12 24Mg 53

127I None

Magnesium-24 Iodine-127

Radioactive Isotopes

12 23Mg 53

125I 92 235U

Magnesium-23 Iodine-125 Uranium-235

12 27Mg 53

131I 92 238U

Magnesium-27 Iodine-131 Uranium-238

Radioactive isotopes are identified by writing the mass number after the element’s name or symbol. Thus, in this example, the isotope is called carbon-14 or C-14. TABLE 16.1 compares some stable, nonradioactive isotopes with some radioactive isotopes.

Mass number (protons and neutrons)

Atomic number (protons)

Symbol of element14 C 6 6 protons (red) 8 neutrons (white)

14 6C carbon-14 C-14

Types of Radiation By emitting radiation, an unstable nucleus forms a more stable, lower-energy nucleus. One type of radiation consists of alpha particles. An alpha particle is identical to a helium (He) nucleus, which has two protons and two neutrons. An alpha particle has a mass number of 4, an atomic number of 2, and a charge of 2 + . The symbol for an alpha particle is the Greek letter alpha (a) or the symbol of a helium nucleus except that the 2 + charge is omitted.

Another type of radiation is a beta particle, which is a high-energy electron, with a charge of 1 - and a mass number of 0. It is represented by the Greek letter beta (b) or by the symbol for the electron including the mass number and the charge (-1

0e).

Beta particle 0-1e or b

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510 CHAPTER 16 Nuclear Chemistry

A positron, similar to a beta particle, has a positive (1 + ) charge with a mass number of 0. It is represented by the Greek letter beta with a 1 + charge, b+, or by the symbol, +1 0e. A positron is an example of antimatter, a term physicists use to describe a particle that is the opposite of another particle, in this case, an electron.

Gamma rays are high-energy radiation, released when an unstable nucleus undergoes a rearrangement of its particles to give a more stable, lower-energy nucleus. Gamma rays are often emitted along with other types of radiation. A gamma ray is written as the Greek letter gamma, g. Because gamma rays are energy only, zeros are used to show that a gamma ray has no mass or charge (0

0g). TABLE 16.2 summarizes the types of radiation we use in nuclear equations.

ENGAGE 16.1 What is the charge and the mass number of an alpha particle emitted by a radioactive atom?

Gamma ray 0 0 g or g

e0+1 orPositron b +

TABLE 16.2 Some Forms of Radiation Type of Radiation Symbol Mass Number Charge

Alpha Particle 2 4He a 4 2 +

Beta Particle -1 0e b 0 1 -

Positron +1 0e b+ 0 1 +

Gamma Ray 0 0g g 0 0

Proton 1 1H p 1 1 +

Neutron 0 1 n n 1 0

PRACTICE PROBLEMS Try Practice Problems 16.1 to 16.10

SAMPLE PROBLEM 16.1 Radiation Particles

TRY IT FIRST

Identify and write the symbol for each of the following types of radiation: a. contains two protons and two neutrons b. has a mass number of 0 and a 1 - charge

SOLUTION

a. An alpha particle, 2 4He or a, has two protons and two neutrons.

b. A beta particle, -1 0e or b, has a mass number of 0 and a 1 - charge.

SELF TEST 16.1

a. Identify and write the symbol for the type of radiation that has a mass number of zero and a 1 + charge.

b. Identify and write the symbol for the type of radiation that has a mass number of 0 and a 0 charge.

ANSWER

a. A positron, +1 0e, has a mass number of 0 and a 1 + charge.

b. Gamma radiation, 0 0g, has a mass number of 0 and a 0 charge.

Biological Effects of Radiation When radiation strikes molecules in its path, electrons may be knocked away, forming unstable ions. If this ionizing radiation passes through the human body, it may interact with water molecules, removing electrons and producing H2O

+, which can cause undesirable chemical reactions.

The cells most sensitive to radiation are the ones undergoing rapid division—those of the bone marrow, skin, reproductive organs, and intestinal lining, as well as all cells of growing children. Damaged cells may lose their ability to produce necessary materials. For example, if radiation damages cells of the bone marrow, red blood cells may no longer be produced. If sperm cells, ova, or the cells of a fetus are damaged, birth defects may result. In contrast, cells of the nerves, muscles, liver, and adult bones are much less sensitive to radiation because they undergo little or no cellular division.

a

b

g

Different types of radiation penetrate the body to different depths.

M16_TIMB8119_06_SE_C16.indd 510 11/26/18 1:09 PM

16.1 Natural Radioactivity 511

Cancer cells are another example of rapidly dividing cells. Because cancer cells are highly sensitive to radiation, large doses of radiation are used to destroy them. The normal tis- sue that surrounds cancer cells divides at a slower rate and suffers less damage from radiation. However, radiation may cause malignant tumors, leukemia, anemia, and genetic mutations.

Radiation Protection Radiation technologists, chemists, doctors, and nurses who work with radioactive isotopes must use proper radiation protection. Proper shielding is necessary to prevent exposure. Alpha particles, which have the largest mass and charge of the radiation particles, travel only a few centimeters in the air before they collide with air molecules, acquire electrons, and become helium atoms. A piece of paper, clothing, and our skin are protection against alpha particles. Lab coats and gloves will also provide sufficient shielding. However, if alpha emitters are ingested or inhaled, the alpha particles they give off can cause serious internal damage.

Beta particles have a very small mass and move much faster and farther than alpha particles, traveling as much as several meters through air. They can pass through paper and penetrate as far as 4 to 5 mm into body tissue. External exposure to beta particles can burn the surface of the skin, but they do not travel far enough to reach the internal organs. Heavy clothing such as lab coats and gloves are needed to protect the skin from beta particles.

Gamma rays travel great distances through the air and pass through many materials, including body tissues. Because gamma rays penetrate so deeply, exposure to gamma rays can be extremely hazardous. Only very dense shielding, such as lead or concrete, will stop them. Syringes used for injections of radioactive materials use shielding made of lead or heavyweight materials such as tungsten and plastic composites.

When working with radioactive materials, medical personnel wear protective clothing and gloves and stand behind a shield (see FIGURE 16.1). Long tongs may be used to pick up vials of radioactive material, keeping them away from the hands and body. TABLE 16.3 summarizes the shielding materials required for the various types of radiation.

FIGURE 16.1 In a nuclear pharmacy, a person working with radioisotopes wears protective clothing and gloves and uses a lead glass shield on a syringe.

TABLE 16.3 Properties of Radiation and Shielding Required Property Alpha (A) Particle Beta (B) Particle Gamma (G) Ray

Travel Distance in Air 2 to 4 cm 200 to 300 cm 500 m

Tissue Depth 0.05 mm 4 to 5 mm 50 cm or more

Shielding Paper, clothing Heavy clothing, lab coats, gloves

Lead, thick concrete

Typical Source Radium-226 Carbon-14 Technetium-99m

ENGAGE 16.2 What types of radiation does a lead shield block?

PRACTICE PROBLEMS

16.1 Natural Radioactivity

16.1 Identify the type of particle or radiation for each of the following:

a. 2 4He b. +1

0e c. 0 0g

16.2 Identify the type of particle or radiation for each of the following:

a. -1 0e b. 1

1H c. 0 1 n

16.3 Naturally occurring potassium consists of three isotopes: potassium-39, potassium-40, and potassium-41.

a. Write the atomic symbol for each isotope. b. In what ways are the isotopes similar, and in what ways do

they differ?

PRACTICE PROBLEMS Try Practice Problems 16.11 and 16.12

People who work in nuclear medicine minimize the time they spend close to radioactive materials to reduce exposure. Remaining in a radioactive area twice as long increases exposure of a person to twice the amount of radiation. The greater the distance from the radioactive source, the lower the intensity of radiation received. By doubling the distance from the radiation source, the intensity of radiation drops to 11222 or one-fourth of its previous value.

M16_TIMB8119_06_SE_C16.indd 511 11/26/18 1:09 PM

512 CHAPTER 16 Nuclear Chemistry

16.4 Naturally occurring iodine is iodine-127. Medically, radioactive isotopes of iodine-125 and iodine-131 are used.

a. Write the atomic symbol for each isotope. b. In what ways are the isotopes similar, and in what ways do

they differ?

16.5 Identify each of the following: a. -1

0X b. 2 4X c. 0

1X

d. 18 38X e. 6

14X

16.6 Identify each of the following: a. 1

1X b. 35 81X c. 0

0X

d. 26 59 X e. +1

0X

Applications 16.7 Write the atomic symbol for each of the following isotopes

used in nuclear medicine: a. copper-64 b. selenium-75 c. sodium-24 d. nitrogen-15

16.8 Write the atomic symbol for each of the following isotopes used in nuclear medicine:

a. indium-111 b. palladium-103 c. barium-131 d. rubidium-82

16.9 Supply the missing information in the following table:

Medical Use Atomic Symbol

Mass Number

Number of Protons

Number of Neutrons

Heart imaging 81 201Tl

Radiation therapy 60 27

Abdominal scan 31 36

Hyperthyroidism 53 131I

Leukemia treatment 32 17

16.10 Supply the missing information in the following table:

Medical Use Atomic Symbol

Mass Number

Number of Protons

Number of Neutrons

Cancer treatment 55 131Cs

Brain scan 43 56

Blood flow 141 58

Bone scan 85 47

Lung function 54 133Xe

16.11 Match the type of radiation (1 to 3) with each of the following statements: 1. alpha particle 2. beta particle 3. gamma radiation

a. does not penetrate skin b. shielding protection includes lead or thick concrete c. can be very harmful if ingested

16.12 Match the type of radiation (1 to 3) with each of the following statements: 1. alpha particle 2. beta particle 3. gamma radiation

a. penetrates farthest into skin and body tissues b. shielding protection includes lab coats and gloves c. travels only a short distance in air

16.2 Nuclear Reactions LEARNING GOAL Write a balanced nuclear equation for radioactive decay, showing mass numbers and atomic numbers.

In a process called radioactive decay, a nucleus spontaneously breaks down by emitting radiation. This process is shown by writing a nuclear equation with the atomic symbols of the original radioactive nucleus on the left, an arrow, and the new nucleus and the type of radiation emitted on the right.

Radioactive nucleus h new nucleus + radiation (a, b, b+, g)

In a nuclear equation, the sum of the mass numbers and the sum of the atomic numbers on one side of the arrow must equal the sum of the mass numbers and the sum of the atomic numbers on the other side.

Alpha Decay An unstable nucleus may emit an alpha particle, which consists of two protons and two neutrons. Thus, the mass number of the radioactive nucleus decreases by 4, and its atomic number decreases by 2. For example, when uranium-238 emits an alpha particle, the new nucleus that forms has a mass number of 234 and an atomic number of 90. Uranium is transformed into a different element, thorium, an example of transmutation.

REVIEW Using Positive and Negative

Numbers in Calculations (1.4)

Solving Equations (1.4)

Counting Protons and Neutrons (4.4)

CORE CHEMISTRY SKILL Writing Nuclear Equations

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16.2 Nuclear Reactions 513

ENGAGE 16.3 What happens to the U-238 nucleus when an alpha particle is emitted?

+U Th He238 4234

Radioactive uranium nucleus

Thorium-234 nucleus

Alpha particle

Radiation

New nucleus

Neutron

146 neutrons 92 protons

Proton

238 92U

234 90Th

4 2He

144 neutrons 90 protons

2 neutrons 2 protons

Radioactive nucleus

New nucleus

Alpha particle

92 90 2

In the nuclear equation for alpha decay, the mass number of the new nucleus decreases by 4 and its atomic number decreases by 2.

We can look at writing a balanced nuclear equation for americium-241, which under- goes alpha decay as shown in Sample Problem 16.2.

SAMPLE PROBLEM 16.2 Writing a Nuclear Equation for Alpha Decay

TRY IT FIRST

Smoke detectors that are used in homes contain americium-241, which undergoes alpha decay. When alpha particles collide with air molecules, charged particles are produced that generate an electrical current. If smoke particles enter the detector, they interfere with the formation of charged particles in the air, and the electrical current is interrupted. This causes the alarm to sound and warns the occupants of the danger of fire. Write the balanced nuclear equation for the alpha decay of americium-241.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

Am-241, alpha decay

balanced nuclear equation

mass number, atomic number of new nucleus

STEP 1 Write the incomplete nuclear equation.

95 241Am h ? + 24He

STEP 2 Determine the missing mass number. In the equation, the mass number, 241, is equal to the sum of the mass numbers of the new nucleus and the alpha particle.

241 = ? + 4 ? = 241 - 4 = 237 (mass number of new nucleus)

STEP 3 Determine the missing atomic number. The atomic number, 95, must equal the sum of the atomic numbers of the new nucleus and the alpha particle.

95 = ? + 2 ? = 95 - 2 = 93 (atomic number of new nucleus)

STEP 4 Determine the symbol of the new nucleus and complete the nuclear equation. On the periodic table, the element that has atomic number 93 is neptunium, Np. The atomic symbol for this isotope of Np is written 93

237Np.

95 241Am h 93

237Np + 24He Pu

4 2He

241 95Am

237 93Np

A smoke detector contains the radioactive isotope americium-241, which undergoes alpha decay.

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514 CHAPTER 16 Nuclear Chemistry

PRACTICE PROBLEMS Try Practice Problems 16.13 and 16.14

SELF TEST 16.2

a. Write the balanced nuclear equation for the alpha decay of Po-214. b. Bismuth-213, used to treat leukemia, emits alpha particles that destroy cancer cells.

Write the balanced nuclear equation for the alpha decay of Bi-213.

ANSWER

a. 84 214Po h 82

210Pb + 24He b. 83213Bi h 81209Tl + 24He

Chemistry Link to Health Radon in Our Homes

The presence of radon gas has become an environmental and health issue because of the radiation danger it poses. Radioactive isotopes such as radium-226 are naturally present in many types of rocks and soils. Radium-226 emits an alpha particle and is converted into radon gas, which diffuses out of the rocks and soil.

88 226Ra h 86

222Rn + 24He

Outdoors, radon gas poses little danger because it disperses in the air. However, if the radioactive source is under a house or building, the radon gas can enter the house through cracks in the foundation or other openings. Those who live or work there may inhale the radon. In the lungs, radon-222 emits alpha particles to form polonium-218, which is known to cause lung cancer.

86 222Rn h 84

218Po + 24He

The U.S. Environmental Protection Agency (EPA) estimates that radon causes about 20 000 lung cancer deaths in one year. The EPA recommends that the maximum level of radon not exceed 4 pico- curies (pCi) per liter of air in a home. One picocurie (pCi) is equal to 10-12 curies (Ci); curies are described in Section 16.3. The EPA estimates that about one out of every 15 homes in the United States have radon levels that exceed this maximum.

A radon test kit is used to determine radon levels in buildings.

Beta Decay A beta particle forms as the result of the breakdown of a neutron into a proton and an elec- tron (beta particle). Because the proton remains in the nucleus, the number of protons increases by one, whereas the number of neutrons decreases by one. Thus, in a nuclear equation for beta decay, the mass number of the radioactive nucleus and the mass number of the new nucleus are the same. However, the atomic number of the new nucleus increases by one. For example, the beta decay of a carbon-14 nucleus produces a nitrogen-14 nucleus.

ENGAGE 16.4 What happens to a C-14 nucleus when a beta particle is emitted?

8 neutrons 6 protons

7 neutrons 7 protons

0 neutrons 0 protons 1- charge

14 6

Radioactive carbon nucleus

Radiation

New nucleus

Neutron

Proton

Beta particle

Stable nitrogen-14 nucleus

+ Radioactive

nucleus New

nucleus Beta

particle

eC N147 0

-1

N147

e0-1

14 6C

In the nuclear equation for beta decay, the mass number of the new nucleus remains the same, and its atomic number increases by 1.

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16.2 Nuclear Reactions 515

A radioisotope is injected into the joint to relieve the pain caused by arthritis.

0 -1e

90 39Y

90 40Zr

PRACTICE PROBLEMS Try Practice Problems 16.15 and 16.16

SAMPLE PROBLEM 16.3 Writing a Nuclear Equation for Beta Decay

TRY IT FIRST

The radioactive isotope yttrium-90, a beta emitter, is used in cancer treatment and as a colloidal injection into joints to relieve arthritis pain. Write the balanced nuclear equation for the beta decay of yttrium-90.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

Y-90, beta decay balanced nuclear equation

mass number, atomic number of new nucleus

STEP 1 Write the incomplete nuclear equation.

0 -1e

90 39Y ? +

STEP 2 Determine the missing mass number. In the equation, the mass number, 90, is equal to the sum of the mass numbers of the new nucleus and the beta particle.

90 = ? + 0 ? = 90 - 0 = 90 (mass number of new nucleus)

STEP 3 Determine the missing atomic number. The atomic number, 39, must equal the sum of the atomic numbers of the new nucleus and the beta particle.

39 = ? - 1 ? = 39 + 1 = 40 (atomic number of new nucleus)

STEP 4 Determine the symbol of the new nucleus and complete the nuclear equation. On the periodic table, the element that has atomic number 40 is zirconium, Zr. The atomic symbol for this isotope of Zr is written 40

90 Zr.

0 -1e

90 39Y

90 40Zr +

SELF TEST 16.3

a. Write the balanced nuclear equation for the beta decay of chromium-51. b. Write the balanced nuclear equation for the beta decay of Cu-64, used in cancer therapy.

ANSWER

a. 24 51Cr h 25

51Mn + -1 0e b. 29

64Cu h 30 64Zn + -1 0e

Positron Emission In positron emission, a proton in an unstable nucleus is converted to a neutron and a positron. The neutron remains in the nucleus, but the positron is emitted from the nucleus. In a nuclear equation for positron emission, the mass number of the radioactive nucleus and the mass number of the new nucleus are the same. However, the atomic number of the new nucleus decreases by one, indicating a change of one element into another. For example, an aluminum-24 nucleus under- goes positron emission to produce a magnesium-24 nucleus. The atomic number of magnesium (12) and the charge of the positron (1 + ) give the atomic number of aluminum (13).

24 13Al h

24 12Mg + 0+1e

Positron

Proton in the

nucleus

Positron emitted

New neutron remains in the nucleus

+

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516 CHAPTER 16 Nuclear Chemistry

Gamma Emission Pure gamma emitters are rare, although gamma radiation accompanies most alpha and beta radiation. In radiology, one of the most commonly used gamma emitters is technetium (Tc). The unstable isotope of technetium is written as the metastable (symbol m) isotope technetium-99m, Tc-99m, or 43

99mTc. By emitting energy in the form of gamma rays, the nucleus becomes more stable.

43 99mTc h 43

99Tc + 00g

FIGURE 16.2 summarizes the changes in the nucleus for alpha, beta, positron, and gamma radiation.

ENGAGE 16.5 Why does the atomic number but not the mass number change in positron emission?

PRACTICE PROBLEMS Try Practice Problems 16.17 and 16.18

SAMPLE PROBLEM 16.4 Writing a Nuclear Equation for Positron Emission

TRY IT FIRST

Write the balanced nuclear equation for the positron emission of manganese-49.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

Mn-49, positron emission

balanced nuclear equation

mass number, atomic number of new nucleus

STEP 1 Write the incomplete nuclear equation.

25 49Mn h ? + +1 0e

STEP 2 Determine the missing mass number. In the equation, the mass number, 49, is equal to the sum of the mass numbers of the new nucleus and the positron.

49 = ? + 0

? = 49 - 0 = 49 (mass number of new nucleus)

STEP 3 Determine the missing atomic number. The atomic number, 25, must equal the sum of the atomic numbers of the new nucleus and the positron.

25 = ? + 1

? = 25 - 1 = 24 (atomic number of new nucleus)

STEP 4 Determine the symbol of the new nucleus and complete the nuclear equation. On the periodic table, the element that has atomic number 24 is chromium, Cr. The atomic symbol for this isotope of Cr is written 24

49Cr.

25 49Mn h 24

49Cr + +1 0e

SELF TEST 16.4

a. Write the balanced nuclear equation for the positron emission of xenon-118. b. Write the balanced nuclear equation for the positron emission of arsenic-74, used in

cancer imaging.

ANSWER

a. 54 118Xe h 53

118I + +1 0e b. 33

74As h 32 74Ge + +1 0e

0 +1e

49 24Cr

49 25Mn

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16.2 Nuclear Reactions 517

INTERACTIVE VIDEO Writing Equations for an Isotope Produced by Bombardment

Producing Radioactive Isotopes Today, many radioisotopes are produced in small amounts by bombarding stable, nonra- dioactive isotopes with high-speed particles such as alpha particles, protons, neutrons, and small nuclei. When one of these particles is absorbed, the stable nucleus is converted to a radioactive isotope and usually some type of radiation particle.

FIGURE 16.2 When the nuclei of alpha, beta, positron, and gamma emitters emit radiation, new, more stable nuclei are produced.

Radiation Source

Alpha emitter

Radiation

He

New Nucleus

New element

4 2 +

Mass number - 4 Atomic number - 2

Beta emitter

e New

element 0

-1 +

Mass number same Atomic number + 1

Positron emitter +

Mass number same Atomic number - 1

Gamma emitter

Stable nucleus of the same element

+

Mass number same Atomic number same

g

e0+1

0 0

New element

PRACTICE PROBLEMS Try Practice Problems 16.19 and 16.20

1

Neutron

13

New radioactive nucleus

10

Stable nucleus

4 + + Bombarding

particle

+

2He 5B 7N 0 n When nonradioactive B-10 is bombarded by an alpha particle, the products are radioactive N-13 and a neutron.

All elements that have an atomic number greater than 92 have been produced by bom- bardment. Most have been produced in only small amounts and exist for only a short time, making it difficult to study their properties. For example, when californium-249 is bombarded with nitrogen-15, the radioactive element dubnium-260 and four neutrons are produced.

7 15N + 98249Cf h 105260Db + 401n

Technetium-99m is a radioisotope used in nuclear medicine for several diagnostic pro- cedures, including brain tumor detection and liver and spleen examinations. The source of technetium-99m is molybdenum-99, which is produced in a nuclear reactor by neutron bombardment of molybdenum-98.

0 1n + 4298Mo h 4299Mo

Many radiology laboratories have small generators containing molybdenum-99, which decays to the technetium-99m radioisotope.

42 99Mo h 43

99mTc + -1 0e

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518 CHAPTER 16 Nuclear Chemistry

The technetium-99m radioisotope decays by emitting gamma rays. Gamma emission is desirable for diagnostic work because the gamma rays pass through the body to the detec- tion equipment.

43 99mTc h 43

99Tc + 00g

A generator is used to prepare technetium-99m.

PRACTICE PROBLEMS Try Practice Problems 16.21 and 16.22

SAMPLE PROBLEM 16.5 Writing a Nuclear Equation for an Isotope Produced by Bombardment

TRY IT FIRST

Write the balanced nuclear equation for the bombardment of nickel-58 by a proton, 1 1H,

which produces a radioactive isotope and an alpha particle.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

Ni-58, proton bombardment, to produce alpha particle

new isotope, balanced

nuclear equation

mass number, atomic number of new nucleus

STEP 1 Write the incomplete nuclear equation.

1 1H + 2858Ni h ? + 24He

STEP 2 Determine the missing mass number. In the equation, the sum of the mass numbers of the proton, 1, and the nickel, 58, must equal the sum of the mass numbers of the new nucleus and the alpha particle.

1 + 58 = ? + 4 ? = 59 - 4 = 55 (mass number of new nucleus)

STEP 3 Determine the missing atomic number. The sum of the atomic numbers of the proton, 1, and nickel, 28, must equal the sum of the atomic numbers of the new nucleus and the alpha particle.

1 + 28 = ? + 2 ? = 29 - 2 = 27 (atomic number of new nucleus)

STEP 4 Determine the symbol of the new nucleus, and complete the nuclear equation. On the periodic table, the element that has atomic number 27 is cobalt, Co. The atomic symbol for this isotope of Co is written 27

55Co.

1 1H + 2858Ni h 2755Co + 24He

SELF TEST 16.5

a. The first radioactive isotope was produced in 1934 by the bombardment of aluminum-27 by an alpha particle to produce a radioactive isotope and one neutron. Write the balanced nuclear equation for this bombardment.

b. When iron-54 is bombarded by an alpha particle, the products are a new isotope and two protons. Write the balanced nuclear equation for this bombardment.

ANSWER

a. 2 4He + 1327Al h 1530P + 01n

b. 2 4He + 2654Fe h 2656Fe + 211H

55 27Co

58 28Ni

1 1H

4 2He

PRACTICE PROBLEMS

16.2 Nuclear Reactions

16.13 Write a balanced nuclear equation for the alpha decay of each of the following radioactive isotopes:

a. 84 208 Po b. 90

232 Th c. 102 251 No d. radon-220

16.14 Write a balanced nuclear equation for the alpha decay of each of the following radioactive isotopes:

a. curium-243 b. 99 252 Es c. 98

251 Cf d. 107 261 Bh

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16.3 Radiation Measurement 519

16.3 Radiation Measurement LEARNING GOAL Describe how radiation is detected and measured.

One of the most common instruments for detecting beta and gamma radiation is the Geiger counter. It consists of a metal tube filled with a gas such as argon. When radiation enters a window on the end of the tube, it forms charged particles in the gas, which produce an electrical current. Each burst of current is amplified to give a click and a reading on a meter.

Ar + radiation h Ar+ + e-

Measuring Radiation Radiation is measured in several different ways. When a radiology laboratory obtains a radioisotope, the activity of the sample is measured in terms of the number of nuclear disintegrations per second. The curie (Ci), the original unit of activity, was defined as the number of disintegrations that occur in 1 s for 1 g of radium, which is equal to 3.7 * 1010 disintegrations/s. The unit was named for Polish scientist Marie Curie, who along with her husband, Pierre, discovered the radioactive elements radium and polonium. The SI unit of radiation activity is the becquerel (Bq), which is 1 disintegration/s.

Radiation levels in workers at the Fukushima Daiichi nuclear power plant are measured.

Radiation

Meter

High-voltage power supply

Electrons

A Geiger counter detects alpha particles, beta particles, and gamma rays using the ionization effect produced in a Geiger–Müller tube.

16.15 Write a balanced nuclear equation for the beta decay of each of the following radioactive isotopes:

a. 11 25 Na b. 8

20 O c. strontium-92 d. iron-60

16.16 Write a balanced nuclear equation for the beta decay of each of the following radioactive isotopes:

a. 19 44 K b. iron-59 c. potassium-42 d. 56

141 Ba

16.17 Write a balanced nuclear equation for the positron emission of each of the following radioactive isotopes:

a. silicon-26 b. cobalt-54 c. 37 77 Rb d. 45

93 Rh

16.18 Write a balanced nuclear equation for the positron emission of each of the following radioactive isotopes:

a. boron-8 b. 8 15 O c. 19

40 K d. nitrogen-13

16.19 Complete each of the following nuclear equations, and describe the type of radiation:

a. 13 28 Al h ? + -1 0e b. 73180m Ta h 73180 Ta + ?

c. 29 66 Cu h 30

66 Zn + ? d. ? h 90234 Th + 24 He e. 80

188 Hg h ? + +1 0e

16.20 Complete each of the following nuclear equations, and describe the type of radiation:

a. 6 11 C h 5

11 B + ? b. 1635 S h ? + -1 0e c. ? h 39

90 Y + -1 0e d. 83210 Bi h ? + 24 He e. ? h 39

89 Y + +1 0e

16.21 Complete each of the following bombardment reactions: a. 0

1 n + 49 Be h ? b. 0

1 n + 52131 Te h ? + -1 0e c. 0

1 n + ? h 1124 Na + 24 He d. 2

4 He + 714 N h ? + 11 H

16.22 Complete each of the following bombardment reactions: a. ? + 1840 Ar h 1943 K + 11 H b. 0

1 n + 92238 U h ? c. 0

1 n + ? h 614 C + 11 H d. ? + 2864 Ni h 111272 Rg + 01 n

The rad (radiation absorbed dose) is a unit that measures the amount of radiation absorbed by a gram of material such as body tissue. The SI unit for absorbed dose is the gray (Gy), which is defined as the joules of energy absorbed by 1 kg of body tissue. The gray is equal to 100 rad.

The rem (radiation equivalent in humans) is a unit that measures the biological effects of different kinds of radiation. Although alpha particles do not penetrate the skin,

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520 CHAPTER 16 Nuclear Chemistry

if they should enter the body by some other route, they can cause extensive damage within a short distance in tissue. High-energy radiation, such as beta particles, high-energy pro- tons, and neutrons that travel into tissue, causes more damage. Gamma rays are damaging because they travel a long way through body tissue.

To determine the equivalent dose or rem dose, the absorbed dose (rad) is multiplied by a factor that adjusts for biological damage caused by a particular form of radiation. For beta and gamma radiation the factor is 1, so the biological damage in rems is the same as the absorbed radiation (rad). For high-energy protons and neutrons, the factor is about 10, and for alpha particles it is 20.

Biological damage (rem) = Absorbed dose (rad) * Factor Often, the measurement for an equivalent dose will be in units of millirems (mrem). One

rem is equal to 1000 mrem. The SI unit is the sievert (Sv). One sievert is equal to 100 rem. TABLE 16.4 summarizes the units used to measure radiation.

ENGAGE 16.6 What is the difference between becquerel and rem?

TABLE 16.4 Units of Radiation Measurement Measurement Common Unit SI Unit Relationship

Activity curie (Ci) 1 Ci = 3.7 * 1010 disintegrations/s

becquerel (Bq) 1 Bq = 1 disintegration/s

1 Ci = 3.7 * 1010 Bq

Absorbed Dose rad gray (Gy) 1 Gy = 1 J/kg of tissue

1 Gy = 100 rad

Biological Damage rem sievert (Sv) 1 Sv = 100 rem

Chemistry Link to Health Radiation and Food

Foodborne illnesses caused by pathogenic bacteria such as Salmonella, Listeria, and Escherichia coli (E. coli) have become a major health concern in the United States. E. coli has been respon- sible for outbreaks of illness from contaminated ground beef, fruit juices, lettuce, and alfalfa sprouts.

The U.S. Food and Drug Administration (FDA) has approved the use of 0.3 to 1 kGy of radiation produced by cobalt-60 or cesium-137 for the treatment of foods. The irradiation technology is much like that used to sterilize medical supplies. Radioactive cobalt pellets are placed in stainless steel tubes, which are arranged in racks. When food moves through the series of racks, the gamma rays pass through the food and kill the bacteria.

It is important for consumers to understand that when food is irradiated, it never comes in contact with the radioactive source. The gamma rays pass through the food to kill bacteria, but that does not make the food radioactive. The radiation kills bacteria because it stops their ability to divide and grow. We cook or heat food thoroughly for the same purpose. Radiation, as well as heat, has little effect on the food itself because its cells are no longer dividing or growing. Thus irradiated food is not harmed, although a small amount of vitamins B1 and C may be lost.

Currently, tomatoes, blueberries, strawberries, and mushrooms are being irradiated to allow them to be harvested when completely ripe and extend their shelf life (see FIGURE 16.3). The FDA has also approved the irradiation of pork, poultry, and beef to decrease potential infections and to extend shelf life. Currently, irradiated

vegetable and meat products are available in retail markets in more than 40 countries. In the United States, irradiated foods such as tropical fruits, spinach, and ground meats are found in some stores. Apollo 17 astronauts ate irradiated foods on the Moon, and some U.S. hospitals and nursing homes now use irradiated poultry to reduce the possibility of salmonella infections among residents. The extended shelf life of irradiated food also makes it useful for campers and military personnel. Soon, consumers concerned about food safety will have a choice of irradiated meats, fruits, and vegetables at the market.

FIGURE 16.3 Irradiation of food

The FDA requires this symbol to appear on irradiated retail foods.

After two weeks, the irradiated strawberries on the right show no spoilage. Mold is growing on the nonirradiated ones on the left.

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16.3 Radiation Measurement 521

A dosimeter measures radiation exposure.

People who work in radiology laboratories wear dosimeters attached to their clothing to determine any exposure to radiation such as X-rays, gamma rays, or beta particles. A dosimeter can be thermoluminescent (TLD), optically stimulated luminescence (OSL), or electronic personal (EPD). Dosimeters provide real-time radiation levels measured by moni- tors in the work area.

SAMPLE PROBLEM 16.6 Radiation Measurement

TRY IT FIRST

One treatment for bone pain involves intravenous administration of the radioisotope phosphorus-32, which is incorporated into bone. A typical dose of 7.0 mCi can produce up to 450 rad in the bone. What is the difference between the units of mCi and rad?

SOLUTION

The millicuries (mCi) indicate the activity of the P-32 in terms of the number of nuclei that break down in 1 s. The radiation absorbed dose (rad) is a measure of the amount of radiation absorbed by the bone.

SELF TEST 16.6

a. What is the absorbed dose of 450 rad in grays (Gy)? b. What is the activity of 7.0 mCi in becquerels (Bq)?

ANSWER

a. 4.5 Gy b. 2.6 * 108 Bq

PRACTICE PROBLEMS Try Practice Problems 16.23 to 16.28

TABLE 16.5 Average Annual Radiation Received by a Person in the United States

Source Dose (mSv)

Natural

Ground 0.2

Air, water, food 0.3

Cosmic rays 0.4

Wood, concrete, brick 0.5

Medical

Chest X-ray 0.2

Dental X-ray 0.2

Mammogram 0.4

Hip X-ray 0.6

Lumbar spine X-ray 0.7

Upper gastrointestinal tract X-ray

2

Other

Nuclear power plants 0.001

Television 0.2

Air travel 0.1

Radon 2*

*Varies widely.

Exposure to Radiation Every day, we are exposed to low levels of radiation from naturally occurring radioactive isotopes in the buildings where we live and work, in our food and water, and in the air we breathe. For example, potassium-40, a naturally occurring radioactive isotope, is present in any potassium-containing food. Other naturally occurring radioisotopes in air and food are carbon-14, radon-222, strontium-90, and iodine-131. The average person in the United States is exposed to about 3.6 mSv of radiation annually. Medical sources of radiation, includ- ing dental, hip, spine, and chest X-rays and mammograms, add to our radiation exposure. TABLE 16.5 lists some common sources of radiation.

Another source of background radiation is cosmic radiation produced in space by the Sun. People who live at high altitudes or travel by airplane receive a greater amount of cosmic radiation because there are fewer molecules in the atmosphere to absorb the radia- tion. For example, a person living in Denver receives about twice the cosmic radiation as a person living in Los Angeles. A person living close to a nuclear power plant normally does not receive much additional radiation, perhaps 0.001 mSv in 1 yr. However, in the accident at the Chernobyl nuclear power plant in 1986 in Ukraine, it is estimated that people in a nearby town received as much as 10 mSv/h.

Radiation Sickness The larger the dose of radiation received at one time, the greater the effect on the body. Exposure to radiation of less than 0.25 Sv usually cannot be detected. Whole-body exposure of 1 Sv produces a temporary decrease in the number of white blood cells. If the exposure to radiation is greater than 1 Sv, a person may suffer the symptoms of radiation sickness: nausea, vomiting, fatigue, and a reduction in white-cell count. A whole-body dosage greater than 3 Sv can decrease the white-cell count to zero. The person suffers diarrhea, hair loss, and infection. Exposure to radiation of 5 Sv is expected to cause death in 50% of the people receiving that dose. This amount of radiation to the whole body is called the lethal dose for one-half the population, or the LD50. The LD50 varies for different life forms, as TABLE 16.6 shows. Whole - body radiation of 6 Sv or greater would be fatal to all humans within a few weeks.

TABLE 16.6 Lethal Doses of Radiation for Some Life Forms

Life Form LD50 (Sv)

Insect 1000

Bacterium 500

Rat 8

Human 5

Dog 3

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522 CHAPTER 16 Nuclear Chemistry

PRACTICE PROBLEMS

16.3 Radiation Measurement

16.23 Match each property (1 to 3) with its unit of measurement. 1. activity 2. absorbed dose 3. biological damage

a. rad b. mrem c. mCi d. Gy

16.24 Match each property (1 to 3) with its unit of measurement. 1. activity 2. absorbed dose 3. biological damage

a. mrad b. gray c. becquerel d. Sv

Applications 16.25 Two technicians in a nuclear laboratory were accidentally

exposed to radiation. If one was exposed to 8 mGy and the other to 5 rad, which technician received more radiation?

16.26 Two samples of a radioisotope were spilled in a nuclear laboratory. The activity of one sample was 8 kBq and the other 15 mCi. Which sample produced the higher amount of radiation?

16.27 a. The recommended dosage of iodine-131 is 4.20 mCi/kg of body mass. How many microcuries of iodine-131 are needed for a 70.0-kg person with hyperthyroidism?

b. A person receives 50 rad of gamma radiation. What is that amount in grays?

16.28 a. The dosage of technetium-99m for a lung scan is 20. mCi/kg of body mass. How many millicuries of technetium-99m should be given to a 50.0-kg person (1 mCi = 1000 mCi)?

b. Suppose a person absorbed 50 mrad of alpha radiation. What would be the equivalent dose in millisieverts?

16.4 Half-Life of a Radioisotope LEARNING GOAL Given the half-life of a radioisotope, calculate the amount of radioisotope remaining after one or more half-lives.

The half-life of a radioisotope is the amount of time it takes for one-half of a sample to decay. For example, 53

131I has a half-life of 8.0 days. As 53 131I decays, it produces the non-

radioactive isotope 54 131Xe and a beta particle.

53 131I h 54

131Xe + -1 0e

Suppose we have a sample that initially contains 20. mg of 53 131I. In 8.0 days, one-half (10. mg)

of all the I-131 nuclei in the sample will decay, which leaves 10. mg of I-131. After 16 days (two half-lives), 5.0 mg of the remaining I-131 decays, which leaves 5.0 mg of I-131. After 24 days (three half-lives), 2.5 mg of the remaining I-131 decays, which leaves 2.5 mg of I-131 nuclei still capable of producing radiation.

REVIEW Interpreting Graphs (1.4)

Using Conversion Factors (2.6)

ENGAGE 16.7 If a 24-mg sample of Tc-99m has a half-life of 6.0 h, why are only 3.0 mg of Tc-99m radioactive after 18 h?

I

1 half-life

8.0 days 16 days 24 days

Xe

2 half-lives 3 half-lives

131 131 Xe131 Xe131

I131

I131

I131

131 53

53

53

53

53

54 54 54

I 13153I 131 53I

131 53I

1 half-life

8.0 days 20. mg of

2 half-lives 10. mg of

3 half-lives 5.0 mg of 2.5 mg of

16 days 24 days

In one half-life, the activity of an isotope decreases by half.

A decay curve is a diagram of the decay of a radioactive isotope. FIGURE 16.4 shows such a curve for the 53

131I we have discussed.

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16.4 Half-Life of a Radioisotope 523

FIGURE 16.4 The decay curve for iodine-131 shows that one-half of the radioactive sample decays and one-half remains radioactive after each half-life of 8.0 days.

1 half-life

2 half-lives 3 half-lives 4 half-lives

5 half-livesA m

ou nt

o f

I- 13

1 (m

g)

20

15

10

5

0 0 8 16 24 32 40

Time (days)

SAMPLE PROBLEM 16.7 Using Half-Lives of a Radioisotope in Medicine

TRY IT FIRST

Phosphorus-32, a radioisotope used in the treatment of leukemia, has a half-life of 14.3 days. If a sample contains 8.0 mg of phosphorus-32, how many milligrams of phosphorus-32 remain after 42.9 days?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

8.0 mg of P-32, 42.9 days elapsed, half@life = 14.3 days

milligrams of P-32 remaining

number of half-lives

STEP 2 Write a plan to calculate the unknown quantity.

Half-life number of half-livesdays

Number of half-lives

32 15 P

32 15 Pmilligrams of remainingmilligrams of

STEP 3 Write the half-life equality and conversion factors.

14.3 days and

1 half-life = 14.3 days

1 half-life

1 half-life

14.3 days

STEP 4 Set up the problem to calculate the needed quantity. First, we determine the number of half-lives in the amount of time that has elapsed.

= 42.9 daysnumber of half-lives * 1 half-life

14.3 days = 3.00 half-lives

Now we can determine how much of the sample decays in three half-lives and how many milligrams of the phosphorus remain.

1 half-life 8.0 mg of

2 half-lives32 15P 4.0 mg of

32 15P 2.0 mg of

32 15P 1.0 mg of

32 15P

3 half-lives

14.3 days 28.6 days 42.9 days

SELF TEST 16.7

a. Iron-59 has a half-life of 44 days. If a nuclear laboratory receives a sample of 8.0 mg of iron-59, how many micrograms of iron-59 are still active after 176 days?

b. Rubidium-82, used to diagnose myocardial disease, has a half-life of 1.27 min. If the initial dose is 1480 MBq, how many half-lives have elapsed, and what is the activity of Rb-82 after 5.08 min?

ANSWER

a. 0.50 mg of iron-59 b. 4 half-lives; 92.5 MBq

CORE CHEMISTRY SKILL Using Half-Lives

INTERACTIVE VIDEO

Half-Lives

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524 CHAPTER 16 Nuclear Chemistry

TABLE 16.7 Half-Lives of Some Radioisotopes Element Radioisotope Half-Life Type of Radiation

Naturally Occurring Radioisotopes

Carbon-14 6 14C 5730 yr Beta

Potassium-40 19 40K 1.3 * 109 yr Beta, gamma

Radium-226 88 226Ra 1600 yr Alpha

Strontium-90 38 90Sr 38.1 yr Alpha

Uranium-238 92 238U 4.5 * 109 yr Alpha

Some Medical Radioisotopes

Carbon-11 6 11C 20. min Positron

Chromium-51 24 51Cr 28 days Gamma

Iodine-131 53 131I 8.0 days Gamma

Oxygen-15 8 15O 2.0 min Positron

Iron-59 26 59Fe 44 days Beta, gamma

Radon-222 86 222Rn 3.8 days Alpha

Technetium-99m 43 99mTc 6.0 h Gamma

PRACTICE PROBLEMS Try Practice Problems 16.29 to 16.34

Naturally occurring isotopes of the elements usually have long half-lives, as shown in TABLE 16.7. They disintegrate slowly and produce radiation over a long period of time, even hundreds or millions of years. In contrast, the radioisotopes used in nuclear medicine have much shorter half-lives. They disintegrate rapidly and produce almost all their radiation in a short period of time. For example, technetium-99m emits half of its radiation in the first 6 h. This means that a small amount of the radioisotope given to a patient is essentially gone within two days. The decay products of technetium-99m are totally eliminated by the body.

ENGAGE 16.8 Why do radioisotopes used in nuclear medicine have short half-lives?

Chemistry Link to the Environment Dating Ancient Objects

Radiological dating is a technique used by geologists, archaeologists, and historians to determine the age of ancient objects. The age of an object derived from plants or animals (such as wood, fiber, natural pigments, bone, and cotton or woolen clothing) is determined by measuring the amount of carbon-14, a naturally occurring radioactive form of carbon. In 1960, Willard Libby received the Nobel Prize for the work he did developing carbon-14 dating techniques during the 1940s. Carbon-14 is produced in the upper atmosphere by the bom- bardment of 7

14N by high-energy neutrons from cosmic rays.

1 0n + 14 7N h 14 6C + 11H

Neutron from Nitrogen in Radioactive Proton cosmic rays atmosphere carbon-14

The carbon-14 reacts with oxygen to form radioactive carbon dioxide, 6

14CO2. Living plants continuously absorb carbon dioxide, which incorporates carbon-14 into the plant material. The uptake of carbon-14 stops when the plant dies. As the carbon-14 undergoes beta decay, the amount of radioactive carbon-14 in the plant material steadily decreases.

6 14C h 7

14N + -1 0e

In a process called carbon dating, scientists use the half-life of carbon-14 (5730 yr) to calculate the length of time since the plant died. For example, a wooden beam found in an ancient dwelling might have one-half of the carbon-14 found in living plants today.

Because one half-life of carbon-14 is 5730 yr, the dwelling was con- structed about 5730 yr ago. Carbon-14 dating was used to determine that the Dead Sea Scrolls are about 2000 yr old.

A radiological dating method used for determining the age of much older items is based on the radioisotope uranium-238, which decays through a series of reactions to lead-206. The uranium-238 isotope has an incredibly long half-life, about 4 * 109 yr. Measure- ments of the amounts of uranium-238 and lead-206 enable geologists to determine the age of rock samples. The older rocks will have a higher percentage of lead-206 because more of the uranium-238 has decayed. The age of rocks brought back from the Moon by the Apollo missions, for example, was determined using uranium-238. They were found to be about 4 * 109 yr old, approximately the same age calculated for Earth.

The age of the Dead Sea Scrolls was determined using carbon-14.

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16.4 Half-Life of a Radioisotope 525

SAMPLE PROBLEM 16.8 Dating Using Half-Lives

TRY IT FIRST

The bones of humans and animals assimilate carbon until death. Using radiocarbon dating, the number of half-lives of carbon-14 from a bone sample determines the age of the bone. Suppose a sample is obtained from a prehistoric animal and used for radiocarbon dating. We can calculate the age of the bone or the years elapsed since the animal died by using the half-life of carbon-14, which is 5730 yr. A bone sample from the skeleton of a prehistoric animal has 25% of the activity of C-14 found in a living animal. How many years ago did the prehistoric animal die?

SOLUTION

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM

Given Need Connect

half-life = 5730 yr, 25% of initial C-14 activity

years elapsed number of half-lives

STEP 2 Write a plan to calculate the unknown quantity.

1.0 half-life Activity: 100%

(initial)

2.0 half-lives 50% 25%

STEP 3 Write the half-life equality and conversion factors.

5730 yr 1 half-life

and 1 half-life 5730 yr

1 half-life = 5730 yr

STEP 4 Set up the problem to calculate the needed quantity.

2.0 half-lives * = 11 000 yr 1 half-life

5730 yr Years elapsed =

We would estimate that the animal died 11 000 yr ago.

SELF TEST 16.8

a. Suppose that a piece of wood found in a cave had one-eighth of its original carbon-14 activity. About how many years ago was the wood part of a living tree?

b. How many micrograms of C-14 were initially in an organism if 4.0 mcg remain after 11 460 yr?

ANSWER

a. 17 000 yr b. 16 mcg

PRACTICE PROBLEMS

16.4 Half-Life of a Radioisotope

16.29 For each of the following, indicate if the number of half-lives elapsed is: 1. one half-life 2. two half-lives 3. three half-lives

a. a sample of Pd-103 with a half-life of 17 days after 34 days b. a sample of C-11 with a half-life of 20 min after 20 min c. a sample of At-211 with a half-life of 7 h after 21 h

16.30 For each of the following, indicate if the number of half-lives elapsed is: 1. one half-life 2. two half-lives 3. three half-lives

a. a sample of Ce-141 with a half-life of 32.5 days after 32.5 days b. a sample of F-18 with a half-life of 110 min after 330 min c. a sample of Au-198 with a half-life of 2.7 days after 5.4 days

The age of a bone sample from a skeleton can be determined by carbon dating.

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526 CHAPTER 16 Nuclear Chemistry

Applications 16.31 Technetium-99m is an ideal radioisotope for scanning organs

because it has a half-life of 6.0 h and is a pure gamma emitter. Suppose that 80.0 mg were prepared in the technetium generator this morning. How many milligrams of technetium-99m would remain after each of the following intervals?

a. one half-life b. two half-lives c. 18 h d. 1.5 days

16.32 A sample of sodium-24 with an activity of 12 mCi is used to study the rate of blood flow in the circulatory system. If sodium-24 has a half-life of 15 h, what is the activity after each of the following intervals?

a. one half-life b. 30 h c. three half-lives d. 2.5 days

16.33 Strontium-85, used for bone scans, has a half-life of 65 days. a. How long will it take for the radiation level of strontium-85

to drop to one-fourth of its original level? b. How long will it take for the radiation level of strontium-85

to drop to one-eighth of its original level?

16.34 Fluorine-18, which has a half-life of 110 min, is used in PET scans.

a. If 100. mg of fluorine-18 is shipped at 8:00 a.m., how many milligrams of the radioisotope are still active after 110 min?

b. If 100. mg of fluorine-18 is shipped at 8:00 a.m., how many milligrams of the radioisotope are still active when the sample arrives at the radiology laboratory at 1:30 p.m.?

16.5 Medical Applications Using Radioactivity LEARNING GOAL Describe the use of radioisotopes in medicine.

The first medical radioactive isotope was used to treat a person with leukemia at the University of California at Berkeley. In 1946, radioactive iodine was successfully used to diagnose thy- roid function and to treat hyperthyroidism and thyroid cancer. Radioactive isotopes are now used to produce images of organs including the liver, spleen, thyroid, kidneys, brain, and heart.

To determine the condition of an organ in the body, a radiation technologist may use a radioisotope that concentrates in that organ. The cells in the body do not differentiate between a nonradioactive atom and a radioactive one, so these radioisotopes are easily incorporated. Then the radioactive atoms are detected because they emit radiation. Some radioisotopes used in nuclear medicine are listed in TABLE 16.8.

TABLE 16.8 Medical Applications of Radioisotopes Isotope Half-Life Radiation Medical Application

Au-198 2.7 days Beta Liver imaging; treatment of abdominal carcinoma

Bi-213 46 min Alpha Treatment of leukemia

Ce-141 32.5 days Beta Gastrointestinal tract diagnosis; measuring blood flow to the heart

Cs-131 9.7 days Gamma Prostate brachytherapy

F-18 110 min Positron Positron emission tomography (PET)

Ga-67 78 h Gamma Abdominal imaging; tumor detection

Ga-68 68 min Positron Detection of pancreatic cancer

I-123 13.2 h Gamma Treatment of thyroid, brain, and prostate cancer

I-131 8.0 days Beta Treatment of Graves’ disease, goiter, hyperthyroid- ism, thyroid and prostate cancer

Ir-192 74 days Gamma Treatment of breast and prostate cancer

P-32 14.3 days Beta Treatment of leukemia, excess red blood cells, and pancreatic cancer

Pd-103 17 days Gamma Prostate brachytherapy

Sm-153 46 h Beta Treatment of bone cancer

Sr-85 65 days Gamma Detection of bone lesions; brain scans

Tc-99m 6.0 h Gamma Imaging of skeleton and heart muscle, brain, liver, heart, lungs, bone, spleen, kidney, and thyroid; most widely used radioisotope in nuclear medicine

Xe-133 5.2 days Beta Pulmonary function diagnosis

Y-90 2.7 days Beta Treatment of liver cancer

Ir-192, breast

Sr-85, bone

Tc-99m, brain, bone

Bi-213, P-32, leukemia

Ga-68, pancreas

I-131, thyroid

Au-198, liver

Ce-141, GI tract

Cs-131, Pd-103, prostate

Different radioisotopes are used to diagnose and treat a number of diseases.

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16.5 Medical Applications Using Radioactivity 527

Scans with Radioisotopes After a person receives a radioisotope, the radiation technologist determines the level and location of radioactivity emitted by the radioisotope. An apparatus called a scanner is used to produce an image of the organ. The scanner moves slowly across the body above the region where the organ containing the radioisotope is located. The gamma rays emitted from the radioisotope in the organ can be used to expose a photographic plate, producing a scan of the organ. On a scan, an area of decreased or increased radiation can indicate conditions such as a disease of the organ, a tumor, a blood clot, or edema.

A common method of determining thyroid function is the use of radioactive iodine uptake. Taken orally, the radioisotope iodine-131 mixes with the iodine already present in the thyroid. Twenty-four hours later, the amount of iodine taken up by the thyroid is deter- mined. A detection tube held up to the area of the thyroid gland detects the radiation coming from the iodine-131 that has located there (see FIGURE 16.5).

A person with a hyperactive thyroid will have a higher than normal level of radioac- tive iodine, whereas a person with a hypoactive thyroid will have lower values. If a person has hyperthyroidism, treatment is begun to lower the activity of the thyroid. One treatment involves giving a therapeutic dosage of radioactive iodine, which has a higher radiation level than the diagnostic dose. The radioactive iodine goes to the thyroid where its radiation destroys some of the thyroid cells. The thyroid produces less thyroid hormone, bringing the hyperthyroid condition under control.

Positron Emission Tomography Positron emitters with short half-lives such as carbon-11, oxygen-15, nitrogen-13, and fluorine-18 are used in an imaging method called positron emission tomography (PET). A positron-emitting isotope such as fluorine-18 combined with substances in the body such as glucose is used to study brain function, metabolism, and blood flow.

9 18F h 8

18O + +1 0e As positrons are emitted, they combine with electrons to produce gamma rays that are

detected by computerized equipment to create a three-dimensional image of the organ (see FIGURE 16.6).

Nonradioactive Imaging Computed Tomography Another imaging method used to scan organs such as the brain, lungs, and heart is computed tomography (CT). A computer monitors the absorption of 30 000 X-ray beams directed at successive layers of the target organ. Based on the densities of the tissues and fluids in the organ, the differences in absorption of the X-rays provide a series of images of the organ. This technique is successful in the identification of hemorrhages, tumors, and atrophy.

Magnetic Resonance Imaging Magnetic resonance imaging (MRI) is a powerful imaging technique that does not involve nuclear radiation. It is the least invasive imaging method available. MRI is based on the absorption of energy when the protons in hydrogen atoms are excited by a strong magnetic field. The difference in energy between the two states is released, which produces the

FIGURE 16.5 Detecting radiation from an organ

A scanner detects radiation from a radioisotope in an organ.

A thyroid scan uses radioactive iodine-131.

FIGURE 16.6 These PET scans of the brain show a normal brain on the left and a brain affected by Alzheimer’s disease on the right.

A CT scan shows a tumor (purple) in the brain.

An MRI scan provides images of the heart and lungs.

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528 CHAPTER 16 Nuclear Chemistry

electromagnetic signal that the scanner detects. These signals are sent to a computer system, where a color image of the body is generated. MRI is particularly useful in obtaining images of soft tissues, which contain large amounts of hydrogen atoms.

Titanium “seeds” filled with a radioactive isotope are implanted in the body to treat cancer.

SAMPLE PROBLEM 16.9 Medical Applications of Radioactivity

TRY IT FIRST

In the treatment of abdominal carcinoma, a person is treated with “seeds” of gold-198, which is a beta emitter. Write the balanced nuclear equation for the beta decay of gold-198.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

gold-198, beta decay

balanced nuclear equation

mass number, atomic number of new nucleus

STEP 1 Write the incomplete nuclear equation.

79 198Au h ? + -10e

STEP 2 Determine the missing mass number. In beta decay, the mass number, 198, does not change.

79 198Au h 198? + -10e

STEP 3 Determine the missing atomic number. The atomic number of the new isotope increases by one.

79 198Au h 198 80? + -10e

STEP 4 Determine the symbol of the new nucleus. On the periodic table, mercury, Hg, has the atomic number of 80.

STEP 5 Complete the nuclear equation.

79 198Au h 198 80Hg + -10e

SELF TEST 16.9

a. In an experimental treatment, a person is given boron-10, which is taken up by malignant tumors. When bombarded with neutrons, boron-10 decays by emitting alpha particles that destroy the surrounding tumor cells. Write the balanced nuclear equation for the reaction for this experimental procedure.

b. Fluorine-18, a source of positrons for PET scans, is synthesized by bombarding oxygen-18 with protons. Write the balanced nuclear equation for this reaction.

ANSWER

a. 10n + 10 5B h 73Li + 42He b. 11H + 18 8O h 18 9F + 10n

Chemistry Link to Health Brachytherapy

The process called brachytherapy, or seed implantation, is an internal form of radiation therapy. The prefix brachy is from the Greek word for short distance. With internal radiation, a high dose of radiation is delivered to a cancerous area, whereas normal tissue sustains minimal damage. Because higher doses are used, fewer treatments of shorter duration are needed. Conventional external treatment delivers a lower dose per treatment, but requires six to eight weeks of treatment.

Permanent Brachytherapy One of the most common forms of cancer in males is prostate can- cer. In addition to surgery and chemotherapy, one treatment option is to place 40 or more titanium capsules, or “seeds,” in the malignant area. Each seed, which is the size of a grain of rice, contains radio- active iodine-125, palladium-103, or cesium-131, which decay by gamma emission. The radiation from the seeds destroys the cancer by

PRACTICE PROBLEMS Try Practice Problems 16.35 to 16.40

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16.6 Nuclear Fission and Fusion 529

interfering with the reproduction of cancer cells with minimal damage to adjacent normal tissues. Ninety percent (90%) of the radioisotopes decay within a few months because they have short half-lives.

Isotope I-125 Pd-103 Cs-131

Radiation Gamma Gamma Gamma

Half-Life 60 days 17 days 10 days

Time Required to Deliver 90% of Radiation

7 months 2 months 1 month

Almost no radiation passes out of the patient’s body. The amount of radiation received by a family member is no greater than that received on a long plane flight. Because the radioisotopes decay to products that are not radioactive, the inert titanium capsules can be left in the body.

Temporary Brachytherapy In another type of treatment for prostate cancer, long needles con- taining iridium-192 are placed in the tumor. However, the needles are removed after 5 to 10 min, depending on the activity of the iridium isotope. Compared to permanent brachytherapy, temporary brachy- therapy can deliver a higher dose of radiation over a shorter time. The procedure may be repeated in a few days.

Brachytherapy is also used following breast cancer lumpectomy. An iridium-192 isotope is inserted into the catheter implanted in the space left by the removal of the tumor. Radiation is delivered primarily to the tissue surrounding the cavity that contained the tumor and where the cancer is most likely to recur. The procedure is repeated twice a day for five days to give an absorbed dose of 34 Gy (3400 rad). The catheter is removed, and no radioactive material remains in the body.

In conventional external beam therapy for breast cancer, a patient is given 2 Gy once a day for six to seven weeks, which gives a total absorbed dose of about 80 Gy or 8000 rad. The external beam therapy irradiates the entire breast, including the tumor cavity.

A catheter is placed temporarily in the breast for radiation from Ir-192.

PRACTICE PROBLEMS

16.5 Medical Applications Using Radioactivity

Applications 16.35 Bone and bony structures contain calcium and phosphorus.

a. Why would the radioisotopes calcium-47 and phosphorus-32 be used in the diagnosis and treatment of bone diseases?

b. During nuclear tests, scientists were concerned that strontium-85, a radioactive product, would be harmful to the growth of bone in children. Explain.

16.36 a. Technetium-99m emits only gamma radiation. Why would this type of radiation be used in diagnostic imaging rather than an isotope that also emits beta or alpha radiation?

b. A person with polycythemia vera (excess production of red blood cells) receives radioactive phosphorus-32. Why would this treatment reduce the production of red blood cells in the bone marrow of the patient?

16.37 In a diagnostic test for leukemia, a person receives 4.0 mL of a solution containing selenium-75. If the activity of the selenium-75 is 45 mCi/mL, what dose, in microcuries, does the patient receive?

16.38 A vial contains radioactive iodine-131 with an activity of 2.0 mCi/mL. If a thyroid test requires 3.0 mCi in an “atomic cocktail,” how many milliliters are used to prepare the iodine-131 solution?

16.39 Gallium-68 is taken up by tumors; the emission of positrons allows the tumors to be located.

a. Write an equation for the positron emission of Ga-68. b. If the half-life is 68 min, how much of a 64-mcg sample is

active after 136 min?

16.40 Xenon-133 is used to test lung function; it decays by emitting a beta particle.

a. Write an equation for the beta decay of Xe-133. b. If the half-life of Xe-133 is 5.2 h, how much of a 20.-mCi

sample is still active after 15.6 h?

16.6 Nuclear Fission and Fusion LEARNING GOAL Describe the processes of nuclear fission and fusion.

During the 1930s, scientists bombarding uranium-235 with neutrons discovered that the U-235 nucleus splits into two smaller nuclei and produces a great amount of energy. This was the discovery of nuclear fission. The energy generated by splitting the atom was called atomic energy.

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530 CHAPTER 16 Nuclear Chemistry

Unstable

235 92U

235 92U

236 92U

236 92U

91 36Kr

91 36Kr

142 56 Ba

142 56 Ba energy+ + ++10n 310n

310n 1 0n

If we could determine the mass of the products krypton, barium, and three neutrons with great accuracy, we would find that their total mass is slightly less than the mass of the starting materials. The missing mass has been converted into an enormous amount of energy, consistent with the famous equation derived by Albert Einstein:

E = mc2

where E is the energy released, m is the mass lost, and c is the speed of light, 3 * 108 m/s. Even though the mass loss is very small, when it is multiplied by the speed of light squared, the result is a large value for the energy released. The fission of 1 g of uranium-235 produces about as much energy as the burning of 3 tons of coal.

Chain Reaction Fission begins when a neutron collides with the nucleus of a uranium atom. The resulting nucleus is unstable and splits into smaller nuclei. This fission process also releases several neutrons and large amounts of gamma radiation and energy. The neutrons emitted have high energies and bombard other uranium-235 nuclei. In a chain reaction, there is a rapid increase in the number of high-energy neutrons available to react with more uranium. To sus- tain a nuclear chain reaction, sufficient quantities of uranium-235 must be brought together to provide a critical mass in which almost all the neutrons immediately collide with more uranium-235 nuclei. So much heat and energy build up that an atomic explosion can occur (see FIGURE 16.7).

Nuclear Fusion In fusion, two small nuclei combine to form a larger nucleus. Mass is lost, and a tremendous amount of energy is released, even more than the energy released from nuclear fission. However, a fusion reaction requires a temperature of 100 000 000 °C to overcome the repulsion of the hydrogen nuclei and cause them to undergo fusion. Fusion reactions occur continuously in the Sun and other stars, providing us with heat and light. The huge amounts of energy produced by our sun come from the fusion of 6 * 1011 kg of hydrogen every second. In a fusion reaction, isotopes of hydrogen combine to form helium and large amounts of energy.

PRACTICE PROBLEMS Try Practice Problems 16.41 to 16.44

energy+ + +

+ + +

4 2He

1 0n

2 1H

3 1H

Hydrogen isotopes combine in a fusion reaction to produce helium, a neutron, and energy.

A typical equation for nuclear fission is

ENGAGE 16.9 Why is a critical mass of uranium-235 necessary to sustain a nuclear chain reaction?

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16.6 Nuclear Fission and Fusion 531

Neutron (n)

235 92U

235 92U

235 92U

235 92U

91 36Kr

91 36Kr

91 36Kr

91 36Kr

3

142 56Ba

142 56Ba

142 56Ba

142 56Ba

1 0n

310n

310n

310n

1 0n

FIGURE 16.7 In a nuclear chain reaction, the fission of each uranium-235 atom produces three neutrons that cause the nuclear fission of more and more uranium-235 atoms.

Scientists expect less radioactive waste with shorter half-lives from fusion reactors. However, fusion is still in the experimental stage because the extremely high temperatures needed have been difficult to reach and even more difficult to maintain.

SAMPLE PROBLEM 16.10 Identifying Fission and Fusion

TRY IT FIRST

Indicate whether each of the following is characteristic of the fission or fusion process, or both:

a. A large nucleus breaks apart to produce smaller nuclei. b. Large amounts of energy are released. c. Extremely high temperatures are needed for reaction.

SOLUTION

a. When a large nucleus breaks apart to produce smaller nuclei, the process is fission. b. Large amounts of energy are released in both the fusion and fission processes. c. An extremely high temperature is required for fusion.

SELF TEST 16.10

Classify each of the following nuclear equations as fission or fusion:

a. 2 3He + 23He h 24He + 211H b. 01 n + 92235U h 54140Xe + 3894Sr + 201 n

ANSWER

a. fusion b. fission

PRACTICE PROBLEMS Try Practice Problems 16.45 and 16.46

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532 CHAPTER 16 Nuclear Chemistry

PRACTICE PROBLEMS

16.6 Nuclear Fission and Fusion

16.41 What is nuclear fission?

16.42 How does a chain reaction occur in nuclear fission?

16.43 Complete the following fission reaction:

0 1 n + 92235U h 50131Sn + ? + 201 n + energy

16.44 In another fission reaction, uranium-235 bombarded with a neutron produces strontium-94, another smaller nucleus, and three neutrons. Write the balanced nuclear equation for the fission reaction.

16.45 Indicate whether each of the following is characteristic of the fission or fusion process, or both:

a. Neutrons bombard a nucleus. b. The nuclear process occurs in the Sun. c. A large nucleus splits into smaller nuclei. d. Small nuclei combine to form larger nuclei.

16.46 Indicate whether each of the following is characteristic of the fission or fusion process, or both:

a. Very high temperatures are required to initiate the reaction. b. Less radioactive waste is produced. c. Hydrogen nuclei are the reactants. d. Large amounts of energy are released when the nuclear

reaction occurs.

UPDATE Cardiac Imaging Using a Radioisotope

As part of her nuclear stress test, Simone starts to walk on the treadmill. When she reaches the maximum level, Pauline injects a radioactive isotope containing Tl-201 with an activity of 74 MBq. The radiation emitted from areas of the heart is detected by a scanner and produces

images of her heart muscle. A thallium stress test can determine how effectively coronary arteries provide blood to the heart. If Simone has any damage to her coronary arteries, reduced blood flow during stress would show a narrowing of an artery or a blockage. After Simone rests for 3 h, Pauline injects more Tl-201, and she is placed under the scanner again. A second set of images of her heart muscle at rest are taken. When Simone’s doctor reviews her scans, he assures her that she had normal blood flow to her heart muscle, both at rest and under stress.

Applications 16.47 What is the activity of the radioactive isotope injection for

Simone a. in curies? b. in millicuries?

16.48 If the half-life of Tl-201 is 3.0 days, what is its activity, in megabecquerels

a. after 3.0 days? b. after 6.0 days?

16.49 How many days will it take until the activity of the Tl-201 in Simone’s body is one-eighth of the initial activity?

16.50 Radiation from Tl-201 to Simone’s kidneys can be 24 mGy. What is this amount of radiation in rads?

A thallium stress test can show narrowing of an artery during stress.

Nuclear Equations

GammaAlpha Beta Positron

Activity

Biological Effect

are measured by their

measured in

from the nucleus as

RemGrayRad

as absorbed as exposure

or

Half-Life

Nuclear Medicine

Fission Fusion

and in

and are used in

emit

as shown in

decay by one-half in one

Sievert

Curie

Becquerel

NUCLEAR CHEMISTRY

Radiation and Energy

Radioisotopes

CONCEPT MAP

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Key Terms 533

CHAPTER REVIEW

16.1 Natural Radioactivity LEARNING GOAL Describe alpha, beta, positron, and gamma radiation. • Radioactive isotopes have unstable nuclei that break down (decay),

spontaneously emitting alpha (a), beta (b), positron (b+), and gamma (g) radiation.

• Because radiation can damage the cells in the body, shielding is used to protect from radiation.

16.2 Nuclear Reactions LEARNING GOAL Write a balanced nuclear equation for radioactive decay, showing mass numbers and atomic numbers. • A balanced nuclear

equation is used to represent the changes that take place in the nuclei of the reactants and products.

• The new isotopes and the type of radiation emitted can be determined from the symbols that show the mass numbers and atomic numbers of the isotopes in the nuclear equation.

• A radioisotope is produced artificially when a nonradioactive isotope is bombarded by a small particle.

16.3 Radiation Measurement LEARNING GOAL Describe how radiation is detected and measured. • In a Geiger counter, radiation produces

charged particles in the gas contained in a tube, which generates an electrical current.

• The curie (Ci) and the becquerel (Bq) measure the activity, which is the number of nuclear disintegrations per second.

• The amount of radiation absorbed by a substance is measured in the rad or the gray (Gy).

• The rem and the sievert (Sv) are units used to measure the biological damage from the different types of radiation.

16.4 Half-Life of a Radioisotope LEARNING GOAL Given the half-life of a radioisotope, calculate the amount of radioisotope remaining after one or more half-lives. • Every radioisotope has

its own rate of emitting radiation.

• The time it takes for one-half of a radioactive sample to decay is called its half-life.

• For many medical radioisotopes, such as Tc-99m and I-131, half- lives are short.

• For other isotopes, usually naturally occurring ones such as C-14, Ra-226, and U-238, half-lives are extremely long.

16.5 Medical Applications Using Radioactivity LEARNING GOAL Describe the use of radioisotopes in medicine. • In nuclear medicine, radioisotopes that go

to specific sites in the body are given to the patient.

• By detecting the radiation they emit, an evaluation can be made about the location and extent of an injury, disease, tumor, or the level of function of a particular organ.

• Higher levels of radiation are used to treat or destroy tumors.

16.6 Nuclear Fission and Fusion LEARNING GOAL Describe the processes of nuclear fission and fusion. • In fission, the bombardment

of a large nucleus breaks it apart into smaller nuclei, releasing one or more types of radiation and a great amount of energy.

• In fusion, small nuclei combine to form larger nuclei while great amounts of energy are released.

0 n 1

235 92U

3

91 36Kr

142 56 Ba

1 0 n

1 half-life

2 half-lives 3 half-lives

4 half-lives 5 half-lives

A m

ou nt

o f

I- 13

1 (m

g)

20

15

10

5

0 0 8 16 24 32 40

Time (days)

14

Radioactive carbon nucleus

Radiation

New nucleus

Beta particle 0

Stable nitrogen-14 nucleus

-1e

6C

14 7N

2 4 He or Alpha particle

a

alpha particle A nuclear particle identical to a helium nucleus, symbol a or 2

4He. becquerel (Bq) A unit of activity of a radioactive sample equal to

one disintegration per second. beta particle A particle identical to an electron, symbol -1

0e or b, that forms in the nucleus when a neutron changes to a proton and an electron.

chain reaction A fission reaction that will continue once it has been initiated by a high-energy neutron bombarding a heavy nucleus such as uranium-235.

curie (Ci) A unit of activity of a radioactive sample equal to 3.7 * 1010 disintegrations/s.

decay curve A diagram of the decay of a radioactive element. equivalent dose The measure of biological damage from an

absorbed dose that has been adjusted for the type of radiation.

fission A process in which large nuclei are split into smaller nuclei, releasing large amounts of energy.

fusion A reaction in which large amounts of energy are released when small nuclei combine to form larger nuclei.

gamma ray High-energy radiation, symbol 0 0g, emitted by an unstable

nucleus. gray (Gy) A unit of absorbed dose equal to 100 rad. half-life The length of time it takes for one-half of a radioactive

sample to decay. positron A particle of radiation with no mass and a positive charge,

symbol b+ or +1 0e, produced when a proton is transformed into a

neutron and a positron. rad (radiation absorbed dose) A measure of an amount of radiation

absorbed by the body. radiation Energy or particles released by radioactive atoms.

KEY TERMS

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534 CHAPTER 16 Nuclear Chemistry

radioactive decay The process by which an unstable nucleus breaks down with the release of high-energy radiation.

radioisotope A radioactive atom of an element.

rem (radiation equivalent in humans) A measure of the biological damage caused by radiation (rad * radiation biological factor).

sievert (Sv) A unit of biological damage (equivalent dose) equal to 100 rem.

The chapter section containing each Core Chemistry Skill is shown in parentheses at the end of each heading.

Writing Nuclear Equations (16.2) • A nuclear equation is written with the atomic symbols of the origi-

nal radioactive nucleus on the left, an arrow, and the new nucleus and the type of radiation emitted on the right.

• The sum of the mass numbers and the sum of the atomic numbers on one side of the arrow must equal the sum of the mass numbers and the sum of the atomic numbers on the other side.

• When an alpha particle is emitted, the mass number of the new nucleus decreases by 4, and its atomic number decreases by 2.

• When a beta particle is emitted, there is no change in the mass number of the new nucleus, but its atomic number increases by one.

• When a positron is emitted, there is no change in the mass number of the new nucleus, but its atomic number decreases by one.

• In gamma emission, there is no change in the mass number or the atomic number of the new nucleus.

Example: a. Write a balanced nuclear equation for the alpha decay of Po-210.

b. Write a balanced nuclear equation for the beta decay of Co-60.

Answer: a. When an alpha particle is emitted, we calculate the decrease of 4 in the mass number (210) of the polonium, and a decrease of 2 in its atomic number.

84 210Po h 82

206? + 24He

CORE CHEMISTRY SKILLS Because lead has atomic number 82, the new nucleus must be an isotope of lead.

84 210Po h 82

206Pb + 24He b. When a beta particle is emitted, there is no change

in the mass number (60) of the cobalt, but there is an increase of 1 in its atomic number.

27 60Co h 28

60? + -1 0e Because nickel has atomic number 28, the new nucleus must be an isotope of nickel.

27 60Co h 28

60Ni + -1 0e

Using Half-Lives (16.4) • The half-life of a radioisotope is the amount of time it takes for

one-half of a sample to decay. • The remaining amount of a radioisotope is calculated by dividing

its quantity or activity by one-half for each half-life that has elapsed.

Example: Co-60 has a half-life of 5.3 yr. If the initial sample of Co-60 has an activity of 1200 mCi, what is its activity after 15.9 yr?

Answer:

number of half@lives = 15.9 yr * 1 half@life

5.3 yr = 3.0 half@lives

1200 mCi iih 1 half@life

600 mCi iih 2 half@lives

300 mCi iih 3 half@lives

150 mCi

In 15.9 yr, three half-lives have passed. The activity was reduced from 1200 mCi to 150 mCi.

UNDERSTANDING THE CONCEPTS

The chapter sections to review are shown in parentheses at the end of each problem.

In problems 16.51 to 16.54, a nucleus is shown with protons and neutrons.

16.53 Draw the nucleus of the isotope that is bombarded in the following: (16.2)

+ +

+ positron

+ beta particle

proton neutron

16.51 Draw the new nucleus when this isotope emits a positron to complete the following: (16.2)

16.52 Draw the nucleus that emits a beta particle to complete the following: (16.2)

16.54 Complete the bombardment reaction by drawing the nucleus of the new isotope that is produced in the following: (16.2)

++

16.55 Carbon dating of small bits of charcoal used in cave paintings has determined that some of the paintings are from 10 000 to 30 000 yr old. Carbon-14 has a half-life of 5730 yr. In a 1@mg sample of carbon from a live tree, the activity of

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Additional Practice Problems 535

carbon-14 is 6.4 mCi. If researchers determine that 1 mg of charcoal from a prehistoric cave painting in France has an activity of 0.80 mCi, what is the age of the painting? (16.4)

16.56 Use the following decay curve for iodine-131 to answer problems a to c: (16.4)

a. Complete the values for the mass of radioactive iodine-131 on the vertical axis.

b. Complete the number of days on the horizontal axis. c. What is the half-life, in days, of iodine-131?

The technique of carbon dating is used to determine the age of ancient cave paintings.

M as

s of

r ad

io ac

ti ve

(m g)

80

20

0 8 16 Days

53 I

13 1

ADDITIONAL PRACTICE PROBLEMS

16.57 Determine the number of protons and number of neutrons in the nucleus of each of the following: (16.1)

a. sodium-25 b. nickel-61 c. rubidium-84 d. silver-110

16.58 Determine the number of protons and number of neutrons in the nucleus of each of the following: (16.1)

a. boron-10 b. zinc-72 c. iron-59 d. gold-198

16.59 Identify each of the following as alpha decay, beta decay, positron emission, or gamma emission: (16.2)

a. 13 27mAl h 13

27 Al + 00 g b. 58 B h 48 Be + +1 0e c. 86

220 Rn h 84 216 Po + 24 He

16.60 Identify each of the following as alpha decay, beta decay, positron emission, or gamma emission: (16.2)

a. 55 127 Cs h 54

127 Xe + +1 0e b. 3890 Sr h 3990 Y + -1 0e c. 85

218 At h 83 214 Bi + 24 He

16.61 Write the balanced nuclear equation for each of the following: (16.2)

a. Th-225 (a decay) b. Bi-210 (a decay) c. cesium-137 (b decay) d. tin-126 (b decay) e. F-18 (b+ emission)

16.62 Write the balanced nuclear equation for each of the following: (16.2)

a. potassium-40 (b decay) b. sulfur-35 (b decay) c. platinum-190 (a decay) d. Ra-210 (a decay) e. In-113m (g emission)

16.63 Complete each of the following nuclear equations: (16.2) a. 2

4 He + 714 N h ? + 11 H b. 2

4 He + 1327 Al h 1430 Si + ? c. 0

1 n + 92235 U h 3890 Sr + 301 n + ? d. 12

23m Mg h ? + 00 g

16.64 Complete each of the following nuclear equations: (16.2) a. ? + 2759 Co h 2556 Mn + 24 He b. ? h 7

14 N + -1 0e c. -1

0e + 3676 Kr h ? d. 2

4 He + 95241 Am h ? + 201 n

16.65 Write the balanced nuclear equation for each of the following: (16.2)

a. When two oxygen-16 atoms collide, one of the products is an alpha particle.

b. When californium-249 is bombarded by oxygen-18, a new element, seaborgium-263, and four neutrons are produced.

c. Radon-222 undergoes alpha decay. d. An atom of strontium-80 emits a positron.

16.66 Write the balanced nuclear equation for each of the following: (16.2)

a. Actinium-225 decays to give francium-221. b. Bismuth-211 emits an alpha particle. c. A radioisotope emits a positron to form titanium-48. d. An atom of germanium-69 emits a positron.

16.67 A 120-mg sample of technetium-99m is used for a diagnostic test. If technetium-99m has a half-life of 6.0 h, how many milligrams of the technetium-99m sample remains active 24 h after the test? (16.4)

16.68 The half-life of oxygen-15 is 124 s. If a sample of oxygen-15 has an activity of 4000 Bq, how many minutes will elapse before it has an activity of 500 Bq? (16.4)

16.69 What is the difference between fission and fusion? (16.6)

16.70 What are the products in the fission of uranium-235 that make possible a nuclear chain reaction? (16.6)

16.71 Where does fusion occur naturally? (16.6)

16.72 Why are scientists continuing to try to build a fusion reactor even though the very high temperatures it requires have been difficult to reach and maintain? (16.6)

Applications 16.73 The activity of K-40 in a 70.-kg human body is estimated to

be 120 nCi. What is this activity in becquerels? (16.3)

16.74 The activity of C-14 in a 70.-kg human body is estimated to be 3.7 kBq. What is this activity in microcuries? (16.3)

16.75 If the amount of radioactive phosphorus-32, used to treat leukemia, in a sample decreases from 1.2 mg to 0.30 mg in 28.6 days, what is the half-life of phosphorus-32? (16.4)

16.76 If the amount of radioactive iodine-123, used to treat thyroid cancer, in a sample decreases from 0.4 mg to 0.1 mg in 26.4 h, what is the half-life of iodine-123? (16.4)

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536 CHAPTER 16 Nuclear Chemistry

16.77 Calcium-47, used to evaluate disorders in calcium metabo- lism, has a half-life of 4.5 days. (16.2, 16.4)

a. Write the balanced nuclear equation for the beta decay of calcium-47.

b. How many milligrams of a 16-mg sample of calcium-47 remain after 18 days?

c. How many days have passed if 4.8 mg of calcium-47 decayed to 1.2 mg of calcium-47?

16.78 Cesium-137, used in cancer treatment, has a half-life of 30 yr. (16.2, 16.4)

a. Write the balanced nuclear equation for the beta decay of cesium-137.

b. How many milligrams of a 16-mg sample of cesium-137 remain after 90 yr?

c. How many years are required for 28 mg of cesium-137 to decay to 3.5 mg of cesium-137?

The following problems are related to the topics in this chapter. However, they do not all follow the chapter order, and they require you to combine concepts and skills from several sections. These problems will help you increase your critical thinking skills and prepare for your next exam.

16.79 Write the balanced nuclear equation for each of the following radioactive emissions: (16.2)

a. an alpha particle from Hg-180 b. a beta particle from Au-198 c. a positron from Rb-82

16.80 Write the balanced nuclear equation for each of the following radioactive emissions: (16.2)

a. an alpha particle from Gd-148 b. a beta particle from Sr-90 c. a positron from Al-25

16.81 All the elements beyond uranium, the transuranium elements, have been prepared by bombardment and are not naturally occurring elements. The first transuranium element neptunium, Np, was prepared by bombarding U-238 with neutrons to form a neptunium atom and a beta particle. Complete the following equation: (16.2)

0 1n + 92238U h ? + ?

16.82 One of the most recently produced transuranium elements, oganesson-294 (Og-294), atomic number 118, was prepared by bombarding californium-249 with another isotope. Complete the following equation for the preparation of this new element: (16.2)

? + 98249Cf h 118294Og + 301n 16.83 A 64@mCi sample of Tl-201 decays to 4.0 mCi in 12 days.

What is the half-life, in days, of Tl-201? (16.3, 16.4)

16.84 A wooden object from the site of an ancient temple has a carbon-14 activity of 10 counts/min compared with a

reference piece of wood cut today that has an activity of 40 counts/min. If the half-life for carbon-14 is 5730 yr, what is the age of the ancient wood object? (16.3, 16.4)

16.85 Element 114 was recently named flerovium, symbol Fl. The reaction for its synthesis involves bombarding Pu-244 with Ca-48. Write the balanced nuclear equation for the synthesis of flerovium. (16.2)

16.86 Element 116 was recently named livermorium, symbol Lv. The reaction for its synthesis involves bombarding Cm-248 with Ca-48. Write the balanced nuclear equation for the synthesis of livermorium. (16.2)

Applications 16.87 The half-life for the radioactive decay of calcium-47 is

4.5 days. If a sample has an activity of 1.0 mCi after 27 days, what was the initial activity, in microcuries, of the sample? (16.3, 16.4)

16.88 The half-life for the radioactive decay of Ce-141 is 32.5 days. If a sample has an activity of 4.0 mCi after 130. days have elapsed, what was the initial activity, in microcuries, of the sample? (16.3, 16.4)

16.89 A nuclear technician was accidentally exposed to potassium-42 while doing brain scans for possible tumors. The error was not discovered until 36 h later when the activity of the potassium-42 sample was 2.0 mCi. If potassium-42 has a half-life of 12 h, what was the activity of the sample at the time the technician was exposed? (16.3, 16.4)

16.90 The radioisotope sodium-24 is used to determine the levels of electrolytes in the body. A 16@mg sample of sodium-24 decays to 2.0 mg in 45 h. What is the half-life, in hours, of sodium-24? (16.4)

CHALLENGE PROBLEMS

ANSWERS TO ENGAGE QUESTIONS 16.6 The becquerel measures the number of disintegrations/s,

whereas the rem measures the biological effect of radiation.

16.7 After 18 h, the Tc-99m sample has had 3.0 half-lives. The amount remaining is 1/8 (½ * ½ * ½) of the original 24 mg, or 3.0 mg.

16.8 Radioisotopes used in nuclear medicine have short half-lives, which allows them to be eliminated quickly from the body.

16.9 If there is less than a critical mass of U-235 available, too many neutrons escape, and a chain reaction cannot be sustained.

16.1 An alpha particle has a charge of 2 + and a mass number of 4. 16.2 Lead shielding will block alpha, beta, and gamma radiation.

16.3 When an alpha particle is emitted by a U-238 nucleus, a thorium-234 nucleus is formed.

16.4 When a beta particle is emitted by a C-14 nucleus, a nitrogen-14 nucleus is formed.

16.5 When a positron is emitted, a proton in the nucleus is changed to a neutron. The atomic number decreases but the mass number, the total number of protons plus neutrons, stays the same.

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Answers to Selected Problems 537

ANSWERS TO SELECTED PROBLEMS 16.51 16.1 a. alpha particle b. positron c. gamma radiation

16.3 a. 19 39K 19

40K 19 41K

b. They all have 19 protons and 19 electrons, but they differ in the number of neutrons.

16.5 a. -1 0e or b b. 2

4He or a c. 0 1n or n

d. 18 38Ar e. 6

14C

16.7 a. 29 64Cu b. 34

75Se c. 11 24Na d. 7

15N

16.9

Medical Use Atomic Symbol

Mass Number

Number of Protons

Number of Neutrons

Heart imaging 81 201Tl 201 81 120

Radiation therapy 27 60Co 60 27 33

Abdominal scan 31 67Ga 67 31 36

Hyperthyroidism 53 131I 131 53 78

Leukemia treatment 15 32P 32 15 17

16.11 a. 1, alpha particle b. 3, gamma radiation c. 1, alpha particle

16.13 a. 84 208 Po h 82

204 Pb + 24 He b. 90232 Th h 88228 Ra + 24 He c. 102

251 No h 100 247 Fm + 24 He d. 86220 Rn h 84216 Po + 24 He

16.15 a. 11 25 Na h 12

25 Mg + -1 0e b. 820 O h 920 F + -1 0e c. 38

92 Sr h 39 92 Y + -1 0e d. 2660 Fe h 2760 Co + -1 0e

16.17 a. 14 26 Si h 13

26 Al + +1 0e b. 2754 Co h 2654 Fe + +1 0e c. 37

77 Rb h 36 77 Kr + +1 0e d. 4593 Rh h 4493 Ru + +1 0e

16.19 a. 14 28 Si, beta decay b. 0

0 g, gamma emission

c. -1 0e, beta decay d. 92

238 U, alpha decay

e. 79 188 Au, positron emission

16.21 a. 4 10 Be b. 53

132 I c. 13 27 Al d. 8

17 O

16.23 a. 2, absorbed dose b. 3, biological damage c. 1, activity d. 2, absorbed dose

16.25 The technician exposed to 5 rad received more radiation.

16.27 a. 294 mCi b. 0.5 Gy

16.29 a. two half-lives b. one half-life c. three half-lives

16.31 a. 40.0 mg b. 20.0 mg c. 10.0 mg d. 1.25 mg

16.33 a. 130 days b. 195 days

16.35 a. The radioactive isotopes of Ca and P will become part of the bony structures of the body, where their radiation can be used to diagnose or treat bone diseases.

b. The body will accumulate radioactive strontium in bones in the same way that it incorporates calcium. Radioactive strontium is harmful to children because its radiation causes more damage in cells that are dividing rapidly.

16.37 180 mCi

16.39 a. 31 68 Ga h 30

68 Mn + +1 0e b. 16 mcg 16.41 Nuclear fission is the splitting of a large atom into smaller

fragments with the release of a large amount of energy.

16.43 42 103Mo

16.45 a. fission b. fusion c. fission d. fusion

16.47 a. 2.0 * 10-3 Ci b. 2.0 mCi 16.49 9.0 days

+ positron

16.53

++

16.55 17 000 yr old

16.57 a. 11 protons and 14 neutrons b. 28 protons and 33 neutrons c. 37 protons and 47 neutrons

d. 47 protons and 63 neutrons

16.59 a. gamma emission b. positron emission c. alpha decay

16.61 a. 90 225 Th h 88

221 Ra + 24 He b. 83210 Bi h 81206 Tl + 24 He c. 55

137 Cs h 56 137 Ba + -1 0e d. 50126 Sn h 51126 Sb + -1 0e

e. 9 18F h 8

18O + +1 0e 16.63 a. 8

17 O b. 1 1 H c. 54

143 Xe d. 12 23 Mg

16.65 a. 8 16 O + 816 O h 1428 Si + 24 He

b. 8 18 O + 98249 Cf h 106263 Sg + 401 n

c. 86 222 Rn h 84

218 PO + 24 He d. 38

80 Sr h 37 80 Rb + +1 0e

16.67 7.5 mg of Tc-99m

16.69 In the fission process, an atom splits into smaller nuclei. In fusion, small nuclei combine (fuse) to form a larger nucleus.

16.71 Fusion occurs naturally in the Sun and other stars.

16.73 4.4 * 103 Bq 16.75 14.3 days

16.77 a. 20 47 Ca h 21

47 Sc + -1 0e b. 1.0 mg of Ca-47 c. 9.0 days

16.79 a. 80 180 Hg h 78

176 Pt + 24 He b. 79198 Au h 80198 Hg + -1 0e c. 37

82 Rb h 36 82 Kr + +1 0e

16.81 0 1 n + 92238 U h 93239 Np + -1 0e

16.83 3.0 days

16.85 20 48 Ca + 94244 Pu h 114292 Fl

16.87 64 mCi

16.89 16 mCi

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CI.33 Consider the reaction of sodium oxalate (Na2C2O4) and potassium permanganate (KMnO4) in acidic solution. The unbalanced equation is the following: (9.2, 9.3, 12.4, 12.5, 15.2)

MnO4 -(aq) + C2O4 2-(aq) h Mn2+(aq) + CO2(g)

c. How many joules were released by the reaction of the magnesium? Assume the density of the HCl solution is 1.00 g/mL and the specific heat of the HCl solution is the same as that of water.

d. What is the heat of reaction for magnesium in joules/gram? in kilojoules/mol?

CI.36 The iceman known as Ötzi was discovered in a high mountain pass on the Austrian–Italian border. Samples of his hair and bones had carbon-14 activity that was 50% of that present in new hair or bone. Carbon-14 undergoes beta decay and has a half-life of 5730 yr. (16.2, 16.4)

COMBINING IDEAS from Chapters 15 and 16

In an oxidation–reduction titration, the KMnO4 from the buret reacts with Na2C2O4.

Magnesium metal reacts vigorously with hydrochloric acid.

The mummified remains of Ötzi were discovered in 1991.

a. What is the balanced oxidation half-reaction? b. What is the balanced reduction half-reaction? c. What is the balanced ionic equation for the reaction? d. If 24.6 mL of a KMnO4 solution is needed to titrate a

solution containing 0.758 g of sodium oxalate (Na2C2O4), what is the molarity of the KMnO4 solution?

CI.34 A strip of magnesium metal dissolves rapidly in 6.00 mL of a 0.150 M HCl solution, producing hydrogen gas and magnesium chloride. (9.2, 9.3, 12.4, 12.6, 14.6, 15.1, 15.2)

HCl(aq) + Mg(s) h H2(g) + MgCl2(aq) Unbalanced

a. Assign oxidation numbers to all of the elements in the reactants and products.

b. What is the balanced chemical equation for the reaction? c. What is the oxidizing agent? d. What is the reducing agent? e. What is the pH of the 0.150 M HCl solution? f. How many grams of magnesium can dissolve in the HCl

solution?

CI.35 A piece of magnesium with a mass of 0.121 g is added to 50.0 mL of a 1.00 M HCl solution at a temperature of 22 °C. When the magnesium dissolves, the solution reaches a temperature of 33 °C. For the equation, see your answer to problem CI.34b. (9.2, 9.3, 9.4, 9.6, 12.4, 12.6)

a. What is the limiting reactant? b. What volume, in milliliters, of hydrogen gas would be

produced at 33 °C when the pressure is 750. mmHg?

a. How long ago did Ötzi live? b. Write a balanced nuclear equation for the decay of

carbon-14.

CI.37 Some of the isotopes of silicon are listed in the following table: (4.4, 5.4, 10.1, 10.3, 16.2, 16.4)

Isotope % Natural Abundance

Atomic Mass

Half-Life (radioactive) Radiation

27 14Si 26.987 4.9 s Positron 28 14Si 92.230 27.977 Stable None 29 14Si 4.683 28.976 Stable None 30 14Si 3.087 29.974 Stable None 31 14Si 30.975 2.6 h Beta

a. In the following table, indicate the number of protons, neutrons, and electrons for each isotope listed:

Isotope Number of Protons

Number of Neutrons

Number of Electrons

27 14 Si 28 14 Si 29 14 Si 30 14 Si 31 14 Si

b. What are the electron configuration and the abbreviated electron configuration of silicon?

c. Calculate the atomic mass for silicon using the isotopes that have a natural abundance.

d. Write the balanced nuclear equations for the positron emission of Si-27 and the beta decay of Si-31.

e. How many hours are needed for a sample of Si-31 with an activity of 16 mCi to decay to 2.0 mCi?

f. Draw the Lewis structures and predict the shape of SiCl4.

538

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Combining Ideas from Chapters 15 and 16 539

ANSWERS

b. 1s22s22p63s23p2; [Ne]3s23p2

c. 28.085 amu d. 2714Si h

27 13Al + 0+1e

31 14Si h

31 15P + 0-1e

e. 7.8 h

CI.33 a. C2O4 2-(aq) h

2CO2(g) + 2 e-

b. 5 e- + 8H+(aq) + MnO4 -(aq) h Mn2+(aq) + 4H2O(l) c. 16H+(aq) + 2MnO4 -(aq) + 5C2O4 2-(aq) h 10CO2(g) + 2Mn2+(aq) + 8H2O(l) d. 0.0920 M KMnO4 solution

CI.35 a. Mg is the limiting reactant. b. 127 mL of H2(g) c. 2.3 * 103 J d. 1.9 * 104 J/g; 460 kJ/mol

CI.40 Radon levels in a home can be measured with a home radon-detection kit. Environmental agencies have set the maximum level of radon-222 in a home at 4 picocuries per liter (pCi/L) of air. (11.7, 16.2, 16.3, 16.4)

CI.38 K+ is an electrolyte required by the human body and found in many foods as well as salt substitutes. One of the isotopes of potassium is potassium-40, which has a natural abundance of 0.012% and a half-life of 1.3 * 109 yr. The isotope potassium-40 decays to calcium-40 or to argon-40. A typical activity for potassium-40 is 7.0 mCi per gram of potassium. (16.2, 16.3, 16.4)

Potassium chloride is used as a salt substitute.

A home detection kit is used to measure the level of radon-222.

a. Write a balanced nuclear equation for each type of decay, and identify the particle emitted.

b. A shaker of salt substitute contains 1.6 oz of K. What is the activity, in millicuries and becquerels, of the potassium-40 in the shaker?

CI.39 Uranium-238 decays in a series of nuclear changes until stable lead-206 is produced. Complete the following nuclear equations that are part of the uranium-238 decay series: (16.2, 16.3, 16.4)

a. 23892 U h 234 90Th + ? b. 23490Th h

? + 0-1e

c. ? h 222

86Rn + 42He

a. Write the balanced nuclear equation for the decay of Ra-226.

b. Write the balanced nuclear equation for the decay of Rn-222.

c. If a room contains 24 000 atoms of radon-222, how many atoms of radon-222 remain after 15.2 days?

d. Suppose a room has a volume of 72 000 L (7.2 * 104 L). If the radon level is the maximum allowed (4 pCi/L), how many alpha particles are emitted from Rn-222 in one day? (1 Ci = 3.7 * 1010 disintegrations per second)

CI.39 a. 23892U h 234

90Th + 42Ηe b. 23490Th h

234 91Pa + 0-1e

c. 22688Ra h 222

86Rn + 42ΗeIsotope Number of Protons

Number of Neutrons

Number of Electrons

27 14Si 14 13 14 28 14Si 14 14 14 29 14Si 14 15 14 30 14Si 14 16 14 31 14Si 14 17 14

CI.37 a.

Si tetrahedralCl Cl

Cl

Cl

Si ClCl Cl

Cl f.

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540

At 4:35 a.m., a rescue crew responds to a call about a house fire. At the scene, Jack, a firefighter/ emergency medical technician (EMT), finds Diane lying in the front yard of her house. In his assessment, Jack reports that Diane has second- and third-degree burns over 40% of her body as well as a broken leg. He places an oxygen re-breather mask on Diane to provide a high concentration of oxygen. Another firefighter/EMT, Nancy, begins dressing the burns with sterile water and cling film, a first aid material made of polyvinyl chloride, which does not stick to the skin and is protective. Jack and his crew transport Diane to the burn center for further treatment.

At the scene of the fire, arson investigators use trained dogs to search for traces of accelerants and fuel. Gasoline, which is often found at arson scenes, is a mixture of organic molecules called alkanes. Alkanes or hydrocarbons are chains of carbon and hydrogen atoms. The alkanes present in gasoline consist of a mixture of compounds with five to eight carbon atoms in a chain. Alkanes are extremely combustible; they react with oxygen to form carbon dioxide, water, and large amounts of heat. Because alkanes undergo combustion reactions, they can be used to start arson fires.

CAREER

Firefighter/Emergency Medical Technician Firefighters/emergency medical technicians are first responders to fires, accidents, and other emergency situations. They are required to have an emergency medical technician certification in order to be able to treat seriously injured people. By combining the skills of a firefighter and an emergency medical technician, they increase the survival rates of the injured. The physical demands of firefighters are extremely high as they fight, extinguish, and prevent fires while wearing heavy protective clothing. They also train for and participate in firefighting drills, and maintain fire equipment so that it is always working and ready. Firefighters must also be knowledgeable about fire codes, arson, and the handling and disposal of hazardous materials. Because firefighters also provide emergency care for sick and injured people, they need to be aware of emergency medical and rescue procedures, as well as the proper methods for controlling the spread of infectious disease.

Organic Chemistry

UPDATE Diane’s Treatment in the Burn Unit

When Diane arrives at the hospital, she is diagnosed with second- and third-degree burns. You can see Diane’s treatment in the UPDATE Diane’s Treatment in the Burn Unit, page 580, and see the results of the arson investigation into the house fire.

17

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17.1 Alkanes 541

TABLE 17.1 Some Properties of Organic and Inorganic Compounds Property Organic Example: C3H8 Inorganic Example: NaCl

Elements Present

C and H, sometimes O, S, N, P, or Cl (F, Br, I)

C and H Most metals and nonmetals

Na and Cl

Particles Molecules C3H8 Mostly ions Na+ and Cl-

Bonding Mostly covalent Covalent Many are ionic, some covalent

Ionic

Polarity of Bonds

Nonpolar unless a strongly electronegative atom is present

Nonpolar Most are ionic or polar covalent, a few are nonpolar covalent

Ionic

Melting Point Usually low - 188 °C Usually high 801 °C Boiling Point Usually low - 42 °C Usually high 1413 °C Flammability High Burns in air Low Does not burn

Solubility in Water

Not soluble unless a polar group is present

No Most are soluble unless nonpolar

Yes

LOOKING AHEAD

17.1 Alkanes 541 17.2 Alkenes, Alkynes, and

Polymers 551 17.3 Aromatic Compounds 557 17.4 Alcohols and Ethers 560 17.5 Aldehydes and

Ketones 564 17.6 Carboxylic Acids and

Esters 568 17.7 Amines and Amides 575

17.1 Alkanes LEARNING GOAL Identify the properties of organic or inorganic compounds. Write the IUPAC names and draw the condensed structural and line-angle formulas for alkanes.

Organic chemistry is the study of carbon compounds. The element carbon has a special role because many carbon atoms can bond together to give a vast array of molecular compounds. Organic compounds always contain carbon and hydrogen, and sometimes other nonmetals such as oxygen, sulfur, nitrogen, phosphorus, or a halogen. We find organic compounds in many common products we use every day, such as gasoline, medicines, shampoos, plastics, and perfumes. The food we eat is composed of organic compounds such as carbohydrates, fats, and proteins that supply us with fuel for energy and the carbon atoms needed to build and repair the cells of our bodies.

The formulas of organic compounds are written with carbon first, followed by hydrogen, and then any other elements. Organic compounds typically have low melting and boiling points, are not soluble in water, and are less dense than water. For example, vegetable oil, which is a mixture of organic compounds, does not dissolve in water but floats on top. Many organic compounds undergo combustion and burn vigorously in air. By contrast, many inorganic compounds have high melting and boiling points. Inorganic compounds that are ionic are usually soluble in water, and most do not burn in air. TABLE 17.1 contrasts some of the properties associated with organic and inorganic compounds, such as propane, C3H8, and sodium chloride, NaCl (see FIGURE 17.1).

REVIEW Balancing a Chemical Equation

(8.2)

Drawing Lewis Structures (10.1)

Predicting Shape (10.3)

Vegetable oil, a mixture of organic compounds, is not soluble in water.

CH3 CH2 CH3

Na+

Cl- FIGURE 17.1 Propane, C3H8, is an organic compound, whereas sodium chloride, NaCl, is an inorganic compound.

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542 CHAPTER 17 Organic Chemistry

Representations of Carbon Compounds Hydrocarbons are organic compounds that consist of only carbon and hydrogen. In organic molecules, every carbon atom has four bonds. In the simplest hydrocarbon, methane (CH4), the carbon atom forms an octet by sharing its four valence electrons with four hydrogen atoms.

C 4H+ =C H

H H H

Methane

H

H

H

C H

The most accurate representation of methane is the three-dimensional space-filling model (a) in which spheres show the relative size and shape of all the atoms. Another type of three-dimensional representation is the ball-and-stick model (b), where the atoms are shown as balls and the bonds between them are shown as sticks. In the ball-and-stick model of methane, CH4, the covalent bonds from the carbon atom to each hydrogen atom are directed to the corners of a tetrahedron with bond angles of 109°. In the wedge–dash model (c), the three-dimensional shape is represented by symbols of the atoms with lines for bonds in the plane of the page, wedges for bonds that project out from the page, and dashes for bonds that are behind the page.

ENGAGE 17.1 Why does methane have a tetrahedral shape?

C

H

H

H H

109°

CH4

H H H

H

C

Three-Dimensional and Two-Dimensional Representations of Methane

(a) Space-filling model

(b) Ball-and-stick model

(c) Wedge–dash model

(d) Expanded structural formula

(e) Condensed structural formula

PRACTICE PROBLEMS Try Practice Problems 17.1 to 17.4

SAMPLE PROBLEM 17.1 Properties of Organic Compounds

TRY IT FIRST

Indicate whether the following properties are more typical of organic or inorganic compounds:

a. is not soluble in water b. has a high melting point c. burns in air

SOLUTION

a. Many organic compounds are not soluble in water. b. Inorganic compounds are more likely to have high melting points. c. Organic compounds are more likely to burn in air.

SELF TEST 17.1

a. What elements are always found in organic compounds? b. If butane in a lighter is used to start a fire, is butane an inorganic or organic compound?

ANSWER

a. C and H b. organic compound

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17.1 Alkanes 543

However, the three-dimensional models are awkward to draw and view for more com- plex molecules. Therefore, it is more practical to use their corresponding two-dimensional formulas. The expanded structural formula (d) shows all of the atoms and the bonds connected to each atom. A condensed structural formula (e) shows the carbon atoms each grouped with the attached number of hydrogen atoms.

The hydrocarbon ethane with two carbon atoms and six hydrogen atoms can be repre- sented by a similar set of three- and two-dimensional models and formulas in which each carbon atom is bonded to another carbon and three hydrogen atoms. As in methane, each carbon atom in ethane retains a tetrahedral shape. A hydrocarbon is referred to as a saturated hydrocarbon when all the bonds in the molecule are single bonds.

Three-Dimensional and Two-Dimensional Representations of Ethane

(a) Space-filling model

(b) Ball-and-stick model

(c) Wedge–dash model

(d) Expanded structural formula

(e) Condensed structural formula

C

H

H

H C

H

H

HC C

H

H

H H

H H CH3CH3

Naming Alkanes More than 90% of the compounds in the world are organic compounds. The large number of carbon compounds is possible because the covalent bond between carbon atoms (C ¬ C) is very strong, allowing carbon atoms to form long, stable chains.

The alkanes are a type of hydrocarbon in which the carbon atoms are connected only by single bonds. One of the most common uses of alkanes is as fuels. Methane, used in gas heaters and gas cooktops, is an alkane with one carbon atom. The alkanes ethane, propane, and butane contain two, three, and four carbon atoms, respectively, connected in a row or a continuous chain. As we can see, the names for alkanes end in ane. Such names are part of the IUPAC system (International Union of Pure and Applied Chemistry) used by chemists to name organic compounds. Alkanes with five or more carbon atoms in a chain are named using Greek prefixes: pent (5), hex (6), hept (7), oct (8), non (9), and dec (10) (see TABLE 17.2).

TABLE 17.2 IUPAC Names and Formulas of the First 10 Alkanes Number of Carbon Atoms

IUPAC Name

Molecular Formula Condensed Structural Formula Line-Angle Formula

1 Methane CH4 CH4 2 Ethane C2H6 CH3 ¬ CH3 3 Propane C3H8 CH3 ¬ CH2 ¬ CH3 4 Butane C4H10 CH3 ¬ CH2 ¬ CH2 ¬ CH3 5 Pentane C5H12 CH3 ¬ CH2 ¬ CH2 ¬ CH2 ¬ CH3 6 Hexane C6H14 CH3 ¬ CH2 ¬ CH2 ¬ CH2 ¬ CH2 ¬ CH3 7 Heptane C7H16 CH3 ¬ CH2 ¬ CH2 ¬ CH2 ¬ CH2 ¬ CH2 ¬ CH3 8 Octane C8H18 CH3 ¬ CH2 ¬ CH2 ¬ CH2 ¬ CH2 ¬ CH2 ¬ CH2 ¬ CH3 9 Nonane C9H20 CH3 ¬ CH2 ¬ CH2 ¬ CH2 ¬ CH2 ¬ CH2 ¬ CH2 ¬ CH2 ¬ CH3 10 Decane C10H22 CH3 ¬ CH2 ¬ CH2 ¬ CH2 ¬ CH2 ¬ CH2 ¬ CH2 ¬ CH2 ¬ CH2 ¬ CH3

PRACTICE PROBLEMS Try Practice Problems 17.5 and 17.6

CORE CHEMISTRY SKILL Naming and Drawing Alkanes

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544 CHAPTER 17 Organic Chemistry

Condensed Structural and Line-Angle Formulas In a condensed structural formula, each carbon atom and its attached hydrogen atoms are written as a group. A subscript indicates the number of hydrogen atoms bonded to each carbon atom.

= CH3 = CH2

H

H

CCH

H

H Expanded Condensed CondensedExpanded

When an organic molecule consists of a chain of three or more carbon atoms, the carbon atoms do not lie in a straight line. Rather, they are arranged in a zigzag pattern.

A simplified formula called the line-angle formula shows a zigzag line in which carbon atoms are represented as the ends of each line and as corners. For example, in the line-angle formula of pentane, each line in the zigzag drawing represents a single bond. The carbon atoms on the ends are bonded to three hydrogen atoms. However, the carbon atoms in the middle of the carbon chain are each bonded to two carbons and two hydrogen atoms as shown in Sample Problem 17.2.

ENGAGE 17.2 How does a line-angle formula represent an organic compound of carbon and hydrogen with single bonds?

SAMPLE PROBLEM 17.2 Drawing Formulas for an Alkane

TRY IT FIRST

Draw the expanded, condensed structural, and line-angle formulas for pentane.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

pentane expanded, condensed structural, and line-angle formulas

carbon chain, zigzag line

STEP 1 Draw the carbon chain. A molecule of pentane has five carbon atoms in a continuous chain.

C C C CC

STEP 2 Draw the expanded structural formula by adding the hydrogen atoms using single bonds to each of the carbon atoms.

C C C C HH

H

H

H

H

H

H

C

H

H

H

H

STEP 3 Draw the condensed structural formula by combining the H atoms with each C atom.

Expanded structural formula

Condensed structural formula

C C C C HH

CH3 CH2 CH2 CH3

H

H

H

H

H

H

C

CH2

H

H

H

H

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17.1 Alkanes 545

STEP 4 Draw the line-angle formula as a zigzag line in which the ends and corners represent C atoms.

Condensed structural formula

Line-angle formula

CH3 CH2 CH2 CH3CH2

SELF TEST 17.2

Draw the condensed structural formula and write the name for each of the following line-angle formulas:

a. b.

ANSWER

a. CH2CH3 CH2 CH2 CH2 CH2 CH3 heptane

b. CH3 ¬ CH2 ¬ CH2 ¬ CH3 butane

Structural Isomers When an alkane has four or more carbon atoms, the atoms can be arranged so that a side group called a branch or substituent is attached to a carbon chain. For example, FIGURE 17.2 shows different models and formulas for two compounds that have the same molecular formula, C4H10. One model has a chain of four carbon atoms. In the other model, a carbon atom is attached as a branch or substituent to a carbon in a chain of three atoms. An alkane with at least one branch is called a branched alkane. When the two com- pounds have the same molecular formula but different arrangements of atoms, they are structural isomers.

FIGURE 17.2 The structural isomers of C4H10 have the same number and type of atoms but are bonded in a different order.

CH3 CH2 CH2 CH3

CH3

CH3 CH CH3

Structural Isomers of C4H10 ENGAGE 17.3 How can two or more structural isomers have the same molecular formula?

In another example, we can draw the condensed structural and line-angle formulas for three different structural isomers with the molecular formula C5H12 as follows:

Structural Isomers of C5H12

Condensed Structural Formula

Line-Angle Formula

CH3 CH2 CH2 CH2 CH3

CH3

CH3 CH CH2 CH3

CH3

CH3

CH3 C CH3

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546 CHAPTER 17 Organic Chemistry

TABLE 17.3 Formulas and Names of Some Common Substituents

Formula

Name

Formula

Name

Formula

Name

Formula

Name

CH3 CH2 ethyl

CH3 methyl

CH3 CH CH3 isopropyl

CH3 CH2 CH2 propyl

CH3 CH CH2

CH3

isobutyl

CH3 CH2 CH2CH2 butyl

Br bromo

I iodo

F fluoro

Cl chloro

PRACTICE PROBLEMS Try Practice Problems 17.7 and 17.8

SAMPLE PROBLEM 17.3 Structural Isomers

TRY IT FIRST

Identify each pair of formulas as structural isomers or the same molecule.

a. CH2 CH2

CH3 CH3

CH2 CH2 CH3and

CH3

b. and

SOLUTION

a. Both have the same molecular formula C4H10. Both have continuous four-carbon chains even though the ¬ CH3 ends are drawn above or below the chain. Thus, both formulas represent the same molecule.

b. Both have the same molecular formula C6H14. The formula on the left has a five-carbon chain with a ¬ CH3 substituent on the second carbon of the chain. The formula on the right has a four-carbon chain with two ¬ CH3 substituents. Thus, there is a different order of bonding of atoms, which represents structural isomers.

SELF TEST 17.3

Why do each of the following line-angle formulas represent a different structural isomer of the molecules in Sample Problem 17.3, part b?

a. b.

ANSWER

a. This isomer with the molecular formula C6H14 has a different arrangement of carbon atoms with a ¬ CH3 group on the third carbon of a five-carbon chain.

b. This isomer with the molecular formula C6H14 has a different arrangement of carbon atoms with two ¬ CH3 groups on the second carbon of a four-carbon chain.

Substituents in Alkanes In the IUPAC names for alkanes, a carbon branch is named as an alkyl group, which is an alkane that is missing one hydrogen atom. The alkyl group is named by replacing the ane ending of the corresponding alkane name with yl. Alkyl groups cannot exist on their own: They must be attached to a carbon chain. When a halogen atom is attached to a carbon chain, it is named as a halo group: fluoro, chloro, bromo, or iodo. Some of the common groups attached to carbon chains are illustrated in TABLE 17.3.

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17.1 Alkanes 547

Naming Alkanes with Substituents In the IUPAC system of naming, a carbon chain is numbered to give the location of the substituents. Let’s take a look at how we use the IUPAC system to name the alkane shown in Sample Problem 17.4.

INTERACTIVE VIDEO

Naming Alkanes

SAMPLE PROBLEM 17.4 Writing IUPAC Names for Alkanes with Substituents

TRY IT FIRST

Write the IUPAC name for the following alkane:

CH3

CH3 CH CH2

Br

CH3

CH2 C CH3

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

six-carbon chain, two methyl groups, one bromo group

IUPAC name position of substituents on the carbon chain

STEP 1 Write the alkane name for the longest chain of carbon atoms. The longest chain contains six carbon atoms; it is hexane.

CH3

CH

Br

CH3

CCH2 CH3CH2CH3 hexane

STEP 2 Number the carbon atoms from the end nearer a substituent. The first substituent is attached to carbon 2 counting from the left, and two more substituents are on carbon 4.

CH3

CH

Br

CH3

CCH2 CH3CH2CH3 hexane

1 2 4 5 63

STEP 3 Give the location and name for each substituent (alphabetical order) as a prefix to the name of the main chain. The substituents are listed in alphabeti- cal order (bromo first, then methyl). A hyphen is placed between the number and the substituent name. When there are two or more of the same substituent, a prefix (di, tri, tetra) is used, and commas separate the numbers. However, prefixes are not used to determine the alphabetical order of the substituents.

4-bromo-2,4-dimethylhexane

CH3

CH

Br

CH3

CCH2 CH3CH2CH3

1 2 4 5 63

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548 CHAPTER 17 Organic Chemistry

SELF TEST 17.4

Write the IUPAC name for each of the following compounds:

a. b.

CH3

CH

CH3

CH3

CCH2 CH3CH2CH3

ANSWER

a. 4-isopropylheptane b. 2,4,4-trimethylhexane

Drawing Structural Formulas for Alkanes with Substituents The IUPAC name gives all the information needed to draw the condensed structural for- mula for an alkane. Suppose you are asked to draw the condensed structural formula for 2,3-dimethylbutane. The alkane name gives the number of carbon atoms in the longest chain. The other names indicate the substituents and where they are attached. We can break down the name in the following way:

2,3-Dimethylbutane

2,3- Di methyl but ane

Substituents on carbons 2 and 3

two identical groups

¬ CH3 alkyl groups

four C atoms in the main chain

single (C ¬ C) bonds

SAMPLE PROBLEM 17.5 Drawing Condensed Structural and Line-Angle Formulas from IUPAC Names

TRY IT FIRST

Draw the condensed structural and line-angle formulas for 2,3-dimethylbutane.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

2,3-dimethylbutane condensed structural and line-angle formulas

four-carbon chain, two methyl groups

STEP 1 Draw the main chain of carbon atoms. For butane, we draw a chain and a zigzag line of four carbon atoms.

C ¬ C ¬ C ¬ C

STEP 2 Number the chain and place the substituents on the carbons indicated by the numbers. The first part of the name indicates two methyl groups ¬ CH3: one on carbon 2 and one on carbon 3.

MethylMethyl

CH3

C C C

CH3

C 1 2 3 4 1 2 3 4

STEP 3 For the condensed structural formula, add the correct number of hydrogen atoms to give four bonds to each C atom.

CH3

CH3 CH CH3

CH3

CH

PRACTICE PROBLEMS Try Practice Problems 17.9 and 17.10

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17.1 Alkanes 549

SELF TEST 17.5

Draw the condensed structural and line-angle formulas for each of the following: a. 2-bromo-4-methylpentane b. 2-chloro-2-methylpropane

ANSWER

a. CH3 CH2CH

Br Br

CH3

CH3

CH b. CH3 CH3

CH3

C

Cl Cl

PRACTICE PROBLEMS Try Practice Problems 17.11 to 17.14

Uses of Alkanes The first four alkanes—methane, ethane, propane, and butane—are gases at room tempera- ture and are widely used as heating fuels.

Alkanes having five to eight carbon atoms (pentane, hexane, heptane, and octane) are liquids at room temperature. They are highly volatile, which makes them useful in fuels such as gasoline.

Liquid alkanes with 9 to 17 carbon atoms have higher boiling points and are found in kerosene, diesel, and jet fuels. Motor oil is a mixture of high-molecular-weight liquid hydrocarbons and is used to lubricate the internal components of engines. Mineral oil is a mixture of liquid hydrocarbons and is used as a laxative and a lubricant. Alkanes with 18 or more carbon atoms are waxy solids at room temperature. Known as paraffins, they are used in waxy coatings added to fruits and vegetables to retain moisture, inhibit mold, and enhance appearance. Petrolatum jelly, or Vaseline, is a semisolid mixture of hydrocarbons with more than 25 carbon atoms used in ointments and cosmetics and as a lubricant.

Combustion of Alkanes The carbon–carbon single bonds in alkanes are difficult to break, which makes them the least reactive family of organic compounds. However, alkanes burn readily in oxygen to produce carbon dioxide, water, and energy.

Propane is the gas used in portable heaters and gas barbecues (see FIGURE 17.3). The equation for the combustion of propane (C3H8) is written:

C3H8(g) + 5O2(g) h ∆

3CO2(g) + 4H2O(g) + energy Propane

Solubility and Density Alkanes are nonpolar, which makes them insoluble in water. However, they are soluble in nonpolar solvents. Alkanes have densities from 0.62 g/mL to about 0.79 g/mL, which is less than the density of water (1.0 g/mL).

If there is an oil spill in the ocean, the alkanes in the oil, which do not mix with water, form a thin layer on the surface that spreads over a large area (see FIGURE 17.4).

FIGURE 17.3 The propane fuel in the tank undergoes combustion, which provides energy.

FIGURE 17.4 In oil spills, large quantities of oil spread out to form a thin layer on top of the ocean surface.

The solid alkanes that make up waxy coatings on fruits and vegetables help retain moisture, inhibit mold, and enhance appearance.

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550 CHAPTER 17 Organic Chemistry

In April 2010, an explosion on an oil-drilling rig in the Gulf of Mexico caused the largest oil spill in U.S. history. At its maximum, an estimated 10 million liters of oil was leaked every day. If the crude oil reaches land, there can be considerable damage to beaches, shellfish, birds, and wildlife habitats. When animals such as birds are covered with oil, they must be cleaned quickly because ingestion of the hydrocarbons when they try to clean themselves is fatal.

ENGAGE 17.4 What physical properties cause oil to remain on the surface of water?

PRACTICE PROBLEMS Try Practice Problems 17.15 to 17.18

PRACTICE PROBLEMS

17.1 Alkanes

17.1 Identify each of the following as an organic or inorganic compound:

a. KCl b. C3H7Cl c. is soluble in water d. has a low boiling point e. contains carbon and hydrogen f. contains ionic bonds

17.2 Identify each of the following as an organic or inorganic compound:

a. contains Li and F b. K3PO4 c. contains covalent bonds d. produces ions in water e. is a gas at room temperature f. C4H9Br

17.3 Match each of the following physical and chemical properties with ethane, C2H6, or sodium bromide, NaBr:

a. boils at - 89 °C b. burns vigorously in air c. is a solid at 250 °C d. dissolves in water

17.4 Match each of the following physical and chemical properties with hexane, C6H14 or calcium nitrate, Ca(NO3)2:

a. melts at 500 °C b. is insoluble in water c. does not burn in air d. is a liquid at room temperature

17.5 Write the IUPAC name for each of the following alkanes:

a. CH2 CH3CH2CH2CH2CH3

b.

c.

17.6 Write the IUPAC name for each of the following alkanes: a. CH3 ¬ CH2 ¬ CH2 ¬ CH3 b.

c.

17.7 Indicate whether each of the following pairs of formulas represents structural isomers or the same molecule:

a. and

CH3

CH CH3CH3 CH3

CH3

CH

CH3

b.

andCH2

CH3CH3

CH CH3CH2 CH

CH3 CH3

CH CH3CH3

andCH2

CH3CH3

CH CH3CH2 CH

CH3 CH3

CH CH3CH3

c. and

17.8 Indicate whether each of the following pairs of formulas represents structural isomers or the same molecule:

a. andCH3

CH3

CCH3 CH2

CH3

CH

CH3CH3

CH3

b. andCH

CH3 CH3CH3

CH CH2CH3

CH3

CH CH2 CH

CH3

CH3CH3

andCH

CH3 CH3CH3

CH CH2CH3

CH3

CH CH2 CH

CH3

CH3CH3

c.

and

17.9 Write the IUPAC name for each of the following:

a. CH3 CH2

F

CH CH3 b. CH3 CH3C

CH3

CH3

c.

Cl

17.10 Write the IUPAC name for each of the following:

a. CH3 CH2

CH3

CH CH2 CH2 Br

b. CH3 CH

CH3 CH3

CH CH2 CH2 CH3

c.

ClBr

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17.2 Alkenes, Alkynes, and Polymers 551

17.11 Draw the condensed structural formula for each of the following alkanes:

a. 3,3-dimethylpentane b. 2,3,5-trimethylhexane c. 1-bromo-2-chloroethane

17.12 Draw the condensed structural formula for each of the following alkanes:

a. 3-ethylpentane b. 4-isopropyl-3-methylheptane c. 2-bromopropane

17.13 Draw the line-angle formula for each of the following: a. 3-methylheptane b. 1-chloro-3-methylpentane c. 2,3-dichlorohexane

17.14 Draw the line-angle formula for each of the following: a. 1-bromo-2-methylpentane b. 2,2,3-trimethylbutane c. 3-ethylhexane

17.15 Heptane, used as a solvent for rubber cement, has a density of 0.68 g/mL and boils at 98 °C.

a. Draw the condensed structural and line-angle formulas for heptane.

b. Is heptane a solid, liquid, or gas at room temperature?

c. Is heptane soluble in water? d. Will heptane float on water or sink? e. Write the balanced chemical equation for the complete

combustion of heptane.

17.16 Nonane, found in kerosene, has a density of 0.79 g/mL and boils at 151 °C.

a. Draw the condensed structural and line-angle formulas for nonane.

b. Is nonane a solid, liquid, or gas at room temperature? c. Is nonane soluble in water? d. Will nonane float on water or sink? e. Write the balanced chemical equation for the complete

combustion of nonane.

Applications

17.17 Write the balanced chemical equation for the complete combustion of each of the following:

a. methane b. 3-methylpentane

17.18 Write the balanced chemical equation for the complete combustion of each of the following:

a. pentane b. methylpropane

17.2 Alkenes, Alkynes, and Polymers LEARNING GOAL Write the IUPAC names and draw the condensed structural and line- angle formulas for alkenes and alkynes.

We organize organic compounds into classes or families by their functional groups, which are groups of specific atoms. Compounds that contain the same functional group have similar physical and chemical properties. Identifying functional groups allows us to classify organic compounds according to their structure, to name compounds within each family, predict their chemical reactions, and draw the structures for their products. A list of the common functional groups in organic compounds is shown in TABLE 17.4.

Alkenes and alkynes are classes of unsaturated hydrocarbons that contain double and triple bonds, respectively. The functional group of an alkene contains at least one double bond between carbons. The double bond forms when two adjacent carbon atoms share two pairs of valence electrons. The simplest alkene is ethene, C2H4, which is often called by its common name, ethylene. In ethene, each carbon atom is attached to two H atoms and the other carbon atom in the double bond. The resulting molecule has a flat geometry because the carbon and hydrogen atoms all lie in the same plane. The functional group of an alkyne contains a triple bond, which occurs when two carbon atoms share three pairs of valence electrons.

CC CCH H

Ethene Ethyne

H

H

H

H

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552 CHAPTER 17 Organic Chemistry

TABLE 17.4 Classes of Organic Compounds Class Functional Group Example/Name Occurrence/Use

Alkene C C H2C “ CH2 Ethene (ethylene)

Ripening of fruit, used to make polyethylene

Alkyne ¬ C ‚ C ¬ H ¬ C ‚ C ¬ H Ethyne (acetylene)

Welding fuel

Alcohol ¬ OH CH3 ¬ CH2 ¬ OH Ethanol (ethyl alcohol)

Solvent

Ether ¬ O ¬ CH3 ¬ CH2 ¬ O ¬ CH2 ¬ CH3 Diethyl ether

Solvent

Aldehyde C H

O

CH3 C H

O

Ethanal (acetaldehyde)

Preparation of acetic acid

Ketone C

O

CH3 CH3C

O

Propanone (acetone)

Solvent, paint and fingernail polish remover

Carboxylic acid C OH

O

CH3 C OH

O

Ethanoic acid (acetic acid)

Component of vinegar

Ester C O

O

CH3 CH3C O

O

Methyl ethanoate (methyl acetate)

Rum flavoring

Amine N CH3 ¬ NH2 Methylamine

Odor of fish

Amide C N

O

CH3 C NH2

O

Ethanamide (acetamide)

Odor of mice

ENGAGE 17.5 How does the functional group in an alcohol differ from the functional group in a ketone?

Naming Alkenes and Alkynes The IUPAC names for alkenes and alkynes are similar to those of alkanes. Using the alkane name with the same number of carbon atoms, the ane ending is replaced with ene for an alkene and yne for an alkyne (see TABLE 17.5).

TABLE 17.5 Comparison of Names for Alkanes, Alkenes, and Alkynes Alkane Alkene Alkyne

CH3 ¬ CH3 H2C “ CH2 HC CH Ethane Ethene (ethylene) Ethyne (acetylene)

CH3 ¬ CH2 ¬ CH2 CH3 ¬ CH “ CH2 C CHCH3

Propane Propene Propyne

PRACTICE PROBLEMS Try Practice Problems 17.19 and 17.20

Fruit is ripened with ethene, a plant hormone.

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17.2 Alkenes, Alkynes, and Polymers 553

Hydrogenation In a reaction called hydrogenation, H atoms add to each of the carbon atoms in a double bond of an alkene. During hydrogenation, the double bonds are converted to single bonds in alkanes. A catalyst such as finely divided platinum (Pt), nickel (Ni), or palladium (Pd) is used to speed up the reaction. The equation for the hydrogenation of 2-butene is written as follows:

+ Pt

CH3 CH CH CH3 CH3 CH3H2 CH2 CH2 Butane2-Butene

CORE CHEMISTRY SKILL Writing Equations for

Hydrogenation and Polymerization

PRACTICE PROBLEMS Try Practice Problems 17.21 to 17.24

SAMPLE PROBLEM 17.6 Naming Alkenes and Alkynes

TRY IT FIRST

Write the IUPAC name for the following:

CH3 CH

CH3

CH CH CH3

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

five-carbon chain, double bond, methyl group

IUPAC name replace the ane of the alkane name with ene

STEP 1 Name the longest carbon chain that contains the double bond. There are five carbon atoms in the longest carbon chain containing the double bond. Replace the ane in the corresponding alkane name with ene to give pentene.

CH3

CH3 CH CH CH CH3 pentene

STEP 2 Number the carbon chain starting from the end nearer the double bond. Place the number of the first carbon in the double bond in front of the alkene name.

CH3

CH3 CH3CHCHCH 2-pentene 5 4 3 2 1

Alkenes with two or three carbons do not need numbers.

STEP 3 Give the location and name for each substituent (alphabetical order) as a prefix to the alkene name.

CH3

CH3 CH3CHCHCH 4-methyl-2-pentene 5 4 3 2 1

SELF TEST 17.6

Draw the condensed structural formula for each of the following: a. 1-chloro-3-hexyne b. 1-bromo-1-pentene

ANSWER

a. CH3CH2CH2CH2Cl CC b. CH3CHCHBr CH2CH2

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554 CHAPTER 17 Organic Chemistry

PRACTICE PROBLEMS Try Practice Problems 17.25 and 17.26

SAMPLE PROBLEM 17.7 Writing an Equation for Hydrogenation

TRY IT FIRST

Draw the condensed structural formula for the product of the following hydrogenation reaction:

Pt+CH3 CH2 H2CH

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

alkene + H2 structural formula of product

add 2H to double bond

CH3 ¬ CH2 ¬ CH3

SELF TEST 17.7

a. Draw the condensed structural formula for the product of the hydrogenation of 2-methyl- 1-butene, using a platinum catalyst.

b. Draw the line-angle formula for the product of the hydrogenation of 2-pentene using a palladium catalyst.

ANSWER

a. CH3 CH2 CH3CH

CH3

b.

Chemistry Link to Health Hydrogenation of Unsaturated Fats

Vegetable oils such as corn oil or safflower oil are unsaturated fats composed of fatty acids that contain double bonds. The process of hydrogenation is used commercially to convert the double bonds in the unsaturated fats in vegetable oils to saturated fats such as marga- rine, which are more solid. Adjusting the amount of added hydrogen produces partially hydrogenated fats such as soft margarine, solid margarine and shortenings, which are used in cooking. For example, oleic acid is a typical unsaturated fatty acid in olive oil and has a double bond at carbon 9. When oleic acid is hydrogenated, it is con- verted to stearic acid, a saturated fatty acid.

Oleic acid (found in olive oil and other unsaturated fats)

Stearic acid (found in saturated fats)

O

OH

O

OH

Double bond Single bond

The unsaturated fats in vegetable oils are converted to saturated fats to make a more solid product.

Polymerization A polymer is a large molecule that consists of small repeating units called monomers. In the past hundred years, the plastics industry has made synthetic polymers that are in many of the materials we use every day, such as carpeting, plastic wrap, nonstick pans, plastic cups, and rain gear. In medicine, synthetic polymers are used to replace diseased or damaged body parts such as hip joints, teeth, heart valves, and blood vessels. There are about 100 billion kg of plastics produced every year, which is about 15 kg for every person on Earth.

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17.2 Alkenes, Alkynes, and Polymers 555

TABLE 17.6 Some Alkenes and Their Polymers Recycling Code Monomer Polymer Section Common Uses

2

HDPE

4

LDPE

H2C “ CH2 Ethene (ethylene)

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

Polyethylene (PE)

Plastic bottles, film, insulation materials, shopping bags, plastic furniture

3

V

CH Chloroethene

(vinyl chloride)

Cl

H2C

H

H

C

Cl

H

C

H

H

C

Cl

H

C

H

H

C

Cl

H

C

Polyvinyl chloride (PVC)

Plastic pipes and tubing, garden hoses, garbage bags, shower curtains, credit cards, medical containers, film for packaging foods, drink bottles, blister packs for pills

5

PP

CH

CH3

H2C Propene

(propylene)

H

H

C

CH3

H

C

H

H

C

CH3

H

C

H

H

C

CH3

H

C

Polypropylene (PP)

Ski and hiking clothing, carpets, artificial joints, plastic bottles, food containers, medical face masks, tubing, plastic bags

F F

F

C

F

C Tetrafluoroethene

F

F

C

F

F

C

F

F

C

F

F

C

F

F

C

F

F

C

Polytetrafluoroethylene

Nonstick coatings, Teflon, Gore-Tex rainwear

6

PS CHH2C Phenylethene

(styrene)

CH2 CH CH2 CH Polystyrene (PS)

Plastic coffee cups and cartons, insulation, plastic tableware

Many of the synthetic polymers are made by addition reactions of small alkene monomers. In an addition reaction, a polymer grows longer as each monomer is added at the end of the chain. A polymer may contain as many as 1000 monomers. Many polymerization reactions require high temperature, a catalyst, and high pressure (over 1000 atm). Polyethylene, a polymer made from ethylene monomers, is used in plastic bottles, film, and plastic dinnerware. More polyethylene is produced worldwide than any other polymer. Low-density polyethylene (LDPE) is flexible, breakable, less dense, and more branched than high-density polyethylene (HDPE). High-density polyethylene is stronger, is more dense, and melts at a higher temperature than LDPE.

many

C C

H

H

H

H

H

H

H

H

H

H

H

H

+

H

H

C C C

H

H

CC C +

H

H

C

H

H

C

H

H

C

H

H

C

Ethene (ethylene) monomers Polyethylene section

Monomer unit repeats

TABLE 17.6 shows the recycling code, the alkene monomer, a polymer section, and common uses for some polymers. The alkane-like nature of these plastic synthetic polymers makes them unreactive. Thus, they do not decompose easily (they are not biodegradable). As a result, they have become significant contributors to pollution, on land and in the oceans. Efforts are being made to make them more degradable.

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556 CHAPTER 17 Organic Chemistry

You can identify the type of polymer used to manufacture a plastic item by looking for the recycling code (arrows in a triangle) found on the label or on the bottom of the plastic container. For example, the number 5 or the letters PP inside the triangle is the code for a polypropylene plastic.

PRACTICE PROBLEMS Try Practice Problems 17.27 to 17.32

SAMPLE PROBLEM 17.8 Polymers

TRY IT FIRST

A firefighter/EMT arrives at a home where a premature baby has been delivered. To prevent hypothermia during transport to the neonatal facility, she wraps the baby in clear wrap, which is polyvinyl chloride.

a. Draw the monomer unit and give its IUPAC name. b. Draw a portion of the polymer formed from three monomer units.

SOLUTION

a. CH

Cl

H2C Chloroethene

b.

H

H

C

Cl

H

C

H

H

C

Cl

H

C

H

H

C

Cl

H

C

SELF TEST 17.8

a. Draw the condensed structural formula for the monomer used in the manufacturing of polypropylene (PP), used for plastic dishes.

b. What is the name of the polymer that contains monomers of phenylethene?

ANSWER

a. CH

CH3

H2C b. polystyrene

PRACTICE PROBLEMS

17.2 Alkenes, Alkynes, and Polymers

17.19 Identify each of the following as an alkane, alkene, or alkyne:

a. CH CH2CH3

b. C CHCH2CH3

c.

17.20 Identify each of the following as an alkane, alkene, or alkyne:

a.

b.

CH3

CH3

CH3CH3 C C

c.

CH3

CH3CH3 CH2 CH

17.21 Write the IUPAC name for each of the following:

a. CH2H2C

b.

CH3

CH2CH3 C

c. CH3CCH2 CCH3 17.22 Write the IUPAC name for each of the following:

a. CHH2C CH2 CH3

b. CH3 CH3

CH3

CHCH2C C

c. CH3 CH2 CH3CH CH

17.23 Draw the condensed structural formula for each of the following compounds:

a. propene b. 1-hexyne c. 2-methyl-1-butene

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17.3 Aromatic Compounds 557

17.3 Aromatic Compounds LEARNING GOAL Describe the bonding in benzene; name aromatic compounds and draw their line-angle formulas.

In 1825, Michael Faraday isolated a hydrocarbon called benzene, which consists of a ring of six carbon atoms with one hydrogen atom attached to each carbon. In benzene, each carbon atom uses three valence electrons to bond to the hydrogen atom and two adjacent carbons. That leaves one valence electron, which scientists first thought was shared in a double bond with an adjacent carbon. In 1865, August Kekulé proposed that the carbon atoms in benzene were arranged in a flat ring with alternating single and double bonds between the adjacent carbon atoms. There are two possible structural representations of benzene in which the double bonds can form between two different carbon atoms. If there were double bonds as in alkenes, then benzene should be much more reactive than it is.

However, unlike the alkenes and alkynes, aromatic hydrocarbons do not easily undergo addition reactions. If their reaction behavior is quite different, they must also differ in how the atoms are bonded in their structures. Today, we know that the six electrons are shared equally among the six carbon atoms. This unique feature of benzene makes it especially stable. Benzene is most often represented as a line-angle formula, which shows a hexagon with a circle in the center. Some of the ways to represent benzene are shown as follows:

17.24 Draw the condensed structural formula for each of the following compounds:

a. 1-pentene b. 3-methyl-1-butyne c. 3,4-dimethyl-1-pentene

17.25 Draw the condensed structural formula for the product in each of the following:

a.

Pt+CH3 CH2 CH2 CH2 H2CH

b.

Ni+CH3 CH3CH CH H2

c.

Pt

H2+

17.26 Draw the condensed structural formula for the product in each of the following:

a.

Pt+CH3 CH2 CH2 H2CH

b.

Pt+CH3 CH2 H2CH3CHC

CH3

c.

H2+

Pt

17.27 What is a polymer?

17.28 What is a monomer?

17.29 Write an equation that represents the formation of a portion of polypropylene from three of its monomers.

17.30 Write an equation that represents the formation of a portion of polyvinyl fluoride (PVF) from three monomers of fluoroethene.

Applications

17.31 The plastic polyvinylidene difluoride, PVDF, is made from monomers of 1,1-difluoroethene. Draw the expanded structural formula for a portion of the polymer formed from three monomers of 1,1-difluoroethene.

17.32 The polymer polyacrylonitrile, PAN, used in the fabric material Orlon is made from monomers of acrylonitrile. Draw the expanded structural formula for a portion of the polymer formed from three monomers of acrylonitrile.

Acrylonitrile

H2C

CN

CH

Benzene Equivalent structures for benzene Structural formulas for benzene

H

H

H H

H H

H

H

H H

H H

H

H

H H

H H

Because many compounds containing benzene had fragrant odors, the family of benzene compounds became known as aromatic compounds. Some common examples of aromatic compounds that we use for f lavor are anisole from anise, estragole from tarragon, and thymol from thyme.

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558 CHAPTER 17 Organic Chemistry

Anisole (anise)

O Estragole (tarragon) Thymol (thyme)

OH O

The aroma and flavor of the herbs anise, tarragon, and thyme are because of aromatic compounds.

Naming Aromatic Compounds Many compounds containing benzene have been important in chemistry for years and still use their common names. Toluene is benzene with a methyl group ( ¬ CH3), aniline is ben- zene with an amino group ( ¬ NH2), and phenol is benzene with a hydroxyl group ( ¬ OH). The names toluene, aniline, and phenol are allowed by IUPAC rules.

CH3

Toluene Line-angle formula for toluene

NH2

Aniline

OH

Phenol

When benzene has only one substituent, the ring is not numbered. When there are two substituents, the benzene ring is numbered to give the lowest numbers to the substituents. In common names, when the benzene ring is a substituent, it is named as a phenyl group. When a name such as toluene, phenol, or aniline is used, the carbon atom attached to the methyl, hydroxyl, or amine group is numbered as carbon 1.

Cl

Cl

Cl

Cl 1,4-Dichlorobenzene

Cl

1,3-Dichlorobenzene

Cl

1,2-Dichlorobenzene

1 2

1

3

1

4

NH2

Br Br

OH

Br

3-Bromoaniline 4-Bromophenol 2-Bromotoluene

1

3

1

4

1 2

ENGAGE 17.6 How is the formula of toluene similar to and different from that of phenol?

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17.3 Aromatic Compounds 559

SAMPLE PROBLEM 17.9 Naming Aromatic Compounds

TRY IT FIRST

Write the IUPAC name for the following:

Cl

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

structural formula IUPAC name name of aromatic compound, number substituent

STEP 1 Write the name for the aromatic compound. A benzene ring with a methyl group is named toluene.

STEP 2 Number and name a substituent as a prefix. The methyl group of toluene is attached to carbon 1, and the ring is numbered to give the lower number to the chlorine substituent.

Cl

1

3

3-chlorotoluene

SELF TEST 17.9

Write the IUPAC name for each of the following:

a. b.

OH

Cl

ANSWER

a. 1,3-diethylbenzene b. 2-chlorophenol

Chemistry Link to Health Some Common Aromatic Compounds

Aromatic compounds are common in nature and in medicine. Toluene is used as a reactant to make drugs, dyes, and explosives such as TNT (trinitrotoluene). The benzene ring is found in pain relievers such as aspirin and acetaminophen, and in flavorings such as vanillin.

O2N NO2

NO2

TNT (2,4,6-trinitrotoluene) Aspirin

O

O

O

OH

N

OH

Acetaminophen

O

HO

O

Vanillin

H

O H

PRACTICE PROBLEMS Try Practice Problems 17.33 to 17.36

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560 CHAPTER 17 Organic Chemistry

Chemistry Link to Health Polycyclic Aromatic Hydrocarbons (PAHs)

Large aromatic compounds known as polycyclic aromatic hydrocarbons (PAHs) are formed by fusing together two or more benzene rings edge to edge. In a fused-ring compound, neighboring benzene rings share two carbon atoms. Naphthalene, with two benzene rings, is well known for its use in mothballs; anthracene, with three rings, is used in the manu- facture of dyes.

alterations in the cells. Benzo[a]pyrene, a product of combustion, has been identified in coal tar, tobacco smoke, barbecued meats, and automobile exhaust.

Naphthalene Anthracene Phenanthrene

When a polycyclic compound contains phenanthrene, it may act as a carcinogen, a substance known to cause cancer.

Compounds containing five or more fused benzene rings such as benzo[a]pyrene are potent carcinogens. The molecules interact with the DNA in the cells, causing abnormal cell growth and cancer. Increased exposure to carcinogens increases the chance of DNA

Aromatic compounds such as benzo[a]pyrene are strongly associated with lung cancers.

Benzo[a]pyrene

PRACTICE PROBLEMS

17.3 Aromatic Compounds

17.33 Write the IUPAC name for each of the following:

a.

Cl

b. c.

OH

17.34 Write the IUPAC name for each of the following:

a.

Br

b. c.

Cl

Cl

17.35 Draw the line-angle formula for each of the following compounds:

a. aniline b. 1-chloro-4-fluorobenzene c. 2-ethyltoluene

17.36 Draw the line-angle formula for each of the following compounds:

a. 1-bromo-2-chlorobenzene b. 4-chlorotoluene c. propylbenzene

17.4 Alcohols and Ethers LEARNING GOAL Write the IUPAC and common names for alcohols and the common names for ethers; draw the condensed structural and line-angle formulas when given their names.

In the functional group of an alcohol, a hydroxyl group ( ¬ OH) replaces a hydrogen atom in a hydrocarbon. In a phenol, the hydroxyl group replaces a hydrogen atom attached to a benzene ring. Molecules of alcohols and phenols have a bent shape around the oxygen atom. In the functional group of an ether, an oxygen atom is attached by single bonds to two carbon atoms (see FIGURE 17.5).

ENGAGE 17.7 How does the functional group of an ether differ from that of an alcohol?

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17.4 Alcohols and Ethers 561

FIGURE 17.5 An alcohol or a phenol has a hydrogen atom replaced by a hydroxyl group ( ¬ OH); an ether contains an oxygen atom ( ¬ O ¬ ) bonded to two carbon groups.PhenolMethanol Dimethyl ether

H O

OH OH

CH3CH3 O

O

HCH3 O

Ball-and-Stick Model

Condensed Structural Formula

Line-Angle Formula

Name

Naming Alcohols In the IUPAC system, an alcohol is named by replacing the final e in the corresponding alkane name with ol. The common name of a simple alcohol uses the name of the alkyl group followed by alcohol.

CH3 H CH3 OH CH3 CH2 H CH3 CH2 OH Methane Methanol

(methyl alcohol) Ethanol

(ethyl alcohol) Ethane

Alcohols with one or two carbon atoms do not require a number for the hydroxyl group. When an alcohol consists of a chain with three or more carbon atoms, the chain is numbered to give the position of the ¬ OH group and any substituents on the chain.

CH3 CH3 CH3CHCH2 CH2 OH

OH

3 32 21 1

1-Propanol (propyl alcohol)

2-Propanol (isopropyl alcohol)

We can also draw the line-angle formulas for alcohols as shown for 2-propanol and 2-butanol.

2-Propanol 2-Butanol

OHOH

SAMPLE PROBLEM 17.10 Naming Alcohols

TRY IT FIRST

Write the IUPAC name for the following:

CH3

CHCH2 CH3CHCH3

OH

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

five carbon chain, hydroxyl group, methyl group

IUPAC name position of methyl and hydroxyl groups, replace e in alkane name with ol

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562 CHAPTER 17 Organic Chemistry

PRACTICE PROBLEMS Try Practice Problems 17.37 and 17.38

STEP 1 Name the longest carbon chain attached to the ¬ OH group by replacing the e in the corresponding alkane name with ol. To name the alcohol, the e in the alkane name pentane is replaced by ol.

CH3

OHCH3

CH3CH CH2 CH pentanol

STEP 2 Number the chain starting at the end nearer to the ¬ OH group. This carbon chain is numbered from right to left to give the position of the ¬ OH group as carbon 2, which is shown as a prefix in the name 2-pentanol.

CH3

OHCH3

CH3CH CH2 CH 2-pentanol 5 4 3 2 1

STEP 3 Give the location and name of each substituent.

CH3

OHCH3

CH3CH CH2 CH 5 4 3 2 1

4-methyl-2-pentanol

SELF TEST 17.10

Write the IUPAC name for each of the following:

a. OH

Cl

b.

OH

ANSWER

a. 3-chloro-1-butanol b. 4,6-dimethyl-2-heptanol

Naming Ethers An ether consists of an oxygen atom that is attached by single bonds to two carbon groups that are alkyl or aromatic groups. In the common name of an ether, the names of the alkyl or aromatic groups attached to the oxygen atom are written in alphabetical order, followed by the word ether.

CH3 CH2 CH3CH2O

Methyl group

Common name: methyl propyl ether

Propyl group

OPRACTICE PROBLEMS Try Practice Problems 17.39 to 17.42

Chemistry Link to Health Some Important Alcohols, Phenols, and Ethers

Methanol (methyl alcohol), the simplest alcohol, is found in many solvents and paint removers. If ingested, methanol is oxidized to formalde- hyde, which can cause headaches, blindness, and death. Methanol is used to make plastics, medicines, and fuels. In car racing, it is used as a fuel because it is less f lammable and has a higher octane rating than does gasoline.

Ethanol (ethyl alcohol) has been known since prehistoric times as an intoxicating product formed by the fermentation of grains, sugars, and starches. Recent interest in biofuels has led to increased production of ethanol by fermentation. “Gasohol” is a mixture of ethanol and gasoline used as a fuel.

C6H12O6 2CH3 CH2 OH Fermentation + 2CO2

Glucose Ethanol

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17.4 Alcohols and Ethers 563

Ethanol is also used in hand sanitizers. In an alcohol-containing sanitizer, the amount of ethanol can be as high as 85% (v/v). This amount of ethanol can make hand sanitizers a fire hazard in the home because ethanol is highly flammable. It is recommended that sanitiz- ers containing ethanol be placed in storage areas that are away from heat sources in the home.

Today, ethanol for commercial use is produced by reacting ethene and water at high temperatures and pressures.

300 °C, 200 atm, H+ H2C CH2 + H2O CH3 CH2 OH

Ethene Ethanol

1,2-Ethanediol (ethylene glycol) is used as antifreeze in heating and cooling systems. It is also a solvent for paints, inks, and plastics, and it is used in the production of synthetic fibers such as Dacron. If ingested, it is extremely toxic. In the body, it is oxidized to oxalic acid, which forms insoluble salts in the kidneys that cause renal dam- age, convulsions, and death. Because its sweet taste is attractive to pets and children, ethylene glycol solutions must be carefully stored.

1,2-Ethanediol (ethylene glycol)

Oxalic acid

HO

O O

HO OH [O]

C CCH2 CH2 OH

1,2,3-Propanetriol (glycerol or glycerin), a trihydroxy alcohol, is a viscous liquid obtained from oils and fats during the produc- tion of soaps. The presence of several polar ¬ OH groups makes it strongly attracted to water, a feature that makes glycerol useful as a skin softener in products such as skin lotions, cosmetics, shaving creams, and liquid soaps.

OH

1,2,3-Propanetriol (glycerol)

CH2 CH2CH OHHO

Antifreeze (ethylene glycol) raises the boiling point and decreases the freezing point of water in a radiator.

OH

Vanillin

Isoeugenol

Eugenol

OH

Thymol

OH Nutmeg

Thyme

Cloves Vanilla

OH

OH

O

O

O

HO

Phenols found in essential oils of plants produce the odor or flavor of the plant.

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564 CHAPTER 17 Organic Chemistry

PRACTICE PROBLEMS

17.4 Alcohols and Ethers

17.37 Write the IUPAC name for each of the following: a. CH3 ¬ CH2 ¬ OH

b.

OH

CH3 CH2 CH CH3

c.

OH

17.38 Write the IUPAC name for each of the following:

a.

OH

b.

CH3 CH3

CH3 CH2 CH CH CH2 OH

c.

OH

17.39 Write the common name for each of the following: a. CH3 ¬ O ¬ CH2 ¬ CH3 b. CH3 ¬ CH2 ¬ CH2 ¬ O ¬ CH2 ¬ CH2 ¬ CH3

c. O

17.40 Write the common name for each of the following: a. CH3 ¬ CH2 ¬ O ¬ CH2 ¬ CH2 ¬ CH3

b. O

c. CH3 ¬ O ¬ CH3 17.41 Draw the condensed structural and line-angle formulas for each

of the following: a. 1-propanol b. ethyl propyl ether c. diethyl ether d. 2-methyl-2-butanol

17.42 Draw the condensed structural and line-angle formulas for each of the following:

a. ethyl methyl ether b. 3-methyl-1-butanol c. 2,4-dichloro-3-hexanol d. butyl ethyl ether

Ethers as Anesthetics Anesthesia is the loss of sensation and consciousness. The term ether has been associated with anesthesia because diethyl ether was the most widely used anesthetic for more than a hundred years. Although it is easy to administer, ether is very volatile and highly flammable. A small spark in the operating room could cause an explosion. Since the 1950s, anesthetics such as Forane (isoflurane), Ethrane (enflurane), Suprane (desflurane), and Sevoflurane have been developed that are

Forane (isoflurane)

Ethrane (enflurane)

F F FC C C C C ClCO O

ClF F FFF

F H H F HH Suprane

(desflurane)

F HC C CO

FF F

F H F

F FC C CO

HF H

F F FC

F Sevoflurane

H

not as f lammable. Most of these anesthetics retain the ether group, but the addition of halogen atoms reduces the volatility and flamma- bility of the ethers.

Isoflurane (Forane) is an inhaled anesthetic.

17.5 Aldehydes and Ketones LEARNING GOAL Write the IUPAC and common names for aldehydes and ketones; draw the condensed structural and line-angle formulas when given their names.

Aldehydes and ketones contain a carbonyl group (C “ O) that has a carbon–oxygen double bond with two groups of atoms attached to the carbon atom at angles of 120°. Because the oxygen atom in the carbonyl group is much more electronegative than the carbon atom, the carbonyl group has a dipole with a partial negative charge (d-) on the oxygen and a partial

ENGAGE 17.8 If aldehydes and ketones both contain a carbonyl group, how can you differentiate between compounds from each family?

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17.5 Aldehydes and Ketones 565

FIGURE 17.7 In the structures of aldehydes, the carbonyl group is always the end carbon.

The carbonyl carbon is at the end of the chain

Methanal (formaldehyde)

IUPAC Common

Ethanal (acetaldehyde)

Propanal (propionaldehyde)

Butanal (butyraldehyde)

H H

H H H

CH3

H

O

H

OOO

O

C H

O

C CH2CH3 H

O

C CH2CH3CH3 H

O

C

FIGURE 17.6 The carbonyl group is found in aldehydes and ketones.

Carbonyl group

Aldehyde Ketone

C

O

CH3 H C

O

CH3 CH3

H

O O

d-

d+

ENGAGE 17.9 Why is the carbon in the carbonyl group in aldehydes always at the end of the chain?

CORE CHEMISTRY SKILL Naming Aldehydes and Ketones

positive charge (d+) on the carbon. The polarity of the carbonyl group strongly influences the physical and chemical properties of aldehydes and ketones.

In an aldehyde, the carbon of the carbonyl group is bonded to at least one hydrogen atom. That carbon may also be bonded to another hydrogen atom, a carbon of an alkyl group, or an aromatic ring (see FIGURE 17.6). In a ketone, the carbonyl group is bonded to two alkyl groups or aromatic rings.

Naming Aldehydes In the IUPAC system, an aldehyde is named by replacing the e in the corresponding alkane name with al. No number is needed for the aldehyde group because it always appears at the end of the chain. The aldehydes with carbon chains of one to four carbon atoms are often referred to by their common names, which end in aldehyde (see FIGURE 17.7). The roots (form, acet, propion, and butyr) of these common names are derived from Latin or Greek words.

The IUPAC system names the aldehyde of benzene as benzaldehyde. Any substituents are numbered from the carbonyl group, which is attached to carbon 1.

O

H

Benzaldehyde

SAMPLE PROBLEM 17.11 Naming Aldehydes

TRY IT FIRST

Write the IUPAC name for the following:

CH2CH3

CH3

CH CH2 HC

O

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

five-carbon chain, methyl substituent

IUPAC name position of methyl group, replace e in alkane name with al

STEP 1 Name the longest carbon chain by replacing the e in the alkane name with al. The longest carbon chain has five carbon atoms.

O

CH2CH3 CH CH2 C H

CH3

pentanal

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566 CHAPTER 17 Organic Chemistry

Naming Ketones In the IUPAC system, the name of a ketone is obtained by replacing the e in the correspond- ing alkane name with one. Carbon chains with five carbon atoms or more are numbered from the end nearer the carbonyl group.

Because ketones play a major role in organic chemistry, they are still often referred to by their common names. In the common names for unbranched ketones, the alkyl groups bonded to the carbonyl group are named as substituents and are listed alphabetically, fol- lowed by ketone. Acetone, which is another name for propanone, has been retained by the IUPAC system.

C

O

CH3CH3CH3 CH2 C

OO

CH3CH3

O

CCH2 CH2 CH3

Propanone (dimethyl ketone; acetone)

Butanone (ethyl methyl ketone)

3-Pentanone (diethyl ketone)

OO

ENGAGE 17.10 Why is ethyl propyl ketone the same compound as 3-hexanone?

STEP 2 Name and number any substituents by counting the carbonyl group as carbon 1. Counting from the right, the methyl substituent is on carbon 3.

O

CH2CH3 CH CH2 C H

CH3

3-methylpentanal 12345

SELF TEST 17.11

a. What is the IUPAC name of the following compound?

O

H

b. Draw the line-angle formula for 4-chlorobenzaldehyde.

ANSWER

a. 5-methylhexanal b.

Cl

H

O

SAMPLE PROBLEM 17.12 Naming Ketones

TRY IT FIRST

Write the IUPAC name for the following ketone:

O

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

five-carbon chain, methyl substituent

IUPAC name position of methyl and carbonyl groups, replace e in alkane name with one

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17.5 Aldehydes and Ketones 567

STEP 1 Name the longest carbon chain by replacing the e in the alkane name with one. The longest chain has five carbon atoms, which is named pentanone.

O

pentanone

STEP 2 Number the carbon chain starting from the end nearer the carbonyl group and indicate its location. Counting from the right, the carbonyl group is on carbon 2.

2-pentanone 5 4 3 2 1

O

STEP 3 Name and number any substituents on the carbon chain. Counting from the right, the methyl group is on carbon 4. The IUPAC name is 4-methyl-2-pentanone.

4-methyl-2-pentanone 5 4 3 2 1

O

SELF TEST 17.12

a. What is the IUPAC name of the following compound?

O

b. Draw the line-angle formula for 3-chlorobutanone.

ANSWER

a. 3-hexanone b.

O

Cl PRACTICE PROBLEMS

Try Practice Problems 17.43 to 17.48

Chemistry Link to Health Some Important Aldehydes and Ketones

Formaldehyde, the simplest aldehyde, is a colorless gas with a pungent odor. An aqueous solution called formalin, which contains 40% formaldehyde, is used as a germicide and to preserve biological specimens. Industrially, it is a reactant in the synthesis of polymers used to make fabrics, insulation materials, carpeting, pressed wood products such as plywood, and plastics for kitchen counters. Exposure to formaldehyde fumes can irritate the eyes, nose, and upper respiratory tract and cause skin rashes, headaches, dizziness, and general fatigue.

Several naturally occurring aromatic aldehydes are used to f la- vor food and as fragrances in perfumes. Benzaldehyde is found in almonds, vanillin in vanilla beans, and cinnamaldehyde in cinnamon.

H

O HO

Benzaldehyde (almond)

Vanillin (vanilla)

Cinnamaldehyde (cinnamon)

OOO

H

The simplest ketone, known as acetone or propanone (dimethyl ketone), is a colorless liquid with a mild odor that has wide use as a solvent in cleaning fluids, paint and nail polish removers, and rubber cement.

Methanal (formaldehyde)

H C

O

H

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568 CHAPTER 17 Organic Chemistry

Acetone is extremely flammable, and care must be taken when using it. In the body, acetone may be produced in uncontrolled diabetes, fasting, and high-protein diets when large amounts of fats are metabolized for energy.

Muscone is a ketone used to make musk perfumes, and oil of spearmint contains carvone.

O

O

Carvone (spearmint oil)

Muscone (musk)

Acetone is used as a solvent in paint and nail polish removers.

PRACTICE PROBLEMS

17.5 Aldehydes and Ketones

17.43 Write the common name for each of the following:

a. CH3 C

O

H b.

O

c. H C

O

H

17.44 Write the common name for each of the following:

a. CH3 CH3CH2

O

C b.

O

c. CH3 CH2 C

O

H

17.45 Write the IUPAC name for each of the following:

a. CH3 CH2 C

O

H b.

O

c. H

OCl

17.46 Write the IUPAC name for each of the following:

a.

O

CH2 CH2 C HCH3 b.

O

c.

Cl

H

O

17.47 Draw the condensed structural formula for a and b and the line-angle formula for c.

a. acetaldehyde b. 2-pentanone c. butyl methyl ketone

17.48 Draw the condensed structural formula for a and b and the line-angle formula for c.

a. propionaldehyde b. 3,4-dimethylhexanal c. 4-bromobutanone

17.6 Carboxylic Acids and Esters LEARNING GOAL Write the IUPAC and common names for carboxylic acids and esters; draw the condensed structural and line-angle formulas when given their names.

In the functional group of a carboxylic acid, the carbon atom of a carbonyl group is attached to a hydroxyl group ( ¬ OH), which forms a carboxyl group. Some ways to represent the carboxyl group in carboxylic acids are shown for propanoic acid.

CH2CH3 C CH2CH3 COOHOH

O

OH

O

Propanoic acid (propionic acid)

O

C OH

Carboxyl group

Hydroxyl group

Carbonyl group

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17.6 Carboxylic Acids and Esters 569

IUPAC Names of Carboxylic Acids The IUPAC name of a carboxylic acid replaces the e in the corresponding alkane name with oic acid. If there are substituents, the carbon chain is numbered beginning with the carboxyl carbon.

Methanoic acid

H C

O CH3

OH 2-Methylpropanoic acid

CH3 CH C

O

OH

Cl

3-Chlorobutanoic acid

CH3 CH C

O

CH2 OH

As with the aldehydes, carboxylic acids with one to four carbon atoms have common names, which are derived from their natural sources. The common names use the prefixes of form, acet, propion, and butyr; these common names are derived from Latin or Greek words (see TABLE 17.7).

CORE CHEMISTRY SKILL Naming Carboxylic Acids

The sour taste of vinegar is due to ethanoic acid (acetic acid).

TABLE 17.7 IUPAC and Common Names of Selected Carboxylic Acids Condensed Structural Formula Line-Angle Formula IUPAC Name Common Name Ball-and-Stick Model

H OHC

O

OHH

O Methanoic acid Formic acid

CH3 C OH

O O

OH

Ethanoic acid Acetic acid

CH3 CH2

O

C OH

O

OH

Propanoic acid Propionic acid

CH3 CH2

O

CH2 C OH OH

O Butanoic acid Butyric acid

SAMPLE PROBLEM 17.13 Naming Carboxylic Acids

TRY IT FIRST

Write the IUPAC name for the following:

OH

O

The simplest aromatic carboxylic acid is benzoic acid. The carbon of the carboxyl group is bonded to carbon 1 in the ring, and the ring is numbered to give the lowest numbers for any substituent.

Cl

Cl

Benzoic acid 3,4-Dichlorobenzoic acid

1 1

2 3

4 OH

O

2-Bromobenzoic acid

OH

Br

O

OH

O

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570 CHAPTER 17 Organic Chemistry

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

four-carbon chain, methyl substituent

IUPAC name position of methyl group, replace e in alkane name with oic acid

STEP 1 Identify the longest carbon chain and replace the e in the alkane name with oic acid. The IUPAC name of a carboxylic acid with four carbons is butanoic acid.

butanoic acid OH

O

STEP 2 Name and number any substituents by counting the carboxyl carbon as 1. Counting from the right, the methyl substituent is on carbon 2. The IUPAC name is 2-methylbutanoic acid.

OH

4 3 2 1

2-methylbutanoic acid

O

SELF TEST 17.13

a. Write the IUPAC name for the following:

CH2 CH2 CH2 OHC

O

CH3

b. Draw the line-angle formula for 3-chlorobenzoic acid.

ANSWER

a. pentanoic acid b. Cl OH

O

Chemistry Link to Health Carboxylic Acids in Metabolism

Several carboxylic acids are part of the metabolic processes within our cells. For example, during glycolysis, a molecule of glucose is broken down into two molecules of pyruvic acid, or actually, its carboxylate ion, pyruvate. During strenuous exercise when oxygen levels are low (anaerobic), pyruvic acid is reduced to give lactic acid or the lactate ion.

Reduction

Pyruvic acid Lactic acid

O

OH 2H+C

O

CCH3

OOH

OHCH CCH3

In the citric acid cycle, also called the Krebs cycle, di- and tricarboxylic acids are oxidized and decarboxylated (loss of CO2) to produce energy for the cells of the body. These carboxylic acids

During exercise, pyruvic acid is converted to lactic acid in the muscles.

PRACTICE PROBLEMS Try Practice Problems 17.49 to 17.52

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17.6 Carboxylic Acids and Esters 571

Esters In the functional group of an ester, the ¬ H of the carboxylic acid is replaced by an alkyl group. Fats and oils in our diets contain esters of glycerol and fatty acids, which are long- chain carboxylic acids. The pleasant aromas and flavors of many fruits including bananas, oranges, and strawberries are due to esters.

CH3 C OH CH3 C O CH3

Ethanoic acid (acetic acid)

Methyl ethanoate (methyl acetate)

O O

Carboxylic Acid Ester

O

O

OH

O

Esterification In a reaction called esterification, an ester is produced when a carboxylic acid and an alcohol react in the presence of an acid catalyst (usually H2SO4) and heat. An excess of the alcohol reactant is used to shift the equilibrium in the direction of the formation of the ester product. In this esterification reaction, the ¬ OH removed from the carboxylic acid and the ¬ H removed from the alcohol combine to form water.

Methanol (methyl alcohol)

H+, heat

Ethanoic acid (acetic acid)

Methyl ethanoate (methyl acetate)

CH3 C OH H O CH3+ H OH+

O

CH3 C O CH3

O

are normally referred to by their common names. At the start of the citric acid cycle, citric acid with six carbons is converted to five-carbon a-ketoglutaric acid. Citric acid is also the acid that gives the sour taste to citrus fruits such as lemons and grapefruits.

COOH

CH2

CH2

C

COOH

Citric acid a-Ketoglutaric acid

COOH

COOH

C O

CH2

CH2 CO2+

COOH

HO [O]

The citric acid cycle continues as a-ketoglutaric acid loses CO2 to give a four-carbon succinic acid. Succinic acid is then oxidized to fumaric acid. We see that some of the functional groups we have studied along with reactions such as oxidation are part of the meta- bolic processes that take place in our cells.

[O]

COOH

COOH

CH2

CH2

COOH

COOH

CH

C

Succinic acid Fumaric acid

H

Citric acid gives the sour taste to citrus fruits.

At the pH of the aqueous environment in the cells, the carboxylic acids are ionized, which means it is actually the carboxylate ions that take part in the reactions of the citric acid cycle. For example, in water, succinic acid is in equilibrium with its carboxylate ion, succinate.

COO-

COO-

CH2

CH2

COOH

COOH

CH2

CH2

Succinic acid Succinate ion

+ 2H2O + 2H3O +

CORE CHEMISTRY SKILL Forming Esters

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572 CHAPTER 17 Organic Chemistry

FIGURE 17.8 The ester methyl ethanoate (methyl acetate) is made from methyl alcohol and ethanoic acid (acetic acid).

Methyl ethanoate (methyl acetate)

PRACTICE PROBLEMS Try Practice Problems 17.53 to 17.56

Naming Esters The name of an ester consists of two words that are derived from the names of the alcohol and the acid in that ester. The first word indicates the alkyl part from the alcohol. The second word is the carboxylate part from the carboxylic acid. The IUPAC names of esters use the IUPAC names for the carbon chain of the acid, while the common names of esters use the common names of the acids. Let’s take a look at the following ester, which has a fruity odor. We start by separating the ester into two parts, which gives us the alkyl of the alcohol and the carboxylate of the acid. Then we name the ester as an alkyl carboxylate (see FIGURE 17.8).

ENGAGE 17.11 What is the IUPAC name of the ester formed from propanoic acid and ethanol?

Butanoic acid (butyric acid)

Methanol (methyl alcohol)

Methyl butanoate (methyl butyrate)

CH3 C OCH2CH2CH3 C OH H O CH3 H2O+ H+, heat

CH3CH2 CH2 +

O O

SELF TEST 17.14

a. What are the IUPAC names of the carboxylic acid and alcohol that are needed to form the following ester, which has the odor of apples?

CH3 CH2 CH2 CH2 CH2 CH2 CH3C

O

O

b. The ester that smells like plums can be synthesized from methanoic acid and 1-butanol. Write the balanced chemical equation, using line-angle formulas, for the formation of this ester.

ANSWER

a. propanoic acid and 1-pentanol

b.

+ H2O+OHHO

O

OHH

O H+, heat

SAMPLE PROBLEM 17.14 Writing Esterification Equations

TRY IT FIRST

An ester that has the smell of pineapples can be synthesized from butanoic acid and methanol. Write the balanced chemical equation for the formation of this ester.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

butanoic acid, methanol

esterification equation ester and H2O

SAMPLE PROBLEM 17.15 Naming Esters

TRY IT FIRST

Write the IUPAC and common names for the following:

O

CH3 CH2 C CH2 CH2 CH3O

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17.6 Carboxylic Acids and Esters 573

Esters in Plants Many of the fragrances of perfumes and flowers and the flavors of fruits are due to esters. Small esters are volatile, so we can smell them, and soluble in water, so we can taste them. Several esters and their flavors and odors are listed in TABLE 17.8.

PRACTICE PROBLEMS Try Practice Problems 17.57 to 17.60

The odor of grapes is due to an ester.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

ester IUPAC name, common name

write the alkyl name for the alcohol, change ic acid to ate

STEP 1 Write the name for the carbon chain from the alcohol as an alkyl group. The alcohol that is used for the ester is propanol, which is named as the alkyl group propyl.

O

CH3 CH2 C CH2 CH2 CH3O propyl

STEP 2 Change the ic acid of the acid name to ate. The carboxylic acid with three carbon atoms is propanoic acid. Replacing the ic acid with ate gives the IUPAC name propyl propanoate. The common name of propionic acid gives the common name for the ester of propyl propionate.

O

CH3 CH2 C CH2 CH2 CH3O propyl propanoate

(propyl propionate)

SELF TEST 17.15

a. Write the IUPAC name for the following ester, which gives the odor and flavor to grapes.

O

O

b. Draw the line-angle formula for ethyl methanoate, which is present in bee stings.

ANSWER

a. ethyl heptanoate b.

O

OH

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574 CHAPTER 17 Organic Chemistry

PRACTICE PROBLEMS

17.6 Carboxylic Acids and Esters

17.49 Write the IUPAC and common name (if any) for each of the following carboxylic acids:

a.

C OHCH3

O

b.

C OHCH3 CH2

O

c.

O

OH

d.

Br

OH

O

17.50 Write the IUPAC and common name (if any) for each of the following carboxylic acids:

a.

O

H C OH b.

O

OH

Br

c.

OH

O

Cl

d.

OCH3

CH3 CH2 CH C OH

17.51 Draw the condensed structural formulas for a and b and the line-angle formulas for c and d.

a. butyric acid b. 2-chloroethanoic acid c. benzoic acid d. 3-bromopropanoic acid

17.52 Draw the condensed structural formulas for a and b and the line-angle formulas for c and d.

a. 2-methylhexanoic acid b. 3-ethylbenzoic acid c. 4-chloropentanoic acid d. 2,4-dibromobutanoic acid

17.53 Draw the condensed structural formula for the ester formed when each of the following reacts with ethyl alcohol:

a. acetic acid b. butyric acid

17.54 Draw the condensed structural formula for the ester formed when each of the following reacts with methyl alcohol:

a. formic acid b. propionic acid

17.55 Draw the line-angle formula for the ester formed in each of the following reactions.

a. OH + OH H+, heat

O

b.

+ H+, heat

HO

O

OH

17.56 Draw the line-angle formula for the ester formed in each of the following reactions.

a. OH + OH H+, heat

O

b.

+ OH H+, heat

OH

O

TABLE 17.8 Some Esters in Fruits and Flavorings Condensed Structural Formula and Name Flavor/Odor

Propyl ethanoate (propyl acetate)

CH3 CH2 CH2 CH3

O

C O Pears

Pentyl ethanoate (pentyl acetate)

CH3 CH2 CH2 CH2 CH2 CH3

O

C O Bananas

Octyl ethanoate (octyl acetate)

CH3 CH2 CH2 CH2 CH2 CH2 CH3CH2 CH2

O

C O Oranges

Ethyl butanoate (ethyl butyrate)

CH2CH2CH3 CH2 CH3

O

C O Pineapples

Pentyl butanoate (pentyl butyrate)

CH2CH2CH3 CH2 CH2 CH2 CH2 CH3

O

C O Apricots

Esters such as ethyl butanoate give the odor and flavor to many fruits.

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17.7 Amines and Amides 575

17.57 Write the IUPAC and common names, if any, for each of the following:

a. H O CH3

O

C

b.

O

O

c. CH3 CH2 CH2 CH3C O

O

17.58 Write the IUPAC and common names, if any, for each of the following:

a.

O

O

b. OCH3 CH2 CH2 CH2 CH2 CH3

O

C

c. OCH3 CH2 CH2 CH3

O

C

17.59 Draw the condensed structural formulas for a and b and the line-angle formulas for c and d.

a. methyl acetate b. butyl formate c. ethyl pentanoate d. propyl propanoate

17.60 Draw the condensed structural formulas for a and b and the line-angle formulas for c and d.

a. hexyl acetate b. ethyl formate c. ethyl hexanoate d. methyl benzoate

17.7 Amines and Amides LEARNING GOAL Write the common names for amines and the IUPAC and common names for amides; draw the condensed structural and line-angle formulas when given their names.

In the functional group of an amine, a nitrogen atom is attached to one or more carbon atoms. In methylamine, a methyl group replaces one hydrogen atom in ammonia. The bond- ing of two methyl groups gives dimethylamine. In trimethylamine, methyl groups replace all three hydrogen atoms attached to the nitrogen atom (see FIGURE 17.9).

Ammonia Methylamine Dimethylamine Trimethylamine

NCH3

CH3

CH3NCH3

H

CH3NCH3

H

HNH

H

H

FIGURE 17.9 Amines have one or more carbon atoms bonded to the N atom.

ENGAGE 17.12 What is the difference between diethylamine and triethylamine?

Naming Amines Several systems are used for naming amines. For simple amines, the common names are often used. In the common name, the alkyl groups bonded to the nitrogen atom are listed in alphabetical order. The prefixes di and tri are used to indicate two and three identical substituents.

Aromatic Amines The aromatic amines use the name aniline, which is approved by IUPAC.

NH2

Aniline

NH2

Br 4-Bromoaniline

1

4

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576 CHAPTER 17 Organic Chemistry

Aniline is used to make many dyes, which give color to wool, cotton, and silk fibers, as well as blue jeans. It is also used to make the polymer polyurethane and in the synthesis of the pain reliever acetaminophen.

O

O Indigo

N

N

H

H

Indigo used in blue dyes can be obtained from tropical plants such as Indigofera tinctoria.

PRACTICE PROBLEMS Try Practice Problems 17.61 and 17.62

SAMPLE PROBLEM 17.16 Naming Amines

TRY IT FIRST

Write the common name for each of the following amines:

a. CH3 ¬ CH2 ¬ NH2 b. CH3 CH3N

CH3

SOLUTION

a. This amine has one ethyl group attached to the nitrogen atom; its name is ethylamine. b. The common name for an amine with three methyl groups attached to the nitrogen

atom is trimethylamine.

SELF TEST 17.16

What is the name of each of the following amines?

a.

NH2

b.

CH3

H

NCH3 CH2

ANSWER

a. aniline b. ethylmethylamine

Chemistry Link to the Environment Alkaloids: Amines in Plants

Alkaloids are physiologically active nitrogen-containing compounds produced by plants. The term alkaloid refers to the “alkali-like” or basic characteristics of amines. Certain alkaloids are used in anes- thetics, in antidepressants, and as stimulants, and many are habit forming.

As a stimulant, nicotine increases the level of adrenaline in the blood, which increases the heart rate and blood pressure. Nicotine is addictive because it activates pleasure centers in the brain. Coniine, which is obtained from hemlock, is extremely toxic.

N

N H

N

Nicotine Coniine

Caffeine is a central nervous system stimulant. Present in coffee, tea, soft drinks, energy drinks, chocolate, and cocoa, caffeine increases alertness, but it may cause nervousness and insomnia. Caffeine is also used in certain pain relievers to counteract the drowsiness caused by an antihistamine.

O

O

Caffeine

N N

N N

Caffeine is a stimulant found in coffee, tea, energy drinks, and chocolate.

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17.7 Amines and Amides 577

For many centuries, morphine and codeine, alkaloids found in the oriental poppy plant, have been used as effective painkillers. Codeine, which is structurally similar to morphine, is used in some prescription painkillers and cough syrups. Heroin, obtained by a chemical modification of morphine, is strongly addicting and is not used medically. The structure of the prescription drug OxyContin (oxycodone) used to relieve severe pain is similar to heroin. Today, there are an increasing number of deaths from OxyContin abuse because its physiological effects are also similar to those of heroin.

O

HO

HO

O

HO

O

O

Morphine Codeine

O

OH O

OxyContin

O

O N

N N

N

O

Heroin

O

O

The green, unripe poppy seed capsule contains a milky sap (opium) that is the source of the alkaloids morphine and codeine.

Amides In the functional group of an amide, the hydroxyl group of a carboxylic acid is replaced by a nitrogen atom (see FIGURE 17.10).

FIGURE 17.10 Amides are derivatives of carboxylic acids in which a nitrogen atom replaces the hydroxyl group ( ¬ OH).

Ethanamide (acetamide)

Ethanoic acid (acetic acid)

Carboxylic Acid Amide

Preparation of Amides An amide is produced in a reaction called amidation, in which a carboxylic acid reacts with ammonia or an amine. A molecule of water is eliminated, and the fragments of the carboxylic acid and amine molecules join to form the amide, much like the formation of esters.

C OH + H H

O

NCH3 CH2

H

C H + H2OCH3 CH2 N

O H Heat

Propanoic acid (propionic acid)

Ammonia

Amide bond

Propanamide (propionamide)

CORE CHEMISTRY SKILL Forming Amides

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578 CHAPTER 17 Organic Chemistry

C OH + H CH3

O

NCH3 CH2

H

C CH3 + H2OCH3 CH2 N

O H Heat

Propanoic acid (propionic acid)

Methylamine N-Methylpropanamide (N-methylpropionamide)

SAMPLE PROBLEM 17.17 Formation of Amides

TRY IT FIRST

Draw the condensed structural formula for the amide product in the following reaction:

OH H2N+

O

CCH3 CH3CH2 Heat

SOLUTION

The condensed structural formula for the amide product can be drawn by attaching the carbonyl group from the acid to the nitrogen atom of the amine. The ¬ OH group removed from the acid and ¬ H from the amine combine to form water.

CH3 C

O

CH2 CH3N

H

SELF TEST 17.17

Draw a. the condensed structural formula and b. the line-angle formula for the amide formed from the reaction of propanoic acid and ethylamine.

ANSWER

a.

CH2CH3 C

O

CH2 CH3N

H

b.

O

N

H

PRACTICE PROBLEMS Try Practice Problems 17.63 and 17.64

Naming Amides In the IUPAC and common names for amides, the oic acid or ic acid from the carboxylic acid name is replaced with amide. The amide of benzoic acid is benzamide. We can diagram the name of an amide in the following way:

CH2 C

O

CH3 CH2 NH2

From butanoic acid (butyric acid) From ammonia

Butanamide (butyramide)

IUPAC Common

BenzamideMethanamide (formamide)

Ethanamide (acetamide)

NH2

O

CCH3NH2

O

CH NH2

O ENGAGE 17.13

Why is the product formed by the amidation of pentanoic acid with ammonia named pentanamide?

M17_TIMB8119_06_SE_C17.indd 578 11/30/18 8:08 AM

17.7 Amines and Amides 579

PRACTICE PROBLEMS

17.7 Amines and Amides

17.61 Write the common name for each of the following:

a. CH3 ¬ NH2 b.

H

N

c. CH2 N CH2 CH3CH3

CH3

17.62 Write the common name for each of the following:

a. CH3 ¬ CH2 ¬ CH2 ¬ NH2 b. CH3 N CH2 CH3

CH3

c.

NH2

17.63 Draw the condensed structural or line-angle formula for the amide formed in each of the following reactions:

a.

+ NH3 Heat

OH

O

b. C OH

O

Heat CH3 CH3CH2H2N+

c.

O

OH H2N+ Heat

17.64 Draw the condensed structural or line-angle formula for the amide formed in each of the following reactions:

a. C OH

O

CH3 CH2 Heat

CH2CH2 NH3+

b. C OH

O

CH3 CH H2N CH2 CH3

CH3

+ Heat

c. OH + Heat

H2N

O

17.65 Write the IUPAC and common name (if any) for each of the following amides:

a.

CH3 C

O

NH2

b.

O

Cl

NH2

c.

NH2C

O

H

17.66 Write the IUPAC and common name (if any) for each of the following amides:

a.

CH3 NH2C

O

CH

Br

b.

NH2

O

c.

O

NH2

SAMPLE PROBLEM 17.18 Naming Amides

TRY IT FIRST

Write the IUPAC and common names for the following amide:

CH2 CCH3

O

NH2

SOLUTION

The IUPAC name of the carboxylic acid is propanoic acid; the common name is propionic acid. Replacing the oic acid or ic acid ending with amide gives the IUPAC name of propanamide and common name of propionamide.

SELF TEST 17.18

Write the IUPAC name for each of the following:

a.

NH2

O

NH2

O

Br

b.

NH2

O

NH2

O

Br

ANSWER

a. hexanamide b. 2-bromobenzamide PRACTICE PROBLEMS

Try Practice Problems 17.65 to 17.68

M17_TIMB8119_06_SE_C17.indd 579 11/30/18 8:08 AM

580 CHAPTER 17 Organic Chemistry

17.67 Draw the condensed structural formulas for a and b and the line-angle formulas for c and d.

a. propionamide b. 2-methylpentanamide c. ethanamide d. 4-chlorobenzamide

17.68 Draw the condensed structural formulas for a and b and the line-angle formulas for c and d.

a. heptanamide b. 3-methylbutanamide c. benzamide d. 3-bromopentanamide

UPDATE Diane’s Treatment in the Burn Unit

When Diane arrives at the hospital, she is transferred to the ICU burn unit. She has second- degree burns that cause damage to the underlying layers of skin and third-degree burns that cause damage to all the layers of the skin. The most common complications of burns are

related to infection. To prevent infection, her skin is covered with a topical antibiotic. Over a period of 3 months, grafts of Diane’s unburned skin are used to cover burned areas. She remains in the burn unit for 3 months and then is discharged. However, Diane returns to the hospital for more skin grafts and plastic surgery.

The arson investigators determine that gasoline was the primary accelerant used to start the fire at Diane’s house. Because there was a lot of paper and dry wood in the area, the fire spread quickly. Some of the hydrocarbons found in gasoline include hexane, heptane, octane, nonane, decane, and toluene.

Applications

17.69 The medications for Diane are contained in blister packs made of polychlorotrifluoroethylene (PCTFE). Draw the expanded structural formula for a portion of PCTFE polymer from three monomers of chlorotrifluoroethylene:

17.70 New polymers have been synthesized to replace PVC in IV bags and cling film. One of these is poly (ethylene-vinyl acetate) (PEVA). Draw the expanded structural formula for a portion of PEVA using an alternating sequence that has two of each monomer shown below:

Ethylene

Vinyl acetate

C C

H

HH

H

C C

H

HH

O

C

CH3

O

17.71 Write the balanced chemical equation for the complete combustion of each of the following hydrocarbons found in gasoline:

a. hexane b. toluene

17.72 Write the balanced chemical equation for the complete combustion of each of the following hydrocarbons found in gasoline:

a. nonane b. octane

Chlorotrifluoroethylene

C C

F

FCl

F

Tablets are contained in blister packs made from polychlorotrifluoroethylene (PCTFE).

M17_TIMB8119_06_SE_C17.indd 580 11/30/18 8:08 AM

Chapter Review 581

CHAPTER REVIEW

17.1 Alkanes LEARNING GOAL Identify the properties of organic or inorganic compounds. Write the IUPAC names and draw the condensed structural and line-angle formulas for alkanes. • Organic compounds contain C and H, have covalent bonds, most

form nonpolar molecules, have low melting points and low boiling points, are not very soluble in water, dissolve as molecules in solu- tions, and burn vigorously in air.

• Inorganic compounds are often ionic or contain polar covalent bonds and form polar molecules, have high melting and boiling points, are usually soluble in water, produce ions in water, and do not burn in air.

• Alkanes are hydrocarbons that have only C ¬ C single bonds. • In the expanded structural formula, a separate line is drawn for

every bonded atom. • A condensed structural formula depicts groups composed of each

carbon atom and its attached hydrogen atoms. • A line-angle formula represents the carbon skeleton as ends and

corners of a zigzag line. • Substituents such as alkyl groups and halogen atoms (named as fluoro,

chloro, bromo, or iodo) can replace hydrogen atoms on the main chain. • The IUPAC system is used to name organic compounds by

indicating the number of carbon atoms and the position of any substituents.

17.2 Alkenes, Alkynes, and Polymers LEARNING GOAL Write the IUPAC names and draw the condensed structural and line-angle formulas for alkenes and alkynes. • Alkenes are unsaturated hydrocarbons that contain carbon–carbon

double bonds (C “ C). • Alkynes contain a carbon–carbon triple bond (C ‚ C).

• The IUPAC names of alkenes end with ene; alkyne names end with yne.

• The main chain is numbered from the end nearer the double or triple bond.

• Alkenes react with hydrogen using a metal catalyst to form alkanes. • Polymers are long-chain molecules that consist of many repeating

units of smaller carbon molecules called monomers.

17.3 Aromatic Compounds LEARNING GOAL Describe the bonding in benzene; name aromatic compounds, and draw their line-angle formulas. • Most aromatic compounds contain benzene, C6H6, a cyclic

structure containing six carbon atoms and six hydrogen atoms. • The structure of benzene is represented as a hexagon with a circle

in the center. • The IUPAC system uses the names of benzene, toluene, aniline,

and phenol. • In the IUPAC name, two substituents are numbered and listed in

alphabetical order.

17.4 Alcohols and Ethers LEARNING GOAL Write the IUPAC and common names for alcohols and the common names for ethers; draw the condensed structural and line-angle formulas when given their names. • The functional group of an alcohol is the

hydroxyl group ( ¬ OH) bonded to a carbon chain.

CONCEPT MAP

Triple Bond

C C

Double BondSingle Bond

C C

Alkanes Alkenes Alkynes Aromatic Compounds

Benzene Ring

Organic Compounds

Carbon Atoms

Alcohols Amines

Esters Amides

Condensed Structural and Line-Angle Formulas

are

involves

which contain

Carbon–Oxygen Single Bonds

Ethers Alcohols

with

are in the

Carbon–Oxygen Double Bonds

Aldehydes

Carboxylic Acids

Carboxyl Group

Ketones

with

are in the

have a

Nitrogen Atom

have a

drawn as

contain a

react with

to form to formwith awith a

C C

with a

ORGANIC CHEMISTRY

Methanol

HCH3 O

M17_TIMB8119_06_SE_C17.indd 581 11/30/18 8:08 AM

582 CHAPTER 17 Organic Chemistry

• In a phenol, the hydroxyl group is bonded to an aromatic ring. • In the IUPAC system, the names of alcohols have ol endings, and

the location of the ¬ OH group is given by numbering the carbon chain.

• In an ether, an oxygen atom ( ¬ O ¬ ) is connected by single bonds to two alkyl or aromatic groups.

• In the common names of ethers, the alkyl groups are listed alphabetically followed by the word ether.

17.5 Aldehydes and Ketones LEARNING GOAL Write the IUPAC and common names for aldehydes and ketones; draw the condensed structural and line-angle formulas when given their names. • Aldehydes and ketones

contain a carbonyl group (C “ O), which is strongly polar.

• In aldehydes, the carbonyl group appears at the end of the carbon chain attached to at least one hydrogen atom.

• In ketones, the carbonyl group occurs between two carbon atoms.

• In the IUPAC system, the e in the alkane name is replaced with al for aldehydes and one for ketones.

• For ketones with more than four carbon atoms in the main chain, the carbonyl group is numbered to show its location.

• Many of the simple aldehydes and ketones use common names.

17.6 Carboxylic Acids and Esters LEARNING GOAL Write the IUPAC and common names for carboxylic acids and esters; draw the condensed structural and line-angle formulas when given their names.

• A carboxylic acid contains the carboxyl functional group, which is a hydroxyl group connected to a carbonyl group.

• The IUPAC name of a carboxylic acid is obtained by replacing the e in the alkane name with oic acid.

• The common names of carboxylic acids with one to four carbon atoms are formic acid, acetic acid, propionic acid, and butyric acid.

• In an ester, a carbon atom replaces the H of the hydroxyl group of a carboxylic acid.

• When heated in the presence of a strong acid, a carboxylic acid reacts with an alcohol to produce an ester. A molecule of water is removed: ¬ OH from the carboxylic acid and ¬ H from the alco- hol molecule.

• The names of esters consist of two words, the alkyl group from the alcohol and the name of the carboxylate obtained by replacing the ic acid ending with ate.

17.7 Amines and Amides LEARNING GOAL Write the common names for amines and the IUPAC and common names for amides; draw the condensed structural and line-angle formulas when given their names. • A nitrogen atom attached to one or more

carbon atoms forms an amine. • In the common names of simple amines, the alkyl groups are listed

alphabetically followed by amine. • Amides are derivatives of carboxylic acids in which the hydroxyl

group is replaced by a nitrogen atom. Amides are named by replac- ing the ic acid or oic acid ending with amide.

• Amides are formed by the reaction of a carboxylic acid with ammonia or an amine.

Carbonyl group

Aldehyde Ketone

C

O

CH3 H C

O

CH3 CH3

Dimethylamine

SUMMARY OF NAMING Family Condensed Structural Formula IUPAC Name Common Name

Alkane CH3 ¬ CH2 ¬ CH3 Propane

Alkane with a substituent CH3 CH3CH

CH3

Methylpropane

Alkene CH3 ¬ CH “ CH2 Propene Propylene Alkyne CH3 ¬ C ‚ CH Propyne

Aromatic Benzene

Alcohol CH3 ¬ OH Methanol Methyl alcohol Ether CH3 ¬ O ¬ CH3 Dimethyl ether

Aldehyde CH H

O

Methanal Formaldehyde

Ketone CCH3 CH3

O

Propanone Acetone; dimethyl ketone

M17_TIMB8119_06_SE_C17.indd 582 11/30/18 8:08 AM

SUMMARY OF REACTIONS

The chapter sections to review are shown after the name of each reaction.

Combustion (17.1)

+CH3 CH3(g) 5O2(g) +CH2 3CO2(g) 4H2O(g) + energy ¢

Propane

+H2C CH3 H2CH CH3 CH2 CH3 Pt

Propene Propane

Hydrogenation (17.2)

+CH3 CH3(g) 5O2(g) +CH2 3CO2(g) 4H2O(g) + energy ¢

Propane

+H2C CH3 H2CH CH3 CH2 CH3 Pt

Propene Propane

Polymerization (17.2)

Ethene monomers Polyethylene

+CH2H2C CH2 CH2

Heat, pressure, catalyst

CH2H2C + CH2H2C CH2 CH2 CH2 CH2

Esterification (17.6)

Ethanoic acid (acetic acid)

Methanol (methyl alcohol)

Methyl ethanoate (methyl acetate)

O

C OH HO+ CH3CH3

O

C O H2O+CH3CH3 H+, heat

Amidation (17.7)

Propanoic acid (propionic acid)

Ammonia Propanamide (propionamide)

Heat

O O HH

C OH N HH+CH2CH3 C N H2O+HCH2CH3

Family Condensed Structural Formula IUPAC Name Common Name

Carboxylic acid CCH3 OH

O

Ethanoic acid Acetic acid

Ester CCH3 CH3O

O

Methyl ethanoate Methyl acetate

Amine CH3 ¬ CH2 ¬ NH2 Ethylamine

Amide CCH3 NH2

O

Ethanamide Acetamide

alcohol An organic compound that contains a hydroxyl group ( ¬ OH) attached to a carbon chain.

aldehyde An organic compound that contains a carbonyl group (C “ O) bonded to at least one hydrogen atom.

alkane A type of hydrocarbon in which the carbon atoms are connected only by single bonds.

alkene A type of hydrocarbon that contains a carbon–carbon double bond (C “ C).

alkyl group An alkane minus one hydrogen atom. Alkyl groups are named like the corresponding alkanes except a yl ending replaces ane.

alkyne A type of hydrocarbon that contains a carbon–carbon triple bond (C ‚ C).

amide An organic compound in which the hydroxyl group of a carboxylic acid is replaced by a nitrogen atom.

KEY TERMS

amine An organic compound that contains a nitrogen atom bonded to one or more carbon atoms.

aromatic compound A compound that contains the ring structure of benzene.

benzene A ring of six carbon atoms each of which is attached to one hydrogen atom, C6H6.

carboxylic acid An organic compound that contains the carboxyl group ( ¬ COOH).

condensed structural formula A formula that shows the carbon atoms grouped with the attached number of hydrogen atoms.

ester An organic compound in which the ¬ H of a carboxyl group is replaced by a carbon atom.

ether An organic compound in which an oxygen atom is bonded to two carbon groups that are alkyl or aromatic.

Key Terms 583

M17_TIMB8119_06_SE_C17.indd 583 11/30/18 8:08 AM

584 CHAPTER 17 Organic Chemistry

expanded structural formula A formula that shows all of the atoms and the bonds connected to each atom.

functional group A group of atoms that determines the physical and chemical properties and naming of a class of organic compounds.

hydrocarbon A type of organic compound that contains only carbon and hydrogen.

IUPAC system A naming system used for organic compounds. ketone An organic compound in which a carbonyl group (C “ O) is

bonded to two alkyl or aromatic groups.

line-angle formula A simplified structure that shows a zigzag line in which carbon atoms are represented as the ends of each line and as corners.

monomer The small organic molecule that is repeated many times in a polymer.

polymer A very large molecule that is composed of many small, repeating structural units that are identical.

structural isomers Two compounds that have the same molecular formula but different arrangements of atoms.

substituent A group of atoms such as an alkyl group or a halogen bonded to the main chain of carbon atoms.

The chapter section containing each Core Chemistry Skill is shown in parentheses at the end of each heading.

Naming and Drawing Alkanes (17.1) • The alkanes ethane, propane, and butane contain two, three, and

four carbon atoms, respectively, connected in a row or a continuous chain.

• Alkanes with five or more carbon atoms in a chain are named using the prefixes pent (5), hex (6), hept (7), oct (8), non (9), and dec (10).

• In the condensed structural formula, the carbon and hydrogen atoms on the ends are written as ¬ CH3 and the carbon and hydro- gen atoms in the middle are written as ¬ CH2 ¬ .

Example: a. What is the name of CH3 ¬ CH2 ¬ CH2 ¬ CH3? b. Draw the condensed structural formula for pentane.

Answer: a. An alkane with a four-carbon chain is named butane. b. Pentane is an alkane with a five-carbon chain. The

carbon atoms on the ends are attached to three H atoms each, and the carbon atoms in the middle are attached to two H each. CH3 ¬ CH2 ¬ CH2 ¬ CH2 ¬ CH3

Writing Equations for Hydrogenation and Polymerization (17.2)

• Hydrogenation adds hydrogen atoms to the double bond of an alkene, using a metal catalyst, to form an alkane.

• In polymerization, many small molecules (monomers) join together to form a polymer.

Example: a. Draw the condensed structural formula for 2-methyl- 2-butene.

b. Draw the condensed structural formula for the product of the hydrogenation of 2-methyl-2-butene.

c. Draw the condensed structural formula of a portion of the polymer polybutene (polybutylene), used in syn- thetic rubber, formed from three monomers of 1-butene.

Answer: a. CH3 C CH

CH3

CH3 b. When H2 adds to a double bond, the product is an

alkane with the same number of carbon atoms.

CH3 CH CH2

CH3

CH3

c.

CH3

CH CH2 CH

CH2

CH2CH2

CH3

CH2

CH3

CH

CH2

CH2

CORE CHEMISTRY SKILLS

Naming Aldehydes and Ketones (17.5)

• In the IUPAC system, an aldehyde is named by replacing the e in the alkane name with al and a ketone by replacing the e with one.

• The position of a substituent on an aldehyde is indicated by num- bering the carbon chain from the carbonyl group and given in front of the name.

• For a ketone, the carbon chain is numbered from the end nearer the carbonyl group.

Example: Give the IUPAC name for the following:

O

Answer: 5-methyl-3-hexanone

Naming Carboxylic Acids (17.6)

• The IUPAC name of a carboxylic acid is obtained by replacing the e in the corresponding alkane name with oic acid.

• The common names of carboxylic acids with one to four carbon atoms are formic acid, acetic acid, propionic acid, and butyric acid.

Example: Write the IUPAC name for the following:

CH3 CH CH2 OH

Br O

C

Answer: 3-bromobutanoic acid

Forming Esters (17.6) • Esters are formed when carboxylic acids react with alcohols in the

presence of a strong acid and heat.

Example: Draw the condensed structural formula for the ester prod- uct of the reaction of propanoic acid and methanol.

Answer: CH3 CH2

O

C CH3O

Forming Amides (17.7) • Amides are formed when carboxylic acids react with ammonia or

amines in the presence of heat.

Example: Draw the condensed structural formula for the amide pro- duced by the reaction of methanoic acid and ethylamine.

Answer: H CH3

O H

C CH2N

M17_TIMB8119_06_SE_C17.indd 584 11/30/18 8:08 AM

Additional Practice Problems 585

UNDERSTANDING THE CONCEPTS

The chapter sections to review are shown in parentheses at the end of each problem.

17.73 Draw a portion of the polymer Teflon, which is made from 1,1,2,2-tetrafluoroethene (use three monomers). (17.2)

Citronellal has the following structure: OCH3

C CH CH2 CH2CH3 CH CH2 C H

CH3

a. Complete the IUPAC name for citronellal: ________, ________-di ________-________-octenal.

b. What does the en in octenal signify? c. What does the al in octenal signify?

17.76 Draw the condensed structural formula for each of the following: (17.5)

a. 2-heptanone, an alarm pheromone of bees b. 2,6-dimethyl-3-heptanone, a communication pheromone of

bees

Teflon is used as a nonstick coating on cooking pans.

17.74 A garden hose is made of polyvinyl chloride (PVC), a polymer of chloroethene (vinyl chloride). Draw a portion of the polymer (use three monomers) for PVC. (17.2)

A plastic garden hose is made of PVC.

17.75 Citronellal, found in oil of citronella, lemon, and lemon grass, is used in perfumes and as an insect repellent. (17.2, 17.5)

An insect-repelling candle contains citronellal, which is found in natural sources.

Bees emit chemicals called pheromones to communicate.

ADDITIONAL PRACTICE PROBLEMS

17.77 Draw the condensed structural formula for each of the following: (17.1, 17.2)

a. 3-ethylhexane b. 2-pentene c. 2-hexyne

17.78 Draw the condensed structural formula for each of the following: (17.1, 17.2)

a. 1-chloro-2-butyne b. 2,3-dimethylpentane c. 3-hexene

17.79 Write the IUPAC name for each of the following: (17.1, 17.2) a.

b. CH3 ¬ CH2 ¬ C ‚ CH c. CH CH2CH3 CH3CH

17.80 Write the IUPAC name for each of the following: (17.1, 17.2)

a.

CH3

H2C C CH2 CH2 CH3

b. CH3 CH2 Cl Cl

c.

BrCH2

CH CHCH2CH3

CH3

CH2 CH3 17.81 Classify each of the following according to its functional group:

(17.2)

a. CH3 CH2 CH2 CH3C

O

b. CH3 ¬ CH “ CH2

c.

O

O d. CH3 ¬ CH2 ¬ CH2 ¬ NH2

M17_TIMB8119_06_SE_C17.indd 585 11/30/18 8:08 AM

586 CHAPTER 17 Organic Chemistry

17.82 Classify each of the following according to its functional group: (17.2)

a. NH2

O

b. CH3 CH2 CH2 C OH

O

c. CH3 CH2

CH3

CH OH

d. CH3 CH

CH3

C H

O

17.83 Select the class of organic compound that matches each of the fol- lowing definitions (a to d): alkane, alkene, alkyne, alcohol, ether, aldehyde, ketone, carboxylic acid, ester, amine, or amide. (17.2)

a. contains a hydroxyl group b. contains one or more carbon–carbon double bonds c. contains a carbonyl group bonded to a hydrogen d. contains only carbon–carbon single bonds

17.84 Select the class of organic compound that matches each of the fol- lowing definitions (a to d): alkane, alkene, alkyne, alcohol, ether, aldehyde, ketone, carboxylic acid, ester, amine, or amide. (17.2)

a. contains a carboxyl group in which the hydrogen atom is replaced by a carbon atom

b. contains a carbonyl group bonded to a hydroxyl group c. contains a nitrogen atom bonded to one or more carbon atoms d. contains a carbonyl group bonded to two carbon atoms

17.85 Classify each of the following according to its functional group(s): (17.2)

a.

Almonds

H

O

b.

Cinnamon sticksCinnamon sticks

H

O

17.86 Classify each of the following according to its functional group(s): (17.2)

a. BHA is an antioxidant used as a preservative in foods such as baked goods, butter, meats, and snack foods.

b. CH3 CH3C

O

C

O

Butter

17.87 Name each of the following aromatic compounds: (17.3)

a.

Br

b.

Cl

Cl

c.

17.88 Name each of the following aromatic compounds: (17.3)

a.

NH2

Br

b.

Cl

c.

F

F

17.89 Draw the structural formula for each of the following: (17.3) a. ethylbenzene b. 4-chlorotoluene c. 1,4-dibromobenzene

17.90 Draw the structural formula for each of the following: (17.3) a. 3-fluorotoluene b. 2-methylaniline c. 1,2-dimethylbenzene

17.91 Write the IUPAC name for a and b and the common name for c. (17.4)

a. CH

CH3

OHCH2CH3

OH

O

Baked goods contain BHA as a preservative.

b.

OH

c.

O CH2CH2 CH2 CH3CH2CH3

17.92 Write the IUPAC name for a and b and the common name for c. (17.4)

a.

Cl

OH

b. CH2

CH3

OHCH CH2CH3

c. O

17.93 Draw the condensed structural formula for each of the following: (17.4)

a. dibutyl ether b. 2-methyl-3-pentanol c. 2-methyl-2-propanol

17.94 Draw the condensed structural formula for each of the following: (17.4)

a. 3-hexanol b. 2-pentanol c. ethyl propyl ether

M17_TIMB8119_06_SE_C17.indd 586 11/30/18 8:08 AM

Challenge Problems 587

17.95 Write the IUPAC name for each of the following: (17.5)

a.

H

O

Cl

b.

O

CH2CH2Cl C H

c.

Cl O

CHCH3 CH3C CH2

17.96 Write the IUPAC name for each of the following: (17.5)

a.

O

C CH3

CH3

CHCH3

b. H

O

Cl

c.

O

C OHCH2 CH2

CH3

CHCH3 17.97 Draw the condensed structural formulas for a and b and the

line-angle formula for c. (17.5) a. ethyl methyl ketone b. 3-chloropropionaldehyde c. 2-bromobenzaldehyde

17.98 Draw the condensed structural formulas for a and b and the line-angle formula for c. (17.5)

a. butyraldehyde b. 2-bromobutanal c. 3,5-dimethyl-2-hexanone

17.99 Write the IUPAC name for each of the following: (17.6)

a.

O

C OHCH2

CH3

CHCH3

b.

O

O

c.

O

CCH3 CH2 O CH2 CH3

17.100 Write the IUPAC name for each of the following: (17.6)

a.

CH3

CH2CH CH2

O

C OHCH3

b.

OH

O

Cl

c.

O

O

17.101 Draw the condensed structural formula for each of the following: (17.7)

a. ethylamine b. hexanamide c. triethylamine

17.102 Draw the condensed structural formula for each of the following: (17.7)

a. formamide b. ethylpropylamine c. diethylmethylamine

17.103 Write the name for each of the following: (17.7)

a. N

b. CH2CH3

O

C NH2

c. NH2CH2CH2CH2CH2CH3

17.104 Write the name for each of the following: (17.7)

a. CH2CH2 CH2

O

C NH2Cl

b.

O

C NH2CH2CH2CH2CH3

c.

CH3

CH2

CH2

CH3NCH2 CH2CH3

The following problems are related to the topics in this chapter. However, they do not all follow the chapter order, and they require you to combine concepts and skills from several sections. These problems will help you increase your critical thinking skills and prepare for your next exam.

17.105 Draw the condensed structural formulas and write the IUPAC names for all eight alcohols that have the molecular formula C5H12O. (17.4)

17.106 There are four amine isomers with the molecular formula C3H9N. Draw their condensed structural formulas. (17.7)

17.107 The insect repellent DEET can be made from an amidation reaction of 3-methylbenzoic acid and diethylamine. Draw the line-angle formula for DEET. (17.7)

CHALLENGE PROBLEMS

Many insect repellents contain DEET, which is an amide.

M17_TIMB8119_06_SE_C17.indd 587 11/30/18 8:08 AM

588 CHAPTER 17 Organic Chemistry

17.110 Complete and balance each of the following reactions: (17.1, 17.2)

a. CH2H2C H2 Pt+

b. CH2 CHCH2 CH2CH3 O2CH2 + ¢

17.108 One of the compounds that give blackberries their odor and flavor can be made by heating hexanoic acid and 1-propanol with an acid catalyst. Draw the condensed structural formula for this compound. (17.6)

17.109 Complete and balance each of the following reactions: (17.1, 17.2)

a. + ¢

CHCH2 CCH3 O2

b. CH CH2CH2 CHCH3 H2CH3 Pt+

ANSWERS TO ENGAGE QUESTIONS 17.8 Both aldehydes and ketones contain carbonyl functional

groups. In an aldehyde, the carbonyl group is bonded to at least one hydrogen atom. In a ketone, the carbonyl group is bonded to two carbon atoms.

17.9 In an aldehyde, the carbonyl group is bonded to at least one hydrogen atom, which means that the carbonyl group must be at the end of the chain.

17.10 Ethyl propyl ketone and 3-hexanone each have six carbons with a two-carbon group and a three-carbon group attached to a carbonyl group on the third carbon. Thus, they are the same compound.

17.11 The ester formed from propanoic acid and ethanol is ethyl propanoate.

17.12 Diethylamine has two ethyl groups attached to the nitrogen atom, whereas triethylamine has three ethyl groups attached to the nitrogen atom.

17.13 When an amide is formed, the product is named by changing the oic acid in the name of the carboxylic acid to amide.

17.1 The central carbon atom in methane has four electron groups. To minimize repulsion, the groups are arranged in a tetrahe- dral shape.

17.2 In a line-angle formula, each line is a single bond, and carbon atoms are represented as the corners and ends of each line.

17.3 Structural isomers have the same molecular formula, but the atoms in the formulas are bonded in different ways.

17.4 Oil, which is a mixture of hydrocarbons, is not soluble in water and is less dense than water. Thus, oil will remain on the surface when added to water.

17.5 The functional group in an alcohol is an ¬ OH group bonded to carbon. The functional group in a ketone is an oxygen atom bonded to a carbon by a double bond. This carbon is bonded to two other carbon atoms.

17.6 In toluene, a methyl group is bonded to benzene. In phenol, an ¬ OH group is bonded to benzene.

17.7 The functional group in an ether is an oxygen atom bonded by single bonds to two carbon atoms. The functional group in an alcohol is an ¬ OH group bonded to carbon.

ANSWERS TO SELECTED PROBLEMS

17.13 a. b. Cl

c.

Cl

Cl

17.15 a.

CH2CH2 CH2 CH2 CH2 CH3CH3

b. liquid c. no d. float

e. C7H16(l) + 11O2(g) h ∆

7CO2(g) + 8H2O(g) + energy

17.17 a. CH4(g) + 2O2(g) h ∆

CO2(g) + 2H2O(g) + energy

b. 2C6H14(l) + 19O2(g) h ∆

12CO2(g) + 14H2O(g) + energy 17.19 a. alkene b. alkyne c. alkene

17.21 a. ethene b. 2-methylpropene c. 2-pentyne

17.23 a. CH3 ¬ CH “ CH2 b. HC ‚ C ¬ CH2 ¬ CH2 ¬ CH2 ¬ CH3

c.

CH3

C CH2 CH3H2C

17.1 a. inorganic b. organic c. inorganic d. organic e. organic f. inorganic

17.3 a. ethane b. ethane c. NaBr d. NaBr

17.5 a. hexane b. heptane c. pentane 17.7 a. same molecule b. structural isomers c. structural isomers

17.9 a. 2-fluorobutane b. dimethylpropane c. 2-chloro-3-methylpentane

17.11 a. C CH2 CH3

CH3

CH3

CH2CH3

b. CH CH2 CH CH3

CH3 CH3 CH3

CHCH3

c. Br CH2 CH2 Cl

M17_TIMB8119_06_SE_C17.indd 588 11/30/18 8:08 AM

Answers to Selected Problems 589

17.25 a. CH3 ¬ CH2 ¬ CH2 ¬ CH2 ¬ CH3 b. CH3 ¬ CH2 ¬ CH2 ¬ CH3

c.

17.27 A polymer is a large molecule composed of small units that are repeated many times.

17.29

CH3

CH C C C

H

H

CH3

H

C

CH3H

H

C

H

H

C

CH3

HH

3H2C

17.31

C C C

F

F

H

H

C

HF

H

C

F

F

C

H

HF

17.33 a. 2-chlorotoluene b. ethylbenzene c. phenol

17.35 a.

NH2

b.

Cl

F

c.

17.37 a. ethanol b. 2-butanol c. 2-methyl-2-pentanol

17.39 a. ethyl methyl ether b. dipropyl ether c. butyl propyl ether

17.41 a. OH

CH2CH2 OHCH3

b.

O CH2 CH2 CH3CH2CH3 O

c.

O CH2CH2 CH3CH3

O

d.

OH

CH3

CCH3 CH2 CH3

OH

17.43 a. acetaldehyde b. methyl propyl ketone c. formaldehyde

17.45 a. propanal b. 2-methyl-3-pentanone c. 2-chlorobenzaldehyde

17.47 a.

O

CCH3 H

b.

O

CCH3 CH2 CH2 CH3

c.

O

17.49 a. ethanoic acid (acetic acid) b. propanoic acid (propionic acid) c. 3-methylhexanoic acid d. 3-bromobenzoic acid

17.51 a.

O

CCH2 CH2CH3 OH

b.

O

CCl CH2 OH

c.

O

OH

d.

O

OHBr

17.53 a.

O

CCH3 O CH2 CH3

b.

O

CCH2 CH2 O CH2 CH3CH3

17.55 a. O

O

b.

O

O

17.57 a. methyl methanoate (methyl formate) b. methyl butanoate (methyl butyrate) c. ethyl propanoate (ethyl propionate)

17.59 a.

O

CCH3 O CH3

b.

O

CH O CH2 CH2 CH2 CH3

c.

O

O

d.

O

O

17.61 a. methylamine b. methylpropylamine c. diethylmethylamine

M17_TIMB8119_06_SE_C17.indd 589 11/30/18 8:08 AM

590 CHAPTER 17 Organic Chemistry

17.63

a.

NH2

O

b.

O

CCH3

H

N CH2 CH3

c.

H

O

N

17.65 a. ethanamide (acetamide) b. 2-chlorobutanamide c. methanamide (formamide)

17.67 a.

C NH2

O

CH3 CH2

b.

C NH2

O

CH2 CHCH2CH3

CH3

c.

O

NH2

d.

O

NH2

Cl

17.69

C C C

Cl

F

F

F

C

FCl

F

C

Cl

F

C

F

FF

17.71 a. 2C6H14(l) 19O2(g)+ 12CO2(g) 14H2O(g)+ energy+ ¢

2C6H14(l) 19O2(g)+ 12CO2(g) 14H2O(g)+ energy+ ¢

b. C7H8(l) 9O2(g)+ 7CO2(g) 4H2O(g)+ energy+ ¢

17.73 C C C C C C

F F F F F F

F F F F F F

17.75 a. 3,7-dimethyl-6-octenal b. The en signifies that a double bond is present. c. The al signifies that an aldehyde is present.

17.77 a. CH3 CH2 CH

CH2 CH3

CH2 CH3CH2

b. CH3 CH CH CH2 CH3

c. CH2C CH2CH3 C CH3

17.79 a. 2,2-dimethylbutane b. 1-butyne c. 2-pentene

17.81 a. ketone b. alkene c. ester d. amine

17.83 a. alcohol b. alkene c. aldehyde d. alkane

17.85 a. aromatic, aldehyde b. aromatic, aldehyde, alkene

17.87 a. 3-bromotoluene b. 1,2-dichlorobenzene c. 3-ethyltoluene

17.89 a.

b.

Cl

c.

Br

Br

17.91 a. 2-butanol b. 3-methyl-2-pentanol c. butyl ethyl ether

17.93 a. OCH3 CH2 CH2 CH2 CH2 CH2 CH2 CH3

b. CH CH

CH3 OH

CH2 CH3CH3

c. CH3 C

OH

CH3

CH3

17.95 a. 4-chlorobenzaldehyde b. 3-chloropropanal c. 2-chloro-3-pentanone

17.97 a.

O

CCH3 CH2 CH3

b.

O

CCH2 CH2Cl H

c.

Br O

H

17.99 a. 3-methylbutanoic acid b. methyl benzoate c. ethyl propanoate

17.101 a. CH3 NH2CH2

b. CH2

O

CH2CH2 CH2CH3 NH2C

c. CH3 CH2 CH3CH2

CH3CH2

N

M17_TIMB8119_06_SE_C17.indd 590 11/30/18 8:08 AM

17.107

N

O

17.109 a. + 8CO2 6H2O+ energy+ ¢

CHCH2 C2CH3 11O2

+ 8CO2 6H2O+ energy+ ¢

CHCH2 C2CH3 11O2

17.103 a. butyldimethylamine b. propanamide c. pentylamine

17.105

OH

CH3 CH2 CH2 CH2 CH2 OH

CH3 CH CH2 CH2 CH3

CH3 CH2 CH CH2 CH3

HO CH2 CH CH2 CH3

HO CH2 CH2 CH

CH3 C CH2 CH3

CH3 C CH2 OH

CH3 CH CH CH3

CH3

OH

CH3

CH3

CH3

CH3

OH

CH3

CH3

OH

2-Pentanol

1-Pentanol

3-Pentanol

2-Methyl-1-butanol

3-Methyl-1-butanol

2-Methyl-2-butanol

3-Methyl-2-butanol

2,2-Dimethyl-1-propanol

b. CH CH2CH2 CHCH3 H2CH3

CH2 CH2CH2 CH2CH3 CH3

Pt+

Answers to Selected Problems 591

M17_TIMB8119_06_SE_C17.indd 591 11/30/18 8:08 AM

592

UPDATE Kate’s Program for Type 2 Diabetes

Now Kate uses a glucose meter to check her blood glucose level. You can read about changes she makes to her lifestyle in the UPDATE Kate’s Program for Type 2 Diabetes, pages 633–634, and see how a change in eating habits and an increase in exercise help Kate to decrease her blood sugar and lose weight.

During her annual physical examination, Kate, a 64-year-old woman, complains that her vision is blurry, she feels a frequent need to urinate, and she has gained 22 lb over the past year. She tried to lose weight and increase her exercise for the past six months without success. Her diet is high in carbohydrates. For dinner, Kate typically eats two cups of pasta and three to four slices of bread with butter or olive oil. She also eats eight to ten pieces of fresh fruit per day at meals and as snacks. A lab test shows that her fasting blood glucose level is 178 mg/dL, indicating type 2 diabetes. She is referred to the diabetes specialty clinic, where she meets Paula, a diabetes nurse. Paula explains to Kate that foods like pasta, bread, and fruit can raise blood glucose levels because they contain large amounts of molecules called carbohydrates, which are broken down in the body into glucose molecules.

CAREER

Diabetes Nurse Diabetes nurses teach patients about diabetes so they can self-manage and control their condition. This includes education on proper diets and nutrition for both diabetic and pre- diabetic patients. Diabetes nurses help patients learn to monitor their medication and blood sugar levels and look for symptoms like diabetic nerve damage and vision loss. Diabetes nurses may also work with patients who have been hospitalized because of complications from their disease. Working with patients with diabetes requires a thorough knowledge of the endocrine system, the system that regulates metabolism, so there is some overlap between diabetes and endocrinology nursing.

Biochemistry 18

M18_TIMB8119_06_SE_C18.indd 592 28/11/18 3:08 PM

18.1 Carbohydrates 593

18.1 Carbohydrates LEARNING GOAL Classify a monosaccharide as an aldose or a ketose; draw the open chain and Haworth structures for monosaccharides.

We have many carbohydrates in our food. There are polysaccharides called starches in bread and pasta. The table sugar used to sweeten cereal, tea, or coffee is sucrose, a disaccharide that consists of two simple sugars, glucose and fructose. Carbohydrates such as table sugar, lactose in milk, and cellulose are all made of carbon, hydrogen, and oxygen. Simple sugars, which have formulas of Cn(H2O)n, were once thought to be hydrates of carbon, thus the name carbohydrate.

Glucose (C6H12O6) is the most important simple carbohydrate in metabolism. In a series of reactions called photosynthesis, energy from the Sun is used to combine the carbon atoms from carbon dioxide (CO2) and the hydrogen and oxygen atoms of water (H2O) to form glucose.

Photosynthesis

Respiration Glucose

energy6H2O6CO2 ++ + C6H12O6 6O2

In the body, glucose is oxidized in a series of metabolic reactions known as respiration to release chemical energy to do work in the cells. Carbon dioxide and water are produced and returned to the atmosphere. The combination of photosynthesis and respiration is called the carbon cycle, in which energy from the Sun is stored in plants by photosynthesis and made available to us when the carbohydrates in our diets are metabolized (see FIGURE 18.1).

Monosaccharides The simplest carbohydrates are the monosaccharides. A monosaccharide cannot be split into smaller carbohydrates. A monosaccharide has a chain of three to seven carbon atoms, one in a carbonyl group and all the others are attached to hydroxyl groups ( ¬ OH). In an aldose, the carbonyl group is on the first carbon ( ¬ CHO); a ketose contains the carbonyl group on the second carbon atom as a ketone (C “ O).

FIGURE 18.1 During photosynthesis, energy from the Sun combines CO2 and H2O to form glucose (C6H12O6) and O2. During respiration in the body, carbohydrates are oxidized to CO2 and H2O, while energy is released.

P H

OTOSY NT H

E S

IS

R E

SP IRATION

Energy

Energy

Carbohydrates + O2

CO2 + H2O

LOOKING AHEAD

18.1 Carbohydrates 593 18.2 Disaccharides and

Polysaccharides 598 18.3 Lipids 605 18.4 Amino Acids and

Proteins 612 18.5 Protein Structure 617 18.6 Proteins as Enzymes 622 18.7 Nucleic Acids 624 18.8 Protein Synthesis 629

Carbohydrates contained in foods such as pasta and bread provide energy for the body.

ENGAGE 18.1 What are the reactants and products of respiration?

Erythrose, an aldose

Erythrulose, a ketose

Aldehyde

Ketone

C

C OHH

C OHH

OH

CH2OH

CH2OH

C O

C OHH

CH2OH

Monosaccharides are also classified by the number of carbon atoms. A monosaccharide with three carbon atoms is a triose, one with four carbon atoms is a tetrose; a pentose has five carbons, and a hexose contains six carbons. We can use both classification systems to indicate the aldehyde or ketone group and the number of carbon atoms. An aldopentose is a five-carbon monosaccharide that is an aldehyde; a ketohexose is a six-carbon monosaccharide that is a ketone.

M18_TIMB8119_06_SE_C18.indd 593 28/11/18 3:08 PM

594 CHAPTER 18 Biochemistry

PRACTICE PROBLEMS Try Practice Problems 18.1 to 18.4

ENGAGE 18.2 Why is ribulose classified as a ketopentose and glucose as an aldohexose?

Ribulose

C

C OHH

C OHH

CH2OH

CH2OH

O

Glucose

C

C OHH

OHH

C OHH

C

C

OH H

CH2OH

OH

Glyceraldehyde (aldotriose)

Threose (aldotetrose)

Ribose (aldopentose)

Fructose (ketohexose)

C

C OHH

OH

CH2OH

C

C

OH H

C OHH

OH

CH2OH

C

C OHH

C

C OHH

OHH

OH

CH2OH

C

C OHH

C

C

OH H

OHH

CH2OH

CH2OH

O

Open-Chain Structures of Some Important Monosaccharides Glucose, galactose, and fructose are the most important monosaccharides. They are all hexoses with the molecular formula C6H12O6. In nature and the cells of the body, their most common form is the d isomer, which has the ¬ OH group attached to carbon 5 on the right side of the chain.

H C

C

OH H C C OOH

OH

H H HC C COH

H C OH

OH

H C OHH C

HO

HO H CHO H CHO H

d-Glucose d-Galactose d-Fructose

1

2

3

4

5

6 CH2OH

OH

C OH

C 1

2

3

4

5

6 CH2OH

1

2

3

4

5

6 CH2OH

CH2OH

The most common hexose, d-glucose, C6H12O6, an aldohexose also known as dextrose and blood sugar, is found in fruits, vegetables, corn syrup, and honey. It is a building block of the disaccharides sucrose, lactose, and maltose, and polysaccharides such as starch, cellulose, and glycogen.

d-Galactose is an aldohexose that does not occur in the free form in nature. It is obtained from the disaccharide lactose, a sugar found in milk and milk products. d-Galactose is important in the cellular membranes of the brain and nervous system. The only difference in the structures of d-glucose and d-galactose is the arrangement of the ¬ OH group on carbon 4.

In contrast to glucose and galactose, d-fructose is a ketohexose. The structure of d-fructose differs from d-glucose at carbons 1 and 2 by the location of the carbonyl group. d-Fructose is the sweetest of the carbohydrates, twice as sweet as sucrose (table sugar). This makes d-fructose popular with dieters because less d-fructose and, therefore, fewer calories are needed to provide a pleasant taste. d-Fructose is found in fruit juices and honey.

ENGAGE 18.3 What are the differences in the open-chain structural formulas of d-galactose and d-glucose?

PRACTICE PROBLEMS Try Practice Problems 18.9 to 18.12

The sweet taste of honey is due to the monosaccharides d-glucose and d-fructose.

M18_TIMB8119_06_SE_C18.indd 594 28/11/18 3:08 PM

18.1 Carbohydrates 595

Chemistry Link to Health Hyperglycemia and Hypoglycemia

Kate’s doctor ordered an oral glucose tolerance test (OGTT) to evaluate her body’s ability to return to normal glucose concentrations (70 to 99 mg/dL) in response to the ingestion of a specified amount of glucose (dextrose). After Kate fasts for 12 h, she drinks a solution containing 75 g of glucose. A blood sample is taken immediately, followed by more blood samples each half-hour for 2 h, and then every hour for a total of 5 h. After her test, Kate is told that her blood glucose was 178 mg/dL, which indicates hyperglycemia. The prefix hyper means above or over; hypo means below or under. The term glyc or gluco refers to “sugar.” Thus, the blood sugar level in hyperglycemia is above normal and in hypoglycemia it is below normal.

A disease that can cause hyperglycemia is type 2 diabetes, which occurs when the pancreas is unable to produce sufficient quantities of insulin. As a result, glucose levels in the body f luids can rise as high as 350 mg/dL. Kate’s symptoms of type 2 diabetes include thirst, excessive urination, and increased appetite. In older adults, type 2 diabetes is sometimes a consequence of excessive weight gain.

B lo

od g

lu co

se l

ev el

( m

g/ dL

)

Time following ingestion of glucose (h)

0 1 2

300

200

100

3 4 5

Hyperglycemia

Normal

Hypoglycemia

A glucose solution is given to determine blood glucose levels.

When a person is hypoglycemic, the blood glucose level rises and then decreases rapidly to levels as low as 40 mg/dL. In some cases, hypoglycemia is caused by overproduction of insulin by the pancreas. Low blood glucose can cause dizziness, general weak- ness, and muscle tremors. A diet may be prescribed that consists of several small meals high in protein and low in carbohydrate. Some hypoglycemic patients are finding success with diets that include more complex carbohydrates rather than simple sugars.

CORE CHEMISTRY SKILL Drawing Haworth Structures

Haworth Structures of Monosaccharides Up until now, we have drawn the structures for monosaccharides such as d-glucose as open chains. However, the most stable form of hexoses is a six-atom ring, known as a Haworth structure. We can draw the Haworth structure for d-glucose as shown in Sample Problem 18.1.

SOLUTION

STEP 1 Turn the open-chain structure clockwise by 90°. The ¬ H and ¬ OH groups on the right of the vertical carbon chain are below the horizontal carbon chain. Those on the left of the open chain are above the horizontal carbon chain.

SAMPLE PROBLEM 18.1 Drawing the Haworth Structure for d-Glucose

TRY IT FIRST

d-Glucose has the following open-chain structure. Draw the Haworth structures for d-glucose.

d-Glucose

C

C

C

C

C

CH2OH

H

H

H OH

OH

HHO

OH

OH

M18_TIMB8119_06_SE_C18.indd 595 28/11/18 3:08 PM

596 CHAPTER 18 Biochemistry

CCCCC

O

H

OH

H

H

OH

H

OH

H

OH

123456 HOCH2

H C

C

OH

OHH C

OHH C

HO H

d-Glucose (open chain)

1

2

3

4

5

6 CH2OH

C OH

=

STEP 2 Fold the horizontal carbon chain into a hexagon, rotate the groups on carbon 5, and bond the O on carbon 5 to carbon 1. With carbons 2 and 3 as the base of a hexagon, move the remaining carbons upward. Rotate the groups on carbon 5 so that the ¬ OH group is close to carbon 1. To complete the Haworth structure, draw a bond between the oxygen of the ¬ OH group on carbon 5 to carbon 1. By conven- tion, the carbon atoms in the ring are drawn as corners.

OH 5 6

4

23

1

5 6

4

23

HO

H H

H

H OH

OH

CH2OH

Carbon-5 oxygen bonds to carbon 1

H H

H H

H

HO

OH

OH

OH

O

CH2OH

Cyclic structure

New hydroxyl group on carbon 1

O

H

C 1

CH2OH5 6

4

23

HO

H OH

H

H OH

OH

H

Rotation of groups on carbon 5

O

H

C 1

STEP 3 Draw the new ¬ OH group on carbon 1 below the ring to give the A form or above the ring to give the B form. Because the new ¬ OH group can form above or below the plane of the Haworth structure, there are two forms of d-glucose, which differ only by the position of the ¬ OH group at carbon 1.

a-d-Glucose

a Form

b Form

b-d-Glucose

HO

H H

H

H H H

H H

HH OH

OH OH

O

CH2OH

HO

OH

OH

OH

O

CH2OH

SELF TEST 18.1

a. Draw the Haworth structure for a-d-galactose.

H C OH

HC

OHH C

HO

CHO H

d-Galactose

OH

C

CH2OH

C

C

OHH

OHH

C

CHO H

HO H

d-Mannose

OH

C

CH2OH

b. Draw the Haworth structure for b-d-mannose.

INTERACTIVE VIDEO Haworth Structures of Monosaccharides

M18_TIMB8119_06_SE_C18.indd 596 28/11/18 3:08 PM

18.1 Carbohydrates 597

In contrast to d-glucose and d-galactose, d-fructose is a ketohexose. It forms the five- atom ring when the hydroxyl group on carbon 5 reacts with the carbon of the ketone group. The new hydroxyl group is on carbon 2.

ANSWER

a. CH2OH

OHH

OH H OHH

H HHO

O

b. CH2OH

OH

H

OH

H

OHH H

HHO

O

PRACTICE PROBLEMS Try Practice Problems 18.5 to 18.8

H

OH H

OH H OH

CH2OHHOCH2 O

2 1

5

4 3

6

a-d-Fructose

H

OH H

OH

H OH CH2OH

HOCH2 O

2

1

5

4 3

6

b-d-Fructose

PRACTICE PROBLEMS

18.1 Carbohydrates

18.1 What functional groups are found in all monosaccharides?

18.2 What is the difference between an aldose and a ketose?

18.3 What are the functional groups and number of carbons in a ketopentose?

18.4 What are the functional groups and number of carbons in an aldohexose?

18.5 What are the kind and number of atoms in the ring portion of the Haworth structure of d-glucose?

18.6 What are the kind and number of atoms in the ring portion of the Haworth structure of d-fructose?

18.7 Identify each of the following Haworth structures as the a or b form:

a.

Applications 18.9 Classify each of the following monosaccharides as an aldopen-

tose, aldohexose, ketopentose, or ketohexose:

OH H

H H OH

OH

CH2OHHOCH2 O

CH2OH

OHH

OH H H

OH

H

H HO

O

18.8 Identify each of the following Haworth structures as the a or b form:

a.

b. CH2OH

HO

H

H

H H

H OH

OH OH

O

CH2OH

OHH

OH H OHH

H HHO

O

b. C

C

H C OH

CH2OH

HOH

H OH

C

OH

Xylose

H OH

CH2OH

HOH

HOH C

C

C

C O

CH2OH

Tagatose

18.10 Classify each of the following monosaccharides as an aldopen- tose, aldohexose, ketopentose, or ketohexose:

a. A solution of xylose is given to test its absorption by the intestines.

b. Tagatose, found in fruit, is similar in sweetness to sugar.

a. Psicose is present in low levels in foods.

b. Lyxose is a component of bacterial lipids.

C

C

C

H C OH

O

CH2OH

CH2OH

H OH

H OH

Psicose

H OH

CH2OH

HOH

HOH C

C

C

Lyxose

C

OH

18.11 An infant with galactosemia can utilize d-glucose in milk but not d-galactose. How does the open-chain structure of d-galactose differ from that of d-glucose?

18.12 d-Fructose is the sweetest monosaccharide. How does the open-chain structure of d-fructose differ from that of d-glucose?

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598 CHAPTER 18 Biochemistry

18.2 Disaccharides and Polysaccharides LEARNING GOAL Describe the monosaccharide units and linkages in disaccharides and polysaccharides.

A disaccharide is composed of two monosaccharides linked together. The most common disaccharides are maltose, lactose, and sucrose.

Maltose, or malt sugar, is a dissacharide obtained from starch and is found in germinating grains. Maltose is used in cereals, candies, and the brewing of beverages. Maltose has a glycosidic bond between two glucose molecules. To form maltose, the ¬OH group on carbon 1 in the first glucose forms a bond with the ¬ OH group on carbon 4 in a second glucose molecule, which is designated as an a(1S 4) linkage. Because the second glucose molecule in maltose has a free ¬ OH group on carbon 1, there are a and b forms of maltose.

+

OHH H

H

H H

H H H

H

H H H

H H

OH OH

H

OHHO

O

CH2OH

CH2OH CH2OH

CH2OH

H H

OH

OH HO

O

OHH H

OH HO

O

OH

OH H OH

O

O + H2O

a-d-Glucose a-d-Glucose

a(1S 4)-Glycosidic bond

a-Maltose, a disaccharide

a Form

1 4

1 4

Lactose, milk sugar, is a disaccharide found in milk and milk products (see FIGURE 18.2). The bond in lactose is a b (1S 4)-glycosidic bond because the b form of galactose links to the hydroxyl group on carbon 4 of glucose. It makes up 6 to 8% of human milk and about 4 to 5% of cow’s milk. Some people do not produce sufficient quantities of the enzyme needed to break down lactose, and the sugar remains undigested and undergoes fermentation, producing gas that causes abdominal cramps and diarrhea. In some commercial milk products, an enzyme

ENGAGE 18.4 How does a-maltose differ from b-maltose?

CH2OH

CH2OH

CH2OH

+ H2O

+

b-d-Galactose OH

OH

OHHO

H H

H H

H H

H

H

H H

H

H H

H H

H

H

HH

O CH2OH

a-d-Glucose OH

OH OHHO

O

OHH

OH OH

O

a Form

O

b(1S 4)-Glycosidic bondOH

OH

HO O

1 4

1

4

a-Lactose, a disaccharide FIGURE 18.2 Lactose is a disaccharide found in milk and milk products.

ENGAGE 18.5 What type of glycosidic bond links galactose and glucose in lactose?

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18.2 Disaccharides and Polysaccharides 599

called lactase is added to break down lactose. Because the glucose molecule in lactose has a free ¬ OH group on carbon 1, there are a and b forms of lactose.

Sucrose, ordinary table sugar, is the most abundant disaccharide in the world. Sucrose consists of an a-d-glucose and a b-d-fructose molecule joined by an a,b11S 22-glycosidic bond. The glycosidic bond in sucrose is between carbon 1 of glucose and carbon 2 of fruc- tose. Most of the sucrose for table sugar comes from sugar cane (20% by mass) or sugar beets (15% by mass) (see FIGURE 18.3). Both the raw and refined forms of sugar are sucrose. Some estimates indicate that each person in the United States consumes an average of 68 kg (150 lb) of sucrose every year either by itself or in a variety of food products.

ENGAGE 18.6 What monosaccharides form sucrose?

OHH

OH

H H

H H

HH H

H

H

H

H H

H

H

H HO

O

OH

OH

OH

O

HO

O

OH

OH

O

OH

OH

OH

O

a-d-Glucose

a,b(1S 2)-Glycosidic bond

2

Sucrose, a disaccharide

2

b-d-Fructose

1 1

CH2OHCH2OH

CH2OHCH2OH

HOCH2 HOCH2

++ H2O

FIGURE 18.3 Sucrose, a disaccharide obtained from sugar beets and sugar cane, contains glucose and fructose.

SAMPLE PROBLEM 18.2 Glycosidic Bonds in Disaccharides

TRY IT FIRST

Melibiose is a disaccharide that is 30 times sweeter than sucrose.

OH

OH H

H H

H H

HO H

H

H

O

CH2OH

CH2O

OHH

OH H OHHO

O

Melibiose

14

14

6

a. What are the monosaccharide units in melibiose? b. What type of glycosidic bond links the monosaccharides? c. Identify the structure as a@ or b@melibiose.

SOLUTION

a. First monosaccharide (left) When the ¬ OH group on carbon 4 is above the plane, it is d-galactose. When the ¬ OH group on carbon 1 is below the plane, it is a-d-galactose.

Second monosaccharide (right)

When the ¬ OH group on carbon 4 is below the plane, it is a-d-glucose.

b. Type of glycosidic bond When the ¬ OH group at carbon 1 of a-d-galactose bonds with the ¬ OH group on carbon 6 of glucose, it is an a(1S 6)-glycosidic bond.

c. Name of disaccharide The ¬ OH group on carbon 1 of glucose is below the plane, which is a-melibiose.

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600 CHAPTER 18 Biochemistry

SELF TEST 18.2

a. Isomaltulose is a sugar substitute that is digested more slowly than sucrose. What are the monosaccharide units in isomaltulose, and what is the glycosidic link in isomaltulose?

O

Isomaltulose

1

6

H

H

OH O

OH

OH HO

H

H

H

H

CH2

OH

OH CH2OHH

O

CH2OH

H

b. Cellobiose is a disaccharide composed of two d-glucose molecules connected by a b(1S 4)@glycosidic linkage. Draw the Haworth structure for b@cellobiose.

ANSWER

a. The two monosaccharides in isomaltulose are glucose and fructose. They are linked by an a(1S 6)@glycosidic bond.

b.

OH

OH

OH O

O

OH

OH HO

H

H

H H

H H

H

H H

O

CH2OH

CH2OH

H

Chemistry Link to Health How Sweet is My Sweetener?

Although many of the monosaccharides and disaccharides taste sweet, they differ considerably in their degree of sweetness. Dietetic foods contain sweeteners that are noncarbohydrate or carbohydrates that are sweeter than sucrose. Some examples of sweeteners com- pared with sucrose are shown in TABLE 18.1.

Sucralose, which is known as Splenda, is made from sucrose by replacing some of the hydroxyl groups with chlorine atoms.

Aspartame, which is marketed as NutraSweet and Equal, is used in a large number of sugar-free products. It is a noncarbohydrate sweetener made of aspartate and a methyl ester of phenylalanine. It does have some caloric value, but it is so sweet that only a very small quantity is needed. However, phenylalanine, one of the breakdown products, poses a danger to anyone who cannot metabolize it prop- erly, a condition called phenylketonuria (PKU).

OH

OH H

H H

H

H

H H

HCl

CH2OH

O

OH

OH

O

CH2Cl

ClCH2

O

Sucralose (Splenda)

From aspartate From phenylalanine

Methyl ester

Aspartame (NutraSweet)

-O

ON

H O

O

O

H3N +

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18.2 Disaccharides and Polysaccharides 601

Another artificial sweetener, Neotame, is a modification of the aspartame structure. The addition of a large alkyl group to the amine group prevents enzymes from breaking the amide bond between aspar- tate and phenylalanine. Thus, phenylalanine is not produced when Neotame is used as a sweetener. Very small amounts of Neotame are needed because it is about 10 000 times sweeter than sucrose.

From aspartateLarge alkyl group to modify Aspartame

From phenylalanine

Neotame

N

-O

ON

H

H

O

O

O

TABLE 18.1 Relative Sweetness of Sugars and Artificial Sweeteners

Sweetness Relative to Sucrose (=100)

Monosaccharides

Galactose 30

Glucose 75

Fructose 175

Disaccharides

Lactose 16

Maltose 33

Sucrose 100

Sugar Alcohols

Sorbitol 60

Maltitol 80

Xylitol 100

Noncarbohydrate Sweeteners

Stevia 15 000

Aspartame 18 000

Saccharin 45 000

Sucralose 60 000

Neotame 1 000 000

Advantame 2 000 000

Steviol

OHO

OH

Artificial sweeteners are used as sugar substitutes.

S N H

O

OO Saccharin (Sweet N Low)

Stevia is a sugar substitute obtained from the leaves of a plant Stevia rebaudiana. It is composed of steviol glycosides that are about 150 times sweeter than sucrose. Stevia has been used to sweeten tea and medicines in South America for 1500 years.

Polysaccharides A polysaccharide is a polymer of many monosaccharides joined together. Four biologically important polysaccharides—amylose, amylopectin, glycogen, and cellulose—are polymers of d-glucose that differ only in the type of glycosidic bonds and the amount of branching in the molecule.

Starch, a storage form of glucose in plants, is found as insoluble granules in rice, wheat, potatoes, beans, and cereals. Starch is composed of two kinds of polysaccharides, amylose and amylopectin. Amylose, which makes up about 20% of starch, consists of 250 to 4000 a-d-glucose molecules connected by a(1S 4)-glycosidic bonds in a continuous chain. Some- times called a straight-chain polymer, polymers of amylose are actually coiled in helical fashion.

Amylopectin, which makes up as much as 80% of plant starch, is a branched-chain polysaccharide. Like amylose, the glucose molecules are connected by a(1S 4)-glycosidic bonds. However, at about every 25 glucose units, there is a branch of glucose molecules attached by an a(1S 6)-glycosidic bond between carbon 1 of the branch and carbon 6 in the main chain (see FIGURE 18.4).

ENGAGE 18.7 What types of glycosidic bonds link glucose molecules in amylose and amylopectin?

Advantame

N

O

O

O

N

H

H

HO

O

O-

O

Saccharin, which is marketed as Sweet’N Low, has been used as a noncarbohydrate artificial sweetener for about 100 years.

Advantame is another modification of aspartame with a com- plex organic group bonded to the amine. Advantame is 20 000 times sweeter than sucrose.

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602 CHAPTER 18 Biochemistry

a(1S 4)-Glycosidic bond

Amylose (20%)

Amylopectin, glycogen (80%)

Glucose monomers

CH2OH

OH H H

H H

H H

H H

H H

H H

H H

H H

H H

H H

H H

H H H H H H H H

H H H

H H H H H H

H H H

HHHHHHHH

HHHH

H

O

OH

CH2OH

OH

O

OH

CH2OH

OH

O

OH

CH2OH

OH

O

OH

OO O O O

a(1S 4)-Glycosidic bond

a(1S 6)-Glycosidic bond to branch

CH2OH

OH

O

OH

CH2OH

OH

O

OH

CH2OH

OH

O

OH

CH2

OH

O

O

OH

OO O

CH2OH

OH

O

OH

CH2OH

OH

O

OH

CH2OH

OH

O

OH

OO O

O O

4

1

1 4

6

1

Amylose is an unbranched polysaccharide of glucose molecules.

Amylopectin is a branched chain of glucose molecules.

FIGURE 18.4 Polysaccharides in starch

Starches hydrolyze easily in water and acid to give smaller saccharides called dextrins, which then hydrolyze to maltose and finally glucose. In our bodies, these complex carbo- hydrates are digested by the enzymes amylase (in saliva) and maltase (in the intestine). The glucose obtained provides about 50% of our nutritional calories.

dextrinsAmylose, amylopectin H+ or amylase

maltose H+ or amylase

many d-glucose units H+ or maltase

Glycogen, or animal starch, is a polymer of glucose that is stored in the liver and muscle of animals. It is used in our cells at a rate that maintains the blood level of glucose and provides energy between meals. The structure of glycogen is very similar to that of amylopectin, found in plants, except that glycogen is more highly branched.

Cellulose is the major structural material of wood and plants. Cotton is almost pure cellulose. In cellulose, glucose molecules form a long unbranched chain similar to that of amylose. However, the glucose units in cellulose are linked by b(1S 4)-glycosidic bonds. The cellulose chains do not form coils like amylose but are aligned in parallel rows held in place by hydrogen bonds between hydroxyl groups in adjacent chains, making cellulose insoluble in water. This gives a rigid structure to the cell wells in wood and fiber that is more resistant to hydrolysis than are the starches (see FIGURE 18.5).

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18.2 Disaccharides and Polysaccharides 603

b(1 4)-Glycosidic bond

CH2OH

OH

O

OH

CH2OH

OH

O

OH

O

Cellulose

CH2OH

OH

H

H

H

H

H

O

OHH

H

H

H

H

H

H

H

H

H

H

H

H

H

O

O

O

O

CH2OH

OH

O

OHH

1

4

O

O

O

O

O

O

O

O

O

O

O

O

O

O

O

FIGURE 18.5 The polysaccharide cellulose is composed of glucose units linked by b(1S 4)-glycosidic bonds.

Humans have enzymes in saliva and pancreatic juices that hydrolyze the a(1S 4)- glycosidic bonds of starches, but not the b(1S 4)-glycosidic bonds of cellulose. Thus, humans cannot digest cellulose. Animals such as horses, cows, and goats can obtain glucose from cellulose because their digestive systems contain bacteria that provide enzymes to hydrolyze b(1S 4)-glycosidic bonds.

ENGAGE 18.8 Why are humans unable to digest cellulose?

PRACTICE PROBLEMS Try Practice Problems 18.13 to 18.20

SAMPLE PROBLEM 18.3 Structures of Polysaccharides

TRY IT FIRST

Give the name of one or more polysaccharides described by each of the following:

a. a polysaccharide that is stored in the liver and muscle tissues b. an unbranched polysaccharide containing b(1S 4)-glycosidic bonds c. a polysaccharide containing a(1S 4)- and a(1S 6)-glycosidic bonds

SOLUTION

a. glycogen b. cellulose c. amylopectin, glycogen

SELF TEST 18.3

a. Cellulose and amylose are both unbranched glucose polymers. How do they differ? b. Amylose and glycogen are both glucose polymers. How do they differ?

ANSWER

a. Cellulose contains glucose units connected by b(1S 4)-glycosidic bonds, whereas the glucose units in amylose are connected by a(1S 4)-glycosidic bonds.

b. Amylose is an unbranched polymer of glucose units, whereas glycogen is a branched polymer of glucose.

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604 CHAPTER 18 Biochemistry

PRACTICE PROBLEMS

18.2 Disaccharides and Polysaccharides

18.13 For each of the following, state the monosaccharide units, the type of glycosidic bond, and the name of the disaccharide, including the a or b form:

OH

OH HO

H H

H H H

H

H OH

H

H

O CH2OH CH2OH

OH

OH

H O

O

b.

18.14 For each of the following, state the monosaccharide units, the type of glycosidic bond, and the name of the disaccharide, including the a or b form:

b.

OH

OH

O

HO

H

H H

H

H H

H

HO

OH

OH

O

CH2OH

HOCH2

CH2OH

OH

H

OH

OH OHO

HO

H H

H

H

H O

CH2OH

OH

OH

H H

H

H

O

CH2

Isomaltose

Applications

18.15 Isomaltose, obtained from the breakdown of starch, has the following Haworth structure:

a. Is isomaltose a mono-, di-, or polysaccharide? b. What are the monosaccharides in isomaltose? c. What is the glycosidic link in isomaltose? d. Is this the a or b form of isomaltose?

18.16 Sophorose, found in certain types of beans, has the following Haworth structure:

Sophorose

OH

O

HO OH

H H

H

H

HO

H

CH2OH

OH

OH HO

H H

H

H

O CH2OH

a. Is sophorose a mono-, di-, or polysaccharide? b. What are the monosaccharides in sophorose? c. What is the glycosidic link in sophorose? d. Is this the a or b form of sophorose?

18.17 Identify the disaccharide that fits each of the following descriptions:

a. ordinary table sugar b. found in milk and milk products c. also called malt sugar d. contains galactose and glucose

18.18 Identify the disaccharide that fits each of the following descriptions:

a. used in brewing b. composed of two glucose units c. also called milk sugar d. contains glucose and fructose

18.19 Give the name of one or more polysaccharides that matches each of the following descriptions:

a. not digestible by humans b. the storage form of carbohydrates in plants c. contains only a(1S 4)-glycosidic bonds d. the most highly branched polysaccharide

18.20 Give the name of one or more polysaccharides that matches each of the following descriptions:

a. the storage form of carbohydrates in animals b. contains only b(1S 4)-glycosidic bonds c. contains both a(1S 4)- and a(1S 6)-glycosidic bonds d. produces only maltose during digestion

OH

OH HO

H H

H H H

H

H H

H

H

O CH2OH CH2OH

OH

OH

OH O

O

a.

OH

OH

OH O

O

OH

OH

HO H

H

H H

H H

H

H H

H

O CH2OH

CH2OH a.

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18.3 Lipids 605

18.3 Lipids LEARNING GOAL Draw the condensed structural or line-angle formula for a fatty acid, a triacylglycerol, and the products of hydrogenation or saponification. Identify the steroid nucleus.

Lipids are a family of biomolecules that have the common property of being soluble in organic solvents but not very soluble in water. The word lipid comes from the Greek word lipos, meaning “fat” or “lard.” Within the lipid family, there are certain structures that distinguish the different types of lipids. Lipids that contain fatty acids include waxes and triacylglycerols, commonly known as fats and oils, which are esters of glycerol and fatty acids. Lipids that are steroids do not contain fatty acids but are characterized by the steroid nucleus of four fused carbon rings.

REVIEW Naming Carboxylic Acids (17.6)

Forming Esters (17.6)

Lipids

Waxes

Triacylglycerols

Long-chain alcohol Fatty acid

G l y c e r o l

Fatty acid

Fatty acid

Fatty acid

Fatty acids Steroids

OH

O

Lipids are naturally occurring biomolecules in cells and tissues, which are soluble in organic solvents but not in water.

CORE CHEMISTRY SKILL Identifying Fatty Acids

Fatty Acids A fatty acid contains a long unbranched chain of carbon atoms, usually 12 to 20, with a carboxylic acid group at one end. The long carbon chain makes fatty acids insoluble in water. An example is lauric acid, a 12-carbon acid found in coconut oil, which has a structure that can be drawn in several forms as follows:

Ball-and-stick model of lauric acid

O

Line-angle formula of lauric acid

CH3 (CH2)10 C OH CH3 (CH2)10 COOH

CH3 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 C

O

Condensed structural formulas of lauric acid OH

OH

O

Saturated fatty acids, such as lauric acid, contain only single bonds between carbons. Monounsaturated fatty acids have one double bond in the carbon chain, and polyunsaturated fatty acids have two or more double bonds. TABLE 18. 2 lists some of the typical fatty acids in lipids.

ENGAGE 18.9 Why is stearic acid classified as a saturated fatty acid and oleic acid as a monounsaturated fatty acid?

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606 CHAPTER 18 Biochemistry

TABLE 18.2 Structures and Melting Points of Common Fatty Acids

Name Carbon Atoms: Double Bonds Present in

Melting Point (°C) Structures

Saturated Fatty Acids

Lauric acid 12:0 Coconut 44 CH3 ¬ (CH2)10 ¬ COOH

OH

O

Myristic acid

14:0 Nutmeg 55 CH3 ¬ (CH2)12 ¬ COOH

OH

O

Palmitic acid

16:0 Palm 63 CH3 ¬ (CH2)14 ¬ COOH

OH

O

Stearic acid 18:0 Animal fat 69 CH3 ¬ (CH2)16 ¬ COOH

OH

O

Monounsaturated Fatty Acids

Palmitoleic acid

16:1 Butter 0 CH3 ¬ (CH2)5 ¬ CH “ CH ¬ (CH2)7 ¬ COOH

OH

O

Oleic acid 18:1 Olives, pecan, grapeseed

14 CH3 ¬ (CH2)7 ¬ CH “ CH ¬ (CH2)7 ¬ COOH

OH

O

Polyunsaturated Fatty Acids

Linoleic acid

18:2 Soybean, safflower, sunflower

- 5 CH3 ¬ (CH2)4 ¬ CH “ CH ¬ CH2 ¬ CH “ CH ¬ (CH2)7 ¬ COOH

OH

O

Linolenic acid

18:3 Corn - 11 CH3 ¬ CH2 ¬ CH “ CH ¬ CH2 ¬ CH “ CH ¬ CH2 ¬ CH “ CH ¬ (CH2)7 ¬ COOH

OH

O

SAMPLE PROBLEM 18.4 Structures and Properties of Fatty Acids

TRY IT FIRST

The line-angle formula for vaccenic acid, a fatty acid found in dairy products and human milk, is shown below.

OH

O

a. Why is this substance an acid? b. How many carbon atoms are in vaccenic acid? c. Is the fatty acid saturated, monounsaturated, or polyunsaturated? d. Would it be soluble in water?

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18.3 Lipids 607

O

CH3 (CH2)5 (CH2)7 C OHCH CH

SOLUTION

a. Vaccenic acid contains a carboxylic acid group. b. It contains 18 carbon atoms. c. It is a monounsaturated fatty acid. d. No; its long hydrocarbon chain makes it insoluble in water.

SELF TEST 18.4

Palmitoleic acid is a fatty acid with the following condensed structural formula:

a. How many carbon atoms are in palmitoleic acid? b. Is the fatty acid saturated, monounsaturated, or polyunsaturated? c. Would it be soluble in water?

ANSWER

a. 16 b. monounsaturated c. no

Waxes Waxes are found in many plants and animals. A wax is an ester of a long-chain fatty acid and a long-chain alcohol. The formulas of some common waxes are given in TABLE 18.3. Beeswax obtained from honeycombs and carnauba wax from palm trees are used to give a protective coating to furniture, cars, and floors. Jojoba wax is used in making candles and cosmetics such as lipstick. Lanolin, a mixture of waxes obtained from wool, is used in hand and facial lotions to aid retention of water, which softens the skin.

CORE CHEMISTRY SKILL Drawing Structures for

Triacylglycerols

TABLE 18.3 Some Typical Waxes Type Condensed Structural Formula Source Uses

Beeswax

O

CH3 (CH2)14 (CH2)29 CH3C O Honeycomb Candles, shoe polish

Carnauba wax

O

CH3 (CH2)24 (CH2)29 CH3C O Brazilian palm tree

Waxes for furniture, cars, floors, shoes

Jojoba wax

O

CH3 (CH2)18 (CH2)19 CH3C O Jojoba bush Candles, soaps, cosmetics

Triacylglycerols In the body, fatty acids are stored as triacylglycerols, also called triglycerides, which are triesters of glycerol (a trihydroxy alcohol) and fatty acids. The general structure of a triacylglycerol follows:

Honeycomb (beeswax) is an ester of a saturated fatty acid and a long-chain alcohol.

CH2

O

O

O

CH

CH2

O

O

O

G l y c e r o l

TristearinTriacylglycerol

Fatty acid

Fatty acid

Fatty acid

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608 CHAPTER 18 Biochemistry

In a triacylglycerol, three hydroxyl groups of glycerol form ester bonds with the carboxyl groups of three fatty acids. For example, glycerol and three molecules of stearic acid form glyceryl tristearate (tristearin).

O

Glycerol 3 Stearic acid molecules

Glyceryl tristearate (tristearin)

CH2 CH3C (CH2)16HO HO+

+ O

CH2 CH3C (CH2)16HO HO+

O

CH CH3C (CH2)16HO HO+

O

O CH2 CH3(CH2)16

CH3(CH2)16 3H2O

O C

O

CH3(CH2)16O CCH2

CH O C

Ester bond

Triacylglycerols are the major form of energy storage for animals. Animals that hibernate eat large quantities of plants, seeds, and nuts that are high in calories. Prior to hibernation, these animals, such as polar bears, gain as much as 14 kg a week. As the external tempera- ture drops, the animal goes into hibernation. The body temperature drops to nearly freezing, and cellular activity, respiration, and heart rate are drastically reduced. Animals that live in extremely cold climates will hibernate for 4 to 7 months. During this time, stored fat is the only source of energy.

Melting Points of Fats and Oils A fat is a triacylglycerol that is solid at room temperature, and comes from animal sources such as meat, whole milk, butter, and cheese. An oil is a triacylglycerol commonly containing one or more double bonds that is usually liquid at room temperature, and is obtained from a plant source (see FIGURE 18.6).

Glyceryl trioleate (triolein)

CH2 O

O

O

O

O

O

CH

CH2

FIGURE 18.6 Vegetable oils such as olive oil contain unsaturated fats.

Prior to hibernation, a polar bear eats food with a high caloric content.

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18.3 Lipids 609

FIGURE 18.7 Vegetable oils are liquids at room temperature because they have a higher percentage of unsaturated fatty acids than do animal fats.

Percent (m/m) Saturated and Unsaturated Fatty Acids in Selected Fats and Oils

0 10 20 30 40 50 60 70 80 90 100

Polyunsaturated Monounsaturated Saturated

Corn

Sunflower

Safflower

Soybean

Canola

Olive

Coconut

Butter

Beef tallow

Solid

Liquid

Triacylglycerols are used to thicken creams and lotions.

ENGAGE 18.10 Why is butter a solid at room temperature, whereas canola oil is a liquid?

The amounts of saturated, monounsaturated, and polyunsaturated fatty acids in some typical fats and oils are shown in FIGURE 18.7. Saturated fatty acids have higher melting points than unsaturated fatty acids because they pack together more tightly. Animal fats usually contain more saturated fatty acids than do vegetable oils. Therefore, the melting points of animal fats are higher than those of vegetable oils.

SAMPLE PROBLEM 18.5 Drawing the Structure for a Triacylglycerol

TRY IT FIRST

Draw the condensed structural formula for glyceryl tripalmitoleate (tripalmitolein), which is used in cosmetic creams and lotions.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

glyceryl tripalmitoleate (tripalmitolein)

condensed structural formula

ester of glycerol and three fatty acids

STEP 1 Draw the condensed structural formulas for glycerol and the fatty acids.

CH2 OH +

CH2 OH

CH OH

HO

HO

HO

O

C (CH2)7 (CH2)5 CH3CH CH

O

C (CH2)7 (CH2)5 CH3CH CH

O

C (CH2)7 (CH2)5 CH3CH CH

Glycerol

Palmitoleic acid

Palmitoleic acid

Palmitoleic acid

+

+

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610 CHAPTER 18 Biochemistry

Reactions of Triacylglycerols The double bonds in unsaturated fatty acids react with hydrogen to produce saturated fatty acids. For example, when hydrogen is added to glyceryl trioleate (triolein) using a nickel catalyst, the product is the saturated fat glyceryl tristearate (tristearin).

Ni

O

Glyceryl trioleate (triolein)

Double bonds Single bonds

Glyceryl tristearate (tristearin)

CH2 O C CH CH(CH2)7 (CH2)7 CH3 O

CH O C CH CH(CH2)7 (CH2)7 CH3 3H2+ O

CH2 O C CH CH(CH2)7 (CH2)7 CH3

O

CH2 O C (CH2)7 (CH2)7 CH3 O

CH O C (CH2)7 (CH2)7 CH3 O

CH2 O C (CH2)7 (CH2)7 CH3CH2

CH2

CH2 CH2

CH2

CH2

STEP 2 Form ester bonds between the hydroxyl groups on glycerol and the carboxyl group on each fatty acid.

O

CH2 O C (CH2)7 (CH2)5 CH3CH CH

O

CH2 O C (CH2)7 (CH2)5 CH3CH CH

O

CH O C (CH2)7 (CH2)5 CH3CH CH

Glyceryl tripalmitoleate (tripalmitolein)

b.

SELF TEST 18.5

a. Draw the line-angle formula for the triacylglycerol containing three molecules of myristic acid (14:0).

b. Draw the condensed structural formula for the triacylglycerol containing three molecules of linoleic acid (18:2).

ANSWER

a.

CH2

CH

CH2

(CH2)7

(CH2)7

(CH2)7

CH2

CH2

CH2

O

O

O

CH

CH

CH

(CH2)4

(CH2)4

(CH2)4

CH3

CH3

CH3

CH

CH

CH

CH

CH

CH

CH

CH

CH

C

C

C

O

O

O

CH2

CH

CH2

O

O

O

O

O

O

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18.3 Lipids 611

Saponification occurs when a fat is heated with a strong base such as sodium hydroxide to give glycerol and the sodium salts of the fatty acids, which is soap. When NaOH is used, a solid soap is produced that can be molded into a desired shape; KOH produces a softer, liquid soap. Oils that are polyunsaturated produce softer soaps. Names like “coconut soap” or “avocado shampoo” tell you the sources of the oil used in the reaction.

Glyceryl tripalmitate (tripalmitin)

Glycerol 3 Sodium palmitate (soap)

O

CH2 O C

C

(CH2)14 CH3

Fat or oil strong base glycerol salts of fatty acids (soap) + +

O

CH2 O C (CH2)14 CH3

O

CH O (CH2)14 CH3

CH2 OH

C

CH2 OH

O

CH OH (CH2)14 CH33NaOH

C

O

(CH2)14 CH3 -O

C

O

(CH2)14 CH3

+ +

Na+

-ONa+

-ONa+

Heat

Heat

PRACTICE PROBLEMS Try Practice Problems 18.21 to 18.30

A cross section of a normal, open artery shows no buildup of plaque.

FIGURE 18.8 Excess cholesterol forms plaque that can block an artery.

A cross section of an artery that is almost completely clogged by atherosclerotic plaque.

PRACTICE PROBLEMS Try Practice Problems 18.31 and 18.32

Steroids: Cholesterol Steroids are compounds containing the steroid nucleus, which consists of four carbon rings fused together. Although they are lipids, steroids do not contain fatty acids.

Steroid nucleus Cholesterol

HO

Attaching other atoms and groups of atoms to the steroid nucleus forms a wide variety of steroid compounds. Cholesterol, which is one of the most important and abun- dant steroids in the body, is a sterol because it contains an oxygen atom as a hydroxyl group ( ¬ OH). Like many steroids, cholesterol has methyl groups, a double bond, and a carbon side chain. In other steroids, the hydroxyl group is replaced by a carbonyl group (C “ O). Cholesterol is obtained from eating meats, milk, and eggs, and it is also synthesized by the liver. There is no cholesterol in vegetable and plant products.

Cholesterol in the Body If a diet is high in cholesterol, the liver produces less cholesterol. A typical daily American diet includes 400 to 500 mg of cholesterol, one of the highest in the world. The American Heart Association has recommended that we consume no more than 300 mg of cholesterol a day. Researchers suggest that saturated fats and cholesterol are associated with diseases such as diabetes; cancers of the breast, pancreas, and colon; and atherosclerosis. In atherosclerosis, deposits of a protein–lipid complex (plaque) accumulate in the coronary blood vessels, restricting the flow of blood to the tissue and causing necrosis (death) of the tissue (see FIGURE 18.8). In the heart, plaque accumula- tion could result in a myocardial infarction (heart attack). Other factors that may also increase the risk of heart disease are family history, lack of exercise, smoking, obesity, diabetes, gender, and age.

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612 CHAPTER 18 Biochemistry

PRACTICE PROBLEMS

18.3 Lipids

18.21 Classify each of the following fatty acids as saturated, monoun- saturated, or polyunsaturated (see Table 18.2):

a. lauric acid b. linolenic acid c. stearic acid

18.22 Classify each of the following fatty acids as saturated, monoun- saturated, or polyunsaturated (see Table 18.2):

a. linoleic acid b. palmitoleic acid c. myristic acid

18.23 Caprylic acid is an 8-carbon saturated fatty acid that occurs in coconut oil (10%) and palm kernel oil (4%). Draw the line-angle formula for glyceryl tricaprylate (tricaprylin).

18.24 Draw the condensed structural formula for glyceryl trilaurate (trilaurin).

Applications 18.25 Safflower oil is polyunsaturated, whereas olive oil is monoun-

saturated. Why would safflower oil have a lower melting point than olive oil?

18.26 Olive oil is monounsaturated, whereas butter fat is saturated. Why does olive oil have a lower melting point than butter fat?

18.27 Use condensed structural formulas to write the balanced chemi- cal equation for the hydrogenation of glyceryl tripalmitoleate, a fat containing glycerol and three palmitoleic acid molecules.

18.28 Use condensed structural formulas to write the balanced chemi- cal equation for the hydrogenation of glyceryl trilinolenate, a fat containing glycerol and three linolenic acid molecules.

18.29 Use condensed structural formulas to write the balanced chemi- cal equation for the NaOH saponification of glyceryl trimyristate, a fat containing glycerol and three myristic acid molecules.

18.30 Use condensed structural formulas to write the balanced chemi- cal equation for the NaOH saponification of glyceryl trioleate, a fat containing glycerol and three oleic acid molecules.

18.31 Draw the structure for the steroid nucleus.

18.32 What are the functional groups on the cholesterol molecule?

18.4 Amino Acids and Proteins LEARNING GOAL Describe protein functions and draw structures for amino acids and peptides.

Proteins perform different functions in the body. Some proteins form structural components such as cartilage, muscles, hair, and nails. Wool, silk, feathers, and horns in animals are made of proteins. Proteins that function as enzymes regulate biological reactions such as digestion and cellular metabolism. Other proteins, such as hemoglobin and myoglobin, transport oxygen in the blood and muscle. TABLE 18.4 gives examples of proteins that are classified by their functions in biological systems.

REVIEW Forming Amides (17.7)

TABLE 18.4 Classification of Some Proteins and Their Functions Class of Protein Function Examples

Structural Provide structural components

Collagen is in tendons and cartilage. Keratin is in hair, skin, wool, horns, and nails.

Contractile Make muscles move Myosin and actin contract muscle fibers.

Transport Carry essential substances throughout the body

Hemoglobin transports oxygen. Lipoproteins transport lipids.

Storage Store nutrients Casein stores protein in milk. Ferritin stores iron in the spleen and liver.

Hormone Regulate body metabolism and the nervous system

Insulin regulates blood glucose level. Growth hormone regulates body growth.

Enzyme Catalyze biochemical reactions in the cells

Sucrase catalyzes the hydrolysis of sucrose. Trypsin catalyzes the hydrolysis of proteins.

Protection Recognize and destroy foreign substances

Immunoglobulins stimulate immune responses.

The horns of animals are made of proteins.

Amino Acids Proteins are composed of molecular building blocks called amino acids. Every amino acid, which is ionized in biological environments, has a central alpha carbon atom (a carbon) bonded to an ammonium group ( ¬ NH3 +), a carboxylate group ( ¬ COO-), a hydrogen

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18.4 Amino Acids and Proteins 613

atom, and an R group. The differences in the 20 a-amino acids present in human proteins are due to the unique characteristics of the R groups.

N

C

C C

O

O +

-

Ball-and-stick model of alanine

O-

R group

Carboxylate group

Ammonium group

CNH C

CH3 O

H

H

H

+

a Carbon

Ionized form of alanine

Classification of Amino Acids We classify amino acids using their specific R groups, which determine their properties in aqueous solution. The nonpolar amino acids have hydrogen, alkyl, or aromatic R groups, which make them hydrophobic (“water fearing”). The polar amino acids have R groups that interact with water, which makes them hydrophilic (“water loving”). There are three groups of polar amino acids. The R groups of polar neutral amino acids contain hydroxyl ( ¬ OH), thiol ( ¬ SH), or amide ( ¬ CONH2) R groups. The R group of a polar acidic amino acid contains a carboxylate group ( ¬ COO-). The R group of a polar basic amino acid contains an ammonium group ( ¬ NH3+). The names and structures of the 20 alpha amino acids commonly found in proteins, along with their three-letter and one-letter abbreviations, are shown at physiological pH (7.4) in TABLE 18.5.

CORE CHEMISTRY SKILL Drawing the Structure for an Amino Acid at Physiological pH

ENGAGE 18.11 Why is valine classified as a nonpolar amino acid whereas histidine is classified as a polar amino acid?

TABLE 18.5 Structures, Names, and Abbreviations of 20 Common Amino Acids at Physiological pH (7.4) Nonpolar Amino Acids (Hydrophobic)

+

H

H3N C COO - +

H

H3N C COO - +

H

H3N C COO - +

H

H3N C COO -

Glycine (Gly, G) Alanine (Ala, A) Valine (Val, V) Leucine (Leu, L) Isoleucine (Ile, I)

+

CH3 CH3

CH2H

H

H3N C

CH

COO-

CH3 CH3

CH CH2

CHCH3

CH3

CH3

+

H

H3N C COO - +

H

H3N C COO - +

H

H3N C COO -+

H

H2N C COO -

Phenylalanine (Phe, F) Methionine (Met, M) Proline (Pro, P)a Tryptophan (Trp, W)

CH2 CH2 CH2H2C

N

H

CH2

CH2CH2

S

CH3

The gold boxes contain the R groups that are the unique parts of each amino acid.

The blue boxes contain the atoms and ions that are common to all amino acids.

(continued)

INTERACTIVE VIDEO Amino Acids at Physiological pH

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614 CHAPTER 18 Biochemistry

Polar Neutral Amino Acids (Hydrophilic)

Serine (Ser, S) Threonine (Thr, T) Tyrosine (Tyr, Y) Cysteine (Cys, C) Asparagine (Asn, N) Glutamine (Gln, Q)

O NH2

O NH2

CH2

OH CH3

CH2 CH2 CH2

OH

SH C

CH2

CH2

C

+

H

H3N C COO - +

H

H3N C COO - +

H

H3N C COO - +

H

H3N C COO - +

H

H3N C COO -

OH

+

H

H3N C COO -

CH

Polar Acidic Amino Acids (Hydrophilic) Polar Basic Amino Acids (Hydrophilic)

Aspartate (Asp, D) Glutamate (Glu, E)

H

H3N C COO -

CH2

C

O O-

H

H3N C COO -

CH2

CH2

C

O O-

+ +

Histidine (His, H)b Lysine (Lys, K)

H

H3N C COO -

CH2

H

H3N C COO -

CH2

CH2

CH2

CH2

NH3

+

N H +

+

+

Arginine (Arg, R)

H

H3N C COO -

CH2

CH2

CH2

NH

NH2

C NH2 +

+

N H

aProline is considered to be a cyclic amino acid because its R group bonds to the nitrogen atom attached to the a carbon. bAt physiological pH, some histidine molecules have a positively charged R group, while other histidine molecules have a neutral R group. The molecule with the positively charged R group is shown here.

SAMPLE PROBLEM 18.6 Structural Formula and Polarity of an Amino Acid

TRY IT FIRST

Draw the structure for the amino acid serine at physiological pH, and write the three-letter and one-letter abbreviations.

SOLUTION

The structure of a specific amino acid is drawn by attaching the R group to the central carbon atom of the general structure of an amino acid.

C

O

H3N C

CH2 +

H

OH R group

O-

The abbreviations for serine are Ser and S.

SELF TEST 18.6

a. Classify serine as polar or nonpolar and hydrophobic or hydrophilic. b. Classify valine as polar or nonpolar and hydrophobic or hydrophilic.

ANSWER

a. polar, hydrophilic b. nonpolar, hydrophobic PRACTICE PROBLEMS

Try Practice Problems 18.33 to 18.38

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18.4 Amino Acids and Proteins 615

Chemistry Link to Health Essential Amino Acids and Complete Proteins

Of the 20 common amino acids used to build the proteins in the body, only 11 can be synthesized in the body. The other 9 amino acids, listed in TABLE 18.6, are essential amino acids that must be obtained from the proteins in the diet.

TABLE 18.6 Essential Amino Acids Histidine (His, H) Phenylalanine (Phe, F)

Isoleucine (Ile, I) Threonine (Thr, T)

Leucine (Leu, L) Tryptophan (Trp, W)

Lysine (Lys, K) Valine (Val, V)

Methionine (Met, M)

Complete proteins, which contain all of the essential amino acids, are found in most animal products such as eggs, milk, meat, fish, and poultry. However, gelatin and plant proteins such as grains, beans, and nuts are incomplete proteins because they are deficient in one or more of the essential amino acids. Diets that rely on plant foods for protein must contain a variety of protein sources to obtain all the essential amino acids. Some examples of complementary protein sources include rice and beans, or a peanut butter sandwich on whole grain bread. Rice contains the amino acid methionine that is deficient in beans, whereas peanuts are rich in lysine that is lacking in grains (see TABLE 18.7).

TABLE 18.7 Amino Acid Deficiencies in Selected Vegetables and Grains

Food Source Amino Acid Deficiency

Eggs, milk, meat, fish, poultry None

Wheat, rice, oats Lysine

Corn Lysine, tryptophan

Beans Methionine, tryptophan

Peas, peanuts Methionine

Almonds, walnuts Lysine, tryptophan

Soy Methionine

Complete proteins such as eggs, milk, meat, and fish contain all of the essential amino acids. Incomplete proteins from plants such as grains, beans, and nuts are deficient in one or more essential amino acids.

ENGAGE 18.12 In the dipeptide Val–Thr, how do you know that valine is at the N-terminus and threonine is at the C-terminus?

PRACTICE PROBLEMS Try Practice Problems 18.41 and 18.42

Peptides A peptide bond is an amide bond that forms when the ¬ COO- group of one amino acid reacts with the ¬ NH3 + group of the next amino acid. We can write the formation of the dipeptide glycylalanine (Gly–Ala, GA) between glycine and alanine. The amino acid writ- ten on the left, glycine has a free (unbonded) NH3

+ group. Therefore, it is the amino acid at the N-terminus of the peptide. The amino acid written on the right, alanine, has a free (unbonded) ¬ COO- group. Therefore, it is the amino acid at the C-terminus of the peptide. In the name of a peptide, each amino acid beginning from the N-terminus has the ine, an, or ate replaced by yl. The last amino acid at the C-terminus uses its full name.

CNH C

OH

H

O- CNH C

OH

H

CH3 CH3

O- CNH C

O

H

H

H

N C H2OC

O

O-+ +

Glycine (Gly, G) Alanine (Ala, A) Glycylalanine (Gly–Ala, GA)

Peptide bond C-terminusN-terminus

+

H

H

H

H HH

+ +

A peptide bond between the ionized structures of glycine and alanine forms the dipeptide glycylalanine.

A summary of the classification of amino acids follows:

Type of Amino Acid Number Type of R Groups Interaction with Water

Nonpolar 9 Nonpolar Hydrophobic

Polar, neutral 6 Contain O and S atoms, but no charge Hydrophilic

Polar, acidic 2 Contain carboxylate groups, negative charge Hydrophilic

Polar, basic 3 Contain ammonium groups, positive charge Hydrophilic

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616 CHAPTER 18 Biochemistry

SAMPLE PROBLEM 18.7 Drawing a Peptide

TRY IT FIRST

Draw the structure and give the name for the tripeptide Gly–Ser–Met.

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

Gly–Ser–Met tripeptide structure, name

peptide bonds, amino acid names

STEP 1 Draw the structure for each amino acid in the peptide, starting from the N-terminus.

Gly, G

H

O

N

H

H C

HH

C O-

Ser, S

H

O

N

H

H C

CH2

OH

H

C O-

Met, M

H

O

N

H

H C

CH2

CH2

CH3

S

H

C O- +++

N-terminus C-terminus

STEP 2 Remove the O atom from the carboxylate group of the N-terminal amino acid and two H atoms from the ammonium group in the adjacent amino acid. Use peptide bonds to connect the amino acids. Repeat this process until the C-terminal amino acid is reached.

Gly–Ser–Met, GSMN-terminus C-terminus

H

O

N

H

H C

HH

C

H

O

N C

CH2

OH

H

C

H

O

N C

CH2

CH2

CH3

S

H

C O- +

The tripeptide is named by replacing the last syllable of each amino acid name with yl, starting at the N-terminus. The amino acid at the C-terminus retains its complete name.

N-terminus glycine is named glycyl serine is named seryl

C-terminus methionine keeps its complete name

The tripeptide is named glycylserylmethionine.

SELF TEST 18.7

a. Draw the structure and give the name for Phe–Thr, a section in glucagon, which is a peptide hormone that increases blood glucose levels.

b. Draw the structure and give the name for Gln–Asn, a section in oxytocin, a nonapeptide used to induce labor.

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18.5 Protein Structure 617

ANSWER

a. b. N C

OH

O-

CH3

H3N +

C

HH

CHCH2 O

C

O

C

H Phenylalanylthreonine

N C O-H3N +

C

HH

CH2CH2

CH2

NH2

C

O

O

NH2

C

O

C

O

C

H Glutaminylasparagine

PRACTICE PROBLEMS Try Practice Problems 18.39 and 18.40, 18.43 and 18.44

PRACTICE PROBLEMS

18.4 Amino Acids and Proteins

18.33 Draw the condensed structural formula for each of the following amino acids at physiological pH:

a. glycine b. threonine c. phenylalanine

18.34 Draw the condensed structural formula for each of the following amino acids at physiological pH:

a. tyrosine b. leucine c. methionine

18.35 Classify each of the amino acids in problem 18.33 as polar or nonpolar. If polar, indicate if the R group is neutral, acidic, or basic. Indicate if each would be hydrophobic or hydrophilic.

18.36 Classify each of the amino acids in problem 18.34 as polar or nonpolar. If polar, indicate if the R group is neutral, acidic, or basic. Indicate if each would be hydrophobic or hydrophilic.

18.37 Give the name of the amino acid represented by each of the following abbreviations:

a. Ala b. V c. Lys d. C

18.38 Give the name of the amino acid represented by each of the following abbreviations:

a. Trp b. M c. Pro d. G

18.39 Draw the condensed structural formula for each of the following peptides, and give its three-letter and one-letter abbreviations:

a. alanylcysteine b. serylphenylalanine c. glycylalanylvaline

18.40 Draw the condensed structural formula for each of the following peptides, and give its three-letter and one-letter abbreviations:

a. prolylaspartate b. threonylleucine c. methionylglutaminyllysine

Applications 18.41 Explain why each of the following pairs are complementary

proteins: a. corn and peas b. rice and soy

18.42 Explain why each of the following pairs are complementary proteins:

a. beans and oats b. almonds and peanuts

18.43 Peptides isolated from rapeseed that may lower blood pressure have the following sequence of amino acids. Draw the structure for each peptide and write the one-letter abbreviations.

a. Arg–Ile–Tyr b. Val–Trp–Ile–Ser

18.44 Peptides from sweet potato with antioxidant properties have the following sequence of amino acids. Draw the structure for each peptide and write the one-letter abbreviations.

a. Asp–Cys–Gly–Tyr b. Asn–Tyr–Asp–Glu–Tyr

18.5 Protein Structure LEARNING GOAL Identify the levels of structure of a protein.

A protein is a polypeptide of 50 or more amino acids that has biological activity. Each pro- tein in our cells has a unique sequence of amino acids that determines its three-dimensional structure and biological function.

Primary Structure The primary structure of a protein is the particular sequence of amino acids held together by peptide bonds. The first protein to have its primary structure determined was insulin, which was accomplished by Frederick Sanger in 1953. Since that time, scientists have determined the amino acid sequences of thousands of proteins. Insulin is a hormone that regulates the glucose level in the blood. In the primary structure of human insulin, there are two polypeptide chains. In chain A, there are 21 amino acids, and in chain B there are 30 amino acids. The polypeptide chains are held together by disulfide bonds formed by the thiol groups of the cysteine amino acids in each of the chains (see FIGURE 18.9). Today, human insulin, for the treatment of diabetes, with this exact structure is produced in large quantities through genetic engineering.

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618 CHAPTER 18 Biochemistry

Gly

Ile

Val

Glu

Gln

Cys

Cys

Thr

Ser

Ile

Cys

Ser

Leu

Tyr

Gln

Leu

Glu

Asn

Tyr

Cys

Asn

Phe

Val

Asn

Gln

His

Leu

Cys

Gly

Ser

His

Leu

Val

Glu

Ala

Leu

Tyr

Leu

Val

Cys

Gly

Glu

Arg

Gly

Phe

Phe

Tyr

Thr

Pro

Lys

Thr

N-terminus A chain

C-terminus A chain

N-terminus B chain

C-terminus B chain

S S

S S

S

S

FIGURE 18.9 The sequence of amino acids in human insulin is the primary structure.

Peptide backbone of primary structure

C-terminus

N-terminus

Hydrogen bonds of

secondary structure

Carbon

Oxygen

Nitrogen

R group

Hydrogen

C

H

N

C

H

NC

O

H

N

C

O

H

N

C

O

H

N

C

O

H

N

C

O

H

N

O O

FIGURE 18.10 The a (alpha) helix acquires a coiled shape from hydrogen bonds between the N ¬ H of the peptide bond in one turn of the polypeptide and the C “ O of the peptide bond in the next turn.

Secondary Structure The secondary structure of a protein describes the structure that forms when amino acids form hydrogen bonds between the atoms in the backbone within a single polypeptide chain or between polypeptide chains. The most common types of secondary structures are the alpha helix and the beta-pleated sheet.

In an alpha helix (a helix), hydrogen bonds form between each N ¬ H group and the oxygen of a C “ O group in the next turn of the a helix (see FIGURE 18.10). Because there are many hydrogen bonds along the peptide, it has a helical or coiled shape.

In the secondary structure known as the beta-pleated sheet (b-pleated sheet), hydrogen bonds hold polypeptide chains together side by side. The hydrogen bonds holding the sheets tightly in place account for the strength and durability of proteins such as silk (see FIGURE 18.11).

Collagen Collagen, the most abundant protein in the body, makes up as much as one-third of all the protein in vertebrates. It is found in connective tissue, blood vessels, skin, tendons, liga- ments, the cornea of the eye, and cartilage. The strong structure of collagen is a result of three polypeptides woven together by hydrogen bonds to form a triple helix. When several triple helices wrap together, they form the fibrils that make up connective tissues and ten- dons. In a young person, collagen is elastic. However, as a person ages, additional bonds form between the fibrils, which make collagen less elastic. Cartilage and tendons become more brittle, and wrinkles are seen in the skin.

Tertiary Structure The tertiary structure of a protein involves attractions and repulsions between the R groups of the amino acids in the polypeptide chain. It is the unique three-dimensional shape of the tertiary structure that determines the biological function of the molecule. TABLE 18.8 lists the interactions that stabilize the tertiary structures of proteins.

Triple helix 3 peptide chains

Collagen fibers are triple helices of polypeptide chains held together by hydrogen bonds.

PRACTICE PROBLEMS Try Practice Problems 18.45 to 18.48

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18.5 Protein Structure 619

FIGURE 18.11 In a b (beta)-pleated sheet secondary structure, hydrogen bonds form between side-by-side sections of the peptide chains.

Carbon

Oxygen

Nitrogen

R group

Hydrogen

Hydrogen bonds between peptide

backbonesAlpha helix

Beta-pleated sheet Protein with alpha helices and beta-pleated sheets

The ribbon model of a protein shows regions of alpha helices and beta-pleated sheets.

TABLE 18.8 Some Interactions That Stabilize Tertiary Structures Interaction Nature of Bonding

Hydrophobic Interactions between nonpolar groups

Hydrophilic Attractions between polar groups and water

Salt bridges (ionic bonds) Ionic interactions between acidic and basic amino acids

Hydrogen bonds Attractions between H and O or N

Disulfide bonds Strong covalent links between sulfur atoms of two cysteine amino acids

Hydrogen bond

OH

C O

H N

H

O-

OH

CH2

CH2

NH3

CH2

S

S

S

S

CH3

CH3

-O C

O

CH2

CH2

OH

O

H

+

b-Pleated sheet Hydrogen bonds

a Helix

Hydrophobic interaction

Hydrophilic interaction with water

Hydrogen bond

Salt bridge Disulfide

bonds

Interactions between amino acid R groups fold a polypeptide into a specific three-dimensional shape called its tertiary structure.

ENGAGE 18.13 Why is the interaction between arginine and aspartate in a tertiary structure called a salt bridge?

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620 CHAPTER 18 Biochemistry

PRACTICE PROBLEMS Try Practice Problems 18.49 and 18.50

CORE CHEMISTRY SKILL Identifying the Primary, Secondary, Tertiary, and Quaternary Structures

of Proteins

SAMPLE PROBLEM 18.8 Interactions That Stabilize Tertiary Structures

TRY IT FIRST

What type of interaction would you expect between the R groups of the following amino acids?

a. cysteine and cysteine b. glutamate and lysine c. tyrosine and serine

SOLUTION

ANALYZE THE PROBLEM

Type of R Groups Type of Interaction Connect

identify R groups on amino acids

determine type of interaction between R groups

polarity of R groups

a. Two cysteines, each with an R group containing ¬ SH, will form a disulfide bond. b. The interaction of the ¬ COO- in the R group of glutamate and the ¬ NH3 + in the

R group of lysine will form a salt bridge. c. The R groups in tyrosine and serine both have ¬ OH groups that can form hydrogen bonds.

SELF TEST 18.8

a. What type of interaction would you expect between the R groups of valine and leucine? b. What type of interaction would you expect between the R groups of threonine and

serine?

ANSWER

a. Both are nonpolar and would have hydrophobic interactions. b. Both contain an ¬ OH group and would form hydrogen bonds.

Quaternary Structure When a biologically active protein consists of two or more polypeptide subunits, the struc- tural level is referred to as a quaternary structure. Hemoglobin, a protein that transports oxygen in blood, consists of four polypeptide chains or subunits (see FIGURE 18.12). The subunits are held together in the quaternary structure by the same kinds of interactions that stabilize the tertiary structures. The hemoglobin molecule can bind and transport four mole- cules of oxygen. TABLE 18.9 and FIGURE 18.13 summarize the structural levels of proteins.

TABLE 18.9 Summary of Structural Levels in Proteins Structural Level Characteristics

Primary Peptide bonds join amino acids in a specific sequence in a polypeptide.

Secondary Hydrogen bonds along or between peptide chains form a helix, or a b-pleated sheet.

Tertiary A protein folds into a compact, three-dimensional shape stabilized by interactions between R groups of amino acids.

Quaternary Two or more protein subunits combine to form a biologically active protein stabilized by the same interactions as in the tertiary structure.

SAMPLE PROBLEM 18.9 Identifying Protein Structure

TRY IT FIRST

Indicate whether the following interactions are responsible for the primary, secondary, tertiary, or quaternary structures of proteins:

a. disulfide bonds between portions of a protein chain b. peptide bonds that form a chain of amino acids c. hydrogen bonds between the O of a peptide bond and the H of a peptide bond four

amino acids away

ENGAGE 18.14 What is the difference between a tertiary structure and a quaternary structure?

FIGURE 18.12 In the ribbon model of hemoglobin, the quaternary structure is made up of four polypeptide subunits.

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18.5 Protein Structure 621

Primary structure

Secondary structure

Tertiary structure

Quaternary structure

FIGURE 18.13 Structural levels for proteins

SOLUTION

ANALYZE THE PROBLEM

Given Need Connect

interactions between amino acids

structural level identify characteristics

a. Disulfide bonds help to stabilize the tertiary and quaternary levels of protein structure. b. Peptide bonds form between amino acids in the primary structure of a protein. c. Hydrogen bonding between peptide bonds in the protein backbone forms the secondary

level of protein structure.

SELF TEST 18.9

Identify the structural level represented by each of the following:

a. the grouping of two polypeptides to make an active enzyme b. the folding of a protein into a compact shape

ANSWER

a. quaternary b. tertiary

PRACTICE PROBLEMS Try Practice Problems 18.51 and 18.52

INTERACTIVE VIDEO

Different Levels of Protein Structure

PRACTICE PROBLEMS

18.5 Protein Structure

18.45 Two peptides each contain one molecule of valine and two molecules of serine. Write the three-letter and one-letter abbreviations for their possible primary structures.

18.46 Two peptides each contain one molecule of alanine, one molecule of glycine, and one molecule of isoleucine. Write the three-letter and one-letter abbreviations for their possible primary structures.

18.47 What is the difference in bonding between an a helix and a b-pleated sheet?

18.48 How is the structure of a b-pleated sheet different from that of a triple helix?

18.49 What type of interaction would you expect between the R groups of the following amino acids in a tertiary structure?

a. cysteine and cysteine b. glutamate and arginine c. serine and aspartate d. leucine and leucine

18.50 What type of interaction would you expect between the R groups of the following amino acids in a tertiary structure?

a. phenylalanine and isoleucine b. aspartate and histidine c. asparagine and tyrosine d. alanine and proline

M18_TIMB8119_06_SE_C18.indd 621 28/11/18 3:08 PM

622 CHAPTER 18 Biochemistry

18.6 Proteins as Enzymes LEARNING GOAL Describe the role of an enzyme in an enzyme-catalyzed reaction.

Every second, thousands of chemical reactions occur in our cells. For example, reactions occur to digest the food we eat, convert the digestion products to chemical energy, and synthesize proteins and other molecules in our cells. Although these reactions can occur outside the body, they occur at rates that are too slow to meet our physiological and metabolic needs. In our bod- ies, catalysts known as enzymes increase the rates at which most biological reactions occur.

An enzyme has a unique three-dimensional shape that recognizes and binds a small group of reacting molecules called substrates. In a catalyzed reaction, an enzyme must first bind to a substrate in a way that favors catalysis.

A typical enzyme is much larger than its substrate. However, within its tertiary structure is a region called the active site that binds a substrate or substrates and catalyzes the reac- tion. This active site is often a small pocket within the larger tertiary structure that closely fits the substrate (see FIGURE 18.14).

C

NH3 + O-O

HO

CH2 OH Substrate in

active site

Val

Ala

Lys

Leu

Val

Ser

Enzyme– Substrate Complex

Hydrophobic pocket

Hydrogen bond

Salt bridge

Substrate Products

Enzyme

Active site

FIGURE 18.14 On the surface of an enzyme, a small region called the active site binds a substrate and catalyzes a reaction of that substrate.

Enzyme-Catalyzed Reaction The combination of an enzyme (E) and a substrate (S) within the active site forms an enzyme–substrate (ES) complex that provides an alternative pathway for the reaction with lower activation energy. Within the active site, the amino acid R groups catalyze the reaction to give an enzyme–product (EP) complex. Then the products are released, and the enzyme is available to bind to another substrate molecule.

E + S vh ES complex h EP complex h E + P Enzyme and Substrate Enzyme–Substrate Complex Enzyme–Product Complex Enzyme and Product

18.51 Indicate whether each of the following statements describes the primary, secondary, tertiary, or quaternary protein structure:

a. R groups interact to form disulfide bonds or salt bridges. b. Peptide bonds join the amino acids in a polypeptide chain. c. Several polypeptides in a b-pleated sheet are held together

by hydrogen bonds between adjacent chains.

18.52 Indicate whether each of the following statements describes the primary, secondary, tertiary, or quaternary protein structure:

a. Hydrogen bonding between amino acid R groups in the same polypeptide gives a coiled shape to the protein.

b. Hydrophilic amino acids move to the polar aqueous environment outside the protein.

c. An active protein contains four subunits.

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18.6 Proteins as Enzymes 623

PRACTICE PROBLEMS Try Practice Problems 18.53 to 18.56

Models of Enzyme Action An early theory of enzyme action, called the lock-and-key model, described the active site as having a rigid, nonflexible shape. According to the lock-and-key model, the shape of the active site was analogous to a lock, and its substrate was the key that specifically fit that lock. How- ever, this model was a static one that did not allow for the flexibility of the tertiary shape of an enzyme and the way we now know that the active site can adjust to the shape of a substrate.

In the dynamic model of enzyme action, called the induced-fit model, the flexibility of the active site allows it to adapt to the shape of the substrate. At the same time, the shape of the substrate may be modified to better fit the geometry of the active site. As a result, the fit of both the active site and the substrate provides the best alignment for the catalysis of the reaction of the substrate. In the induced-fit model, substrate and enzyme work together to acquire a geometrical arrangement that lowers the activation energy.

In the hydrolysis of the disaccharide lactose by the enzyme lactase, a molecule of lac- tose binds to the active site of lactase. As the lactose binds to the enzyme, both the active site and the substrate lactose change shape. In this ES complex, the glycosidic bond of lactose is in a position that is favorable for hydrolysis, which is the splitting by water of a large molecule into smaller parts. The R groups on the amino acids in the active site then catalyze the hydrolysis of lactose which produces the monosaccharides glucose and galac- tose. Because the structures of the products are no longer attracted to the active site, they are released, which allows lactase to react with another lactose molecule (see FIGURE 18.15).

PRACTICE PROBLEMS

18.6 Proteins as Enzymes

18.53 Match the terms, (1) enzyme, (2) enzyme–substrate complex, and (3) substrate, with each of the following:

a. has a tertiary structure that recognizes the substrate b. has a structure that fits the active site of an enzyme c. the combination of an enzyme with the substrate

18.54 Match the terms, (1) active site, (2) lock-and-key model, and (3) induced-fit model, with each of the following:

a. the portion of an enzyme where catalytic activity occurs b. an active site that adapts to the shape of a substrate c. an active site that has a rigid shape

18.55 a. Write an equation that represents an enzyme-catalyzed reaction.

b. How is the active site different from the whole enzyme structure?

18.56 a. How does an enzyme speed up the reaction of a substrate? b. After the products have formed, what happens to the

enzyme?

Substrate (S) (lactose)

Products (P)

H2O

Active site

Enzyme (E) (lactase)

OH

Enzyme–Substrate Complex (ES)

Enzyme–Product Complex (EP)

Galactose Glucose

O OH

OH O

O

OH OH OH O

OH

OH O

O

OH OHO

O

OH OH OO

The glycosidic bond of lactose is aligned within the active site for hydrolysis.

The products are released and the enzyme binds to another lactose.

Lactose is hydrolyzed by the enzyme lactase to glucose and galactose.FIGURE 18.15 The induced-fit

model for enzyme action

M18_TIMB8119_06_SE_C18.indd 623 28/11/18 3:08 PM

624 CHAPTER 18 Biochemistry

18.7 Nucleic Acids LEARNING GOAL Describe the structure of the nucleic acids in DNA and RNA.

Nucleic acids are large molecules, found in the nuclei of cells, that store information and direct activities for cellular growth and reproduction. There are two closely related types of nucleic acids: deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). Deoxyribonucleic acid, the genetic material in the nucleus of a cell, contains all the information needed for the development of a complete living organism. Ribonucleic acid interprets the genetic information in DNA for the synthesis of protein.

Both DNA and RNA are unbranched chains of repeating monomer units known as nucleotides. Each nucleotide has three components: a base that contains nitrogen, a five- carbon sugar, and a phosphate group (see FIGURE 18.16). A DNA molecule may contain several million nucleotides; smaller RNA molecules may contain up to several thousand.

Bases The nitrogen-containing bases in nucleic acids are derivatives of pyrimidine or purine. A pyrimidine has a single ring with two nitrogen atoms, and a purine has two rings each with two nitrogen atoms. In DNA, the pyrimidine bases with single rings are cytosine (C) and thymine (T), and the purine bases with double rings are adenine (A) and guanine (G). RNA contains the same bases, except thymine (5-methyluracil) is replaced by uracil (U) (see FIGURE 18.17).

FIGURE 18.16 The general structure of a nucleotide includes a nitrogen-containing base, a sugar, and a phosphate group.

Base

O

OP CH2

O- Sugar

O N

N 5¿

4¿ 3¿ 2¿

1¿

O-

PRACTICE PROBLEMS Try Practice Problems 18.57 and 18.58

FIGURE 18.17 DNA contains the bases A, G, C, and T; RNA contains A, G, C, and U.

Pyrimidine Bases in Nucleic Acids

Purine Bases in Nucleic Acids

N

N

Adenine (A) (DNA and RNA)

NH2

Cytosine (C) (DNA and RNA)

N

N

O

NH2

H Uracil (U)

(RNA only)

H

N

N

O

O

H

Guanine (G) (DNA and RNA)

H

N

N

H

N

N

O

NH2 N

N

H

N

NN

N

H

N

N

Thymine (T) (DNA only)

H

N

H

CH3 N

O

O

Pyrimidine

Purine

Pentose Sugars In RNA, the five-carbon sugar is ribose, which gives the letter R in the abbreviation RNA. The atoms in the pentose sugars are numbered with primes (1′, 2′, 3′, 4′, and 5′) to dif- ferentiate them from the atoms in the bases. In DNA, the five-carbon sugar is deoxyribose, which is similar to ribose except that there is no hydroxyl group ( ¬ OH) on C2′. The deoxy prefix means “without oxygen” and provides the D in DNA (see FIGURE 18.18).

Nucleosides and Nucleotides A nucleoside is produced when a pyrimidine or purine forms a glycosidic bond to C1′ of a sugar, either ribose or deoxyribose. For example, adenine, a purine, and ribose form a nucleoside called adenosine.

FIGURE 18.18 The five-carbon pentose sugar found in RNA is ribose, and in DNA, deoxyribose.

OH

Ribose in RNA

H H H

OH

H

OH

OHHO

Deoxyribose in DNA

H H H

OH

H

H No oxygen is bonded

to this carbon

Pentose Sugars in RNA and DNA

5¿

4¿

3¿ 2¿

5¿

4¿

3¿ 2¿

1¿

1¿

CH2

HO CH2

O

O

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18.7 Nucleic Acids 625

Nucleotides are produced when the C5′ ¬ OH group of ribose or deoxyribose in a nucleoside forms a phosphate ester. All the nucleotides in RNA and DNA are shown in FIGURE 18.19.

ENGAGE 18.15 What are two differences in the nucleotides of RNA and DNA?

FIGURE 18.19 The nucleotides of RNA (shown in black) are similar to those of DNA (shown in magenta), except in DNA the sugar is deoxyribose and deoxythymidine replaces uridine.

Deoxythymidine monophosphate (dTMP)

CH2

H H H

OH

O

OP-O

O-

Uridine monophosphate (UMP)

H

O O

OH (H)

Cytidine monophosphate (CMP) Deoxycytidine monophosphate (dCMP)

Guanosine monophosphate (GMP) Deoxyguanosine monophosphate (dGMP)

Adenosine monophosphate (AMP) Deoxyadenosine monophosphate (dAMP)

5¿ CH2

H H H

OH

O

OP-O

O- H

OH (H)

5¿

O CH2

H H H

OH

O

OP-O

O- H

H

5¿

O CH2

H H H

OH

O

OP-O

O- H

OH

5¿

O CH2

H H H

OH

O

OP-O

O- H

OH (H)

5¿ N

N

O

NH2 H

N

N

O

O

H

N

N N

N

O

NH2

NH2

N

N N

N

H

N

CH3 N

O

O

Phosphoester bond

H2O

+

+

Sugar + +Base Nucleoside H2O

Adenine

Ribose

N-Glycosidic bond

CH2

H H H

OH

HO

H

O

OH

1¿

Adenosine

CH2

H H H

OH

HO

H

O

OH

1¿ OH

NH2

N

N

H

N

N NH2

N

N N

N

A base forms an N-glycosidic bond with ribose or deoxyribose to form a nucleoside.

The name of a nucleoside that contains a purine ends with osine, whereas a nucleo- side that contains a pyrimidine ends with idine. The names of nucleosides of DNA add deoxy to the beginning of their names. The corresponding nucleotides in RNA and DNA are named by adding monophosphate to the end of the nucleoside name. Although the letters A, G, C, U, and T represent the bases, they are often used in the abbreviations of the respective nucleotides.

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626 CHAPTER 18 Biochemistry

FIGURE 18.20 In the primary structure of RNA, the nucleotides are connected by phosphodiester linkages.

CH2

H H H

O

H

O

5¿

O CH2

H H H

O

H

5¿

O P

O-

O-

O P

O-

O

O P

O-

O

O P

O-

O

O CH2

H H H

O

H

5¿

O CH2

H H H

OH

O

H

OH

5¿

3¿

3¿

3¿

3¿

Phosphodiester linkage

3¿

5¿ Adenine (A)

Cytosine (C)

Guanine (G)

Uracil (U)

Free 5¿ end

Free 3¿ end

OH

OH

OH

N

N

O

NH2

H

N

N

O

O

H

N

N N

N

O

NH2

NH2

N

N N

N

A

T (U)

S

S

S

S

P

P

P

P

C

G

Free 5¿ end

Free 3¿ end

Phosphodiester linkage

5¿

3¿

5¿

3¿

5¿

3¿

5¿

3¿

In the primary structure of nucleic acids, each sugar in a sugar– phosphate backbone is attached to a base.

DNA Double Helix: A Secondary Structure In 1953, James Watson and Francis Crick proposed that DNA was a double helix that consists of two polynucleotide strands winding about each other like a spiral staircase. The sugar–phosphate backbones are analogous to the outside railings, with the bases arranged like steps along the inside.

Complementary Base Pairs Each of the bases along one polynucleotide strand forms hydrogen bonds to a specific base on the opposite DNA strand. Adenine only bonds to thymine, and guanine only bonds to cyto- sine (see FIGURE 18.21). The pairs AT and GC are called complementary base pairs. The specific pairing of the bases occurs because adenine and thymine form two hydrogen bonds, while cytosine and guanine form three hydrogen bonds. No other stable base pairs occur.

CORE CHEMISTRY SKILL Writing the Complementary DNA

Strand

Structure of Nucleic Acids The nucleic acids are unbranched chains of many nucleotides in which the 3′ hydroxyl group of the sugar in one nucleotide bonds to the phosphate group on the 5′ carbon atom in the sugar of the next nucleotide. This connection between the sugars in adjacent nucleotides is referred to as a phosphodiester linkage. As more nucleotides are added, a backbone forms that consists of alternating sugar and phosphate groups. The bases, which are attached to each sugar, extend out from the sugar–phosphate backbone.

In any nucleic acid, the sugar at one end has a free 5′ phosphate group, and the sugar at the other end has a free 3′ hydroxyl group. A nucleic acid sequence is read from the free 5′ phosphate end to the free 3′ hydroxyl end using only the letters of the bases. For example, the nucleotide sequence in the section of RNA shown in FIGURE 18.20 is ¬ A C G U ¬ .

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18.7 Nucleic Acids 627

FIGURE 18.21 In the model shown, the sugar– phosphate backbone is represented by a ribbon with hydrogen bonds between complementary base pairs.

C

T O

N

H

H

H

H

H

N

N

N

N

N

O

N N

Guanine – Cytosine Base Pair (three hydrogen bonds)

A

AT

A

G

A

CG

CG

T

A

Adenine Thymine

T

A

GC

CG

N

N

N

N

N O

O

N H

H H

N

CH3

Adenine–Thymine Base Pair (two hydrogen bonds)

5¿ 3¿

Guanine Cytosine

5¿ 3¿

Sugar–phosphate backbone

SAMPLE PROBLEM 18.10 Complementary Base Pairs

TRY IT FIRST

Write the complementary base sequence for the following segment of a strand of DNA:

A C G A T C T

SOLUTION

In the complementary strand of DNA, the complementary base pairs are AT and CG. Original segment of DNA: A C G A T C T

f f f f f f f Complementary segment: T G C T A G A

SELF TEST 18.10

Write the complementary base sequence for each of the following DNA segments: a. G G T T A A C C b. G A T T A C A C G

ANSWER

a. C C A A T T G G b. C T A A T G T G C

M18_TIMB8119_06_SE_C18.indd 627 28/11/18 3:08 PM

628 CHAPTER 18 Biochemistry

FIGURE 18.22 In DNA replication, the separate strands of the parent DNA are the templates for the synthesis of complementary strands, which produces two exact copies of DNA.

C

A T

AT

C

G

G

A T

A T

GC

GC

GC

GC

GC

AT

AT

A T

A T

A T

C G

CG

CG

CG

G

A

GC

AT

AT

A T

C

C

T AAT

New phosphodiester linkages form

Parent DNA

Nucleotides

New DNA strand

New DNA strand

Daughter DNA strands

Eventually, the entire double helix of the parent DNA is copied. In each new DNA molecule, one strand of the double helix is from the original DNA and one is a newly synthesized strand. This process produces two new DNAs called daughter DNA that are identical to each other and exact copies of the original parent DNA. In DNA replication, complementary base pairing ensures the correct placement of bases in the daughter DNA strands.

PRACTICE PROBLEMS Try Practice Problems 18.59 to 18.64

DNA Replication In DNA replication, the strands in the original or parent DNA separate, which allows the synthesis of complementary strands of DNA. The process begins when an enzyme catalyzes the unwinding of a portion of the double helix by breaking the hydro- gen bonds between the complementary bases. These resulting strands or parent DNA now act as templates for the synthesis of new complementary strands of DNA (see FIGURE 18.22).

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18.8 Protein Synthesis 629

18.8 Protein Synthesis LEARNING GOAL Describe the synthesis of protein from mRNA.

Ribonucleic acid, RNA, which makes up most of the nucleic acid found in the cell, is involved with transmitting the genetic information needed to operate the cell. Similar to DNA, RNA mol- ecules are polymers of nucleotides. However, RNA differs from DNA in several important ways:

1. The sugar in RNA is ribose rather than the deoxyribose found in DNA. 2. In RNA, the base uracil replaces thymine. 3. RNA molecules are single stranded, not double stranded. 4. RNA molecules are much smaller than DNA molecules.

Types of RNA There are three major types of RNA in the cells: messenger RNA, ribosomal RNA, and transfer RNA, which are classified according to their function, as shown in TABLE 18.10.

TABLE 18.10 Types of RNA Molecules in Humans

Type Abbreviation Function in the Cell Percentage of Total RNA

Messenger RNA mRNA Carries information for protein synthesis from the DNA to the ribosomes

5

Ribosomal RNA rRNA Major component of the ribosomes; site of protein synthesis

80

Transfer RNA tRNA Brings specific amino acids to the site of protein synthesis

15

In replication, the genetic information in DNA is reproduced by making identical cop- ies of DNA. In transcription, the information contained in DNA is transferred to mRNA molecules. In translation, the genetic information now present in the mRNA is used to build the sequence of amino acids of the desired protein (see FIGURE 18.23).

PRACTICE PROBLEMS

18.7 Nucleic Acids

18.57 Identify each of the following bases as a component of RNA, DNA, or both:

a. thymine b.

18.59 How are the two strands of nucleic acid in DNA held together?

18.60 What is meant by complementary base pairing?

18.61 Write the base sequence in a complementary DNA segment if each original segment has the following base sequence:

a. A A A A A A b. G G G G G G c. A G T C C A G G T d. C T G T A T A C G T T

18.62 Write the base sequence in a complementary DNA segment if each original segment has the following base sequence:

a. T T T T T T b. C C C C C C C C C c. A T G G C A d. A T A T G C G C T A

18.63 What process ensures that the replication of DNA produces identical copies?

18.64 What is daughter DNA?

NH2

N

N

N

N

H

O

H

NH2

N

N

18.58 Identify each of the following bases as a component of RNA, DNA, or both:

a. guanine b.

ENGAGE 18.16 What is the difference between transcription and translation?

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630 CHAPTER 18 Biochemistry

Transcription begins when the section of DNA to be copied unwinds. One strand of DNA acts as a template for the synthesis of mRNA as bonds are formed to each comple- mentary base: C is paired with G, T pairs with A, and A pairs with U (not T).

CORE CHEMISTRY SKILL Writing the mRNA Segment for a

DNA Template

PRACTICE PROBLEMS Try Practice Problems 18.65 and 18.66

SAMPLE PROBLEM 18.11 RNA Synthesis

TRY IT FIRST

The sequence of bases in a segment of the DNA template strand for mRNA is C G A T C A. What corresponding mRNA is produced?

SOLUTION

To form the mRNA, each base in the DNA template strand is paired with its complemen- tary base: G with C, C with G, T with A, and A with U. DNA template strand: C G A T C A

Transcription

4

4

4

4

4

4

Complementary base sequence in mRNA: G C U A G U

SELF TEST 18.11

Write the DNA base sequence that codes for each of the following mRNA segments:

a. G G G U U U A A A b. G A U U A C A C G

ANSWER

a. C C C A A A T T T b. C T A A T G T G C

The Genetic Code The genetic code consists of a series of three nucleotides (triplets) in mRNA, called codons that specify the amino acids and their sequence in a protein. Early work on protein synthesis showed that repeating triplets of uracil (UUU) produced a polypeptide that contained only phenylalanine. Therefore, a sequence of UUU UUU UUU codes for three phenylalanines.

Codons in mRNA: UUU UUU UUU4

4

4

Amino acid sequence: Phe ¬ Phe ¬ Phe

INTERACTIVE VIDEO

Protein Synthesis

Transcription

Protein

Translation

DNA mRNA

Ala Phe

Asp

Thr

Cys

FIGURE 18.23 The genetic information in DNA is replicated in cell division and used to produce messenger RNAs that code for the amino acids needed for protein synthesis.

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18.8 Protein Synthesis 631

TABLE 18.11 mRNA Codons: The Genetic Code for Amino Acids

First Letter

Second Letter Third LetterU C A G

U

UUU UUC

fPhe (F) UUA UUG

fLeu (L)

UCU UCC UCA UCG

∂Ser (S) UAU UAC

fTyr (Y) UAA STOPb

UAG STOPb

UGU UGC

fCys (C) UGA STOPb

UGG Trp (W)

U

C

A

G

C

CUU CUC CUA CUG

∂Leu (L) CCU CCC CCA CCG

∂Pro (P) CAU CAC

fHis (H) CAA CAG

fGln (Q)

CGU CGC CGA CGG

∂Arg (R) U

C

A

G

A

AUU AUC AUA

¶Ile (I)

AUG STARTa/ Met (M)

ACU ACC ACA ACG

∂Thr (T) AAU AAC

fAsn (N) AAA AAG

fLys (K)

AGU AGC

fSer (S) AGA AGG

fArg (R)

U

C

A

G

G

GUU GUC GUA GUG

∂Val (V) GCU GCC GCA GCG

∂Ala (A) GAU GAC

fAsp (D) GAA GAG

fGlu (E)

GGU GGC GGA GGG

∂ Gly (G) U

C

A

G

STARTa codon signals the initiation of a peptide chain. STOPb codons signal the end of a peptide chain.

Codons have now been determined for all 20 amino acids. A total of 64 codons is possible from the triplet combinations of A, G, C, and U. Three of these, UGA, UAA, and UAG, are stop signals that code for the termination of protein synthesis. All the other codons shown in TABLE 18.11 specify amino acids. Thus, one amino acid can have several codons. For example, glycine has four codons: GGU, GGC, GGA, and GGG. The triplet AUG has two roles in protein synthesis. At the beginning of an mRNA, the codon AUG signals the start of protein synthesis. In the middle of a series of codons, AUG codes for the amino acid methionine.

Protein Synthesis: Translation Once the mRNA is synthesized, it migrates out of the nucleus into the cytoplasm to the ribosomes. At the ribosomes, the translation process converts the codons on mRNA into amino acids to make a protein.

Protein synthesis begins when the mRNA combines with a ribosome. There, tRNA molecules, which carry amino acids, align with mRNA, and a peptide bond forms between the amino acids. After the first tRNA detaches from the ribosome, the ribosome shifts to the next codon on the mRNA. Each time the ribosome shifts and the next tRNA aligns with the mRNA, a peptide bond joins the new amino acid to the growing polypeptide chain. After all the amino acids for a particular protein have been linked together by peptide bonds, the ribosome encounters a stop codon. Because there are no tRNAs to complement the termina- tion codon, protein synthesis ends, and the completed polypeptide chain is released from the ribosome. Then interactions between the amino acids in the chain form the protein into the three-dimensional structure that makes the polypeptide into a biologically active protein (see FIGURE 18.24).

CORE CHEMISTRY SKILL Writing the Amino Acid for an

mRNA Codon

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632 CHAPTER 18 Biochemistry

FIGURE 18.24 In the translation process, the mRNA synthesized by transcription attaches to a ribosome, and tRNAs pick up their amino acids, bind to the appropriate codon, and place them in a growing peptide chain.

CC C C CAA AUUUU UGGG

T T

T

T

T

A

A A

A

G

G

G

C

C

U U

A

AT A

A A U

A A

G G C

C

GG

G

C

CC

C G G

GA A

AU U

T

Nucleus (site of transcription)

DNA

mRNA is made from a DNA template

mRNA leaves the nucleus, attaches to a ribosome, and translation begins

Correct amino acid is attached to each tRNA

tRNA reenters the cytoplasmic pool of free tRNA, ready to be reactivated with a new amino acid

As the ribosome moves along the mRNA, a new amino acid forms a peptide bond to the growing protein chain Incoming tRNA

hydrogen bonds to a complementary mRNA sequence (codon)

mRNA

DNA

Amino acids

Growing polypeptide chain

Nuclear membrane

Cys

Ala Ile Phe

Asp

Thr

Direction of ribosome advance

Portion of mRNA already

translated

Codon

Ribosome

Cytoplasm (site of translation)

Peptide bond

mRNA

M18_TIMB8119_06_SE_C18.indd 632 28/11/18 3:08 PM

Update 633

PRACTICE PROBLEMS Try Practice Problems 18.67 to 18.70

SAMPLE PROBLEM 18.12 Protein Synthesis: Translation

TRY IT FIRST

Use three-letter and one-letter abbreviations to write the amino acid sequence for the peptide from the mRNA sequence of UCA AAA GCC CUU.

SOLUTION

Each of the codons specifies a particular amino acid. Using Table 18.11, we write the codons and the amino acids in the peptide.

ANALYZE THE PROBLEM

Given Need Connect

mRNA UCA AAA GCC CUU amino acid sequence

genetic code

mRNA codons: UCA AAA GCC CUU4

4

4

4

Amino acid sequence: Ser ¬ Lys ¬ Ala ¬ Leu, SKAL

SELF TEST 18.12

Write three-letter and one-letter abbreviations of the amino acid sequence for the peptide from each of the following mRNA sequences:

a. GGG AGC AGU GAG GUU b. UAU GAU UCC ACG CCG

ANSWER

a. Gly–Ser–Ser–Glu–Val, GSSEV b. Tyr–Asp–Ser–Thr–Pro, YDSTP

PRACTICE PROBLEMS

18.8 Protein Synthesis

18.65 Write the corresponding section of mRNA produced from the following section of DNA template:

C C G A A G G T T C A C

18.66 Write the corresponding section of mRNA produced from the following section of DNA template:

T A C G G A A G C T A

18.67 What amino acid is coded for by each of the following RNA codons?

a. CUG b. CCU c. GGU d. AGG

18.68 What amino acid is coded for by each of the following RNA codons?

a. AAG b. GUC c. CGG d. GCA

Applications 18.69 The following is a segment of the DNA template that codes for

human insulin:

TTT GTG AAC CAA CAC CTG

a. Write the corresponding mRNA segment. b. Write the three-letter and one-letter abbreviations for the

corresponding peptide segment.

18.70 The following is a segment of the DNA template that codes for human insulin:

TGC GGC TCA CAC CTG GTG

a. Write the corresponding mRNA segment. b. Write the three-letter and one-letter abbreviations for the

corresponding peptide segment.

UPDATE Kate’s Program for Type 2 Diabetes

At Kate’s next appointment, Paula shows Kate how to use a glucose meter. She instructs Kate to measure her blood glucose level before and after breakfast and dinner each day. Paula explains to Kate that her pre- meal blood glucose level should be 110 mg/dL or less, and if it increases

by more than 50 mg/dL after the meal, she needs to lower the amount of carbohydrates she consumes.

Kate and Paula plan several meals. They combine fruits and vegetables that have high and low levels of carbohydrates in the same meal to stay within the recommended range of about 45 to 60 g of carbohydrate per meal. Kate and Paula also discuss the fact that complex carbohydrates in the body take longer to break down into glucose and, therefore, raise the blood sugar level more gradually.

M18_TIMB8119_06_SE_C18.indd 633 28/11/18 3:08 PM

634 CHAPTER 18 Biochemistry

CONCEPT MAP

Carbohydrates

Monosaccharides: Glucose,

Galactose, Fructose

Lipids Proteins Nucleic Acids

BIOCHEMISTRY

and in

undergo

consist of consist of

Disaccharides: Maltose, Lactose, Sucrose

Polysaccharides: Amylose,

Amylopectin, Glycogen, Cellulose

in

Hydrogenation

undergoundergo

found in

with glycerol

Fatty Acids

Triacylglycerols

Saponification

Saturated Unsaturated

Fats Oils

areare

Sterols

of the

Steroid Nucleus

Cholesterol

such as

in a specific order as

Amino Acids

Primary Structure

Secondary Structure

contain peptide bonds between

and hydrogen bond as

Tertiary and Quaternary Structures

with attractions that give

contains

RNA

Nucleotides A,U,G,C

contains

areconsist of

found in

are

arranged as a

with

DNA

Double Helix

Deoxynucleotides A,T,G,C

Complementary Base Pairs

AT, GC

Protein Synthesis

mRNA

with three types

triplets of bases

are

place the

proper

rRNA

and

tRNA

and

Codons

uses

involved in

Protein

produce a produce

a

Amino Acids

specific for

After her appointment, Kate increases her exercise to walking 30 minutes twice a day. She begins to change her diet by eating six small meals a day consisting mostly of small amounts of fruits, vegetables without starch such as green beans and broccoli, small servings of whole grains, and chicken or fish. Kate also decreases the amounts of breads and pasta she eats because she knows carbohydrates will raise her blood sugar.

After three months, Kate reports that she lost 10 lb and that her blood glucose dropped to 146 mg/dL. Her blurry vision improved, and her need to urinate decreased.

Applications

18.71 Kate’s blood volume is 3.9 L. Before treatment, if her blood glucose was 178 mg/dL, how many grams of glucose were in her blood?

18.72 Kate’s blood volume is 3.9 L. After three months of diet and exercise, if her blood glucose is 146 mg/dL, how many grams of glucose are in her blood?

18.73 For breakfast, Kate had 1 cup of orange juice (23 g carbohydrate), 2 slices of wheat toast (24 g carbohydrate), 2 tablespoons of grape jam (26 g carbohydrate), and coffee with sugar substitute (0 g carbohydrate).

a. Has Kate remained within the limit of 45 to 60 g of carbohydrate?

b. Using the energy value of 4 kcal/g for carbohydrate, calculate the total kilocalories from carbohydrates in Kate’s breakfast, rounded to the tens place.

18.74 The next day, Kate had 1 cup of cereal (15 g carbohydrate) with skim milk (7 g carbohydrate), 1 banana (17 g carbohydrate), and 1/2 cup of orange juice (12 g carbohydrate) for breakfast.

a. Has Kate remained within the limit of 45 to 60 g of carbohydrate?

b. Using the energy value of 4 kcal/g for carbohydrate, calculate the total kilocalories from carbohydrates in Kate’s breakfast, rounded to the tens place.

M18_TIMB8119_06_SE_C18.indd 634 28/11/18 3:08 PM

Chapter Review 635

CHAPTER REVIEW

18.1 Carbohydrates LEARNING GOAL Classify a carbohydrate as an aldose or a ketose; draw the open-chain and Haworth structures for monosaccharides. • Carbohydrates are composed of carbon,

hydrogen, and oxygen. • Monosaccharides are polyhydroxy aldehydes

(aldoses) or ketones (ketoses). • Monosaccharides are also classified by their

number of carbon atoms: triose, tetrose, pen- tose, or hexose. Important monosaccharides are glucose, galactose, and fructose.

• The predominant form of monosaccharides is the cyclic form of five or six atoms.

• The Haworth structure forms when an ¬ OH group (usually the one on carbon 5 in hexoses) reacts with the carbonyl group of the same molecule.

18.2 Disaccharides and Polysaccharides LEARNING GOAL Describe the monosaccharide units and linkages in disaccharides and polysaccharides. • Disaccharides are two monosaccharide

units joined together by a glycosidic bond. • In the most common disaccharides malt-

ose, lactose, and sucrose, there is at least one glucose unit.

• Polysaccharides are polymers of monosaccharide units. Amylose is an unbranched polymer of glucose, and amylopectin is a branched polymer of glucose. Glycogen, the storage form of glucose in ani- mals, is similar to amylopectin with more branching. Cellulose is also a polymer of glucose, but in cellulose the glycosidic bonds are b bonds rather than a bonds.

18.3 Lipids LEARNING GOAL Draw the condensed structural or line-angle formula for a fatty acid, a triacylglycerol, and the products of hydrogenation or saponification. Identify the steroid nucleus. • Lipids are nonpolar compounds that are not

soluble in water. • Classes of lipids include waxes, fats, oils, and steroids. • Fatty acids are long-chain carboxylic acids that may be saturated or

unsaturated. • Triacylglycerols are esters of glycerol with three fatty acids. • Fats contain more saturated fatty acids and have higher melting

points than most vegetable oils. • The hydrogenation of unsaturated fatty acids converts double

bonds to single bonds. • In saponification, a fat heated with a strong base produces glycerol

and the salts of the fatty acids (soap). • Steroids are lipids containing the steroid nucleus, which is a fused

structure of four rings.

18.4 Amino Acids and Proteins LEARNING GOAL Describe protein functions and draw structures for amino acids and peptides. • Some proteins are enzymes or hormones, whereas others are

important in structure, transport, protection, storage, and muscle contraction.

C

C

C

C

C

OH

HO H

H OH

H OH

H OH

CH2OH

• A group of 20 amino acids provides the molecular building blocks of proteins.

• Attached to the central (alpha) carbon of each amino acid are an ammonium group, a carboxylate group, a hydrogen atom, and a unique R group.

• Peptides form when an amide bond links the carboxylate group of one amino acid and the ammonium group of a second amino acid.

18.5 Protein Structure LEARNING GOAL Identify the levels of structure of a protein. • Long chains of amino acids that are

biologically active are called proteins. • The primary structure of a protein is its

sequence of amino acids joined by peptide bonds.

• In the secondary structure, hydrogen bonds between atoms in the peptide bonds produce a characteristic shape such as an a helix or a b-pleated sheet.

• The most abundant protein in the body is collagen, which is composed of fibrils of triple helices that are hydrogen-bonded.

• A tertiary structure is stabilized by interactions between R groups of amino acids in one region of the polypeptide chain with R groups in different regions of the protein.

• In a quaternary structure, two or more subunits combine for biological activity.

18.6 Proteins as Enzymes LEARNING GOAL Describe the role of an enzyme in an enzyme-catalyzed reaction. • Enzymes are proteins that act as biologi-

cal catalysts by accelerating the rate of cellular reactions.

• Within the tertiary structure of an enzyme, a small pocket called the active site binds the substrate.

• In the lock-and-key model, an early theory of enzyme action, a substrate precisely fits the shape of the active site.

• In the induced-fit model, both the active site and the substrate undergo changes in their shapes to give the best fit for efficient catalysis. In the enzyme–substrate complex, catalysis takes place when the amino acid R groups in the active site of an enzyme react with a substrate.

• When the products of catalysis are released, the enzyme can bind another substrate molecule.

18.7 Nucleic Acids LEARNING GOAL Describe the structure of the nucleic acids in DNA and RNA. • Nucleic acids, such as deoxyribonu-

cleic acid (DNA) and ribonucleic acid (RNA), are polymers of nucleotides.

• A nucleotide is composed of three parts: a base, a pentose sugar, and a phosphate group.

• In DNA, the sugar is deoxyribose, and the base can be adenine, thymine,

Enzyme–Substrate Complex (ES)

O

OH OH OO

C

T A

A

G

A

CG

CG

T

A

Sugar–phosphate backbone

T

A

GC

CG

5¿ 3¿

M18_TIMB8119_06_SE_C18.indd 635 28/11/18 3:08 PM

636 CHAPTER 18 Biochemistry

UCA AAA GCC CUU

Ser Lys Ala Leu

• The bases in the mRNA are complementary to the bases in DNA, except A in DNA is paired with U in RNA.

• The genetic code consists of a series of codons, which are sequences of three bases that specify the amino acids in a protein.

• Proteins are synthesized at the ribosomes. • During translation, tRNAs bring the appropriate amino acids to the

mRNA at the ribosome and peptide bonds form. • When the polypeptide is released, it takes on its secondary and

tertiary structures and becomes a functional protein in the cell.

active site A pocket in the tertiary enzyme structure that binds substrate and catalyzes a reaction.

amino acid The building block of proteins, consisting of a hydrogen atom, an ammonium group, a carboxylate group, and a unique R group attached to the alpha carbon.

amylopectin A branched-chain polymer of starch composed of glucose units joined by a(1S 4)@ and a(1S 6)@glycosidic bonds.

amylose An unbranched polymer of starch composed of glucose units joined by a(1S 4)@glycosidic bonds.

carbohydrate A simple or complex sugar composed of carbon, hydrogen, and oxygen.

cellulose An unbranched polysaccharide composed of glucose units linked by b(1S 4)@glycosidic bonds that cannot be hydrolyzed by the human digestive system.

codon A sequence of three bases in mRNA that specifies a certain amino acid to be placed in a protein. A few codons signal the start or stop of protein synthesis.

complementary base pairs In DNA, adenine is always paired with thymine (A and T or T and A), and guanine is always paired with cytosine (G and C or C and G). In forming RNA, adenine is paired with uracil (A and U).

disaccharide A carbohydrate composed of two monosaccharides joined by a glycosidic bond.

DNA Deoxyribonucleic acid; the genetic material of all cells contain- ing nucleotides with deoxyribose, phosphate, and the four bases: adenine, thymine, guanine, and cytosine.

double helix The helical shape of the double chain of DNA that is like a spiral staircase with a sugar–phosphate backbone on the outside and base pairs like stair steps on the inside.

enzyme A protein that catalyzes a biological reaction. fat A triacylglycerol that is solid at room temperature and usually

comes from animal sources. fatty acid A long-chain carboxylic acid found in many lipids. fructose A ketohexose which is combined with glucose in sucrose. galactose An aldohexose that is combined with glucose in lactose. genetic code The sequence of codons in mRNA that specifies the

amino acid order for the synthesis of protein. glucose An aldohexose found in fruits, vegetables, corn syrup, and

honey that is also known as blood sugar and dextrose. The most prevalent monosaccharide in the diet. Most polysaccharides are polymers of glucose.

glycogen A polysaccharide formed in the liver and muscles for the storage of glucose as an energy reserve. It is composed of glucose in a highly branched polymer joined by a(1S 4)@ and a(1S 6)-glycosidic bonds.

Haworth structure The ring structure of a monosaccharide.

induced-fit model A model of enzyme action in which the shape of a substrate and the active site of the enzyme adjust to give an optimal fit.

lactose A disaccharide consisting of glucose and galactose found in milk and milk products.

lipids A family of compounds that is nonpolar in nature and not soluble in water; includes fats, oils, waxes, and steroids.

maltose A disaccharide consisting of two glucose units; it is obtained from the hydrolysis of starch.

monosaccharide A polyhydroxy compound that contains an alde- hyde or a ketone group.

nucleic acid A large molecule composed of nucleotides; found as a double helix in DNA and as the single strands of RNA.

nucleoside The combination of a pentose sugar and a base. nucleotide Building block of a nucleic acid consisting of a base, a

pentose sugar (ribose or deoxyribose), and a phosphate group. oil A triacylglycerol that is usually a liquid at room temperature and

is obtained from a plant source. polysaccharide A polymer of many monosaccharide units, usually

glucose. Polysaccharides differ in the types of glycosidic bonds and the amount of branching in the polymer.

primary structure The specific sequence of the amino acids in a protein.

protein A term used for biologically active polypeptides that have many amino acids linked together by peptide bonds.

quaternary structure A protein structure in which two or more protein subunits form an active protein.

replication The process of duplicating DNA by pairing the bases on each parent strand with their complementary bases.

RNA Ribonucleic acid; a type of nucleic acid that is a single strand of nucleotides containing ribose, phosphate, and the four bases: adenine, cytosine, guanine, and uracil.

secondary structure The formation of an a helix or b-pleated sheet by hydrogen bonds.

steroids Types of lipid composed of a multicyclic ring system. sucrose A disaccharide composed of glucose and fructose;

commonly called table sugar or “sugar.” tertiary structure The folding of the secondary structure of a protein

into a compact structure that is stabilized by the interactions of R groups such as salt bridges and disulfide bonds.

transcription The transfer of genetic information from DNA by the formation of mRNA.

translation The interpretation of the codons in mRNA as amino acids in a peptide.

triacylglycerols A family of lipids composed of three fatty acids bonded through ester bonds to glycerol, a trihydroxy alcohol.

KEY TERMS

guanine, or cytosine. In RNA, the sugar is ribose and uracil replaces thymine.

• Each nucleic acid has its own unique sequence of bases. • A DNA molecule consists of two strands of nucleotides that are

wound around each other like a spiral staircase. • The two strands are held together by hydrogen bonds between

complementary base pairs AT and GC.

18.8 Protein Synthesis LEARNING GOAL Describe the synthesis of protein from mRNA.

M18_TIMB8119_06_SE_C18.indd 636 28/11/18 3:08 PM

Core Chemistry Skills 637

The chapter section containing each Core Chemistry Skill is shown in parentheses at the end of each heading.

Drawing Haworth Structures (18.1) • The Haworth structure shows the ring structure of a

monosaccharide. • Groups on the right side of the open-chain structure of the mono-

saccharide are below the plane of the ring, those on the left are above the plane.

• The new ¬ OH group is the a form if it is below the plane of the ring, or the b form if it is above the plane.

Example: Draw the Haworth structure for b-d-idose.

CORE CHEMISTRY SKILLS

H

H H OH

H

H

OH

OH

HO

CH2OH

O

OH

O

H

H

OH

OH

OH

HOH C

C

C

H

CH2OH

C

C

OH

d-Idose

Answer: In the b-form, the new ¬ OH group is above the plane of the ring.

Identifying Fatty Acids (18.3) • Fatty acids are unbranched carboxylic acids that typically contain

an even number (12 to 20) of carbon atoms. • Fatty acids may be saturated, monounsaturated with one double

bond, or polyunsaturated with two or more double bonds.

Example: State the number of carbon atoms, saturated or unsaturated, and name of the following:

Answer: 16 carbon atoms, saturated, palmitic acid

Drawing Structures for Triacylglycerols (18.3) • Triacylglycerols are esters of glycerol with three long-chain fatty

acids.

Example: Draw the line-angle formula of and name the triacylglycerol formed from glycerol and palmitic acid (16:0).

Answer:

CH2

O

CH

CH2

O

Glyceryl tripalmitate (tripalmitin)

O

O O

O

H3N C

H

CH2

SH

O

C O- +

Drawing the Structure for an Amino Acid at Physiological pH (18.4) • The central a carbon of each amino acid is bonded to an

ammonium group ( ¬ NH3 +), a carboxylate group ( ¬ COO-), a hydrogen atom, and a unique R group.

• The R group gives an amino acid the property of being nonpolar, polar, acidic, or basic.

Example: Draw the condensed structural formula for cysteine at physiological pH.

Answer:

Identifying the Primary, Secondary, Tertiary, and Quaternary Structures of Proteins (18.5) • The primary structure of a protein is the sequence of amino acids

joined by peptide bonds. • In the secondary structures of proteins, hydrogen bonds between

atoms in the peptide bonds produce an alpha helix or a beta-pleated sheet.

• The tertiary structure of a protein is stabilized by R groups that have hydrophobic interactions, attractions between groups and water, hydrogen bonds, disulfide bonds, and salt bridges.

• In a quaternary structure, two or more tertiary subunits are combined for biological activity, held by the same interactions found in tertiary structures.

Example: Identify the following as characteristic of the primary, secondary, tertiary, or quaternary structure of a protein:

a. The R groups of two amino acids interact to form a salt bridge.

b. Eight amino acids form peptide bonds. c. A polypeptide forms an alpha helix. d. Two amino acids with hydrophobic R groups interact. e. A protein with biological activity contains four

polypeptide subunits.

Answer: a. tertiary, quaternary b. primary c. secondary d. tertiary, quaternary e. quaternary

M18_TIMB8119_06_SE_C18.indd 637 28/11/18 3:08 PM

638 CHAPTER 18 Biochemistry

Writing the Complementary DNA Strand (18.7) • During DNA replication, new DNA strands are made along each of

the original DNA strands. • The new strand of DNA is made by forming hydrogen bonds with

the bases in the template strand: A with T and T with A; C with G and G with C.

Example: Write the complementary base sequence for the following DNA segment: A T T C G G T A C

Answer: T A A G C C A T G

Writing the mRNA Segment for a DNA Template (18.8) • Transcription is the process that produces mRNA from one strand

of DNA. • The bases in the mRNA are complementary to the DNA except that

A in DNA is paired with U in RNA.

Example: Write the corresponding section of mRNA produced from the following section of DNA template: C G C A T G T C A

Answer: G C G U A C A G U

Writing the Amino Acid for an mRNA Codon (18.8) • The genetic code consists of a sequence of three bases (codons)

that specifies the order for the amino acids in a protein. • The codon AUG signals the start of transcription and codons UAG,

UGA, and UAA signal it to stop.

Example: Use three-letter and one-letter abbreviations to write the amino acid sequence for the peptide from the mRNA sequence CCG UAU GGG.

Answer: Pro9Tyr9Gly, PYG

UNDERSTANDING THE CONCEPTS

The chapter sections to review are shown in parentheses at the end of each problem.

18.75 Melezitose, a carbohydrate secreted by insects, has the following Haworth structure: (18.1, 18.2)

HO

HO

O H

H

H

OH

OH O

CH2OH

HOCH2

H

O

H

OH

H

H

O

H

O H

H

H

OH

OH

CH2OH

H

H

CH2OH

Melezitose

a. Is melezitose a mono-, di-, tri-, or polysaccharide? b. What ketohexose and aldohexose are present in melezitose?

18.76 What are the disaccharides and polysaccharides present in each of the following? (18.1, 18.2)

(a) (b)

(c)

18.77 One of the triacylglycerols in palm oil is glyceryl tripalmitate. Draw the condensed structural formula for glyceryl tripalmitate. (18.3)

The fruit from palm trees are a source of palm oil.

18.78 Jojoba wax in candles consists of a 20-carbon saturated fatty acid and a 20-carbon saturated alcohol. Draw the condensed structural formula for jojoba wax. (18.3)

Candles contain jojoba wax.

18.79 Identify each of the following as a saturated, monounsaturated, or polyunsaturated fatty acid: (18.3)

a. CH3 ¬ (CH2)4 ¬ (CH “ CH ¬ CH2)2 ¬ (CH2)6 ¬ COOH b. lauric acid

Salmon is a good source of unsaturated fatty acids.

(d)

18.80 Identify each of the following as a saturated, monounsaturated, or polyunsaturated fatty acid: (18.3)

a. CH3 ¬ (CH2)14 ¬ COOH b. CH3 ¬ (CH2)7 ¬ CH “ CH ¬ (CH2)7 ¬ COOH

M18_TIMB8119_06_SE_C18.indd 638 28/11/18 3:08 PM

Additional Practice Problems 639

18.81 Seeds and vegetables are often deficient in one or more essential amino acids. The table below shows which essential amino acids are present in each food. (18.4)

Source Lysine Tryptophan Methionine

Oatmeal No Yes Yes

Rice No Yes Yes

Garbanzo beans Yes No Yes

Lima beans Yes No No

Cornmeal No No Yes

Use the table to decide if each food combination provides the essential amino acids lysine, tryptophan, and methionine.

a. rice and garbanzo beans b. lima beans and cornmeal c. garbanzo beans and lima beans

18.82 Seeds and vegetables are often deficient in one or more essential amino acids. Using the table in problem 18.81, state whether each combination provides the essential amino acids lysine, tryptophan, and methionine. (18.4)

b. (CH2)4H3NCH2 C O - and

O +

18.84 Identify the amino acids and type of interaction that occurs between the following R groups in tertiary protein structures: (18.4, 18.5)

b. CH3CH2 CH3

CH3

CH and

CH2CH2 HSSH and a.

CH2HOCH2 C NH2 and

O

a.

18.85 Answer the following questions for the given section of DNA: (18.7, 18.8)

a. Complete the bases in the parent and new strands.

A

C GG C C

T G T

New strand:

Parent strand:

b. Using the new strand as a template, write the mRNA sequence.

c. Write the 3-letter abbreviations for the amino acids that would go into the peptide from the mRNA you wrote in part b.

18.86 Answer the following questions for the given section of DNA: (18.7, 18.8)

a. Complete the bases in the parent and new strands.

A

C G C G

G T C T

New strand:

Parent strand:

b. Using the new strand as a template, write the mRNA sequence.

c. Write the 3-letter abbreviations for the amino acids that would go into the peptide from the mRNA you wrote in part b.

Oatmeal is deficient in the essential amino acid lysine.

a. rice and lima beans b. rice and oatmeal c. oatmeal and lima beans

18.83 Identify the amino acids and type of interaction that occurs between the following R groups in tertiary protein structures: (18.4, 18.5)

ADDITIONAL PRACTICE PROBLEMS

18.87 What are the structural differences in d-glucose and d-galactose? (18.1)

18.88 What are the structural differences in d-glucose and d-fructose? (18.1)

18.89 Draw the Haworth structure for a@d- and b@d-gulose, whose open-chain structure is shown at right. (18.1)

d-Gulose

C

C OHH

H

OH

HO

C OHH

C

CH

OH

CH2OH

M18_TIMB8119_06_SE_C18.indd 639 28/11/18 3:08 PM

640 CHAPTER 18 Biochemistry

The following problems are related to the topics in this chapter. However, they do not all follow the chapter order, and they require you to combine concepts and skills from several sections. These problems will help you increase your critical thinking skills and prepare for your next exam.

18.101 Raffinose is a trisaccharide found in green vegetables such as cabbage, asparagus, and broccoli. It is composed of three different monosaccharides. Identify the monosaccharides in raffinose. (18.1, 18.2)

CHALLENGE PROBLEMS

HO O

H

H

H

OH

OH

CH2OH

HO

O H H

H

H

O

H

OH

OH OH

H

H

CH2OH

O

O H

H

CH2OH

H

OH

CH2

Raffinose

18.95 Identify the base and sugar in each of the following nucleosides: (18.7)

a. deoxythymidine b. adenosine c. cytidine d. deoxyguanosine

18.96 Identify the base and sugar in each of the following nucleotides: (18.7)

a. CMP b. dAMP c. dGMP d. UMP

18.97 Write the complementary base sequence for each of the following parent DNA segments: (18.7)

a. G A C T T A G G C b. T G C A A A C T A G C T c. A T C G A T C G A T C G

18.98 Write the complementary base sequence for each of the following parent DNA segments: (18.7)

a. T T A C G G A C C G C b. A T A G C C C T T A C T G G c. G G C C T A C C T T A A C G

18.99 Match the following statements with rRNA, mRNA, or tRNA: (18.8)

a. carries genetic information from the nucleus to the ribosomes

b. combines with proteins to form ribosomes

18.100 Match the following statements with rRNA, mRNA, or tRNA: (18.8)

a. acts as a template for protein synthesis b. brings amino acids to the ribosomes for protein synthesis

18.90 From the compounds shown, select those that match the following: (18.1)

b. an aldopentose a. an aldohexose c. a ketohexose

1 2

C

H

HO

C OH

H

H C OH

CH2OH

C

HO

HO

C H

C H

OH

H C OH

H

CH2OH

C

OH C

OH

3

C O

C

HO

HO

C H

C H

OHH

CH2OH

CH2OH

18.91 Gentiobiose, a carbohydrate found in saffron, contains two glucose molecules linked by a b(1S 6)-glycosidic bond. Draw the Haworth structure for a-gentiobiose. (18.1, 18.2)

18.92 b-Cellobiose is a disaccharide obtained from the hydrolysis of cellulose. It contains two glucose molecules linked by a b(1S 4)@glycosidic bond. Draw the Haworth structure for b-cellobiose. (18.1, 18.2)

18.93 Draw the condensed structural formula for Ser–Lys–Asp at physiological pH. (18.4)

18.94 Draw the condensed structural formula for Val–Ala–Leu at physiological pH. (18.4)

18.102 The disaccharide trehalose found in mushrooms is composed of two a-d-glucose molecules joined by an a(1S 1)-glycosidic bond. Draw the Haworth structure for trehalose. (18.1, 18.2)

18.103 Sunflower seed oil can be used to make margarine. A triacyl- glycerol in sunflower seed oil contains two linoleic acids and one oleic acid. (18.3)

Sunflower oil is obtained from the seeds of the sunflower.

a. Draw the condensed structural formulas for two isomers of the triacylglycerol in sunflower oil.

b. Using one of the isomers, write the reaction that takes place when sunflower seed oil is used to make solid margarine.

18.104 What type of interaction would you expect between the following amino acids in a tertiary structure? (18.4, 18.5)

a. threonine and glutamine b. valine and alanine c. arginine and glutamate

18.105 What are some differences between each of the following pairs? (18.4, 18.5)

a. secondary and tertiary protein structures b. essential and nonessential amino acids c. polar and nonpolar amino acids

18.106 What are some differences between each of the following pairs? (18.4, 18.5)

a. a salt bridge and a disulfide bond b. an a helix and a b-pleated sheet c. tertiary and quaternary structures of proteins

M18_TIMB8119_06_SE_C18.indd 640 28/11/18 3:09 PM

Answers to Engage Questions 641

Applications 18.107 Endorphins are polypeptides that reduce pain. Use three-

letter abbreviations to write the amino acid sequence for the endorphin leucine enkephalin (leu-enkephalin), which has the following mRNA: (18.4, 18.5, 18.8)

AUG UAC GGU GGA UUU CUA UAA

18.108 Endorphins are polypeptides that reduce pain. Use three- letter abbreviations to write the amino acid sequence for the endorphin methionine enkephalin (met-enkephalin), which has the following mRNA: (18.4, 18.5, 18.8)

AUG UAC GGU GGA UUU AUG UAA

18.109 Aspartame, which is used in artificial sweeteners, contains the following dipeptide: (18.4, 18.5)

18.110 The tripeptide shown has a strong attraction for Cu2+ and is present in human blood and saliva. (18.4, 18.5)

C

H H HH H

H3N +

H CH2

N N

H

N H

C C

O

O-C

OCH2

CH2

CH2

CH2

NH3

CC

O

+

N +

ANSWERS TO ENGAGE QUESTIONS 18.9 Stearic acid has only carbon–carbon single bonds; it is

saturated. Oleic acid has one carbon–carbon double bond; it is monounsaturated.

18.10 Butter contains over 60% saturated fatty acids and is a solid at room temperature. Canola oil contains only about 5% saturated fatty acids and is a liquid at room temperature.

18.11 Valine has a hydrocarbon R group: it is a nonpolar amino acid. Histidine has a charged R group: it is a polar amino acid.

18.12 Valine is at the N-terminus because it is written first. Threo- nine is at the C-terminus because it is written last.

18.13 Arginine is a basic amino acid, while aspartate is an acidic amino acid. The ionic interaction between acidic and basic amino acids is called a salt bridge.

18.14 The tertiary structure is the compact shape of a protein. The quaternary structure is the combination of protein subunits.

18.15 The nucleotides in DNA contain the sugar deoxyribose and the bases adenine, guanine, cytosine, and thymine. The nucleotides in RNA contain the sugar ribose and the bases adenine, guanine, cytosine, and uracil.

18.16 In transcription, genetic information is transferred to mRNA. In translation, the genetic information is used to form a protein.

18.1 The reactants of respiration are glucose and oxygen; the prod- ucts are carbon dioxide, water, and energy.

18.2 Ribulose has five carbons and contains a carbonyl group bonded to two other carbons. Ribulose is a ketopentose. Glu- cose has six carbons and contains a carbonyl group bonded to one carbon and one hydrogen. Glucose is an aldohexose.

18.3 The open-chain structure of d-galactose has the hydroxyl group on carbon 4 on the left. The open-chain structure of d-glucose has the hydroxyl group on carbon 4 on the right.

18.4 In a-maltose, the ¬ OH group on carbon 1 is below the plane of the ring. In b-maltose, the ¬ OH group on carbon 1 is above the plane of the ring.

18.5 In lactose, the galactose and glucose are linked by a b(1 S 4) glycosidic bond.

18.6 Sucrose is composed of glucose and fructose.

18.7 In amylose and amylopectin, the glucose molecules are linked by a(1 S 4) glycosidic bonds. In amylopectin, there are also branches of glucose molecules linked by a(1S 6) glycosidic bonds.

18.8 Humans do not have the enzymes needed to break the b(1 S 4) glycosidic bonds in cellulose.

a. What are the amino acids in the dipeptide? b. What is the name of the dipeptide in aspartame? c. Give the three-letter and one-letter abbreviations for the

dipeptide in aspartame.

a. What are the amino acids in the tripeptide? b. What is the name of this tripeptide? c. Give the three-letter and one-letter abbreviations for the

tripeptide.

18.111 The following sequence is a portion of a DNA template: (18.4, 18.5, 18.8)

GCT TTT CAA AAA a. Write the corresponding mRNA segment. b. Write the three-letter and one-letter abbreviations for the

corresponding peptide segment.

18.112 The following sequence is a portion of a DNA template: (18.4, 18.5, 18.8)

TGT GGG GTT ATT a. Write the corresponding mRNA segment. b. Write the three-letter and one-letter abbreviations for the

corresponding peptide segment.

N C

O-

O-

O

H3N +

C

CH2 CH2

C

O

C

O

C

HH H

Some artificial sweeteners contain aspartame, which is an ester of a dipeptide.

M18_TIMB8119_06_SE_C18.indd 641 28/11/18 3:09 PM

642 CHAPTER 18 Biochemistry

ANSWERS TO SELECTED PROBLEMS 18.29 18.1 Hydroxyl groups are found in all monosaccharides along with

a carbonyl on the first or second carbon.

18.3 A ketopentose contains hydroxyl and ketone functional groups and has five carbon atoms.

18.5 In the ring portion of the Haworth structure of glucose, there are five carbon atoms and an oxygen atom.

18.7 a. a form b. a form

18.9 a. ketohexose b. aldopentose

18.11 In galactose, the ¬ OH group on carbon 4 extends to the left. In glucose, this ¬ OH group extends to the right.

18.13 a. one molecule of glucose and one molecule of galactose; b(1S 4)-glycosidic bond; b-lactose

b. two molecules of glucose; a(1S 4)-glycosidic bond; a-maltose

18.15 a. disaccharide b. two molecules of glucose c. a(1S 6)-glycosidic bond d. a 18.17 a. sucrose b. lactose c. maltose d. lactose

18.19 a. cellulose b. amylose, amylopectin c. amylose d. glycogen

18.21 a. saturated b. polyunsaturated c. saturated

18.23

CH2

O

CH

CH2

O

O

O

O

O

18.25 Safflower oil contains polyunsaturated fatty acids; olive oil contains a large amount of monounsaturated fatty acids. Molecules with more unsaturated fatty acids have lower melting points.

O

O

O

O

OCH

CH2 O

CH2 O C (CH2)12 CH3

C (CH2)12 CH3

C (CH2)12 CH3

CH3CH OH

OHCH2

OHCH2

C-O (CH2)12

3NaOH+

3Na++

Heat

18.31

18.33 a. +

H O

H

H3N C C O - b.

+

H

H3N C

CH

C O-

O

CH3 OH

c. +

H

H3N C

CH2 O

C O-

18.35 a. nonpolar, hydrophobic b. polar, neutral, hydrophilic c. nonpolar, hydrophobic

18.37 a. alanine b. valine c. lysine d. cysteine

CH2 O C (CH2)7 (CH2)5 CH3CH CH

O

CH2 O C (CH2)7 (CH2)5 CH3CH CH

O CH O C (CH2)7 (CH2)5 CH3 3H2+CH CH

O

OCH Ni

CH2 O

CH2 O C (CH2)14 CH3

C (CH2)14 CH3

C (CH2)14 CH3

O

O

O

18.27

M18_TIMB8119_06_SE_C18.indd 642 28/11/18 3:09 PM

Answers to Selected Problems 643

18.49 a. disulfide bond b. salt bridge c. hydrogen bond d. hydrophobic interaction

18.51 a. tertiary and quaternary b. primary c. secondary

18.53 a. (1) enzyme b. (3) substrate c. (2) enzyme–substrate complex

18.55 a. E + S vh ES h EP h E + P b. The active site is a region or pocket within the tertiary

structure of an enzyme that accepts the substrate, aligns the substrate for reaction, and catalyzes the reaction.

18.57 a. DNA b. both DNA and RNA

18.59 The two DNA strands are held together by hydrogen bonds between the complementary bases in each strand.

18.61 a. T T T T T T b. C C C C C C c. T C A G G T C C A d. G A C A T A T G C A A

18.63 The DNA strands separate to allow each of the bases to pair with its complementary base, which produces two exact copies of the original DNA.

18.65 G G C U U C C A A G U G

18.67 a. leucine (Leu) b. proline (Pro) c. glycine (Gly) d. arginine (Arg)

18.69 a. AAA CAC UUG GUU GUG GAC b. Lys–His–Leu–Val–Val–Asp, KHLVVD

18.71 6.9 g of glucose

18.73 a. Kate’s breakfast had 73 g of carbohydrate. She still needs to cut down the amount of carbohydrate.

b. 290 kcal

18.75 a. Melezitose is a trisaccharide. b. Melezitose contains two glucose molecules and a fructose

molecule.

+ H3N

HH

N C

SH

CH2

H

C C

O

C

OCH3

O-

Ala–Cys, AC

18.39 a.

b.

CH2 O

C

H

N O-

CH2

C

H H

OH

C

O

C +

H3N

Ser–Phe, SF

c. O- +

H3N

CH3

CH3 CH3

O

H H H

H

H H

N C C

O

N C C

OCH

CC

Gly–Ala–Val, GAV

18.41 a. Corn contains methionine which is lacking in peas, while peas contain lysine and tryptophan, lacking in corn.

b. Rice contains methionine which is lacking in soy, while soy contain lysine, lacking in rice.

18.43 a.

OH

CH2CHCH3CH2

CH2

CH2

CH2

CH3

NH

C

NH2

NH2

+ H3N

+

C

H HH H H

C N C N CC

OO

O-C

O

RIY

b.

+ H3N C C N C N C CC

O

NC

OO

O-C

OCH

H H HHHH H

CH2

OHCH3 CH3

H

N

CH2 CHCH3

CH2

CH3

VWIS

18.45 Val–Ser–Ser, VSS; Ser–Val–Ser, SVS; or Ser–Ser–Val, SSV

18.47 In the a helix, hydrogen bonds form between the oxygen atom in the C “ O group and hydrogen in the N ¬ H group in the next turn of the chain. In the b-pleated sheet, side-by-side hydrogen bonds occur between parallel peptides or across sections of a long polypeptide chain.

18.79 a. polyunsaturated b. saturated

18.81 a. yes b. no c. no

18.83 a. asparagine and serine; hydrogen bond b. aspartate and lysine; salt bridge

G A C G T

TC A GG C CC ANew strand:

T C G GParent strand:

18.85 a.

b.

G A C G UU C G G

c. Asp–Pro–Trp

18.87 They differ only at carbon 4 where the ¬ OH group in glucose is on the right side and in galactose it is on the left side.

OCH

CH2 O

CH2 O C (CH2)14 CH3

C (CH2)14 CH3

C (CH2)14 CH3

O

O

O

Glyceryl tripalmitate

18.77

M18_TIMB8119_06_SE_C18.indd 643 28/11/18 3:09 PM

644 CHAPTER 18 Biochemistry

18.95 a. thymine and deoxyribose b. adenine and ribose c. cytosine and ribose d. guanine and deoxyribose

18.97 a. C T G A A T G G C b. A C G T T T G A T G A c. T A G C T A G C T A G C

18.99 a. mRNA b. rRNA

18.101 galactose, glucose, and fructose

18.89

a-D-Gulose b-D-Gulose

OH

OHH H

H H H H H H

H

H

OH

HO

CH2OH

O

OH

OH

OH

HO

CH2OH

O

18.91

O

CH2

OH

HHO

OH

H

H

OH H

OH H HO

H

H

CH2OH

O

OH

H

H

H O

18.93 C

OH

H

O

C +

+

H3N N

H

O

C N

CH2

C

CH2

CH2

C

CH2

CH2

NH3

CH2

O-

O-O

C C

O

H H H

18.103 a. O

CH2 O C (CH2)7 (CH2)4CH2 CH3CH CH CH CH

O

CH2 O C (CH2)7 CH2 (CH2)4 CH3CH CH CH CH

O

CH O C (CH2)7 (CH2)7 CH3CH CH

b.

5H2 Ni

+

CH2 O C (CH2)16 CH3

O

CH2 O C (CH2)16 CH3

O CH O C (CH2)16 CH3

O

O

CH2 O C (CH2)7 (CH2)4CH2 CH3CH CH CH CH

O

CH2 O C (CH2)7 CH2 (CH2)4 CH3CH CH CH CH

O

CH O C (CH2)7 (CH2)7 CH3CH CH

18.105 a. The secondary structure of a protein depends on hydrogen bonds to form an a a helix or a b-pleated sheet. The tertiary structure is determined by the interactions of R groups such as disulfide bonds, hydrogen bonds, and salt bridges.

b. Nonessential amino acids are synthesized by the body, but essential amino acids must be supplied by the diet.

c. Polar amino acids have hydrophilic R groups; nonpolar amino acids have hydrophobic R groups.

18.107 START - Tyr - Gly - Gly - Phe - Leu - STOP 18.109 a. aspartate and phenylalanine b. aspartylphenylalanine c. Asp–Phe, DF

18.111 a. CGA AAA GUU UUU b. Arg–Lys–Val–Phe, RKVF

O

CH2 O C (CH2)7 (CH2)4CH2 CH3CH CH CH CH

O

CH2 O C (CH2)7 (CH2)7 CH3CH CH

O

CH O C (CH2)7 (CH2)4CH2 CH3CH CH CH CH

M18_TIMB8119_06_SE_C18.indd 644 28/11/18 3:09 PM

645

CI.41 A compound called butylated hydroxytoluene, or BHT, with a molecular formula C15H24O, is added to preserve foods such as cereal. As an antioxidant, BHT reacts with oxygen in the cereal container, which protects the food from spoilage. (2.5, 2.6, 7.1, 7.2, 15.1, 17.4)

CI.44 In response to signals from the nervous system, the hypothalamus secretes a polypeptide hormone known as gonadotropin-releasing factor (GnRF), which stimulates the pituitary gland to release other hormones into the bloodstream.

COMBINING IDEAS from Chapters 17 and 18

OH

BHT is an antioxidant added to preserve foods such as cereal.

a. What functional group is present in BHT? b. Why is BHT referred to as an “antioxidant”? c. The U.S. Food and Drug Administration (FDA) allows a

maximum of 50. ppm of BHT added to cereal. How many milligrams of BHT could be added to a box of cereal that contains 15 oz of dry cereal?

CI.42 Olive oil contains a high percentage of glyceryl trioleate (triolein). (7.2, 11.7, 12.6, 18.3)

One of the triacylglycerols in olive oil is glyceryl trioleate (triolein).

a. Draw the condensed structural formula for glyceryl trioleate (triolein).

b. How many liters of H2 gas at STP are needed to com- pletely react with the double bonds in 100. g of triolein?

c. How many milliliters of a 6.00 M NaOH solution are needed to completely saponify 100. g of triolein?

CI.43 A sink drain can become clogged with solid fat such as glyceryl tristearate (tristearin). (8.2, 12.6, 18.3)

A sink drain can become clogged with saturated fats.

a. How would adding lye (NaOH) to the sink drain remove the blockage?

b. Use condensed structural formulas to write a balanced equation for the reaction that occurs.

c. How many milliliters of a 0.500 M NaOH solution are needed to completely saponify 10.0 g of tristearin?

Two of the hormones are known as gonadotropins, which are the luteinizing hormone (LH) in males, and the follicle-stimulating hormone (FSH) in females. GnRF is a decapeptide with the following primary structure: (18.4, 18.5)

Glu9His9Trp9Ser9Tyr9Gly9Leu9Arg9Pro9Gly

a. What is the N-terminal amino acid in GnRF? b. What is the C-terminal amino acid in GnRF? c. Which amino acids in GnRF are nonpolar? d. Draw the condensed structural formulas at physiological

pH for the acidic and basic amino acids in GnRF. e. Draw the condensed structural formulas at physiological

pH for the primary structure of the first three amino acids starting from the N-terminus of GnRF.

CI.45 The plastic known as PETE (polyethyleneterephthalate) is a polymer of terephthalic acid and ethylene glycol. PETE is used to make plastic soft drink bottles and containers for salad dressings, shampoos, and dishwashing liquids. Today, PETE is the most widely recycled of all the plastics; in a single year, 1.80 * 109 lb of PETE are recycled. After PETE is separated from other plastics, it can be used in polyester fabric, fill for sleeping bags, door mats, and tennis ball containers. The density of PETE is 1.38 g/mL. (2.5, 2.6, 17.6)

Pituitary gland

Hypothalamus

Gonadotropin-releasing factor (GnRF) is secreted by the hypothalamus.

Terephthalic acid Ethylene glycol

OHHO

OO

OH HO

Plastic bottles made of PETE are ready to be recycled.

M18_TIMB8119_06_SE_C18.indd 645 28/11/18 8:03 PM

646 CHAPTER 18 Biochemistry

a. Write the complementary (template) strand for this normal DNA segment.

b. Write the mRNA sequence using the (template) strand in part a.

c. What amino acids are placed in the beta chain using the portion of mRNA in part b?

d. What is the order of nucleotides after T is deleted? e. Write the template strand for the mutated DNA segment. f. Write the mRNA sequence from the mutated DNA seg-

ment using the template strand in part e. g. What amino acids are placed in the beta chain by the

mutated DNA segment? h. How might the properties of this segment of the beta chain

be different from the properties of the normal protein? i. How might the level of structure in hemoglobin be affected

if beta chains are not produced?

a. Draw the line-angle formula for the ester formed from one molecule of terephthalic acid and one molecule of ethylene glycol.

b. Draw the line-angle formula for the product formed when a second molecule of ethylene glycol reacts with the ester you drew for the answer in part a.

c. How many kilograms of PETE are recycled in one year? d. What volume, in liters, of PETE is recycled in one year? e. Suppose a landfill with an area of a football field and a

depth of 5.0 m holds 2.7 * 107 L of recycled PETE. If all of the PETE that is recycled in a year were placed in land- fills, how many would it fill?

CI.46 Thalassemia is an inherited genetic mutation that limits the production of hemoglobin. If less hemoglobin is produced, there is a shortage of red blood cells (anemia). As a result, the body does not have sufficient amounts of oxygen. In one form of thalassemia, thymine (T) is deleted from section 91 (bold) in the following segment of normal DNA: (18.4, 18.5, 18.8)

89 90 91 92 93 94

AGT CAG CTG CAC TGT GAC A. c

CI.45 a. CI.41 a. The ¬ OH group in BHT is bonded to a carbon atom in an aromatic ring, which means BHT has a phenol func- tional group.

b. BHT is referred to as an “antioxidant” because it reacts with oxygen in the food container, rather than the food, thus preventing or retarding spoilage of the food.

c. 21 mg of BHT

CI.43 a. Adding NaOH will saponify the glyceryl tristearate (fat), breaking it up into fatty acid salts and glycerol, which are soluble and will wash down the drain.

b.

ANSWERS

O

O

O

O

-O

OCH

CH2 O

CH2 O C (CH2)16 CH3

C (CH2)16 CH3

C (CH2)16 CH3 +

+ CH3CH OH

OHCH2

OHCH2

C (CH2)16

3NaOH

3Na+

c. 67.3 mL of a 0.500 M NaOH solution

OHO

OO

OH

b.

OO

OO

OHHO

c. 8.18 * 108 kg of PETE d. 5.93 * 108 L of PETE e. 22 landfills

M18_TIMB8119_06_SE_C18.indd 646 28/11/18 3:09 PM

C-1

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Chapter 10 p. 269 fotostorm/Getty Images p. 291 top: FOTOSEARCH RM/AGE Fotostock p. 291 bottom: John A. Rizzo/Getty Images p. 293 top, right: johnnyscriv/Getty Images p. 295 top, right: Wsphotos/Getty Images p. 297 Nikkytok/Fotolia p. 298 fotostorm/Getty Images p. 303 top: Mark Dahners/AP Images p. 303 bottom: Vasily Pindyurin/Getty Images p. 309 top: a-wrangler/Getty Images p. 309 bottom: Eric Schrader/Pearson Education, Inc. p. 310 left: jane/Getty Images p. 310 right: gvictoria/Getty Images

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Credits C-3

Chapter 11 p. 311 Adam Gault/Science Photo Library/Alamy p. 313 NASA p. 315 top: GybasDigiPhoto/Shutterstock p. 315 bottom: Kenneth William Caleno/Shutterstock p. 316 LeventeGyori/Shutterstock p. 317 Levent Konuk/Shutterstock p. 320 Steve Bower/Shutterstock p. 323 top, right: Prasit Rodphan/Shutterstock p. 325 Andrey Nekrasov/Alamy Stock Photo p. 328 Tomasz Wojnarowicz/Fotolia p. 330 yuri10b/Getty Images p. 331 top: Jaral Lertjamekorn/Shutterstock p. 331 right: LoloStock/Shutterstock p. 335 Eric Schrader/Fundamental Photographs, NYC p. 340 Mic Smith/Alamy Stock Photo p. 341 Adam Gault/Science Photo Library/Alamy p. 346 Library of Congress

Chapter 12 p. 350 AJPhoto/Science Source p. 353 top: Thinkstock Images/Getty Images p. 356 Comstock Images/Getty Images p. 358 left: Dr. Marazzi/Science Source p. 358 right: Remik44992/Shutterstock p. 359 Mega Pixel/Shutterstock p. 360 bottom: CNRI/Science Source p. 379 Ecologix Environmental Systems, LLC p. 380 top: Photowind/Shutterstock p. 380 bottom: jane/Getty Images p. 382 Florida Images/Alamy Stock Photo p. 385 Thinkstock Images/Getty Images p. 387 Picsfive/Shutterstock p. 388 AJPhoto/Science Source p. 392 top: Professor25/Getty Images p. 392 bottom: Brand X Pictures/Getty Images

Chapter 13 p. 398 Dante Fenolio/Science Source p. 406 Gudellaphoto/Fotolia p. 417 Incamerastock/Alamy Stock Photo p. 419 top: Sirtravelalot/Shutterstock p. 419 bottom: Photobac/Shutterstock p. 421 top: Scimat/Science Source p. 421 bottom: Taras Verkhovynets/Shutterstock p. 423 A.P. Fitzpatrick Art Materials/Dorling Kindersley p. 424 Brian Lasenby/Fotolia

Chapter 14 p. 431 Anna Ivanova/123rf p. 432 Magicinfoto/Shutterstock p. 433 top: Kul Bhatia/Science Source p. 433 bottom: Lukas Gojda/Shutterstock

p. 438 Magone/123RF p. 439 bottom: Richard Megna/Fundamental Photographs,

NYC. p. 440 Lana Langlois/Shutterstock p. 448 vinegar: George Tsartsianidis/123RF p. 448 can: Fotofermer/Shutterstock p. 448 detergent: PeterG/Shutterstock p. 448 bleach: Luisa Leal/123RF p. 449 top: Alexander Gospodinov/Fotolia p. 449 bottom: Wayne Hutchinson/Alamy Stock Photo p. 450 Eric Schrader/Pearson Education, Inc. p. 451 Ana Bokan/Shutterstock p. 453 Yenyu Shih/Shutterstock p. 454 Richard Megna/Fundamental Photographs, NYC. p. 455 Eric Schrader/Pearson Education, Inc. p. 456 Gerald Bernard/Shutterstock p. 463 Anna Ivanova/123rf p. 468 geotrac/Getty Images p. 470 Universal Images Group/SuperStock p. 473 top, left: Oleksandr Kalinichenko/Shutterstock p. 473 center, left: Patsy Michaud/Shutterstock p. 473 bottom: a-wrangler/Getty Images p. 473 center: isifa Image Service s.r.o./Alamy Stock Photo p. 473 center, right: Rostislav Sedlacek/Fotolia p. 474 top: Charles D. Winters/Science Source p. 474 bottom: Picsfive/Shutterstock

Chapter 15 p. 476 Craig Holmes Premium/Alamy Stock Photo p. 477 Martyn F. Chillmaid/Science Source p. 482 Andrew Lambert Photography/Science Source p. 485 Richard Megna/Fundamental Photographs p. 488 Charles D. Winters/Science Source p. 493 top: Terekhov igor/Shutterstock p. 493 bottom: Paul Orr/Shutterstock p. 494 top: Yenes/Shutterstock p. 494 bottom: alexlmx/123RF p. 495 Olivier DIGOIT/Alamy Stock Photo p. 498 left: Jim Barber/Shutterstock p. 498 right: karandaev/123RF p. 499 top: Craig Holmes Premium/Alamy Stock Photo p. 499 bottom: Artproem/Shutterstock p. 500 left: Andrew Lambert Photography/Science Source p. 500 top, right: Richard Megna/Fundamental Photographs p. 500 bottom, right: Charles D. Winters/Science Source p. 502 left: Michal Baranski/123RF p. 502 right: Andrew Lambert/Science Source

Chapter 16 p. 508 Tyler Olson/Shutterstock p. 510 Celig/Shutterstock p. 511 Josh Sher/Science Source p. 513 Uberphotos/Getty Images p. 514 Photo courtesy of Spruce Environmental

Technologies, Inc.

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C-4 Credits

p. 515 Laurent/B. HOP AME/BSIP SA/Alamy Stock Photo p. 518 Australian Nuclear Science and Technology

Organization p. 519 left: Kyodo/Newscom p. 519 right: Don Farrall/Getty Images p. 520 left: U.S. Food & Drug Administration (FDA) p. 521 Jürgen Schulzki/Alamy Stock Photo p. 524 Library of Congress (Photo duplication) p. 525 Newscom p. 527 top, right: Burger/Phanie068689/SuperStock p. 527 center right: Pasieka/Science Source p. 527 bottom, left: Mehau Kulyk/Science Source p. 527 bottom, center: GJLP/Science Source p. 527 bottom, right: Lawrence Berkeley National Library/

Getty Images p. 528 Editorial Image,LLC/Alamy p. 529 left: Karen C. Timberlake p. 529 right: Theodore R. Allen, Ph.D.,Esp/Cytyc Hologic

Corporation p. 532 left: CNRI/Science Source p. 532 right: Tyler Olson/Shutterstock p. 535 robertharding/Alamy Stock Photo p. 538 top, left: Andrew Lambert Photography/SPL/Science

Source p. 538 bottom, left: Richard Megna/Fundamental

Photographs, NYC. p. 538 top, right: Sven Hoppe/picture-alliance/dpa/AP

Images p. 539 left: Danny Smythe/Shutterstock

Chapter 17 p. 540 Corepics/Fotolia p. 542 left: lillisphotography/Getty Images p. 549 bottom, left: Roza/Fotolia p. 552 Tim Hall/Getty Images p. 555 Siede Preis/Getty Images p. 558 left: Alexandr KanÃμkin/Shutterstock p. 558 center: Vblinov/Shutterstock p. 558 right: dirkr/Getty Images p. 560 Mediscan/Alamy Stock Photo p. 562 ASA studio/Shutterstock

p. 563 right: jane/Getty Images p. 564 Renata Kazakova/Shutterstock p. 568 Robert Przybysz/Shutterstock p. 570 Jose Manuel Ribeiro/Thomson Reuters (Markets)

LLC p. 574 Lynn Watson/Shutterstock p. 576 left: blickwinkel/Koenig/Alamy Stock Photo p. 576 right: Harvey Male/Alamy Stock Photo p. 577 Arterra Picture Library/De Meester Johan/Alamy

Stock Photo p. 580 bottom: Sumroeng/Fotolia p. 585 top, left: Parpalea Catalin/Shutterstock p. 585 right: Ivan Marjanovic/Shutterstock p. 585 center, left: Siede Preis/Getty Images p. 586 almonds: Pavlo Kucherov/Fotolia p. 586 cinnamon: C Squared Studios/Getty Images p. 586 muffins: riderfoot/Fotolia p. 586 butter: Multiart/Shutterstock

Chapter 18 p. 592 Rolf Bruderer/Blend Images/Alamy Stock Photo p. 593 bottom: AZP Worldwide/Shutterstock p. 595 Coprid/Shutterstock p. 601 Tetra Images/Getty Images p. 603 Danny E Hooks/Shutterstock p. 607 Lotus_studio/Shutterstock p. 608 left: Dawn Wilson Photography/Getty Images p. 609 KMNPhoto/Shutterstock p. 611 National Heart, Lung, and Blood Institute p. 612 Fuse/Getty Images p. 618 Laurence Mouton/PhotoAlto sas/Alamy Stock Photo p. 633 Rolf Bruderer/Blend Images/Alamy Stock Photo p. 638 Ekkachai/Shutterstock p. 638 3355m/Fotolia p. 639 Mara Zemgaliete/Fotolia p. 640 margo555/Fotolia p. 641 Scott A. Frangos/Shutterstock p. 645 bottom, left: Lauree Feldman/Getty Images p. 645 top, right: Martin M. Rotker/Science Source p. 645 bottom, right: David Zaitz/Alamy Stock Photo

Z01_TIMB8119_06_SE_CRED.indd 4 11/27/18 12:21 PM

I-1

A Abbreviated electron configuration, 136 Absolute zero, 77 Acetylsalicylic acid (aspirin), 450 Acid A substance that dissolves in water and produces

hydrogen ions (H+), according to the Arrhenius theory. All acids are hydrogen ion donors, according to the Brønsted–Lowry theory, 432

Acid–base titration, 457–458 endpoint, 457 indicator, 457

Acid dissociation constant, Ka The numerical value of the product of the ions from the dissociation of a weak acid divided by the concentration of the weak acid, 442

Acid dissociation expression, 442 Acidosis, 462 Acid reflux, 463 Acid reflux disease, 463 Acids and bases, 431–475 Acids and bases, reactions of, 454–457

acids and carbonates or bicarbonates, 454 acids and metals, 454 neutralization, 455, 456t neutralization equations balancing, 455–456

Acids and bases, strengths of, 437–442, 438t diprotic acids, 439–440 direction of reaction, 441 dissociation, 437 strong acids, 437 strong bases, 440 weak acids, 438 weak bases, 440

Acids and carbonates or bicarbonates, 454 Acids and metals, 454 Activation energy The energy that must be provided by a

collision to break apart the bonds of the reacting molecules, 399

Active site A pocket in the tertiary enzyme structure that binds substrate and catalyzes a reaction, 622

Activity series A table of half-reactions with the metals that oxidize most easily at the top, and the metals that do not oxidize easily at the bottom, 488, 489t

Actual yield The actual amount of product produced by a reaction, 252

Adding significant zeroes, 36 Addition and subtraction with measured numbers, 37 Air, as mixture of gas, 336–337, 336t Alcohol An organic compound that contains a hydroxyl group

( ¬ OH) attached to a carbon chain, 560 Alcohols and ethers, 560–564

alcohol, 560 ether, 560 naming alcohols, 561 naming ethers, 562

Alcohols, phenols, and ethers use, 562–564 Aldehyde An organic compound that contains a carbonyl

group (C “ O) bonded to at least one hydrogen atom, 564–567

Aldehydes and ketones, 564–568 aldehyde, 565 ketones, 565 naming aldehydes, 565 naming ketones, 566

Alkali metal An element in Group 1A (1), except hydrogen, that is a soft, shiny metal with one electron in its outermost energy level, 104

Alkaline earth metal An element in Group 2A (2) that has two electrons in its outermost energy level, 104

Alkaloids, amines in plants, 576–577 Alkane A type of hydrocarbon in which the carbon atoms are

connected only by single bonds, 541–551 combustion, 549 condensed structural and line-angle formulas, 544–545 condensed structural formula, 542 drawing structural formulas for alkanes with substituents,

548–549 expanded structural formula, 542 hydrocarbons, 542 IUPAC system of naming, 543 naming alkanes with substituents, 547–548 representations of carbon compounds, 542–543 solubility and density, 549–550 structural isomers, 545 substituent, 545 substituents in alkanes, 546, 546t uses, 549

Alkene A type of hydrocarbon that contains a carbon–carbon double bond (C “ C), 551

Alkenes, alkynes, and polymers, 551–557 alkene, 551 alkyne, 551 monomer, 554 naming, 552, 552t polymer, 554 polymerization, 554–557, 555t recycling code, 555t

Alkyl group An alkane minus one hydrogen atom. Alkyl groups are named like the corresponding alkanes except a yl ending replaces ane, 546

Alkyne A type of hydrocarbon that contains a carbon–carbon triple bond (C ‚ C), 555

Glossary/Index

Z02_TIMB8119_06_SE_IDX.indd 1 11/30/18 11:23 AM

I-2 Glossary/Index

Alpha decay, 512–514 Alpha helix, 618 Alpha particle A nuclear particle identical to a helium

nucleus, symbol a or 2 4He, 509

Amide An organic compound in which the hydroxyl group of a carboxylic acid is replaced by a nitrogen atom, 577

Amine An organic compound that contains a nitrogen atom bonded to one or more carbon atoms, 575

Amino acid The building block of proteins, consisting of a hydrogen atom, an ammonium group, a carboxylate group, and a unique R group attached to the alpha carbon, 612–613, 613t

codons for, 630 essential, 615 N- and C-terminus, 615 polar and nonpolar, 613 structure of, 613

Amino acids and proteins, 612–617 amino acids, 612–613 peptides, 615–617 protein classifications, 612t

Amphoteric Substances that can act as either an acid or a base in water, 436

Amylopectin A branched-chain polymer of starch com- posed of glucose units joined by a(1S 4)@ and a(1S 6)@glycosidic bonds, 601

Amylose An unbranched polymer of starch composed of glucose units joined by a(1S 4)@glycosidic bonds, 601

Aniline, 575 Anion A negatively charged ion such as Cl-, O2-, or SO4

2-, 158 Anode The electrode where oxidation takes place, 489 Antacids, 456 Aromatic compound A compound that contains the ring

structure of benzene, 557–560 benzene, 557 naming, 558

Atmosphere (atm) A unit equal to the pressure exerted by a column of mercury 760 mm high, 314

Atmospheric pressure The pressure exerted by the atmosphere, 312

Atom The smallest particle of an element that retains the characteristics of the element, 108–110

Dalton’s atomic theory, 108 electrical charge, 108

Atomic mass The weighted average mass of all the naturally occurring isotopes of an element, 115–117, 116t

calculation, 115–117 weighted average analogy, 115

Atomic mass unit (amu) A small mass unit used to describe the mass of extremely small particles such as atoms and subatomic particles; 1 amu is equal to one-twelfth the mass of a 6

12C atom, 109–110 Atomic number A number that is equal to the number of pro-

tons in an atom, 111 Atomic size The distance between the outermost electrons and

the nucleus, 145

Atomic spectrum A series of lines specific for each element produced by photons emitted by electrons dropping to lower energy levels, 129

changes in energy level, 130–131 electron energy levels, 129–130 photons, 129

Atomic symbol An abbreviation used to indicate the mass number and atomic number of an isotope, 113–114, 114t

Avogadro’s law A gas law stating that the volume of a gas is directly related to the number of moles of gas when pressure and temperature do not change, 327

Avogadro’s number The number of items in a mole, equal to 6.022 * 1023, 184–186

B Balanced equation The final form of a chemical equation that

shows the same number of atoms of each element in the reactants and products, 215

Balanced equation information, 240t Balancing equations, 217–222

polyatomic ions, 220–222 whole-number coefficients, 219–220

Base A substance that dissolves in water and produces hydrox- ide ions (OH-), according to the Arrhenius theory. All bases are hydrogen ion acceptors, according to the Brønsted–Lowry theory, 433

Base dissociation constant, Kb The numerical value of the product of the ions from the dissociation of a weak base divided by the concentration of the weak base, 443

Bases in nucleic acids, 624 Batteries, 492–496

dry-cell batteries, 493–494 lead storage battery 494–495 lithium-ion batteries, 494–495

Becquerel (Bq) A unit of activity of a radioactive sample equal to one disintegration per second, 519

Bent The shape of a molecule with two bonded atoms and one lone pair or two lone pairs, 280

Beta decay, 514–515 Beta particle A particle identical to an electron, symbol -1

0e or b, that forms in the nucleus when a neutron changes to a proton and an electron, 509

Beta pleated sheet, 619 Benzene A ring of six carbon atoms each of which is attached

to one hydrogen atom, C6H6, 557 Biochemistry, 592–646 Biological effects of radiation, 510–511 Biological reactions to UV light, 128 Blood gases, 337 Blood pressure measurement, 316 Boiling The formation of bubbles of gas throughout a

liquid, 293 Boiling point (bp) The temperature at which a liquid changes

to gas (boils) and gas changes to liquid (condenses), 293 Boiling point elevation, 383–384, 383t Bonding, 269–310

Z02_TIMB8119_06_SE_IDX.indd 2 11/30/18 11:23 AM

Glossary/Index I-3

diabetes nurse, 592 dialysis nurse, 350 dietitian 68 environmental scientist, 239 exercise physiologist, 213 farmer, 100 firefighter/emergency medical technician, 540 forensic scientist, 1 histologist, 269 materials engineer, 125 medical laboratory technologist, 431 pharmacist, 156 radiation technologist, 508 registered nurse, 27 respiratory therapist, 311 veterinarian, 183

Catalyst A substance that increases the rate of reaction by lowering the activation energy, 401

and effect on equilibrium, 416 and rates of reaction, 401–402

Cathode The electrode where reduction takes place, 489 Cation A positively charged ion such as Na+, Mg2+, Al3+, or

NH4 +, 158

Cellulose An unbranched polysaccharide composed of glucose units linked by b(1S 4)@glycosidic bonds that cannot be hydrolyzed by the human digestive system, 602

Celsius (°C) temperature scale A temperature scale on which water has a freezing point of 0 °C and a boiling point of 100 °C, 29

Centimeter (cm) A unit of length in the metric system; there are 2.54 cm in 1 in., 29

Central atoms with four electron groups (tetrahedral), 280–283 Central atoms with three electron groups (trigonal), 279–280 Central atoms with two electron groups (linear), 279 Chain reaction A fission reaction that will continue once it

has been initiated by a high-energy neutron bombarding a heavy nucleus such as uranium-235, 530

Change in temperature and effect on equilibrium, 418–419 Change of state The transformation of one state of matter to

another, for example, solid to liquid, liquid to solid, liq- uid to gas, 291–298

Charles’s law A gas law stating that the volume of a gas is directly related to the Kelvin temperature when pressure and moles of the gas do not change, 320

Chemical A substance that always has the same composition and properties wherever it is found, 1

Chemical change A change during which the original sub- stance is converted into a new substance that has a different composition and new physical and chemical properties, 73

Chemical equation A shorthand way to represent a chemi- cal reaction using chemical formulas to indicate the reactants and products and coefficients to show reacting ratios, 214–217, 214t

balanced equations, 215–216 products, 215 reactants, 215 symbols, 215t

double bond, 276 electronegativity and bond polarity, 283–287 intermolecular forces, 288–291 Lewis structures for molecules and polyatomic ions,

270–276 octet rule exceptions, 276 octet rule, 157 polarity of molecules, 287–288 resonance structures, 276–279 shapes of molecules and polyatomic atoms, 279–283 triple bond, 276

Boyle’s law A gas law stating that the pressure of a gas is inversely related to the volume when temperature and moles of the gas do not change, 317

Brachytherapy, 528–529 Breathing, pressure–volume relationship, 318 Brønsted–Lowry acids and bases An acid is a hydrogen ion

donor; a base is a hydrogen ion acceptor, 434–437 amphoteric substance, 436 conjugate acid–base pairs, 435–437 hydronium ion, 434

Buffers, 458–463 buffer solution, 458 pH of a buffer calculation, 460–462, 462t

Buffer solution A solution of a weak acid and its conjugate base or a weak base and its conjugate acid that maintains the pH by neutralizing added acid or base, 459

C Calculations, 34–38

adding significant zeroes, 36 addition and subtraction with measured numbers, 37 multiplication and division with measured numbers, 36 rounding off, 35–36, 35t

Calculator operations, 11–12 Calculators and scientific notation, 19 Calorie (cal) The amount of heat energy that raises the tem-

perature of exactly 1 g of water by exactly 1°C, 80, 87 Carbohydrate A simple or complex sugar composed of car-

bon, hydrogen, and oxygen, 593–597 fructose, 594 galactose, 594 glucose, 594 Haworth structures of monosaccharides, 595–597 monosaccharides, 593–594 open-chain structures, 594

Carbon-14 dating, 524 Carbon monoxide toxicity, 227 Carboxylic acid An organic compound that contains the car-

boxyl group ( ¬ COOH), 568 Carboxylic acids and esters, 568–575

carboxylic acid, 568 naming carboxylic acids, 569–570, 569t

Carboxylic acids in metabolism, 570–571 Careers

chemical oceanographer, 398 dentist, 476

Z02_TIMB8119_06_SE_IDX.indd 3 11/30/18 11:23 AM

I-4 Glossary/Index

sweeteners, 600–601, 600t toxicology and risk–benefit assessment, 52 water in the body, 352–353 weight, losing and gaining, 89

Cholesterol, 611 Chromatography, 73 Codon A sequence of three bases in mRNA that specifies

a certain amino acid to be placed in a protein. A few codons signal the start or stop of protein synthesis, 630–631, 631t

Coefficients Whole numbers placed in front of the formulas to balance the number of atoms or moles of atoms of each element on both sides of an equation, 215

Cold packs and hot packs, 257 Collagen, 618 Collision theory A model for a chemical reaction stating that

molecules must collide with sufficient energy and proper orientation to form products, 399

Colloid A mixture having particles that are moderately large. Colloids pass through filters but cannot pass through semipermeable membranes, 378

Combination reaction A chemical reaction in which reactants combine to form a single product, 223

Combined gas law A relationship that combines several gas laws relating pressure, volume, and temperature when the amount of gas does not change, 325–326

Combining energy calculations, 296–297 Combustion reaction A chemical reaction in which a fuel

containing carbon reacts with oxygen to produce CO2, H2O, and energy, 225

Common ions, formulas and names, 159t Complementary base pairs In DNA, adenine is always

paired with thymine (A and T or T and A), and guanine is always paired with cytosine (G and C or C and G). In forming RNA, adenine is paired with uracil (A and U), 626

Compound A pure substance consisting of two or more elements, with a definite composition, that can be broken down into simpler substances only by chemical methods, 69

Computed tomography (CT), 527 Concentration A measure of the amount of solute that is

dissolved in a specified amount of solution, 363–371, 363t

conversion factors from concentrations, 370t mass percent (m/m), 364 mass/volume percent (m/v), 366–368 molarity (M), 368–371 volume percent (v/v), 365–366

Concentration changes and effect on equilibrium, 415–416 of reactants and rates of reaction, 401

Conclusion An explanation of an observation that has been validated by repeated experiments that support a hypothesis, 4

Condensation The change of state from a gas to a liquid, 293–294

Chemical equilibrium The point at which the rate of forward and reverse reactions are equal so that no further change in concentrations of reactants and products takes place, 403–406

equilibrium, 403–404 reversible reaction, 403

Chemical formula The group of symbols and subscripts that represents the atoms or ions in a compound, 162

Chemical formulas for ionic compounds, 162 Chemical oceanographer, 398 Chemical properties The properties that indicate the ability

of a substance to change into a new substance, 73 Chemical reactions, 213–238 Chemical symbol An abbreviation that represents the name of

an element, 101–102, 101t Chemistry The study of the composition, structure, properties

and reactions of matter, 2 Chemistry Links to the Environment

alkaloids: amines in plants, 576–577 corrosion: oxidation of metals, 494–495 dating ancient objects, 524 fertilizers, 196 fuel cells: clean energy for the future, 496

Chemistry Links to Health alcohols, phenols, and ethers, 562–564 aldehydes and ketones, 567–568 antacids, 456 aromatic compounds, 559 biological reactions to UV light, 128 blood gases, 337 blood pressure measurement, 316 body temperature variation, 79 bone density, 56 brachytherapy, 528–529 breathing mixtures, 70 breathing, pressure–volume relationship, 318 buffers in blood plasma, 462 carbon monoxide toxicity, 227 carboxylic acids in metabolism, 570–571 cold packs and hot packs, 257 electrolytes in body fluids, 356 elements essential to health, 106, 106t, 107t essential amino acids and proteins, 615 gout and kidney stones, 358 hemodialysis and the artificial kidney, 387–388 homeostasis and regulation of body temperature, 419 hydrogenation of unsaturated fats, 554 hyperbaric chambers, 340 hyperglycemia and hypoglycemia, 595 ions in the body, 161, 161t oxygen–hemoglobin equilibrium and hypoxia, 417 Paracelsus (early chemist), 4 polycyclic aromatic hydrocarbons (PAHs), 560 radiation and food, 520 radon in homes, 514 steam burns, 297 stomach acid, HCl, 453

Z02_TIMB8119_06_SE_IDX.indd 4 11/30/18 11:23 AM

Glossary/Index I-5

Dialysis by kidneys and artificial kidney, 387–388 Dialysis nurse, 350 Dietitian, 68 Dilution A process by which water (solvent) is added to a

solution to increase the volume and decrease (dilute) the solute concentration, 371–374

Dipole The separation of positive and negative charge in a polar bond indicated by an arrow that is drawn from the more positive atom to the more negative atom, 284

Dipole–dipole attractions Intermolecular forces between oppositely charged ends of polar molecules, 289

Diprotic acids, 439–440 Direction of reaction, 441 Direct relationship A relationship in which two properties

increase or decrease together, 320 Disaccharide A carbohydrate composed of two monosaccha-

rides joined by a glycosidic bond, 598 Dispersion forces Weak dipole bonding that results from a

momentary polarization of nonpolar molecules, 289 Dissociation The separation of an acid or a base into ions in

water, 437 Dissociation of water, 444–447

neutral solution, 445 using Kw to calculate hydronium ion concentration in a

solution, 446 water dissociation constant, Kw, 445 water dissociation expression, 445

Dissociation of weak acids and bases, 442–444 acid dissociation constant, Ka, 442 writing expressions for, 442–444

DNA Deoxyribonucleic acid; the genetic material of all cells containing nucleotides with deoxyribose, phosphate, and the four bases: adenine, thymine, guanine, and cytosine, 624

Dosage problems conversion factors, 44 Double bond A sharing of two pairs of electrons by two

atoms, 276 Double helix The helical shape of the double chain of DNA

that is like a spiral staircase with a sugar–phosphate backbone on the outside and base pairs like stair steps on the inside, 626

Double replacement reaction A reaction in which the positive ions in the reacting compounds exchange places, 225

Drawing Lewis structures, 272–274, 272t Dry-cell batteries, 493–494

E Effect on equilibrium

catalyst, 416 change in temperature, 418–419 concentration changes, 415–416 volume change, 416–417

Electrical energy from oxidation–reduction reactions, 488–496 activity series, 488, 489t anode, 490 batteries, 492–496

Condensed structural formula A formula that shows the carbon atoms grouped with the attached number of hydrogen atoms, 542

Condensed structural formula of alkanes, 542 Condition changes and effect on equilibrium, 418t Conservation of mass, 240–241

balanced equation information, 240t Law of Conservation of Mass, 240

Conjugate acid–base pair An acid and a base that differ by one H+. When an acid donates a hydrogen ion, the prod- uct is its conjugate base, which is capable of accepting a hydrogen ion in the reverse reaction, 435

Conversion factor A ratio in which the numerator and denom- inator are quantities from an equality or given relation- ship, 42–47, 43t

from dosage problems, 44 metric conversion factors, 43 metric–U.S. system conversion factors, 43 from percent, ppm, and ppb, 44 with powers, 45

Conversion of units in a fraction, 51 Cooling curve A diagram that illustrates temperature changes

and changes of state for a substance as heat is removed, 295

Corrosion, 494–495 Covalent bond A sharing of valence electrons by atoms, 157,

172 Cubic centimeter (cm3, cc) The volume of a cube that has

1-cm sides; 1 cm3 is equal to 1 mL, 41 Curie (Ci) A unit of activity of a radioactive sample equal to

3.7 * 1010 disintegrations/s, 519

D Dalton’s law A gas law stating that the total pressure exerted

by a mixture of gases in a container is the sum of the partial pressures that each gas would exert alone, 335–341

d block The 10 elements in Groups 3B (3) to 2B (12) in which electrons fill the five d orbitals, 140

Decay curve A diagram of the decay of a radioactive element, 522, 524t

Decomposition reaction A reaction in which a single reactant splits into two or more simpler substances, 223

Density The relationship of the mass of an object to its volume expressed as grams per cubic centimeter (g/cm3), grams per milliliter (g/mL), or grams per liter (g/L), 53–58, 54t

calculating, 54–55 problem solving with, 55–57 volume displacement, 55

Dentist, 476 Deposition The change of a gas directly to a solid; the reverse

of sublimation, 292 Diabetes nurse, 592 Dialysis A process in which water and small solute particles

pass through a semipermeable membrane, 387

Z02_TIMB8119_06_SE_IDX.indd 5 11/30/18 11:23 AM

I-6 Glossary/Index

examples, 197t Endothermic reaction A reaction in which the energy of the

products is higher than that of the reactants, 255 Endpoint The point at which an indicator changes color. For

the indicator phenolphthalein, the color change occurs when the number of moles of OH- is equal to the number of moles of H3O

+ in the sample, 457 Energy The ability to do work, 79–83, 80t

heat and, 80 kinetic and potential, 80 units, 80

Energy in chemical reactions, 254–260 calculations of heat in reactions, 256–257 endothermic reactions, 255 energy units used for chemical reactions, 254 exothermic reactions, 254 heat of reaction, 254 Hess’s law, 257–259

Energy level A group of electrons with similar energy, 129 Energy units used for chemical reactions, 254 Energy value The kilocalories (or kilojoules) obtained per

gram of the food types: carbohydrate, fat, and protein, 88–90, 88t

Environmental scientist, 239 Enzyme A protein that catalyzes a biological reaction, 622 Enzyme-catalyzed reaction, 622 Equality A relationship between two units that measure the

same quantity, 40 Equilibrium, 403–404 Equilibrium constant calculation, 407–409 Equilibrium constant, Kc The numerical value obtained

by substituting the equilibrium concentrations of the components into the equilibrium expression, 406–410, 408t

equilibrium constant calculation, 407–409 equilibrium expression, 406–407 heterogeneous equilibrium, 408 homogeneous equilibrium, 408

Equilibrium expression The ratio of the concentrations of products to the concentrations of reactants, with each component raised to an exponent equal to the coefficient of that compound in the balanced chemical equation, 406–407

Equilibrium in saturated solutions, 420–424 molar solubility, S, 423–424 solubility product constant (Ksp), 421 solubility product expression, 421

Equivalent dose The measure of biological damage from an absorbed dose that has been adjusted for the type of radiation, 520

Ester An organic compound in which the ¬ H of a carboxyl group is replaced by a carbon atom, 571

naming esters, 572 Esterification, 571–572 Esters in plants, 573–574, 573t

cathode, 490 voltaic cell, 490–492

Electrolysis The use of electrical energy to run a nonspontane- ous oxidation–reduction reaction in an electrolytic cell, 497

Electrolysis of sodium chloride, 497 Electrolyte A substance that produces ions when dissolved in

water; its solution conducts electricity, 355 Electrolytes and nonelectrolytes, 355–357

classification of solutes in aqueous solutions, 355t electrolytes, 355 nonelectrolytes, 355 strong electrolyte, 355 weak electrolytes, 355

Electrolytes in body fluids, 356 Electrolytic cell A cell in which electrical energy is used to

make a nonspontaneous oxidation–reduction reaction happen, 497

Electromagnetic radiation Forms of energy such as visible light, microwaves, radio waves, infrared, ultraviolet light, and X-rays that travel as waves at the speed of light, 126

electromagnetic spectrum, 126–128 frequency, 126 wavelength, 126

Electromagnetic spectrum The arrangement of types of radiation from long wavelengths to short wavelengths, 126

Electroplating, 498 Electron A negatively charged subatomic particle having a

minute mass that is usually ignored in mass calculations; its symbol is e-, 108

Electron configuration A list of the number of electrons in each sublevel within an atom, arranged by increasing energy, 139

Electronegativity The relative ability of an element to attract electrons in a bond, 283–287

dipole, 284 nonpolar covalent bond, 284 polar covalent bond, 284 polarity, 284 variations in bonding, 285–286

Electron configurations and the periodic table, 139–143 d block, 140 exceptions in sublevel block order, 142–143 f block, 140 p block, 140 Period 4 and above, 141 s block, 140

Element A pure substance containing only one type of matter, which cannot be broken down by chemical methods, 69, 101, 101t

Empirical formula The simplest or smallest whole-number ratio of the atoms in a formula, 197–201, 197t

calculating 197 conversion of decimal numbers to whole numbers, 199t

Z02_TIMB8119_06_SE_IDX.indd 6 11/30/18 11:23 AM

Glossary/Index I-7

Gases, 311–349 Gay-Lussac’s law A gas law stating that the pressure of a gas

is directly related to the Kelvin temperature when the number of moles of a gas and its volume do not change, 322

Genetic code The sequence of codons in mRNA that specifies the amino acid order for the synthesis of protein, 630–631, 631t

Glucose An aldohexose found in fruits, vegetables, corn syrup, and honey that is also known as blood sugar and dextrose. The most prevalent monosaccharide in the diet. Most polysaccharides are polymers of glucose, 594

Glycogen A polysaccharide formed in the liver and muscles for the storage of glucose as an energy reserve. It is composed of glucose in a highly branched polymer joined by a(1S 4)@ and a(1S 6)@glycosidic bonds, 602

Gout and kidney stones, 358 Gram (g) The metric unit used in measurements of mass, 29 Graph interpretation, 14–16 Gray (Gy) A unit of absorbed dose equal to 100 rad, 519 Group A vertical column in the periodic table that contains

elements having similar physical and chemical properties, 103

Group number A number that appears at the top of each vertical column (group) in the periodic table and indicates the number of electrons in the outermost energy level, 103

H Half-life The length of time it takes for one-half of a

radioactive sample to decay, 522 Half-life of a radioisotope, 522–526, 524t

decay curve, 522, 524t Half-reaction method A method of balancing oxidation–

reduction reactions in which the half-reactions are balanced separately and then combined to give the complete reaction, 483

Half-reactions, 477–478 Halogen An element in Group 7A (17)—fluorine, chlorine,

bromine, iodine, astatine, and tennessine—that has seven electrons in its outermost energy level, 104

Haworth structure The ring structure of a monosaccharide, 595–597

Heat The energy associated with the motion of particles in a substance, 80

Heat equation A relationship that calculates heat (q) given the mass, specific heat, and temperature change for a substance, 83–85

Heating curve A diagram that illustrates the temperature changes and changes of state of a substance as it is heated, 295

Heat of fusion The energy required to melt exactly 1 g of a substance at its melting point. For water, 334 J is needed to melt 1 g of ice; 334 J is released when 1 g of water freezes, 292

Ether An organic compound in which an oxygen atom is bonded to two carbon groups that are alkyl or aromatic, 560

Evaporation The formation of a gas (vapor) by the escape of high-energy molecules from the surface of a liquid, 293

Exact number A number obtained by counting or by definition, 32–33, 32t

Exceptions in sublevel block order, 142–143 Exercise physiologist, 213 Exothermic reaction A reaction in which the energy of the

products is lower than that of the reactants, 254 Expanded structural formula A formula that shows all of the

atoms and the bonds connected to each atom, 542 Experiment A procedure that tests the validity of a hypothesis, 4

F Fahrenheit scale (°F) A scale where water freezes at 32° and

boils at 212°, 29 Farmer, 100 Fat A triacylglycerol that is solid at room temperature and

usually comes from animal sources, 608 Fatty acid A long-chain carboxylic acid found in many lipids,

605–606, 606t f block The 14 elements in the rows at the bottom of the

periodic table in which electrons fill the seven 4f and 5f orbitals, 140

Filtration, 73 Firefighter/emergency medical technician, 540 Fission A process in which large nuclei are split into smaller

nuclei, releasing large amounts of energy, 529 Forensic scientist, 1 Freezing A change of state from liquid to solid, 292 Freezing point (fp) The temperature at which a liquid changes

to a solid (freezes) and a solid changes to a liquid (melts), 292

Freezing point, lowering, 381–382 Frequency The number of times the crests of a wave pass a

point in 1 s, 126 Fructose A ketohexose which is combined with glucose in

sucrose, 594 Fuel cells: clean energy for the future, 496 Functional group A group of atoms that determines the

physical and chemical properties and naming of a class of organic compounds, 551

Fusion A reaction in which large amounts of energy are released when small nuclei combine to form larger nuclei, 530

G Galactose An aldohexose that is combined with glucose in

lactose, 594 Gamma ray High-energy radiation, symbol

0

0g, emitted by an unstable nucleus, 510

Gas A state of matter that does not have a definite shape or volume, 72

Z02_TIMB8119_06_SE_IDX.indd 7 11/30/18 11:23 AM

I-8 Glossary/Index

size, mass, and melting and boiling points, 290–291 International System of Units (SI) The official system of

measurement throughout the world, except for the United States, that modifies the metric system, 28

Inverse relationship A relationship in which two properties change in opposite directions, 317

Ion An atom or group of atoms having an electrical charge because of a loss or gain of electrons, 157

Ionic and molecular compounds, 156–182 Ionic and polar solutes, 354 Ionic bond The attraction between a positive ion and a nega-

tive when electrons are transferred from a metal to a nonmetal, 157

Ionic charge The difference between the number of protons (positive) and the number of electrons (negative) written in the upper right corner of the symbol for the element or polyatomic ion, 158

Ionic charges from group numbers, 159 Ionic compound A compound of positive and negative ions

held together by ionic bonds, 161 Ionic compounds, 161–164

chemical formulas, 162 properties, 161–162 subscripts in formulas, 163 writing ionic formulas from ionic charges, 163–164

Ionization energy The energy needed to remove the least tightly bound electron from the outermost energy level of an atom, 146

Ions and transfer of electrons, 157–161 common ions, formulas and names, 159t ionic charges from group numbers, 159 monatomic ions and nearest noble gases, 159t negative ions, 158–159 positive ions, 157–158

Ions in the body, 161, 161t Isotonic solutions, 385–386 Isotope An atom that differs only in mass number from

another atom of the same element. Isotopes have the same atomic number (number of protons), but different numbers of neutrons, 114, 114t

IUPAC system A naming system used for organic compounds, 543, 545t

J Joule (J) The SI unit of heat energy; 4.184 J = 1 cal, 80

K Kelvin (K) temperature scale A temperature scale on which

the lowest possible temperature is 0 K, 29 Ketone An organic compound in which a carbonyl group

(C “ O) is bonded to two alkyl or aromatic groups, 565 Kilogram (kg) A metric mass of 1000 g, equal to 2.205 lb.

The kilogram is the SI standard unit of mass, 29 Kinetic energy The energy of moving particles, 80 Kinetic molecular theory of gases A model used to explain

the behavior of gases, 312

Heat of reaction The heat (symbol ∆H) absorbed or released when a reaction takes place at constant pressure, 256

Heat of vaporization The energy required to vaporize 1 g of a substance at its boiling point. For water, 2260 J is needed to vaporize exactly 1 g of water; 1 g of steam gives off 2260 J when it condenses, 293–294

Hemodialysis, 387 Henry’s law The solubility of a gas in a liquid is directly

related to the pressure of that gas above the liquid, 359 Hess’s law Heat can be absorbed or released in a single chem-

ical reaction or in several steps, 257–259 Heterogeneous equilibrium An equilibrium system in which

the components are in different states, 408 Heterogeneous mixture, 70 Histologist, 269 Homeostasis and regulation of body temperature, 419 Homogeneous equilibrium An equilibrium system in which

all components are in the same state, 408 Homogeneous mixture, 70 Hydration The process of surrounding dissolved ions by

water molecules, 354 Hydrocarbon A type of organic compound that contains only

carbon and hydrogen, 542 Hydrogenation, 553–554 Hydrogenation of unsaturated fats, 554 Hydrogen bond The attraction between a partially positive H

atom and a strongly electronegative atom of N, O, or F, 289

Hydronium ion, H3O + The ion formed by the attraction of a

hydrogen ion, H+, to a water molecule, 434 Hydronium calculation from pH, 452–453

Hyperbaric chambers, 340 Hyperglycemia and hypoglycemia, 595 Hypothesis An unverified explanation of a natural

phenomenon, 4 Hypotonic and hypertonic solutions, 386–387

I Ideal gas constant, R A numerical value that relates the

quantities P, V, n, and T in the ideal gas law, PV = nRT, 330

Ideal gas law A law that combines the four measured properties of a gas PV = nRT, 329–333

ideal gas constant R, 330 molar mass of a gas, 332

Indicator A substance added to a titration sample that changes color when the pH of the solution changes, 457

Induced-fit model A model of enzyme action in which the shape of a substrate and the active site of the enzyme adjust to give an optimal fit, 623

Inorganic compounds, 541, 541t Intermolecular forces, 288–291

dipole–dipole attractions, 289 dispersion forces, 289, 290t hydrogen bond, 289 intermolecular forces and melting points, 289, 290t

Z02_TIMB8119_06_SE_IDX.indd 8 11/30/18 11:23 AM

Glossary/Index I-9

Mass number The total number of protons and neutrons in the nucleus of an atom, 111

Mass of product calculation from limiting reactant, 250–252 Mass percent (m/m) The grams of solute in 100 g of solution,

364 Mass percent composition The percent by mass of the

elements in a formula, 194–197 Mass/volume percent (m/v) The grams of solute in 100. mL

of solution, 366 Materials engineer, 125 Math skills, 9–16 Matter The material that makes up a substance and has mass

and occupies space, 1, 69–71 chemical properties and changes, 73–74, 74t classifications, 69–71 physical properties and changes, 73, 74t states and properties, 72–75, 72t

Measured number A number obtained when a quantity is determined by using a measuring device, 31

Measurements, 27–67 Medical applications using radioactivity, 526–529, 526t

computed tomography (CT), 527 magnetic resonance imaging (MRI), 527 positron emission tomography (PET), 527 scans with radioisotopes, 527

Medical laboratory technologist, 431 Melting The change of state from a solid to a liquid, 292 Melting point (mp) The temperature at which a solid becomes

a liquid (melts). It is the same temperature as the freezing point, 292

Metal An element that is shiny, malleable, ductile, and a good conductor of heat and electricity. The metals are located to the left of the heavy zigzag line on the periodic table, 104

Metallic character A measure of how easily an element loses a valence electron, 147

Metalloid An element with properties of both metals and nonmetals located along the heavy zigzag line on the periodic table, 105

Meter (m) The metric unit for length that is slightly longer than a yard. The meter is the SI standard unit of length, 29

Metric conversion factors, 43 Metric system A system of measurement used by scientists

and in most countries of the world, 28, 28t Metric–U.S. system conversion factors, 43 Milliliter (mL) A metric unit of volume equal to one-thousandth

of a liter (0.001 L), 28 Mixture The physical combination of two or more substances

that are not chemically combined, 69–70 heterogeneous mixture, 70 homogeneous mixture, 70 solution, 70

Molality (m) The number of moles of solute particles in exactly 1 kg of solvent, 380

Molality and freezing point lowering/boiling point elevation, 378–84

Molarity (M) The number of moles of solute in exactly 1 L of solution, 368

L Lactose A disaccharide consisting of glucose and galactose

found in milk and milk products, 598 Law of Conservation of Mass In a chemical reaction, the

total mass of the reactants is equal to the total mass of the products; matter is neither lost nor gained, 240

Lead storage battery 494–495 Le Châtelier’s principle When a stress is placed on a system

at equilibrium, the equilibrium shifts to relieve that stress, 414–420

effect of catalyst on equilibrium, 416 effect of change in temperature on equilibrium, 418–419 effect of concentration changes on equilibrium, 415–416 effect of volume change on equilibrium, 416–417 effects of condition changes on equilibrium, 418t

Length, 29, 40 Lewis structure A structure drawn in which the valence elec-

trons of all the atoms are arranged to give octets except two electrons for hydrogen, 272

Lewis structures for molecules and polyatomic ions, 270–276

for atoms, 270–271, 270t for ionic compounds, 271 for molecular compounds, 271–272, 272t sharing electrons between atoms of different elements,

272, 272t Lewis symbol The representation of an atom that shows the

valence electrons as dots placed around the symbol of an element, 270

Limiting reactant The reactant used up during a chemical reaction, which limits the amount of product that can form, 247–252

calculating mass of product from limiting reactant, 250–252 calculating moles of product from limiting reactant,

248–249 Line-angle formula A simplified structure that shows a zigzag

line in which carbon atoms are represented as the ends of each line and as corners, 544–545

Linear The shape of a molecule that has two bonded atoms and no lone pairs, 279

Lipids A family of compounds that is nonpolar in nature and not soluble in water; includes fats, oils, waxes, and steroids, 605–612

Liquid A state of matter that takes the shape of its container but has a definite volume, 72

Liter (L) The metric unit for volume that is slightly larger than a quart, 28

Lithium-ion batteries, 494–495 Lock-and-key model of enzyme action, 623

M Magnetic resonance imaging (MRI), 527 Maltose A disaccharide consisting of two glucose units; it is

obtained from the hydrolysis of starch, 598 Mass A measure of the quantity of material in an object, 29 Mass calculation for chemical reactions, 245–247

Z02_TIMB8119_06_SE_IDX.indd 9 11/30/18 11:23 AM

I-10 Glossary/Index

beta particle, 509 biological effects of radiation, 510–511 gamma rays, 510 positron, 510 radiation, 509 radiation protection, 511–512, 511t radioisotope, 509 symbols, 509

Negative ions, 158–159 Neutral The term that describes a solution with equal

concentrations of H3O + and OH-, 445

Neutralization A reaction between an acid and a base to form water and a salt, 455

Neutralization equations, balancing, 455–456 Neutral solution, 445 Neutron A neutral subatomic particle having a mass of about

1 amu and found in the nucleus of an atom; its symbol is n or n0, 109

Nitrogen narcosis, 70 Noble gas An element in Group 8A (18) of the periodic table,

generally unreactive and seldom found in combination with other elements, that has eight electrons (helium has two electrons) in its outermost energy level, 104

Nonelectrolyte A substance that dissolves in water as molecules; its solution does not conduct an electrical current, 355

Nonmetal An element with little or no luster that is a poor conductor of heat and electricity. The nonmetals are located to the right of the heavy zigzag line on the periodic table, 104

Nonpolar covalent bond A covalent bond in which the electrons are shared equally between atoms, 284

Nonpolar molecule A molecule that has only nonpolar bonds or in which the bond dipoles cancel, 287

Nonpolar solutes, 354 Nuclear chemistry, 508–539 Nuclear fission and fusion, 529–532

chain reaction, 530 fission, 529 fusion, 530

Nuclear reactions, 512–519 alpha decay, 512–514 beta decay, 514–515 gamma emission, 516–517 radioactive decay, 512 radioactive isotope production, 517–518

Nucleic acid A large molecule composed of nucleotides; found as a double helix in DNA and as the single strands of RNA, 624–629

bases, 624 complementary base pairs, 626 DNA, 624 double helix, 626 nucleic acid structure, 626 nucleosides, 624–625 nucleotides, 624–625

Molar mass The mass in grams of 1 mol of an element equal numerically to its atomic mass. The molar mass of a compound is equal to the sum of the atomic masses of the elements in the formula, 188–190

calculations using, 190–194 of compound, 188–190

Molar mass of a gas, 332 Molar solubility, S, 423–424 Molar volume A volume of 22.4 L occupied by 1 mol of a gas

at STP conditions of 0 °C (273 K) and 1 atm, 328 Mole A group of atoms, molecules, or formula units that

contains 6.022 * 1023 of these items, 184–188 Avogadro’s number, 184–186 conversion factors using chemical formula, 186–187 of elements in chemical compound, 186

Molecular compound A combination of atoms in which stable electron configurations are attained by sharing electrons, 172

Molecular compounds and sharing electrons, 172–176 names and formulas of, 173–174, 173t writing formulas from the names of, 174–176

Molecular formula The actual formula that gives the number of atoms of each type of element in the compound, 197, 201–203

calculating, 202–203 vs. empirical formula, 201–202, 201t

Molecule The smallest unit of two or more atoms held together by covalent bonds, 287

Mole–mole factor A conversion factor that relates the number of moles of two compounds in an equation derived from their coefficients, 242

Mole–mole factors in calculations, 242–244 Mole relationships in chemical equations,

241–244 mole–mole factors, 242

Moles of product from limiting reactant calculation, 248–249 Monomer The small organic molecule that is repeated many

times in a polymer, 554 Monosaccharide A polyhydroxy compound that contains an

aldehyde or a ketone group, 593–594 Multiplication and division with measured

numbers, 36

N Names and formulas of molecular compounds,

173–174, 173t Naming and writing ionic formulas, 164–168, 164t

determination of variable charge, 165–166, 166t metals with variable charges, 165, 165t writing formulas from the name of an ionic

compound, 167–168 Naming ionic compounds with polyatomic

ions, 171–172, 171t Natural radioactivity, 509–512, 510t

alpha particle, 509

Z02_TIMB8119_06_SE_IDX.indd 10 11/30/18 11:23 AM

Glossary/Index I-11

in acidic solution, 485–486 in basic solution, 486–488 using half-reactions, 483–484

Oxidation–reduction reaction A reaction in which the oxidation of one reactant is always accompanied by the reduction of another reactant, 228–231, 477

in biological systems, 229–230 characteristics, 230t

Oxidation–reduction reactions requiring electricity, 497–499 electrolysis, 497 electrolysis of sodium chloride, 497 electrolytic cell, 497 electroplating, 498

Oxidizing agent The reactant that gains electrons and is reduced, 481

Oxidizing and reducing agents, 481–482 Oxygen–hemoglobin equilibrium and hypoxia, 417

P Partial pressure The pressure exerted by a single gas in a gas

mixture, 335–340 air, as mixture of gas, 336–337, 336t Dalton’s law, 336 gases collected over water, 338–340

Particles in solution, 380 p block The elements in Groups 3A (13) to 8A (18) in which

electrons fill the p orbitals, 140 Pentose sugars, 624 Peptides, 615–617 Percentage calculation, 12–13 Percent, ppm, and ppb conversion factors, 44 Percent yield The ratio of the actual yield for a reaction to the

theoretical yield possible for the reaction, 252–254 Period A horizontal row of elements in the periodic table, 103

Period 1: hydrogen and helium, 136 Period 2: lithium to neon, 136–137 Period 3: sodium to argon, 137–138 Period 4 and above, 141

Periodic table An arrangement of elements by increasing atomic number such that elements having similar chemi- cal behavior are grouped in vertical columns, 103–107

groups and periods, 103–104 names of groups, 104

pH A measure of the [H3O +] in a solution; pH = - log[H3O+],

449 buffer calculation, 460–462, 462t calculation of solutions, 449–450 scale, 447–454

Pharmacist, 156 Photon A packet of energy that has both particle and

wave characteristics and travels at the speed of light, 129

pH scale, 447–454 calculation of hydronium from pH, 452–453 calculation of pH of solutions, 449–450

pentose sugars, 624 replication, 628–629 RNA, 624 structure, 626

Nucleoside The combination of a pentose sugar and a base, 624–625

Nucleotide Building block of a nucleic acid consisting of a base, a pentose sugar (ribose or deoxyribose), and a phosphate group, 624–625

Nucleus The compact, extremely dense center of an atom, containing the protons and neutrons of the atom, 109

Number of electrons in sublevels, 134 Nutrition, 87–90, 88t

O Observation Information determined by noting and recording

a natural phenomenon, 3–4 Octet rule Elements in Groups 1A to 7A (1, 2, 13 to 17) react

with other elements by forming ionic or covalent bonds to produce a stable electron configuration, 157

Oil A triacylglycerol that is usually a liquid at room tempera- ture and is obtained from a plant source, 608

Open-chain structures, 594 Orbital The region around the nucleus of an atom where

electrons of certain energy are most likely to be found: s orbitals are spherical; p orbitals have two lobes, 132

Orbital capacity and electron spin, 134 Orbital diagram A diagram that shows the distribution of

electrons in the orbitals of the energy levels, 135 Orbital diagrams and electron configurations, 135–139

Period 1: hydrogen and helium, 136 Period 2: lithium to neon, 136–137 Period 3: sodium to argon, 137–138

Organic chemistry, 541 Organic compounds, 541 Organic compound classes, 552t Osmosis The flow of a solvent, usually water, through a semi-

permeable membrane into a solution of higher solute concentration, 385–388

dialysis, 387 hypotonic and hypertonic solutions, 386–387 isotonic solutions, 385–386 osmotic pressure, 385

Osmotic pressure The pressure that prevents the flow of water into the more concentrated solution, 385

Oxidation The loss of electrons by a substance. Biological oxidation may involve the addition of oxygen or the loss of hydrogen, 228, 477

Oxidation and reduction, 476–507 Oxidation number A number equal to zero in an element or

the charge of a monatomic ion; in molecular compounds and polyatomic ions, oxidation numbers are assigned using a set of rules, 478

Oxidation–reduction equation balancing, 483–488

Z02_TIMB8119_06_SE_IDX.indd 11 11/30/18 11:23 AM

I-12 Glossary/Index

Products The substances formed as a result of a chemical reaction, 215

Protein A term used for biologically active polypeptides that have many amino acids linked together by peptide bonds, 617

Protein classifications, 612t Proteins as enzymes, 622–623

active site, 622 enzyme-catalyzed reaction, 622 enzymes, 622 induced-fit model of enzyme action, 623 lock-and-key model of enzyme action, 623

Protein structure, 617–622, 620t collagen, 618 primary structure, 617–618 quaternary structure, 620 secondary structure, 618 tertiary structure, 618–619, 619t

Protein synthesis, 629–634 codons, 630, 631t genetic code, 630–631 RNA types, 629t transcription, 629, 631–633 translation, 629

Proton A positively charged subatomic particle having a mass of about 1 amu and found in the nucleus of an atom; its symbol is p or p+, 108

Pure substance A type of matter that has a definite composition, 69

Q Quaternary structure A protein structure in which two or

more protein subunits form an active protein, 620

R Rad (radiation absorbed dose) A measure of an amount of

radiation absorbed by the body, 519 Radiation Energy or particles released by radioactive atoms,

509 Radiation and food, 520 Radiation measurement, 519–522

becquerel (Bq), 519 curie (Ci), 519 equivalent dose, 520 exposure to radiation, 521, 521t gray (Gy), 519 rad (radiation absorbed dose), 519 sievert (Sv), 520 units, radiation measurement, 520t

Radiation protection, 511–512, 511t Radiation sickness, 521, 521t Radiation technologist, 508 Radioactive decay The process by which an unstable nucleus

breaks down with the release of high-energy radiation, 512 Radioactive isotope production, 517–518 Radioisotope A radioactive atom of an element, 509, 509t

Physical change A change in which the physical properties of a substance change but its identity stays the same, 73, 74t

Physical properties The properties that can be observed or measured without affecting the identity of a substance, 73, 74t

Place value identification, 10 pOH A measure of the [OH-] in a solution;

pOH = - log[OH-], 451 Polar covalent bond A covalent bond in which the electrons

are shared unequally between atoms, 284 Polar molecule A molecule containing bond dipoles that do

not cancel, 287 Polarity A measure of the unequal sharing of electrons,

indicated by the difference in electronegativities, 284 Polarity of molecules, 287–288 Polyatomic ion A group of covalently bonded nonmetal atoms

that has an overall electrical charge, 168–172 names, 169, 169t naming ionic compounds with, 171–172, 171t writing formulas for compounds with, 170–171

Polymer A very large molecule that is composed of many small, repeating structural units that are identical, 554

Polymerization, 554–557, 555t Polysaccharide A polymer of many monosaccharide units,

usually glucose. Polysaccharides differ in the types of glycosidic bonds and the amount of branching in the polymer, 601–604

amylopectin, 601 amylose, 601 cellulose, 602 glycogen, 602

Positive and negative numbers, 11 Positive ions, 157–158 Positron A particle of radiation with no mass and a positive

charge, symbol b+ or +1 0e, produced when a proton is

transformed into a neutron and a positron, 510 Positron emission tomography (PET), 527 Potential energy A type of energy related to position or

composition of a substance, 80 Powers conversion factors, 45 Prefix The part of the name of a metric unit that precedes the

base unit and specifies the size of the measurement. All prefixes are related on a decimal scale, 39, 39

Pressure The force exerted by gas particles that hit the walls of a container, 312

Pressure and volume, 317–319 Boyle’s law, 317 inverse relationship, 317

Primary structure The specific sequence of the amino acids in a protein, 617–618

Principal quantum number (n) The number (n = 1, n = 2, ...) assigned to an energy level, 129

Problem solving, 47–53 conversion of units in a fraction, 51 two or more conversion factors, 49–50

Z02_TIMB8119_06_SE_IDX.indd 12 11/30/18 11:23 AM

Glossary/Index I-13

Second (s) A unit of time used in both SI and metric systems, 29

Secondary structure The formation of an a helix or b-pleated sheet by hydrogen bonds, 618

Shapes of molecules and polyatomic atoms, 279–283 bent, 280 central atom with four electron groups (tetrahedral),

280–283 central atoms with three electron groups (trigonal),

279–280 central atoms with two electron groups (linear), 279 trigonal pyramidal, 280–281, 281t

Shapes of orbitals, 132–133 Sharing electrons between atoms of different

elements, 272, 272t Sievert (Sv) A unit of biological damage equal to

100 rem, 520 Significant figures (SFs) The numbers recorded in a

measurement, 31, 31t, 34–38 Significant zeroes and scientific notation, 32 Single replacement reaction A reaction in which an element

replaces a different element in a compound, 224 Solid A state of matter that has its own shape and

volume, 72 Soluble and insoluble ionic compounds, 359–363 Solubility The maximum amount of solute that can dissolve in

100. g of solvent, usually water, at a given temperature, 357–363

effects of temperature, 359 saturated solution, 357 soluble and insoluble ionic compounds, 359–363 solubility rules, 360t unsaturated solution, 357

Solubility product constant, Ksp The product of the concentrations of the ions in a saturated solution of a slightly soluble ionic compound, with each concentration raised to a power equal to its coefficient in the balanced equilibrium equation, 421

Solubility product expression The product of the ion concentrations, with each concentration raised to an exponent equal to the coefficient in the balanced chemical equation, 421

Solubility rules A set of guidelines that states whether an ionic compound is soluble or insoluble in water, 359

Solute The component in a solution that is present in the lesser amount, 351

Solution A homogeneous mixture in which the solute is made up of small particles (ions or molecules) that can pass through filters and semipermeable membranes, 351

Solvent The substance in which the solute dissolves; usually the component present in greater amount, 351

Solving equations, 13–14 Specific gravity (sp gr) A relationship between the density of

a substance and the density of water, 57 Specific heat (SH) A quantity of heat that changes the

temperature of exactly 1 g of a substance by exactly 1 °C, 82–87

heat equation, 83–85

Radon in homes, 514 Rate of reaction The speed at which reactants form products,

399–403 Reactants The initial substances that undergo change in a

chemical reaction, 215 Reaction types summary, 226t Recycling code Arrows in a triangle found on the label or on

the bottom of a container, 555 Reducing agent The reactant that loses electrons and is

oxidized, 481 Reduction The gain of electrons by a substance. Biological

reduction may involve the loss of oxygen or the gain of hydrogen, 228, 477

Registered nurse, 27 Rem (radiation equivalent in humans) A measure of the

biological damage caused by the various kinds of radiation (rad * radiation biological factor), 519

Replication The process of duplicating DNA by pairing the bases on each parent strand with their complementary bases, 628–629

Representative element An element in the first two columns on the left of the periodic table and the last six columns on the right that has a group number of 1A through 8A or 1, 2, and 13 through 18, 103

Resonance structures Two or more Lewis structures that can be drawn for a molecule or polyatomic ion by placing a multiple bond between different atoms, 276–279

Respiratory therapist, 311 Reversible reaction A reaction in which a forward reaction

occurs from reactants to products and a reverse reaction occurs from products back to reactants, 403

RNA Ribonucleic acid; a type of nucleic acid that is a single strand of nucleotides containing ribose, phosphate, and the four bases: adenine, cytosine, guanine, and uracil, 624

RNA types, 629t Rounding off, 35–36, 35t

S Salt An ionic compound that contains a metal ion or NH4

+ and a nonmetal or polyatomic ion other than OH-, 454

Saturated solution A solution containing the maxi- mum amount of solute that can dissolve at a given temperature. Any additional solute will remain undissolved in the container, 357

s block The elements in Groups 1A (1) and 2A (2) in which electrons fill the s orbitals, 140

Scientific method The process of making observations, proposing a hypothesis, and testing the hypothesis; after repeated experiments validate the hypothesis, a conclusion may be drawn as to its validity, 3–5

Scientific notation A form of writing large and small numbers using a coefficient that is at least 1 but less than 10, followed by a power of 10, 17–20, 18t

calculators and, 19 conversion to a standard number, 19–20 powers of 10, 18t

Z02_TIMB8119_06_SE_IDX.indd 13 11/30/18 11:23 AM

I-14 Glossary/Index

vapor pressure and boiling point, 323 vapor pressure and boiling point of water, 324t vapor pressure of water, 323t

Temperature and rate of reaction, 401 Temperature and volume, 320–322

Charles’s law, 320 direct relationship, 320

Tertiary structure The folding of the secondary structure of a protein into a compact structure that is stabilized by the interactions of R groups such as salt bridges and disul- fide bonds, 618–619, 619t

Tetrahedral The shape of a molecule with four bonded atoms, 280

Theoretical yield The maximum amount of product that a reaction can produce from a given amount of reactant, 252

Titration The addition of base to an acid sample to determine the concentration of the acid, 457

Transcription The transfer of genetic information from DNA by the formation of mRNA, 629, 631–633

Transition element An element in the center of the periodic table that is designated with the letter “B” or the group number of 3 through 12, 103

Translation The interpretation of the codons in mRNA as amino acids in a peptide, 629

Trends, periodic properties, 143–147 atomic size, 145–146 group number and valence electrons, 144 ionization energy, 146 metallic character, 147

Triacylglycerols A family of lipids composed of three fatty acids bonded through ester bonds to glycerol, a trihy- droxy alcohol, 607–608, 610–611

Trigonal planar The shape of a molecule with three bonded atoms and no lone pairs, 279

Trigonal pyramidal The shape of a molecule that has three bonded atoms and one lone pair, 280

Triple bond A sharing of three pairs of electrons by two atoms, 276

Two or more conversion factors, 49–50

U Unsaturated solution A solution that contains less solute than

can be dissolved, 357

V Valence electrons The electrons in the highest energy level of

an atom, 144 Valence shell electron-pair repulsion (VSEPR) theory A the-

ory that predicts the shape of a molecule by moving the electron groups on a central atom as far apart as possible to minimize the mutual repulsion of the electrons, 279

Vapor pressure The pressure exerted by the particles of vapor above a liquid, 323

heat exchange, 85–86 Standard number, 19–20 States of matter Three forms of matter: solid, liquid, and gas,

72, 72t Steam burns, 297 Steroids Types of lipid composed of a multicyclic ring system,

611 Stomach acid, HCl, 453 STP Standard conditions of exactly 0 °C (273 K) temperature

and 1 atm pressure used for the comparison of gases, 328

Strategies to improve learning and understanding, 6–9 study habits, 6–7 study plan, 8–9

Strong acid An acid that completely dissociates in water, 437 Strong base A base that completely dissociates in water, 440 Strong electrolyte A compound that ionizes completely when

it dissolves in water; its solution is a good conductor of electricity, 355

Structural isomers Two compounds that have the same molecular formula but different arrangements of atoms, 545

Study habits, 6–7 Study plan, 8–9 Subatomic particle A particle within an atom; protons,

neutrons, and electrons are subatomic particles, 108 Sublevel A group of orbitals of equal energy within energy

levels. The number of sublevels in each energy level is the same as the principal quantum number (n), 131

Sublevel blocks, 140 Sublevels and orbitals, 131–134

number of electrons in sublevels, 134, 134t orbital capacity and electron spin, 134 shapes of orbitals, 132–133

Sublimation The change of state in which a solid is trans- formed directly to a gas without forming a liquid first, 292

Subscripts in formulas for ionic compounds, 163 Substituent A group of atoms such as an alkyl group or a

halogen bonded to the main chain of carbon atoms, 545 Substituents in alkanes, 546, 546t Sucrose A disaccharide composed of glucose and fructose;

commonly called table sugar or “sugar,” 599 Suspension A mixture in which the solute particles are large

enough and heavy enough to settle out and be retained by both filters and semipermeable membranes, 379

T Temperature An indicator of the hotness or coldness of an

object, 29, 75–78, 77t Celsius scale, 29 Fahrenheit scale, 29 Kelvin scale, 29

Temperature and pressure (Gay-Lussac’s law), 322–324 Gay-Lussac’s law, 322

Z02_TIMB8119_06_SE_IDX.indd 14 11/30/18 11:23 AM

Glossary/Index I-15

Water dissociation expression The product of [H3O +] and

[OH-] in solution; Kw = [H3O +][OH-], 445

Water in the body, 352–353 Wavelength The distance between adjacent crests or troughs

in a wave, 126 Waxes, 607, 607t Weak acid An acid that is a poor donor of H+ and dissociates

only slightly in water, 438 Weak base A base that is a poor acceptor of H+ and produces

only a small number of ions in water, 440 Weak electrolyte A substance that produces only a

few ions along with many molecules when it dissolves in water; its solution is a weak conductor of electricity, 355

Weight, 29 Weighted average, 115 Whole-number coefficients, 219–220 Writing formulas for compounds with polyatomic

ions, 170–171 Writing formulas from the names of molecular

compounds, 174–176 Writing ionic formulas from ionic charges, 163–164

boiling point, 323 boiling point of water, 324t water, 323t

Veterinarian, 183 Voltaic cell A type of cell with two compartments that uses

spontaneous oxidation–reduction reactions to produce electrical energy, 489

Volume (V) The amount of space occupied by a substance, 28 Volume and moles, 327–329

Avogadro’s law, 327 molar volume, 328 STP, 328

Volume change and effect on equilibrium, 416–417 Volume displacement, 55 Volume percent (v/v) A percent concentration that relates the

volume of the solute in 100. mL of solution, 365

W Water as solvent, 351 Water dissociation constant, Kw The numerical value

of the product of [H3O +] and [OH-] in solution;

Kw = 1.0 * 10-14, 445

Z02_TIMB8119_06_SE_IDX.indd 15 11/30/18 11:23 AM

Length SI Unit Meter (m) Volume SI Unit Cubic Meter (m3) Mass SI Unit Kilogram (kg)

1 meter (m) = 100 centimeters (cm) 1 liter (L) = 1000 milliliters (mL) 1 kilogram (kg) = 1000 grams (g) 1 mg = 1000 mcg 1 m = 1000 millimeters (mm) 1 mL = 1 cm3 1 dl = 100 mL 1 g = 1000 milligrams (mg) 1 lb = 16 oz 1 cm = 10 mm 1 L = 1.057 quart (qt) 1 mL = 1cc 1 kg = 2.205 lb 1 kilometer (km) = 0.6214 mile (mi) 1 qt = 946.4 mL 1 lb = 453.6 g 1 inch (in.) = 2.54 cm (exact) 1 pint (pt) = 473.2 mL 1 mol = 6.022 * 1023 particles

1 gal = 3.785 L Water 1 tsp = 5 mL 1 tbsp (T) = 15 mL

density = 1.00 g/mL (at 4 °C)

Temperature SI Unit Kelvin (K) Pressure SI Unit Pascal (Pa) Energy SI Unit Joule (J)

TF = 1.8(TC) + 32 1 atm = 760 mmHg 1 calorie (cal) = 4.184 J (exact)

TC = TF - 32

1.8 1 atm = 101.325 kPa 1 kcal = 1000 cal

TK = TC + 273 1 atm = 760 Torr 1 mol of gas = 22.4 L (STP) R = 0.0821 L # atm / mol # K R = 62.4 mmHg # atm / mol # K

Water Heat of fusion = 334 J/g Heat of vaporization = 2260 J/g Specific heat (SH) = 4.184 J/g °C; 1.00 cal/g °C

Metric and SI Prefixes

Prefix Symbol Power of Ten

Prefixes That Increase the Size of the Unit

peta P 1015

tera T 1012

giga G 109

mega M 106

kilo k 103

Prefixes That Decrease the Size of the Unit deci d 10-1

centi c 10-2

milli m 10-3

micro m (mc) 10-6

nano n 10-9

pico p 10-12

femto f 10-15

Formulas and Molar Masses of Some Typical Compounds

Name Formula Molar Mass

(g/mol)

Ammonia NH3 17.03 Ammonium chloride NH4Cl 53.49 Ammonium sulfate (NH4)2SO4 132.15 Bromine Br2 159.80 Butane C4H10 58.12 Calcium carbonate CaCO3 100.09 Calcium chloride CaCl2 110.98 Calcium hydroxide Ca(OH)2 74.10 Calcium oxide CaO 56.08 Carbon dioxide CO2 44.01 Chlorine Cl2 70.90 Copper(II) sulfide CuS 95.62 Hydrogen H2 2.016

Metric and SI Units and Some Useful Conversion Factors

Name Formula Molar Mass

(g/mol)

Hydrogen chloride HCl 36.46 Iron(III) oxide Fe2O3 159.70 Magnesium oxide MgO 40.31 Methane CH4 16.04 Nitrogen N2 28.02 Oxygen O2 32.00 Potassium carbonate K2CO3 138.21 Potassium nitrate KNO3 101.11 Propane C3H8 44.09 Sodium chloride NaCl 58.44 Sodium hydroxide NaOH 40.00 Sulfur trioxide SO3 80.07 Water H2O 18.02

CVR_TIMB8119_06_SE_BEP.indd 2 11/29/18 1:24 PM

Formulas and Charges of Some Common Cations

Cations (Fixed Charge)

1+ 2+ 3+

Li+ Lithium Mg2+ Magnesium Al3+ Aluminum

Na+ Sodium Ca2+ Calcium

K+ Potassium Sr2+ Strontium

NH4 + Ammonium Ba2+ Barium

H3O + Hydronium Zn2+ Zinc

Ag + Silver Cd2+ Cadmium

Cations (Variable Charge)

1+ or 2+ 1+ or 3+

Cu+ Copper(I) Cu2+ Copper(II) Au+ Gold(I) Au3+ Gold(III)

Hg2 2+ Mercury(I) Hg2+ Mercury(II)

2+ or 3+ 2+ or 4+

Fe2+ Iron(II) Fe3+ Iron(III) Sn2+ Tin(II) Sn4+ Tin(IV)

Co2+ Cobalt(II) Co3+ Cobalt(III) Pb2+ Lead(II) Pb4+ Lead(IV)

Cr2+ Chromium(II) Cr3+ Chromium(III)

Mn2+ Manganese(II) Mn3+ Manganese(III)

Ni2+ Nickel(II) Ni3+ Nickel(III)

3+ or 5+

Bi3+ Bismuth(III) Bi5+ Bismuth(V)

Formulas and Charges of Some Common Anions

Monatomic Ions

F - Fluoride Br- Bromide O2- Oxide N3- Nitride

Cl- Chloride I- Iodide S2- Sulfide P3- Phosphide

Polyatomic Ions

HCO3 - Hydrogen carbonate (bicarbonate) CO3

2- Carbonate C2H3O2

- Acetate CN- Cyanide NO3

- Nitrate NO2 - Nitrite

H2PO4 - Dihydrogen phosphate HPO4

2- Hydrogen phosphate PO4 3- Phosphate

H2PO3 - Dihydrogen phosphite HPO3

2- Hydrogen phosphite PO3 3- Phosphite

HSO4 - Hydrogen sulfate (bisulfate) SO4

2- Sulfate

HSO3 - Hydrogen sulfite (bisulfite) SO3

2- Sulfite ClO4

- Perchlorate ClO3 - Chlorate

ClO2 - Chlorite ClO- Hypochlorite

OH- Hydroxide CrO4 2- Chromate

MnO4 - Permanganate Cr2O7

2- Dichromate

SCN- Thiocyanate

Functional Groups in Organic Compounds

Type Functional Group

Haloalkane ¬ F, ¬ Cl, ¬ Br, or ¬ I Alkene ¬ CH “ CH ¬ Alkyne ¬ C ‚ C ¬ Aromatic Benzene ring Alcohol ¬ OH Ether ¬ O ¬

Aldehyde

O

C H

Ketone

O

C

Type Functional Group

Carboxylic acid

O

C OH

Ester

O

C O

Amine ¬ NH2

Amide

O

C NH2

CVR_TIMB8119_06_SE_BEP.indd 3 11/29/18 1:24 PM

First published in July 2021.

New Enterprise House St Helens Street Derby DE1 3GY UK

email: [email protected]

Copyright © 2021 David Icke

No part of this book may be reproduced in any form without permission from the

Publisher, except for the quotation of brief passages in criticism

Cover Design: Gareth Icke

Book Design: Neil Hague

British Library Cataloguing-in Publication Data

A catalogue record for this book is

available from the British Library

eISBN 978-18384153-1-0

Dedication:

To Freeeeeedom!

Renegade:

Adjective

‘Having rejected tradition: Unconventional.’

Merriam-Webster Dictionary

Acquiescence to tyranny is the death of the spirit

You may be 38 years old, as I happen to be. And one day,

some great opportunity stands before you and calls you to

stand up for some great principle, some great issue, some

great cause. And you refuse to do it because you are afraid

… You refuse to do it because you want to live longer …

You’re afraid that you will lose your job, or you are afraid

that you will be criticised or that you will lose your

popularity, or you’re afraid that somebody will stab you, or

shoot at you or bomb your house; so you refuse to take the

stand.

Well, you may go on and live until you are 90, but you’re just

as dead at 38 as you would be at 90. And the cessation of

breathing in your life is but the belated announcement of an

earlier death of the spirit.

Martin Luther King

How the few control the many and always have – the many do whatever they’re told

‘Forward, the Light Brigade!’

Was there a man dismayed?

Not though the soldier knew

Someone had blundered.

Theirs not to make reply,

Theirs not to reason why,

Theirs but to do and die.

Into the valley of Death

Rode the six hundred.

Cannon to right of them,

Cannon to le� of them,

Cannon in front of them

Volleyed and thundered;

Stormed at with shot and shell,

Boldly they rode and well,

Into the jaws of Death,

Into the mouth of hell

Rode the six hundred

Alfred Lord Tennyson (1809-1892)

The mist is li�ing slowly

I can see the way ahead

And I’ve le� behind the empty streets

That once inspired my life

And the strength of the emotion

Is like thunder in the air

’Cos the promise that we made each other

Haunts me to the end

The secret of your beauty

And the mystery of your soul

I’ve been searching for in everyone I meet

And the times I’ve been mistaken

It’s impossible to say

And the grass is growing

Underneath our feet

The words that I remember

From my childhood still are true

That there’s none so blind

As those who will not see

And to those who lack the courage

And say it’s dangerous to try

Well they just don’t know

That love eternal will not be denied

I know you’re out there somewhere

Somewhere, somewhere

I know you’re out there somewhere

Somewhere you can hear my voice

I know I’ll find you somehow

Somehow, somehow

I know I’ll find you somehow

And somehow I’ll return again to you

The Moody Blues

Are you a gutless wonder - or a Renegade Mind?

Monuments put from pen to paper,

Turns me into a gutless wonder,

And if you tolerate this,

Then your children will be next.

Gravity keeps my head down,

Or is it maybe shame ...

Manic Street Preachers

Rise like lions a�er slumber

In unvanquishable number.

Shake your chains to earth like dew

Which in sleep have fallen on you.

Ye are many – they are few.

Percy Shelley

CHAPTER 1

CHAPTER 2

CHAPTER 3

CHAPTER 4

CHAPTER 5

CHAPTER 6

CHAPTER 7

CHAPTER 8

CHAPTER 9

CHAPTER 10

CHAPTER 11

CHAPTER 12

Postscript

APPENDIX

BIBLIOGRAPHY

INDEX

Contents

‘I’m thinking’ – Oh, but are you? Renegade perception The Pushbacker sting ‘Covid’: The calculated catastrophe There is no ‘virus’ Sequence of deceit War on your mind ‘Reframing’ insanity We must have it? So what is it? Human 2.0 Who controls the Cult? Escaping Wetiko     Cowan-Kaufman-Morell Statement on Virus Isolation    

F

CHAPTER ONE

I’m thinking’ – Oh, but are you?

Think for yourself and let others enjoy the privilege of doing so too

Voltaire

rench-born philosopher, mathematician and scientist René

Descartes became famous for his statement in Latin in the 17th

century which translates into English as: ‘I think, therefore I am.’

On the face of it that is true. Thought reflects perception and

perception leads to both behaviour and self-identity. In that sense

‘we’ are what we think. But who or what is doing the thinking and is

thinking the only route to perception? Clearly, as we shall see, ‘we’

are not always the source of ‘our’ perception, indeed with regard to

humanity as a whole this is rarely the case; and thinking is far from

the only means of perception. Thought is the village idiot compared

with other expressions of consciousness that we all have the

potential to access and tap into. This has to be true when we are

those other expressions of consciousness which are infinite in nature.

We have forgo�en this, or, more to the point, been manipulated to

forget.

These are not just the esoteric musings of the navel. The whole

foundation of human control and oppression is control of

perception. Once perception is hijacked then so is behaviour which

is dictated by perception. Collective perception becomes collective

behaviour and collective behaviour is what we call human society.

Perception is all and those behind human control know that which is

why perception is the target 24/7 of the psychopathic manipulators

that I call the Global Cult. They know that if they dictate perception

they will dictate behaviour and collectively dictate the nature of

human society. They are further aware that perception is formed

from information received and if they control the circulation of

information they will to a vast extent direct human behaviour.

Censorship of information and opinion has become globally Nazi-

like in recent years and never more blatantly than since the illusory

‘virus pandemic’ was triggered out of China in 2019 and across the

world in 2020. Why have billions submi�ed to house arrest and

accepted fascistic societies in a way they would have never believed

possible? Those controlling the information spewing from

government, mainstream media and Silicon Valley (all controlled by

the same Global Cult networks) told them they were in danger from

a ‘deadly virus’ and only by submi�ing to house arrest and

conceding their most basic of freedoms could they and their families

be protected. This monumental and provable lie became the

perception of the billions and therefore the behaviour of the billions. In

those few words you have the whole structure and modus operandi

of human control. Fear is a perception – False Emotion Appearing

Real – and fear is the currency of control. In short … get them by the

balls (or give them the impression that you have) and their hearts

and minds will follow. Nothing grips the dangly bits and freezes the

rear-end more comprehensively than fear.

World number 1

There are two ‘worlds’ in what appears to be one ‘world’ and the

prime difference between them is knowledge. First we have the mass

of human society in which the population is maintained in coldly-

calculated ignorance through control of information and the

‘education’ (indoctrination) system. That’s all you really need to

control to enslave billions in a perceptual delusion in which what are

perceived to be their thoughts and opinions are ever-repeated

mantras that the system has been downloading all their lives

through ‘education’, media, science, medicine, politics and academia

in which the personnel and advocates are themselves

overwhelmingly the perceptual products of the same repetition.

Teachers and academics in general are processed by the same

programming machine as everyone else, but unlike the great

majority they never leave the ‘education’ program. It gripped them

as students and continues to grip them as programmers of

subsequent generations of students. The programmed become the

programmers – the programmed programmers. The same can

largely be said for scientists, doctors and politicians and not least

because as the American writer Upton Sinclair said: ‘It is difficult to

get a man to understand something when his salary depends upon

his not understanding it.’ If your career and income depend on

thinking the way the system demands then you will – bar a few free-

minded exceptions – concede your mind to the Perceptual

Mainframe that I call the Postage Stamp Consensus. This is a tiny

band of perceived knowledge and possibility ‘taught’ (downloaded)

in the schools and universities, pounded out by the mainstream

media and on which all government policy is founded. Try thinking,

and especially speaking and acting, outside of the ‘box’ of consensus

and see what that does for your career in the Mainstream Everything

which bullies, harasses, intimidates and ridicules the population into

compliance. Here we have the simple structure which enslaves most

of humanity in a perceptual prison cell for an entire lifetime and I’ll

go deeper into this process shortly. Most of what humanity is taught

as fact is nothing more than programmed belief. American science

fiction author Frank Herbert was right when he said: ‘Belief can be

manipulated. Only knowledge is dangerous.’ In the ‘Covid’ age

belief is promoted and knowledge is censored. It was always so, but

never to the extreme of today.

World number 2

A ‘number 2’ is slang for ‘doing a poo’ and how appropriate that is

when this other ‘world’ is doing just that on humanity every minute

of every day. World number 2 is a global network of secret societies

and semi-secret groups dictating the direction of society via

governments, corporations and authorities of every kind. I have

spent more than 30 years uncovering and exposing this network that

I call the Global Cult and knowing its agenda is what has made my

books so accurate in predicting current and past events. Secret

societies are secret for a reason. They want to keep their hoarded

knowledge to themselves and their chosen initiates and to hide it

from the population which they seek through ignorance to control

and subdue. The whole foundation of the division between World 1

and World 2 is knowledge. What number 1 knows number 2 must not.

Knowledge they have worked so hard to keep secret includes (a) the

agenda to enslave humanity in a centrally-controlled global

dictatorship, and (b) the nature of reality and life itself. The la�er (b)

must be suppressed to allow the former (a) to prevail as I shall be

explaining. The way the Cult manipulates and interacts with the

population can be likened to a spider’s web. The ‘spider’ sits at the

centre in the shadows and imposes its will through the web with

each strand represented in World number 2 by a secret society,

satanic or semi-secret group, and in World number 1 – the world of

the seen – by governments, agencies of government, law

enforcement, corporations, the banking system, media

conglomerates and Silicon Valley (Fig 1 overleaf). The spider and the

web connect and coordinate all these organisations to pursue the

same global outcome while the population sees them as individual

entities working randomly and independently. At the level of the

web governments are the banking system are the corporations are the

media are Silicon Valley are the World Health Organization working

from their inner cores as one unit. Apparently unconnected

countries, corporations, institutions, organisations and people are on

the same team pursuing the same global outcome. Strands in the web

immediately around the spider are the most secretive and exclusive

secret societies and their membership is emphatically restricted to

the Cult inner-circle emerging through the generations from

particular bloodlines for reasons I will come to. At the core of the

core you would get them in a single room. That’s how many people

are dictating the direction of human society and its transformation

through the ‘Covid’ hoax and other means. As the web expands out

from the spider we meet the secret societies that many people will be

aware of – the Freemasons, Knights Templar, Knights of Malta, Opus

Dei, the inner sanctum of the Jesuit Order, and such like. Note how

many are connected to the Church of Rome and there is a reason for

that. The Roman Church was established as a revamp, a rebranding,

of the relocated ‘Church’ of Babylon and the Cult imposing global

tyranny today can be tracked back to Babylon and Sumer in what is

now Iraq.

Figure 1: The global web through which the few control the many. (Image Neil Hague.)

Inner levels of the web operate in the unseen away from the public

eye and then we have what I call the cusp organisations located at

the point where the hidden meets the seen. They include a series of

satellite organisations answering to a secret society founded in

London in the late 19th century called the Round Table and among

them are the Royal Institute of International Affairs (UK, founded in

1920); Council on Foreign Relations (US, 1921); Bilderberg Group

(worldwide, 1954); Trilateral Commission (US/worldwide, 1972); and

the Club of Rome (worldwide, 1968) which was created to exploit

environmental concerns to justify the centralisation of global power

to ‘save the planet’. The Club of Rome instigated with others the

human-caused climate change hoax which has led to all the ‘green

new deals’ demanding that very centralisation of control. Cusp

organisations, which include endless ‘think tanks’ all over the world,

are designed to coordinate a single global policy between political

and business leaders, intelligence personnel, media organisations

and anyone who can influence the direction of policy in their own

sphere of operation. Major players and regular a�enders will know

what is happening – or some of it – while others come and go and

are kept overwhelmingly in the dark about the big picture. I refer to

these cusp groupings as semi-secret in that they can be publicly

identified, but what goes on at the inner-core is kept very much ‘in

house’ even from most of their members and participants through a

fiercely-imposed system of compartmentalisation. Only let them

know what they need to know to serve your interests and no more.

The structure of secret societies serves as a perfect example of this

principle. Most Freemasons never get higher than the bo�om three

levels of ‘degree’ (degree of knowledge) when there are 33 official

degrees of the Sco�ish Rite. Initiates only qualify for the next higher

‘compartment’ or degree if those at that level choose to allow them.

Knowledge can be carefully assigned only to those considered ‘safe’.

I went to my local Freemason’s lodge a few years ago when they

were having an ‘open day’ to show how cuddly they were and when

I cha�ed to some of them I was astonished at how li�le the rank and

file knew even about the most ubiquitous symbols they use. The

mushroom technique – keep them in the dark and feed them bullshit

– applies to most people in the web as well as the population as a

whole. Sub-divisions of the web mirror in theme and structure

transnational corporations which have a headquarters somewhere in

the world dictating to all their subsidiaries in different countries.

Subsidiaries operate in their methodology and branding to the same

centrally-dictated plan and policy in pursuit of particular ends. The

Cult web functions in the same way. Each country has its own web

as a subsidiary of the global one. They consist of networks of secret

societies, semi-secret groups and bloodline families and their job is

to impose the will of the spider and the global web in their particular

country. Subsidiary networks control and manipulate the national

political system, finance, corporations, media, medicine, etc. to

ensure that they follow the globally-dictated Cult agenda. These

networks were the means through which the ‘Covid’ hoax could be

played out with almost every country responding in the same way.

The ‘Yessir’ pyramid

Compartmentalisation is the key to understanding how a tiny few

can dictate the lives of billions when combined with a top-down

sequence of imposition and acquiescence. The inner core of the Cult

sits at the peak of the pyramidal hierarchy of human society (Fig 2

overleaf). It imposes its will – its agenda for the world – on the level

immediately below which acquiesces to that imposition. This level

then imposes the Cult will on the level below them which acquiesces

and imposes on the next level. Very quickly we meet levels in the

hierarchy that have no idea there even is a Cult, but the sequence of

imposition and acquiescence continues down the pyramid in just the

same way. ‘I don’t know why we are doing this but the order came

from “on-high” and so we be�er just do it.’ Alfred Lord Tennyson

said of the cannon fodder levels in his poem The Charge of the Light

Brigade: ‘Theirs not to reason why; theirs but to do and die.’ The next

line says that ‘into the valley of death rode the six hundred’ and they

died because they obeyed without question what their perceived

‘superiors’ told them to do. In the same way the population

capitulated to ‘Covid’. The whole hierarchical pyramid functions

like this to allow the very few to direct the enormous many.

Eventually imposition-acquiescence-imposition-acquiescence comes

down to the mass of the population at the foot of the pyramid. If

they acquiesce to those levels of the hierarchy imposing on them

(governments/law enforcement/doctors/media) a circuit is

completed between the population and the handful of super-

psychopaths in the Cult inner core at the top of the pyramid.

Without a circuit-breaking refusal to obey, the sequence of

imposition and acquiescence allows a staggeringly few people to

impose their will upon the entirety of humankind. We are looking at

the very sequence that has subjugated billions since the start of 2020.

Our freedom has not been taken from us. Humanity has given it

away. Fascists do not impose fascism because there are not enough

of them. Fascism is imposed by the population acquiescing to

fascism. Put another way allowing their perceptions to be

programmed to the extent that leads to the population giving their

freedom away by giving their perceptions – their mind – away. If this

circuit is not broken by humanity ceasing to cooperate with their

own enslavement then nothing can change. For that to happen

people have to critically think and see through the lies and window

dressing and then summon the backbone to act upon what they see.

The Cult spends its days working to stop either happening and its

methodology is systematic and highly detailed, but it can be

overcome and that is what this book is all about.

Figure 2: The simple sequence of imposition and compliance that allows a handful of people at the peak of the pyramid to dictate the lives of billions.

The Life Program

Okay, back to world number 1 or the world of the ‘masses’. Observe

the process of what we call ‘life’ and it is a perceptual download

from cradle to grave. The Cult has created a global structure in

which perception can be programmed and the program continually

topped-up with what appears to be constant confirmation that the

program is indeed true reality. The important word here is ‘appears’.

This is the structure, the fly-trap, the Postage Stamp Consensus or

Perceptual Mainframe, which represents that incredibly narrow

band of perceived possibility delivered by the ‘education’ system,

mainstream media, science and medicine. From the earliest age the

download begins with parents who have themselves succumbed to

the very programming their children are about to go through. Most

parents don’t do this out of malevolence and mostly it is quite the

opposite. They do what they believe is best for their children and

that is what the program has told them is best. Within three or four

years comes the major transition from parental programming to full-

blown state (Cult) programming in school, college and university

where perceptually-programmed teachers and academics pass on

their programming to the next generations. Teachers who resist are

soon marginalised and their careers ended while children who resist

are called a problem child for whom Ritalin may need to be

prescribed. A few years a�er entering the ‘world’ children are under

the control of authority figures representing the state telling them

when they have to be there, when they can leave and when they can

speak, eat, even go to the toilet. This is calculated preparation for a

lifetime of obeying authority in all its forms. Reflex-action fear of

authority is instilled by authority from the start. Children soon learn

the carrot and stick consequences of obeying or defying authority

which is underpinned daily for the rest of their life. Fortunately I

daydreamed through this crap and never obeyed authority simply

because it told me to. This approach to my alleged ‘be�ers’ continues

to this day. There can be consequences of pursuing open-minded

freedom in a world of closed-minded conformity. I spent a lot of time

in school corridors a�er being ejected from the classroom for not

taking some of it seriously and now I spend a lot of time being

ejected from Facebook, YouTube and Twi�er. But I can tell you that

being true to yourself and not compromising your self-respect is far

more exhilarating than bowing to authority for authority’s sake. You

don’t have to be a sheep to the shepherd (authority) and the sheep

dog (fear of not obeying authority).

The perceptual download continues throughout the formative

years in school, college and university while script-reading

‘teachers’, ‘academics’ ‘scientists’, ‘doctors’ and ‘journalists’ insist

that ongoing generations must be as programmed as they are.

Accept the program or you will not pass your ‘exams’ which confirm

your ‘degree’ of programming. It is tragic to think that many parents

pressure their offspring to work hard at school to download the

program and qualify for the next stage at college and university. The

late, great, American comedian George Carlin said: ‘Here’s a bumper

sticker I’d like to see: We are proud parents of a child who has

resisted his teachers’ a�empts to break his spirit and bend him to the

will of his corporate masters.’ Well, the best of luck finding many of

those, George. Then comes the moment to leave the formal

programming years in academia and enter the ‘adult’ world of work.

There you meet others in your chosen or prescribed arena who went

through the same Postage Stamp Consensus program before you

did. There is therefore overwhelming agreement between almost

everyone on the basic foundations of Postage Stamp reality and the

rejection, even contempt, of the few who have a mind of their own

and are prepared to use it. This has two major effects. Firstly, the

consensus confirms to the programmed that their download is really

how things are. I mean, everyone knows that, right? Secondly, the

arrogance and ignorance of Postage Stamp adherents ensure that

anyone questioning the program will have unpleasant consequences

for seeking their own truth and not picking their perceptions from

the shelf marked: ‘Things you must believe without question and if

you don’t you’re a dangerous lunatic conspiracy theorist and a

harebrained nu�er’.

Every government, agency and corporation is founded on the

same Postage Stamp prison cell and you can see why so many

people believe the same thing while calling it their own ‘opinion’.

Fusion of governments and corporations in pursuit of the same

agenda was the definition of fascism described by Italian dictator

Benito Mussolini. The pressure to conform to perceptual norms

downloaded for a lifetime is incessant and infiltrates society right

down to family groups that become censors and condemners of their

own ‘black sheep’ for not, ironically, being sheep. We have seen an

explosion of that in the ‘Covid’ era. Cult-owned global media

unleashes its propaganda all day every day in support of the Postage

Stamp and targets with abuse and ridicule anyone in the public eye

who won’t bend their mind to the will of the tyranny. Any response

to this is denied (certainly in my case). They don’t want to give a

platform to expose official lies. Cult-owned-and-created Internet

giants like Facebook, Google, YouTube and Twi�er delete you for

having an unapproved opinion. Facebook boasts that its AI censors

delete 97-percent of ‘hate speech’ before anyone even reports it.

Much of that ‘hate speech’ will simply be an opinion that Facebook

and its masters don’t want people to see. Such perceptual oppression

is widely known as fascism. Even Facebook executive Benny

Thomas, a ‘CEO Global Planning Lead’, said in comments secretly

recorded by investigative journalism operation Project Veritas that

Facebook is ‘too powerful’ and should be broken up:

I mean, no king in history has been the ruler of two billion people, but Mark Zuckerberg is … And he’s 36. That’s too much for a 36-year-old ... You should not have power over two billion people. I just think that’s wrong.

Thomas said Facebook-owned platforms like Instagram, Oculus, and

WhatsApp needed to be separate companies. ‘It’s too much power

when they’re all one together’. That’s the way the Cult likes it,

however. We have an executive of a Cult organisation in Benny

Thomas that doesn’t know there is a Cult such is the

compartmentalisation. Thomas said that Facebook and Google ‘are

no longer companies, they’re countries’. Actually they are more

powerful than countries on the basis that if you control information

you control perception and control human society.

I love my oppressor

Another expression of this psychological trickery is for those who

realise they are being pressured into compliance to eventually

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convince themselves to believe the official narratives to protect their

self-respect from accepting the truth that they have succumbed to

meek and subservient compliance. Such people become some of the

most vehement defenders of the system. You can see them

everywhere screaming abuse at those who prefer to think for

themselves and by doing so reminding the compliers of their own

capitulation to conformity. ‘You are talking dangerous nonsense you

Covidiot!!’ Are you trying to convince me or yourself? It is a potent

form of Stockholm syndrome which is defined as: ‘A psychological

condition that occurs when a victim of abuse identifies and a�aches,

or bonds, positively with their abuser.’ An example is hostages

bonding and even ‘falling in love’ with their kidnappers. The

syndrome has been observed in domestic violence, abused children,

concentration camp inmates, prisoners of war and many and various

Satanic cults. These are some traits of Stockholm syndrome listed at

goodtherapy.org:

Positive regard towards perpetrators of abuse or captor [see

‘Covid’].

Failure to cooperate with police and other government authorities

when it comes to holding perpetrators of abuse or kidnapping

accountable [or in the case of ‘Covid’ cooperating with the police

to enforce and defend their captors’ demands].

Li�le or no effort to escape [see ‘Covid’].

Belief in the goodness of the perpetrators or kidnappers [see

‘Covid’].

Appeasement of captors. This is a manipulative strategy for

maintaining one’s safety. As victims get rewarded – perhaps with

less abuse or even with life itself – their appeasing behaviours are

reinforced [see ‘Covid’].

Learned helplessness. This can be akin to ‘if you can’t beat ‘em,

join ‘em’. As the victims fail to escape the abuse or captivity, they

may start giving up and soon realize it’s just easier for everyone if

they acquiesce all their power to their captors [see ‘Covid’].

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Feelings of pity toward the abusers, believing they are actually

victims themselves. Because of this, victims may go on a crusade

or mission to ‘save’ [protect] their abuser [see the venom

unleashed on those challenging the official ‘Covid’ narrative].

Unwillingness to learn to detach from their perpetrators and heal.

In essence, victims may tend to be less loyal to themselves than to

their abuser [ definitely see ‘Covid’].

Ponder on those traits and compare them with the behaviour of

great swathes of the global population who have defended

governments and authorities which have spent every minute

destroying their lives and livelihoods and those of their children and

grandchildren since early 2020 with fascistic lockdowns, house arrest

and employment deletion to ‘protect’ them from a ‘deadly virus’ that

their abusers’ perceptually created to bring about this very outcome.

We are looking at mass Stockholm syndrome. All those that agree to

concede their freedom will believe those perceptions are originating

in their own independent ‘mind’ when in fact by conceding their

reality to Stockholm syndrome they have by definition conceded any

independence of mind. Listen to the ‘opinions’ of the acquiescing

masses in this ‘Covid’ era and what gushes forth is the repetition of

the official version of everything delivered unprocessed, unfiltered

and unquestioned. The whole programming dynamic works this

way. I must be free because I’m told that I am and so I think that I

am.

You can see what I mean with the chapter theme of ‘I’m thinking –

Oh, but are you?’ The great majority are not thinking, let alone for

themselves. They are repeating what authority has told them to

believe which allows them to be controlled. Weaving through this

mentality is the fear that the ‘conspiracy theorists’ are right and this

again explains the o�en hysterical abuse that ensues when you dare

to contest the official narrative of anything. Denial is the mechanism

of hiding from yourself what you don’t want to be true. Telling

people what they want to hear is easy, but it’s an infinitely greater

challenge to tell them what they would rather not be happening.

One is akin to pushing against an open door while the other is met

with vehement resistance no ma�er what the scale of evidence. I

don’t want it to be true so I’ll convince myself that it’s not. Examples

are everywhere from the denial that a partner is cheating despite all

the signs to the reflex-action rejection of any idea that world events

in which country a�er country act in exactly the same way are

centrally coordinated. To accept the la�er is to accept that a force of

unspeakable evil is working to destroy your life and the lives of your

children with nothing too horrific to achieve that end. Who the heck

wants that to be true? But if we don’t face reality the end is duly

achieved and the consequences are far worse and ongoing than

breaking through the walls of denial today with the courage to make

a stand against tyranny.

Connect the dots – but how?

A crucial aspect of perceptual programming is to portray a world in

which everything is random and almost nothing is connected to

anything else. Randomness cannot be coordinated by its very nature

and once you perceive events as random the idea they could be

connected is waved away as the rantings of the tinfoil-hat brigade.

You can’t plan and coordinate random you idiot! No, you can’t, but

you can hide the coldly-calculated and long-planned behind the

illusion of randomness. A foundation manifestation of the Renegade

Mind is to scan reality for pa�erns that connect the apparently

random and turn pixels and dots into pictures. This is the way I

work and have done so for more than 30 years. You look for

similarities in people, modus operandi and desired outcomes and

slowly, then ever quicker, the picture forms. For instance: There

would seem to be no connection between the ‘Covid pandemic’ hoax

and the human-caused global-warming hoax and yet they are masks

(appropriately) on the same face seeking the same outcome. Those

pushing the global warming myth through the Club of Rome and

other Cult agencies are driving the lies about ‘Covid’ – Bill Gates is

an obvious one, but they are endless. Why would the same people be

involved in both when they are clearly not connected? Oh, but they

are. Common themes with personnel are matched by common goals.

The ‘solutions’ to both ‘problems’ are centralisation of global power

to impose the will of the few on the many to ‘save’ humanity from

‘Covid’ and save the planet from an ‘existential threat’ (we need

‘zero Covid’ and ‘zero carbon emissions’). These, in turn, connect

with the ‘dot’ of globalisation which was coined to describe the

centralisation of global power in every area of life through incessant

political and corporate expansion, trading blocks and superstates

like the European Union. If you are the few and you want to control

the many you have to centralise power and decision-making. The

more you centralise power the more power the few at the centre will

have over the many; and the more that power is centralised the more

power those at the centre have to centralise even quicker. The

momentum of centralisation gets faster and faster which is exactly

the process we have witnessed. In this way the hoaxed ‘pandemic’

and the fakery of human-caused global warming serve the interests

of globalisation and the seizure of global power in the hands of the

Cult inner-circle which is behind ‘Covid’, ‘climate change’ and

globalisation. At this point random ‘dots’ become a clear and

obvious picture or pa�ern.

Klaus Schwab, the classic Bond villain who founded the Cult’s

Gates-funded World Economic Forum, published a book in 2020, The

Great Reset, in which he used the ‘problem’ of ‘Covid’ to justify a

total transformation of human society to ‘save’ humanity from

‘climate change’. Schwab said: ‘The pandemic represents a rare but

narrow window of opportunity to reflect, reimagine, and reset our

world.’ What he didn’t mention is that the Cult he serves is behind

both hoaxes as I show in my book The Answer. He and the Cult don’t

have to reimagine the world. They know precisely what they want

and that’s why they destroyed human society with ‘Covid’ to ‘build

back be�er’ in their grand design. Their job is not to imagine, but to

get humanity to imagine and agree with their plans while believing

it’s all random. It must be pure coincidence that ‘The Great Reset’

has long been the Cult’s code name for the global imposition of

fascism and replaced previous code-names of the ‘New World

Order’ used by Cult frontmen like Father George Bush and the ‘New

Order of the Ages’ which emerged from Freemasonry and much

older secret societies. New Order of the Ages appears on the reverse

of the Great Seal of the United States as ‘Novus ordo seclorum’

underneath the Cult symbol used since way back of the pyramid and

all seeing-eye (Fig 3). The pyramid is the hierarchy of human control

headed by the illuminated eye that symbolises the force behind the

Cult which I will expose in later chapters. The term ‘Annuit Coeptis’

translates as ‘He favours our undertaking’. We are told the ‘He’ is

the Christian god, but ‘He’ is not as I will be explaining.

Figure 3: The all-seeing eye of the Cult ‘god’ on the Freemason-designed Great Seal of the United States and also on the dollar bill.

Having you on

Two major Cult techniques of perceptual manipulation that relate to

all this are what I have called since the 1990s Problem-Reaction-

Solution (PRS) and the Totalitarian Tiptoe (TT). They can be

uncovered by the inquiring mind with a simple question: Who

benefits? The answer usually identifies the perpetrators of a given

action or happening through the concept of ‘he who most benefits

from a crime is the one most likely to have commi�ed it’. The Latin

‘Cue bono?’ – Who benefits? – is widely a�ributed to the Roman

orator and statesman Marcus Tullius Cicero. No wonder it goes back

so far when the concept has been relevant to human behaviour since

history was recorded. Problem-Reaction-Solution is the technique

used to manipulate us every day by covertly creating a problem (or

the illusion of one) and offering the solution to the problem (or the

illusion of one). In the first phase you create the problem and blame

someone or something else for why it has happened. This may relate

to a financial collapse, terrorist a�ack, war, global warming or

pandemic, anything in fact that will allow you to impose the

‘solution’ to change society in the way you desire at that time. The

‘problem’ doesn’t have to be real. PRS is manipulation of perception

and all you need is the population to believe the problem is real.

Human-caused global warming and the ‘Covid pandemic’ only have

to be perceived to be real for the population to accept the ‘solutions’ of

authority. I refer to this technique as NO-Problem-Reaction-Solution.

Billions did not meekly accept house arrest from early 2020 because

there was a real deadly ‘Covid pandemic’ but because they

perceived – believed – that to be the case. The antidote to Problem-

Reaction-Solution is to ask who benefits from the proposed solution.

Invariably it will be anyone who wants to justify more control

through deletion of freedom and centralisation of power and

decision-making.

The two world wars were Problem-Reaction-Solutions that

transformed and realigned global society. Both were manipulated

into being by the Cult as I have detailed in books since the mid-

1990s. They dramatically centralised global power, especially World

War Two, which led to the United Nations and other global bodies

thanks to the overt and covert manipulations of the Rockefeller

family and other Cult bloodlines like the Rothschilds. The UN is a

stalking horse for full-blown world government that I will come to

shortly. The land on which the UN building stands in New York was

donated by the Rockefellers and the same Cult family was behind

Big Pharma scalpel and drug ‘medicine’ and the creation of the

World Health Organization as part of the UN. They have been

stalwarts of the eugenics movement and funded Hitler’s race-purity

expert’ Ernst Rudin. The human-caused global warming hoax has

been orchestrated by the Club of Rome through the UN which is

manufacturing both the ‘problem’ through its Intergovernmental

Panel on Climate Change and imposing the ‘solution’ through its

Agenda 21 and Agenda 2030 which demand the total centralisation

of global power to ‘save the world’ from a climate hoax the United

Nations is itself perpetrating. What a small world the Cult can be

seen to be particularly among the inner circles. The bedfellow of

Problem-Reaction-Solution is the Totalitarian Tiptoe which became

the Totalitarian Sprint in 2020. The technique is fashioned to hide the

carefully-coordinated behind the cover of apparently random events.

You start the sequence at ‘A’ and you know you are heading for ‘Z’.

You don’t want people to know that and each step on the journey is

presented as a random happening while all the steps strung together

lead in the same direction. The speed may have quickened

dramatically in recent times, but you can still see the incremental

approach of the Tiptoe in the case of ‘Covid’ as each new imposition

takes us deeper into fascism. Tell people they have to do this or that

to get back to ‘normal’, then this and this and this. With each new

demand adding to the ones that went before the population’s

freedom is deleted until it disappears. The spider wraps its web

around the flies more comprehensively with each new diktat. I’ll

highlight this in more detail when I get to the ‘Covid’ hoax and how

it has been pulled off. Another prime example of the Totalitarian

Tiptoe is how the Cult-created European Union went from a ‘free-

trade zone’ to a centralised bureaucratic dictatorship through the

Tiptoe of incremental centralisation of power until nations became

mere administrative units for Cult-owned dark suits in Brussels.

The antidote to ignorance is knowledge which the Cult seeks

vehemently to deny us, but despite the systematic censorship to that

end the Renegade Mind can overcome this by vociferously seeking

out the facts no ma�er the impediments put in the way. There is also

a method of thinking and perceiving – knowing – that doesn’t even

need names, dates, place-type facts to identify the pa�erns that

reveal the story. I’ll get to that in the final chapter. All you need to

know about the manipulation of human society and to what end is

still out there – at the time of writing – in the form of books, videos

and websites for those that really want to breach the walls of

programmed perception. To access this knowledge requires the

abandonment of the mainstream media as a source of information in

the awareness that this is owned and controlled by the Cult and

therefore promotes mass perceptions that suit the Cult. Mainstream

media lies all day, every day. That is its function and very reason for

being. Where it does tell the truth, here and there, is only because the

truth and the Cult agenda very occasionally coincide. If you look for

fact and insight to the BBC, CNN and virtually all the rest of them

you are asking to be conned and perceptually programmed.

Know the outcome and you’ll see the journey

Events seem random when you have no idea where the world is

being taken. Once you do the random becomes the carefully

planned. Know the outcome and you’ll see the journey is a phrase I

have been using for a long time to give context to daily happenings

that appear unconnected. Does a problem, or illusion of a problem,

trigger a proposed ‘solution’ that further drives society in the

direction of the outcome? Invariably the answer will be yes and the

random – abracadabra – becomes the clearly coordinated. So what is

this outcome that unlocks the door to a massively expanded

understanding of daily events? I will summarise its major aspects –

the fine detail is in my other books – and those new to this

information will see that the world they thought they were living in

is a very different place. The foundation of the Cult agenda is the

incessant centralisation of power and all such centralisation is

ultimately in pursuit of Cult control on a global level. I have

described for a long time the planned world structure of top-down

dictatorship as the Hunger Games Society. The term obviously

comes from the movie series which portrayed a world in which a

few living in military-protected hi-tech luxury were the overlords of

a population condemned to abject poverty in isolated ‘sectors’ that

were not allowed to interact. ‘Covid’ lockdowns and travel bans

anyone? The ‘Hunger Games’ pyramid of structural control has the

inner circle of the Cult at the top with pre�y much the entire

population at the bo�om under their control through dependency

for survival on the Cult. The whole structure is planned to be

protected and enforced by a military-police state (Fig 4).

Here you have the reason for the global lockdowns of the fake

pandemic to coldly destroy independent incomes and livelihoods

and make everyone dependent on the ‘state’ (the Cult that controls

the ‘states’). I have warned in my books for many years about the

plan to introduce a ‘guaranteed income’ – a barely survivable

pi�ance – designed to impose dependency when employment was

destroyed by AI technology and now even more comprehensively at

great speed by the ‘Covid’ scam. Once the pandemic was played and

lockdown consequences began to delete independent income the

authorities began to talk right on cue about the need for a

guaranteed income and a ‘Great Reset’. Guaranteed income will be

presented as benevolent governments seeking to help a desperate

people – desperate as a direct result of actions of the same

governments. The truth is that such payments are a trap. You will

only get them if you do exactly what the authorities demand

including mass vaccination (genetic manipulation). We have seen

this theme already in Australia where those dependent on

government benefits have them reduced if parents don’t agree to

have their children vaccinated according to an insane health-

destroying government-dictated schedule. Calculated economic

collapse applies to governments as well as people. The Cult wants

rid of countries through the creation of a world state with countries

broken up into regions ruled by a world government and super

states like the European Union. Countries must be bankrupted, too,

to this end and it’s being achieved by the trillions in ‘rescue

packages’ and furlough payments, trillions in lost taxation, and

money-no-object spending on ‘Covid’ including constant all-

medium advertising (programming) which has made the media

dependent on government for much of its income. The day of

reckoning is coming – as planned – for government spending and

given that it has been made possible by printing money and not by

production/taxation there is inflation on the way that has the

potential to wipe out monetary value. In that case there will be no

need for the Cult to steal your money. It just won’t be worth

anything (see the German Weimar Republic before the Nazis took

over). Many have been okay with lockdowns while ge�ing a

percentage of their income from so-called furlough payments

without having to work. Those payments are dependent, however,

on people having at least a theoretical job with a business considered

non-essential and ordered to close. As these business go under

because they are closed by lockdown a�er lockdown the furlough

stops and it will for everyone eventually. Then what? The ‘then

what?’ is precisely the idea.

Figure 4: The Hunger Games Society structure I have long warned was planned and now the ‘Covid’ hoax has made it possible. This is the real reason for lockdowns.

Hired hands

Between the Hunger Games Cult elite and the dependent population

is planned to be a vicious military-police state (a fusion of the two

into one force). This has been in the making for a long time with

police looking ever more like the military and carrying weapons to

match. The pandemic scam has seen this process accelerate so fast as

lockdown house arrest is brutally enforced by carefully recruited

fascist minds and gormless system-servers. The police and military

are planned to merge into a centrally-directed world army in a

global structure headed by a world government which wouldn’t be

elected even by the election fixes now in place. The world army is

not planned even to be human and instead wars would be fought,

primarily against the population, using robot technology controlled

by artificial intelligence. I have been warning about this for decades

and now militaries around the world are being transformed by this

very AI technology. The global regime that I describe is a particular

form of fascism known as a technocracy in which decisions are not

made by clueless and co-opted politicians but by unelected

technocrats – scientists, engineers, technologists and bureaucrats.

Cult-owned-and-controlled Silicon Valley giants are examples of

technocracy and they already have far more power to direct world

events than governments. They are with their censorship selecting

governments. I know that some are calling the ‘Great Reset’ a

Marxist communist takeover, but fascism and Marxism are different

labels for the same tyranny. Tell those who lived in fascist Germany

and Stalinist Russia that there was a difference in the way their

freedom was deleted and their lives controlled. I could call it a fascist

technocracy or a Marxist technocracy and they would be equally

accurate. The Hunger Games society with its world government

structure would oversee a world army, world central bank and single

world cashless currency imposing its will on a microchipped

population (Fig 5). Scan its different elements and see how the

illusory pandemic is forcing society in this very direction at great

speed. Leaders of 23 countries and the World Health Organization

(WHO) backed the idea in March, 2021, of a global treaty for

‘international cooperation’ in ‘health emergencies’ and nations

should ‘come together as a global community for peaceful

cooperation that extends beyond this crisis’. Cut the Orwellian

bullshit and this means another step towards global government.

The plan includes a cashless digital money system that I first warned

about in 1993. Right at the start of ‘Covid’ the deeply corrupt Tedros

Adhanom Ghebreyesus, the crooked and merely gofer ‘head’ of the

World Health Organization, said it was possible to catch the ‘virus’

by touching cash and it was be�er to use cashless means. The claim

was ridiculous nonsense and like the whole ‘Covid’ mind-trick it

was nothing to do with ‘health’ and everything to do with pushing

every aspect of the Cult agenda. As a result of the Tedros lie the use

of cash has plummeted. The Cult script involves a single world

digital currency that would eventually be technologically embedded

in the body. China is a massive global centre for the Cult and if you

watch what is happening there you will know what is planned for

everywhere. The Chinese government is developing a digital

currency which would allow fines to be deducted immediately via

AI for anyone caught on camera breaking its fantastic list of laws

and the money is going to be programmable with an expiry date to

ensure that no one can accrue wealth except the Cult and its

operatives.

Figure 5: The structure of global control the Cult has been working towards for so long and this has been enormously advanced by the ‘Covid’ illusion.

Serfdom is so smart

The Cult plan is far wider, extreme, and more comprehensive than

even most conspiracy researchers appreciate and I will come to the

true depths of deceit and control in the chapters ‘Who controls the

Cult?’ and ‘Escaping Wetiko’. Even the world that we know is crazy

enough. We are being deluged with ever more sophisticated and

controlling technology under the heading of ‘smart’. We have smart

televisions, smart meters, smart cards, smart cars, smart driving,

smart roads, smart pills, smart patches, smart watches, smart skin,

smart borders, smart pavements, smart streets, smart cities, smart

communities, smart environments, smart growth, smart planet ...

smart everything around us. Smart technologies and methods of

operation are designed to interlock to create a global Smart Grid

connecting the entirety of human society including human minds to

create a centrally-dictated ‘hive’ mind. ‘Smart cities’ is code for

densely-occupied megacities of total surveillance and control

through AI. Ever more destructive frequency communication

systems like 5G have been rolled out without any official testing for

health and psychological effects (colossal). 5G/6G/7G systems are

needed to run the Smart Grid and each one becomes more

destructive of body and mind. Deleting independent income is

crucial to forcing people into these AI-policed prisons by ending

private property ownership (except for the Cult elite). The Cult’s

Great Reset now openly foresees a global society in which no one

will own any possessions and everything will be rented while the

Cult would own literally everything under the guise of government

and corporations. The aim has been to use the lockdowns to destroy

sources of income on a mass scale and when the people are destitute

and in unrepayable amounts of debt (problem) Cult assets come

forward with the pledge to write-off debt in return for handing over

all property and possessions (solution). Everything – literally

everything including people – would be connected to the Internet

via AI. I was warning years ago about the coming Internet of Things

(IoT) in which all devices and technology from your car to your

fridge would be plugged into the Internet and controlled by AI.

Now we are already there with much more to come. The next stage

is the Internet of Everything (IoE) which is planned to include the

connection of AI to the human brain and body to replace the human

mind with a centrally-controlled AI mind. Instead of perceptions

being manipulated through control of information and censorship

those perceptions would come direct from the Cult through AI.

What do you think? You think whatever AI decides that you think.

In human terms there would be no individual ‘think’ any longer. Too

incredible? The ravings of a lunatic? Not at all. Cult-owned crazies

in Silicon Valley have been telling us the plan for years without

explaining the real motivation and calculated implications. These

include Google executive and ‘futurist’ Ray Kurzweil who highlights

the year 2030 for when this would be underway. He said:

Our thinking ... will be a hybrid of biological and non-biological thinking ... humans will be able to extend their limitations and ‘think in the cloud’ ... We’re going to put gateways to the cloud in our brains ... We’re going to gradually merge and enhance ourselves ... In my view, that’s the nature of being human – we transcend our limitations.

As the technology becomes vastly superior to what we are then the small proportion that is still human gets smaller and smaller and smaller until it’s just utterly negligible.

The sales-pitch of Kurzweil and Cult-owned Silicon Valley is that

this would make us ‘super-human’ when the real aim is to make us

post-human and no longer ‘human’ in the sense that we have come

to know. The entire global population would be connected to AI and

become the centrally-controlled ‘hive-mind’ of externally-delivered

perceptions. The Smart Grid being installed to impose the Cult’s will

on the world is being constructed to allow particular locations – even

one location – to control the whole global system. From these prime

control centres, which absolutely include China and Israel, anything

connected to the Internet would be switched on or off and

manipulated at will. Energy systems could be cut, communication

via the Internet taken down, computer-controlled driverless

autonomous vehicles driven off the road, medical devices switched

off, the potential is limitless given how much AI and Internet

connections now run human society. We have seen nothing yet if we

allow this to continue. Autonomous vehicle makers are working

with law enforcement to produce cars designed to automatically pull

over if they detect a police or emergency vehicle flashing from up to

100 feet away. At a police stop the car would be unlocked and the

window rolled down automatically. Vehicles would only take you

where the computer (the state) allowed. The end of petrol vehicles

and speed limiters on all new cars in the UK and EU from 2022 are

steps leading to electric computerised transport over which

ultimately you have no control. The picture is far bigger even than

the Cult global network or web and that will become clear when I

get to the nature of the ‘spider’. There is a connection between all

these happenings and the instigation of DNA-manipulating

‘vaccines’ (which aren’t ‘vaccines’) justified by the ‘Covid’ hoax. That

connection is the unfolding plan to transform the human body from

a biological to a synthetic biological state and this is why synthetic

biology is such a fast-emerging discipline of mainstream science.

‘Covid vaccines’ are infusing self-replicating synthetic genetic

material into the cells to cumulatively take us on the Totalitarian

Tiptoe from Human 1.0 to the synthetic biological Human 2.0 which

will be physically and perceptually a�ached to the Smart Grid to one

hundred percent control every thought, perception and deed.

Humanity needs to wake up and fast.

This is the barest explanation of where the ‘outcome’ is planned to

go but it’s enough to see the journey happening all around us. Those

new to this information will already see ‘Covid’ in a whole new

context. I will add much more detail as we go along, but for the

minutiae evidence see my mega-works, The Answer, The Trigger and

Everything You Need to Know But Have Never Been Told.

Now – how does a Renegade Mind see the ‘world’?

A

CHAPTER TWO

Renegade Perception

It is one thing to be clever and another to be wise

George R.R. Martin

simple definition of the difference between a programmed

mind and a Renegade Mind would be that one sees only dots

while the other connects them to see the picture. Reading reality

with accuracy requires the observer to (a) know the planned

outcome and (b) realise that everything, but everything, is connected.

The entirety of infinite reality is connected – that’s its very nature –

and with human society an expression of infinite reality the same

must apply. Simple cause and effect is a connection. The effect is

triggered by the cause and the effect then becomes the cause of

another effect. Nothing happens in isolation because it can’t. Life in

whatever reality is simple choice and consequence. We make choices

and these lead to consequences. If we don’t like the consequences we

can make different choices and get different consequences which

lead to other choices and consequences. The choice and the

consequence are not only connected they are indivisible. You can’t

have one without the other as an old song goes. A few cannot

control the world unless those being controlled allow that to happen

– cause and effect, choice and consequence. Control – who has it and

who doesn’t – is a two-way process, a symbiotic relationship,

involving the controller and controlled. ‘They took my freedom

away!!’ Well, yes, but you also gave it to them. Humanity is

subjected to mass control because humanity has acquiesced to that

control. This is all cause and effect and literally a case of give and

take. In the same way world events of every kind are connected and

the Cult works incessantly to sell the illusion of the random and

coincidental to maintain the essential (to them) perception of dots

that hide the picture. Renegade Minds know this and constantly

scan the world for pa�erns of connection. This is absolutely pivotal

in understanding the happenings in the world and without that

perspective clarity is impossible. First you know the planned

outcome and then you identify the steps on the journey – the day-by-

day apparently random which, when connected in relation to the

outcome, no longer appear as individual events, but as the

proverbial chain of events leading in the same direction. I’ll give you

some examples:

Political puppet show

We are told to believe that politics is ‘adversarial’ in that different

parties with different beliefs engage in an endless tussle for power.

There may have been some truth in that up to a point – and only a

point – but today divisions between ‘different’ parties are rhetorical

not ideological. Even the rhetorical is fusing into one-speak as the

parties eject any remaining free thinkers while others succumb to the

ever-gathering intimidation of anyone with the ‘wrong’ opinion. The

Cult is not a new phenomenon and can be traced back thousands of

years as my books have documented. Its intergenerational initiates

have been manipulating events with increasing effect the more that

global power has been centralised. In ancient times the Cult secured

control through the system of monarchy in which ‘special’

bloodlines (of which more later) demanded the right to rule as kings

and queens simply by birthright and by vanquishing others who

claimed the same birthright. There came a time, however, when

people had matured enough to see the unfairness of such tyranny

and demanded a say in who governed them. Note the word –

governed them. Not served them – governed them, hence government

defined as ‘the political direction and control exercised over the

actions of the members, citizens, or inhabitants of communities,

societies, and states; direction of the affairs of a state, community,

etc.’ Governments exercise control over rather than serve just like the

monarchies before them. Bizarrely there are still countries like the

United Kingdom which are ruled by a monarch and a government

that officially answers to the monarch. The UK head of state and that

of Commonwealth countries such as Canada, Australia and New

Zealand is ‘selected’ by who in a single family had unprotected sex

with whom and in what order. Pinch me it can’t be true. Ouch! Shit,

it is. The demise of monarchies in most countries offered a potential

vacuum in which some form of free and fair society could arise and

the Cult had that base covered. Monarchies had served its interests

but they couldn’t continue in the face of such widespread opposition

and, anyway, replacing a ‘royal’ dictatorship that people could see

with a dictatorship ‘of the people’ hiding behind the concept of

‘democracy’ presented far greater manipulative possibilities and

ways of hiding coordinated tyranny behind the illusion of ‘freedom’.

Democracy is quite wrongly defined as government selected by

the population. This is not the case at all. It is government selected

by some of the population (and then only in theory). This ‘some’

doesn’t even have to be the majority as we have seen so o�en in first-

past-the-post elections in which the so-called majority party wins

fewer votes than the ‘losing’ parties combined. Democracy can give

total power to a party in government from a minority of the votes

cast. It’s a sleight of hand to sell tyranny as freedom. Seventy-four

million Trump-supporting Americans didn’t vote for the

‘Democratic’ Party of Joe Biden in the distinctly dodgy election in

2020 and yet far from acknowledging the wishes and feelings of that

great percentage of American society the Cult-owned Biden

government set out from day one to destroy them and their right to a

voice and opinion. Empty shell Biden and his Cult handlers said

they were doing this to ‘protect democracy’. Such is the level of

lunacy and sickness to which politics has descended. Connect the

dots and relate them to the desired outcome – a world government

run by self-appointed technocrats and no longer even elected

politicians. While operating through its political agents in

government the Cult is at the same time encouraging public distain

for politicians by pu�ing idiots and incompetents in theoretical

power on the road to deleting them. The idea is to instil a public

reaction that says of the technocrats: ‘Well, they couldn’t do any

worse than the pathetic politicians.’ It’s all about controlling

perception and Renegade Minds can see through that while

programmed minds cannot when they are ignorant of both the

planned outcome and the manipulation techniques employed to

secure that end. This knowledge can be learned, however, and fast if

people choose to get informed.

Politics may at first sight appear very difficult to control from a

central point. I mean look at the ‘different’ parties and how would

you be able to oversee them all and their constituent parts? In truth,

it’s very straightforward because of their structure. We are back to

the pyramid of imposition and acquiescence. Organisations are

structured in the same way as the system as a whole. Political parties

are not open forums of free expression. They are hierarchies. I was a

national spokesman for the British Green Party which claimed to be

a different kind of politics in which influence and power was

devolved; but I can tell you from direct experience – and it’s far

worse now – that Green parties are run as hierarchies like all the

others however much they may try to hide that fact or kid

themselves that it’s not true. A very few at the top of all political

parties are directing policy and personnel. They decide if you are

elevated in the party or serve as a government minister and to do

that you have to be a yes man or woman. Look at all the maverick

political thinkers who never ascended the greasy pole. If you want to

progress within the party or reach ‘high-office’ you need to fall into

line and conform. Exceptions to this are rare indeed. Should you

want to run for parliament or Congress you have to persuade the

local or state level of the party to select you and for that you need to

play the game as dictated by the hierarchy. If you secure election and

wish to progress within the greater structure you need to go on

conforming to what is acceptable to those running the hierarchy

from the peak of the pyramid. Political parties are perceptual gulags

and the very fact that there are party ‘Whips’ appointed to ‘whip’

politicians into voting the way the hierarchy demands exposes the

ridiculous idea that politicians are elected to serve the people they

are supposed to represent. Cult operatives and manipulation has

long seized control of major parties that have any chance of forming

a government and at least most of those that haven’t. A new party

forms and the Cult goes to work to infiltrate and direct. This has

reached such a level today that you see video compilations of

‘leaders’ of all parties whether Democrats, Republicans,

Conservative, Labour and Green parroting the same Cult mantra of

‘Build Back Be�er’ and the ‘Great Reset’ which are straight off the

Cult song-sheet to describe the transformation of global society in

response to the Cult-instigated hoaxes of the ‘Covid pandemic’ and

human-caused ‘climate change’. To see Caroline Lucas, the Green

Party MP that I knew when I was in the party in the 1980s, speaking

in support of plans proposed by Cult operative Klaus Schwab

representing the billionaire global elite is a real head-shaker.

Many parties – one master

The party system is another mind-trick and was instigated to change

the nature of the dictatorship by swapping ‘royalty’ for dark suits

that people believed – though now ever less so – represented their

interests. Understanding this trick is to realise that a single force (the

Cult) controls all parties either directly in terms of the major ones or

through manipulation of perception and ideology with others. You

don’t need to manipulate Green parties to demand your

transformation of society in the name of ‘climate change’ when they

are obsessed with the lie that this is essential to ‘save the planet’. You

just give them a platform and away they go serving your interests

while believing they are being environmentally virtuous. America’s

political structure is a perfect blueprint for how the two or multi-

party system is really a one-party state. The Republican Party is

controlled from one step back in the shadows by a group made up of

billionaires and their gofers known as neoconservatives or Neocons.

I have exposed them in fine detail in my books and they were the

driving force behind the policies of the imbecilic presidency of Boy

George Bush which included 9/11 (see The Trigger for a

comprehensive demolition of the official story), the subsequent ‘war

on terror’ (war of terror) and the invasions of Afghanistan and Iraq.

The la�er was a No-Problem-Reaction-Solution based on claims by

Cult operatives, including Bush and British Prime Minister Tony

Blair, about Saddam Hussein’s ‘weapons of mass destruction’ which

did not exist as war criminals Bush and Blair well knew.

Figure 6: Different front people, different parties – same control system.

The Democratic Party has its own ‘Neocon’ group controlling

from the background which I call the ‘Democons’ and here’s the

penny-drop – the Neocons and Democons answer to the same

masters one step further back into the shadows (Fig 6). At that level

of the Cult the Republican and Democrat parties are controlled by

the same people and no ma�er which is in power the Cult is in

power. This is how it works in almost every country and certainly in

Britain with Conservative, Labour, Liberal Democrat and Green

parties now all on the same page whatever the rhetoric may be in

their feeble a�empts to appear different. Neocons operated at the

time of Bush through a think tank called The Project for the New

American Century which in September, 2000, published a document

entitled Rebuilding America’s Defenses: Strategies, Forces, and Resources

For a New Century demanding that America fight ‘multiple,

simultaneous major theatre wars’ as a ‘core mission’ to force regime-

change in countries including Iraq, Libya and Syria. Neocons

arranged for Bush (‘Republican’) and Blair (‘Labour Party’) to front-

up the invasion of Iraq and when they departed the Democons

orchestrated the targeting of Libya and Syria through Barack Obama

(‘Democrat’) and British Prime Minister David Cameron

(‘Conservative Party’). We have ‘different’ parties and ‘different’

people, but the same unfolding script. The more the Cult has seized

the reigns of parties and personnel the more their policies have

transparently pursued the same agenda to the point where the

fascist ‘Covid’ impositions of the Conservative junta of Jackboot

Johnson in Britain were opposed by the Labour Party because they

were not fascist enough. The Labour Party is likened to the US

Democrats while the Conservative Party is akin to a British version

of the Republicans and on both sides of the Atlantic they all speak

the same language and support the direction demanded by the Cult

although some more enthusiastically than others. It’s a similar story

in country a�er country because it’s all centrally controlled. Oh, but

what about Trump? I’ll come to him shortly. Political ‘choice’ in the

‘party’ system goes like this: You vote for Party A and they get into

government. You don’t like what they do so next time you vote for

Party B and they get into government. You don’t like what they do

when it’s pre�y much the same as Party A and why wouldn’t that be

with both controlled by the same force? Given that only two,

sometimes three, parties have any chance of forming a government

to get rid of Party B that you don’t like you have to vote again for

Party A which … you don’t like. This, ladies and gentlemen, is what

they call ‘democracy’ which we are told – wrongly – is a term

interchangeable with ‘freedom’.

The cult of cults

At this point I need to introduce a major expression of the Global

Cult known as Sabbatian-Frankism. Sabbatian is also spelt as

Sabbatean. I will summarise here. I have published major exposés

and detailed background in other works. Sabbatian-Frankism

combines the names of two frauds posing as ‘Jewish’ men, Sabbatai

Zevi (1626-1676), a rabbi, black magician and occultist who

proclaimed he was the Jewish messiah; and Jacob Frank (1726-1791),

the Polish ‘Jew’, black magician and occultist who said he was the

reincarnation of ‘messiah’ Zevi and biblical patriarch Jacob. They

worked across two centuries to establish the Sabbatian-Frankist cult

that plays a major, indeed central, role in the manipulation of human

society by the Global Cult which has its origins much further back in

history than Sabbatai Zevi. I should emphasise two points here in

response to the shrill voices that will scream ‘anti-Semitism’: (1)

Sabbatian-Frankists are NOT Jewish and only pose as such to hide

their cult behind a Jewish façade; and (2) my information about this

cult has come from Jewish sources who have long realised that their

society and community has been infiltrated and taken over by

interloper Sabbatian-Frankists. Infiltration has been the foundation

technique of Sabbatian-Frankism from its official origin in the 17th

century. Zevi’s Sabbatian sect a�racted a massive following

described as the biggest messianic movement in Jewish history,

spreading as far as Africa and Asia, and he promised a return for the

Jews to the ‘Promised Land’ of Israel. Sabbatianism was not Judaism

but an inversion of everything that mainstream Judaism stood for. So

much so that this sinister cult would have a feast day when Judaism

had a fast day and whatever was forbidden in Judaism the

Sabbatians were encouraged and even commanded to do. This

included incest and what would be today called Satanism. Members

were forbidden to marry outside the sect and there was a system of

keeping their children ignorant of what they were part of until they

were old enough to be trusted not to unknowingly reveal anything

to outsiders. The same system is employed to this day by the Global

Cult in general which Sabbatian-Frankism has enormously

influenced and now largely controls.

Zevi and his Sabbatians suffered a setback with the intervention

by the Sultan of the Islamic O�oman Empire in the Middle East and

what is now the Republic of Turkey where Zevi was located. The

Sultan gave him the choice of proving his ‘divinity’, converting to

Islam or facing torture and death. Funnily enough Zevi chose to

convert or at least appear to. Some of his supporters were

disillusioned and dri�ed away, but many did not with 300 families

also converting – only in theory – to Islam. They continued behind

this Islamic smokescreen to follow the goals, rules and rituals of

Sabbatianism and became known as ‘crypto-Jews’ or the ‘Dönmeh’

which means ‘to turn’. This is rather ironic because they didn’t ‘turn’

and instead hid behind a fake Islamic persona. The process of

appearing to be one thing while being very much another would

become the calling card of Sabbatianism especially a�er Zevi’s death

and the arrival of the Satanist Jacob Frank in the 18th century when

the cult became Sabbatian-Frankism and plumbed still new depths

of depravity and infiltration which included – still includes – human

sacrifice and sex with children. Wherever Sabbatians go paedophilia

and Satanism follow and is it really a surprise that Hollywood is so

infested with child abuse and Satanism when it was established by

Sabbatian-Frankists and is still controlled by them? Hollywood has

been one of the prime vehicles for global perceptual programming

and manipulation. How many believe the version of ‘history’

portrayed in movies when it is a travesty and inversion (again) of the

truth? Rabbi Marvin Antelman describes Frankism in his book, To

Eliminate the Opiate, as ‘a movement of complete evil’ while Jewish

professor Gershom Scholem said of Frank in The Messianic Idea in

Judaism: ‘In all his actions [he was] a truly corrupt and degenerate

individual ... one of the most frightening phenomena in the whole of

Jewish history.’ Frank was excommunicated by traditional rabbis, as

was Zevi, but Frank was undeterred and enjoyed vital support from

the House of Rothschild, the infamous banking dynasty whose

inner-core are Sabbatian-Frankists and not Jews. Infiltration of the

Roman Church and Vatican was instigated by Frank with many

Dönmeh ‘turning’ again to convert to Roman Catholicism with a

view to hijacking the reins of power. This was the ever-repeating

modus operandi and continues to be so. Pose as an advocate of the

religion, culture or country that you want to control and then

manipulate your people into the positions of authority and influence

largely as advisers, administrators and Svengalis for those that

appear to be in power. They did this with Judaism, Christianity

(Christian Zionism is part of this), Islam and other religions and

nations until Sabbatian-Frankism spanned the world as it does

today.

Sabbatian Saudis and the terror network

One expression of the Sabbatian-Frankist Dönmeh within Islam is

the ruling family of Saudi Arabia, the House of Saud, through which

came the vile distortion of Islam known as Wahhabism. This is the

violent creed followed by terrorist groups like Al-Qaeda and ISIS or

Islamic State. Wahhabism is the hand-chopping, head-chopping

‘religion’ of Saudi Arabia which is used to keep the people in a

constant state of fear so the interloper House of Saud can continue to

rule. Al-Qaeda and Islamic State were lavishly funded by the House

of Saud while being created and directed by the Sabbatian-Frankist

network in the United States that operates through the Pentagon,

CIA and the government in general of whichever ‘party’. The front

man for the establishment of Wahhabism in the middle of the 18th

century was a Sabbatian-Frankist ‘crypto-Jew’ posing as Islamic

called Muhammad ibn Abd al-Wahhab. His daughter would marry

the son of Muhammad bin Saud who established the first Saudi state

before his death in 1765 with support from the British Empire. Bin

Saud’s successors would establish modern Saudi Arabia in league

with the British and Americans in 1932 which allowed them to seize

control of Islam’s major shrines in Mecca and Medina. They have

dictated the direction of Sunni Islam ever since while Iran is the

major centre of the Shiite version and here we have the source of at

least the public conflict between them. The Sabbatian network has

used its Wahhabi extremists to carry out Problem-Reaction-Solution

terrorist a�acks in the name of ‘Al-Qaeda’ and ‘Islamic State’ to

justify a devastating ‘war on terror’, ever-increasing surveillance of

the population and to terrify people into compliance. Another

insight of the Renegade Mind is the streetwise understanding that

just because a country, location or people are a�acked doesn’t mean

that those apparently representing that country, location or people

are not behind the a�ackers. O�en they are orchestrating the a�acks

because of the societal changes that can be then justified in the name

of ‘saving the population from terrorists’.

I show in great detail in The Trigger how Sabbatian-Frankists were

the real perpetrators of 9/11 and not ‘19 Arab hijackers’ who were

blamed for what happened. Observe what was justified in the name

of 9/11 alone in terms of Middle East invasions, mass surveillance

and control that fulfilled the demands of the Project for the New

American Century document published by the Sabbatian Neocons.

What appear to be enemies are on the deep inside players on the

same Sabbatian team. Israel and Arab ‘royal’ dictatorships are all

ruled by Sabbatians and the recent peace agreements between Israel

and Saudi Arabia, the United Arab Emirates (UAE) and others are

only making formal what has always been the case behind the

scenes. Palestinians who have been subjected to grotesque tyranny

since Israel was bombed and terrorised into existence in 1948 have

never stood a chance. Sabbatian-Frankists have controlled Israel (so

the constant theme of violence and war which Sabbatians love) and

they have controlled the Arab countries that Palestinians have

looked to for real support that never comes. ‘Royal families’ of the

Arab world in Saudi Arabia, Bahrain, UAE, etc., are all Sabbatians

with allegiance to the aims of the cult and not what is best for their

Arabic populations. They have stolen the oil and financial resources

from their people by false claims to be ‘royal dynasties’ with a

genetic right to rule and by employing vicious militaries to impose

their will.

Satanic ‘illumination’

The Satanist Jacob Frank formed an alliance in 1773 with two other

Sabbatians, Mayer Amschel Rothschild (1744-1812), founder of the

Rothschild banking dynasty, and Jesuit-educated fraudulent Jew,

Adam Weishaupt, and this led to the formation of the Bavarian

Illuminati, firstly under another name, in 1776. The Illuminati would

be the manipulating force behind the French Revolution (1789-1799)

and was also involved in the American Revolution (1775-1783)

before and a�er the Illuminati’s official creation. Weishaupt would

later become (in public) a Protestant Christian in archetypal

Sabbatian style. I read that his name can be decoded as Adam-Weis-

haupt or ‘the first man to lead those who know’. He wasn’t a leader

in the sense that he was a subordinate, but he did lead those below

him in a crusade of transforming human society that still continues

today. The theme was confirmed as early as 1785 when a horseman

courier called Lanz was reported to be struck by lighting and

extensive Illuminati documents were found in his saddlebags. They

made the link to Weishaupt and detailed the plan for world takeover.

Current events with ‘Covid’ fascism have been in the making for a

very long time. Jacob Frank was jailed for 13 years by the Catholic

Inquisition a�er his arrest in 1760 and on his release he headed for

Frankfurt, Germany, home city and headquarters of the House of

Rothschild where the alliance was struck with Mayer Amschel

Rothschild and Weishaupt. Rothschild arranged for Frank to be

given the title of Baron and he became a wealthy nobleman with a

big following of Jews in Germany, the Austro-Hungarian Empire

and other European countries. Most of them would have believed he

was on their side.

The name ‘Illuminati’ came from the Zohar which is a body of

works in the Jewish mystical ‘bible’ called the Kabbalah. ‘Zohar’ is

the foundation of Sabbatian-Frankist belief and in Hebrew ‘Zohar’

means ‘splendour’, ‘radiance’, ‘illuminated’, and so we have

‘Illuminati’. They claim to be the ‘Illuminated Ones’ from their

knowledge systematically hidden from the human population and

passed on through generations of carefully-chosen initiates in the

global secret society network or Cult. Hidden knowledge includes

an awareness of the Cult agenda for the world and the nature of our

collective reality that I will explore later. Cult ‘illumination’ is

symbolised by the torch held by the Statue of Liberty which was

gi�ed to New York by French Freemasons in Paris who knew exactly

what it represents. ‘Liberty’ symbolises the goddess worshipped in

Babylon as Queen Semiramis or Ishtar. The significance of this will

become clear. Notice again the ubiquitous theme of inversion with

the Statue of ‘Liberty’ really symbolising mass control (Fig 7). A

mirror-image statute stands on an island in the River Seine in Paris

from where New York Liberty originated (Fig 8). A large replica of

the Liberty flame stands on top of the Pont de l’Alma tunnel in Paris

where Princess Diana died in a Cult ritual described in The Biggest

Secret. Lucifer ‘the light bringer’ is related to all this (and much more

as we’ll see) and ‘Lucifer’ is a central figure in Sabbatian-Frankism

and its associated Satanism. Sabbatians reject the Jewish Torah, or

Pentateuch, the ‘five books of Moses’ in the Old Testament known as

Genesis, Exodus, Leviticus, Numbers, and Deuteronomy which are

claimed by Judaism and Christianity to have been dictated by ‘God’

to Moses on Mount Sinai. Sabbatians say these do not apply to them

and they seek to replace them with the Zohar to absorb Judaism and

its followers into their inversion which is an expression of a much

greater global inversion. They want to delete all religions and force

humanity to worship a one-world religion – Sabbatian Satanism that

also includes worship of the Earth goddess. Satanic themes are being

more and more introduced into mainstream society and while

Christianity is currently the foremost target for destruction the

others are planned to follow.

Figure 7: The Cult goddess of Babylon disguised as the Statue of Liberty holding the flame of Lucifer the ‘light bringer’.

Figure 8: Liberty’s mirror image in Paris where the New York version originated.

Marx brothers

Rabbi Marvin Antelman connects the Illuminati to the Jacobins in To

Eliminate the Opiate and Jacobins were the force behind the French

Revolution. He links both to the Bund der Gerechten, or League of

the Just, which was the network that inflicted communism/Marxism

on the world. Antelman wrote:

The original inner circle of the Bund der Gerechten consisted of born Catholics, Protestants and Jews [Sabbatian-Frankist infiltrators], and those representatives of respective subdivisions formulated schemes for the ultimate destruction of their faiths. The heretical Catholics laid plans which they felt would take a century or more for the ultimate destruction of the church; the apostate Jews for the ultimate destruction of the Jewish religion.

Sabbatian-created communism connects into this anti-religion

agenda in that communism does not allow for the free practice of

religion. The Sabbatian ‘Bund’ became the International Communist

Party and Communist League and in 1848 ‘Marxism’ was born with

the Communist Manifesto of Sabbatian assets Karl Marx and

Friedrich Engels. It is absolutely no coincidence that Marxism, just a

different name for fascist and other centrally-controlled tyrannies, is

being imposed worldwide as a result of the ‘Covid’ hoax and nor

that Marxist/fascist China was the place where the hoax originated.

The reason for this will become very clear in the chapter ‘Covid: The

calculated catastrophe’. The so-called ‘Woke’ mentality has hijacked

traditional beliefs of the political le� and replaced them with far-

right make-believe ‘social justice’ be�er known as Marxism. Woke

will, however, be swallowed by its own perceived ‘revolution’ which

is really the work of billionaires and billionaire corporations feigning

being ‘Woke’. Marxism is being touted by Wokers as a replacement

for ‘capitalism’ when we don’t have ‘capitalism’. We have cartelism

in which the market is stitched up by the very Cult billionaires and

corporations bankrolling Woke. Billionaires love Marxism which

keeps the people in servitude while they control from the top.

Terminally naïve Wokers think they are ‘changing the world’ when

it’s the Cult that is doing the changing and when they have played

their vital part and become surplus to requirements they, too, will be

targeted. The Illuminati-Jacobins were behind the period known as

‘The Terror’ in the French Revolution in 1793 and 1794 when Jacobin

Maximillian de Robespierre and his Orwellian ‘Commi�ee of Public

Safety’ killed 17,000 ‘enemies of the Revolution’ who had once been

‘friends of the Revolution’. Karl Marx (1818-1883), whose Sabbatian

creed of Marxism has cost the lives of at least 100 million people, is a

hero once again to Wokers who have been systematically kept

ignorant of real history by their ‘education’ programming. As a

result they now promote a Sabbatian ‘Marxist’ abomination destined

at some point to consume them. Rabbi Antelman, who spent decades

researching the Sabbatian plot, said of the League of the Just and

Karl Marx:

Contrary to popular opinion Karl Marx did not originate the Communist Manifesto. He was paid for his services by the League of the Just, which was known in its country of origin, Germany, as the Bund der Geaechteten.

Antelman said the text a�ributed to Marx was the work of other

people and Marx ‘was only repeating what others already said’.

Marx was ‘a hired hack – lackey of the wealthy Illuminists’. Marx

famously said that religion was the ‘opium of the people’ (part of the

Sabbatian plan to demonise religion) and Antelman called his books,

To Eliminate the Opiate. Marx was born Jewish, but his family

converted to Christianity (Sabbatian modus operandi) and he

a�acked Jews, not least in his book, A World Without Jews. In doing

so he supported the Sabbatian plan to destroy traditional Jewishness

and Judaism which we are clearly seeing today with the vindictive

targeting of orthodox Jews by the Sabbatian government of Israel

over ‘Covid’ laws. I don’t follow any religion and it has done much

damage to the world over centuries and acted as a perceptual

straightjacket. Renegade Minds, however, are always asking why

something is being done. It doesn’t ma�er if they agree or disagree

with what is happening – why is it happening is the question. The

‘why?’ can be answered with regard to religion in that religions

create interacting communities of believers when the Cult wants to

dismantle all discourse, unity and interaction (see ‘Covid’

lockdowns) and the ultimate goal is to delete all religions for a one-

world religion of Cult Satanism worshipping their ‘god’ of which

more later. We see the same ‘why?’ with gun control in America. I

don’t have guns and don’t want them, but why is the Cult seeking to

disarm the population at the same time that law enforcement

agencies are armed to their molars and why has every tyrant in

history sought to disarm people before launching the final takeover?

They include Hitler, Stalin, Pol Pot and Mao who followed

confiscation with violent seizing of power. You know it’s a Cult

agenda by the people who immediately race to the microphones to

exploit dead people in multiple shootings. Ultra-Zionist Cult lackey

Senator Chuck Schumer was straight on the case a�er ten people

were killed in Boulder, Colorado in March, 2121. Simple rule … if

Schumer wants it the Cult wants it and the same with his ultra-

Zionist mate the wild-eyed Senator Adam Schiff. At the same time

they were calling for the disarmament of Americans, many of whom

live a long way from a police response, Schumer, Schiff and the rest

of these pampered clowns were si�ing on Capitol Hill behind a

razor-wired security fence protected by thousands of armed troops

in addition to their own armed bodyguards. Mom and pop in an

isolated home? They’re just potential mass shooters.

Zion Mainframe

Sabbatian-Frankists and most importantly the Rothschilds were

behind the creation of ‘Zionism’, a political movement that

demanded a Jewish homeland in Israel as promised by Sabbatai

Zevi. The very symbol of Israel comes from the German meaning of

the name Rothschild. Dynasty founder Mayer Amschel Rothschild

changed the family name from Bauer to Rothschild, or ‘Red-Shield’

in German, in deference to the six-pointed ‘Star of David’ hexagram

displayed on the family’s home in Frankfurt. The symbol later

appeared on the flag of Israel a�er the Rothschilds were centrally

involved in its creation. Hexagrams are not a uniquely Jewish

symbol and are widely used in occult (‘hidden’) networks o�en as a

symbol for Saturn (see my other books for why). Neither are

Zionism and Jewishness interchangeable. Zionism is a political

movement and philosophy and not a ‘race’ or a people. Many Jews

oppose Zionism and many non-Jews, including US President Joe

Biden, call themselves Zionists as does Israel-centric Donald Trump.

America’s support for the Israel government is pre�y much a gimme

with ultra-Zionist billionaires and corporations providing fantastic

and dominant funding for both political parties. Former

Congresswoman Cynthia McKinney has told how she was

approached immediately she ran for office to ‘sign the pledge’ to

Israel and confirm that she would always vote in that country’s best

interests. All American politicians are approached in this way.

Anyone who refuses will get no support or funding from the

enormous and all-powerful Zionist lobby that includes organisations

like mega-lobby group AIPAC, the American Israel Public Affairs

Commi�ee. Trump’s biggest funder was ultra-Zionist casino and

media billionaire Sheldon Adelson while major funders of the

Democratic Party include ultra-Zionist George Soros and ultra-

Zionist financial and media mogul, Haim Saban. Some may reel back

at the suggestion that Soros is an Israel-firster (Sabbatian-controlled

Israel-firster), but Renegade Minds watch the actions not the words

and everywhere Soros donates his billions the Sabbatian agenda

benefits. In the spirit of Sabbatian inversion Soros pledged $1 billion

for a new university network to promote ‘liberal values and tackle

intolerance’. He made the announcement during his annual speech

at the Cult-owned World Economic Forum in Davos, Switzerland, in

January, 2020, a�er his ‘harsh criticism’ of ‘authoritarian rulers’

around the world. You can only laugh at such brazen mendacity.

How he doesn’t laugh is the mystery. Translated from the Orwellian

‘liberal values and tackle intolerance’ means teaching non-white

people to hate white people and for white people to loathe

themselves for being born white. The reason for that will become

clear.

The ‘Anti-Semitism’ fraud

Zionists support the Jewish homeland in the land of Palestine which

has been the Sabbatian-Rothschild goal for so long, but not for the

benefit of Jews. Sabbatians and their global Anti-Semitism Industry

have skewed public and political opinion to equate opposing the

violent extremes of Zionism to be a blanket a�ack and condemnation

of all Jewish people. Sabbatians and their global Anti-Semitism

Industry have skewed public and political opinion to equate

opposing the violent extremes of Zionism to be a blanket a�ack and

condemnation of all Jewish people. This is nothing more than a

Sabbatian protection racket to stop legitimate investigation and

exposure of their agendas and activities. The official definition of

‘anti-Semitism’ has more recently been expanded to include criticism

of Zionism – a political movement – and this was done to further stop

exposure of Sabbatian infiltrators who created Zionism as we know

it today in the 19th century. Renegade Minds will talk about these

subjects when they know the shit that will come their way. People

must decide if they want to know the truth or just cower in the

corner in fear of what others will say. Sabbatians have been trying to

label me as ‘anti-Semitic’ since the 1990s as I have uncovered more

and more about their background and agendas. Useless, gutless,

fraudulent ‘journalists’ then just repeat the smears without question

and on the day I was writing this section a pair of unquestioning

repeaters called Ben Quinn and Archie Bland (how appropriate)

outright called me an ‘anti-Semite’ in the establishment propaganda

sheet, the London Guardian, with no supporting evidence. The

Sabbatian Anti-Semitism Industry said so and who are they to

question that? They wouldn’t dare. Ironically ‘Semitic’ refers to a

group of languages in the Middle East that are almost entirely

Arabic. ‘Anti-Semitism’ becomes ‘anti-Arab’ which if the

consequences of this misunderstanding were not so grave would be

hilarious. Don’t bother telling Quinn and Bland. I don’t want to

confuse them, bless ‘em. One reason I am dubbed ‘anti-Semitic’ is

that I wrote in the 1990s that Jewish operatives (Sabbatians) were

heavily involved in the Russian Revolution when Sabbatians

overthrew the Romanov dynasty. This apparently made me ‘anti-

Semitic’. Oh, really? Here is a section from The Trigger:

British journalist Robert Wilton confirmed these themes in his 1920 book The Last Days of the Romanovs when he studied official documents from the Russian government to identify the members of the Bolshevik ruling elite between 1917 and 1919. The Central Committee included 41 Jews among 62 members; the Council of the People’s Commissars had 17 Jews out of 22 members; and 458 of the 556 most important Bolshevik positions between 1918 and 1919 were occupied by Jewish people. Only 17 were Russian. Then there were the 23 Jews among the 36 members of the vicious Cheka Soviet secret police established in 1917 who would soon appear all across the country.

Professor Robert Service of Oxford University, an expert on 20th century Russian history, found evidence that [‘Jewish’] Leon Trotsky had sought to make sure that Jews were enrolled in the Red Army and were disproportionately represented in the Soviet civil bureaucracy that included the Cheka which performed mass arrests, imprisonment and executions of ‘enemies of the people’. A US State Department Decimal File (861.00/5339) dated November 13th, 1918, names [Rothschild banking agent in America] Jacob Schiff and a list of ultra-Zionists as funders of the Russian Revolution leading to claims of a ‘Jewish plot’, but the key point missed by all is they were not ‘Jews’ – they were Sabbatian-Frankists.

Britain’s Winston Churchill made the same error by mistake or

otherwise. He wrote in a 1920 edition of the Illustrated Sunday Herald

that those behind the Russian revolution were part of a ‘worldwide

conspiracy for the overthrow of civilisation and for the

reconstitution of society on the basis of arrested development, of

envious malevolence, and impossible equality’ (see ‘Woke’ today

because that has been created by the same network). Churchill said

there was no need to exaggerate the part played in the creation of

Bolshevism and in the actual bringing about of the Russian

Revolution ‘by these international and for the most part atheistical

Jews’ [‘atheistical Jews’ = Sabbatians]. Churchill said it is certainly a

very great one and probably outweighs all others: ‘With the notable

exception of Lenin, the majority of the leading figures are Jews.’ He

went on to describe, knowingly or not, the Sabbatian modus

operandi of placing puppet leaders nominally in power while they

control from the background:

Moreover, the principal inspiration and driving power comes from the Jewish leaders. Thus Tchitcherin, a pure Russian, is eclipsed by his nominal subordinate, Litvinoff, and the influence of Russians like Bukharin or Lunacharski cannot be compared with the power of Trotsky, or of Zinovieff, the Dictator of the Red Citadel (Petrograd), or of Krassin or Radek – all Jews. In the Soviet institutions the predominance of Jews is even more astonishing. And the prominent, if not indeed the principal, part in the system of terrorism applied by the Extraordinary Commissions for Combatting Counter-Revolution has been taken by Jews, and in some notable cases by Jewesses.

What I said about seriously disproportionate involvement in the

Russian Revolution by Jewish ‘revolutionaries’ (Sabbatians) is

provable fact, but truth is no defence against the Sabbatian Anti-

Semitism Industry, its repeater parrots like Quinn and Bland, and

the now breathtaking network of so-called ‘Woke’ ‘anti-hate’ groups

with interlocking leaderships and funding which have the role of

discrediting and silencing anyone who gets too close to exposing the

Sabbatians. We have seen ‘truth is no defence’ confirmed in legal

judgements with the Saskatchewan Human Rights Commission in

Canada decreeing this: ‘Truthful statements can be presented in a

manner that would meet the definition of hate speech, and not all

truthful statements must be free from restriction.’ Most ‘anti-hate’

activists, who are themselves consumed by hatred, are too stupid

and ignorant of the world to know how they are being used. They

are far too far up their own virtue-signalling arses and it’s far too

dark for them to see anything.

The ‘revolution’ game

The background and methods of the ‘Russian’ Revolution are

straight from the Sabbatian playbook seen in the French Revolution

and endless others around the world that appear to start as a

revolution of the people against tyrannical rule and end up with a

regime change to more tyrannical rule overtly or covertly. Wars,

terror a�acks and regime overthrows follow the Sabbatian cult

through history with its agents creating them as Problem-Reaction-

Solutions to remove opposition on the road to world domination.

Sabbatian dots connect the Rothschilds with the Illuminati, Jacobins

of the French Revolution, the ‘Bund’ or League of the Just, the

International Communist Party, Communist League and the

Communist Manifesto of Karl Marx and Friedrich Engels that would

lead to the Rothschild-funded Russian Revolution. The sequence

comes under the heading of ‘creative destruction’ when you advance

to your global goal by continually destroying the status quo to install

a new status quo which you then also destroy. The two world wars

come to mind. With each new status quo you move closer to your

planned outcome. Wars and mass murder are to Sabbatians a

collective blood sacrifice ritual. They are obsessed with death for

many reasons and one is that death is an inversion of life. Satanists

and Sabbatians are obsessed with death and o�en target churches

and churchyards for their rituals. Inversion-obsessed Sabbatians

explain the use of inverted symbolism including the inverted

pentagram and inverted cross. The inversion of the cross has been

related to targeting Christianity, but the cross was a religious symbol

long before Christianity and its inversion is a statement about the

Sabbatian mentality and goals more than any single religion.

Sabbatians operating in Germany were behind the rise of the

occult-obsessed Nazis and the subsequent Jewish exodus from

Germany and Europe to Palestine and the United States a�er World

War Two. The Rothschild dynasty was at the forefront of this both as

political manipulators and by funding the operation. Why would

Sabbatians help to orchestrate the horrors inflicted on Jews by the

Nazis and by Stalin a�er they organised the Russian Revolution?

Sabbatians hate Jews and their religion, that’s why. They pose as

Jews and secure positions of control within Jewish society and play

the ‘anti-Semitism’ card to protect themselves from exposure

through a global network of organisations answering to the

Sabbatian-created-and-controlled globe-spanning intelligence

network that involves a stunning web of military-intelligence

operatives and operations for a tiny country of just nine million.

Among them are Jewish assets who are not Sabbatians but have been

convinced by them that what they are doing is for the good of Israel

and the Jewish community to protect them from what they have

been programmed since childhood to believe is a Jew-hating hostile

world. The Jewish community is just a highly convenient cover to

hide the true nature of Sabbatians. Anyone ge�ing close to exposing

their game is accused by Sabbatian place-people and gofers of ‘anti-

Semitism’ and claiming that all Jews are part of a plot to take over

the world. I am not saying that. I am saying that Sabbatians – the real

Jew-haters – have infiltrated the Jewish community to use them both

as a cover and an ‘anti-Semitic’ defence against exposure. Thus we

have the Anti-Semitism Industry targeted researchers in this way

and most Jewish people think this is justified and genuine. They

don’t know that their ‘Jewish’ leaders and institutions of state,

intelligence and military are not controlled by Jews at all, but cultists

and stooges of Sabbatian-Frankism. I once added my name to a pro-

Jewish freedom petition online and the next time I looked my name

was gone and text had been added to the petition blurb to a�ack me

as an ‘anti-Semite’ such is the scale of perceptual programming.

Moving on America

I tell the story in The Trigger and a chapter called ‘Atlantic Crossing’

how particularly a�er Israel was established the Sabbatians moved

in on the United States and eventually grasped control of

government administration, the political system via both Democrats

and Republicans, the intelligence community like the CIA and

National Security Agency (NSA), the Pentagon and mass media.

Through this seriously compartmentalised network Sabbatians and

their operatives in Mossad, Israeli Defense Forces (IDF) and US

agencies pulled off 9/11 and blamed it on 19 ‘Al-Qaeda hijackers’

dominated by men from, or connected to, Sabbatian-ruled Saudi

Arabia. The ‘19’ were not even on the planes let alone flew those big

passenger jets into buildings while being largely incompetent at

piloting one-engine light aircra�. ‘Hijacker’ Hani Hanjour who is

said to have flown American Airlines Flight 77 into the Pentagon

with a turn and manoeuvre most professional pilots said they would

have struggled to do was banned from renting a small plane by

instructors at the Freeway Airport in Bowie, Maryland, just six weeks

earlier on the grounds that he was an incompetent pilot. The Jewish

population of the world is just 0.2 percent with even that almost

entirely concentrated in Israel (75 percent Jewish) and the United

States (around two percent). This two percent and globally 0.2

percent refers to Jewish people and not Sabbatian interlopers who are

a fraction of that fraction. What a sobering thought when you think

of the fantastic influence on world affairs of tiny Israel and that the

Project for the New America Century (PNAC) which laid out the

blueprint in September, 2000, for America’s war on terror and regime

change wars in Iraq, Libya and Syria was founded and dominated by

Sabbatians known as ‘Neocons’. The document conceded that this

plan would not be supported politically or publicly without a major

a�ack on American soil and a Problem-Reaction-Solution excuse to

send troops to war across the Middle East. Sabbatian Neocons said:

... [The] process of transformation ... [war and regime change] ... is likely to be a long one, absent some catastrophic and catalysing event – like a new Pearl Harbor.

Four months later many of those who produced that document

came to power with their inane puppet George Bush from the long-

time Sabbatian Bush family. They included Sabbatian Dick Cheney

who was officially vice-president, but really de-facto president for

the entirety of the ‘Bush’ government. Nine months a�er the ‘Bush’

inauguration came what Bush called at the time ‘the Pearl Harbor of

the 21st century’ and with typical Sabbatian timing and symbolism

2001 was the 60th anniversary of the a�ack in 1941 by the Japanese

Air Force on Pearl Harbor, Hawaii, which allowed President

Franklin Delano Roosevelt to take the United States into a Sabbatian-

instigated Second World War that he said in his election campaign

that he never would. The evidence is overwhelming that Roosevelt

and his military and intelligence networks knew the a�ack was

coming and did nothing to stop it, but they did make sure that

America’s most essential naval ships were not in Hawaii at the time.

Three thousand Americans died in the Pearl Harbor a�acks as they

did on September 11th. By the 9/11 year of 2001 Sabbatians had

widely infiltrated the US government, military and intelligence

operations and used their compartmentalised assets to pull off the

‘Al-Qaeda’ a�acks. If you read The Trigger it will blow your mind to

see the u�erly staggering concentration of ‘Jewish’ operatives

(Sabbatian infiltrators) in essential positions of political, security,

legal, law enforcement, financial and business power before, during,

and a�er the a�acks to make them happen, carry them out, and then

cover their tracks – and I do mean staggering when you think of that

0.2 percent of the world population and two percent of Americans

which are Jewish while Sabbatian infiltrators are a fraction of that. A

central foundation of the 9/11 conspiracy was the hijacking of

government, military, Air Force and intelligence computer systems

in real time through ‘back-door’ access made possible by Israeli

(Sabbatian) ‘cyber security’ so�ware. Sabbatian-controlled Israel is

on the way to rivalling Silicon Valley for domination of cyberspace

and is becoming the dominant force in cyber-security which gives

them access to entire computer systems and their passcodes across

the world. Then add to this that Zionists head (officially) Silicon

Valley giants like Google (Larry Page and Sergey Brin), Google-

owned YouTube (Susan Wojcicki), Facebook (Mark Zuckerberg and

Sheryl Sandberg), and Apple (Chairman Arthur D. Levinson), and

that ultra-Zionist hedge fund billionaire Paul Singer has a $1 billion

stake in Twi�er which is only nominally headed by ‘CEO’ pothead

Jack Dorsey. As cable news host Tucker Carlson said of Dorsey:

‘There used to be debate in the medical community whether

dropping a ton of acid had permanent effects and I think that debate

has now ended.’ Carlson made the comment a�er Dorsey told a

hearing on Capitol Hill (if you cut through his bullshit) that he

believed in free speech so long as he got to decide what you can hear

and see. These ‘big names’ of Silicon Valley are only front men and

women for the Global Cult, not least the Sabbatians, who are the true

controllers of these corporations. Does anyone still wonder why

these same people and companies have been ferociously censoring

and banning people (like me) for exposing any aspect of the Cult

agenda and especially the truth about the ‘Covid’ hoax which

Sabbatians have orchestrated?

The Jeffrey Epstein paedophile ring was a Sabbatian operation. He

was officially ‘Jewish’ but he was a Sabbatian and women abused by

the ring have told me about the high number of ‘Jewish’ people

involved. The Epstein horror has Sabbatian wri�en all over it and

matches perfectly their modus operandi and obsession with sex and

ritual. Epstein was running a Sabbatian blackmail ring in which

famous people with political and other influence were provided

with young girls for sex while everything was being filmed and

recorded on hidden cameras and microphones at his New York

house, Caribbean island and other properties. Epstein survivors

have described this surveillance system to me and some have gone

public. Once the famous politician or other figure knew he or she

was on video they tended to do whatever they were told. Here we go

again …when you’ve got them by the balls their hearts and minds

will follow. Sabbatians use this blackmail technique on a wide scale

across the world to entrap politicians and others they need to act as

demanded. Epstein’s private plane, the infamous ‘Lolita Express’,

had many well-known passengers including Bill Clinton while Bill

Gates has flown on an Epstein plane and met with him four years

a�er Epstein had been jailed for paedophilia. They subsequently met

many times at Epstein’s home in New York according to a witness

who was there. Epstein’s infamous side-kick was Ghislaine Maxwell,

daughter of Mossad agent and ultra-Zionist mega-crooked British

businessman, Bob Maxwell, who at one time owned the Daily Mirror

newspaper. Maxwell was murdered at sea on his boat in 1991 by

Sabbatian-controlled Mossad when he became a liability with his

business empire collapsing as a former Mossad operative has

confirmed (see The Trigger).

Money, money, money, funny money …

Before I come to the Sabbatian connection with the last three US

presidents I will lay out the crucial importance to Sabbatians of

controlling banking and finance. Sabbatian Mayer Amschel

Rothschild set out to dominate this arena in his family’s quest for

total global control. What is freedom? It is, in effect, choice. The

more choices you have the freer you are and the fewer your choices

the more you are enslaved. In the global structure created over

centuries by Sabbatians the biggest decider and restrictor of choice is

… money. Across the world if you ask people what they would like

to do with their lives and why they are not doing that they will reply

‘I don’t have the money’. This is the idea. A global elite of multi-

billionaires are described as ‘greedy’ and that is true on one level;

but control of money – who has it and who doesn’t – is not primarily

about greed. It’s about control. Sabbatians have seized ever more

control of finance and sucked the wealth of the world out of the

hands of the population. We talk now, a�er all, about the ‘One-

percent’ and even then the wealthiest are a lot fewer even than that.

This has been made possible by a money scam so outrageous and so

vast it could rightly be called the scam of scams founded on creating

‘money’ out of nothing and ‘loaning’ that with interest to the

population. Money out of nothing is called ‘credit’. Sabbatians have

asserted control over governments and banking ever more

completely through the centuries and secured financial laws that

allow banks to lend hugely more than they have on deposit in a

confidence trick known as fractional reserve lending. Imagine if you

could lend money that doesn’t exist and charge the recipient interest

for doing so. You would end up in jail. Bankers by contrast end up in

mansions, private jets, Malibu and Monaco.

Banks are only required to keep a fraction of their deposits and

wealth in their vaults and they are allowed to lend ‘money’ they

don’t have called ‘credit. Go into a bank for a loan and if you succeed

the banker will not move any real wealth into your account. They

will type into your account the amount of the agreed ‘loan’ – say

£100,000. This is not wealth that really exists; it is non-existent, fresh-

air, created-out-of-nothing ‘credit’ which has never, does not, and

will never exist except in theory. Credit is backed by nothing except

wind and only has buying power because people think that it has

buying power and accept it in return for property, goods and

services. I have described this situation as like those cartoon

characters you see chasing each other and when they run over the

edge of a cliff they keep running forward on fresh air until one of

them looks down, realises what’s happened, and they all crash into

the ravine. The whole foundation of the Sabbatian financial system is

to stop people looking down except for periodic moments when they

want to crash the system (as in 2008 and 2020 ongoing) and reap the

rewards from all the property, businesses and wealth their borrowers

had signed over as ‘collateral’ in return for a ‘loan’ of fresh air. Most

people think that money is somehow created by governments when

it comes into existence from the start as a debt through banks

‘lending’ illusory money called credit. Yes, the very currency of

exchange is a debt from day one issued as an interest-bearing loan.

Why don’t governments create money interest-free and lend it to

their people interest-free? Governments are controlled by Sabbatians

and the financial system is controlled by Sabbatians for whom

interest-free money would be a nightmare come true. Sabbatians

underpin their financial domination through their global network of

central banks, including the privately-owned US Federal Reserve

and Britain’s Bank of England, and this is orchestrated by a

privately-owned central bank coordination body called the Bank for

International Se�lements in Basle, Switzerland, created by the usual

suspects including the Rockefellers and Rothschilds. Central bank

chiefs don’t answer to governments or the people. They answer to

the Bank for International Se�lements or, in other words, the Global

Cult which is dominated today by Sabbatians.

Built-in disaster

There are so many constituent scams within the overall banking

scam. When you take out a loan of thin-air credit only the amount of

that loan is theoretically brought into circulation to add to the

amount in circulation; but you are paying back the principle plus

interest. The additional interest is not created and this means that

with every ‘loan’ there is a shortfall in the money in circulation

between what is borrowed and what has to be paid back. There is

never even close to enough money in circulation to repay all

outstanding public and private debt including interest. Coldly

weaved in the very fabric of the system is the certainty that some

will lose their homes, businesses and possessions to the banking

‘lender’. This is less obvious in times of ‘boom’ when the amount of

money in circulation (and the debt) is expanding through more

people wanting and ge�ing loans. When a downturn comes and the

money supply contracts it becomes painfully obvious that there is

not enough money to service all debt and interest. This is less

obvious in times of ‘boom’ when the amount of money in circulation

(and the debt) is expanding through more people wanting and

ge�ing loans. When a downturn comes and the money supply

contracts and it becomes painfully obvious – as in 2008 and currently

– that there is not enough money to service all debt and interest.

Sabbatian banksters have been leading the human population

through a calculated series of booms (more debt incurred) and busts

(when the debt can’t be repaid and the banks get the debtor’s

tangible wealth in exchange for non-existent ‘credit’). With each

‘bust’ Sabbatian bankers have absorbed more of the world’s tangible

wealth and we end up with the One-percent. Governments are in

bankruptcy levels of debt to the same system and are therefore

owned by a system they do not control. The Federal Reserve,

‘America’s central bank’, is privately-owned and American

presidents only nominally appoint its chairman or woman to

maintain the illusion that it’s an arm of government. It’s not. The

‘Fed’ is a cartel of private banks which handed billions to its

associates and friends a�er the crash of 2008 and has been Sabbatian-

controlled since it was manipulated into being in 1913 through the

covert trickery of Rothschild banking agents Jacob Schiff and Paul

Warburg, and the Sabbatian Rockefeller family. Somehow from a

Jewish population of two-percent and globally 0.2 percent (Sabbatian

interlopers remember are far smaller) ultra-Zionists headed the

Federal Reserve for 31 years between 1987 and 2018 in the form of

Alan Greenspan, Bernard Bernanke and Janet Yellen (now Biden’s

Treasury Secretary) with Yellen’s deputy chairman a Israeli-

American duel citizen and ultra-Zionist Stanley Fischer, a former

governor of the Bank of Israel. Ultra-Zionist Fed chiefs spanned the

presidencies of Ronald Reagan (‘Republican’), Father George Bush

(‘Republican’), Bill Clinton (‘Democrat’), Boy George Bush

(‘Republican’) and Barack Obama (‘Democrat’). We should really

add the pre-Greenspan chairman, Paul Adolph Volcker, ‘appointed’

by Jimmy Carter (‘Democrat’) who ran the Fed between 1979 and

1987 during the Carter and Reagan administrations before

Greenspan took over. Volcker was a long-time associate and business

partner of the Rothschilds. No ma�er what the ‘party’ officially in

power the United States economy was directed by the same force.

Here are members of the Obama, Trump and Biden administrations

and see if you can make out a common theme.

Barack Obama (‘Democrat’)

Ultra-Zionists Robert Rubin, Larry Summers, and Timothy Geithner

ran the US Treasury in the Clinton administration and two of them

reappeared with Obama. Ultra-Zionist Fed chairman Alan

Greenspan had manipulated the crash of 2008 through deregulation

and jumped ship just before the disaster to make way for ultra-

Zionist Bernard Bernanke to hand out trillions to Sabbatian ‘too big

to fail’ banks and businesses, including the ubiquitous ultra-Zionist

Goldman Sachs which has an ongoing staff revolving door operation

between itself and major financial positions in government

worldwide. Obama inherited the fallout of the crash when he took

office in January, 2009, and fortunately he had the support of his

ultra-Zionist White House Chief of Staff Rahm Emmanuel, son of a

terrorist who helped to bomb Israel into being in 1948, and his ultra-

Zionist senior adviser David Axelrod, chief strategist in Obama’s two

successful presidential campaigns. Emmanuel, later mayor of

Chicago and former senior fundraiser and strategist for Bill Clinton,

is an example of the Sabbatian policy a�er Israel was established of

migrating insider families to America so their children would be

born American citizens. ‘Obama’ chose this financial team

throughout his administration to respond to the Sabbatian-instigated

crisis:

Timothy Geithner (ultra-Zionist) Treasury Secretary; Jacob J. Lew,

Treasury Secretary; Larry Summers (ultra-Zionist), director of the

White House National Economic Council; Paul Adolph Volcker

(Rothschild business partner), chairman of the Economic Recovery

Advisory Board; Peter Orszag (ultra-Zionist), director of the Office of

Management and Budget overseeing all government spending;

Penny Pritzker (ultra-Zionist), Commerce Secretary; Jared Bernstein

(ultra-Zionist), chief economist and economic policy adviser to Vice

President Joe Biden; Mary Schapiro (ultra-Zionist), chair of the

Securities and Exchange Commission (SEC); Gary Gensler (ultra-

Zionist), chairman of the Commodity Futures Trading Commission

(CFTC); Sheila Bair (ultra-Zionist), chair of the Federal Deposit

Insurance Corporation (FDIC); Karen Mills (ultra-Zionist), head of

the Small Business Administration (SBA); Kenneth Feinberg (ultra-

Zionist), Special Master for Executive [bail-out] Compensation.

Feinberg would be appointed to oversee compensation (with strings)

to 9/11 victims and families in a campaign to stop them having their

day in court to question the official story. At the same time ultra-

Zionist Bernard Bernanke was chairman of the Federal Reserve and

these are only some of the ultra-Zionists with allegiance to

Sabbatian-controlled Israel in the Obama government. Obama’s

biggest corporate donor was ultra-Zionist Goldman Sachs which had

employed many in his administration.

Donald Trump (‘Republican’)

Trump claimed to be an outsider (he wasn’t) who had come to ‘drain

the swamp’. He embarked on this goal by immediately appointing

ultra-Zionist Steve Mnuchin, a Goldman Sachs employee for 17

years, as his Treasury Secretary. Others included Gary Cohn (ultra-

Zionist), chief operating officer of Goldman Sachs, his first Director

of the National Economic Council and chief economic adviser, who

was later replaced by Larry Kudlow (ultra-Zionist). Trump’s senior

adviser throughout his four years in the White House was his

sinister son-in-law Jared Kushner, a life-long friend of Israel Prime

Minister Benjamin Netanyahu. Kushner is the son of a convicted

crook who was pardoned by Trump in his last days in office. Other

ultra-Zionists in the Trump administration included: Stephen Miller,

Senior Policy Adviser; Avrahm Berkowitz, Deputy Adviser to Trump

and his Senior Adviser Jared Kushner; Ivanka Trump, Adviser to the

President, who converted to Judaism when she married Jared

Kushner; David Friedman, Trump lawyer and Ambassador to Israel;

Jason Greenbla�, Trump Organization executive vice president and

chief legal officer, who was made Special Representative for

International Negotiations and the Israeli-Palestinian Conflict; Rod

Rosenstein, Deputy A�orney General; Elliot Abrams, Special

Representative for Venezuela, then Iran; John Eisenberg, National

Security Council Legal Adviser and Deputy Council to the President

for National Security Affairs; Anne Neuberger, Deputy National

Manager, National Security Agency; Ezra Cohen-Watnick, Acting

Under Secretary of Defense for Intelligence; Elan Carr, Special Envoy

to monitor and combat anti-Semitism; Len Khodorkovsky, Deputy

Special Envoy to monitor and combat anti-Semitism; Reed Cordish,

Assistant to the President, Intragovernmental and Technology

Initiatives. Trump Vice President Mike Pence and Secretary of State

Mike Pompeo, both Christian Zionists, were also vehement

supporters of Israel and its goals and ambitions.

Donald ‘free-speech believer’ Trump pardoned a number of

financial and violent criminals while ignoring calls to pardon Julian

Assange and Edward Snowden whose crimes are revealing highly

relevant information about government manipulation and

corruption and the widespread illegal surveillance of the American

people by US ‘security’ agencies. It’s so good to know that Trump is

on the side of freedom and justice and not mega-criminals with

allegiance to Sabbatian-controlled Israel. These included a pardon

for Israeli spy Jonathan Pollard who was jailed for life in 1987 under

the Espionage Act. Aviem Sella, the Mossad agent who recruited

Pollard, was also pardoned by Trump while Assange sat in jail and

Snowden remained in exile in Russia. Sella had ‘fled’ (was helped to

escape) to Israel in 1987 and was never extradited despite being

charged under the Espionage Act. A Trump White House statement

said that Sella’s clemency had been ‘supported by Benjamin

Netanyahu, Ron Dermer, Israel’s US Ambassador, David Friedman,

US Ambassador to Israel and Miriam Adelson, wife of leading

Trump donor Sheldon Adelson who died shortly before. Other

friends of Jared Kushner were pardoned along with Sholom Weiss

who was believed to be serving the longest-ever white-collar prison

sentence of more than 800 years in 2000. The sentence was

commuted of Ponzi-schemer Eliyahu Weinstein who defrauded Jews

and others out of $200 million. I did mention that Assange and

Snowden were ignored, right? Trump gave Sabbatians almost

everything they asked for in military and political support, moving

the US Embassy from Tel Aviv to Jerusalem with its critical symbolic

and literal implications for Palestinian statehood, and the ‘deal of the

Century’ designed by Jared Kushner and David Friedman which

gave the Sabbatian Israeli government the green light to

substantially expand its already widespread program of building

illegal Jewish-only se�lements in the occupied land of the West

Bank. This made a two-state ‘solution’ impossible by seizing all the

land of a potential Palestinian homeland and that had been the plan

since 1948 and then 1967 when the Arab-controlled Gaza Strip, West

Bank, Sinai Peninsula and Syrian Golan Heights were occupied by

Israel. All the talks about talks and road maps and delays have been

buying time until the West Bank was physically occupied by Israeli

real estate. Trump would have to be a monumentally ill-informed

idiot not to see that this was the plan he was helping to complete.

The Trump administration was in so many ways the Kushner

administration which means the Netanyahu administration which

means the Sabbatian administration. I understand why many

opposing Cult fascism in all its forms gravitated to Trump, but he

was a crucial part of the Sabbatian plan and I will deal with this in

the next chapter.

Joe Biden (‘Democrat’)

A barely cognitive Joe Biden took over the presidency in January,

2021, along with his fellow empty shell, Vice-President Kamala

Harris, as the latest Sabbatian gofers to enter the White House.

Names on the door may have changed and the ‘party’ – the force

behind them remained the same as Zionists were appointed to a

stream of pivotal areas relating to Sabbatian plans and policy. They

included: Janet Yellen, Treasury Secretary, former head of the Federal

Reserve, and still another ultra-Zionist running the US Treasury a�er

Mnuchin (Trump), Lew and Geithner (Obama), and Summers and

Rubin (Clinton); Anthony Blinken, Secretary of State; Wendy

Sherman, Deputy Secretary of State (so that’s ‘Biden’s’ Sabbatian

foreign policy sorted); Jeff Zients, White House coronavirus

coordinator; Rochelle Walensky, head of the Centers for Disease

Control; Rachel Levine, transgender deputy health secretary (that’s

‘Covid’ hoax policy under control); Merrick Garland, A�orney

General; Alejandro Mayorkas, Secretary of Homeland Security; Cass

Sunstein, Homeland Security with responsibility for new

immigration laws; Avril Haines, Director of National Intelligence;

Anne Neuberger, National Security Agency cybersecurity director

(note, cybersecurity); David Cohen, CIA Deputy Director; Ronald

Klain, Biden’s Chief of Staff (see Rahm Emanuel); Eric Lander, a

‘leading geneticist’, Office of Science and Technology Policy director

(see Smart Grid, synthetic biology agenda); Jessica Rosenworcel,

acting head of the Federal Communications Commission (FCC)

which controls Smart Grid technology policy and electromagnetic

communication systems including 5G. How can it be that so many

pivotal positions are held by two-percent of the American

population and 0.2 percent of the world population administration

a�er administration no ma�er who is the president and what is the

party? It’s a coincidence? Of course it’s not and this is why

Sabbatians have built their colossal global web of interlocking ‘anti-

hate’ hate groups to condemn anyone who asks these glaring

questions as an ‘anti-Semite’. The way that Jewish people horrifically

abused in Sabbatian-backed Nazi Germany are exploited to this end

is stomach-turning and disgusting beyond words.

Political fusion

Sabbatian manipulation has reversed the roles of Republicans and

Democrats and the same has happened in Britain with the

Conservative and Labour Parties. Republicans and Conservatives

were always labelled the ‘right’ and Democrats and Labour the ‘le�’,

but look at the policy positions now and the Democrat-Labour ‘le�’

has moved further to the ‘right’ than Republicans and Conservatives

under the banner of ‘Woke’, the Cult-created far-right tyranny.

Where once the Democrat-Labour ‘le�’ defended free speech and

human rights they now seek to delete them and as I said earlier

despite the ‘Covid’ fascism of the Jackboot Johnson Conservative

government in the UK the Labour Party of leader Keir Starmer

demanded even more extreme measures. The Labour Party has been

very publicly absorbed by Sabbatians a�er a political and media

onslaught against the previous leader, the weak and inept Jeremy

Corbyn, over made-up allegations of ‘anti-Semitism’ both by him

and his party. The plan was clear with this ‘anti-Semite’ propaganda

and what was required in response was a swi� and decisive ‘fuck

off’ from Corbyn and a statement to expose the Anti-Semitism

Industry (Sabbatian) a�empt to silence Labour criticism of the Israeli

government (Sabbatians) and purge the party of all dissent against

the extremes of ultra-Zionism (Sabbatians). Instead Corbyn and his

party fell to their knees and appeased the abusers which, by

definition, is impossible. Appeasing one demand leads only to a new

demand to be appeased until takeover is complete. Like I say – ‘fuck

off’ would have been a much more effective policy and I have used it

myself with great effect over the years when Sabbatians are on my

case which is most of the time. I consider that fact a great

compliment, by the way. The outcome of the Labour Party

capitulation is that we now have a Sabbatian-controlled

Conservative Party ‘opposed’ by a Sabbatian-controlled Labour

Party in a one-party Sabbatian state that hurtles towards the

extremes of tyranny (the Sabbatian cult agenda). In America the

situation is the same. Labour’s Keir Starmer spends his days on his

knees with his tongue out pointing to Tel Aviv, or I guess now

Jerusalem, while Boris Johnson has an ‘anti-Semitism czar’ in the

form of former Labour MP John Mann who keeps Starmer company

on his prayer mat.

Sabbatian influence can be seen in Jewish members of the Labour

Party who have been ejected for criticism of Israel including those

from families that suffered in Nazi Germany. Sabbatians despise real

Jewish people and target them even more harshly because it is so

much more difficult to dub them ‘anti-Semitic’ although in their

desperation they do try.

I

CHAPTER THREE

The Pushbacker sting

Until you realize how easy it is for your mind to be manipulated, you

remain the puppet of someone else’s game

Evita Ochel

will use the presidencies of Trump and Biden to show how the

manipulation of the one-party state plays out behind the illusion

of political choice across the world. No two presidencies could – on

the face of it – be more different and apparently at odds in terms of

direction and policy.

A Renegade Mind sees beyond the obvious and focuses on

outcomes and consequences and not image, words and waffle. The

Cult embarked on a campaign to divide America between those who

blindly support its agenda (the mentality known as ‘Woke’) and

those who are pushing back on where the Cult and its Sabbatians

want to go. This presents infinite possibilities for dividing and ruling

the population by se�ing them at war with each other and allows a

perceptual ring fence of demonisation to encircle the Pushbackers in

a modern version of the Li�le Big Horn in 1876 when American

cavalry led by Lieutenant Colonel George Custer were drawn into a

trap, surrounded and killed by Native American tribes defending

their land of thousands of years from being seized by the

government. In this modern version the roles are reversed and it’s

those defending themselves from the Sabbatian government who are

surrounded and the government that’s seeking to destroy them. This

trap was set years ago and to explain how we must return to 2016

and the emergence of Donald Trump as a candidate to be President

of the United States. He set out to overcome the best part of 20 other

candidates in the Republican Party before and during the primaries

and was not considered by many in those early stages to have a

prayer of living in the White House. The Republican Party was said

to have great reservations about Trump and yet somehow he won

the nomination. When you know how American politics works –

politics in general – there is no way that Trump could have become

the party’s candidate unless the Sabbatian-controlled ‘Neocons’ that

run the Republican Party wanted that to happen. We saw the proof

in emails and documents made public by WikiLeaks that the

Democratic Party hierarchy, or Democons, systematically

undermined the campaign of Bernie Sanders to make sure that

Sabbatian gofer Hillary Clinton won the nomination to be their

presidential candidate. If the Democons could do that then the

Neocons in the Republican Party could have derailed Trump in the

same way. But they didn’t and at that stage I began to conclude that

Trump could well be the one chosen to be president. If that was the

case the ‘why’ was pre�y clear to see – the goal of dividing America

between Cult agenda-supporting Wokers and Pushbackers who

gravitated to Trump because he was telling them what they wanted

to hear. His constituency of support had been increasingly ignored

and voiceless for decades and profoundly through the eight years of

Sabbatian puppet Barack Obama. Now here was someone speaking

their language of pulling back from the incessant globalisation of

political and economic power, the exporting of American jobs to

China and elsewhere by ‘American’ (Sabbatian) corporations, the

deletion of free speech, and the mass immigration policies that had

further devastated job opportunities for the urban working class of

all races and the once American heartlands of the Midwest.

Beware the forked tongue

Those people collectively sighed with relief that at last a political

leader was apparently on their side, but another trait of the

Renegade Mind is that you look even harder at people telling you

what you want to hear than those who are telling you otherwise.

Obviously as I said earlier people wish what they want to hear to be

true and genuine and they are much more likely to believe that than

someone saying what they don’t want to here and don’t want to be

true. Sales people are taught to be skilled in eliciting by calculated

questioning what their customers want to hear and repeating that

back to them as their own opinion to get their targets to like and

trust them. Assets of the Cult are also sales people in the sense of

selling perception. To read Cult manipulation you have to play the

long and expanded game and not fall for the Vaudeville show of

party politics. Both American parties are vehicles for the Cult and

they exploit them in different ways depending on what the agenda

requires at that moment. Trump and the Republicans were used to

be the focus of dividing America and isolating Pushbackers to open

the way for a Biden presidency to become the most extreme in

American history by advancing the full-blown Woke (Cult) agenda

with the aim of destroying and silencing Pushbackers now labelled

Nazi Trump supporters and white supremacists.

Sabbatians wanted Trump in office for the reasons described by

ultra-Zionist Saul Alinsky (1909-1972) who was promoting the Woke

philosophy through ‘community organising’ long before anyone had

heard of it. In those days it still went by its traditional name of

Marxism. The reason for the manipulated Trump phenomenon was

laid out in Alinsky’s 1971 book, Rules for Radicals, which was his

blueprint for overthrowing democratic and other regimes and

replacing them with Sabbatian Marxism. Not surprisingly his to-do

list was evident in the Sabbatian French and Russian ‘Revolutions’

and that in China which will become very relevant in the next

chapter about the ‘Covid’ hoax. Among Alinsky’s followers have

been the deeply corrupt Barack Obama, House Speaker Nancy Pelosi

and Hillary Clinton who described him as a ‘hero’. All three are

Sabbatian stooges with Pelosi personifying the arrogant corrupt

idiocy that so widely fronts up for the Cult inner core. Predictably as

a Sabbatian advocate of the ‘light-bringer’ Alinsky features Lucifer

on the dedication page of his book as the original radical who gained

his own kingdom (‘Earth’ as we shall see). One of Alinsky’s golden

radical rules was to pick an individual and focus all a�ention, hatred

and blame on them and not to target faceless bureaucracies and

corporations. Rules for Radicals is really a Sabbatian handbook with

its contents repeatedly employed all over the world for centuries and

why wouldn’t Sabbatians bring to power their designer-villain to be

used as the individual on which all a�ention, hatred and blame was

bestowed? This is what they did and the only question for me is how

much Trump knew that and how much he was manipulated. A bit of

both, I suspect. This was Alinsky’s Trump technique from a man

who died in 1972. The technique has spanned history:

Pick the target, freeze it, personalize it, polarize it. Don’t try to attack abstract corporations or bureaucracies. Identify a responsible individual. Ignore attempts to shift or spread the blame.

From the moment Trump came to illusory power everything was

about him. It wasn’t about Republican policy or opinion, but all

about Trump. Everything he did was presented in negative,

derogatory and abusive terms by the Sabbatian-dominated media

led by Cult operations such as CNN, MSNBC, The New York Times

and the Jeff Bezos-owned Washington Post – ‘Pick the target, freeze it,

personalize it, polarize it.’ Trump was turned into a demon to be

vilified by those who hated him and a demi-god loved by those who

worshipped him. This, in turn, had his supporters, too, presented as

equally demonic in preparation for the punchline later down the line

when Biden was about to take office. It was here’s a Trump, there’s a

Trump, everywhere a Trump, Trump. Virtually every news story or

happening was filtered through the lens of ‘The Donald’. You loved

him or hated him and which one you chose was said to define you as

Satan’s spawn or a paragon of virtue. Even supporting some Trump

policies or statements and not others was enough for an assault on

your character. No shades of grey were or are allowed. Everything is

black and white (literally and figuratively). A Californian I knew had

her head u�erly scrambled by her hatred for Trump while telling

people they should love each other. She was so totally consumed by

Trump Derangement Syndrome as it became to be known that this

glaring contradiction would never have occurred to her. By

definition anyone who criticised Trump or praised his opponents

was a hero and this lady described Joe Biden as ‘a kind, honest

gentleman’ when he’s a provable liar, mega-crook and vicious piece

of work to boot. Sabbatians had indeed divided America using

Trump as the fall-guy and all along the clock was ticking on the

consequences for his supporters.

In hock to his masters

Trump gave Sabbatians via Israel almost everything they wanted in

his four years. Ask and you shall receive was the dynamic between

himself and Benjamin Netanyahu orchestrated by Trump’s ultra-

Zionist son-in-law Jared Kushner, his ultra-Zionist Ambassador to

Israel, David Friedman, and ultra-Zionist ‘Israel adviser’, Jason

Greenbla�. The last two were central to the running and protecting

from collapse of his business empire, the Trump Organisation, and

colossal business failures made him forever beholding to Sabbatian

networks that bailed him out. By the start of the 1990s Trump owed

$4 billion to banks that he couldn’t pay and almost $1billion of that

was down to him personally and not his companies. This mega-

disaster was the result of building two new casinos in Atlantic City

and buying the enormous Taj Mahal operation which led to

crippling debt payments. He had borrowed fantastic sums from 72

banks with major Sabbatian connections and although the scale of

debt should have had him living in a tent alongside the highway

they never foreclosed. A plan was devised to li� Trump from the

mire by BT Securities Corporation and Rothschild Inc. and the case

was handled by Wilber Ross who had worked for the Rothschilds for

27 years. Ross would be named US Commerce Secretary a�er

Trump’s election. Another crucial figure in saving Trump was ultra-

Zionist ‘investor’ Carl Icahn who bought the Taj Mahal casino. Icahn

was made special economic adviser on financial regulation in the

Trump administration. He didn’t stay long but still managed to find

time to make a tidy sum of a reported $31.3 million when he sold his

holdings affected by the price of steel three days before Trump

imposed a 235 percent tariff on steel imports. What amazing bits of

luck these people have. Trump and Sabbatian operatives have long

had a close association and his mentor and legal adviser from the

early 1970s until 1986 was the dark and genetically corrupt ultra-

Zionist Roy Cohn who was chief counsel to Senator Joseph

McCarthy’s ‘communist’ witch-hunt in the 1950s. Esquire magazine

published an article about Cohn with the headline ‘Don’t mess with

Roy Cohn’. He was described as the most feared lawyer in New York

and ‘a ruthless master of dirty tricks ... [with] ... more than one Mafia

Don on speed dial’. Cohn’s influence, contacts, support and

protection made Trump a front man for Sabbatians in New York

with their connections to one of Cohn’s many criminal employers,

the ‘Russian’ Sabbatian Mafia. Israel-centric media mogul Rupert

Murdoch was introduced to Trump by Cohn and they started a long

friendship. Cohn died in 1986 weeks a�er being disbarred for

unethical conduct by the Appellate Division of the New York State

Supreme Court. The wheels of justice do indeed run slow given the

length of Cohn’s crooked career.

QAnon-sense

We are asked to believe that Donald Trump with his fundamental

connections to Sabbatian networks and operatives has been leading

the fight to stop the Sabbatian agenda for the fascistic control of

America and the world. Sure he has. A man entrapped during his

years in the White House by Sabbatian operatives and whose biggest

financial donor was casino billionaire Sheldon Adelson who was

Sabbatian to his DNA?? Oh, do come on. Trump has been used to

divide America and isolate Pushbackers on the Cult agenda under

the heading of ‘Trump supporters’, ‘insurrectionists’ and ‘white

supremacists’. The US Intelligence/Mossad Psyop or psychological

operation known as QAnon emerged during the Trump years as a

central pillar in the Sabbatian campaign to lead Pushbackers into the

trap set by those that wished to destroy them. I knew from the start

that QAnon was a scam because I had seen the same scenario many

times before over 30 years under different names and I had wri�en

about one in particular in the books. ‘Not again’ was my reaction

when QAnon came to the fore. The same script is pulled out every

few years and a new name added to the le�erhead. The story always

takes the same form: ‘Insiders’ or ‘the good guys’ in the government-

intelligence-military ‘Deep State’ apparatus were going to instigate

mass arrests of the ‘bad guys’ which would include the Rockefellers,

Rothschilds, Barack Obama, Hillary Clinton, George Soros, etc., etc.

Dates are given for when the ‘good guys’ are going to move in, but

the dates pass without incident and new dates are given which pass

without incident. The central message to Pushbackers in each case is

that they don’t have to do anything because there is ‘a plan’ and it is

all going to be sorted by the ‘good guys’ on the inside. ‘Trust the

plan’ was a QAnon mantra when the only plan was to misdirect

Pushbackers into pu�ing their trust in a Psyop they believed to be

real. Beware, beware, those who tell you what you want to hear and

always check it out. Right up to Biden’s inauguration QAnon was

still claiming that ‘the Storm’ was coming and Trump would stay on

as president when Biden and his cronies were arrested and jailed. It

was never going to happen and of course it didn’t, but what did

happen as a result provided that punchline to the Sabbatian

Trump/QAnon Psyop.

On January 6th, 2021, a very big crowd of Trump supporters

gathered in the National Mall in Washington DC down from the

Capitol Building to protest at what they believed to be widespread

corruption and vote fraud that stopped Trump being re-elected for a

second term as president in November, 2020. I say as someone that

does not support Trump or Biden that the evidence is clear that

major vote-fixing went on to favour Biden, a man with cognitive

problems so advanced he can o�en hardly string a sentence together

without reading the words wri�en for him on the Teleprompter.

Glaring ballot discrepancies included serious questions about

electronic voting machines that make vote rigging a comparative

cinch and hundreds of thousands of paper votes that suddenly

appeared during already advanced vote counts and virtually all of

them for Biden. Early Trump leads in crucial swing states suddenly

began to close and disappear. The pandemic hoax was used as the

excuse to issue almost limitless numbers of mail-in ballots with no

checks to establish that the recipients were still alive or lived at that

address. They were sent to streams of people who had not even

asked for them. Private organisations were employed to gather these

ballots and who knows what they did with them before they turned

up at the counts. The American election system has been

manipulated over decades to become a sick joke with more holes

than a Swiss cheese for the express purpose of dictating the results.

Then there was the criminal manipulation of information by

Sabbatian tech giants like Facebook, Twi�er and Google-owned

YouTube which deleted pro-Trump, anti-Biden accounts and posts

while everything in support of Biden was le� alone. Sabbatians

wanted Biden to win because a�er the dividing of America it was

time for full-on Woke and every aspect of the Cult agenda to be

unleashed.

Hunter gatherer

Extreme Silicon Valley bias included blocking information by the

New York Post exposing a Biden scandal that should have ended his

bid for president in the final weeks of the campaign. Hunter Biden,

his monumentally corrupt son, is reported to have sent a laptop to

be repaired at a local store and failed to return for it. Time passed

until the laptop became the property of the store for non-payment of

the bill. When the owner saw what was on the hard drive he gave a

copy to the FBI who did nothing even though it confirmed

widespread corruption in which the Joe Biden family were using his

political position, especially when he was vice president to Obama,

to make multiple millions in countries around the world and most

notably Ukraine and China. Hunter Biden’s one-time business

partner Tony Bobulinski went public when the story broke in the

New York Post to confirm the corruption he saw and that Joe Biden

not only knew what was going on he also profited from the spoils.

Millions were handed over by a Chinese company with close

connections – like all major businesses in China – to the Chinese

communist party of President Xi Jinping. Joe Biden even boasted at a

meeting of the Cult’s World Economic Forum that as vice president

he had ordered the government of Ukraine to fire a prosecutor. What

he didn’t mention was that the same man just happened to be

investigating an energy company which was part of Hunter Biden’s

corrupt portfolio. The company was paying him big bucks for no

other reason than the influence his father had. Overnight Biden’s

presidential campaign should have been over given that he had lied

publicly about not knowing what his son was doing. Instead almost

the entire Sabbatian-owned mainstream media and Sabbatian-

owned Silicon Valley suppressed circulation of the story. This alone

went a mighty way to rigging the election of 2020. Cult assets like

Mark Zuckerberg at Facebook also spent hundreds of millions to be

used in support of Biden and vote ‘administration’.

The Cult had used Trump as the focus to divide America and was

now desperate to bring in moronic, pliable, corrupt Biden to

complete the double-whammy. No way were they going to let li�le

things like the will of the people thwart their plan. Silicon Valley

widely censored claims that the election was rigged because it was

rigged. For the same reason anyone claiming it was rigged was

denounced as a ‘white supremacist’ including the pathetically few

Republican politicians willing to say so. Right across the media

where the claim was mentioned it was described as a ‘false claim’

even though these excuses for ‘journalists’ would have done no

research into the subject whatsoever. Trump won seven million more

votes than any si�ing president had ever achieved while somehow a

cognitively-challenged soon to be 78-year-old who was hidden away

from the public for most of the campaign managed to win more

votes than any presidential candidate in history. It makes no sense.

You only had to see election rallies for both candidates to witness the

enthusiasm for Trump and the apathy for Biden. Tens of thousands

would a�end Trump events while Biden was speaking in empty car

parks with o�en only television crews a�ending and framing their

shots to hide the fact that no one was there. It was pathetic to see

footage come to light of Biden standing at a podium making

speeches only to TV crews and party fixers while reading the words

wri�en for him on massive Teleprompter screens. So, yes, those

protestors on January 6th had a point about election rigging, but

some were about to walk into a trap laid for them in Washington by

the Cult Deep State and its QAnon Psyop. This was the Capitol Hill

riot ludicrously dubbed an ‘insurrection’.

The spider and the fly

Renegade Minds know there are not two ‘sides’ in politics, only one

side, the Cult, working through all ‘sides’. It’s a stage show, a puppet

show, to direct the perceptions of the population into focusing on

diversions like parties and candidates while missing the puppeteers

with their hands holding all the strings. The Capitol Hill

‘insurrection’ brings us back to the Li�le Big Horn. Having created

two distinct opposing groupings – Woke and Pushbackers – the trap

was about to be sprung. Pushbackers were to be encircled and

isolated by associating them all in the public mind with Trump and

then labelling Trump as some sort of Confederate leader. I knew

immediately that the Capitol riot was a set-up because of two things.

One was how easy the rioters got into the building with virtually no

credible resistance and secondly I could see – as with the ‘Covid’

hoax in the West at the start of 2020 – how the Cult could exploit the

situation to move its agenda forward with great speed. My

experience of Cult techniques and activities over more than 30 years

has showed me that while they do exploit situations they haven’t

themselves created this never happens with events of fundamental

agenda significance. Every time major events giving cultists the

excuse to rapidly advance their plan you find they are manipulated

into being for the specific reason of providing that excuse – Problem-

Reaction-Solution. Only a tiny minority of the huge crowd of

Washington protestors sought to gain entry to the Capitol by

smashing windows and breaching doors. That didn’t ma�er. The

whole crowd and all Pushbackers, even if they did not support

Trump, were going to be lumped together as dangerous

insurrectionists and conspiracy theorists. The la�er term came into

widespread use through a CIA memo in the 1960s aimed at

discrediting those questioning the nonsensical official story of the

Kennedy assassination and it subsequently became widely

employed by the media. It’s still being used by inept ‘journalists’

with no idea of its origin to discredit anyone questioning anything

that authority claims to be true. When you are perpetrating a

conspiracy you need to discredit the very word itself even though

the dictionary definition of conspiracy is merely ‘the activity of

secretly planning with other people to do something bad or illegal‘

and ‘a general agreement to keep silent about a subject for the

purpose of keeping it secret’. On that basis there are conspiracies

almost wherever you look. For obvious reasons the Cult and its

lapdog media have to claim there are no conspiracies even though

the word appears in state laws as with conspiracy to defraud, to

murder, and to corrupt public morals.

Agent provocateurs are widely used by the Cult Deep State to

manipulate genuine people into acting in ways that suit the desired

outcome. By genuine in this case I mean protestors genuinely

supporting Trump and claims that the election was stolen. In among

them, however, were agents of the state wearing the garb of Trump

supporters and QAnon to pump-prime the Capital riot which some

genuine Trump supporters naively fell for. I described the situation

as ‘Come into my parlour said the spider to the fly’. Leaflets

appeared through the Woke paramilitary arm Antifa, the anti-fascist

fascists, calling on supporters to turn up in Washington looking like

Trump supporters even though they hated him. Some of those

arrested for breaching the Capitol Building were sourced to Antifa

and its stable mate Black Lives Ma�er. Both organisations are funded

by Cult billionaires and corporations. One man charged for the riot

was according to his lawyer a former FBI agent who had held top

secret security clearance for 40 years. A�orney Thomas Plofchan said

of his client, 66-year-old Thomas Edward Caldwell:

He has held a Top Secret Security Clearance since 1979 and has undergone multiple Special Background Investigations in support of his clearances. After retiring from the Navy, he

worked as a section chief for the Federal Bureau of Investigation from 2009-2010 as a GS-12 [mid-level employee].

He also formed and operated a consulting firm performing work, often classified, for U.S government customers including the US. Drug Enforcement Agency, Department of Housing and Urban Development, the US Coast Guard, and the US Army Personnel Command.

A judge later released Caldwell pending trial in the absence of

evidence about a conspiracy or that he tried to force his way into the

building. The New York Post reported a ‘law enforcement source‘ as

saying that ‘at least two known Antifa members were spo�ed’ on

camera among Trump supporters during the riot while one of the

rioters arrested was John Earle Sullivan, a seriously extreme Black

Lives Ma�er Trump-hater from Utah who was previously arrested

and charged in July, 2020, over a BLM-Antifa riot in which drivers

were threatened and one was shot. Sullivan is the founder of Utah-

based Insurgence USA which is an affiliate of the Cult-created-and-

funded Black Lives Ma�er movement. Footage appeared and was

then deleted by Twi�er of Trump supporters calling out Antifa

infiltrators and a group was filmed changing into pro-Trump

clothing before the riot. Security at the building was pathetic – as

planned. Colonel Leroy Fletcher Prouty, a man with long experience

in covert operations working with the US security apparatus, once

described the tell-tale sign to identify who is involved in an

assassination. He said:

No one has to direct an assassination – it happens. The active role is played secretly by permitting it to happen. This is the greatest single clue. Who has the power to call off or reduce the usual security precautions?

This principle applies to many other situations and certainly to the

Capitol riot of January 6th, 2021.

The sting

With such a big and potentially angry crowd known to be gathering

near the Capitol the security apparatus would have had a major

police detail to defend the building with National Guard troops on

standby given the strength of feeling among people arriving from all

over America encouraged by the QAnon Psyop and statements by

Donald Trump. Instead Capitol Police ‘security’ was flimsy, weak,

and easily breached. The same number of officers was deployed as

on a regular day and that is a blatant red flag. They were not staffed

or equipped for a possible riot that had been an obvious possibility

in the circumstances. No protective and effective fencing worth the

name was put in place and there were no contingency plans. The

whole thing was basically a case of standing aside and waving

people in. Once inside police mostly backed off apart from one

Capitol police officer who ridiculously shot dead unarmed Air Force

veteran protestor Ashli Babbi� without a warning as she climbed

through a broken window. The ‘investigation’ refused to name or

charge the officer a�er what must surely be considered a murder in

the circumstances. They just li�ed a carpet and swept. The story was

endlessly repeated about five people dying in the ‘armed

insurrection’ when there was no report of rioters using weapons.

Apart from Babbi� the other four died from a heart a�ack, strokes

and apparently a drug overdose. Capitol police officer Brian Sicknick

was reported to have died a�er being bludgeoned with a fire

extinguisher when he was alive a�er the riot was over and died later

of what the Washington Medical Examiner’s Office said was a stroke.

Sicknick had no external injuries. The lies were delivered like rapid

fire. There was a narrative to build with incessant repetition of the lie

until the lie became the accepted ‘everybody knows that’ truth. The

‘Big Lie’ technique of Nazi Propaganda Minister Joseph Goebbels is

constantly used by the Cult which was behind the Nazis and is

today behind the ‘Covid’ and ‘climate change’ hoaxes. Goebbels

said:

If you tell a lie big enough and keep repeating it, people will eventually come to believe it. The lie can be maintained only for such time as the State can shield the people from the political, economic and/or military consequences of the lie. It thus becomes vitally important for the State to use all of its powers to repress dissent, for the truth is the mortal enemy of the lie, and thus by extension, the truth is the greatest enemy of the State.

Most protestors had a free run of the Capitol Building. This

allowed pictures to be taken of rioters in iconic parts of the building

including the Senate chamber which could be used as propaganda

images against all Pushbackers. One Congresswoman described the

scene as ‘the worst kind of non-security anybody could ever

imagine’. Well, the first part was true, but someone obviously did

imagine it and made sure it happened. Some photographs most

widely circulated featured people wearing QAnon symbols and now

the Psyop would be used to dub all QAnon followers with the

ubiquitous fit-all label of ‘white supremacist’ and ‘insurrectionists’.

When a Muslim extremist called Noah Green drove his car at two

police officers at the Capitol Building killing one in April, 2021, there

was no such political and media hysteria. They were just

disappointed he wasn’t white.

The witch-hunt

Government prosecutor Michael Sherwin, an aggressive, dark-eyed,

professional Ro�weiler led the ‘investigation’ and to call it over the

top would be to understate reality a thousand fold. Hundreds were

tracked down and arrested for the crime of having the wrong

political views and people were jailed who had done nothing more

than walk in the building, commi�ed no violence or damage to

property, took a few pictures and le�. They were labelled a ‘threat to

the Republic’ while Biden sat in the White House signing executive

orders wri�en for him that were dismantling ‘the Republic’. Even

when judges ruled that a mother and son should not be in jail the

government kept them there. Some of those arrested have been

badly beaten by prison guards in Washington and lawyers for one

man said he suffered a fractured skull and was made blind in one

eye. Meanwhile a woman is shot dead for no reason by a Capitol

Police officer and we are not allowed to know who he is never mind

what has happened to him although that will be nothing. The Cult’s

QAnon/Trump sting to identify and isolate Pushbackers and then

target them on the road to crushing and deleting them was a

resounding success. You would have thought the Russians had

invaded the building at gunpoint and lined up senators for a firing

squad to see the political and media reaction. Congresswoman

Alexandria Ocasio-Cortez is a child in a woman’s body, a terrible-

twos, me, me, me, Woker narcissist of such proportions that words

have no meaning. She said she thought she was going to die when

‘insurrectionists’ banged on her office door. It turned out she wasn’t

even in the Capitol Building when the riot was happening and the

‘banging’ was a Capitol Police officer. She referred to herself as a

‘survivor’ which is an insult to all those true survivors of violent and

sexual abuse while she lives her pampered and privileged life

talking drivel for a living. Her Woke colleague and fellow mega-

narcissist Rashida Tlaib broke down describing the devastating

effect on her, too, of not being in the building when the rioters were

there. Ocasio-Cortez and Tlaib are members of a fully-Woke group

of Congresswomen known as ‘The Squad’ along with Ilhan Omar

and Ayanna Pressley. The Squad from what I can see can be

identified by its vehement anti-white racism, anti-white men agenda,

and, as always in these cases, the absence of brain cells on active

duty.

The usual suspects were on the riot case immediately in the form

of Democrat ultra-Zionist senators and operatives Chuck Schumer

and Adam Schiff demanding that Trump be impeached for ‘his part

in the insurrection’. The same pair of prats had led the failed

impeachment of Trump over the invented ‘Russia collusion’

nonsense which claimed Russia had helped Trump win the 2016

election. I didn’t realise that Tel Aviv had been relocated just outside

Moscow. I must find an up-to-date map. The Russia hoax was a

Sabbatian operation to keep Trump occupied and impotent and to

stop any rapport with Russia which the Cult wants to retain as a

perceptual enemy to be pulled out at will. Puppet Biden began

a�acking Russia when he came to office as the Cult seeks more

upheaval, division and war across the world. A two-year stage show

‘Russia collusion inquiry’ headed by the not-very-bright former 9/11

FBI chief Robert Mueller, with support from 19 lawyers, 40 FBI

agents plus intelligence analysts, forensic accountants and other

staff, devoured tens of millions of dollars and found no evidence of

Russia collusion which a ten-year-old could have told them on day

one. Now the same moronic Schumer and Schiff wanted a second

impeachment of Trump over the Capitol ‘insurrection’ (riot) which

the arrested development of Schumer called another ‘Pearl Harbor’

while others compared it with 9/11 in which 3,000 died and, in the

case of CNN, with the Rwandan genocide in the 1990s in which an

estimated 500,000 to 600,000 were murdered, between 250, 000 and

500,000 women were raped, and populations of whole towns were

hacked to death with machetes. To make those comparisons purely

for Cult political reasons is beyond insulting to those that suffered

and lost their lives and confirms yet again the callous inhumanity

that we are dealing with. Schumer is a monumental idiot and so is

Schiff, but they serve the Cult agenda and do whatever they’re told

so they get looked a�er. Talking of idiots – another inane man who

spanned the Russia and Capitol impeachment a�empts was Senator

Eric Swalwell who had the nerve to accuse Trump of collusion with

the Russians while sleeping with a Chinese spy called Christine Fang

or ‘Fang Fang’ which is straight out of a Bond film no doubt starring

Klaus Schwab as the bloke living on a secret island and controlling

laser weapons positioned in space and pointing at world capitals.

Fang Fang plays the part of Bond’s infiltrator girlfriend which I’m

sure she would enjoy rather more than sharing a bed with the

brainless Swalwell, lying back and thinking of China. The FBI

eventually warned Swalwell about Fang Fang which gave her time

to escape back to the Chinese dictatorship. How very thoughtful of

them. The second Trump impeachment also failed and hardly

surprising when an impeachment is supposed to remove a si�ing

president and by the time it happened Trump was no longer

president. These people are running your country America, well,

officially anyway. Terrifying isn’t it?

Outcomes tell the story - always

The outcome of all this – and it’s the outcome on which Renegade

Minds focus, not the words – was that a vicious, hysterical and

obviously pre-planned assault was launched on Pushbackers to

censor, silence and discredit them and even targeted their right to

earn a living. They have since been condemned as ‘domestic

terrorists’ that need to be treated like Al-Qaeda and Islamic State.

‘Domestic terrorists’ is a label the Cult has been trying to make stick

since the period of the Oklahoma bombing in 1995 which was

blamed on ‘far-right domestic terrorists’. If you read The Trigger you

will see that the bombing was clearly a Problem-Reaction-Solution

carried out by the Deep State during a Bill Clinton administration so

corrupt that no dictionary definition of the term would even nearly

suffice. Nearly 30, 000 troops were deployed from all over America

to the empty streets of Washington for Biden’s inauguration. Ten

thousand of them stayed on with the pretext of protecting the capital

from insurrectionists when it was more psychological programming

to normalise the use of the military in domestic law enforcement in

support of the Cult plan for a police-military state. Biden’s fascist

administration began a purge of ‘wrong-thinkers’ in the military

which means anyone that is not on board with Woke. The Capitol

Building was surrounded by a fence with razor wire and the Land of

the Free was further symbolically and literally dismantled. The circle

was completed with the installation of Biden and the exploitation of

the QAnon Psyop.

America had never been so divided since the civil war of the 19th

century, Pushbackers were isolated and dubbed terrorists and now,

as was always going to happen, the Cult immediately set about

deleting what li�le was le� of freedom and transforming American

society through a swish of the hand of the most controlled

‘president’ in American history leading (officially at least) the most

extreme regime since the country was declared an independent state

on July 4th, 1776. Biden issued undebated, dictatorial executive

orders almost by the hour in his opening days in office across the

whole spectrum of the Cult wish-list including diluting controls on

the border with Mexico allowing thousands of migrants to illegally

enter the United States to transform the demographics of America

and import an election-changing number of perceived Democrat

voters. Then there were Biden deportation amnesties for the already

illegally resident (estimated to be as high as 20 or even 30 million). A

bill before Congress awarded American citizenship to anyone who

could prove they had worked in agriculture for just 180 days in the

previous two years as ‘Big Ag’ secured its slave labour long-term.

There were the plans to add new states to the union such as Puerto

Rico and making Washington DC a state. They are all parts of a plan

to ensure that the Cult-owned Woke Democrats would be

permanently in power.

Border – what border?

I have exposed in detail in other books how mass immigration into

the United States and Europe is the work of Cult networks fuelled by

the tens of billions spent to this and other ends by George Soros and

his global Open Society (open borders) Foundations. The impact can

be seen in America alone where the population has increased by 100

million in li�le more than 30 years mostly through immigration. I

wrote in The Answer that the plan was to have so many people

crossing the southern border that the numbers become unstoppable

and we are now there under Cult-owned Biden. El Salvador in

Central America puts the scale of what is happening into context. A

third of the population now lives in the United States, much of it

illegally, and many more are on the way. The methodology is to

crush Central and South American countries economically and

spread violence through machete-wielding psychopathic gangs like

MS-13 based in El Salvador and now operating in many American

cities. Biden-imposed lax security at the southern border means that

it is all but open. He said before his ‘election’ that he wanted to see a

surge towards the border if he became president and that was the

green light for people to do just that a�er election day to create the

human disaster that followed for both America and the migrants.

When that surge came the imbecilic Alexandria Ocasio-Cortez said it

wasn’t a ‘surge’ because they are ‘children, not insurgents’ and the

term ‘surge’ (used by Biden) was a claim of ‘white supremacists’.

This disingenuous lady may one day enter the realm of the most

basic intelligence, but it won’t be any time soon.

Sabbatians and the Cult are in the process of destroying America

by importing violent people and gangs in among the genuine to

terrorise American cities and by overwhelming services that cannot

cope with the sheer volume of new arrivals. Something similar is

happening in Europe as Western society in general is targeted for

demographic and cultural transformation and upheaval. The plan

demands violence and crime to create an environment of

intimidation, fear and division and Soros has been funding the

election of district a�orneys across America who then stop

prosecuting many crimes, reduce sentences for violent crimes and

free as many violent criminals as they can. Sabbatians are creating

the chaos from which order – their order – can respond in a classic

Problem-Reaction-Solution. A Freemasonic moto says ‘Ordo Ab

Chao’ (Order out of Chaos) and this is why the Cult is constantly

creating chaos to impose a new ‘order’. Here you have the reason

the Cult is constantly creating chaos. The ‘Covid’ hoax can be seen

with those entering the United States by plane being forced to take a

‘Covid’ test while migrants flooding through southern border

processing facilities do not. Nothing is put in the way of mass

migration and if that means ignoring the government’s own ‘Covid’

rules then so be it. They know it’s all bullshit anyway. Any pushback

on this is denounced as ‘racist’ by Wokers and Sabbatian fronts like

the ultra-Zionist Anti-Defamation League headed by the appalling

Jonathan Greenbla� which at the same time argues that Israel should

not give citizenship and voting rights to more Palestinian Arabs or

the ‘Jewish population’ (in truth the Sabbatian network) will lose

control of the country.

Society-changing numbers

Biden’s masters have declared that countries like El Salvador are so

dangerous that their people must be allowed into the United States

for humanitarian reasons when there are fewer murders in large

parts of many Central American countries than in US cities like

Baltimore. That is not to say Central America cannot be a dangerous

place and Cult-controlled American governments have been making

it so since way back, along with the dismantling of economies, in a

long-term plan to drive people north into the United States. Parts of

Central America are very dangerous, but in other areas the story is

being greatly exaggerated to justify relaxing immigration criteria.

Migrants are being offered free healthcare and education in the

United States as another incentive to head for the border and there is

no requirement to be financially independent before you can enter to

prevent the resources of America being drained. You can’t blame

migrants for seeking what they believe will be a be�er life, but they

are being played by the Cult for dark and nefarious ends. The

numbers since Biden took office are huge. In February, 2021, more

than 100,000 people were known to have tried to enter the US

illegally through the southern border (it was 34,000 in the same

month in 2020) and in March it was 170,000 – a 418 percent increase

on March, 2020. These numbers are only known people, not the ones

who get in unseen. The true figure for migrants illegally crossing the

border in a single month was estimated by one congressman at

250,000 and that number will only rise under Biden’s current policy.

Gangs of murdering drug-running thugs that control the Mexican

side of the border demand money – thousands of dollars – to let

migrants cross the Rio Grande into America. At the same time gun

ba�les are breaking out on the border several times a week between

rival Mexican drug gangs (which now operate globally) who are

equipped with sophisticated military-grade weapons, grenades and

armoured vehicles. While the Capitol Building was being ‘protected’

from a non-existent ‘threat’ by thousands of troops, and others were

still deployed at the time in the Cult Neocon war in Afghanistan, the

southern border of America was le� to its fate. This is not

incompetence, it is cold calculation.

By March, 2021, there were 17,000 unaccompanied children held at

border facilities and many of them are ensnared by people traffickers

for paedophile rings and raped on their journey north to America.

This is not conjecture – this is fact. Many of those designated

children are in reality teenage boys or older. Meanwhile Wokers

posture their self-purity for encouraging poor and tragic people to

come to America and face this nightmare both on the journey and at

the border with the disgusting figure of House Speaker Nancy Pelosi

giving disingenuous speeches about caring for migrants. The

woman’s evil. Wokers condemned Trump for having children in

cages at the border (so did Obama, Shhhh), but now they are sleeping

on the floor without access to a shower with one border facility 729

percent over capacity. The Biden insanity even proposed flying

migrants from the southern border to the northern border with

Canada for ‘processing’. The whole shambles is being overseen by

ultra-Zionist Secretary of Homeland Security, the moronic liar

Alejandro Mayorkas, who banned news cameras at border facilities

to stop Americans seeing what was happening. Mayorkas said there

was not a ban on news crews; it was just that they were not allowed

to film. Alongside him at Homeland Security is another ultra-Zionist

Cass Sunstein appointed by Biden to oversee new immigration laws.

Sunstein despises conspiracy researchers to the point where he

suggests they should be banned or taxed for having such views. The

man is not bonkers or anything. He’s perfectly well-adjusted, but

adjusted to what is the question. Criticise what is happening and

you are a ‘white supremacist’ when earlier non-white immigrants

also oppose the numbers which effect their lives and opportunities.

Black people in poor areas are particularly damaged by uncontrolled

immigration and the increased competition for work opportunities

with those who will work for less. They are also losing voting power

as Hispanics become more dominant in former black areas. It’s a

downward spiral for them while the billionaires behind the policy

drone on about how much they care about black people and

‘racism’. None of this is about compassion for migrants or black

people – that’s just wind and air. Migrants are instead being

mercilessly exploited to transform America while the countries they

leave are losing their future and the same is true in Europe. Mass

immigration may now be the work of Woke Democrats, but it can be

traced back to the 1986 Immigration Reform and Control Act (it

wasn’t) signed into law by Republican hero President Ronald

Reagan which gave amnesty to millions living in the United States

illegally and other incentives for people to head for the southern

border. Here we have the one-party state at work again.

Save me syndrome

Almost every aspect of what I have been exposing as the Cult

agenda was on display in even the first days of ‘Biden’ with silencing

of Pushbackers at the forefront of everything. A Renegade Mind will

view the Trump years and QAnon in a very different light to their

supporters and advocates as the dots are connected. The

QAnon/Trump Psyop has given the Cult all it was looking for. We

may not know how much, or li�le, that Trump realised he was being

used, but that’s a side issue. This pincer movement produced the

desired outcome of dividing America and having Pushbackers

isolated. To turn this around we have to look at new routes to

empowerment which do not include handing our power to other

people and groups through what I will call the ‘Save Me Syndrome’

– ‘I want someone else to do it so that I don’t have to’. We have seen

this at work throughout human history and the QAnon/Trump

Psyop is only the latest incarnation alongside all the others. Religion

is an obvious expression of this when people look to a ‘god’ or priest

to save them or tell them how to be saved and then there are ‘save

me’ politicians like Trump. Politics is a diversion and not a ‘saviour’.

It is a means to block positive change, not make it possible.

Save Me Syndrome always comes with the same repeating theme

of handing your power to whom or what you believe will save you

while your real ‘saviour’ stares back from the mirror every morning.

Renegade Minds are constantly vigilant in this regard and always

asking the question ‘What can I do?’ rather than ‘What can someone

else do for me?’ Gandhi was right when he said: ‘You must be the

change you want to see in the world.’ We are indeed the people we

have been waiting for. We are presented with a constant ra� of

reasons to concede that power to others and forget where the real

power is. Humanity has the numbers and the Cult does not. It has to

use diversion and division to target the unstoppable power that

comes from unity. Religions, governments, politicians, corporations,

media, QAnon, are all different manifestations of this power-

diversion and dilution. Refusing to give your power to governments

and instead handing it to Trump and QAnon is not to take a new

direction, but merely to recycle the old one with new names on the

posters. I will explore this phenomenon as we proceed and how to

break the cycles and recycles that got us here through the mists of

repeating perception and so repeating history.

For now we shall turn to the most potent example in the entire

human story of the consequences that follow when you give your

power away. I am talking, of course, of the ‘Covid’ hoax.

W

CHAPTER FOUR

‘Covid’: Calculated catastrophe

Facts are threatening to those invested in fraud

DaShanne Stokes

e can easily unravel the real reason for the ‘Covid pandemic’

hoax by employing the Renegade Mind methodology that I

have outlined this far. We’ll start by comparing the long-planned

Cult outcome with the ‘Covid pandemic’ outcome. Know the

outcome and you’ll see the journey.

I have highlighted the plan for the Hunger Games Society which

has been in my books for so many years with the very few

controlling the very many through ongoing dependency. To create

this dependency it is essential to destroy independent livelihoods,

businesses and employment to make the population reliant on the

state (the Cult) for even the basics of life through a guaranteed

pi�ance income. While independence of income remained these Cult

ambitions would be thwarted. With this knowledge it was easy to

see where the ‘pandemic’ hoax was going once talk of ‘lockdowns’

began and the closing of all but perceived ‘essential’ businesses to

‘save’ us from an alleged ‘deadly virus’. Cult corporations like

Amazon and Walmart were naturally considered ‘essential’ while

mom and pop shops and stores had their doors closed by fascist

decree. As a result with every new lockdown and new regulation

more small and medium, even large businesses not owned by the

Cult, went to the wall while Cult giants and their frontmen and

women grew financially fa�er by the second. Mom and pop were

denied an income and the right to earn a living and the wealth of

people like Jeff Bezos (Amazon), Mark Zuckerberg (Facebook) and

Sergei Brin and Larry Page (Google/Alphabet) have reached record

levels. The Cult was increasing its own power through further

dramatic concentrations of wealth while the competition was being

destroyed and brought into a state of dependency. Lockdowns have

been instigated to secure that very end and were never anything to

do with health. My brother Paul spent 45 years building up a bus

repair business, but lockdowns meant buses were running at a

fraction of normal levels for months on end. Similar stories can told

in their hundreds of millions worldwide. Efforts of a lifetime coldly

destroyed by Cult multi-billionaires and their lackeys in government

and law enforcement who continued to earn their living from the

taxation of the people while denying the right of the same people to

earn theirs. How different it would have been if those making and

enforcing these decisions had to face the same financial hardships of

those they affected, but they never do.

Gates of Hell

Behind it all in the full knowledge of what he is doing and why is

the psychopathic figure of Cult operative Bill Gates. His puppet

Tedros at the World Health Organization declared ‘Covid’ a

pandemic in March, 2020. The WHO had changed the definition of a

‘pandemic’ in 2009 just a month before declaring the ‘swine flu

pandemic’ which would not have been so under the previous

definition. The same applies to ‘Covid’. The definition had

included… ‘an infection by an infectious agent, occurring

simultaneously in different countries, with a significant mortality

rate relative to the proportion of the population infected’. The new

definition removed the need for ‘significant mortality’. The

‘pandemic’ has been fraudulent even down to the definition, but

Gates demanded economy-destroying lockdowns, school closures,

social distancing, mandatory masks, a ‘vaccination’ for every man,

woman and child on the planet and severe consequences and

restrictions for those that refused. Who gave him this power? The

Cult did which he serves like a li�le boy in short trousers doing

what his daddy tells him. He and his psychopathic missus even

smiled when they said that much worse was to come (what they

knew was planned to come). Gates responded in the ma�er-of-fact

way of all psychopaths to a question about the effect on the world

economy of what he was doing:

Well, it won’t go to zero but it will shrink. Global GDP is probably going to take the biggest hit ever [Gates was smiling as he said this] … in my lifetime this will be the greatest economic hit. But you don’t have a choice. People act as if you have a choice. People don’t feel like going to the stadium when they might get infected … People are deeply affected by seeing these stats, by knowing they could be part of the transmission chain, old people, their parents and grandparents, could be affected by this, and so you don’t get to say ignore what is going on here.

There will be the ability to open up, particularly in rich countries, if things are done well over the next few months, but for the world at large normalcy only returns when we have largely vaccinated the entire population.

The man has no compassion or empathy. How could he when he’s

a psychopath like all Cult players? My own view is that even beyond

that he is very seriously mentally ill. Look in his eyes and you can

see this along with his crazy flailing arms. You don’t do what he has

done to the world population since the start of 2020 unless you are

mentally ill and at the most extreme end of psychopathic. You

especially don’t do it when to you know, as we shall see, that cases

and deaths from ‘Covid’ are fakery and a product of monumental

figure massaging. ‘These stats’ that Gates referred to are based on a

‘test’ that’s not testing for the ‘virus’ as he has known all along. He

made his fortune with big Cult support as an infamously ruthless

so�ware salesman and now buys global control of ‘health’ (death)

policy without the population he affects having any say. It’s a

breathtaking outrage. Gates talked about people being deeply

affected by fear of ‘Covid’ when that was because of him and his

global network lying to them minute-by-minute supported by a

lying media that he seriously influences and funds to the tune of

hundreds of millions. He’s handed big sums to media operations

including the BBC, NBC, Al Jazeera, Univision, PBS NewsHour,

ProPublica, National Journal, The Guardian, The Financial Times, The

Atlantic, Texas Tribune, USA Today publisher Ganne�, Washington

Monthly, Le Monde, Center for Investigative Reporting, Pulitzer

Center on Crisis Reporting, National Press Foundation, International

Center for Journalists, Solutions Journalism Network, the Poynter

Institute for Media Studies, and many more. Gates is everywhere in

the ‘Covid’ hoax and the man must go to prison – or a mental facility

– for the rest of his life and his money distributed to those he has

taken such enormous psychopathic pleasure in crushing.

The Muscle

The Hunger Games global structure demands a police-military state

– a fusion of the two into one force – which viciously imposes the

will of the Cult on the population and protects the Cult from public

rebellion. In that regard, too, the ‘Covid’ hoax just keeps on giving.

O�en unlawful, ridiculous and contradictory ‘Covid’ rules and

regulations have been policed across the world by moronic

automatons and psychopaths made faceless by face-nappy masks

and acting like the Nazi SS and fascist blackshirts and brownshirts of

Hitler and Mussolini. The smallest departure from the rules decreed

by the psychos in government and their clueless gofers were jumped

upon by the face-nappy fascists. Brutality against public protestors

soon became commonplace even on girls, women and old people as

the brave men with the batons – the Face-Nappies as I call them –

broke up peaceful protests and handed out fines like confe�i to

people who couldn’t earn a living let alone pay hundreds of pounds

for what was once an accepted human right. Robot Face-Nappies of

No�ingham police in the English East Midlands fined one group

£11,000 for a�ending a child’s birthday party. For decades I charted

the transformation of law enforcement as genuine, decent officers

were replaced with psychopaths and the brain dead who would

happily and brutally do whatever their masters told them. Now they

were let loose on the public and I would emphasise the point that

none of this just happened. The step-by-step change in the dynamic

between police and public was orchestrated from the shadows by

those who knew where this was all going and the same with the

perceptual reframing of those in all levels of authority and official

administration through ‘training courses’ by organisations such as

Common Purpose which was created in the late 1980s and given a

massive boost in Blair era Britain until it became a global

phenomenon. Supposed public ‘servants’ began to view the

population as the enemy and the same was true of the police. This

was the start of the explosion of behaviour manipulation

organisations and networks preparing for the all-war on the human

psyche unleashed with the dawn of 2020. I will go into more detail

about this later in the book because it is a core part of what is

happening.

Police desecrated beauty spots to deter people gathering and

arrested women for walking in the countryside alone ‘too far’ from

their homes. We had arrogant, clueless sergeants in the Isle of Wight

police where I live posting on Facebook what they insisted the

population must do or else. A schoolmaster sergeant called Radford

looked young enough for me to ask if his mother knew he was out,

but he was posting what he expected people to do while a Sergeant

Wilkinson boasted about fining lads for meeting in a McDonald’s car

park where they went to get a lockdown takeaway. Wilkinson added

that he had even cancelled their order. What a pair of prats these

people are and yet they have increasingly become the norm among

Jackboot Johnson’s Yellowshirts once known as the British police.

This was the theme all over the world with police savagery common

during lockdown protests in the United States, the Netherlands, and

the fascist state of Victoria in Australia under its tyrannical and

again moronic premier Daniel Andrews. Amazing how tyrannical

and moronic tend to work as a team and the same combination

could be seen across America as arrogant, narcissistic Woke

governors and mayors such as Gavin Newsom (California), Andrew

Cuomo (New York), Gretchen Whitmer (Michigan), Lori Lightfoot

(Chicago) and Eric Garce�i (Los Angeles) did their Nazi and Stalin

impressions with the full support of the compliant brutality of their

enforcers in uniform as they arrested small business owners defying

fascist shutdown orders and took them to jail in ankle shackles and

handcuffs. This happened to bistro owner Marlena Pavlos-Hackney

in Gretchen Whitmer’s fascist state of Michigan when police arrived

to enforce an order by a state-owned judge for ‘pu�ing the

community at risk’ at a time when other states like Texas were

dropping restrictions and migrants were pouring across the

southern border without any ‘Covid’ questions at all. I’m sure there

are many officers appalled by what they are ordered to do, but not

nearly enough of them. If they were truly appalled they would not

do it. As the months passed every opportunity was taken to have the

military involved to make their presence on the streets ever more

familiar and ‘normal’ for the longer-term goal of police-military

fusion.

Another crucial element to the Hunger Games enforcement

network has been encouraging the public to report neighbours and

others for ‘breaking the lockdown rules’. The group faced with

£11,000 in fines at the child’s birthday party would have been

dobbed-in by a neighbour with a brain the size of a pea. The

technique was most famously employed by the Stasi secret police in

communist East Germany who had public informants placed

throughout the population. A police chief in the UK says his force

doesn’t need to carry out ‘Covid’ patrols when they are flooded with

so many calls from the public reporting other people for visiting the

beach. Dorset police chief James Vaughan said people were so

enthusiastic about snitching on their fellow humans they were now

operating as an auxiliary arm of the police: ‘We are still ge�ing

around 400 reports a week from the public, so we will respond to

reports …We won’t need to be doing hotspot patrols because people

are very quick to pick the phone up and tell us.’ Vaughan didn’t say

that this is a pillar of all tyrannies of whatever complexion and the

means to hugely extend the reach of enforcement while spreading

distrust among the people and making them wary of doing anything

that might get them reported. Those narcissistic Isle of Wight

sergeants Radford and Wilkinson never fail to add a link to their

Facebook posts where the public can inform on their fellow slaves.

Neither would be self-aware enough to realise they were imitating

the Stasi which they might well never have heard of. Government

psychologists that I will expose later laid out a policy to turn

communities against each other in the same way.

A coincidence? Yep, and I can knit fog

I knew from the start of the alleged pandemic that this was a Cult

operation. It presented limitless potential to rapidly advance the Cult

agenda and exploit manipulated fear to demand that every man,

woman and child on the planet was ‘vaccinated’ in a process never

used on humans before which infuses self-replicating synthetic

material into human cells. Remember the plan to transform the

human body from a biological to a synthetic biological state. I’ll deal

with the ‘vaccine’ (that’s not actually a vaccine) when I focus on the

genetic agenda. Enough to say here that mass global ‘vaccination’

justified by this ‘new virus’ set alarms ringing a�er 30 years of

tracking these people and their methods. The ‘Covid’ hoax officially

beginning in China was also a big red flag for reasons I will be

explaining. The agenda potential was so enormous that I could

dismiss any idea that the ‘virus’ appeared naturally. Major

happenings with major agenda implications never occur without

Cult involvement in making them happen. My questions were

twofold in early 2020 as the media began its campaign to induce

global fear and hysteria: Was this alleged infectious agent released

on purpose by the Cult or did it even exist at all? I then did what I

always do in these situations. I sat, observed and waited to see

where the evidence and information would take me. By March and

early April synchronicity was strongly – and ever more so since then

– pointing me in the direction of there is no ‘virus’. I went public on

that with derision even from swathes of the alternative media that

voiced a scenario that the Chinese government released the ‘virus’ in

league with Deep State elements in the United States from a top-

level bio-lab in Wuhan where the ‘virus’ is said to have first

appeared. I looked at that possibility, but I didn’t buy it for several

reasons. Deaths from the ‘virus’ did not in any way match what they

would have been with a ‘deadly bioweapon’ and it is much more

effective if you sell the illusion of an infectious agent rather than

having a real one unless you can control through injection who has it

and who doesn’t. Otherwise you lose control of events. A made-up

‘virus’ gives you a blank sheet of paper on which you can make it do

whatever you like and have any symptoms or mutant ‘variants’ you

choose to add while a real infectious agent would limit you to what

it actually does. A phantom disease allows you to have endless

ludicrous ‘studies’ on the ‘Covid’ dollar to widen the perceived

impact by inventing ever more ‘at risk’ groups including one study

which said those who walk slowly may be almost four times more

likely to die from the ‘virus’. People are in psychiatric wards for less.

A real ‘deadly bioweapon’ can take out people in the hierarchy

that are not part of the Cult, but essential to its operation. Obviously

they don’t want that. Releasing a real disease means you

immediately lose control of it. Releasing an illusory one means you

don’t. Again it’s vital that people are extra careful when dealing with

what they want to hear. A bioweapon unleashed from a Chinese

laboratory in collusion with the American Deep State may fit a

conspiracy narrative, but is it true? Would it not be far more effective

to use the excuse of a ‘virus’ to justify the real bioweapon – the

‘vaccine’? That way your disease agent does not have to be

transmi�ed and arrives directly through a syringe. I saw a French

virologist Luc Montagnier quoted in the alternative media as saying

he had discovered that the alleged ‘new’ severe acute respiratory

syndrome coronavirus , or SARS-CoV-2, was made artificially and

included elements of the human immunodeficiency ‘virus’ (HIV)

and a parasite that causes malaria. SARS-CoV-2 is alleged to trigger

an alleged illness called Covid-19. I remembered Montagnier’s name

from my research years before into claims that an HIV ‘retrovirus’

causes AIDs – claims that were demolished by Berkeley virologist

Peter Duesberg who showed that no one had ever proved that HIV

causes acquired immunodeficiency syndrome or AIDS. Claims that

become accepted as fact, publicly and medically, with no proof

whatsoever are an ever-recurring story that profoundly applies to

‘Covid’. Nevertheless, despite the lack of proof, Montagnier’s team

at the Pasteur Institute in Paris had a long dispute with American

researcher Robert Gallo over which of them discovered and isolated

the HIV ‘virus’ and with no evidence found it to cause AIDS. You will

see later that there is also no evidence that any ‘virus’ causes any

disease or that there is even such a thing as a ‘virus’ in the way it is

said to exist. The claim to have ‘isolated’ the HIV ‘virus’ will be

presented in its real context as we come to the shocking story – and

it is a story – of SARS-CoV-2 and so will Montagnier’s assertion that

he identified the full SARS-CoV-2 genome.

Hoax in the making

We can pick up the ‘Covid’ story in 2010 and the publication by the

Rockefeller Foundation of a document called ‘Scenarios for the

Future of Technology and International Development’. The inner

circle of the Rockefeller family has been serving the Cult since John

D. Rockefeller (1839-1937) made his fortune with Standard Oil. It is

less well known that the same Rockefeller – the Bill Gates of his day

– was responsible for establishing what is now referred to as ‘Big

Pharma’, the global network of pharmaceutical companies that make

outrageous profits dispensing scalpel and drug ‘medicine’ and are

obsessed with pumping vaccines in ever-increasing number into as

many human arms and backsides as possible. John D. Rockefeller

was the driving force behind the creation of the ‘education’ system

in the United States and elsewhere specifically designed to program

the perceptions of generations therea�er. The Rockefeller family

donated exceptionally valuable land in New York for the United

Nations building and were central in establishing the World Health

Organization in 1948 as an agency of the UN which was created

from the start as a Trojan horse and stalking horse for world

government. Now enter Bill Gates. His family and the Rockefellers

have long been extremely close and I have seen genealogy which

claims that if you go back far enough the two families fuse into the

same bloodline. Gates has said that the Bill and Melinda Gates

Foundation was inspired by the Rockefeller Foundation and why not

when both are serving the same Cult? Major tax-exempt foundations

are overwhelmingly criminal enterprises in which Cult assets fund

the Cult agenda in the guise of ‘philanthropy’ while avoiding tax in

the process. Cult operatives can become mega-rich in their role of

front men and women for the psychopaths at the inner core and

they, too, have to be psychopaths to knowingly serve such evil. Part

of the deal is that a big percentage of the wealth gleaned from

representing the Cult has to be spent advancing the ambitions of the

Cult and hence you have the Rockefeller Foundation, Bill and

Melinda Gates Foundation (and so many more) and people like

George Soros with his global Open Society Foundations spending

their billions in pursuit of global Cult control. Gates is a global

public face of the Cult with his interventions in world affairs

including Big Tech influence; a central role in the ‘Covid’ and

‘vaccine’ scam; promotion of the climate change shakedown;

manipulation of education; geoengineering of the skies; and his

food-control agenda as the biggest owner of farmland in America,

his GMO promotion and through other means. As one writer said:

‘Gates monopolizes or wields disproportionate influence over the

tech industry, global health and vaccines, agriculture and food policy

(including biopiracy and fake food), weather modification and other

climate technologies, surveillance, education and media.’ The almost

limitless wealth secured through Microso� and other not-allowed-

to-fail ventures (including vaccines) has been ploughed into a long,

long list of Cult projects designed to enslave the entire human race.

Gates and the Rockefellers have been working as one unit with the

Rockefeller-established World Health Organization leading global

‘Covid’ policy controlled by Gates through his mouth-piece Tedros.

Gates became the WHO’s biggest funder when Trump announced

that the American government would cease its donations, but Biden

immediately said he would restore the money when he took office in

January, 2021. The Gates Foundation (the Cult) owns through

limitless funding the world health system and the major players

across the globe in the ‘Covid’ hoax.

Okay, with that background we return to that Rockefeller

Foundation document of 2010 headed ‘Scenarios for the Future of

Technology and International Development’ and its ‘imaginary’

epidemic of a virulent and deadly influenza strain which infected 20

percent of the global population and killed eight million in seven

months. The Rockefeller scenario was that the epidemic destroyed

economies, closed shops, offices and other businesses and led to

governments imposing fierce rules and restrictions that included

mandatory wearing of face masks and body-temperature checks to

enter communal spaces like railway stations and supermarkets. The

document predicted that even a�er the height of the Rockefeller-

envisaged epidemic the authoritarian rule would continue to deal

with further pandemics, transnational terrorism, environmental

crises and rising poverty. Now you may think that the Rockefellers

are our modern-day seers or alternatively, and rather more likely,

that they well knew what was planned a few years further on.

Fascism had to be imposed, you see, to ‘protect citizens from risk

and exposure’. The Rockefeller scenario document said:

During the pandemic, national leaders around the world flexed their authority and imposed airtight rules and restrictions, from the mandatory wearing of face masks to body-temperature checks at the entries to communal spaces like train stations and supermarkets. Even after the pandemic faded, this more authoritarian control and oversight of citizens and their activities stuck and even intensified. In order to protect themselves from the spread of increasingly global problems – from pandemics and transnational terrorism to environmental crises and rising poverty – leaders around the world took a firmer grip on power.

At first, the notion of a more controlled world gained wide acceptance and approval. Citizens willingly gave up some of their sovereignty – and their privacy – to more paternalistic states in exchange for greater safety and stability. Citizens were more tolerant, and even eager, for top- down direction and oversight, and national leaders had more latitude to impose order in the ways they saw fit.

In developed countries, this heightened oversight took many forms: biometric IDs for all citizens, for example, and tighter regulation of key industries whose stability was deemed vital to national interests. In many developed countries, enforced cooperation with a suite of new regulations and agreements slowly but steadily restored both order and, importantly, economic growth.

There we have the prophetic Rockefellers in 2010 and three years

later came their paper for the Global Health Summit in Beijing,

China, when government representatives, the private sector,

international organisations and groups met to discuss the next 100

years of ‘global health’. The Rockefeller Foundation-funded paper

was called ‘Dreaming the Future of Health for the Next 100 Years

and more prophecy ensued as it described a dystopian future: ‘The

abundance of data, digitally tracking and linking people may mean

the ‘death of privacy’ and may replace physical interaction with

transient, virtual connection, generating isolation and raising

questions of how values are shaped in virtual networks.’ Next in the

‘Covid’ hoax preparation sequence came a ‘table top’ simulation in

2018 for another ‘imaginary’ pandemic of a disease called Clade X

which was said to kill 900 million people. The exercise was

organised by the Gates-funded Johns Hopkins University’s Center

for Health Security in the United States and this is the very same

university that has been compiling the disgustingly and

systematically erroneous global figures for ‘Covid’ cases and deaths.

Similar Johns Hopkins health crisis scenarios have included the Dark

Winter exercise in 2001 and Atlantic Storm in 2005.

Nostradamus 201

For sheer predictive genius look no further prophecy-watchers than

the Bill Gates-funded Event 201 held only six weeks before the

‘coronavirus pandemic’ is supposed to have broken out in China

and Event 201 was based on a scenario of a global ‘coronavirus

pandemic’. Melinda Gates, the great man’s missus, told the BBC that

he had ‘prepared for years’ for a coronavirus pandemic which told

us what we already knew. Nostradamugates had predicted in a TED

talk in 2015 that a pandemic was coming that would kill a lot of

people and demolish the world economy. My god, the man is a

machine – possibly even literally. Now here he was only weeks

before the real thing funding just such a simulated scenario and

involving his friends and associates at Johns Hopkins, the World

Economic Forum Cult-front of Klaus Schwab, the United Nations,

Johnson & Johnson, major banks, and officials from China and the

Centers for Disease Control in the United States. What synchronicity

– Johns Hopkins would go on to compile the fraudulent ‘Covid’

figures, the World Economic Forum and Schwab would push the

‘Great Reset’ in response to ‘Covid’, the Centers for Disease Control

would be at the forefront of ‘Covid’ policy in the United States,

Johnson & Johnson would produce a ‘Covid vaccine’, and

everything would officially start just weeks later in China. Spooky,

eh? They were even accurate in creating a simulation of a ‘virus’

pandemic because the ‘real thing’ would also be a simulation. Event

201 was not an exercise preparing for something that might happen;

it was a rehearsal for what those in control knew was going to

happen and very shortly. Hours of this simulation were posted on

the Internet and the various themes and responses mirrored what

would soon be imposed to transform human society. News stories

were inserted and what they said would be commonplace a few

weeks later with still more prophecy perfection. Much discussion

focused on the need to deal with misinformation and the ‘anti-vax

movement’ which is exactly what happened when the ‘virus’ arrived

– was said to have arrived – in the West.

Cult-owned social media banned criticism and exposure of the

official ‘virus’ narrative and when I said there was no ‘virus’ in early

April, 2020, I was banned by one platform a�er another including

YouTube, Facebook and later Twi�er. The mainstream broadcast

media in Britain was in effect banned from interviewing me by the

Tony-Blair-created government broadcasting censor Ofcom headed

by career government bureaucrat Melanie Dawes who was

appointed just as the ‘virus’ hoax was about to play out in January,

2020. At the same time the Ickonic media platform was using Vimeo,

another ultra-Zionist-owned operation, while our own player was

being created and they deleted in an instant hundreds of videos,

documentaries, series and shows to confirm their unbelievable

vindictiveness. We had copies, of course, and they had to be restored

one by one when our player was ready. These people have no class.

Sabbatian Facebook promised free advertisements for the Gates-

controlled World Health Organization narrative while deleting ‘false

claims and conspiracy theories’ to stop ‘misinformation’ about the

alleged coronavirus. All these responses could be seen just a short

while earlier in the scenarios of Event 201. Extreme censorship was

absolutely crucial for the Cult because the official story was so

ridiculous and unsupportable by the evidence that it could never

survive open debate and the free-flow of information and opinion. If

you can’t win a debate then don’t have one is the Cult’s approach

throughout history. Facebook’s li�le boy front man – front boy –

Mark Zuckerberg equated ‘credible and accurate information’ with

official sources and exposing their lies with ‘misinformation’.

Silencing those that can see

The censorship dynamic of Event 201 is now the norm with an army

of narrative-supporting ‘fact-checker’ organisations whose entire

reason for being is to tell the public that official narratives are true

and those exposing them are lying. One of the most appalling of

these ‘fact-checkers’ is called NewsGuard founded by ultra-Zionist

Americans Gordon Crovitz and Steven Brill. Crovitz is a former

publisher of The Wall Street Journal, former Executive Vice President

of Dow Jones, a member of the Council on Foreign Relations (CFR),

and on the board of the American Association of Rhodes Scholars.

The CFR and Rhodes Scholarships, named a�er Rothschild agent

Cecil Rhodes who plundered the gold and diamonds of South Africa

for his masters and the Cult, have featured widely in my books.

NewsGuard don’t seem to like me for some reason – I really can’t

think why – and they have done all they can to have me censored

and discredited which is, to quote an old British politician, like being

savaged by a dead sheep. They are, however, like all in the

censorship network, very well connected and funded by

organisations themselves funded by, or connected to, Bill Gates. As

you would expect with anything associated with Gates NewsGuard

has an offshoot called HealthGuard which ‘fights online health care

hoaxes’. How very kind. Somehow the NewsGuard European

Managing Director Anna-Sophie Harling, a remarkably young-

looking woman with no broadcasting experience and li�le hands-on

work in journalism, has somehow secured a position on the ‘Content

Board’ of UK government broadcast censor Ofcom. An executive of

an organisation seeking to discredit dissidents of the official

narratives is making decisions for the government broadcast

‘regulator’ about content?? Another appalling ‘fact-checker’ is Full

Fact funded by George Soros and global censors Google and

Facebook.

It’s amazing how many activists in the ‘fact-checking’, ‘anti-hate’,

arena turn up in government-related positions – people like UK

Labour Party activist Imran Ahmed who heads the Center for

Countering Digital Hate founded by people like Morgan

McSweeney, now chief of staff to the Labour Party’s hapless and

useless ‘leader’ Keir Starmer. Digital Hate – which is what it really is

– uses the American spelling of Center to betray its connection to a

transatlantic network of similar organisations which in 2020

shapeshi�ed from a�acking people for ‘hate’ to a�acking them for

questioning the ‘Covid’ hoax and the dangers of the ‘Covid vaccine’.

It’s just a coincidence, you understand. This is one of Imran Ahmed’s

hysterical statements: ‘I would go beyond calling anti-vaxxers

conspiracy theorists to say they are an extremist group that pose a

national security risk.’ No one could ever accuse this prat of

understatement and he’s including in that those parents who are

now against vaccines a�er their children were damaged for life or

killed by them. He’s such a nice man. Ahmed does the rounds of the

Woke media ge�ing so�-ball questions from spineless ‘journalists’

who never ask what right he has to campaign to destroy the freedom

of speech of others while he demands it for himself. There also

seems to be an overrepresentation in Ofcom of people connected to

the narrative-worshipping BBC. This incredible global network of

narrative-support was super-vital when the ‘Covid’ hoax was played

in the light of the mega-whopper lies that have to be defended from

the spotlight cast by the most basic intelligence.

Setting the scene

The Cult plays the long game and proceeds step-by-step ensuring

that everything is in place before major cards are played and they

don’t come any bigger than the ‘Covid’ hoax. The psychopaths can’t

handle events where the outcome isn’t certain and as li�le as

possible – preferably nothing – is le� to chance. Politicians,

government and medical officials who would follow direction were

brought to illusory power in advance by the Cult web whether on

the national stage or others like state governors and mayors of

America. For decades the dynamic between officialdom, law

enforcement and the public was changed from one of service to one

of control and dictatorship. Behaviour manipulation networks

established within government were waiting to impose the coming

‘Covid’ rules and regulations specifically designed to subdue and

rewire the psyche of the people in the guise of protecting health.

These included in the UK the Behavioural Insights Team part-owned

by the British government Cabinet Office; the Scientific Pandemic

Insights Group on Behaviours (SPI-B); and a whole web of

intelligence and military groups seeking to direct the conversation

on social media and control the narrative. Among them are the

cyberwarfare (on the people) 77th Brigade of the British military

which is also coordinated through the Cabinet Office as civilian and

military leadership continues to combine in what they call the

Fusion Doctrine. The 77th Brigade is a British equivalent of the

infamous Israeli (Sabbatian) military cyberwarfare and Internet

manipulation operation Unit 8200 which I expose at length in The

Trigger. Also carefully in place were the medical and science advisers

to government – many on the payroll past or present of Bill Gates –

and a whole alternative structure of unelected government stood by

to take control when elected parliaments were effectively closed

down once the ‘Covid’ card was slammed on the table. The structure

I have described here and so much more was installed in every

major country through the Cult networks. The top-down control

hierarchy looks like this: The Cult – Cult-owned Gates – the World

Health Organization and Tedros – Gates-funded or controlled chief

medical officers and science ‘advisers’ (dictators) in each country –

political ‘leaders’– law enforcement – The People. Through this

simple global communication and enforcement structure the policy

of the Cult could be imposed on virtually the entire human

population so long as they acquiesced to the fascism. With

everything in place it was time for the bu�on to be pressed in late

2019/early 2020.

These were the prime goals the Cult had to secure for its will to

prevail:

1) Locking down economies, closing all but designated ‘essential’ businesses (Cult-owned

corporations were ‘essential’), and pu�ing the population under house arrest was an

imperative to destroy independent income and employment and ensure dependency on the

Cult-controlled state in the Hunger Games Society. Lockdowns had to be established as the

global blueprint from the start to respond to the ‘virus’ and followed by pre�y much the

entire world.

2) The global population had to be terrified into believing in a deadly ‘virus’ that didn’t

actually exist so they would unquestioningly obey authority in the belief that authority

must know how best to protect them and their families. So�ware salesman Gates would

suddenly morph into the world’s health expert and be promoted as such by the Cult-owned

media.

3) A method of testing that wasn’t testing for the ‘virus’, but was only claimed to be, had to

be in place to provide the illusion of ‘cases’ and subsequent ‘deaths’ that had a very

different cause to the ‘Covid-19’ that would be scribbled on the death certificate.

4) Because there was no ‘virus’ and the great majority testing positive with a test not testing

for the ‘virus’ would have no symptoms of anything the lie had to be sold that people

without symptoms (without the ‘virus’) could still pass it on to others. This was crucial to

justify for the first time quarantining – house arresting – healthy people. Without this the

economy-destroying lockdown of everybody could not have been credibly sold.

5) The ‘saviour’ had to be seen as a vaccine which beyond evil drug companies were

working like angels of mercy to develop as quickly as possible, with all corners cut, to save

the day. The public must absolutely not know that the ‘vaccine’ had nothing to do with a

‘virus’ or that the contents were ready and waiting with a very different motive long before

the ‘Covid’ card was even li�ed from the pack.

I said in March, 2020, that the ‘vaccine’ would have been created

way ahead of the ‘Covid’ hoax which justified its use and the

following December an article in the New York Intelligencer

magazine said the Moderna ‘vaccine’ had been ‘designed’ by

January, 2020. This was ‘before China had even acknowledged that

the disease could be transmi�ed from human to human, more than a

week before the first confirmed coronavirus case in the United

States’. The article said that by the time the first American death was

announced a month later ‘the vaccine had already been

manufactured and shipped to the National Institutes of Health for

the beginning of its Phase I clinical trial’. The ‘vaccine’ was actually

‘designed’ long before that although even with this timescale you

would expect the article to ask how on earth it could have been done

that quickly. Instead it asked why the ‘vaccine’ had not been rolled

out then and not months later. Journalism in the mainstream is truly

dead. I am going to detail in the next chapter why the ‘virus’ has

never existed and how a hoax on that scale was possible, but first the

foundation on which the Big Lie of ‘Covid’ was built.

The test that doesn’t test

Fraudulent ‘testing’ is the bo�om line of the whole ‘Covid’ hoax and

was the means by which a ‘virus’ that did not exist appeared to exist.

They could only achieve this magic trick by using a test not testing

for the ‘virus’. To use a test that was testing for the ‘virus’ would

mean that every test would come back negative given there was no

‘virus’. They chose to exploit something called the RT-PCR test

invented by American biochemist Kary Mullis in the 1980s who said

publicly that his PCR test … cannot detect infectious disease. Yes, the

‘test’ used worldwide to detect infectious ‘Covid’ to produce all the

illusory ‘cases’ and ‘deaths’ compiled by Johns Hopkins and others

cannot detect infectious disease. This fact came from the mouth of the

man who invented PCR and was awarded the Nobel Prize in

Chemistry in 1993 for doing so. Sadly, and incredibly conveniently

for the Cult, Mullis died in August, 2019, at the age of 74 just before

his test would be fraudulently used to unleash fascism on the world.

He was said to have died from pneumonia which was an irony in

itself. A few months later he would have had ‘Covid-19’ on his death

certificate. I say the timing of his death was convenient because had

he lived Mullis, a brilliant, honest and decent man, would have been

vociferously speaking out against the use of his test to detect ‘Covid’

when it was never designed, or able, to do that. I know that to be

true given that Mullis made the same point when his test was used

to ‘detect’ – not detect – HIV. He had been seriously critical of the

Gallo/Montagnier claim to have isolated the HIV ‘virus’ and shown

it to cause AIDS for which Mullis said there was no evidence. AIDS

is actually not a disease but a series of diseases from which people

die all the time. When they die from those same diseases a�er a

positive ‘test’ for HIV then AIDS goes on their death certificate. I

think I’ve heard that before somewhere. Countries instigated a

policy with ‘Covid’ that anyone who tested positive with a test not

testing for the ‘virus’ and died of any other cause within 28 days and

even longer ‘Covid-19’ had to go on the death certificate. Cases have

come from the test that can’t test for infectious disease and the

deaths are those who have died of anything a�er testing positive

with a test not testing for the ‘virus’. I’ll have much more later about

the death certificate scandal.

Mullis was deeply dismissive of the now US ‘Covid’ star Anthony

Fauci who he said was a liar who didn’t know anything about

anything – ‘and I would say that to his face – nothing.’ He said of

Fauci: ‘The man thinks he can take a blood sample, put it in an

electron microscope and if it’s got a virus in there you’ll know it – he

doesn’t understand electron microscopy and he doesn’t understand

medicine and shouldn’t be in a position like he’s in.’ That position,

terrifyingly, has made him the decider of ‘Covid’ fascism policy on

behalf of the Cult in his role as director since 1984 of the National

Institute of Allergy and Infectious Diseases (NIAID) while his record

of being wrong is laughable; but being wrong, so long as it’s the right

kind of wrong, is why the Cult loves him. He’ll say anything the Cult

tells him to say. Fauci was made Chief Medical Adviser to the

President immediately Biden took office. Biden was installed in the

White House by Cult manipulation and one of his first decisions was

to elevate Fauci to a position of even more control. This is a

coincidence? Yes, and I identify as a flamenco dancer called Lola.

How does such an incompetent criminal like Fauci remain in that

pivotal position in American health since the 1980s? When you serve

the Cult it looks a�er you until you are surplus to requirements.

Kary Mullis said prophetically of Fauci and his like: ‘Those guys

have an agenda and it’s not an agenda we would like them to have

… they make their own rules, they change them when they want to,

and Tony Fauci does not mind going on television in front of the

people who pay his salary and lie directly into the camera.’ Fauci has

done that almost daily since the ‘Covid’ hoax began. Lying is in

Fauci’s DNA. To make the situation crystal clear about the PCR test

this is a direct quote from its inventor Kary Mullis:

It [the PCR test] doesn’t tell you that you’re sick and doesn’t tell you that the thing you ended up with was really going to hurt you ...’

Ask yourself why governments and medical systems the world over

have been using this very test to decide who is ‘infected’ with the

SARS-CoV-2 ‘virus’ and the alleged disease it allegedly causes,

‘Covid-19’. The answer to that question will tell you what has been

going on. By the way, here’s a li�le show-stopper – the ‘new’ SARS-

CoV-2 ‘virus’ was ‘identified’ as such right from the start using … the

PCR test not testing for the ‘virus’. If you are new to this and find that

shocking then stick around. I have hardly started yet. Even worse,

other ‘tests’, like the ‘Lateral Flow Device’ (LFD), are considered so

useless that they have to be confirmed by the PCR test! Leaked emails

wri�en by Ben Dyson, adviser to UK ‘Health’ Secretary Ma�

Hancock, said they were ‘dangerously unreliable’. Dyson, executive

director of strategy at the Department of Health, wrote: ‘As of today,

someone who gets a positive LFD result in (say) London has at best a

25 per cent chance of it being a true positive, but if it is a self-

reported test potentially as low as 10 per cent (on an optimistic

assumption about specificity) or as low as 2 per cent (on a more

pessimistic assumption).’ These are the ‘tests’ that schoolchildren

and the public are being urged to have twice a week or more and

have to isolate if they get a positive. Each fake positive goes in the

statistics as a ‘case’ no ma�er how ludicrously inaccurate and the

‘cases’ drive lockdown, masks and the pressure to ‘vaccinate’. The

government said in response to the email leak that the ‘tests’ were

accurate which confirmed yet again what shocking bloody liars they

are. The real false positive rate is 100 percent as we’ll see. In another

‘you couldn’t make it up’ the UK government agreed to pay £2.8

billion to California’s Innova Medical Group to supply the irrelevant

lateral flow tests. The company’s primary test-making centre is in

China. Innova Medical Group, established in March, 2020, is owned

by Pasaca Capital Inc, chaired by Chinese-American millionaire

Charles Huang who was born in Wuhan.

How it works – and how it doesn’t

The RT-PCR test, known by its full title of Polymerase chain reaction,

is used across the world to make millions, even billions, of copies of

a DNA/RNA genetic information sample. The process is called

‘amplification’ and means that a tiny sample of genetic material is

amplified to bring out the detailed content. I stress that it is not

testing for an infectious disease. It is simply amplifying a sample of

genetic material. In the words of Kary Mullis: ‘PCR is … just a

process that’s used to make a whole lot of something out of

something.’ To emphasise the point companies that make the PCR

tests circulated around the world to ‘test’ for ‘Covid’ warn on the

box that it can’t be used to detect ‘Covid’ or infectious disease and is

for research purposes only. It’s okay, rest for a minute and you’ll be

fine. This is the test that produces the ‘cases’ and ‘deaths’ that have

been used to destroy human society. All those global and national

medical and scientific ‘experts’ demanding this destruction to ‘save

us’ KNOW that the test is not testing for the ‘virus’ and the cases and

deaths they claim to be real are an almost unimaginable fraud. Every

one of them and so many others including politicians and

psychopaths like Gates and Tedros must be brought before

Nuremburg-type trials and jailed for the rest of their lives. The more

the genetic sample is amplified by PCR the more elements of that

material become sensitive to the test and by that I don’t mean

sensitive for a ‘virus’ but for elements of the genetic material which

is naturally in the body or relates to remnants of old conditions of

various kinds lying dormant and causing no disease. Once the

amplification of the PCR reaches a certain level everyone will test

positive. So much of the material has been made sensitive to the test

that everyone will have some part of it in their body. Even lying

criminals like Fauci have said that once PCR amplifications pass 35

cycles everything will be a false positive that cannot be trusted for

the reasons I have described. I say, like many proper doctors and

scientists, that 100 percent of the ‘positives’ are false, but let’s just go

with Fauci for a moment.

He says that any amplification over 35 cycles will produce false

positives and yet the US Centers for Disease Control (CDC) and

Food and Drug Administration (FDA) have recommended up to 40

cycles and the National Health Service (NHS) in Britain admi�ed in

an internal document for staff that it was using 45 cycles of

amplification. A long list of other countries has been doing the same

and at least one ‘testing’ laboratory has been using 50 cycles. Have

you ever heard a doctor, medical ‘expert’ or the media ask what level

of amplification has been used to claim a ‘positive’. The ‘test’ comes

back ‘positive’ and so you have the ‘virus’, end of story. Now we can

see how the government in Tanzania could send off samples from a

goat and a pawpaw fruit under human names and both came back

positive for ‘Covid-19’. Tanzania president John Magufuli mocked

the ‘Covid’ hysteria, the PCR test and masks and refused to import

the DNA-manipulating ‘vaccine’. The Cult hated him and an article

sponsored by the Bill Gates Foundation appeared in the London

Guardian in February, 2021, headed ‘It’s time for Africa to rein in

Tanzania’s anti-vaxxer president’. Well, ‘reined in’ he shortly was.

Magufuli appeared in good health, but then, in March, 2021, he was

dead at 61 from ‘heart failure’. He was replaced by Samia Hassan

Suhulu who is connected to Klaus Schwab’s World Economic Forum

and she immediately reversed Magufuli’s ‘Covid’ policy. A sample of

cola tested positive for ‘Covid’ with the PCR test in Germany while

American actress and singer-songwriter Erykah Badu tested positive

in one nostril and negative in the other. Footballer Ronaldo called

the PCR test ‘bullshit’ a�er testing positive three times and being

forced to quarantine and miss matches when there was nothing

wrong with him. The mantra from Tedros at the World Health

Organization and national governments (same thing) has been test,

test, test. They know that the more tests they can generate the more

fake ‘cases’ they have which go on to become ‘deaths’ in ways I am

coming to. The UK government has its Operation Moonshot planned

to test multiple millions every day in workplaces and schools with

free tests for everyone to use twice a week at home in line with the

Cult plan from the start to make testing part of life. A government

advertisement for an ‘Interim Head of Asymptomatic Testing

Communication’ said the job included responsibility for delivering a

‘communications strategy’ (propaganda) ‘to support the expansion

of asymptomatic testing that ‘normalises testing as part of everyday life’.

More tests means more fake ‘cases’, ‘deaths’ and fascism. I have

heard of, and from, many people who booked a test, couldn’t turn

up, and yet got a positive result through the post for a test they’d

never even had. The whole thing is crazy, but for the Cult there’s

method in the madness. Controlling and manipulating the level of

amplification of the test means the authorities can control whenever

they want the number of apparent ‘cases’ and ‘deaths’. If they want

to justify more fascist lockdown and destruction of livelihoods they

keep the amplification high. If they want to give the illusion that

lockdowns and the ‘vaccine’ are working then they lower the

amplification and ‘cases’ and ‘deaths’ will appear to fall. In January,

2021, the Cult-owned World Health Organization suddenly warned

laboratories about over-amplification of the test and to lower the

threshold. Suddenly headlines began appearing such as: ‘Why ARE

“Covid” cases plummeting?’ This was just when the vaccine rollout

was underway and I had predicted months before they would make

cases appear to fall through amplification tampering when the

‘vaccine’ came. These people are so predictable.

Cow vaccines?

The question must be asked of what is on the test swabs being poked

far up the nose of the population to the base of the brain? A nasal

swab punctured one woman’s brain and caused it to leak fluid. Most

of these procedures are being done by people with li�le training or

medical knowledge. Dr Lorraine Day, former orthopaedic trauma

surgeon and Chief of Orthopaedic Surgery at San Francisco General

Hospital, says the tests are really a ‘vaccine’. Cows have long been

vaccinated this way. She points out that masks have to cover the nose

and the mouth where it is claimed the ‘virus’ exists in saliva. Why

then don’t they take saliva from the mouth as they do with a DNA

test instead of pushing a long swab up the nose towards the brain?

The ethmoid bone separates the nasal cavity from the brain and

within that bone is the cribriform plate. Dr Day says that when the

swab is pushed up against this plate and twisted the procedure is

‘depositing things back there’. She claims that among these ‘things’

are nanoparticles that can enter the brain. Researchers have noted

that a team at the Gates-funded Johns Hopkins have designed tiny,

star-shaped micro-devices that can latch onto intestinal mucosa and

release drugs into the body. Mucosa is the thin skin that covers the

inside surface of parts of the body such as the nose and mouth and

produces mucus to protect them. The Johns Hopkins micro-devices

are called ‘theragrippers’ and were ‘inspired’ by a parasitic worm

that digs its sharp teeth into a host’s intestines. Nasal swabs are also

coated in the sterilisation agent ethylene oxide. The US National

Cancer Institute posts this explanation on its website:

At room temperature, ethylene oxide is a flammable colorless gas with a sweet odor. It is used primarily to produce other chemicals, including antifreeze. In smaller amounts, ethylene oxide is used as a pesticide and a sterilizing agent. The ability of ethylene oxide to damage DNA makes it an effective sterilizing agent but also accounts for its cancer-causing activity.

The Institute mentions lymphoma and leukaemia as cancers most

frequently reported to be associated with occupational exposure to

ethylene oxide along with stomach and breast cancers. How does

anyone think this is going to work out with the constant testing

regime being inflicted on adults and children at home and at school

that will accumulate in the body anything that’s on the swab?

Doctors know best

It is vital for people to realise that ‘hero’ doctors ‘know’ only what

the Big Pharma-dominated medical authorities tell them to ‘know’

and if they refuse to ‘know’ what they are told to ‘know’ they are out

the door. They are mostly not physicians or healers, but repeaters of

the official narrative – or else. I have seen alleged professional

doctors on British television make shocking statements that we are

supposed to take seriously. One called ‘Dr’ Amir Khan, who is

actually telling patients how to respond to illness, said that men

could take the birth pill to ‘help slow down the effects of Covid-19’.

In March, 2021, another ridiculous ‘Covid study’ by an American

doctor proposed injecting men with the female sex hormone

progesterone as a ‘Covid’ treatment. British doctor Nighat Arif told

the BBC that face coverings were now going to be part of ongoing

normal. Yes, the vaccine protects you, she said (evidence?) … but the

way to deal with viruses in the community was always going to

come down to hand washing, face covering and keeping a physical

distance. That’s not what we were told before the ‘vaccine’ was

circulating. Arif said she couldn’t imagine ever again going on the

underground or in a li� without a mask. I was just thanking my

good luck that she was not my doctor when she said – in March,

2021 – that if ‘we are behaving and we are doing all the right things’

she thought we could ‘have our nearest and dearest around us at

home … around Christmas and New Year! Her patronising delivery

was the usual school teacher talking to six-year-olds as she repeated

every government talking point and probably believed them all. If

we have learned anything from the ‘Covid’ experience surely it must

be that humanity’s perception of doctors needs a fundamental

rethink. NHS ‘doctor’ Sara Kayat told her television audience that

the ‘Covid vaccine’ would ‘100 percent prevent hospitalisation and

death’. Not even Big Pharma claimed that. We have to stop taking

‘experts’ at their word without question when so many of them are

clueless and only repeating the party line on which their careers

depend. That is not to say there are not brilliants doctors – there are

and I have spoken to many of them since all this began – but you

won’t see them in the mainstream media or quoted by the

psychopaths and yes-people in government.

Remember the name – Christian Drosten

German virologist Christian Drosten, Director of Charité Institute of

Virology in Berlin, became a national star a�er the pandemic hoax

began. He was feted on television and advised the German

government on ‘Covid’ policy. Most importantly to the wider world

Drosten led a group that produced the ‘Covid’ testing protocol for

the PCR test. What a remarkable feat given the PCR cannot test for

infectious disease and even more so when you think that Drosten

said that his method of testing for SARS-CoV-2 was developed

‘without having virus material available’. He developed a test for a

‘virus’ that he didn’t have and had never seen. Let that sink in as you

survey the global devastation that came from what he did. The

whole catastrophe of Drosten’s ‘test’ was based on the alleged

genetic sequence published by Chinese scientists on the Internet. We

will see in the next chapter that this alleged ‘genetic sequence’ has

never been produced by China or anyone and cannot be when there

is no SARS-CoV-2. Drosten, however, doesn’t seem to let li�le details

like that get in the way. He was the lead author with Victor Corman

from the same Charité Hospital of the paper ‘Detection of 2019 novel

coronavirus (2019-nCoV) by real-time PCR‘ published in a magazine

called Eurosurveillance. This became known as the Corman-Drosten

paper. In November, 2020, with human society devastated by the

effects of the Corman-Drosten test baloney, the protocol was publicly

challenged by 22 international scientists and independent

researchers from Europe, the United States, and Japan. Among them

were senior molecular geneticists, biochemists, immunologists, and

microbiologists. They produced a document headed ‘External peer

review of the RTPCR test to detect SARS-Cov-2 Reveals 10 Major

Flaws At The Molecular and Methodological Level: Consequences

•

•

•

•

•

•

For False-Positive Results’. The flaws in the Corman-Drosten test

included the following:

The test is non-specific because of erroneous design

Results are enormously variable

The test is unable to discriminate between the whole ‘virus’ and

viral fragments

It doesn’t have positive or negative controls

The test lacks a standard operating procedure

It is unsupported by proper peer view

The scientists said the PCR ‘Covid’ testing protocol was not

founded on science and they demanded the Corman-Drosten paper

be retracted by Eurosurveillance. They said all present and previous

Covid deaths, cases, and ‘infection rates’ should be subject to a

massive retroactive inquiry. Lockdowns and travel restrictions

should be reviewed and relaxed and those diagnosed through PCR

to have ‘Covid-19’ should not be forced to isolate. Dr Kevin Corbe�,

a health researcher and nurse educator with a long academic career

producing a stream of peer-reviewed publications at many UK

universities, made the same point about the PCR test debacle. He

said of the scientists’ conclusions: ‘Every scientific rationale for the

development of that test has been totally destroyed by this paper. It’s

like Hiroshima/Nagasaki to the Covid test.’ He said that China

hadn’t given them an isolated ‘virus’ when Drosten developed the

test. Instead they had developed the test from a sequence in a gene

bank.’ Put another way … they made it up! The scientists were

supported in this contention by a Portuguese appeals court which

ruled in November, 2020, that PCR tests are unreliable and it is

unlawful to quarantine people based solely on a PCR test. The point

about China not providing an isolated virus must be true when the

‘virus’ has never been isolated to this day and the consequences of

that will become clear. Drosten and company produced this useless

‘protocol’ right on cue in January, 2020, just as the ‘virus’ was said to

be moving westward and it somehow managed to successfully pass

a peer-review in 24 hours. In other words there was no peer-review

for a test that would be used to decide who had ‘Covid’ and who

didn’t across the world. The Cult-created, Gates-controlled World

Health Organization immediately recommended all its nearly 200

member countries to use the Drosten PCR protocol to detect ‘cases’

and ‘deaths’. The sting was underway and it continues to this day.

So who is this Christian Drosten that produced the means through

which death, destruction and economic catastrophe would be

justified? His education background, including his doctoral thesis,

would appear to be somewhat shrouded in mystery and his track

record is dire as with another essential player in the ‘Covid’ hoax,

the Gates-funded Professor Neil Ferguson at the Gates-funded

Imperial College in London of whom more shortly. Drosten

predicted in 2003 that the alleged original SARS ‘virus’ (SARS-1’)

was an epidemic that could have serious effects on economies and an

effective vaccine would take at least two years to produce. Drosten’s

answer to every alleged ‘outbreak’ is a vaccine which you won’t be

shocked to know. What followed were just 774 official deaths

worldwide and none in Germany where there were only nine cases.

That is even if you believe there ever was a SARS ‘virus’ when the

evidence is zilch and I will expand on this in the next chapter.

Drosten claims to be co-discoverer of ‘SARS-1’ and developed a test

for it in 2003. He was screaming warnings about ‘swine flu’ in 2009

and how it was a widespread infection far more severe than any

dangers from a vaccine could be and people should get vaccinated. It

would be helpful for Drosten’s vocal chords if he simply recorded

the words ‘the virus is deadly and you need to get vaccinated’ and

copies could be handed out whenever the latest made-up threat

comes along. Drosten’s swine flu epidemic never happened, but Big

Pharma didn’t mind with governments spending hundreds of

millions on vaccines that hardly anyone bothered to use and many

who did wished they hadn’t. A study in 2010 revealed that the risk

of dying from swine flu, or H1N1, was no higher than that of the

annual seasonal flu which is what at least most of ‘it’ really was as in

the case of ‘Covid-19’. A media investigation into Drosten asked

how with such a record of inaccuracy he could be the government

adviser on these issues. The answer to that question is the same with

Drosten, Ferguson and Fauci – they keep on giving the authorities

the ‘conclusions’ and ‘advice’ they want to hear. Drosten certainly

produced the goods for them in January, 2020, with his PCR protocol

garbage and provided the foundation of what German internal

medicine specialist Dr Claus Köhnlein, co-author of Virus Mania,

called the ‘test pandemic’. The 22 scientists in the Eurosurveillance

challenge called out conflicts of interest within the Drosten ‘protocol’

group and with good reason. Olfert Landt, a regular co-author of

Drosten ‘studies’, owns the biotech company TIB Molbiol

Syntheselabor GmbH in Berlin which manufactures and sells the

tests that Drosten and his mates come up with. They have done this

with SARS, Enterotoxigenic E. coli (ETEC), MERS, Zika ‘virus’,

yellow fever, and now ‘Covid’. Landt told the Berliner Zeitung

newspaper:

The testing, design and development came from the Charité [Drosten and Corman]. We simply implemented it immediately in the form of a kit. And if we don’t have the virus, which originally only existed in Wuhan, we can make a synthetic gene to simulate the genome of the virus. That’s what we did very quickly.

This is more confirmation that the Drosten test was designed

without access to the ‘virus’ and only a synthetic simulation which is

what SARS-CoV-2 really is – a computer-generated synthetic fiction.

It’s quite an enterprise they have going here. A Drosten team decides

what the test for something should be and Landt’s biotech company

flogs it to governments and medical systems across the world. His

company must have made an absolute fortune since the ‘Covid’ hoax

began. Dr Reiner Fuellmich, a prominent German consumer

protection trial lawyer in Germany and California, is on Drosten’s

case and that of Tedros at the World Health Organization for crimes

against humanity with a class-action lawsuit being prepared in the

United States and other legal action in Germany.

Why China?

Scamming the world with a ‘virus’ that doesn’t exist would seem

impossible on the face of it, but not if you have control of the

relatively few people that make policy decisions and the great

majority of the global media. Remember it’s not about changing

‘real’ reality it’s about controlling perception of reality. You don’t have

to make something happen you only have make people believe that

it’s happening. Renegade Minds understand this and are therefore

much harder to swindle. ‘Covid-19’ is not a ‘real’ ‘virus’. It’s a mind

virus, like a computer virus, which has infected the minds, not the

bodies, of billions. It all started, publically at least, in China and that

alone is of central significance. The Cult was behind the revolution

led by its asset Mao Zedong, or Chairman Mao, which established

the People’s Republic of China on October 1st, 1949. It should have

been called The Cult’s Republic of China, but the name had to reflect

the recurring illusion that vicious dictatorships are run by and for

the people (see all the ‘Democratic Republics’ controlled by tyrants).

In the same way we have the ‘Biden’ Democratic Republic of

America officially ruled by a puppet tyrant (at least temporarily) on

behalf of Cult tyrants. The creation of Mao’s merciless

communist/fascist dictatorship was part of a frenzy of activity by the

Cult at the conclusion of World War Two which, like the First World

War, it had instigated through its assets in Germany, Britain, France,

the United States and elsewhere. Israel was formed in 1948; the

Soviet Union expanded its ‘Iron Curtain’ control, influence and

military power with the Warsaw Pact communist alliance in 1955;

the United Nations was formed in 1945 as a Cult precursor to world

government; and a long list of world bodies would be established

including the World Health Organization (1948), World Trade

Organization (1948 under another name until 1995), International

Monetary Fund (1945) and World Bank (1944). Human society was

redrawn and hugely centralised in the global Problem-Reaction-

Solution that was World War Two. All these changes were

significant. Israel would become the headquarters of the Sabbatians

and the revolution in China would prepare the ground and control

system for the events of 2019/2020.

Renegade Minds know there are no borders except for public

consumption. The Cult is a seamless, borderless global entity and to

understand the game we need to put aside labels like borders,

nations, countries, communism, fascism and democracy. These

delude the population into believing that countries are ruled within

their borders by a government of whatever shade when these are

mere agencies of a global power. America’s illusion of democracy

and China’s communism/fascism are subsidiaries – vehicles – for the

same agenda. We may hear about conflict and competition between

America and China and on the lower levels that will be true; but at

the Cult level they are branches of the same company in the way of

the McDonald’s example I gave earlier. I have tracked in the books

over the years support by US governments of both parties for

Chinese Communist Party infiltration of American society through

allowing the sale of land, even military facilities, and the acquisition

of American business and university influence. All this is

underpinned by the infamous stealing of intellectual property and

technological know-how. Cult-owned Silicon Valley corporations

waive their fraudulent ‘morality’ to do business with human-rights-

free China; Cult-controlled Disney has become China’s PR

department; and China in effect owns ‘American’ sports such as

basketball which depends for much of its income on Chinese

audiences. As a result any sports player, coach or official speaking

out against China’s horrific human rights record is immediately

condemned or fired by the China-worshipping National Basketball

Association. One of the first acts of China-controlled Biden was to

issue an executive order telling federal agencies to stop making

references to the ‘virus’ by the ‘geographic location of its origin’.

Long-time Congressman Jerry Nadler warned that criticising China,

America’s biggest rival, leads to hate crimes against Asian people in

the United States. So shut up you bigot. China is fast closing in on

Israel as a country that must not be criticised which is apt, really,

given that Sabbatians control them both. The two countries have

developed close economic, military, technological and strategic ties

which include involvement in China’s ‘Silk Road’ transport and

economic initiative to connect China with Europe. Israel was the first

country in the Middle East to recognise the establishment of Mao’s

tyranny in 1950 months a�er it was established.

Project Wuhan – the ‘Covid’ Psyop

I emphasise again that the Cult plays the long game and what is

happening to the world today is the result of centuries of calculated

manipulation following a script to take control step-by-step of every

aspect of human society. I will discuss later the common force

behind all this that has spanned those centuries and thousands of

years if the truth be told. Instigating the Mao revolution in China in

1949 with a 2020 ‘pandemic’ in mind is not only how they work – the

71 years between them is really quite short by the Cult’s standards of

manipulation preparation. The reason for the Cult’s Chinese

revolution was to create a fiercely-controlled environment within

which an extreme structure for human control could be incubated to

eventually be unleashed across the world. We have seen this happen

since the ‘pandemic’ emerged from China with the Chinese control-

structure founded on AI technology and tyrannical enforcement

sweep across the West. Until the moment when the Cult went for

broke in the West and put its fascism on public display Western

governments had to pay some lip-service to freedom and democracy

to not alert too many people to the tyranny-in-the-making. Freedoms

were more subtly eroded and power centralised with covert

government structures put in place waiting for the arrival of 2020

when that smokescreen of ‘freedom’ could be dispensed with. The

West was not able to move towards tyranny before 2020 anything

like as fast as China which was created as a tyranny and had no

limits on how fast it could construct the Cult’s blueprint for global

control. When the time came to impose that structure on the world it

was the same Cult-owned Chinese communist/fascist government

that provided the excuse – the ‘Covid pandemic’. It was absolutely

crucial to the Cult plan for the Chinese response to the ‘pandemic’ –

draconian lockdowns of the entire population – to become the

blueprint that Western countries would follow to destroy the

livelihoods and freedom of their people. This is why the Cult-

owned, Gates-owned, WHO Director-General Tedros said early on:

The Chinese government is to be congratulated for the extraordinary measures it has taken to contain the outbreak. China is actually setting a new standard for outbreak response and it is not an exaggeration.

Forbes magazine said of China: ‘… those measures protected untold

millions from ge�ing the disease’. The Rockefeller Foundation

‘epidemic scenario’ document in 2010 said ‘prophetically’:

However, a few countries did fare better – China in particular. The Chinese government’s quick imposition and enforcement of mandatory quarantine for all citizens, as well as its instant and near-hermetic sealing off of all borders, saved millions of lives, stopping the spread of the virus far earlier than in other countries and enabling a swifter post-pandemic recovery.

Once again – spooky.

The first official story was the ‘bat theory’ or rather the bat

diversion. The source of the ‘virus outbreak’ we were told was a

‘‘wet market’ in Wuhan where bats and other animals are bought

and eaten in horrifically unhygienic conditions. Then another story

emerged through the alternative media that the ‘virus’ had been

released on purpose or by accident from a BSL-4 (biosafety level 4)

laboratory in Wuhan not far from the wet market. The lab was

reported to create and work with lethal concoctions and

bioweapons. Biosafety level 4 is the highest in the World Health

Organization system of safety and containment. Renegade Minds are

aware of what I call designer manipulation. The ideal for the Cult is

for people to buy its prime narrative which in the opening salvoes of

the ‘pandemic’ was the wet market story. It knows, however, that

there is now a considerable worldwide alternative media of

researchers sceptical of anything governments say and they are o�en

given a version of events in a form they can perceive as credible

while misdirecting them from the real truth. In this case let them

think that the conspiracy involved is a ‘bioweapon virus’ released

from the Wuhan lab to keep them from the real conspiracy – there is

no ‘virus’. The WHO’s current position on the source of the outbreak

at the time of writing appears to be: ‘We haven’t got a clue, mate.’

This is a good position to maintain mystery and bewilderment. The

inner circle will know where the ‘virus’ came from – nowhere. The

bo�om line was to ensure the public believed there was a ‘virus’ and

it didn’t much ma�er if they thought it was natural or had been

released from a lab. The belief that there was a ‘deadly virus’ was all

that was needed to trigger global panic and fear. The population was

terrified into handing their power to authority and doing what they

were told. They had to or they were ‘all gonna die’.

In March, 2020, information began to come my way from real

doctors and scientists and my own additional research which had

my intuition screaming: ‘Yes, that’s it! There is no virus.’ The

‘bioweapon’ was not the ‘virus’; it was the ‘vaccine’ already being

talked about that would be the bioweapon. My conclusion was

further enhanced by happenings in Wuhan. The ‘virus’ was said to

be sweeping the city and news footage circulated of people

collapsing in the street (which they’ve never done in the West with

the same ‘virus’). The Chinese government was building ‘new

hospitals’ in a ma�er of ten days to ‘cope with demand’ such was the

virulent nature of the ‘virus’. Yet in what seemed like no time the

‘new hospitals’ closed – even if they even opened – and China

declared itself ‘virus-free’. It was back to business as usual. This was

more propaganda to promote the Chinese draconian lockdowns in

the West as the way to ‘beat the virus’. Trouble was that we

subsequently had lockdown a�er lockdown, but never business as

usual. As the people of the West and most of the rest of the world

were caught in an ever-worsening spiral of lockdown, social

distancing, masks, isolated old people, families forced apart, and

livelihood destruction, it was party-time in Wuhan. Pictures

emerged of thousands of people enjoying pool parties and concerts.

It made no sense until you realised there never was a ‘virus’ and the

whole thing was a Cult set-up to transform human society out of one

its major global strongholds – China.

How is it possible to deceive virtually the entire world population

into believing there is a deadly virus when there is not even a ‘virus’

let alone a deadly one? It’s nothing like as difficult as you would

think and that’s clearly true because it happened.

Postscript: See end of book Postscript for more on the ‘Wuhan lab virus release’ story which the authorities and media were pushing

heavily in the summer of 2021 to divert a�ention from the truth that

the ‘Covid virus’ is pure invention.

T

CHAPTER FIVE

There is no ‘virus’

You can fool some of the people all of the time, and all of the people

some of the time, but you cannot fool all of the people all of the time

Abraham Lincoln

he greatest form of mind control is repetition. The more you

repeat the same mantra of alleged ‘facts’ the more will accept

them to be true. It becomes an ‘everyone knows that, mate’. If you

can also censor any other version or alternative to your alleged

‘facts’ you are pre�y much home and cooking.

By the start of 2020 the Cult owned the global mainstream media

almost in its entirety to spew out its ‘Covid’ propaganda and ignore

or discredit any other information and view. Cult-owned social

media platforms in Cult-owned Silicon Valley were poised and

ready to unleash a campaign of ferocious censorship to obliterate all

but the official narrative. To complete the circle many demands for

censorship by Silicon Valley were led by the mainstream media as

‘journalists’ became full-out enforcers for the Cult both as

propagandists and censors. Part of this has been the influx of young

people straight out of university who have become ‘journalists’ in

significant positions. They have no experience and a headful of

programmed perceptions from their years at school and university at

a time when today’s young are the most perceptually-targeted

generations in known human history given the insidious impact of

technology. They enter the media perceptually prepared and ready

to repeat the narratives of the system that programmed them to

repeat its narratives. The BBC has a truly pathetic ‘specialist

disinformation reporter’ called Marianna Spring who fits this bill

perfectly. She is clueless about the world, how it works and what is

really going on. Her role is to discredit anyone doing the job that a

proper journalist would do and system-serving hacks like Spring

wouldn’t dare to do or even see the need to do. They are too busy

licking the arse of authority which can never be wrong and, in the

case of the BBC propaganda programme, Panorama, contacting

payments systems such as PayPal to have a donations page taken

down for a film company making documentaries questioning

vaccines. Even the BBC soap opera EastEnders included a

disgracefully biased scene in which an inarticulate white working

class woman was made to look foolish for questioning the ‘vaccine’

while a well-spoken black man and Asian woman promoted the

government narrative. It ticked every BBC box and the fact that the

black and minority community was resisting the ‘vaccine’ had

nothing to do with the way the scene was wri�en. The BBC has

become a disgusting tyrannical propaganda and censorship

operation that should be defunded and disbanded and a free media

take its place with a brief to stop censorship instead of demanding it.

A BBC ‘interview’ with Gates goes something like: ‘Mr Gates, sir, if I

can call you sir, would you like to tell our audience why you are

such a great man, a wonderful humanitarian philanthropist, and

why you should absolutely be allowed as a so�ware salesman to

decide health policy for approaching eight billion people? Thank

you, sir, please sir.’ Propaganda programming has been incessant

and merciless and when all you hear is the same story from the

media, repeated by those around you who have only heard the same

story, is it any wonder that people on a grand scale believe absolute

mendacious garbage to be true? You are about to see, too, why this

level of information control is necessary when the official ‘Covid’

narrative is so nonsensical and unsupportable by the evidence.

Structure of Deceit

The pyramid structure through which the ‘Covid’ hoax has been

manifested is very simple and has to be to work. As few people as

possible have to be involved with full knowledge of what they are

doing – and why – or the real story would get out. At the top of the

pyramid are the inner core of the Cult which controls Bill Gates who,

in turn, controls the World Health Organization through his pivotal

funding and his puppet Director-General mouthpiece, Tedros.

Before he was appointed Tedros was chair of the Gates-founded

Global Fund to ‘fight against AIDS, tuberculosis and malaria’, a

board member of the Gates-funded ‘vaccine alliance’ GAVI, and on

the board of another Gates-funded organisation. Gates owns him

and picked him for a specific reason – Tedros is a crook and worse.

‘Dr’ Tedros (he’s not a medical doctor, the first WHO chief not to be)

was a member of the tyrannical Marxist government of Ethiopia for

decades with all its human rights abuses. He has faced allegations of

corruption and misappropriation of funds and was exposed three

times for covering up cholera epidemics while Ethiopia’s health

minister. Tedros appointed the mass-murdering genocidal

Zimbabwe dictator Robert Mugabe as a WHO goodwill ambassador

for public health which, as with Tedros, is like appointing a

psychopath to run a peace and love campaign. The move was so

ridiculous that he had to drop Mugabe in the face of widespread

condemnation. American economist David Steinman, a Nobel peace

prize nominee, lodged a complaint with the International Criminal

Court in The Hague over alleged genocide by Tedros when he was

Ethiopia’s foreign minister. Steinman says Tedros was a ‘crucial

decision maker’ who directed the actions of Ethiopia’s security forces

from 2013 to 2015 and one of three officials in charge when those

security services embarked on the ‘killing’ and ‘torturing’ of

Ethiopians. You can see where Tedros is coming from and it’s

sobering to think that he has been the vehicle for Gates and the Cult

to direct the global response to ‘Covid’. Think about that. A

psychopathic Cult dictates to psychopath Gates who dictates to

psychopath Tedros who dictates how countries of the world must

respond to a ‘Covid virus’ never scientifically shown to exist. At the

same time psychopathic Cult-owned Silicon Valley information

giants like Google, YouTube, Facebook and Twi�er announced very

early on that they would give the Cult/Gates/Tedros/WHO version

of the narrative free advertising and censor those who challenged

their intelligence-insulting, mendacious story.

The next layer in the global ‘medical’ structure below the Cult,

Gates and Tedros are the chief medical officers and science ‘advisers’

in each of the WHO member countries which means virtually all of

them. Medical officers and arbiters of science (they’re not) then take

the WHO policy and recommended responses and impose them on

their country’s population while the political ‘leaders’ say they are

deciding policy (they’re clearly not) by ‘following the science’ on the

advice of the ‘experts’ – the same medical officers and science

‘advisers’ (dictators). In this way with the rarest of exceptions the

entire world followed the same policy of lockdown, people

distancing, masks and ‘vaccines’ dictated by the psychopathic Cult,

psychopathic Gates and psychopathic Tedros who we are supposed

to believe give a damn about the health of the world population they

are seeking to enslave. That, amazingly, is all there is to it in terms of

crucial decision-making. Medical staff in each country then follow

like sheep the dictates of the shepherds at the top of the national

medical hierarchies – chief medical officers and science ‘advisers’

who themselves follow like sheep the shepherds of the World Health

Organization and the Cult. Shepherds at the national level o�en

have major funding and other connections to Gates and his Bill and

Melinda Gates Foundation which carefully hands out money like

confe�i at a wedding to control the entire global medical system

from the WHO down.

Follow the money

Christopher Whi�y, Chief Medical Adviser to the UK Government at

the centre of ‘virus’ policy, a senior adviser to the government’s

Scientific Advisory Group for Emergencies (SAGE), and Executive

Board member of the World Health Organization, was gi�ed a grant

of $40 million by the Bill and Melinda Gates Foundation for malaria

research in Africa. The BBC described the unelected Whi�y as ‘the

official who will probably have the greatest impact on our everyday

lives of any individual policymaker in modern times’ and so it

turned out. What Gates and Tedros have said Whi�y has done like

his equivalents around the world. Patrick Vallance, co-chair of SAGE

and the government’s Chief Scientific Adviser, is a former executive

of Big Pharma giant GlaxoSmithKline with its fundamental financial

and business connections to Bill Gates. In September, 2020, it was

revealed that Vallance owned a deferred bonus of shares in

GlaxoSmithKline worth £600,000 while the company was

‘developing’ a ‘Covid vaccine’. Move along now – nothing to see

here – what could possibly be wrong with that? Imperial College in

London, a major player in ‘Covid’ policy in Britain and elsewhere

with its ‘Covid-19’ Response Team, is funded by Gates and has big

connections to China while the now infamous Professor Neil

Ferguson, the useless ‘computer modeller’ at Imperial College is also

funded by Gates. Ferguson delivered the dramatically inaccurate

excuse for the first lockdowns (much more in the next chapter). The

Institute for Health Metrics and Evaluation (IHME) in the United

States, another source of outrageously false ‘Covid’ computer

models to justify lockdowns, is bankrolled by Gates who is a

vehement promotor of lockdowns. America’s version of Whi�y and

Vallance, the again now infamous Anthony Fauci, has connections to

‘Covid vaccine’ maker Moderna as does Bill Gates through funding

from the Bill and Melinda Gates Foundation. Fauci is director of the

National Institute of Allergy and Infectious Diseases (NIAID), a

major recipient of Gates money, and they are very close. Deborah

Birx who was appointed White House Coronavirus Response

Coordinator in February, 2020, is yet another with ties to Gates.

Everywhere you look at the different elements around the world

behind the coordination and decision making of the ‘Covid’ hoax

there is Bill Gates and his money. They include the World Health

Organization; Centers for Disease Control (CDC) in the United

States; National Institutes of Health (NIH) of Anthony Fauci;

Imperial College and Neil Ferguson; the London School of Hygiene

where Chris Whi�y worked; Regulatory agencies like the UK

Medicines & Healthcare products Regulatory Agency (MHRA)

which gave emergency approval for ‘Covid vaccines’; Wellcome

Trust; GAVI, the Vaccine Alliance; the Coalition for Epidemic

Preparedness Innovations (CEPI); Johns Hopkins University which

has compiled the false ‘Covid’ figures; and the World Economic

Forum. A Nationalfile.com article said:

Gates has a lot of pull in the medical world, he has a multi-million dollar relationship with Dr. Fauci, and Fauci originally took the Gates line supporting vaccines and casting doubt on [the drug hydroxychloroquine]. Coronavirus response team member Dr. Deborah Birx, appointed by former president Obama to serve as United States Global AIDS Coordinator, also sits on the board of a group that has received billions from Gates’ foundation, and Birx reportedly used a disputed Bill Gates-funded model for the White House’s Coronavirus effort. Gates is a big proponent for a population lockdown scenario for the Coronavirus outbreak.

Another funder of Moderna is the Defense Advanced Research

Projects Agency (DARPA), the technology-development arm of the

Pentagon and one of the most sinister organisations on earth.

DARPA had a major role with the CIA covert technology-funding

operation In-Q-Tel in the development of Google and social media

which is now at the centre of global censorship. Fauci and Gates are

extremely close and openly admit to talking regularly about ‘Covid’

policy, but then why wouldn’t Gates have a seat at every national

‘Covid’ table a�er his Foundation commi�ed $1.75 billion to the

‘fight against Covid-19’. When passed through our Orwellian

Translation Unit this means that he has bought and paid for the Cult-

driven ‘Covid’ response worldwide. Research the major ‘Covid’

response personnel in your own country and you will find the same

Gates funding and other connections again and again. Medical and

science chiefs following World Health Organization ‘policy’ sit atop

a medical hierarchy in their country of administrators, doctors and

nursing staff. These ‘subordinates’ are told they must work and

behave in accordance with the policy delivered from the ‘top’ of the

national ‘health’ pyramid which is largely the policy delivered by

the WHO which is the policy delivered by Gates and the Cult. The

whole ‘Covid’ narrative has been imposed on medical staff by a

climate of fear although great numbers don’t even need that to

comply. They do so through breathtaking levels of ignorance and

include doctors who go through life simply repeating what Big

Pharma and their hierarchical masters tell them to say and believe.

No wonder Big Pharma ‘medicine’ is one of the biggest killers on

Planet Earth.

The same top-down system of intimidation operates with regard

to the Cult Big Pharma cartel which also dictates policy through

national and global medical systems in this way. The Cult and Big

Pharma agendas are the same because the former controls and owns

the la�er. ‘Health’ administrators, doctors, and nursing staff are told

to support and parrot the dictated policy or they will face

consequences which can include being fired. How sad it’s been to see

medical staff meekly repeating and imposing Cult policy without

question and most of those who can see through the deceit are only

willing to speak anonymously off the record. They know what will

happen if their identity is known. This has le� the courageous few to

expose the lies about the ‘virus’, face masks, overwhelmed hospitals

that aren’t, and the dangers of the ‘vaccine’ that isn’t a vaccine. When

these medical professionals and scientists, some renowned in their

field, have taken to the Internet to expose the truth their articles,

comments and videos have been deleted by Cult-owned Facebook,

Twi�er and YouTube. What a real head-shaker to see YouTube

videos with leading world scientists and highly qualified medical

specialists with an added link underneath to the notorious Cult

propaganda website Wikipedia to find the ‘facts’ about the same

subject.

HIV – the ‘Covid’ trial-run

I’ll give you an example of the consequences for health and truth

that come from censorship and unquestioning belief in official

narratives. The story was told by PCR inventor Kary Mullis in his

book Dancing Naked in the Mind Field. He said that in 1984 he

accepted as just another scientific fact that Luc Montagnier of

France’s Pasteur Institute and Robert Gallo of America’s National

Institutes of Health had independently discovered that a ‘retrovirus’

dubbed HIV (human immunodeficiency virus) caused AIDS. They

were, a�er all, Mullis writes, specialists in retroviruses. This is how

the medical and science pyramids work. Something is announced or

assumed and then becomes an everybody-knows-that purely through

repetition of the assumption as if it is fact. Complete crap becomes

accepted truth with no supporting evidence and only repetition of

the crap. This is how a ‘virus’ that doesn’t exist became the ‘virus’

that changed the world. The HIV-AIDS fairy story became a multi-

billion pound industry and the media poured out propaganda

terrifying the world about the deadly HIV ‘virus’ that caused the

lethal AIDS. By then Mullis was working at a lab in Santa Monica,

California, to detect retroviruses with his PCR test in blood

donations received by the Red Cross. In doing so he asked a

virologist where he could find a reference for HIV being the cause of

AIDS. ‘You don’t need a reference,’ the virologist said … ‘Everybody

knows it.’ Mullis said he wanted to quote a reference in the report he

was doing and he said he felt a li�le funny about not knowing the

source of such an important discovery when everyone else seemed

to. The virologist suggested he cite a report by the Centers for

Disease Control and Prevention (CDC) on morbidity and mortality.

Mullis read the report, but it only said that an organism had been

identified and did not say how. The report did not identify the

original scientific work. Physicians, however, assumed (key recurring

theme) that if the CDC was convinced that HIV caused AIDS then

proof must exist. Mullis continues:

I did computer searches. Neither Montagnier, Gallo, nor anyone else had published papers describing experiments which led to the conclusion that HIV probably caused AIDS. I read the papers in Science for which they had become well known as AIDS doctors, but all they had said there was that they had found evidence of a past infection by something which was probably HIV in some AIDS patients.

They found antibodies. Antibodies to viruses had always been considered evidence of past disease, not present disease. Antibodies signaled that the virus had been defeated. The patient had saved himself. There was no indication in these papers that this virus caused a disease. They didn’t show that everybody with the antibodies had the disease. In fact they found some healthy people with antibodies.

Mullis asked why their work had been published if Montagnier

and Gallo hadn’t really found this evidence, and why had they been

fighting so hard to get credit for the discovery? He says he was

hesitant to write ‘HIV is the probable cause of AIDS’ until he found

published evidence to support that. ‘Tens of thousands of scientists

and researchers were spending billions of dollars a year doing

research based on this idea,’ Mullis writes. ‘The reason had to be

there somewhere; otherwise these people would not have allowed

their research to se�le into one narrow channel of investigation.’ He

said he lectured about PCR at numerous meetings where people

were always talking about HIV and he asked them how they knew

that HIV was the cause of AIDS:

Everyone said something. Everyone had the answer at home, in the office, in some drawer. They all knew, and they would send me the papers as soon as they got back. But I never got any papers. Nobody ever sent me the news about how AIDS was caused by HIV.

Eventually Mullis was able to ask Montagnier himself about the

reference proof when he lectured in San Diego at the grand opening

of the University of California AIDS Research Center. Mullis says

this was the last time he would ask his question without showing

anger. Montagnier said he should reference the CDC report. ‘I read

it’, Mullis said, and it didn’t answer the question. ‘If Montagnier

didn’t know the answer who the hell did?’ Then one night Mullis

was driving when an interview came on National Public Radio with

Peter Duesberg, a prominent virologist at Berkeley and a California

Scientist of the Year. Mullis says he finally understood why he could

not find references that connected HIV to AIDS – there weren’t any!

No one had ever proved that HIV causes AIDS even though it had

spawned a multi-billion pound global industry and the media was

repeating this as fact every day in their articles and broadcasts

terrifying the shit out of people about AIDS and giving the

impression that a positive test for HIV (see ‘Covid’) was a death

sentence. Duesberg was a threat to the AIDS gravy train and the

agenda that underpinned it. He was therefore abused and castigated

a�er he told the Proceedings of the National Academy of Sciences

there was no good evidence implicating the new ‘virus’. Editors

rejected his manuscripts and his research funds were deleted. Mullis

points out that the CDC has defined AIDS as one of more than 30

diseases if accompanied by a positive result on a test that detects

antibodies to HIV; but those same diseases are not defined as AIDS

cases when antibodies are not detected:

If an HIV-positive woman develops uterine cancer, for example, she is considered to have AIDS. If she is not HIV positive, she simply has uterine cancer. An HIV-positive man with tuberculosis has AIDS; if he tests negative he simply has tuberculosis. If he lives in Kenya or Colombia, where the test for HIV antibodies is too expensive, he is simply presumed to have the antibodies and therefore AIDS, and therefore he can be treated in the World Health Organization’s clinic. It’s the only medical help available in some places. And it’s free, because the countries that support WHO are worried about AIDS.

Mullis accuses the CDC of continually adding new diseases (see ever

more ‘Covid symptoms’) to the grand AIDS definition and of

virtually doctoring the books to make it appear as if the disease

continued to spread. He cites how in 1993 the CDC enormously

broadened its AIDS definition and county health authorities were

delighted because they received $2,500 per year from the Federal

government for every reported AIDS case. Ladies and gentlemen, I

have just described, via Kary Mullis, the ‘Covid pandemic’ of 2020

and beyond. Every element is the same and it’s been pulled off in the

same way by the same networks.

The ‘Covid virus’ exists? Okay – prove it. Er … still waiting

What Kary Mullis described with regard to ‘HIV’ has been repeated

with ‘Covid’. A claim is made that a new, or ‘novel’, infection has

been found and the entire medical system of the world repeats that

as fact exactly as they did with HIV and AIDS. No one in the

mainstream asks rather relevant questions such as ‘How do you

know?’ and ‘Where is your proof?’ The SARS-Cov-2 ‘virus’ and the

‘Covid-19 disease’ became an overnight ‘everybody-knows-that’.

The origin could be debated and mulled over, but what you could

not suggest was that ‘SARS-Cov-2’ didn’t exist. That would be

ridiculous. ‘Everybody knows’ the ‘virus’ exists. Well, I didn’t for

one along with American proper doctors like Andrew Kaufman and

Tom Cowan and long-time American proper journalist Jon

Rappaport. We dared to pursue the obvious and simple question:

‘Where’s the evidence?’ The overwhelming majority in medicine,

journalism and the general public did not think to ask that. A�er all,

everyone knew there was a new ‘virus’. Everyone was saying so and I

heard it on the BBC. Some would eventually argue that the ‘deadly

virus’ was nothing like as deadly as claimed, but few would venture

into the realms of its very existence. Had they done so they would

have found that the evidence for that claim had gone AWOL as with

HIV causes AIDS. In fact, not even that. For something to go AWOL

it has to exist in the first place and scientific proof for a ‘SARS-Cov-2’

can be filed under nothing, nowhere and zilch.

Dr Andrew Kaufman is a board-certified forensic psychiatrist in

New York State, a Doctor of Medicine and former Assistant

Professor and Medical Director of Psychiatry at SUNY Upstate

Medical University, and Medical Instructor of Hematology and

Oncology at the Medical School of South Carolina. He also studied

biology at the Massachuse�s Institute of Technology (MIT) and

trained in Psychiatry at Duke University. Kaufman is retired from

allopathic medicine, but remains a consultant and educator on

natural healing, I saw a video of his very early on in the ‘Covid’ hoax

in which he questioned claims about the ‘virus’ in the absence of any

supporting evidence and with plenty pointing the other way. I did

everything I could to circulate his work which I felt was asking the

pivotal questions that needed an answer. I can recommend an

excellent pull-together interview he did with the website The Last

Vagabond entitled Dr Andrew Kaufman: Virus Isolation, Terrain Theory

and Covid-19 and his website is andrewkaufmanmd.com. Kaufman is

not only a forensic psychiatrist; he is forensic in all that he does. He

always reads original scientific papers, experiments and studies

instead of second-third-fourth-hand reports about the ‘virus’ in the

media which are repeating the repeated repetition of the narrative.

When he did so with the original Chinese ‘virus’ papers Kaufman

realised that there was no evidence of a ‘SARS-Cov-2’. They had

never – from the start – shown it to exist and every repeat of this

claim worldwide was based on the accepted existence of proof that

was nowhere to be found – see Kary Mullis and HIV. Here we go

again.

Let’s postulate

Kaufman discovered that the Chinese authorities immediately

concluded that the cause of an illness that broke out among about

200 initial patients in Wuhan was a ‘new virus’ when there were no

grounds to make that conclusion. The alleged ‘virus’ was not

isolated from other genetic material in their samples and then shown

through a system known as Koch’s postulates to be the causative

agent of the illness. The world was told that the SARS-Cov-2 ‘virus’

caused a disease they called ‘Covid-19’ which had ‘flu-like’

symptoms and could lead to respiratory problems and pneumonia.

If it wasn’t so tragic it would almost be funny. ‘Flu-like’ symptoms’?

Pneumonia? Respiratory disease? What in CHINA and particularly in

Wuhan, one of the most polluted cities in the world with a resulting

epidemic of respiratory disease?? Three hundred thousand people

get pneumonia in China every year and there are nearly a billion

cases worldwide of ‘flu-like symptoms’. These have a whole range of

causes – including pollution in Wuhan – but no other possibility was

credibly considered in late 2019 when the world was told there was a

new and deadly ‘virus’. The global prevalence of pneumonia and

‘flu-like systems’ gave the Cult networks unlimited potential to re-

diagnose these other causes as the mythical ‘Covid-19’ and that is

what they did from the very start. Kaufman revealed how Chinese

medical and science authorities (all subordinates to the Cult-owned

communist government) took genetic material from the lungs of

only a few of the first patients. The material contained their own

cells, bacteria, fungi and other microorganisms living in their bodies.

The only way you could prove the existence of the ‘virus’ and its

responsibility for the alleged ‘Covid-19’ was to isolate the virus from

all the other material – a process also known as ‘purification’ – and

then follow the postulates sequence developed in the late 19th

century by German physician and bacteriologist Robert Koch which

became the ‘gold standard’ for connecting an alleged causation

agent to a disease:

1. The microorganism (bacteria, fungus, virus, etc.) must be present in every case of the

disease and all patients must have the same symptoms. It must also not be present in healthy

individuals.

2. The microorganism must be isolated from the host with the disease. If the microorganism

is a bacteria or fungus it must be grown in a pure culture. If it is a virus, it must be purified

(i.e. containing no other material except the virus particles) from a clinical sample.

3. The specific disease, with all of its characteristics, must be reproduced when the

infectious agent (the purified virus or a pure culture of bacteria or fungi) is inoculated into a

healthy, susceptible host.

4. The microorganism must be recoverable from the experimentally infected host as in step

2.

Not one of these criteria has been met in the case of ‘SARS-Cov-2’ and

‘Covid-19’. Not ONE. EVER. Robert Koch refers to bacteria and not

viruses. What are called ‘viral particles’ are so minute (hence masks

are useless by any definition) that they could only be seen a�er the

invention of the electron microscope in the 1930s and can still only

be observed through that means. American bacteriologist and

virologist Thomas Milton Rivers, the so-called ‘Father of Modern

Virology’ who was very significantly director of the Rockefeller

Institute for Medical Research in the 1930s, developed a less

stringent version of Koch’s postulates to identify ‘virus’ causation

known as ‘Rivers criteria’. ‘Covid’ did not pass that process either.

Some even doubt whether any ‘virus’ can be isolated from other

particles containing genetic material in the Koch method. Freedom

of Information requests in many countries asking for scientific proof

that the ‘Covid virus’ has been purified and isolated and shown to

exist have all come back with a ‘we don’t have that’ and when this

happened with a request to the UK Department of Health they

added this comment:

However, outside of the scope of the [Freedom of Information Act] and on a discretionary basis, the following information has been advised to us, which may be of interest. Most infectious diseases are caused by viruses, bacteria or fungi. Some bacteria or fungi have the capacity to grow on their own in isolation, for example in colonies on a petri dish. Viruses are different in that they are what we call ‘obligate pathogens’ – that is, they cannot survive or reproduce without infecting a host ...

… For some diseases, it is possible to establish causation between a microorganism and a disease by isolating the pathogen from a patient, growing it in pure culture and reintroducing it to a healthy organism. These are known as ‘Koch’s postulates’ and were developed in 1882. However, as our understanding of disease and different disease-causing agents has advanced, these are no longer the method for determining causation [Andrew Kaufman asks why in that case are there two published articles falsely claiming to satisfy Koch’s postulates].

It has long been known that viral diseases cannot be identified in this way as viruses cannot be grown in ‘pure culture’. When a patient is tested for a viral illness, this is normally done by looking for the presence of antigens, or viral genetic code in a host with molecular biology techniques [Kaufman asks how you could know the origin of these chemicals without having a pure culture for comparison].

For the record ‘antigens’ are defined so:

Invading microorganisms have antigens on their surface that the human body can recognise as being foreign – meaning not belonging to it. When the body recognises a foreign antigen, lymphocytes (white blood cells) produce antibodies, which are complementary in shape to the antigen.

Notwithstanding that this is open to question in relation to ‘SARS-

Cov-2’ the presence of ‘antibodies’ can have many causes and they

are found in people that are perfectly well. Kary Mullis said:

‘Antibodies … had always been considered evidence of past disease,

not present disease.’

‘Covid’ really is a computer ‘virus’ Where the UK Department of Health statement says ‘viruses’ are

now ‘diagnosed’ through a ‘viral genetic code in a host with

molecular biology techniques’, they mean … the PCR test which its

inventor said cannot test for infectious disease. They have no

credible method of connecting a ‘virus’ to a disease and we will see

that there is no scientific proof that any ‘virus’ causes any disease or

there is any such thing as a ‘virus’ in the way that it is described.

Tenacious Canadian researcher Christine Massey and her team made

some 40 Freedom of Information requests to national public health

agencies in different countries asking for proof that SARS-CoV-2 has

been isolated and not one of them could supply that information.

Massey said of her request in Canada: ‘Freedom of Information

reveals Public Health Agency of Canada has no record of ‘SARS-

COV-2’ isolation performed by anyone, anywhere, ever.’ If you

accept the comment from the UK Department of Health it’s because

they can’t isolate a ‘virus’. Even so many ‘science’ papers claimed to

have isolated the ‘Covid virus’ until they were questioned and had

to admit they hadn’t. A reply from the Robert Koch Institute in

Germany was typical: ‘I am not aware of a paper which purified

isolated SARS-CoV-2.’ So what the hell was Christian Drosten and

his gang using to design the ‘Covid’ testing protocol that has

produced all the illusory Covid’ cases and ‘Covid’ deaths when the

head of the Chinese version of the CDC admi�ed there was a

problem right from the start in that the ‘virus’ had never been

isolated/purified? Breathe deeply: What they are calling ‘Covid’ is

actually created by a computer program i.e. they made it up – er, that’s

it. They took lung fluid, with many sources of genetic material, from

one single person alleged to be infected with Covid-19 by a PCR test

which they claimed, without clear evidence, contained a ‘virus’. They

used several computer programs to create a model of a theoretical

virus genome sequence from more than fi�y-six million small

sequences of RNA, each of an unknown source, assembling them

like a puzzle with no known solution. The computer filled in the

gaps with sequences from bits in the gene bank to make it look like a

bat SARS-like coronavirus! A wave of the magic wand and poof, an

in silico (computer-generated) genome, a scientific fantasy, was

created. UK health researcher Dr Kevin Corbe� made the same point

with this analogy:

… It’s like giving you a few bones and saying that’s your fish. It could be any fish. Not even a skeleton. Here’s a few fragments of bones. That’s your fish … It’s all from gene bank and the bits of the virus sequence that weren’t there they made up.

They synthetically created them to fill in the blanks. That’s what genetics is; it’s a code. So it’s ABBBCCDDD and you’re missing some what you think is EEE so you put it in. It’s all

synthetic. You just manufacture the bits that are missing. This is the end result of the geneticization of virology. This is basically a computer virus.

Further confirmation came in an email exchange between British

citizen journalist Frances Leader and the government’s Medicines &

Healthcare Products Regulatory Agency (the Gates-funded MHRA)

which gave emergency permission for untested ‘Covid vaccines’ to

be used. The agency admi�ed that the ‘vaccine’ is not based on an

isolated ‘virus’, but comes from a computer-generated model. Frances

Leader was naturally banned from Cult-owned fascist Twi�er for

making this exchange public. The process of creating computer-

generated alleged ‘viruses’ is called ‘in silico’ or ‘in silicon’ –

computer chips – and the term ‘in silico’ is believed to originate with

biological experiments using only a computer in 1989. ‘Vaccines’

involved with ‘Covid’ are also produced ‘in silico’ or by computer

not a natural process. If the original ‘virus’ is nothing more than a

made-up computer model how can there be ‘new variants’ of

something that never existed in the first place? They are not new

‘variants’; they are new computer models only minutely different to

the original program and designed to further terrify the population

into having the ‘vaccine’ and submi�ing to fascism. You want a ‘new

variant’? Click, click, enter – there you go. Tell the medical

profession that you have discovered a ‘South African variant’, ‘UK

variants’ or a ‘Brazilian variant’ and in the usual HIV-causes-AIDS

manner they will unquestioningly repeat it with no evidence

whatsoever to support these claims. They will go on television and

warn about the dangers of ‘new variants’ while doing nothing more

than repeating what they have been told to be true and knowing that

any deviation from that would be career suicide. Big-time insiders

will know it’s a hoax, but much of the medical community is clueless

about the way they are being played and themselves play the public

without even being aware they are doing so. What an interesting

‘coincidence’ that AstraZeneca and Oxford University were

conducting ‘Covid vaccine trials’ in the three countries – the UK,

South Africa and Brazil – where the first three ‘variants’ were

claimed to have ‘broken out’.

Here’s your ‘virus’ – it’s a unicorn

Dr Andrew Kaufman presented a brilliant analysis describing how

the ‘virus’ was imagined into fake existence when he dissected an

article published by Nature and wri�en by 19 authors detailing

alleged ‘sequencing of a complete viral genome’ of the ‘new SARS-

CoV-2 virus’. This computer-modelled in silico genome was used as a

template for all subsequent genome sequencing experiments that

resulted in the so-called variants which he said now number more

than 6,000. The fake genome was constructed from more than 56

million individual short strands of RNA. Those li�le pieces were

assembled into longer pieces by finding areas of overlapping

sequences. The computer programs created over two million

possible combinations from which the authors simply chose the

longest one. They then compared this to a ‘bat virus’ and the

computer ‘alignment’ rearranged the sequence and filled in the gaps!

They called this computer-generated abomination the ‘complete

genome’. Dr Tom Cowan, a fellow medical author and collaborator

with Kaufman, said such computer-generation constitutes scientific

fraud and he makes this superb analogy:

Here is an equivalency: A group of researchers claim to have found a unicorn because they found a piece of a hoof, a hair from a tail, and a snippet of a horn. They then add that information into a computer and program it to re-create the unicorn, and they then claim this computer re-creation is the real unicorn. Of course, they had never actually seen a unicorn so could not possibly have examined its genetic makeup to compare their samples with the actual unicorn’s hair, hooves and horn.

The researchers claim they decided which is the real genome of SARS-CoV-2 by ‘consensus’, sort of like a vote. Again, different computer programs will come up with different versions of the imaginary ‘unicorn’, so they come together as a group and decide which is the real imaginary unicorn.

This is how the ‘virus’ that has transformed the world was brought

into fraudulent ‘existence’. Extraordinary, yes, but as the Nazis said

the bigger the lie the more will believe it. Cowan, however, wasn’t

finished and he went on to identify what he called the real

blockbuster in the paper. He quotes this section from a paper wri�en

by virologists and published by the CDC and then explains what it

means:

Therefore, we examined the capacity of SARS-CoV-2 to infect and replicate in several common primate and human cell lines, including human adenocarcinoma cells (A549), human liver cells (HUH 7.0), and human embryonic kidney cells (HEK-293T). In addition to Vero E6 and Vero CCL81 cells. ... Each cell line was inoculated at high multiplicity of infection and examined 24h post-infection.

No CPE was observed in any of the cell lines except in Vero cells, which grew to greater than 10 to the 7th power at 24 h post-infection. In contrast, HUH 7.0 and 293T showed only modest viral replication, and A549 cells were incompatible with SARS CoV-2 infection.

Cowan explains that when virologists a�empt to prove infection

they have three possible ‘hosts’ or models on which they can test.

The first was humans. Exposure to humans was generally not done

for ethical reasons and has never been done with SARS-CoV-2 or any

coronavirus. The second possible host was animals. Cowan said that

forge�ing for a moment that they never actually use purified virus

when exposing animals they do use solutions that they claim contain

the virus. Exposure to animals has been done with SARS-CoV-2 in

an experiment involving mice and this is what they found: None of

the wild (normal) mice got sick. In a group of genetically-modified

mice, a statistically insignificant number lost weight and had slightly

bristled fur, but they experienced nothing like the illness called

‘Covid-19’. Cowan said the third method – the one they mostly rely

on – is to inoculate solutions they say contain the virus onto a variety

of tissue cultures. This process had never been shown to kill tissue

unless the sample material was starved of nutrients and poisoned as

part of the process. Yes, incredibly, in tissue experiments designed to

show the ‘virus’ is responsible for killing the tissue they starve the

tissue of nutrients and add toxic drugs including antibiotics and they

do not have control studies to see if it’s the starvation and poisoning

that is degrading the tissue rather than the ‘virus’ they allege to be in

there somewhere. You want me to pinch you? Yep, I understand.

Tom Cowan said this about the whole nonsensical farce as he

explains what that quote from the CDC paper really means:

The shocking thing about the above quote is that using their own methods, the virologists found that solutions containing SARS-CoV-2 – even in high amounts – were NOT, I repeat NOT, infective to any of the three human tissue cultures they tested. In plain English, this means they proved, on their terms, that this ‘new coronavirus’ is not infectious to human beings. It is ONLY infective to monkey kidney cells, and only then when you add two potent drugs (gentamicin and amphotericin), known to be toxic to kidneys, to the mix.

My friends, read this again and again. These virologists, published by the CDC, performed a clear proof, on their terms, showing that the SARS-CoV-2 virus is harmless to human beings. That is the only possible conclusion, but, unfortunately, this result is not even mentioned in their conclusion. They simply say they can provide virus stocks cultured only on monkey Vero cells, thanks for coming.

Cowan concluded: ‘If people really understood how this “science”

was done, I would hope they would storm the gates and demand

honesty, transparency and truth.’ Dr Michael Yeadon, former Vice

President and Chief Scientific Adviser at drug giant Pfizer has been a

vocal critic of the ‘Covid vaccine’ and its potential for multiple harm.

He said in an interview in April, 2021, that ‘not one [vaccine] has the

virus. He was asked why vaccines normally using a ‘dead’ version of

a disease to activate the immune system were not used for ‘Covid’

and instead we had the synthetic methods of the ‘mRNA Covid

vaccine’. Yeadon said that to do the former ‘you’d have to have some

of [the virus] wouldn’t you?’ He added: ‘No-one’s got any –

seriously.’ Yeadon said that surely they couldn’t have fooled the

whole world for a year without having a virus, ‘but oddly enough

ask around – no one’s got it’. He didn’t know why with all the ‘great

labs’ around the world that the virus had not been isolated – ‘Maybe

they’ve been too busy running bad PCR tests and vaccines that

people don’t need.’ What is today called ‘science’ is not ‘science’ at

all. Science is no longer what is, but whatever people can be

manipulated to believe that it is. Real science has been hijacked by the

Cult to dispense and produce the ‘expert scientists’ and contentions

that suit the agenda of the Cult. How big-time this has happened

with the ‘Covid’ hoax which is entirely based on fake science

delivered by fake ‘scientists’ and fake ‘doctors’. The human-caused

climate change hoax is also entirely based on fake science delivered

by fake ‘scientists’ and fake ‘climate experts’. In both cases real

scientists, climate experts and doctors have their views suppressed

and deleted by the Cult-owned science establishment, media and

Silicon Valley. This is the ‘science’ that politicians claim to be

‘following’ and a common denominator of ‘Covid’ and climate are

Cult psychopaths Bill Gates and his mate Klaus Schwab at the Gates-

funded World Economic Forum. But, don’t worry, it’s all just a

coincidence and absolutely nothing to worry about. Zzzzzzzz.

What is a ‘virus’ REALLY?

Dr Tom Cowan is one of many contesting the very existence of

viruses let alone that they cause disease. This is understandable

when there is no scientific evidence for a disease-causing ‘virus’.

German virologist Dr Stefan Lanka won a landmark case in 2017 in

the German Supreme Court over his contention that there is no such

thing as a measles virus. He had offered a big prize for anyone who

could prove there is and Lanka won his case when someone sought

to claim the money. There is currently a prize of more than 225,000

euros on offer from an Isolate Truth Fund for anyone who can prove

the isolation of SARS-CoV-2 and its genetic substance. Lanka wrote

in an article headed ‘The Misconception Called Virus’ that scientists

think a ‘virus’ is causing tissue to become diseased and degraded

when in fact it is the processes they are using which do that – not a

‘virus’. Lanka has done an important job in making this point clear

as Cowan did in his analysis of the CDC paper. Lanka says that all

claims about viruses as disease-causing pathogens are wrong and

based on ‘easily recognisable, understandable and verifiable

misinterpretations.’ Scientists believed they were working with

‘viruses’ in their laboratories when they were really working with

‘typical particles of specific dying tissues or cells …’ Lanka said that

the tissue decaying process claimed to be caused by a ‘virus’ still

happens when no alleged ‘virus’ is involved. It’s the process that does

the damage and not a ‘virus’. The genetic sample is deprived of

nutrients, removed from its energy supply through removal from

the body and then doused in toxic antibiotics to remove any bacteria.

He confirms again that establishment scientists do not (pinch me)

conduct control experiments to see if this is the case and if they did

they would see the claims that ‘viruses’ are doing the damage is

nonsense. He adds that during the measles ‘virus’ court case he

commissioned an independent laboratory to perform just such a

control experiment and the result was that the tissues and cells died

in the exact same way as with alleged ‘infected’ material. This is

supported by a gathering number of scientists, doctors and

researchers who reject what is called ‘germ theory’ or the belief in

the body being infected by contagious sources emi�ed by other

people. Researchers Dawn Lester and David Parker take the same

stance in their highly-detailed and sourced book What Really Makes

You Ill – Why everything you thought you knew about disease is wrong

which was recommended to me by a number of medical

professionals genuinely seeking the truth. Lester and Parker say

there is no provable scientific evidence to show that a ‘virus’ can be

transmi�ed between people or people and animals or animals and

people:

The definition also claims that viruses are the cause of many diseases, as if this has been definitively proven. But this is not the case; there is no original scientific evidence that definitively demonstrates that any virus is the cause of any disease. The burden of proof for any theory lies with those who proposed it; but none of the existing documents provides ‘proof’ that supports the claim that ‘viruses’ are pathogens.

Dr Tom Cowan employs one of his clever analogies to describe the

process by which a ‘virus’ is named as the culprit for a disease when

what is called a ‘virus’ is only material released by cells detoxing

themselves from infiltration by chemical or radiation poisoning. The

tidal wave of technologically-generated radiation in the ‘smart’

modern world plus all the toxic food and drink are causing this to

happen more than ever. Deluded ‘scientists’ misread this as a

gathering impact of what they wrongly label ‘viruses’.

Paper can infect houses

Cowan said in an article for davidicke.com – with his tongue only

mildly in his cheek – that he believed he had made a tremendous

discovery that may revolutionise science. He had discovered that

small bits of paper are alive, ‘well alive-ish’, can ‘infect’ houses, and

then reproduce themselves inside the house. The result was that this

explosion of growth in the paper inside the house causes the house

to explode, blowing it to smithereens. His evidence for this new

theory is that in the past months he had carefully examined many of

the houses in his neighbourhood and found almost no scraps of

paper on the lawns and surrounds of the house. There was an

occasional stray label, but nothing more. Then he would return to

these same houses a week or so later and with a few, not all of them,

particularly the old and decrepit ones, he found to his shock and

surprise they were li�ered with stray bits of paper. He knew then

that the paper had infected these houses, made copies of itself, and

blew up the house. A young boy on a bicycle at one of the sites told

him he had seen a demolition crew using dynamite to explode the

house the previous week, but Cowan dismissed this as the idle

thoughts of silly boys because ‘I was on to something big’. He was

on to how ‘scientists’ mistake genetic material in the detoxifying

process for something they call a ‘virus’. Cowan said of his house

and paper story:

If this sounds crazy to you, it’s because it should. This scenario is obviously nuts. But consider this admittedly embellished, for effect, current viral theory that all scientists, medical doctors and virologists currently believe.

He takes the example of the ‘novel SARS-Cov2’ virus to prove the

point. First they take someone with an undefined illness called

‘Covid-19’ and don’t even a�empt to find any virus in their sputum.

Never mind the scientists still describe how this ‘virus’, which they

have not located a�aches to a cell receptor, injects its genetic

material, in ‘Covid’s’ case, RNA, into the cell. The RNA once inserted

exploits the cell to reproduce itself and makes ‘thousands, nay

millions, of copies of itself … Then it emerges victorious to claim its

next victim’:

If you were to look in the scientific literature for proof, actual scientific proof, that uniform SARS-CoV2 viruses have been properly isolated from the sputum of a sick person, that actual spike proteins could be seen protruding from the virus (which has not been found), you would find that such evidence doesn’t exist.

If you go looking in the published scientific literature for actual pictures, proof, that these spike proteins or any viral proteins are ever attached to any receptor embedded in any cell membrane, you would also find that no such evidence exists. If you were to look for a video or documented evidence of the intact virus injecting its genetic material into the body of the cell, reproducing itself and then emerging victorious by budding off the cell membrane, you would find that no such evidence exists.

The closest thing you would find is electron micrograph pictures of cellular particles, possibly attached to cell debris, both of which to be seen were stained by heavy metals, a process that completely distorts their architecture within the living organism. This is like finding bits of paper stuck to the blown-up bricks, thereby proving the paper emerged by taking pieces of the bricks on its way out.

The Enders baloney

Cowan describes the ‘Covid’ story as being just as make-believe as

his paper story and he charts back this fantasy to a Nobel Prize

winner called John Enders (1897-1985), an American biomedical

scientist who has been dubbed ‘The Father of Modern Vaccines’.

Enders is claimed to have ‘discovered’ the process of the viral

culture which ‘proved’ that a ‘virus’ caused measles. Cowan

explains how Enders did this ‘by using the EXACT same procedure

that has been followed by every virologist to find and characterize

every new virus since 1954’. Enders took throat swabs from children

with measles and immersed them in 2ml of milk. Penicillin (100u/ml)

and the antibiotic streptomycin (50,g/ml) were added and the whole

mix was centrifuged – rotated at high speed to separate large cellular

debris from small particles and molecules as with milk and cream,

for example. Cowan says that if the aim is to find li�le particles of

genetic material (‘viruses’) in the snot from children with measles it

would seem that the last thing you would do is mix the snot with

other material – milk –that also has genetic material. ‘How are you

ever going to know whether whatever you found came from the snot

or the milk?’ He points out that streptomycin is a ‘nephrotoxic’ or

poisonous-to-the-kidney drug. You will see the relevance of that

shortly. Cowan says that it gets worse, much worse, when Enders

describes the culture medium upon which the virus ‘grows’: ‘The

culture medium consisted of bovine amniotic fluid (90%), beef

embryo extract (5%), horse serum (5%), antibiotics and phenol red as

an indicator of cell metabolism.’ Cowan asks incredulously: ‘Did he

just say that the culture medium also contained fluids and tissues

that are themselves rich sources of genetic material?’ The genetic

cocktail, or ‘medium’, is inoculated onto tissue and cells from rhesus

monkey kidney tissue. This is where the importance of streptomycin

comes in and currently-used antimicrobials and other drugs that are

poisonous to kidneys and used in ALL modern viral cultures (e.g.

gentamicin, streptomycin, and amphotericin). Cowan asks: ‘How are

you ever going to know from this witch’s brew where any genetic

material comes from as we now have five different sources of rich

genetic material in our mix?’ Remember, he says, that all genetic

material, whether from monkey kidney tissues, bovine serum, milk,

etc., is made from the exact same components. The same central

question returns: ‘How are you possibly going to know that it was

the virus that killed the kidney tissue and not the toxic antibiotic and

starvation rations on which you are growing the tissue?’ John Enders

answered the question himself – you can’t:

A second agent was obtained from an uninoculated culture of monkey kidney cells. The cytopathic changes [death of the cells] it induced in the unstained preparations could not be distinguished with confidence from the viruses isolated from measles.

The death of the cells (‘cytopathic changes’) happened in exactly

the same manner, whether they inoculated the kidney tissue with the

measles snot or not, Cowan says. ‘This is evidence that the

destruction of the tissue, the very proof of viral causation of illness,

was not caused by anything in the snot because they saw the same

destructive effect when the snot was not even used … the cytopathic,

i.e., cell-killing, changes come from the process of the culture itself,

not from any virus in any snot, period.’ Enders quotes in his 1957

paper a virologist called Ruckle as reporting similar findings ‘and in

addition has isolated an agent from monkey kidney tissue that is so

far indistinguishable from human measles virus’. In other words,

Cowan says, these particles called ‘measles viruses’ are simply and

clearly breakdown products of the starved and poisoned tissue. For

measles ‘virus’ see all ‘viruses’ including the so-called ‘Covid virus’.

Enders, the ‘Father of Modern Vaccines’, also said:

There is a potential risk in employing cultures of primate cells for the production of vaccines composed of attenuated virus, since the presence of other agents possibly latent in primate tissues cannot be definitely excluded by any known method.

Cowan further quotes from a paper published in the journal

Viruses in May, 2020, while the ‘Covid pandemic’ was well

underway in the media if not in reality. ‘EVs’ here refers to particles

of genetic debris from our own tissues, such as exosomes of which

more in a moment: ‘The remarkable resemblance between EVs and

viruses has caused quite a few problems in the studies focused on

the analysis of EVs released during viral infections.’ Later the paper

adds that to date a reliable method that can actually guarantee a

complete separation (of EVs from viruses) DOES NOT EXIST. This

was published at a time when a fairy tale ‘virus’ was claimed in total

certainty to be causing a fairy tale ‘viral disease’ called ‘Covid-19’ – a

fairy tale that was already well on the way to transforming human

society in the image that the Cult has worked to achieve for so long.

Cowan concludes his article:

To summarize, there is no scientific evidence that pathogenic viruses exist. What we think of as ‘viruses’ are simply the normal breakdown products of dead and dying tissues and cells. When we are well, we make fewer of these particles; when we are starved, poisoned, suffocated by wearing masks, or afraid, we make more.

There is no engineered virus circulating and making people sick. People in laboratories all over the world are making genetically modified products to make people sick. These are called vaccines. There is no virome, no ‘ecosystem’ of viruses, viruses are not 8%, 50% or 100 % of our genetic material. These are all simply erroneous ideas based on the misconception called a virus.

What is ‘Covid’? Load of bollocks

The background described here by Cowan and Lanka was

emphasised in the first video presentation that I saw by Dr Andrew

Kaufman when he asked whether the ‘Covid virus’ was in truth a

natural defence mechanism of the body called ‘exosomes’. These are

released by cells when in states of toxicity – see the same themes

returning over and over. They are released ever more profusely as

chemical and radiation toxicity increases and think of the potential

effect therefore of 5G alone as its destructive frequencies infest the

human energetic information field with a gathering pace (5G went

online in Wuhan in 2019 as the ‘virus’ emerged). I’ll have more about

this later. Exosomes transmit a warning to the rest of the body that

‘Houston, we have a problem’. Kaufman presented images of

exosomes and compared them with ‘Covid’ under an electron

microscope and the similarity was remarkable. They both a�ach to

the same cell receptors (claimed in the case of ‘Covid’), contain the

same genetic material in the form of RNA or ribonucleic acid, and

both are found in ‘viral cell cultures’ with damaged or dying cells.

James Hildreth MD, President and Chief Executive Officer of the

Meharry Medical College at Johns Hopkins, said: ‘The virus is fully

an exosome in every sense of the word.’ Kaufman’s conclusion was

that there is no ‘virus’: ‘This entire pandemic is a completely

manufactured crisis … there is no evidence of anyone dying from

[this] illness.’ Dr Tom Cowan and Sally Fallon Morell, authors of The

Contagion Myth, published a statement with Dr Kaufman in

February, 2021, explaining why the ‘virus’ does not exist and you can

read it that in full in the Appendix.

‘Virus’ theory can be traced to the ‘cell theory’ in 1858 of German

physician Rudolf Virchow (1821-1920) who contended that disease

originates from a single cell infiltrated by a ‘virus’. Dr Stefan Lanka

said that findings and insights with respect to the structure, function

and central importance of tissues in the creation of life, which were

already known in 1858, comprehensively refute the cell theory.

Virchow ignored them. We have seen the part later played by John

Enders in the 1950s and Lanka notes that infection theories were

only established as a global dogma through the policies and

eugenics of the Third Reich in Nazi Germany (creation of the same

Sabbatian cult behind the ‘Covid’ hoax). Lanka said: ‘Before 1933,

scientists dared to contradict this theory; a�er 1933, these critical

scientists were silenced’. Dr Tom Cowan’s view is that ill-heath is

caused by too much of something, too li�le of something, or

toxification from chemicals and radiation – not contagion. We must

also highlight as a major source of the ‘virus’ theology a man still

called the ‘Father of Modern Virology’ – Thomas Milton Rivers

(1888-1962). There is no way given the Cult’s long game policy that it

was a coincidence for the ‘Father of Modern Virology’ to be director

of the Rockefeller Institute for Medical Research from 1937 to 1956

when he is credited with making the Rockefeller Institute a leader in

‘viral research’. Cult Rockefellers were the force behind the creation

of Big Pharma ‘medicine’, established the World Health

Organisation in 1948, and have long and close associations with the

Gates family that now runs the WHO during the pandemic hoax

through mega-rich Cult gofer and psychopath Bill Gates.

Only a Renegade Mind can see through all this bullshit by asking

the questions that need to be answered, not taking ‘no’ or

prevarication for an answer, and certainly not hiding from the truth

in fear of speaking it. Renegade Minds have always changed the

world for the be�er and they will change this one no ma�er how

bleak it may currently appear to be.

A

CHAPTER SIX

Sequence of deceit

If you tell the truth, you don’t have to remember anything

Mark Twain

gainst the background that I have laid out this far the sequence

that took us from an invented ‘virus’ in Cult-owned China in

late 2019 to the fascist transformation of human society can be seen

and understood in a whole new context.

We were told that a deadly disease had broken out in Wuhan and

the world media began its campaign (coordinated by behavioural

psychologists as we shall see) to terrify the population into

unquestioning compliance. We were shown images of Chinese

people collapsing in the street which never happened in the West

with what was supposed to be the same condition. In the earliest

days when alleged cases and deaths were few the fear register was

hysterical in many areas of the media and this would expand into

the common media narrative across the world. The real story was

rather different, but we were never told that. The Chinese

government, one of the Cult’s biggest centres of global operation,

said they had discovered a new illness with flu-like and pneumonia-

type symptoms in a city with such toxic air that it is overwhelmed

with flu-like symptoms, pneumonia and respiratory disease. Chinese

scientists said it was a new – ‘novel’ – coronavirus which they called

Sars-Cov-2 and that it caused a disease they labelled ‘Covid-19’.

There was no evidence for this and the ‘virus’ has never to this day

been isolated, purified and its genetic code established from that. It

was from the beginning a computer-generated fiction. Stories of

Chinese whistleblowers saying the number of deaths was being

supressed or that the ‘new disease’ was related to the Wuhan bio-lab

misdirected mainstream and alternative media into cul-de-sacs to

obscure the real truth – there was no ‘virus’.

Chinese scientists took genetic material from the lung fluid of just

a few people and said they had found a ‘new’ disease when this

material had a wide range of content. There was no evidence for a

‘virus’ for the very reasons explained in the last two chapters. The

‘virus’ has never been shown to (a) exist and (b) cause any disease.

People were diagnosed on symptoms that are so widespread in

Wuhan and polluted China and with a PCR test that can’t detect

infectious disease. On this farce the whole global scam was sold to

the rest of the world which would also diagnose respiratory disease

as ‘Covid-19’ from symptoms alone or with a PCR test not testing for

a ‘virus’. Flu miraculously disappeared worldwide in 2020 and into

2021 as it was redesignated ‘Covid-19’. It was really the same old flu

with its ‘flu-like’ symptoms a�ributed to ‘flu-like’ ‘Covid-19’. At the

same time with very few exceptions the Chinese response of

draconian lockdown and fascism was the chosen weapon to respond

across the West as recommended by the Cult-owned Tedros at the

Cult-owned World Health Organization run by the Cult-owned

Gates. All was going according to plan. Chinese scientists –

everything in China is controlled by the Cult-owned government –

compared their contaminated RNA lung-fluid material with other

RNA sequences and said it appeared to be just under 80 percent

identical to the SARS-CoV-1 ‘virus’ claimed to be the cause of the

SARS (severe acute respiratory syndrome) ‘outbreak’ in 2003. They

decreed that because of this the ‘new virus’ had to be related and

they called it SARS-CoV-2. There are some serious problems with

this assumption and assumption was all it was. Most ‘factual’ science

turns out to be assumptions repeated into everyone-knows-that. A

match of under 80-percent is meaningless. Dr Kaufman makes the

point that there’s a 96 percent genetic correlation between humans

and chimpanzees, but ‘no one would say our genetic material is part

of the chimpanzee family’. Yet the Chinese authorities were claiming

that a much lower percentage, less than 80 percent, proved the

existence of a new ‘coronavirus’. For goodness sake human DNA is

60 percent similar to a banana.

You are feeling sleepy

The entire ‘Covid’ hoax is a global Psyop, a psychological operation

to program the human mind into believing and fearing a complete

fantasy. A crucial aspect of this was what appeared to happen in Italy.

It was all very well streaming out daily images of an alleged

catastrophe in Wuhan, but to the Western mind it was still on the

other side of the world in a very different culture and se�ing. A

reaction of ‘this could happen to me and my family’ was still nothing

like as intense enough for the mind-doctors. The Cult needed a

Western example to push people over that edge and it chose Italy,

one of its major global locations going back to the Roman Empire.

An Italian ‘Covid’ crisis was manufactured in a particular area called

Lombardy which just happens to be notorious for its toxic air and

therefore respiratory disease. Wuhan, China, déjà vu. An hysterical

media told horror stories of Italians dying from ‘Covid’ in their

droves and how Lombardy hospitals were being overrun by a tidal

wave of desperately ill people needing treatment a�er being struck

down by the ‘deadly virus’. Here was the psychological turning

point the Cult had planned. Wow, if this is happening in Italy, the

Western mind concluded, this indeed could happen to me and my

family. Another point is that Italian authorities responded by

following the Chinese blueprint so vehemently recommended by the

Cult-owned World Health Organization. They imposed fascistic

lockdowns on the whole country viciously policed with the help of

surveillance drones sweeping through the streets seeking out anyone

who escaped from mass house arrest. Livelihoods were destroyed

and psychology unravelled in the way we have witnessed since in all

lockdown countries. Crucial to the plan was that Italy responded in

this way to set the precedent of suspending freedom and imposing

fascism in a ‘Western liberal democracy’. I emphasised in an

animated video explanation on davidicke.com posted in the summer

of 2020 how important it was to the Cult to expand the Chinese

lockdown model across the West. Without this, and the bare-faced lie

that non-symptomatic people could still transmit a ‘disease’ they

didn’t have, there was no way locking down the whole population,

sick and not sick, could be pulled off. At just the right time and with

no evidence Cult operatives and gofers claimed that people without

symptoms could pass on the ‘disease’. In the name of protecting the

‘vulnerable’ like elderly people, who lockdowns would kill by the

tens of thousands, we had for the first time healthy people told to

isolate as well as the sick. The great majority of people who tested

positive had no symptoms because there was nothing wrong with

them. It was just a trick made possible by a test not testing for the

‘virus’.

Months a�er my animated video the Gates-funded Professor Neil

Ferguson at the Gates-funded Imperial College confirmed that I was

right. He didn’t say it in those terms, naturally, but he did say it.

Ferguson will enter the story shortly for his outrageously crazy

‘computer models’ that led to Britain, the United States and many

other countries following the Chinese and now Italian methods of

response. Put another way, following the Cult script. Ferguson said

that SAGE, the UK government’s scientific advisory group which has

controlled ‘Covid’ policy from the start, wanted to follow the

Chinese lockdown model (while they all continued to work and be

paid), but they wondered if they could possibly, in Ferguson’s

words, ‘get away with it in Europe’. ‘Get away with it’? Who the hell

do these moronic, arrogant people think they are? This appalling

man Ferguson said that once Italy went into national lockdown they

realised they, too, could mimic China:

It’s a communist one-party state, we said. We couldn’t get away with it in Europe, we thought … and then Italy did it. And we realised we could. Behind this garbage from Ferguson is a simple fact: Doing the same as China in every country was the plan from the start and Ferguson’s ‘models’ would play a central role in achieving that. It’s just a coincidence, of course, and absolutely nothing to worry your little head about.

Oops, sorry, our mistake

Once the Italian segment of the Psyop had done the job it was

designed to do a very different story emerged. Italian authorities

revealed that 99 percent of those who had ‘died from Covid-19’ in

Italy had one, two, three, or more ‘co-morbidities’ or illnesses and

health problems that could have ended their life. The US Centers for

Disease Control and Prevention (CDC) published a figure of 94

percent for Americans dying of ‘Covid’ while having other serious

medical conditions – on average two to three (some five or six) other

potential causes of death. In terms of death from an unproven ‘virus’

I say it is 100 percent. The other one percent in Italy and six percent

in the US would presumably have died from ‘Covid’s’ flu-like

symptoms with a range of other possible causes in conjunction with

a test not testing for the ‘virus’. Fox News reported that even more

startling figures had emerged in one US county in which 410 of 422

deaths a�ributed to ‘Covid-19’ had other potentially deadly health

conditions. The Italian National Health Institute said later that the

average age of people dying with a ‘Covid-19’ diagnosis in Italy was

about 81. Ninety percent were over 70 with ten percent over 90. In

terms of other reasons to die some 80 percent had two or more

chronic diseases with half having three or more including

cardiovascular problems, diabetes, respiratory problems and cancer.

Why is the phantom ‘Covid-19’ said to kill overwhelmingly old

people and hardly affect the young? Old people continually die of

many causes and especially respiratory disease which you can re-

diagnose ‘Covid-19’ while young people die in tiny numbers by

comparison and rarely of respiratory disease. Old people ‘die of

Covid’ because they die of other things that can be redesignated

‘Covid’ and it really is that simple.

Flu has flown

The blueprint was in place. Get your illusory ‘cases’ from a test not

testing for the ‘virus’ and redesignate other causes of death as

‘Covid-19’. You have an instant ‘pandemic’ from something that is

nothing more than a computer-generated fiction. With near-on a

billion people having ‘flu-like’ symptoms every year the potential

was limitless and we can see why flu quickly and apparently

miraculously disappeared worldwide by being diagnosed ‘Covid-19’.

The painfully bloody obvious was explained away by the childlike

media in headlines like this in the UK ‘Independent’: ‘Not a single

case of flu detected by Public Health England this year as Covid

restrictions suppress virus’. I kid you not. The masking, social

distancing and house arrest that did not make the ‘Covid virus’

disappear somehow did so with the ‘flu virus’. Even worse the

article, by a bloke called Samuel Love�, suggested that maybe the

masking, sanitising and other ‘Covid’ measures should continue to

keep the flu away. With a ridiculousness that disturbs your breathing

(it’s ‘Covid-19’) the said Love� wrote: ‘With widespread social

distancing and mask-wearing measures in place throughout the UK,

the usual routes of transmission for influenza have been blocked.’

He had absolutely no evidence to support that statement, but look at

the consequences of him acknowledging the obvious. With flu not

disappearing at all and only being relabelled ‘Covid-19’ he would

have to contemplate that ‘Covid’ was a hoax on a scale that is hard to

imagine. You need guts and commitment to truth to even go there

and that’s clearly something Samuel Love� does not have in

abundance. He would never have got it through the editors anyway.

Tens of thousands die in the United States alone every winter from

flu including many with pneumonia complications. CDC figures

record 45 million Americans diagnosed with flu in 2017-2018 of

which 61,000 died and some reports claim 80,000. Where was the

same hysteria then that we have seen with ‘Covid-19’? Some 250,000

Americans are admi�ed to hospital with pneumonia every year with

about 50,000 cases proving fatal. About 65 million suffer respiratory

disease every year and three million deaths makes this the third

biggest cause of death worldwide. You only have to redesignate a

portion of all these people ‘Covid-19’ and you have an instant global

pandemic or the appearance of one. Why would doctors do this? They

are told to do this and all but a few dare not refuse those who must

be obeyed. Doctors in general are not researching their own

knowledge and instead take it direct and unquestioned from the

authorities that own them and their careers. The authorities say they

must now diagnose these symptoms ‘Covid-19’ and not flu, or

whatever, and they do it. Dark suits say put ‘Covid-19’ on death

certificates no ma�er what the cause of death and the doctors do it.

Renegade Minds don’t fall for the illusion that doctors and medical

staff are all highly-intelligent, highly-principled, seekers of medical

truth. Some are, but not the majority. They are repeaters, gofers, and

yes sir, no sir, purveyors of what the system demands they purvey.

The ‘Covid’ con is not merely confined to diseases of the lungs.

Instructions to doctors to put ‘Covid-19’ on death certificates for

anyone dying of anything within 28 days (or much more) of a

positive test not testing for the ‘virus’ opened the floodgates. The

term dying with ‘Covid’ and not of ‘Covid’ was coined to cover the

truth. Whether it was a with or an of they were all added to the death

numbers a�ributed to the ‘deadly virus’ compiled by national

governments and globally by the Gates-funded Johns Hopkins

operation in the United States that was so involved in those

‘pandemic’ simulations. Fraudulent deaths were added to the ever-

growing list of fraudulent ‘cases’ from false positives from a false

test. No wonder Professor Walter Ricciardi, scientific advisor to the

Italian minister of health, said a�er the Lombardy hysteria had done

its job that ‘Covid’ death rates were due to Italy having the second

oldest population in the world and to how hospitals record deaths:

The way in which we code deaths in our country is very generous in the sense that all the people who die in hospitals with the coronavirus are deemed to be dying of the coronavirus. On re-evaluation by the National Institute of Health, only 12 per cent of death certificates have shown a direct causality from coronavirus, while 88 per cent of patients who have died have at least one pre-morbidity – many had two or three.

This is extraordinary enough when you consider the propaganda

campaign to use Italy to terrify the world, but how can they even say

twelve percent were genuine when the ‘virus’ has not been shown to

exist, its ‘code’ is a computer program, and diagnosis comes from a

test not testing for it? As in China, and soon the world, ‘Covid-19’ in

Italy was a redesignation of diagnosis. Lies and corruption were to

become the real ‘pandemic’ fuelled by a pathetically-compliant

medical system taking its orders from the tiny few at the top of their

national hierarchy who answered to the World Health Organization

which answers to Gates and the Cult. Doctors were told – ordered –

to diagnose a particular set of symptoms ‘Covid-19’ and put that on

the death certificate for any cause of death if the patient had tested

positive with a test not testing for the virus or had ‘Covid’ symptoms

like the flu. The United States even introduced big financial

incentives to manipulate the figures with hospitals receiving £4,600

from the Medicare system for diagnosing someone with regular

pneumonia, $13,000 if they made the diagnosis from the same

symptoms ‘Covid-19’ pneumonia, and $39, 000 if they put a ‘Covid’

diagnosed patient on a ventilator that would almost certainly kill

them. A few – painfully and pathetically few – medical

whistleblowers revealed (before Cult-owned YouTube deleted their

videos) that they had been instructed to ‘let the patient crash’ and

put them straight on a ventilator instead of going through a series of

far less intrusive and dangerous methods as they would have done

before the pandemic hoax began and the financial incentives kicked

in. We are talking cold-blooded murder given that ventilators are so

damaging to respiratory systems they are usually the last step before

heaven awaits. Renegade Minds never fall for the belief that people

in white coats are all angels of mercy and cannot be full-on

psychopaths. I have explained in detail in The Answer how what I am

describing here played out across the world coordinated by the

World Health Organization through the medical hierarchies in

almost every country.

Medical scientist calls it

Information about the non-existence of the ‘virus’ began to emerge

for me in late March, 2020, and mushroomed a�er that. I was sent an

email by Sir Julian Rose, a writer, researcher, and organic farming

promotor, from a medical scientist friend of his in the United States.

Even at that early stage in March the scientist was able to explain

how the ‘Covid’ hoax was being manipulated. He said there were no

reliable tests for a specific ‘Covid-19 virus’ and nor were there any

reliable agencies or media outlets for reporting numbers of actual

‘Covid-19’ cases. We have seen in the long period since then that he

was absolutely right. ‘Every action and reaction to Covid-19 is based

on totally flawed data and we simply cannot make accurate

assessments,’ he said. Most people diagnosed with ‘Covid-19’ were

showing nothing more than cold and flu-like symptoms ‘because

most coronavirus strains are nothing more than cold/flu-like

symptoms’. We had farcical situations like an 84-year-old German

man testing positive for ‘Covid-19’ and his nursing home ordered to

quarantine only for him to be found to have a common cold. The

scientist described back then why PCR tests and what he called the

‘Mickey Mouse test kits’ were useless for what they were claimed to

be identifying. ‘The idea these kits can isolate a specific virus like

Covid-19 is nonsense,’ he said. Significantly, he pointed out that ‘if

you want to create a totally false panic about a totally false pandemic

– pick a coronavirus’. This is exactly what the Cult-owned Gates,

World Economic Forum and Johns Hopkins University did with

their Event 201 ‘simulation’ followed by their real-life simulation

called the ‘pandemic’. The scientist said that all you had to do was

select the sickest of people with respiratory-type diseases in a single

location – ‘say Wuhan’ – and administer PCR tests to them. You can

then claim that anyone showing ‘viral sequences’ similar to a

coronavirus ‘which will inevitably be quite a few’ is suffering from a

‘new’ disease:

Since you already selected the sickest flu cases a fairly high proportion of your sample will go on to die. You can then say this ‘new’ virus has a CFR [case fatality rate] higher than the flu and use this to infuse more concern and do more tests which will of course produce more ‘cases’, which expands the testing, which produces yet more ‘cases’ and so on and so on. Before long you have your ‘pandemic’, and all you have done is use a simple test kit trick to convert the worst flu and pneumonia cases into something new that doesn’t ACTUALLY EXIST [my emphasis].

He said that you then ‘just run the same scam in other countries’

and make sure to keep the fear message running high ‘so that people

•

•

•

will feel panicky and less able to think critically’. The only problem

to overcome was the fact there is no actual new deadly pathogen and

only regular sick people. This meant that deaths from the ‘new

deadly pathogen’ were going to be way too low for a real new

deadly virus pandemic, but he said this could be overcome in the

following ways – all of which would go on to happen:

1. You can claim this is just the beginning and more deaths are imminent [you underpin this

with fantasy ‘computer projections’]. Use this as an excuse to quarantine everyone and then

claim the quarantine prevented the expected millions of dead.

2. You can [say that people] ‘minimizing’ the dangers are irresponsible and bully them into

not talking about numbers.

3. You can talk crap about made up numbers hoping to blind people with pseudoscience.

4. You can start testing well people (who, of course, will also likely have shreds of

coronavirus [RNA] in them) and thus inflate your ‘case figures’ with ‘asymptomatic

carriers’ (you will of course have to spin that to sound deadly even though any virologist

knows the more symptom-less cases you have the less deadly is your pathogen).

The scientist said that if you take these simple steps ‘you can have

your own entirely manufactured pandemic up and running in

weeks’. His analysis made so early in the hoax was brilliantly

prophetic of what would actually unfold. Pulling all the information

together in these recent chapters we have this is simple 1, 2, 3, of

how you can delude virtually the entire human population into

believing in a ‘virus’ that doesn’t exist:

A ‘Covid case’ is someone who tests positive with a test not

testing for the ‘virus’.

A ‘Covid death’ is someone who dies of any cause within 28 days

(or much longer) of testing positive with a test not testing for the

‘virus.

Asymptomatic means there is nothing wrong with you, but they

claim you can pass on what you don’t have to justify locking

down (quarantining) healthy people in totality.

The foundations of the hoax are that simple. A study involving ten

million people in Wuhan, published in November, 2020, demolished

the whole lie about those without symptoms passing on the ‘virus’.

They found ‘300 asymptomatic cases’ and traced their contacts to

find that not one of them was detected with the ‘virus’.

‘Asymptomatic’ patients and their contacts were isolated for no less

than two weeks and nothing changed. I know it’s all crap, but if you

are going to claim that those without symptoms can transmit ‘the

virus’ then you must produce evidence for that and they never have.

Even World Health Organization official Dr Maria Van Kerkhove,

head of the emerging diseases and zoonosis unit, said as early as

June, 2020, that she doubted the validity of asymptomatic

transmission. She said that ‘from the data we have, it still seems to

be rare that an asymptomatic person actually transmits onward to a

secondary individual’ and by ‘rare’ she meant that she couldn’t cite

any case of asymptomatic transmission.

The Ferguson factor

The problem for the Cult as it headed into March, 2020, when the

script had lockdown due to start, was that despite all the

manipulation of the case and death figures they still did not have

enough people alleged to have died from ‘Covid’ to justify mass

house arrest. This was overcome in the way the scientist described:

‘You can claim this is just the beginning and more deaths are

imminent … Use this as an excuse to quarantine everyone and then

claim the quarantine prevented the expected millions of dead.’ Enter

one Professor Neil Ferguson, the Gates-funded ‘epidemiologist’ at

the Gates-funded Imperial College in London. Ferguson is Britain’s

Christian Drosten in that he has a dire record of predicting health

outcomes, but is still called upon to advise government on the next

health outcome when another ‘crisis’ comes along. This may seem to

be a strange and ridiculous thing to do. Why would you keep

turning for policy guidance to people who have a history of being

monumentally wrong? Ah, but it makes sense from the Cult point of

view. These ‘experts’ keep on producing predictions that suit the

Cult agenda for societal transformation and so it was with Neil

Ferguson as he revealed his horrific (and clearly insane) computer

model predictions that allowed lockdowns to be imposed in Britain,

the United States and many other countries. Ferguson does not have

even an A-level in biology and would appear to have no formal

training in computer modelling, medicine or epidemiology,

according to Derek Winton, an MSc in Computational Intelligence.

He wrote an article somewhat aghast at what Ferguson did which

included taking no account of respiratory disease ‘seasonality’ which

means it is far worse in the winter months. Who would have thought

that respiratory disease could be worse in the winter? Well, certainly

not Ferguson.

The massively China-connected Imperial College and its bizarre

professor provided the excuse for the long-incubated Chinese model

of human control to travel westward at lightning speed. Imperial

College confirms on its website that it collaborates with the Chinese

Research Institute; publishes more than 600 research papers every

year with Chinese research institutions; has 225 Chinese staff; 2,600

Chinese students – the biggest international group; 7,000 former

students living in China which is the largest group outside the UK;

and was selected for a tour by China’s President Xi Jinping during

his state visit to the UK in 2015. The college takes major donations

from China and describes itself as the UK’s number one university

collaborator with Chinese research institutions. The China

communist/fascist government did not appear phased by the woeful

predictions of Ferguson and Imperial when during the lockdown

that Ferguson induced the college signed a five-year collaboration

deal with China tech giant Huawei that will have Huawei’s indoor

5G network equipment installed at the college’s West London tech

campus along with an ‘AI cloud platform’. The deal includes Chinese

sponsorship of Imperial’s Venture Catalyst entrepreneurship

competition. Imperial is an example of the enormous influence the

Chinese government has within British and North American

universities and research centres – and further afield. Up to 200

academics from more than a dozen UK universities are being

investigated on suspicion of ‘unintentionally’ helping the Chinese

government build weapons of mass destruction by ‘transferring

world-leading research in advanced military technology such as

aircra�, missile designs and cyberweapons’. Similar scandals have

broken in the United States, but it’s all a coincidence. Imperial

College serves the agenda in many other ways including the

promotion of every aspect of the United Nations Agenda 21/2030

(the Great Reset) and produced computer models to show that

human-caused ‘climate change’ is happening when in the real world

it isn’t. Imperial College is driving the climate agenda as it drives the

‘Covid’ agenda (both Cult hoaxes) while Patrick Vallance, the UK

government’s Chief Scientific Adviser on ‘Covid’, was named Chief

Scientific Adviser to the UN ‘climate change’ conference known as

COP26 hosted by the government in Glasgow, Scotland. ‘Covid’ and

‘climate’ are fundamentally connected.

Professor Woeful

From Imperial’s bosom came Neil Ferguson still advising

government despite his previous disasters and it was announced

early on that he and other key people like UK Chief Medical Adviser

Chris Whi�y had caught the ‘virus’ as the propaganda story was

being sold. Somehow they managed to survive and we had Prime

Minister Boris Johnson admi�ed to hospital with what was said to be

a severe version of the ‘virus’ in this same period. His whole policy

and demeanour changed when he returned to Downing Street. It’s a

small world with these government advisors – especially in their

communal connections to Gates – and Ferguson had partnered with

Whi�y to write a paper called ‘Infectious disease: Tough choices to

reduce Ebola transmission’ which involved another scare-story that

didn’t happen. Ferguson’s ‘models’ predicted that up to150, 000

could die from ‘mad cow disease’, or BSE, and its version in sheep if

it was transmi�ed to humans. BSE was not transmi�ed and instead

triggered by an organophosphate pesticide used to treat a pest on

cows. Fewer than 200 deaths followed from the human form. Models

by Ferguson and his fellow incompetents led to the unnecessary

culling of millions of pigs, ca�le and sheep in the foot and mouth

outbreak in 2001 which destroyed the lives and livelihoods of

farmers and their families who had o�en spent decades building

their herds and flocks. Vast numbers of these animals did not have

foot and mouth and had no contact with the infection. Another

‘expert’ behind the cull was Professor Roy Anderson, a computer

modeller at Imperial College specialising in the epidemiology of

human, not animal, disease. Anderson has served on the Bill and

Melinda Gates Grand Challenges in Global Health advisory board

and chairs another Gates-funded organisation. Gates is everywhere.

In a precursor to the ‘Covid’ script Ferguson backed closing

schools ‘for prolonged periods’ over the swine flu ‘pandemic’ in 2009

and said it would affect a third of the world population if it

continued to spread at the speed he claimed to be happening. His

mates at Imperial College said much the same and a news report

said: ‘One of the authors, the epidemiologist and disease modeller

Neil Ferguson, who sits on the World Health Organisation’s

emergency commi�ee for the outbreak, said the virus had “full

pandemic potential”.’ Professor Liam Donaldson, the Chris Whi�y

of his day as Chief Medical Officer, said the worst case could see 30

percent of the British people infected by swine flu with 65,000 dying.

Ferguson and Donaldson were indeed proved correct when at the

end of the year the number of deaths a�ributed to swine flu was 392.

The term ‘expert’ is rather liberally applied unfortunately, not least

to complete idiots. Swine flu ‘projections’ were great for

GlaxoSmithKline (GSK) as millions rolled in for its Pandemrix

influenza vaccine which led to brain damage with children most

affected. The British government (taxpayers) paid out more than £60

million in compensation a�er GSK was given immunity from

prosecution. Yet another ‘Covid’ déjà vu. Swine flu was supposed to

have broken out in Mexico, but Dr Wolfgang Wodarg, a German

doctor, former member of parliament and critic of the ‘Covid’ hoax,

observed ‘the spread of swine flu’ in Mexico City at the time. He

said: ‘What we experienced in Mexico City was a very mild flu

which did not kill more than usual – which killed even fewer people

than usual.’ Hyping the fear against all the facts is not unique to

‘Covid’ and has happened many times before. Ferguson is reported

to have over-estimated the projected death toll of bird flu (H5N1) by

some three million-fold, but bird flu vaccine makers again made a

killing from the scare. This is some of the background to the Neil

Ferguson who produced the perfectly-timed computer models in

early 2020 predicting that half a million people would die in Britain

without draconian lockdown and 2.2 million in the United States.

Politicians panicked, people panicked, and lockdowns of alleged

short duration were instigated to ‘fla�en the curve’ of cases gleaned

from a test not testing for the ‘virus’. I said at the time that the public

could forget the ‘short duration’ bit. This was an agenda to destroy

the livelihoods of the population and force them into mass control

through dependency and there was going to be nothing ‘short’ about

it. American researcher Daniel Horowitz described the consequences

of the ‘models’ spewed out by Gates-funded Ferguson and Imperial

College:

What led our government and the governments of many other countries into panic was a single Imperial College of UK study, funded by global warming activists, that predicted 2.2 million deaths if we didn’t lock down the country. In addition, the reported 8-9% death rate in Italy scared us into thinking there was some other mutation of this virus that they got, which might have come here.

Together with the fact that we were finally testing and had the ability to actually report new cases, we thought we were headed for a death spiral. But again … we can’t flatten a curve if we don’t know when the curve started.

How about it never started?

Giving them what they want

An investigation by German news outlet Welt Am Sonntag (World on

Sunday) revealed how in March, 2020, the German government

gathered together ‘leading scientists from several research institutes

and universities’ and ‘together, they were to produce a [modelling]

paper that would serve as legitimization for further tough political

measures’. The Cult agenda was justified by computer modelling not

based on evidence or reality; it was specifically constructed to justify

the Cult demand for lockdowns all over the world to destroy the

independent livelihoods of the global population. All these

modellers and everyone responsible for the ‘Covid’ hoax have a date

with a trial like those in Nuremberg a�er World War Two when

Nazis faced the consequences of their war crimes. These corrupt-

beyond-belief ‘modellers’ wrote the paper according to government

instructions and it said that that if lockdown measures were li�ed

then up to one million Germans would die from ‘Covid-19’ adding

that some would die ‘agonizingly at home, gasping for breath’

unable to be treated by hospitals that couldn’t cope. All lies. No

ma�er – it gave the Cult all that it wanted. What did long-time

government ‘modeller’ Neil Ferguson say? If the UK and the United

States didn’t lockdown half a million would die in Britain and 2.2

million Americans. Anyone see a theme here? ‘Modellers’ are such a

crucial part of the lockdown strategy that we should look into their

background and follow the money. Researcher Rosemary Frei

produced an excellent article headlined ‘The Modelling-paper

Mafiosi’. She highlights a guy called John Edmunds, a British

epidemiologist, and professor in the Faculty of Epidemiology and

Population Health at the London School of Hygiene & Tropical

Medicine. He studied at Imperial College. Edmunds is a member of

government ‘Covid’ advisory bodies which have been dictating

policy, the New and Emerging Respiratory Virus Threats Advisory

Group (NERVTAG) and the Scientific Advisory Group for

Emergencies (SAGE).

Ferguson, another member of NERVTAG and SAGE, led the way

with the original ‘virus’ and Edmunds has followed in the ‘variant’

stage and especially the so-called UK or Kent variant known as the

‘Variant of Concern’ (VOC) B.1.1.7. He said in a co-wri�en report for

the Centre for Mathematical modelling of Infectious Diseases at the

London School of Hygiene and Tropical Medicine, with input from

the Centre’s ‘Covid-19’ Working Group, that there was ‘a realistic

possibility that VOC B.1.1.7 is associated with an increased risk of

death compared to non-VOC viruses’. Fear, fear, fear, get the

vaccine, fear, fear, fear, get the vaccine. Rosemary Frei reveals that

almost all the paper’s authors and members of the modelling centre’s

‘Covid-19’ Working Group receive funding from the Bill and

Melinda Gates Foundation and/or the associated Gates-funded

Wellcome Trust. The paper was published by e-journal Medr χiv

which only publishes papers not peer-reviewed and the journal was

established by an organisation headed by Facebook’s Mark

Zuckerberg and his missus. What a small world it is. Frei discovered

that Edmunds is on the Scientific Advisory Board of the Coalition for

Epidemic Preparedness Innovations (CEPI) which was established

by the Bill and Melinda Gates Foundation, Klaus Schwab’s Davos

World Economic Forum and Big Pharma giant Wellcome. CEPI was

‘launched in Davos [in 2017] to develop vaccines to stop future

epidemics’, according to its website. ‘Our mission is to accelerate the

development of vaccines against emerging infectious diseases and

enable equitable access to these vaccines for people during

outbreaks.’ What kind people they are. Rosemary Frei reveals that

Public Health England (PHE) director Susan Hopkins is an author of

her organisation’s non-peer-reviewed reports on ‘new variants’.

Hopkins is a professor of infectious diseases at London’s Imperial

College which is gi�ed tens of millions of dollars a year by the Bill

and Melinda Gates Foundation. Gates-funded modelling disaster

Neil Ferguson also co-authors Public Health England reports and he

spoke in December, 2020, about the potential danger of the B.1.1.7.

‘UK variant’ promoted by Gates-funded modeller John Edmunds.

When I come to the ‘Covid vaccines’ the ‘new variants’ will be

shown for what they are – bollocks.

Connections, connections

All these people and modellers are lockdown-obsessed or, put

another way, they demand what the Cult demands. Edmunds said in

January, 2021, that to ease lockdowns too soon would be a disaster

and they had to ‘vaccinate much, much, much more widely than the

elderly’. Rosemary Frei highlights that Edmunds is married to

Jeanne Pimenta who is described in a LinkedIn profile as director of

epidemiology at GlaxoSmithKline (GSK) and she held shares in the

company. Patrick Vallance, co-chair of SAGE and the government’s

Chief Scientific Adviser, is a former executive of GSK and has a

deferred bonus of shares in the company worth £600,000. GSK has

serious business connections with Bill Gates and is collaborating

with mRNA-’vaccine’ company CureVac to make ‘vaccines’ for the

new variants that Edmunds is talking about. GSK is planning a

‘Covid vaccine’ with drug giant Sanofi. Puppet Prime Minister Boris

Johnson announced in the spring of 2021 that up to 60 million

vaccine doses were to be made at the GSK facility at Barnard Castle

in the English North East. Barnard Castle, with a population of just

6,000, was famously visited in breach of lockdown rules in April,

2020, by Johnson aide Dominic Cummings who said that he drove

there ‘to test his eyesight’ before driving back to London. Cummings

would be be�er advised to test his integrity – not that it would take

long. The GSK facility had nothing to do with his visit then although

I’m sure Patrick Vallance would have been happy to arrange an

introduction and some tea and biscuits. Ruthless psychopath Gates

has made yet another fortune from vaccines in collaboration with Big

Pharma companies and gushes at the phenomenal profits to be made

from vaccines – more than a 20-to-1 return as he told one

interviewer. Gates also tweeted in December, 2019, with the

foreknowledge of what was coming: ‘What’s next for our

foundation? I’m particularly excited about what the next year could

mean for one of the best buys in global health: vaccines.’

Modeller John Edmunds is a big promotor of vaccines as all these

people appear to be. He’s the dean of the London School of Hygiene

& Tropical Medicine’s Faculty of Epidemiology and Population

Health which is primarily funded by the Bill and Melinda Gates

Foundation and the Gates-established and funded GAVI vaccine

alliance which is the Gates vehicle to vaccinate the world. The

organisation Doctors Without Borders has described GAVI as being

‘aimed more at supporting drug-industry desires to promote new

products than at finding the most efficient and sustainable means for

fighting the diseases of poverty’. But then that’s why the psychopath

Gates created it. John Edmunds said in a video that the London

School of Hygiene & Tropical Medicine is involved in every aspect of

vaccine development including large-scale clinical trials. He

contends that mathematical modelling can show that vaccines

protect individuals and society. That’s on the basis of shit in and shit

out, I take it. Edmunds serves on the UK Vaccine Network as does

Ferguson and the government’s foremost ‘Covid’ adviser, the grim-

faced, dark-eyed Chris Whi�y. The Vaccine Network says it works

‘to support the government to identify and shortlist targeted

investment opportunities for the most promising vaccines and

vaccine technologies that will help combat infectious diseases with

epidemic potential, and to address structural issues related to the

UK’s broader vaccine infrastructure’. Ferguson is acting Director of

the Imperial College Vaccine Impact Modelling Consortium which

has funding from the Bill and Melina Gates Foundation and the

Gates-created GAVI ‘vaccine alliance’. Anyone wonder why these

characters see vaccines as the answer to every problem? Ferguson is

wildly enthusiastic in his support for GAVI’s campaign to vaccine

children en masse in poor countries. You would expect someone like

Gates who has constantly talked about the need to reduce the

population to want to fund vaccines to keep more people alive. I’m

sure that’s why he does it. The John Edmunds London School of

Hygiene & Tropical Medicine (LSHTM) has a Vaccines

Manufacturing Innovation Centre which develops, tests and

commercialises vaccines. Rosemary Frei writes:

The vaccines centre also performs affiliated activities like combating ‘vaccine hesitancy’. The latter includes the Vaccine Confidence Project. The project’s stated purpose is, among other things, ‘to provide analysis and guidance for early response and engagement with the public to ensure sustained confidence in vaccines and immunisation’. The Vaccine Confidence Project’s director is LSHTM professor Heidi Larson. For more than a decade she’s been researching how to combat vaccine hesitancy.

How the bloody hell can blokes like John Edmunds and Neil

Ferguson with those connections and financial ties model ‘virus’ case

and death projections for the government and especially in a way

that gives their paymasters like Gates exactly what they want? It’s

insane, but this is what you find throughout the world.

‘Covid’ is not dangerous, oops, wait, yes it is

Only days before Ferguson’s nightmare scenario made Jackboot

Johnson take Britain into a China-style lockdown to save us from a

deadly ‘virus’ the UK government website gov.uk was reporting

something very different to Ferguson on a page of official

government guidance for ‘high consequence infectious diseases

(HCID)’. It said this about ‘Covid-19’:

As of 19 March 2020, COVID-19 is no longer considered to be a high consequence infectious diseases (HCID) in the UK [my emphasis]. The 4 nations public health HCID group made an interim recommendation in January 2020 to classify COVID-19 as an HCID. This was based on consideration of the UK HCID criteria about the virus and the disease with information available during the early stages of the outbreak.

Now that more is known about COVID-19, the public health bodies in the UK have reviewed the most up to date information about COVID-19 against the UK HCID criteria. They have determined that several features have now changed; in particular, more information is available about mortality rates (low overall), and there is now greater clinical awareness and a specific and sensitive laboratory test, the availability of which continues to increase. The Advisory Committee on Dangerous Pathogens (ACDP) is also of the opinion that COVID-19 should no longer be classified as an HCID.

Soon a�er the government had been exposed for downgrading the

risk they upgraded it again and everyone was back to singing from

the same Cult hymn book. Ferguson and his fellow Gates clones

indicated that lockdowns and restrictions would have to continue

until a Gates-funded vaccine was developed. Gates said the same

because Ferguson and his like were repeating the Gates script which

is the Cult script. ‘Fla�en the curve’ became an ongoing nightmare of

continuing lockdowns with periods in between of severe restrictions

in pursuit of destroying independent incomes and had nothing to do

with protecting health about which the Cult gives not a shit. Why

wouldn’t Ferguson be pushing a vaccine ‘solution’ when he’s owned

by vaccine-obsessive Gates who makes a fortune from them and

when Ferguson heads the Vaccine Impact Modelling Consortium at

Imperial College funded by the Gates Foundation and GAVI, the

‘vaccine alliance’, created by Gates as his personal vaccine

promotion operation? To compound the human catastrophe that

Ferguson’s ‘models’ did so much to create he was later exposed for

breaking his own lockdown rules by having sexual liaisons with his

married girlfriend Antonia Staats at his home while she was living at

another location with her husband and children. Staats was a

‘climate’ activist and senior campaigner at the Soros-funded Avaaz

which I wouldn’t trust to tell me that grass is green. Ferguson had to

resign as a government advisor over this hypocrisy in May, 2020, but

a�er a period of quiet he was back being quoted by the ridiculous

media on the need for more lockdowns and a vaccine rollout. Other

government-advising ‘scientists’ from Imperial College’ held the fort

in his absence and said lockdown could be indefinite until a vaccine

was found. The Cult script was being sung by the payrolled choir. I

said there was no intention of going back to ‘normal’ when the

‘vaccine’ came because the ‘vaccine’ is part of a very different agenda

that I will discuss in Human 2.0. Why would the Cult want to let the

world go back to normal when destroying that normal forever was

the whole point of what was happening? House arrest, closing

businesses and schools through lockdown, (un)social distancing and

masks all followed the Ferguson fantasy models. Again as I

predicted (these people are so predictable) when the ‘vaccine’

arrived we were told that house arrest, lockdown, (un)social

distancing and masks would still have to continue. I will deal with

the masks in the next chapter because they are of fundamental

importance.

Where’s the ‘pandemic’?

Any mildly in-depth assessment of the figures revealed what was

really going on. Cult-funded and controlled organisations still have

genuine people working within them such is the number involved.

So it is with Genevieve Briand, assistant program director of the

Applied Economics master’s degree program at Johns Hopkins

University. She analysed the impact that ‘Covid-19’ had on deaths

from all causes in the United States using official data from the CDC

for the period from early February to early September, 2020. She

found that allegedly ‘Covid’ related-deaths exceeded those from

heart disease which she found strange with heart disease always the

biggest cause of fatalities. Her research became even more significant

when she noted the sudden decline in 2020 of all non-’Covid’ deaths:

‘This trend is completely contrary to the pa�ern observed in all

previous years … the total decrease in deaths by other causes almost

exactly equals the increase in deaths by Covid-19.’ This was such a

game, set and match in terms of what was happening that Johns

Hopkins University deleted the article on the grounds that it ‘was

being used to support false and dangerous inaccuracies about the

impact of the pandemic’. No – because it exposed the scam from

official CDC figures and this was confirmed when those figures were

published in January, 2021. Here we can see the effect of people

dying from heart a�acks, cancer, road accidents and gunshot

wounds – anything – having ‘Covid-19’ on the death certificate along

with those diagnosed from ‘symptoms’ who had even not tested

positive with a test not testing for the ‘virus’. I am not kidding with

the gunshot wounds, by the way. Brenda Bock, coroner in Grand

County, Colorado, revealed that two gunshot victims tested positive

for the ‘virus’ within the previous 30 days and were therefore

classified as ‘Covid deaths’. Bock said: ‘These two people had tested

positive for Covid, but that’s not what killed them. A gunshot

wound is what killed them.’ She said she had not even finished her

investigation when the state listed the gunshot victims as deaths due

to the ‘virus’. The death and case figures for ‘Covid-19’ are an

absolute joke and yet they are repeated like parrots by the media,

politicians and alleged medical ‘experts’. The official Cult narrative

is the only show in town.

Genevieve Briand found that deaths from all causes were not

exceptional in 2020 compared with previous years and a Spanish

magazine published figures that said the same about Spain which

was a ‘Covid’ propaganda hotspot at one point. Discovery Salud, a

health and medicine magazine, quoted government figures which

showed how 17,000 fewer people died in Spain in 2020 than in 2019

and more than 26,000 fewer than in 2018. The age-standardised

mortality rate for England and Wales when age distribution is taken

into account was significantly lower in 2020 than the 1970s, 80s and

90s, and was only the ninth highest since 2000. Where is the

‘pandemic’?

Post mortems and autopsies virtually disappeared for ‘Covid’

deaths amid claims that ‘virus-infected’ bodily fluids posed a risk to

those carrying out the autopsy. This was rejected by renowned

German pathologist and forensic doctor Klaus Püschel who said that

he and his staff had by then done 150 autopsies on ‘Covid’ patients

with no problems at all. He said they were needed to know why

some ‘Covid’ patients suffered blood clots and not severe respiratory

infections. The ‘virus’ is, a�er all, called SARS or ‘severe acute

respiratory syndrome’. I highlighted in the spring of 2020 this

phenomenon and quoted New York intensive care doctor Cameron

Kyle-Sidell who posted a soon deleted YouTube video to say that

they had been told to prepare to treat an infectious disease called

‘Covid-19’, but that was not what they were dealing with. Instead he

likened the lung condition of the most severely ill patients to what

you would expect with cabin depressurisation in a plane at 30,000

feet or someone dropped on the top of Everest without oxygen or

acclimatisation. I have never said this is not happening to a small

minority of alleged ‘Covid’ patients – I am saying this is not caused

by a phantom ‘contagious virus’. Indeed Kyle-Sidell said that

‘Covid-19’ was not the disease they were told was coming their way.

‘We are operating under a medical paradigm that is untrue,’ he said,

and he believed they were treating the wrong disease: ‘These people

are being slowly starved of oxygen.’ Patients would take off their

oxygen masks in a state of fear and stress and while they were blue

in the face on the brink of death. They did not look like patients

dying of pneumonia. You can see why they don’t want autopsies

when their virus doesn’t exist and there is another condition in some

people that they don’t wish to be uncovered. I should add here that

the 5G system of millimetre waves was being rapidly introduced

around the world in 2020 and even more so now as they fire 5G at

the Earth from satellites. At 60 gigahertz within the 5G range that

frequency interacts with the oxygen molecule and stops people

breathing in sufficient oxygen to be absorbed into the bloodstream.

They are installing 5G in schools and hospitals. The world is not

mad or anything. 5G can cause major changes to the lungs and blood

as I detail in The Answer and these consequences are labelled ‘Covid-

19’, the alleged symptoms of which can be caused by 5G and other

electromagnetic frequencies as cells respond to radiation poisoning.

The ‘Covid death’ scam

Dr Sco� Jensen, a Minnesota state senator and medical doctor,

exposed ‘Covid’ Medicare payment incentives to hospitals and death

certificate manipulation. He said he was sent a seven-page document

by the US Department of Health ‘coaching’ him on how to fill out

death certificates which had never happened before. The document

said that he didn’t need to have a laboratory test for ‘Covid-19’ to

put that on the death certificate and that shocked him when death

certificates are supposed to be about facts. Jensen described how

doctors had been ‘encouraged, if not pressured’ to make a diagnosis

of ‘Covid-19’ if they thought it was probable or ‘presumed’. No

positive test was necessary – not that this would have ma�ered

anyway. He said doctors were told to diagnose ‘Covid’ by symptoms

when these were the same as colds, allergies, other respiratory

problems, and certainly with influenza which ‘disappeared’ in the

‘Covid’ era. A common sniffle was enough to get the dreaded

verdict. Ontario authorities decreed that a single care home resident

with one symptom from a long list must lead to the isolation of the

entire home. Other courageous doctors like Jensen made the same

point about death figure manipulation and how deaths by other

causes were falling while ‘Covid-19 deaths’ were rising at the same

rate due to re-diagnosis. Their videos rarely survive long on

YouTube with its Cult-supporting algorithms courtesy of CEO Susan

Wojcicki and her bosses at Google. Figure-tampering was so glaring

and ubiquitous that even officials were le�ing it slip or outright

saying it. UK chief scientific adviser Patrick Vallance said on one

occasion that ‘Covid’ on the death certificate doesn’t mean ‘Covid’

was the cause of death (so why the hell is it there?) and we had the

rare sight of a BBC reporter telling the truth when she said:

‘Someone could be successfully treated for Covid, in say April,

discharged, and then in June, get run over by a bus and die … That

person would still be counted as a Covid death in England.’ Yet the

BBC and the rest of the world media went on repeating the case and

death figures as if they were real. Illinois Public Health Director Dr

Ngozi Ezike revealed the deceit while her bosses must have been

clenching their bu�ocks:

If you were in a hospice and given a few weeks to live and you were then found to have Covid that would be counted as a Covid death. [There might be] a clear alternate cause, but it is still listed as a Covid death. So everyone listed as a Covid death doesn’t mean that was the cause of the death, but that they had Covid at the time of death.

Yes, a ‘Covid virus’ never shown to exist and tested for with a test

not testing for the ‘virus’. In the first period of the pandemic hoax

through the spring of 2020 the process began of designating almost

everything a ‘Covid’ death and this has continued ever since. I sat in

a restaurant one night listening to a loud conversation on the next

table where a family was discussing in bewilderment how a relative

who had no symptoms of ‘Covid’, and had died of a long-term

problem, could have been diagnosed a death by the ‘virus’. I could

understand their bewilderment. If they read this book they will

know why this medical fraud has been perpetrated the world over.

Some media truth shock

The media ignored the evidence of death certificate fraud until

eventually one columnist did speak out when she saw it first-hand.

Bel Mooney is a long-time national newspaper journalist in Britain

currently working for the Daily Mail. Her article on February 19th,

2021, carried this headline: ‘My dad Ted passed three Covid tests

and died of a chronic illness yet he’s officially one of Britain’s 120,000

victims of the virus and is far from alone ... so how many more are

there?’ She told how her 99-year-old father was in a care home with

a long-standing chronic obstructive pulmonary disease and vascular

dementia. Maybe, but he was still aware enough to tell her from the

start that there was no ‘virus’ and he refused the ‘vaccine’ for that

reason. His death was not unexpected given his chronic health

problems and Mooney said she was shocked to find that ‘Covid-19’

was declared the cause of death on his death certificate. She said this

was a ‘bizarre and unacceptable untruth’ for a man with long-time

health problems who had tested negative twice at the home for the

‘virus’. I was also shocked by this story although not by what she

said. I had been highlighting the death certificate manipulation for

ten months. It was the confirmation that a professional full-time

journalist only realised this was going on when it affected her

directly and neither did she know that whether her dad tested

positive or negative was irrelevant with the test not testing for the

‘virus’. Where had she been? She said she did not believe in

‘conspiracy theories’ without knowing I’m sure that this and

‘conspiracy theorists’ were terms put into widespread circulation by

the CIA in the 1960s to discredit those who did not accept the

ridiculous official story of the Kennedy assassination. A blanket

statement of ‘I don’t believe in conspiracy theories’ is always bizarre.

The dictionary definition of the term alone means the world is

drowning in conspiracies. What she said was even more da� when

her dad had just been affected by the ‘Covid’ conspiracy. Why else

does she think that ‘Covid-19’ was going on the death certificates of

people who died of something else?

To be fair once she saw from personal experience what was

happening she didn’t mince words. Mooney was called by the care

home on the morning of February 9th to be told her father had died

in his sleep. When she asked for the official cause of death what

came back was ‘Covid-19’. Mooney challenged this and was told

there had been deaths from Covid on the dementia floor (confirmed

by a test not testing for the ‘virus’) so they considered it ‘reasonable

to assume’. ‘But doctor,’ Mooney rightly protested, ‘an assumption

isn’t a diagnosis.’ She said she didn’t blame the perfectly decent and

sympathetic doctor – ‘he was just doing his job’. Sorry, but that’s

bullshit. He wasn’t doing his job at all. He was pu�ing a false cause of

death on the death certificate and that is a criminal offence for which

he should be brought to account and the same with the millions of

doctors worldwide who have done the same. They were not doing

their job they were following orders and that must not wash at new

Nuremberg trials any more than it did at the first ones. Mooney’s

doctor was ‘assuming’ (presuming) as he was told to, but ‘just

following orders’ makes no difference to his actions. A doctor’s job is

to serve the patient and the truth, not follow orders, but that’s what

they have done all over the world and played a central part in

making the ‘Covid’ hoax possible with all its catastrophic

consequences for humanity. Shame on them and they must answer

for their actions. Mooney said her disquiet worsened when she

registered her father’s death by telephone and was told by the

registrar there had been very many other cases like hers where ‘the

deceased’ had not tested positive for ‘Covid’ yet it was recorded as

the cause of death. The test may not ma�er, but those involved at

their level think it ma�ers and it shows a callous disregard for

accurate diagnosis. The pressure to do this is coming from the top of

the national ‘health’ pyramids which in turn obey the World Health

Organization which obeys Gates and the Cult. Mooney said the

registrar agreed that this must distort the national figures adding

that ‘the strangest thing is that every winter we record countless

deaths from flu, and this winter there have been none. Not one!’ She

asked if the registrar thought deaths from flu were being

misdiagnosed and lumped together with ‘Covid’ deaths. The answer

was a ‘puzzled yes’. Mooney said that the funeral director said the

same about ‘Covid’ deaths which had nothing to do with ‘Covid’.

They had lost count of the number of families upset by this and

other funeral companies in different countries have had the same

experience. Mooney wrote:

The nightly shroud-waving and shocking close-ups of pain imposed on us by the TV news bewildered and terrified the population into eager compliance with lockdowns. We were invited to ‘save the NHS’ and to grieve for strangers – the real-life loved ones behind those shocking death counts. Why would the public imagine what I now fear, namely that the way Covid-19 death statistics are compiled might make the numbers seem greater than they are?

Oh, just a li�le bit – like 100 percent.

Do the maths

Mooney asked why a country would wish to skew its mortality

figures by wrongly certifying deaths? What had been going on?

Well, if you don’t believe in conspiracies you will never find the

answer which is that it’s a conspiracy. She did, however, describe

what she had discovered as a ‘national scandal’. In reality it’s a

global scandal and happening everywhere. Pillars of this conspiracy

were all put into place before the bu�on was pressed with the

Drosten PCR protocol and high amplifications to produce the cases

and death certificate changes to secure illusory ‘Covid’ deaths.

Mooney notes that normally two doctors were needed to certify a

death, with one having to know the patient, and how the rules were

changed in the spring of 2020 to allow one doctor to do this. In the

same period ‘Covid deaths’ were decreed to be all cases where

Covid-19 was put on the death certificate even without a positive test

or any symptoms. Mooney asked: ‘How many of the 30,851 (as of

January 15) care home resident deaths with Covid-19 on the

certificate (32.4 per cent of all deaths so far) were based on an

assumption, like that of my father? And what has that done to our

national psyche?’All of them is the answer to the first question and it

has devastated and dismantled the national psyche, actually the

global psyche, on a colossal scale. In the UK case and death data is

compiled by organisations like Public Health England (PHE) and the

Office for National Statistics (ONS). Mooney highlights the insane

policy of counting a death from any cause as ‘Covid-19’ if this

happens within 28 days of a positive test (with a test not testing for

the ‘virus’) and she points out that ONS statistics reflect deaths

‘involving Covid’ ‘or due to Covid’ which meant in practice any

death where ‘Covid-19’ was mentioned on the death certificate. She

described the consequences of this fraud:

Most people will accept the narrative they are fed, so panicky governments here and in Europe witnessed the harsh measures enacted in totalitarian China and jumped into lockdown. Headlines about Covid deaths tolled like the knell that would bring doomsday to us all. Fear stalked our empty streets. Politicians parroted the frankly ridiculous aim of ‘zero Covid’ and shut down the economy, while most British people agreed that lockdown was essential and (astonishingly to me, as a patriotic Brit) even wanted more restrictions.

For what? Lies on death certificates? Never mind the grim toll of lives ruined, suicides, schools closed, rising inequality, depression, cancelled hospital treatments, cancer patients in a torture of waiting, poverty, economic devastation, loneliness, families kept apart, and so on. How many lives have been lost as a direct result of lockdown?

She said that we could join in a national chorus of shock and horror

at reaching the 120,000 death toll which was surely certain to have

been totally skewed all along, but what about the human cost of

lockdown justified by these ‘death figures’? The British Medical

Journal had reported a 1,493 percent increase in cases of children

taken to Great Ormond Street Hospital with abusive head injuries

alone and then there was the effect on families:

Perhaps the most shocking thing about all this is that families have been kept apart – and obeyed the most irrational, changing rules at the whim of government – because they believed in the statistics. They succumbed to fear, which his generation rejected in that war fought for freedom. Dad (God rest his soul) would be angry. And so am I.

Another theme to watch is that in the winter months when there

are more deaths from all causes they focus on ‘Covid’ deaths and in

the summer when the British Lung Foundation says respiratory

disease plummets by 80 percent they rage on about ‘cases’. Either

way fascism on population is always the answer.

Nazi eugenics in the 21st century

Elderly people in care homes have been isolated from their families

month a�er lonely month with no contact with relatives and

grandchildren who were banned from seeing them. We were told

that lockdown fascism was to ‘protect the vulnerable’ like elderly

people. At the same time Do Not Resuscitate (DNR) orders were

placed on their medical files so that if they needed resuscitation it

wasn’t done and ‘Covid-19’ went on their death certificates. Old

people were not being ‘protected’ they were being culled –

murdered in truth. DNR orders were being decreed for disabled and

young people with learning difficulties or psychological problems.

The UK Care Quality Commission, a non-departmental body of the

Department of Health and Social Care, found that 34 percent of

those working in health and social care were pressured into placing

‘do not a�empt cardiopulmonary resuscitation’ orders on ‘Covid’

patients who suffered from disabilities and learning difficulties

without involving the patient or their families in the decision. UK

judges ruled that an elderly woman with dementia should have the

DNA-manipulating ‘Covid vaccine’ against her son’s wishes and that

a man with severe learning difficulties should have the jab despite

his family’s objections. Never mind that many had already died. The

judiciary always supports doctors and government in fascist

dictatorships. They wouldn’t dare do otherwise. A horrific video was

posted showing fascist officers from Los Angeles police forcibly

giving the ‘Covid’ shot to women with special needs who were

screaming that they didn’t want it. The same fascists are seen giving

the jab to a sleeping elderly woman in a care home. This is straight

out of the Nazi playbook. Hitler’s Nazis commi�ed mass murder of

the mentally ill and physically disabled throughout Germany and

occupied territories in the programme that became known as Aktion

T4, or just T4. Sabbatian-controlled Hitler and his grotesque crazies

set out to kill those they considered useless and unnecessary. The

Reich Commi�ee for the Scientific Registering of Hereditary and

Congenital Illnesses registered the births of babies identified by

physicians to have ‘defects’. By 1941 alone more than 5,000 children

were murdered by the state and it is estimated that in total the

number of innocent people killed in Aktion T4 was between 275,000

and 300,000. Parents were told their children had been sent away for

‘special treatment’ never to return. It is rather pathetic to see claims

about plans for new extermination camps being dismissed today

when the same force behind current events did precisely that 80

years ago. Margaret Sanger was a Cult operative who used ‘birth

control’ to sanitise her programme of eugenics. Organisations she

founded became what is now Planned Parenthood. Sanger proposed

that ‘the whole dysgenic population would have its choice of

segregation or sterilization’. These included epileptics, ‘feeble-

minded’, and prostitutes. Sanger opposed charity because it

perpetuated ‘human waste‘. She reveals the Cult mentality and if

anyone thinks that extermination camps are a ‘conspiracy theory’

their naivety is touching if breathtakingly stupid.

If you don’t believe that doctors can act with callous disregard for

their patients it is worth considering that doctors and medical staff

agreed to put government-decreed DNR orders on medical files and

do nothing when resuscitation is called for. I don’t know what you

call such people in your house. In mine they are Nazis from the Josef

Mengele School of Medicine. Phenomenal numbers of old people

have died worldwide from the effects of lockdown, depression, lack

of treatment, the ‘vaccine’ (more later) and losing the will to live. A

common response at the start of the manufactured pandemic was to

remove old people from hospital beds and transfer them to nursing

homes. The decision would result in a mass cull of elderly people in

those homes through lack of treatment – not ‘Covid’. Care home

whistleblowers have told how once the ‘Covid’ era began doctors

would not come to their homes to treat patients and they were

begging for drugs like antibiotics that o�en never came. The most

infamous example was ordered by New York governor Andrew

Cuomo, brother of a moronic CNN host, who amazingly was given

an Emmy Award for his handling of the ‘Covid crisis’ by the

ridiculous Wokers that hand them out. Just how ridiculous could be

seen in February, 2021, when a Department of Justice and FBI

investigation began into how thousands of old people in New York

died in nursing homes a�er being discharged from hospital to make

way for ‘Covid’ patients on Cuomo’s say-so – and how he and his

staff covered up these facts. This couldn’t have happened to a nicer

psychopath. Even then there was a ‘Covid’ spin. Reports said that

thousands of old people who tested positive for ‘Covid’ in hospital

were transferred to nursing homes to both die of ‘Covid’ and

transmit it to others. No – they were in hospital because they were ill

and the fact that they tested positive with a test not testing for the

‘virus’ is irrelevant. They were ill o�en with respiratory diseases

ubiquitous in old people near the end of their lives. Their transfer

out of hospital meant that their treatment stopped and many would

go on to die.

They’re old. Who gives a damn?

I have exposed in the books for decades the Cult plan to cull the

world’s old people and even to introduce at some point what they

call a ‘demise pill’ which at a certain age everyone would take and

be out of here by law. In March, 2021, Spain legalised euthanasia and

assisted suicide following the Netherlands, Belgium, Luxembourg

and Canada on the Tiptoe to the demise pill. Treatment of old people

by many ‘care’ homes has been a disgrace in the ‘Covid’ era. There

are many, many, caring staff – I know some. There have, however,

been legions of stories about callous treatment of old people and

their families. Police were called when families came to take their

loved ones home in the light of isolation that was killing them. They

became prisoners of the state. Care home residents in insane, fascist

Ontario, Canada, were not allowed to leave their room once the

‘Covid’ hoax began. UK staff have even wheeled elderly people

away from windows where family members were talking with them.

Oriana Criscuolo from Stockport in the English North West dropped

off some things for her 80-year-old father who has Parkinson’s

disease and dementia and she wanted to wave to him through a

ground-floor window. She was told that was ‘illegal’. When she went

anyway they closed the curtains in the middle of the day. Oriana

said:

It’s just unbelievable. I cannot understand how care home staff – people who are being paid to care – have become so uncaring. Their behaviour is inhumane and cruel. It’s beyond belief.

She was right and this was not a one-off. What a way to end your life

in such loveless circumstances. UK registered nurse Nicky Millen, a

proper old school nurse for 40 years, said that when she started her

career care was based on dignity, choice, compassion and empathy.

Now she said ‘the things that are important to me have gone out of

the window.’ She was appalled that people were dying without their

loved ones and saying goodbye on iPads. Nicky described how a

distressed 89-year-old lady stroked her face and asked her ‘how

many paracetamol would it take to finish me off’. Life was no longer

worth living while not seeing her family. Nicky said she was

humiliated in front of the ward staff and patients for le�ing the lady

stroke her face and giving her a cuddle. Such is the dehumanisation

that the ‘Covid’ hoax has brought to the surface. Nicky worked in

care homes where patients told her they were being held prisoner. ‘I

want to live until I die’, one said to her. ‘I had a lady in tears because

she hadn’t seen her great-grandson.’ Nicky was compassionate old

school meeting psychopathic New Normal. She also said she had

worked on a ‘Covid’ ward with no ‘Covid’ patients. Jewish writer

Shai Held wrote an article in March, 2020, which was headlined ‘The

Staggering, Heartless Cruelty Toward the Elderly’. What he

described was happening from the earliest days of lockdown. He

said ‘the elderly’ were considered a group and not unique

individuals (the way of the Woke). Shai Held said:

Notice how the all-too-familiar rhetoric of dehumanization works: ‘The elderly’ are bunched together as a faceless mass, all of them considered culprits and thus effectively deserving of the suffering the pandemic will inflict upon them. Lost entirely is the fact that the elderly are individual human beings, each with a distinctive face and voice, each with hopes and dreams, memories and regrets, friendships and marriages, loves lost and loves sustained.

‘The elderly’ have become another dehumanised group for which

anything goes and for many that has resulted in cold disregard for

their rights and their life. The distinctive face that Held talks about is

designed to be deleted by masks until everyone is part of a faceless

mass.

‘War-zone’ hospitals myth

Again and again medical professionals have told me what was really

going on and how hospitals ‘overrun like war zones’ according to

the media were virtually empty. The mantra from medical

whistleblowers was please don’t use my name or my career is over.

Citizen journalists around the world sneaked into hospitals to film

evidence exposing the ‘war-zone’ lie. They really were largely empty

with closed wards and operating theatres. I met a hospital worker in

my town on the Isle of Wight during the first lockdown in 2020 who

said the only island hospital had never been so quiet. Lockdown was

justified by the psychopaths to stop hospitals being overrun. At the

same time that the island hospital was near-empty the military

arrived here to provide extra beds. It was all propaganda to ramp up

the fear to ensure compliance with fascism as were never-used

temporary hospitals with thousands of beds known as Nightingales

and never-used make-shi� mortuaries opened by the criminal UK

government. A man who helped to install those extra island beds

a�ributed to the army said they were never used and the hospital

was empty. Doctors and nurses ‘stood around talking or on their

phones, wandering down to us to see what we were doing’. There

were no masks or social distancing. He accused the useless local

island paper, the County Press, of ‘pumping the fear as if our hospital

was overrun and we only have one so it should have been’. He

described ambulances parked up with crews outside in deck chairs.

When his brother called an ambulance he was told there was a two-

hour backlog which he called ‘bullshit’. An old lady on the island fell

‘and was in a bad way’, but a caller who rang for an ambulance was

told the situation wasn’t urgent enough. Ambulance stations were

working under capacity while people would hear ambulances with

sirens blaring driving through the streets. When those living near

the stations realised what was going on they would follow them as

they le�, circulated around an urban area with the sirens going, and

then came back without stopping. All this was to increase levels of

fear and the same goes for the ‘ventilator shortage crisis’ that cost

tens of millions for hastily produced ventilators never to be used.

Ambulance crews that agreed to be exploited in this way for fear

propaganda might find themselves a mirror. I wish them well with

that. Empty hospitals were the obvious consequence of treatment

and diagnoses of non-’Covid’ conditions cancelled and those

involved handed a death sentence. People have been dying at home

from undiagnosed and untreated cancer, heart disease and other life-

threatening conditions to allow empty hospitals to deal with a

‘pandemic’ that wasn’t happening.

Death of the innocent

‘War-zones’ have been laying off nursing staff, even doctors where

they can. There was no work for them. Lockdown was justified by

saving lives and protecting the vulnerable they were actually killing

with DNR orders and preventing empty hospitals being ‘overrun’. In

Britain the mantra of stay at home to ‘save the NHS’ was everywhere

and across the world the same story was being sold when it was all

lies. Two California doctors, Dan Erickson and Artin Massihi at

Accelerated Urgent Care in Bakersfield, held a news conference in

April, 2020, to say that intensive care units in California were ‘empty,

essentially’, with hospitals shu�ing floors, not treating patients and

laying off doctors. The California health system was working at

minimum capacity ‘ge�ing rid of doctors because we just don’t have

the volume’. They said that people with conditions such as heart

disease and cancer were not coming to hospital out of fear of ‘Covid-

19’. Their video was deleted by Susan Wojcicki’s Cult-owned

YouTube a�er reaching five million views. Florida governor Ron

Desantis, who rejected the severe lockdowns of other states and is

being targeted for doing so, said that in March, 2020, every US

governor was given models claiming they would run out of hospital

beds in days. That was never going to happen and the ‘modellers’

knew it. Deceit can be found at every level of the system. Urgent

children’s operations were cancelled including fracture repairs and

biopsies to spot cancer. Eric Nicholls, a consultant paediatrician, said

‘this is obviously concerning and we need to return to normal

operating and to increase capacity as soon as possible’. Psychopaths

in power were rather less concerned because they are psychopaths.

Deletion of urgent care and diagnosis has been happening all over

the world and how many kids and others have died as a result of the

actions of these cold and heartless lunatics dictating ‘health’ policy?

The number must be stratospheric. Richard Sullivan, professor of

cancer and global health at King’s College London, said people

feared ‘Covid’ more than cancer such was the campaign of fear.

‘Years of lost life will be quite dramatic’, Sullivan said, with ‘a huge

amount of avoidable mortality’. Sarah Woolnough, executive

director for policy at Cancer Research UK, said there had been a 75

percent drop in urgent referrals to hospitals by family doctors of

people with suspected cancer. Sullivan said that ‘a lot of services

have had to scale back – we’ve seen a dramatic decrease in the

amount of elective cancer surgery’. Lockdown deaths worldwide has

been absolutely fantastic with the New York Post reporting how data

confirmed that ‘lockdowns end more lives than they save’:

There was a sharp decline in visits to emergency rooms and an increase in fatal heart attacks because patients didn’t receive prompt treatment. Many fewer people were screened for cancer. Social isolation contributed to excess deaths from dementia and Alzheimer’s.

Researchers predicted that the social and economic upheaval would lead to tens of thousands of “deaths of despair” from drug overdoses, alcoholism and suicide. As unemployment surged and mental-health and substance-abuse treatment programs were interrupted, the reported levels of anxiety, depression and suicidal thoughts increased dramatically, as did alcohol sales and fatal drug overdoses.

This has been happening while nurses and other staff had so much

time on their hands in the ‘war-zones’ that Tic-Tok dancing videos

began appearing across the Internet with medical staff dancing

around in empty wards and corridors as people died at home from

causes that would normally have been treated in hospital.

Mentions in dispatches

One brave and truth-commi�ed whistleblower was Louise

Hampton, a call handler with the UK NHS who made a viral

Internet video saying she had done ‘fuck all’ during the ‘pandemic’

which was ‘a load of bollocks’. She said that ‘Covid-19’ was

rebranded flu and of course she lost her job. This is what happens in

the medical and endless other professions now when you tell the

truth. Louise filmed inside ‘war-zone’ accident and emergency

departments to show they were empty and I mean empty as in no

one there. The mainstream media could have done the same and

blown the gaff on the whole conspiracy. They haven’t to their eternal

shame. Not that most ‘journalists’ seem capable of manifesting

shame as with the psychopaths they slavishly repeat without

question. The relative few who were admi�ed with serious health

problems were le� to die alone with no loved ones allowed to see

them because of ‘Covid’ rules and they included kids dying without

the comfort of mum and dad at their bedside while the evil behind

this couldn’t give a damn. It was all good fun to them. A Sco�ish

NHS staff nurse publicly quit in the spring of 2021 saying: ‘I can no

longer be part of the lies and the corruption by the government.’ She

said hospitals ‘aren’t full, the beds aren’t full, beds have been shut,

wards have been shut’. Hospitals were never busy throughout

‘Covid’. The staff nurse said that Nicola Sturgeon, tragically the

leader of the Sco�ish government, was on television saying save the

hospitals and the NHS – ‘but the beds are empty’ and ‘we’ve not

seen flu, we always see flu every year’. She wrote to government and

spoke with her union Unison (the unions are Cult-compromised and

useless, but nothing changed. Many of her colleagues were scared of

losing their jobs if they spoke out as they wanted to. She said

nursing staff were being affected by wearing masks all day and ‘my

head is spli�ing every shi� from wearing a mask’. The NHS is part

of the fascist tyranny and must be dismantled so we can start again

with human beings in charge. (Ironically, hospitals were reported to

be busier again when official ‘Covid’ cases fell in spring/summer of

2021 and many other conditions required treatment at the same time

as the fake vaccine rollout.)

I will cover the ‘Covid vaccine’ scam in detail later, but it is

another indicator of the sickening disregard for human life that I am

highlighting here. The DNA-manipulating concoctions do not fulfil

the definition of a ‘vaccine’, have never been used on humans before

and were given only emergency approval because trials were not

completed and they continued using the unknowing public. The

result was what a NHS senior nurse with responsibility for ‘vaccine’

procedure said was ‘genocide’. She said the ‘vaccines’ were not

‘vaccines’. They had not been shown to be safe and claims about

their effectiveness by drug companies were ‘poetic licence’. She

described what was happening as a ‘horrid act of human

annihilation’. The nurse said that management had instigated a

policy of not providing a Patient Information Leaflet (PIL) before

people were ‘vaccinated’ even though health care professionals are

supposed to do this according to protocol. Patients should also be

told that they are taking part in an ongoing clinical trial. Her

challenges to what is happening had seen her excluded from

meetings and ridiculed in others. She said she was told to ‘watch my

step … or I would find myself surplus to requirements’. The nurse,

who spoke anonymously in fear of her career, said she asked her

NHS manager why he/she was content with taking part in genocide

against those having the ‘vaccines’. The reply was that everyone had

to play their part and to ‘put up, shut up, and get it done’.

Government was ‘leaning heavily’ on NHS management which was

clearly leaning heavily on staff. This is how the global ‘medical’

hierarchy operates and it starts with the Cult and its World Health

Organization.

She told the story of a doctor who had the Pfizer jab and when

questioned had no idea what was in it. The doctor had never read

the literature. We have to stop treating doctors as intellectual giants

when so many are moral and medical pygmies. The doctor did not

even know that the ‘vaccines’ were not fully approved or that their

trials were ongoing. They were, however, asking their patients if

they minded taking part in follow-ups for research purposes – yes,

the ongoing clinical trial. The nurse said the doctor’s ignorance was

not rare and she had spoken to a hospital consultant who had the jab

without any idea of the background or that the ‘trials’ had not been

completed. Nurses and pharmacists had shown the same ignorance.

‘My NHS colleagues have forsaken their duty of care, broken their

code of conduct – Hippocratic Oath – and have been brainwashed

just the same as the majority of the UK public through propaganda

…’ She said she had not been able to recruit a single NHS colleague,

doctor, nurse or pharmacist to stand with her and speak out. Her

union had refused to help. She said that if the genocide came to light

she would not hesitate to give evidence at a Nuremberg-type trial

against those in power who could have affected the outcomes but

didn’t.

And all for what?

To put the nonsense into perspective let’s say the ‘virus’ does exist

and let’s go completely crazy and accept that the official

manipulated figures for cases and deaths are accurate. Even then a

study by Stanford University epidemiologist Dr John Ioannidis

published on the World Health Organization website produced an

average infection to fatality rate of … 0.23 percent! Ioannidis said: ‘If

one could sample equally from all locations globally, the median

infection fatality rate might even be substantially lower than the

0.23% observed in my analysis.’ For healthy people under 70 it was

… 0.05 percent! This compares with the 3.4 percent claimed by the

Cult-owned World Health Organization when the hoax was first

played and maximum fear needed to be generated. An updated

Stanford study in April, 2021, put the ‘infection’ to ‘fatality’ rate at

just 0.15 percent. Another team of scientists led by Megan O’Driscoll

and Henrik Salje studied data from 45 countries and published their

findings on the Nature website. For children and young people the

figure is so small it virtually does not register although authorities

will be hyping dangers to the young when they introduce DNA-

manipulating ‘vaccines’ for children. The O’Driscoll study produced

an average infection-fatality figure of 0.003 for children from birth to

four; 0.001 for 5 to 14; 0.003 for 15 to 19; and it was still only 0.456 up

to 64. To claim that children must be ‘vaccinated’ to protect them

from ‘Covid’ is an obvious lie and so there must be another reason

and there is. What’s more the average age of a ‘Covid’ death is akin

to the average age that people die in general. The average age of

death in England is about 80 for men and 83 for women. The average

age of death from alleged ‘Covid’ is between 82 and 83. California

doctors, Dan Erickson and Artin Massihi, said at their April media

conference that projection models of millions of deaths had been

‘woefully inaccurate’. They produced detailed figures showing that

Californians had a 0.03 chance of dying from ‘Covid’ based on the

number of people who tested positive (with a test not testing for the

‘virus’). Erickson said there was a 0.1 percent chance of dying from

‘Covid’ in the state of New York, not just the city, and a 0.05 percent

chance in Spain, a centre of ‘Covid-19’ hysteria at one stage. The

Stanford studies supported the doctors’ data with fatality rate

estimates of 0.23 and 0.15 percent. How close are these figures to my

estimate of zero? Death-rate figures claimed by the World Health

Organization at the start of the hoax were some 15 times higher. The

California doctors said there was no justification for lockdowns and

the economic devastation they caused. Everything they had ever

learned about quarantine was that you quarantine the sick and not

the healthy. They had never seen this before and it made no medical

sense.

Why in the in the light of all this would governments and medical

systems the world over say that billions must go under house arrest;

lose their livelihood; in many cases lose their mind, their health and

their life; force people to wear masks dangerous to health and

psychology; make human interaction and even family interaction a

criminal offence; ban travel; close restaurants, bars, watching live

sport, concerts, theatre, and any activity involving human

togetherness and discourse; and closing schools to isolate children

from their friends and cause many to commit suicide in acts of

hopelessness and despair? The California doctors said lockdown

consequences included increased child abuse, partner abuse,

alcoholism, depression, and other impacts they were seeing every

day. Who would do that to the entire human race if not mentally-ill

psychopaths of almost unimaginable extremes like Bill Gates? We

must face the reality of what we are dealing with and come out of

denial. Fascism and tyranny are made possible only by the target

population submi�ing and acquiescing to fascism and tyranny. The

whole of human history shows that to be true. Most people naively

and unquestioning believed what they were told about a ‘deadly

virus’ and meekly and weakly submi�ed to house arrest. Those who

didn’t believe it – at least in total – still submi�ed in fear of the

consequences of not doing so. For the rest who wouldn’t submit

draconian fines have been imposed, brutal policing by psychopaths

for psychopaths, and condemnation from the meek and weak who

condemn the Pushbackers on behalf of the very force that has them,

too, in its gunsights. ‘Pathetic’ does not even begin to suffice.

Britain’s brainless ‘Health’ Secretary Ma� Hancock warned anyone

lying to border officials about returning from a list of ‘hotspot’

countries could face a jail sentence of up to ten years which is more

than for racially-aggravated assault, incest and a�empting to have

sex with a child under 13. Hancock is a lunatic, but he has the state

apparatus behind him in a Cult-led chain reaction and the same with

UK ‘Vaccine Minister’ Nadhim Zahawi, a prominent member of the

mega-Cult secret society, Le Cercle, which featured in my earlier

books. The Cult enforces its will on governments and medical

systems; government and medical systems enforce their will on

business and police; business enforces its will on staff who enforce it

on customers; police enforce the will of the Cult on the population

and play their essential part in creating a world of fascist control that

their own children and grandchildren will have to live in their entire

lives. It is a hierarchical pyramid of imposition and acquiescence

and, yes indeedy, of clinical insanity.

Does anyone bright enough to read this book have to ask what the

answer is? I think not, but I will reveal it anyway in the fewest of

syllables: Tell the psychos and their moronic lackeys to fuck off and

let’s get on with our lives. We are many – They are few.

I

CHAPTER SEVEN

War on your mind

One believes things because one has been conditioned to believe

them

Aldous Huxley, Brave New World

have described the ‘Covid’ hoax as a ‘Psyop’ and that is true in

every sense and on every level in accordance with the definition of

that term which is psychological warfare. Break down the ‘Covid

pandemic’ to the foundation themes and it is psychological warfare

on the human individual and collective mind.

The same can be said for the entire human belief system involving

every subject you can imagine. Huxley was right in his contention

that people believe what they are conditioned to believe and this

comes from the repetition throughout their lives of the same

falsehoods. They spew from government, corporations, media and

endless streams of ‘experts’ telling you what the Cult wants you to

believe and o�en believing it themselves (although far from always).

‘Experts’ are rewarded with ‘prestigious’ jobs and titles and as

agents of perceptual programming with regular access to the media.

The Cult has to control the narrative – control information – or they

lose control of the vital, crucial, without-which-they-cannot-prevail

public perception of reality. The foundation of that control today is

the Internet made possible by the Defense Advanced Research

Projects Agency (DARPA), the incredibly sinister technological arm

of the Pentagon. The Internet is the result of military technology.

DARPA openly brags about establishing the Internet which has been

a long-term project to lasso the minds of the global population. I

have said for decades the plan is to control information to such an

extreme that eventually no one would see or hear anything that the

Cult does not approve. We are closing in on that end with ferocious

censorship since the ‘Covid’ hoax began and in my case it started

back in the 1990s in terms of books and speaking venues. I had to

create my own publishing company in 1995 precisely because no one

else would publish my books even then. I think they’re all still

running.

Cult Internet

To secure total control of information they needed the Internet in

which pre-programmed algorithms can seek out ‘unclean’ content

for deletion and even stop it being posted in the first place. The Cult

had to dismantle print and non-Internet broadcast media to ensure

the transfer of information to the appropriate-named ‘Web’ – a

critical expression of the Cult web. We’ve seen the ever-quickening

demise of traditional media and control of what is le� by a tiny

number of corporations operating worldwide. Independent

journalism in the mainstream is already dead and never was that

more obvious than since the turn of 2020. The Cult wants all

information communicated via the Internet to globally censor and

allow the plug to be pulled any time. Lockdowns and forced

isolation has meant that communication between people has been

through electronic means and no longer through face-to-face

discourse and discussion. Cult psychopaths have targeted the bars,

restaurants, sport, venues and meeting places in general for this

reason. None of this is by chance and it’s to stop people gathering in

any kind of privacy or number while being able to track and monitor

all Internet communications and block them as necessary. Even

private messages between individuals have been censored by these

fascists that control Cult fronts like Facebook, Twi�er, Google and

YouTube which are all officially run by Sabbatian place-people and

from the background by higher-level Sabbatian place people.

Facebook, Google, Amazon and their like were seed-funded and

supported into existence with money-no-object infusions of funds

either directly or indirectly from DARPA and CIA technology arm

In-Q-Tel. The Cult plays the long game and prepares very carefully

for big plays like ‘Covid’. Amazon is another front in the

psychological war and pre�y much controls the global market in

book sales and increasingly publishing. Amazon’s limitless funds

have deleted fantastic numbers of independent publishers to seize

global domination on the way to deciding which books can be sold

and circulated and which cannot. Moves in that direction are already

happening. Amazon’s leading light Jeff Bezos is the grandson of

Lawrence Preston Gise who worked with DARPA predecessor

ARPA. Amazon has big connections to the CIA and the Pentagon.

The plan I have long described went like this:

1. Employ military technology to establish the Internet.

2. Sell the Internet as a place where people can freely communicate without censorship and

allow that to happen until the Net becomes the central and irreversible pillar of human

society. If the Internet had been highly censored from the start many would have rejected it.

3. Fund and manipulate major corporations into being to control the circulation of

information on your Internet using cover stories about geeks in garages to explain how they

came about. Give them unlimited funds to expand rapidly with no need to make a profit for

years while non-Cult companies who need to balance the books cannot compete. You know

that in these circumstances your Googles, YouTubes, Facebooks and Amazons are going to

secure near monopolies by either crushing or buying up the opposition.

4. Allow freedom of expression on both the Internet and communication platforms to draw

people in until the Internet is the central and irreversible pillar of human society and your

communication corporations have reached a stage of near monopoly domination.

5. Then unleash your always-planned frenzy of censorship on the basis of ‘where else are

you going to go?’ and continue to expand that until nothing remains that the Cult does not

want its human targets to see.

The process was timed to hit the ‘Covid’ hoax to ensure the best

chance possible of controlling the narrative which they knew they

had to do at all costs. They were, a�er all, about to unleash a ‘deadly

virus’ that didn’t really exist. If you do that in an environment of

free-flowing information and opinion you would be dead in the

water before you could say Gates is a psychopath. The network was

in place through which the Cult-created-and-owned World Health

Organization could dictate the ‘Covid’ narrative and response policy

slavishly supported by Cult-owned Internet communication giants

and mainstream media while those telling a different story were

censored. Google, YouTube, Facebook and Twi�er openly

announced that they would do this. What else would we expect from

Cult-owned operations like Facebook which former executives have

confirmed set out to make the platform more addictive than

cigare�es and coldly manipulates emotions of its users to sow

division between people and groups and scramble the minds of the

young? If Zuckerberg lives out the rest of his life without going to

jail for crimes against humanity, and most emphatically against the

young, it will be a travesty of justice. Still, no ma�er, cause and effect

will catch up with him eventually and the same with Sergey Brin

and Larry Page at Google with its CEO Sundar Pichai who fix the

Google search results to promote Cult narratives and hide the

opposition. Put the same key words into Google and other search

engines like DuckDuckGo and you will see how different results can

be. Wikipedia is another intensely biased ‘encyclopaedia’ which

skews its content to the Cult agenda. YouTube links to Wikipedia’s

version of ‘Covid’ and ‘climate change’ on video pages in which

experts in their field offer a different opinion (even that is

increasingly rare with Wojcicki censorship). Into this ‘Covid’ silence-

them network must be added government media censors, sorry

‘regulators’, such as Ofcom in the UK which imposed tyrannical

restrictions on British broadcasters that had the effect of banning me

from ever appearing. Just to debate with me about my evidence and

views on ‘Covid’ would mean breaking the fascistic impositions of

Ofcom and its CEO career government bureaucrat Melanie Dawes.

Gutless British broadcasters tremble at the very thought of fascist

Ofcom.

Psychos behind ‘Covid’

The reason for the ‘Covid’ catastrophe in all its facets and forms can

be seen by whom and what is driving the policies worldwide in such

a coordinated way. Decisions are not being made to protect health,

but to target psychology. The dominant group guiding and

‘advising’ government policy are not medical professionals. They are

psychologists and behavioural scientists. Every major country has its

own version of this phenomenon and I’ll use the British example to

show how it works. In many ways the British version has been

affecting the wider world in the form of the huge behaviour

manipulation network in the UK which operates in other countries.

The network involves private companies, government, intelligence

and military. The Cabinet Office is at the centre of the government

‘Covid’ Psyop and part-owns, with ‘innovation charity’ Nesta, the

Behavioural Insights Team (BIT) which claims to be independent of

government but patently isn’t. The BIT was established in 2010 and

its job is to manipulate the psyche of the population to acquiesce to

government demands and so much more. It is also known as the

‘Nudge Unit’, a name inspired by the 2009 book by two ultra-

Zionists, Cass Sunstein and Richard Thaler, called Nudge: Improving

Decisions About Health, Wealth, and Happiness. The book, as with the

Behavioural Insights Team, seeks to ‘nudge’ behaviour (manipulate

it) to make the public follow pa�erns of action and perception that

suit those in authority (the Cult). Sunstein is so skilled at this that he

advises the World Health Organization and the UK Behavioural

Insights Team and was Administrator of the White House Office of

Information and Regulatory Affairs in the Obama administration.

Biden appointed him to the Department of Homeland Security –

another ultra-Zionist in the fold to oversee new immigration laws

which is another policy the Cult wants to control. Sunstein is

desperate to silence anyone exposing conspiracies and co-authored a

2008 report on the subject in which suggestions were offered to ban

‘conspiracy theorizing’ or impose ‘some kind of tax, financial or

otherwise, on those who disseminate such theories’. I guess a

psychiatrist’s chair is out of the question?

Sunstein’s mate Richard Thaler, an ‘academic affiliate’ of the UK

Behavioural Insights Team, is a proponent of ‘behavioural

economics’ which is defined as the study of ‘the effects of

psychological, cognitive, emotional, cultural and social factors on the

decisions of individuals and institutions’. Study the effects so they

can be manipulated to be what you want them to be. Other leading

names in the development of behavioural economics are ultra-

Zionists Daniel Kahneman and Robert J. Shiller and they, with

Thaler, won the Nobel Memorial Prize in Economic Sciences for their

work in this field. The Behavioural Insights Team is operating at the

heart of the UK government and has expanded globally through

partnerships with several universities including Harvard, Oxford,

Cambridge, University College London (UCL) and Pennsylvania.

They claim to have ‘trained’ (reframed) 20,000 civil servants and run

more than 750 projects involving 400 randomised controlled trials in

dozens of countries’ as another version of mind reframers Common

Purpose. BIT works from its office in New York with cities and their

agencies, as well as other partners, across the United States and

Canada – this is a company part-owned by the British government

Cabinet Office. An executive order by President Cult-servant Obama

established a US Social and Behavioral Sciences Team in 2015. They

all have the same reason for being and that’s to brainwash the

population directly and by brainwashing those in positions of

authority.

‘Covid’ mind game

Another prime aspect of the UK mind-control network is the

‘independent’ [joke] Scientific Pandemic Insights Group on

Behaviours (SPI-B) which ‘provides behavioural science advice

aimed at anticipating and helping people adhere to interventions

that are recommended by medical or epidemiological experts’. That

means manipulating public perception and behaviour to do

whatever government tells them to do. It’s disgusting and if they

really want the public to be ‘safe’ this lot should all be under lock

and key. According to the government website SPI-B consists of

‘behavioural scientists, health and social psychologists,

anthropologists and historians’ and advises the Whi�y-Vallance-led

Scientific Advisory Group for Emergencies (SAGE) which in turn

advises the government on ‘the science’ (it doesn’t) and ‘Covid’

policy. When politicians say they are being guided by ‘the science’

this is the rabble in each country they are talking about and that

‘science’ is dominated by behaviour manipulators to enforce

government fascism through public compliance. The Behaviour

Insight Team is headed by psychologist David Solomon Halpern, a

visiting professor at King’s College London, and connects with a

national and global web of other civilian and military organisations

as the Cult moves towards its goal of fusing them into one fascistic

whole in every country through its ‘Fusion Doctrine’. The behaviour

manipulation network involves, but is not confined to, the Foreign

Office; National Security Council; government communications

headquarters (GCHQ); MI5; MI6; the Cabinet Office-based Media

Monitoring Unit; and the Rapid Response Unit which ‘monitors

digital trends to spot emerging issues; including misinformation and

disinformation; and identifies the best way to respond’.

There is also the 77th Brigade of the UK military which operates

like the notorious Israeli military’s Unit 8200 in manipulating

information and discussion on the Internet by posing as members of

the public to promote the narrative and discredit those who

challenge it. Here we have the military seeking to manipulate

domestic public opinion while the Nazis in government are fine with

that. Conservative Member of Parliament Tobias Ellwood, an

advocate of lockdown and control through ‘vaccine passports’, is a

Lieutenant Colonel reservist in the 77th Brigade which connects with

the military operation jHub, the ‘innovation centre’ for the Ministry

of Defence and Strategic Command. jHub has also been involved

with the civilian National Health Service (NHS) in ‘symptom

tracing’ the population. The NHS is a key part of this mind control

network and produced a document in December, 2020, explaining to

staff how to use psychological manipulation with different groups

and ages to get them to have the DNA-manipulating ‘Covid vaccine’

that’s designed to cumulatively rewrite human genetics. The

document, called ‘Optimising Vaccination Roll Out – Do’s and Dont’s

for all messaging, documents and “communications” in the widest

sense’, was published by NHS England and the NHS Improvement

Behaviour Change Unit in partnership with Public Health England

and Warwick Business School. I hear the mantra about ‘save the

NHS’ and ‘protect the NHS’ when we need to scrap the NHS and

start again. The current version is far too corrupt, far too anti-human

and totally compromised by Cult operatives and their assets. UK

government broadcast media censor Ofcom will connect into this

web – as will the BBC with its tremendous Ofcom influence – to

control what the public see and hear and dictate mass perception.

Nuremberg trials must include personnel from all these

organisations.

The fear factor

The ‘Covid’ hoax has led to the creation of the UK Cabinet Office-

connected Joint Biosecurity Centre (JBC) which is officially described

as providing ‘expert advice on pandemics’ using its independent [all

Cult operations are ‘independent’] analytical function to provide

real-time analysis about infection outbreaks to identify and respond

to outbreaks of Covid-19’. Another role is to advise the government

on a response to spikes in infections – ‘for example by closing

schools or workplaces in local areas where infection levels have

risen’. Put another way, promoting the Cult agenda. The Joint

Biosecurity Centre is modelled on the Joint Terrorism Analysis

Centre which analyses intelligence to set ‘terrorism threat levels’ and

here again you see the fusion of civilian and military operations and

intelligence that has led to military intelligence producing

documents about ‘vaccine hesitancy’ and how it can be combated.

Domestic civilian ma�ers and opinions should not be the business of

the military. The Joint Biosecurity Centre is headed by Tom Hurd,

director general of the Office for Security and Counter-Terrorism

from the establishment-to-its-fingertips Hurd family. His father is

former Foreign Secretary Douglas Hurd. How coincidental that Tom

•

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Hurd went to the elite Eton College and Oxford University with

Boris Johnson. Imperial College with its ridiculous computer

modeller Neil Ferguson will connect with this gigantic web that will

itself interconnect with similar set-ups in other major and not so

major countries. Compared with this Cult network the politicians, be

they Boris Johnson, Donald Trump or Joe Biden, are bit-part players

‘following the science’. The network of psychologists was on the

‘Covid’ case from the start with the aim of generating maximum fear

of the ‘virus’ to ensure compliance by the population. A government

behavioural science group known as SPI-B produced a paper in

March, 2020, for discussion by the main government science

advisory group known as SAGE. It was headed ‘Options for

increasing adherence to social distancing measures’ and it said the

following in a section headed ‘Persuasion’:

A substantial number of people still do not feel sufficiently

personally threatened; it could be that they are reassured by the

low death rate in their demographic group, although levels of

concern may be rising. Having a good understanding of the risk

has been found to be positively associated with adoption of

COVID-19 social distancing measures in Hong Kong.

The perceived level of personal threat needs to be increased

among those who are complacent, using hard-hi�ing evaluation

of options for increasing social distancing emotional messaging.

To be effective this must also empower people by making clear

the actions they can take to reduce the threat.

Responsibility to others: There seems to be insufficient

understanding of, or feelings of responsibility about, people’s role

in transmi�ing the infection to others … Messaging about actions

need to be framed positively in terms of protecting oneself and

the community, and increase confidence that they will be effective.

Some people will be more persuaded by appeals to play by the

rules, some by duty to the community, and some to personal risk.

All these different approaches are needed. The messaging also

needs to take account of the realities of different people’s lives.

Messaging needs to take account of the different motivational

levers and circumstances of different people.

All this could be achieved the SPI-B psychologists said by using the

media to increase the sense of personal threat which translates as terrify

the shit out of the population, including children, so they all do what

we want. That’s not happened has it? Those excuses for ‘journalists’

who wouldn’t know journalism if it bit them on the arse (the great

majority) have played their crucial part in serving this Cult-

government Psyop to enslave their own kids and grandkids. How

they live with themselves I have no idea. The psychological war has

been underpinned by constant government ‘Covid’ propaganda in

almost every television and radio ad break, plus the Internet and

print media, which has pounded out the fear with taxpayers footing

the bill for their own programming. The result has been people

terrified of a ‘virus’ that doesn’t exist or one with a tiny fatality rate

even if you believe it does. People walk down the street and around

the shops wearing face-nappies damaging their health and

psychology while others report those who refuse to be that naïve to

the police who turn up in their own face-nappies. I had a cameraman

come to my flat and he was so frightened of ‘Covid’ he came in

wearing a mask and refused to shake my hand in case he caught

something. He had – naïveitis – and the thought that he worked in

the mainstream media was both depressing and made his behaviour

perfectly explainable. The fear which has gripped the minds of so

many and frozen them into compliance has been carefully cultivated

by these psychologists who are really psychopaths. If lives get

destroyed and a lot of young people commit suicide it shows our

plan is working. SPI-B then turned to compulsion on the public to

comply. ‘With adequate preparation, rapid change can be achieved’,

it said. Some countries had introduced mandatory self-isolation on a

wide scale without evidence of major public unrest and a large

majority of the UK’s population appeared to be supportive of more

coercive measures with 64 percent of adults saying they would

support pu�ing London under a lockdown (watch the ‘polls’ which

are designed to make people believe that public opinion is in favour

or against whatever the subject in hand).

For ‘aggressive protective measures’ to be effective, the SPI-B

paper said, special a�ention should be devoted to those population

groups that are more at risk. Translated from the Orwellian this

means making the rest of population feel guilty for not protecting

the ‘vulnerable’ such as old people which the Cult and its agencies

were about to kill on an industrial scale with lockdown, lack of

treatment and the Gates ‘vaccine’. Psychopath psychologists sold

their guilt-trip so comprehensively that Los Angeles County

Supervisor Hilda Solis reported that children were apologising (from

a distance) to their parents and grandparents for bringing ‘Covid’

into their homes and ge�ing them sick. ‘… These apologies are just

some of the last words that loved ones will ever hear as they die

alone,’ she said. Gut-wrenchingly Solis then used this childhood

tragedy to tell children to stay at home and ‘keep your loved ones

alive’. Imagine heaping such potentially life-long guilt on a kid when

it has absolutely nothing to do with them. These people are deeply

disturbed and the psychologists behind this even more so.

Uncivil war – divide and rule

Professional mind-controllers at SPI-B wanted the media to increase

a sense of responsibility to others (do as you’re told) and promote

‘positive messaging’ for those actions while in contrast to invoke

‘social disapproval’ by the unquestioning, obedient, community of

anyone with a mind of their own. Again the compliant Goebbels-like

media obliged. This is an old, old, trick employed by tyrannies the

world over throughout human history. You get the target population

to keep the target population in line – your line. SPI-B said this could

‘play an important role in preventing anti-social behaviour or

discouraging failure to enact pro-social behaviour’. For ‘anti-social’

in the Orwellian parlance of SPI-B see any behaviour that

government doesn’t approve. SPI-B recommendations said that

‘social disapproval’ should be accompanied by clear messaging and

promotion of strong collective identity – hence the government and

celebrity mantra of ‘we’re all in this together’. Sure we are. The mind

doctors have such contempt for their targets that they think some

clueless comedian, actor or singer telling them to do what the

government wants will be enough to win them over. We have had

UK comedian Lenny Henry, actor Michael Caine and singer Elton

John wheeled out to serve the propagandists by urging people to

have the DNA-manipulating ‘Covid’ non-’vaccine’. The role of

Henry and fellow black celebrities in seeking to coax a ‘vaccine’

reluctant black community into doing the government’s will was

especially stomach-turning. An emotion-manipulating script and

carefully edited video featuring these black ‘celebs’ was such an

insult to the intelligence of black people and where’s the self-respect

of those involved selling their souls to a fascist government agenda?

Henry said he heard black people’s ‘legitimate worries and

concerns’, but people must ‘trust the facts’ when they were doing

exactly that by not having the ‘vaccine’. They had to include the

obligatory reference to Black Lives Ma�er with the line … ‘Don’t let

coronavirus cost even more black lives – because we ma�er’. My

god, it was pathetic. ‘I know the vaccine is safe and what it does.’

How? ‘I’m a comedian and it says so in my script.’

SPI-B said social disapproval needed to be carefully managed to

avoid victimisation, scapegoating and misdirected criticism, but they

knew that their ‘recommendations’ would lead to exactly that and

the media were specifically used to stir-up the divide-and-conquer

hostility. Those who conform like good li�le baa, baas, are praised

while those who have seen through the tidal wave of lies are

‘Covidiots’. The awake have been abused by the fast asleep for not

conforming to fascism and impositions that the awake know are

designed to endanger their health, dehumanise them, and tear

asunder the very fabric of human society. We have had the curtain-

twitchers and morons reporting neighbours and others to the face-

nappied police for breaking ‘Covid rules’ with fascist police

delighting in posting links and phone numbers where this could be

done. The Cult cannot impose its will without a compliant police

and military or a compliant population willing to play their part in

enslaving themselves and their kids. The words of a pastor in Nazi

Germany are so appropriate today:

First they came for the socialists and I did not speak out because I was not a socialist.

Then they came for the trade unionists and I did not speak out because I was not a trade unionist.

Then they came for the Jews and I did not speak out because I was not a Jew.

Then they came for me and there was no one left to speak for me.

Those who don’t learn from history are destined to repeat it and so

many are.

‘Covid’ rules: Rewiring the mind

With the background laid out to this gigantic national and global

web of psychological manipulation we can put ‘Covid’ rules into a

clear and sinister perspective. Forget the claims about protecting

health. ‘Covid’ rules are about dismantling the human mind,

breaking the human spirit, destroying self-respect, and then pu�ing

Humpty Dumpty together again as a servile, submissive slave. Social

isolation through lockdown and distancing have devastating effects

on the human psyche as the psychological psychopaths well know

and that’s the real reason for them. Humans need contact with each

other, discourse, closeness and touch, or they eventually, and

literarily, go crazy. Masks, which I will address at some length,

fundamentally add to the effects of isolation and the Cult agenda to

dehumanise and de-individualise the population. To do this while

knowing – in fact seeking – this outcome is the very epitome of evil

and psychologists involved in this are the epitome of evil. They must

like all the rest of the Cult demons and their assets stand trial for

crimes against humanity on a scale that defies the imagination.

Psychopaths in uniform use isolation to break enemy troops and

agents and make them subservient and submissive to tell what they

know. The technique is rightly considered a form of torture and

torture is most certainly what has been imposed on the human

population.

Clinically-insane American psychologist Harry Harlow became

famous for his isolation experiments in the 1950s in which he

separated baby monkeys from their mothers and imprisoned them

for months on end in a metal container or ‘pit of despair’. They soon

began to show mental distress and depression as any idiot could

have predicted. Harlow put other monkeys in steel chambers for

three, six or twelve months while denying them any contact with

animals or humans. He said that the effects of total social isolation

for six months were ‘so devastating and debilitating that we had

assumed initially that twelve months of isolation would not produce

any additional decrement’; but twelve months of isolation ‘almost

obliterated the animals socially’. This is what the Cult and its

psychopaths are doing to you and your children. Even monkeys in

partial isolation in which they were not allowed to form

relationships with other monkeys became ‘aggressive and hostile,

not only to others, but also towards their own bodies’. We have seen

this in the young as a consequence of lockdown. UK government

psychopaths launched a public relations campaign telling people not

to hug each other even a�er they received the ‘Covid-19 vaccine’

which we were told with more lies would allow a return to ‘normal

life’. A government source told The Telegraph: ‘It will be along the

lines that it is great that you have been vaccinated, but if you are

going to visit your family and hug your grandchildren there is a

chance you are going to infect people you love.’ The source was

apparently speaking from a secure psychiatric facility. Janet Lord,

director of Birmingham University’s Institute of Inflammation and

Ageing, said that parents and grandparents should avoid hugging

their children. Well, how can I put it, Ms Lord? Fuck off. Yep, that’ll

do.

Destroying the kids – where are the parents?

Observe what has happened to people enslaved and isolated by

lockdown as suicide and self-harm has soared worldwide,

particularly among the young denied the freedom to associate with

their friends. A study of 49,000 people in English-speaking countries

concluded that almost half of young adults are at clinical risk of

mental health disorders. A national survey in America of 1,000

currently enrolled high school and college students found that 5

percent reported a�empting suicide during the pandemic. Data from

the US CDC’s National Syndromic Surveillance Program from

January 1st to October 17th, 2020, revealed a 31 percent increase in

mental health issues among adolescents aged 12 to 17 compared

with 2019. The CDC reported that America in general suffered the

biggest drop in life expectancy since World War Two as it fell by a

year in the first half of 2020 as a result of ‘deaths of despair’ –

overdoses and suicides. Deaths of despair have leapt by more than

20 percent during lockdown and include the highest number of fatal

overdoses ever recorded in a single year – 81,000. Internet addiction

is another consequence of being isolated at home which lowers

interest in physical activities as kids fall into inertia and what’s the

point? Children and young people are losing hope and giving up on

life, sometimes literally. A 14-year-old boy killed himself in

Maryland because he had ‘given up’ when his school district didn’t

reopen; an 11-year-old boy shot himself during a zoom class; a

teenager in Maine succumbed to the isolation of the ‘pandemic’

when he ended his life a�er experiencing a disrupted senior year at

school. Children as young as nine have taken their life and all these

stories can be repeated around the world. Careers are being

destroyed before they start and that includes those in sport in which

promising youngsters have not been able to take part. The plan of

the psycho-psychologists is working all right. Researchers at

Cambridge University found that lockdowns cause significant harm

to children’s mental health. Their study was published in the

Archives of Disease in Childhood, and followed 168 children aged

between 7 and 11. The researchers concluded:

During the UK lockdown, children’s depression symptoms have increased substantially, relative to before lockdown. The scale of this effect has direct relevance for the continuation of different elements of lockdown policy, such as complete or partial school closures …

… Specifically, we observed a statistically significant increase in ratings of depression, with a medium-to-large effect size. Our findings emphasise the need to incorporate the potential impact of lockdown on child mental health in planning the ongoing response to the global pandemic and the recovery from it.

Not a chance when the Cult’s psycho-psychologists were ge�ing

exactly what they wanted. The UK’s Royal College of Paediatrics and

Child Health has urged parents to look for signs of eating disorders

in children and young people a�er a three to four fold increase.

Specialists say the ‘pandemic’ is a major reason behind the rise. You

don’t say. The College said isolation from friends during school

closures, exam cancellations, loss of extra-curricular activities like

sport, and an increased use of social media were all contributory

factors along with fears about the virus (psycho-psychologists

again), family finances, and students being forced to quarantine.

Doctors said young people were becoming severely ill by the time

they were seen with ‘Covid’ regulations reducing face-to-face

consultations. Nor is it only the young that have been devastated by

the psychopaths. Like all bullies and cowards the Cult is targeting

the young, elderly, weak and infirm. A typical story was told by a

British lady called Lynn Parker who was not allowed to visit her

husband in 2020 for the last ten and half months of his life ‘when he

needed me most’ between March 20th and when he died on

December 19th. This vacates the criminal and enters the territory of

evil. The emotional impact on the immune system alone is immense

as are the number of people of all ages worldwide who have died as

a result of Cult-demanded, Gates-demanded, lockdowns.

Isolation is torture

The experience of imposing solitary confinement on millions of

prisoners around the world has shown how a large percentage

become ‘actively psychotic and/or acutely suicidal’. Social isolation

has been found to trigger ‘a specific psychiatric syndrome,

characterized by hallucinations; panic a�acks; overt paranoia;

diminished impulse control; hypersensitivity to external stimuli; and

difficulties with thinking, concentration and memory’. Juan Mendez,

a United Nations rapporteur (investigator), said that isolation is a

form of torture. Research has shown that even a�er isolation

prisoners find it far more difficult to make social connections and I

remember cha�ing to a shop assistant a�er one lockdown who told

me that when her young son met another child again he had no idea

how to act or what to do. Hannah Flanagan, Director of Emergency

Services at Journey Mental Health Center in Dane County,

Wisconsin, said: ‘The specificity about Covid social distancing and

isolation that we’ve come across as contributing factors to the

suicides are really new to us this year.’ But they are not new to those

that devised them. They are ge�ing the effect they want as the

population is psychologically dismantled to be rebuilt in a totally

different way. Children and the young are particularly targeted.

They will be the adults when the full-on fascist AI-controlled

technocracy is planned to be imposed and they are being prepared

to meekly submit. At the same time older people who still have a

memory of what life was like before – and how fascist the new

normal really is – are being deleted. You are going to see efforts to

turn the young against the old to support this geriatric genocide.

Hannah Flanagan said the big increase in suicide in her county

proved that social isolation is not only harmful, but deadly. Studies

have shown that isolation from others is one of the main risk factors

in suicide and even more so with women. Warnings that lockdown

could create a ‘perfect storm’ for suicide were ignored. A�er all this

was one of the reasons for lockdown. Suicide, however, is only the

most extreme of isolation consequences. There are many others. Dr

Dhruv Khullar, assistant professor of healthcare policy at Weill

Cornell Medical College, said in a New York Times article in 2016 long

before the fake ‘pandemic’:

A wave of new research suggests social separation is bad for us. Individuals with less social connection have disrupted sleep patterns, altered immune systems, more inflammation and higher levels of stress hormones. One recent study found that isolation increases the risk of heart disease by 29 percent and stroke by 32 percent. Another analysis that pooled data from 70 studies and 3.4 million people found that socially isolated individuals had a 30 percent higher risk of dying in the next seven years, and that this effect was largest in middle age.

Loneliness can accelerate cognitive decline in older adults, and isolated individuals are twice as likely to die prematurely as those with more robust social interactions. These effects start early: Socially isolated children have significantly poorer health 20 years later, even after controlling for other factors. All told, loneliness is as important a risk factor for early death as obesity and smoking.

There you have proof from that one article alone four years before

2020 that those who have enforced lockdown, social distancing and

isolation knew what the effect would be and that is even more so

with professional psychologists that have been driving the policy

across the globe. We can go back even further to the years 2000 and

2003 and the start of a major study on the effects of isolation on

health by Dr Janine Gronewold and Professor Dirk M. Hermann at

the University Hospital in Essen, Germany, who analysed data on

4,316 people with an average age of 59 who were recruited for the

long-term research project. They found that socially isolated people

are more than 40 percent more likely to have a heart a�ack, stroke,

or other major cardiovascular event and nearly 50 percent more

likely to die from any cause. Given the financial Armageddon

unleashed by lockdown we should note that the study found a

relationship between increased cardiovascular risk and lack of

financial support. A�er excluding other factors social isolation was

still connected to a 44 percent increased risk of cardiovascular

problems and a 47 percent increased risk of death by any cause. Lack

of financial support was associated with a 30 percent increase in the

risk of cardiovascular health events. Dr Gronewold said it had been

known for some time that feeling lonely or lacking contact with close

friends and family can have an impact on physical health and the

study had shown that having strong social relationships is of high

importance for heart health. Gronewold said they didn’t understand

yet why people who are socially isolated have such poor health

outcomes, but this was obviously a worrying finding, particularly

during these times of prolonged social distancing. Well, it can be

explained on many levels. You only have to identify the point in the

body where people feel loneliness and missing people they are

parted from – it’s in the centre of the chest where they feel the ache

of loneliness and the ache of missing people. ‘My heart aches for

you’ … ‘My heart aches for some company.’ I will explain this more

in the chapter Escaping Wetiko, but when you realise that the body

is the mind – they are expressions of each other – the reason why

state of the mind dictates state of the body becomes clear.

American psychologist Ranjit Powar was highlighting the effects

of lockdown isolation as early as April, 2020. She said humans have

evolved to be social creatures and are wired to live in interactive

groups. Being isolated from family, friends and colleagues could be

unbalancing and traumatic for most people and could result in short

or even long-term psychological and physical health problems. An

increase in levels of anxiety, aggression, depression, forgetfulness

and hallucinations were possible psychological effects of isolation.

‘Mental conditions may be precipitated for those with underlying

pre-existing susceptibilities and show up in many others without

any pre-condition.’ Powar said personal relationships helped us cope

with stress and if we lost this outlet for le�ing off steam the result

can be a big emotional void which, for an average person, was

difficult to deal with. ‘Just a few days of isolation can cause

increased levels of anxiety and depression’ – so what the hell has

been the effect on the global population of 18 months of this at the

time of writing? Powar said: ‘Add to it the looming threat of a

dreadful disease being repeatedly hammered in through the media

and you have a recipe for many shades of mental and physical

distress.’ For those with a house and a garden it is easy to forget that

billions have had to endure lockdown isolation in tiny overcrowded

flats and apartments with nowhere to go outside. The psychological

and physical consequences of this are unimaginable and with lunatic

and abusive partners and parents the consequences have led to

tremendous increases in domestic and child abuse and alcoholism as

people seek to shut out the horror. Ranjit Powar said:

Staying in a confined space with family is not all a rosy picture for everyone. It can be extremely oppressive and claustrophobic for large low-income families huddled together in small single-room houses. Children here are not lucky enough to have many board/electronic games or books to keep them occupied.

Add to it the deep insecurity of running out of funds for food and basic necessities. On the other hand, there are people with dysfunctional family dynamics, such as domineering, abusive or alcoholic partners, siblings or parents which makes staying home a period of trial. Incidence of suicide and physical abuse against women has shown a worldwide increase. Heightened anxiety and depression also affect a person’s immune system, making them more susceptible to illness.

To think that Powar’s article was published on April 11th, 2020.

Six-feet fantasy

Social (unsocial) distancing demanded that people stay six feet or

two metres apart. UK government advisor Robert Dingwall from the

New and Emerging Respiratory Virus Threats Advisory Group said

in a radio interview that the two-metre rule was ‘conjured up out of

nowhere’ and was not based on science. No, it was not based on

medical science, but it didn’t come out of nowhere. The distance

related to psychological science. Six feet/two metres was adopted in

many countries and we were told by people like the criminal

Anthony Fauci and his ilk that it was founded on science. Many

schools could not reopen because they did not have the space for six-

feet distancing. Then in March, 2021, a�er a year of six-feet ‘science’,

a study published in the Journal of Infectious Diseases involving more

than 500,000 students and almost 100,000 staff over 16 weeks

revealed no significant difference in ‘Covid’ cases between six feet

and three feet and Fauci changed his tune. Now three feet was okay.

There is no difference between six feet and three inches when there is

no ‘virus’ and they got away with six feet for psychological reasons

for as long as they could. I hear journalists and others talk about

‘unintended consequences’ of lockdown. They are not unintended at

all; they have been coldly-calculated for a specific outcome of human

control and that’s why super-psychopaths like Gates have called for

them so vehemently. Super-psychopath psychologists have

demanded them and psychopathic or clueless, spineless, politicians

have gone along with them by ‘following the science’. But it’s not

science at all. ‘Science’ is not what is; it’s only what people can be

manipulated to believe it is. The whole ‘Covid’ catastrophe is

founded on mind control. Three word or three statement mantras

issued by the UK government are a well-known mind control

technique and so we’ve had ‘Stay home/protect the NHS/save lives’,

‘Stay alert/control the virus/save lives’ and ‘hands/face/space’. One

of the most vocal proponents of extreme ‘Covid’ rules in the UK has

been Professor Susan Michie, a member of the British Communist

Party, who is not a medical professional. Michie is the director of the

Centre for Behaviour Change at University College London. She is a

behavioural psychologist and another filthy rich ‘Marxist’ who praised

China’s draconian lockdown. She was known by fellow students at

Oxford University as ‘Stalin’s nanny’ for her extreme Marxism.

Michie is an influential member of the UK government’s Scientific

Advisory Group for Emergencies (SAGE) and behavioural

manipulation groups which have dominated ‘Covid’ policy. She is a

consultant adviser to the World Health Organization on ‘Covid-19’

and behaviour. Why the hell are lockdowns anything to do with her

when they are claimed to be about health? Why does a behavioural

psychologist from a group charged with changing the behaviour of

the public want lockdown, human isolation and mandatory masks?

Does that question really need an answer? Michie absolutely has to

explain herself before a Nuremberg court when humanity takes back

its world again and even more so when you see the consequences of

masks that she demands are compulsory. This is a Michie classic:

The benefits of getting primary school children to wear masks is that regardless of what little degree of transmission is occurring in those age groups it could help normalise the practice. Young children wearing masks may be more likely to get their families to accept masks.

Those words alone should carry a prison sentence when you

ponder on the callous disregard for children involved and what a

statement it makes about the mind and motivations of Susan Michie.

What a lovely lady and what she said there encapsulates the

mentality of the psychopaths behind the ‘Covid’ horror. Let us

compare what Michie said with a countrywide study in Germany

published at researchsquare.com involving 25,000 school children

and 17,854 health complaints submi�ed by parents. Researchers

found that masks are harming children physically, psychologically,

and behaviourally with 24 health issues associated with mask

wearing. They include: shortness of breath (29.7%); dizziness

(26.4%); increased headaches (53%); difficulty concentrating (50%);

drowsiness or fatigue (37%); and malaise (42%). Nearly a third of

children experienced more sleep issues than before and a quarter

developed new fears. Researchers found health issues and other

impairments in 68 percent of masked children covering their faces

for an average of 4.5 hours a day. Hundreds of those taking part

experienced accelerated respiration, tightness in the chest, weakness,

and short-term impairment of consciousness. A reminder of what

Michie said again:

The benefits of getting primary school children to wear masks is that regardless of what little degree of transmission is occurring in those age groups it could help normalise the practice. Young children wearing masks may be more likely to get their families to accept masks.

Psychopaths in government and psychology now have children and

young people – plus all the adults – wearing masks for hours on end

while clueless teachers impose the will of the psychopaths on the

young they should be protecting. What the hell are parents doing?

Cult lab rats

We have some schools already imposing on students microchipped

buzzers that activate when they get ‘too close’ to their pals in the

way they do with lab rats. How apt. To the Cult and its brain-dead

servants our children are lab rats being conditioned to be

unquestioning, dehumanised slaves for the rest of their lives.

Children and young people are being weaned and frightened away

from the most natural human instincts including closeness and

touch. I have tracked in the books over the years how schools were

banning pupils from greeting each other with a hug and the whole

Cult-induced Me Too movement has terrified men and boys from a

relaxed and natural interaction with female friends and work

colleagues to the point where many men try never to be in a room

alone with a woman that’s not their partner. Airhead celebrities have

as always played their virtue-signalling part in making this happen

with their gross exaggeration. For every monster like Harvey

Weinstein there are at least tens of thousands of men that don’t treat

women like that; but everyone must be branded the same and policy

changed for them as well as the monster. I am going to be using the

word ‘dehumanise’ many times in this chapter because that is what

the Cult is seeking to do and it goes very deep as we shall see. Don’t

let them kid you that social distancing is planned to end one day.

That’s not the idea. We are seeing more governments and companies

funding and producing wearable gadgets to keep people apart and

they would not be doing that if this was meant to be short-term. A

tech start-up company backed by GCHQ, the British Intelligence and

military surveillance headquarters, has created a social distancing

wrist sensor that alerts people when they get too close to others. The

CIA has also supported tech companies developing similar devices.

The wearable sensor was developed by Tended, one of a number of

start-up companies supported by GCHQ (see the CIA and DARPA).

The device can be worn on the wrist or as a tag on the waistband and

will vibrate whenever someone wearing the device breaches social

distancing and gets anywhere near natural human contact. The

company had a lucky break in that it was developing a distancing

sensor when the ‘Covid’ hoax arrived which immediately provided a

potentially enormous market. How fortunate. The government in

big-time Cult-controlled Ontario in Canada is investing $2.5 million

in wearable contact tracing technology that ‘will alert users if they

may have been exposed to the Covid-19 in the workplace and will

beep or vibrate if they are within six feet of another person’.

Facedrive Inc., the technology company behind this, was founded in

2016 with funding from the Ontario Together Fund and obviously

they, too, had a prophet on the board of directors. The human

surveillance and control technology is called TraceSCAN and would

be worn by the human cyborgs in places such as airports,

workplaces, construction sites, care homes and … schools.

I emphasise schools with children and young people the prime

targets. You know what is planned for society as a whole if you keep

your eyes on the schools. They have always been places where the

state program the next generation of slaves to be its compliant

worker-ants – or Woker-ants these days; but in the mist of the

‘Covid’ madness they have been transformed into mind laboratories

on a scale never seen before. Teachers and head teachers are just as

programmed as the kids – o�en more so. Children are kept apart

from human interaction by walk lanes, classroom distancing,

staggered meal times, masks, and the rolling-out of buzzer systems.

Schools are now physically laid out as a laboratory maze for lab-rats.

Lunatics at a school in Anchorage, Alaska, who should be

prosecuted for child abuse, took away desks and forced children to

kneel (know your place) on a mat for five hours a day while wearing

a mask and using their chairs as a desk. How this was supposed to

impact on a ‘virus’ only these clinically insane people can tell you

and even then it would be clap-trap. The school banned recess

(interaction), art classes (creativity), and physical exercise (ge�ing

body and mind moving out of inertia). Everyone behind this outrage

should be in jail or be�er still a mental institution. The behavioural

manipulators are all for this dystopian approach to schools.

Professor Susan Michie, the mind-doctor and British Communist

Party member, said it was wrong to say that schools were safe. They

had to be made so by ‘distancing’, masks and ventilation (si�ing all

day in the cold). I must ask this lady round for dinner on a night I

know I am going to be out and not back for weeks. She probably

wouldn’t be able to make it, anyway, with all the visits to her own

psychologist she must have block-booked.

Masking identity

I know how shocking it must be for you that a behaviour

manipulator like Michie wants everyone to wear masks which have

long been a feature of mind-control programs like the infamous

MKUltra in the United States, but, there we are. We live and learn. I

spent many years from 1996 to right across the millennium

researching mind control in detail on both sides of the Atlantic and

elsewhere. I met a large number of mind-control survivors and

many had been held captive in body and mind by MKUltra. MK

stands for mind-control, but employs the German spelling in

deference to the Nazis spirited out of Germany at the end of World

War Two by Operation Paperclip in which the US authorities, with

help from the Vatican, transported Nazi mind-controllers and

engineers to America to continue their work. Many of them were

behind the creation of NASA and they included Nazi scientist and

SS officer Wernher von Braun who swapped designing V-2 rockets to

bombard London with designing the Saturn V rockets that powered

the NASA moon programme’s Apollo cra�. I think I may have

mentioned that the Cult has no borders. Among Paperclip escapees

was Josef Mengele, the Angel of Death in the Nazi concentration

camps where he conducted mind and genetic experiments on

children o�en using twins to provide a control twin to measure the

impact of his ‘work’ on the other. If you want to observe the Cult

mentality in all its extremes of evil then look into the life of Mengele.

I have met many people who suffered mercilessly under Mengele in

the United States where he operated under the name Dr Greene and

became a stalwart of MKUltra programming and torture. Among his

locations was the underground facility in the Mojave Desert in

California called the China Lake Naval Weapons Station which is

almost entirely below the surface. My books The Biggest Secret,

Children of the Matrix and The Perception Deception have the detailed

background to MKUltra.

The best-known MKUltra survivor is American Cathy O’Brien. I

first met her and her late partner Mark Phillips at a conference in

Colorado in 1996. Mark helped her escape and deprogram from

decades of captivity in an offshoot of MKUltra known as Project

Monarch in which ‘sex slaves’ were provided for the rich and

famous including Father George Bush, Dick Cheney and the

Clintons. Read Cathy and Mark’s book Trance-Formation of America

and if you are new to this you will be shocked to the core. I read it in

1996 shortly before, with the usual synchronicity of my life, I found

myself given a book table at the conference right next to hers.

MKUltra never ended despite being very publicly exposed (only a

small part of it) in the 1970s and continues in other guises. I am still

in touch with Cathy. She contacted me during 2020 a�er masks

became compulsory in many countries to tell me how they were

used as part of MKUltra programming. I had been observing ‘Covid

regulations’ and the relationship between authority and public for

months. I saw techniques that I knew were employed on individuals

in MKUltra being used on the global population. I had read many

books and manuals on mind control including one called Silent

Weapons for Quiet Wars which came to light in the 1980s and was a

guide on how to perceptually program on a mass scale. ‘Silent

Weapons’ refers to mind-control. I remembered a line from the

manual as governments, medical authorities and law enforcement

agencies have so obviously talked to – or rather at – the adult

population since the ‘Covid’ hoax began as if they are children. The

document said:

If a person is spoken to by a T.V. advertiser as if he were a twelve-year-old, then, due to suggestibility, he will, with a certain probability, respond or react to that suggestion with the uncritical response of a twelve-year-old and will reach in to his economic reservoir and deliver its energy to buy that product on impulse when he passes it in the store.

That’s why authority has spoken to adults like children since all this

began.

Why did Michael Jackson wear masks?

Every aspect of the ‘Covid’ narrative has mind-control as its central

theme. Cathy O’Brien wrote an article for davidicke.com about the

connection between masks and mind control. Her daughter Kelly

who I first met in the 1990s was born while Cathy was still held

captive in MKUltra. Kelly was forced to wear a mask as part of her

programming from the age of two to dehumanise her, target her

sense of individuality and reduce the amount of oxygen her brain

and body received. Bingo. This is the real reason for compulsory

masks, why they have been enforced en masse, and why they seek to

increase the number they demand you wear. First one, then two,

with one disgraceful alleged ‘doctor’ recommending four which is

nothing less than a death sentence. Where and how o�en they must

be worn is being expanded for the purpose of mass mind control

and damaging respiratory health which they can call ‘Covid-19’.

Canada’s government headed by the man-child Justin Trudeau, says

it’s fine for children of two and older to wear masks. An insane

‘study’ in Italy involving just 47 children concluded there was no

problem for babies as young as four months wearing them. Even a�er

people were ‘vaccinated’ they were still told to wear masks by the

criminal that is Anthony Fauci. Cathy wrote that mandating masks

is allowing the authorities literally to control the air we breathe

which is what was done in MKUltra. You might recall how the

singer Michael Jackson wore masks and there is a reason for that. He

was subjected to MKUltra mind control through Project Monarch

and his psyche was scrambled by these simpletons. Cathy wrote:

In MKUltra Project Monarch mind control, Michael Jackson had to wear a mask to silence his voice so he could not reach out for help. Remember how he developed that whisper voice when he wasn’t singing? Masks control the mind from the outside in, like the redefining of words is doing. By controlling what we can and cannot say for fear of being labeled racist or beaten, for example, it ultimately controls thought that drives our words and ultimately actions (or lack thereof).

Likewise, a mask muffles our speech so that we are not heard, which controls voice … words … mind. This is Mind Control. Masks are an obvious mind control device, and I am disturbed so many people are complying on a global scale. Masks depersonalize while making a person feel as though they have no voice. It is a barrier to others. People who would never choose to comply but are forced to wear a mask in order to keep their job, and ultimately their family fed, are compromised. They often feel shame and are subdued. People have stopped talking with each other while media controls the narrative.

The ‘no voice’ theme has o�en become literal with train

passengers told not to speak to each other in case they pass on the

‘virus’, singing banned for the same reason and bonkers California

officials telling people riding roller coasters that they cannot shout

and scream. Cathy said she heard every day from healed MKUltra

survivors who cannot wear a mask without flashing back on ways

their breathing was controlled – ‘from ball gags and penises to water

boarding’. She said that through the years when she saw images of

people in China wearing masks ‘due to pollution’ that it was really

to control their oxygen levels. ‘I knew it was as much of a population

control mechanism of depersonalisation as are burkas’, she said.

Masks are another Chinese communist/fascist method of control that

has been swept across the West as the West becomes China at

lightning speed since we entered 2020.

Mask-19

There are other reasons for mandatory masks and these include

destroying respiratory health to call it ‘Covid-19’ and stunting brain

development of children and the young. Dr Margarite Griesz-

Brisson MD, PhD, is a Consultant Neurologist and

Neurophysiologist and the Founder and Medical Director of the

London Neurology and Pain Clinic. Her CV goes down the street

and round the corner. She is clearly someone who cares about people

and won’t parrot the propaganda. Griesz-Brisson has a PhD in

pharmacology, with special interest in neurotoxicology,

environmental medicine, neuroregeneration and neuroplasticity (the

way the brain can change in the light of information received). She

went public in October, 2020, with a passionate warning about the

effects of mask-wearing laws:

The reinhalation of our exhaled air will without a doubt create oxygen deficiency and a flooding of carbon dioxide. We know that the human brain is very sensitive to oxygen deprivation. There are nerve cells for example in the hippocampus that can’t be longer than 3 minutes without oxygen – they cannot survive. The acute warning symptoms are headaches, drowsiness, dizziness, issues in concentration, slowing down of reaction time – reactions of the cognitive system.

Oh, I know, let’s tell bus, truck and taxi drivers to wear them and

people working machinery. How about pilots, doctors and police?

Griesz-Brisson makes the important point that while the symptoms

she mentions may fade as the body readjusts this does not alter the

fact that people continue to operate in oxygen deficit with long list of

potential consequences. She said it was well known that

neurodegenerative diseases take years or decades to develop. ‘If

today you forget your phone number, the breakdown in your brain

would have already started 20 or 30 years ago.’ She said

degenerative processes in your brain are ge�ing amplified as your

oxygen deprivation continues through wearing a mask. Nerve cells

in the brain are unable to divide themselves normally in these

circumstances and lost nerve cells will no longer be regenerated.

‘What is gone is gone.’ Now consider that people like shop workers

and schoolchildren are wearing masks for hours every day. What in

the name of sanity is going to be happening to them? ‘I do not wear

a mask, I need my brain to think’, Griesz-Brisson said, ‘I want to

have a clear head when I deal with my patients and not be in a

carbon dioxide-induced anaesthesia’. If you are told to wear a mask

anywhere ask the organisation, police, store, whatever, for their risk

assessment on the dangers and negative effects on mind and body of

enforcing mask-wearing. They won’t have one because it has never

been done not even by government. All of them must be subject to

class-action lawsuits as the consequences come to light. They don’t

do mask risk assessments for an obvious reason. They know what

the conclusions would be and independent scientific studies that

have been done tell a horror story of consequences.

‘Masks are criminal’

Dr Griesz-Brisson said that for children and adolescents, masks are

an absolute no-no. They had an extremely active and adaptive

immune system and their brain was incredibly active with so much

to learn. ‘The child’s brain, or the youth’s brain, is thirsting for

oxygen.’ The more metabolically active an organ was, the more

oxygen it required; and in children and adolescents every organ was

metabolically active. Griesz-Brisson said that to deprive a child’s or

adolescent’s brain of oxygen, or to restrict it in any way, was not only

dangerous to their health, it was absolutely criminal. ‘Oxygen

deficiency inhibits the development of the brain, and the damage

that has taken place as a result CANNOT be reversed.’ Mind

manipulators of MKUltra put masks on two-year-olds they wanted

to neurologically rewire and you can see why. Griesz-Brisson said a

child needs the brain to learn and the brain needs oxygen to

function. ‘We don’t need a clinical study for that. This is simple,

indisputable physiology.’ Consciously and purposely induced

oxygen deficiency was an absolutely deliberate health hazard, and

an absolute medical contraindication which means that ‘this drug,

this therapy, this method or measure should not be used, and is not

allowed to be used’. To coerce an entire population to use an

absolute medical contraindication by force, she said, there had to be

definite and serious reasons and the reasons must be presented to

competent interdisciplinary and independent bodies to be verified

and authorised. She had this warning of the consequences that were

coming if mask wearing continued:

When, in ten years, dementia is going to increase exponentially, and the younger generations couldn’t reach their god-given potential, it won’t help to say ‘we didn’t need the masks’. I know how damaging oxygen deprivation is for the brain, cardiologists know how damaging it is for the heart, pulmonologists know how damaging it is for the lungs. Oxygen deprivation damages every single organ. Where are our health departments, our health insurance, our medical associations? It would have been their duty to be vehemently against the lockdown and to stop it and stop it from the very beginning.

Why do the medical boards issue punishments to doctors who give people exemptions? Does the person or the doctor seriously have to prove that oxygen deprivation harms people? What kind of medicine are our doctors and medical associations representing? Who is responsible for this crime? The ones who want to enforce it? The ones who let it happen and play along, or the ones who don’t prevent it?

All of the organisations and people she mentions there either

answer directly to the Cult or do whatever hierarchical levels above

them tell them to do. The outcome of both is the same. ‘It’s not about

masks, it’s not about viruses, it’s certainly not about your health’,

Griesz-Brisson said. ‘It is about much, much more. I am not

participating. I am not afraid.’ They were taking our air to breathe

and there was no unfounded medical exemption from face masks.

Oxygen deprivation was dangerous for every single brain. It had to

be the free decision of every human being whether they want to

wear a mask that was absolutely ineffective to protect themselves

from a virus. She ended by rightly identifying where the

responsibility lies for all this:

The imperative of the hour is personal responsibility. We are responsible for what we think, not the media. We are responsible for what we do, not our superiors. We are responsible for our health, not the World Health Organization. And we are responsible for what happens in our country, not the government.

Halle-bloody-lujah.

But surgeons wear masks, right?

Independent studies of mask-wearing have produced a long list of

reports detailing mental, emotional and physical dangers. What a

definition of insanity to see police officers imposing mask-wearing

on the public which will cumulatively damage their health while the

police themselves wear masks that will cumulatively damage their

health. It’s u�er madness and both public and police do this because

‘the government says so’ – yes a government of brain-donor idiots

like UK Health Secretary Ma� Hancock reading the ‘follow the

science’ scripts of psychopathic, lunatic psychologists. The response

you get from Stockholm syndrome sufferers defending the very

authorities that are destroying them and their families is that

‘surgeons wear masks’. This is considered the game, set and match

that they must work and don’t cause oxygen deficit. Well, actually,

scientific studies have shown that they do and oxygen levels are

monitored in operating theatres to compensate. Surgeons wear

masks to stop spi�le and such like dropping into open wounds – not

to stop ‘viral particles’ which are so miniscule they can only be seen

through an electron microscope. Holes in the masks are significantly

bigger than ‘viral particles’ and if you sneeze or cough they will

breach the mask. I watched an incredibly disingenuous ‘experiment’

that claimed to prove that masks work in catching ‘virus’ material

from the mouth and nose. They did this with a slow motion camera

and the mask did block big stuff which stayed inside the mask and

•

•

•

against the face to be breathed in or cause infections on the face as

we have seen with many children. ‘Viral particles’, however, would

never have been picked up by the camera as they came through the

mask when they are far too small to be seen. The ‘experiment’ was

therefore disingenuous and useless.

Studies have concluded that wearing masks in operating theatres

(and thus elsewhere) make no difference to preventing infection

while the opposite is true with toxic shite building up in the mask

and this had led to an explosion in tooth decay and gum disease

dubbed by dentists ‘mask mouth’. You might have seen the Internet

video of a furious American doctor urging people to take off their

masks a�er a four-year-old patient had been rushed to hospital the

night before and nearly died with a lung infection that doctors

sourced to mask wearing. A study in the journal Cancer Discovery

found that inhalation of harmful microbes can contribute to

advanced stage lung cancer in adults and long-term use of masks

can help breed dangerous pathogens. Microbiologists have said

frequent mask wearing creates a moist environment in which

microbes can grow and proliferate before entering the lungs. The

Canadian Agency for Drugs and Technologies in Health, or CADTH,

a Canadian national organisation that provides research and

analysis to healthcare decision-makers, said this as long ago as 2013

in a report entitled ‘Use of Surgical Masks in the Operating Room: A

Review of the Clinical Effectiveness and Guidelines’. It said:

No evidence was found to support the use of surgical face masks

to reduce the frequency of surgical site infections

No evidence was found on the effectiveness of wearing surgical

face masks to protect staff from infectious material in the

operating room.

Guidelines recommend the use of surgical face masks by staff in

the operating room to protect both operating room staff and

patients (despite the lack of evidence).

We were told that the world could go back to ‘normal’ with the

arrival of the ‘vaccines’. When they came, fraudulent as they are, the

story changed as I knew that it would. We are in the midst of

transforming ‘normal’, not going back to it. Mary Ramsay, head of

immunisation at Public Health England, echoed the words of US

criminal Anthony Fauci who said masks and other regulations must

stay no ma�er if people are vaccinated. The Fauci idiot continued to

wear two masks – different colours so both could be clearly seen –

a�er he claimed to have been vaccinated. Senator Rand Paul told

Fauci in one exchange that his double-masks were ‘theatre’ and he

was right. It’s all theatre. Mary Ramsay back-tracked on the vaccine-

return-to-normal theme when she said the public may need to wear

masks and social-distance for years despite the jabs. ‘People have got

used to those lower-level restrictions now, and [they] can live with

them’, she said telling us what the idea has been all along. ‘The

vaccine does not give you a pass, even if you have had it, you must

continue to follow all the guidelines’ said a Public Health England

statement which reneged on what we had been told before and

made having the ‘vaccine’ irrelevant to ‘normality’ even by the

official story. Spain’s fascist government trumped everyone by

passing a law mandating the wearing of masks on the beach and

even when swimming in the sea. The move would have devastated

what’s le� of the Spanish tourist industry, posed potential breathing

dangers to swimmers and had Northern European sunbathers

walking around with their forehead brown and the rest of their face

white as a sheet. The ruling was so crazy that it had to be retracted

a�er pressure from public and tourist industry, but it confirmed

where the Cult wants to go with masks and how clinically insane

authority has become. The determination to make masks permanent

and hide the serious dangers to body and mind can be seen in the

censorship of scientist Professor Denis Rancourt by Bill Gates-

funded academic publishing website ResearchGate over his papers

exposing the dangers and uselessness of masks. Rancourt said:

ResearchGate today has permanently locked my account, which I have had since 2015. Their reasons graphically show the nature of their attack against democracy, and their corruption of

science … By their obscene non-logic, a scientific review of science articles reporting on harms caused by face masks has a ‘potential to cause harm’. No criticism of the psychological device (face masks) is tolerated, if the said criticism shows potential to influence public policy.

This is what happens in a fascist world.

Where are the ‘greens’ (again)?

Other dangers of wearing masks especially regularly relate to the

inhalation of minute plastic fibres into the lungs and the deluge of

discarded masks in the environment and oceans. Estimates

predicted that more than 1.5 billion disposable masks will end up in

the world’s oceans every year polluting the water with tons of plastic

and endangering marine wildlife. Studies project that humans are

using 129 billion face masks each month worldwide – about three

million a minute. Most are disposable and made from plastic, non-

biodegradable microfibers that break down into smaller plastic

particles that become widespread in ecosystems. They are li�ering

cities, clogging sewage channels and turning up in bodies of water. I

have wri�en in other books about the immense amounts of

microplastics from endless sources now being absorbed into the

body. Rolf Halden, director of the Arizona State University (ASU)

Biodesign Center for Environmental Health Engineering, was the

senior researcher in a 2020 study that analysed 47 human tissue

samples and found microplastics in all of them. ‘We have detected

these chemicals of plastics in every single organ that we have

investigated’, he said. I wrote in The Answer about the world being

deluged with microplastics. A study by the Worldwide Fund for

Nature (WWF) found that people are consuming on average every

week some 2,000 tiny pieces of plastic mostly through water and also

through marine life and the air. Every year humans are ingesting

enough microplastics to fill a heaped dinner plate and in a life-time

of 79 years it is enough to fill two large waste bins. Marco

Lambertini, WWF International director general said: ‘Not only are

plastics polluting our oceans and waterways and killing marine life –

it’s in all of us and we can’t escape consuming plastics,’ American

geologists found tiny plastic fibres, beads and shards in rainwater

samples collected from the remote slopes of the Rocky Mountain

National Park near Denver, Colorado. Their report was headed: ‘It is

raining plastic.’ Rachel Adams, senior lecturer in Biomedical Science

at Cardiff Metropolitan University, said that among health

consequences are internal inflammation and immune responses to a

‘foreign body’. She further pointed out that microplastics become

carriers of toxins including mercury, pesticides and dioxins (a

known cause of cancer and reproductive and developmental

problems). These toxins accumulate in the fa�y tissues once they

enter the body through microplastics. Now this is being

compounded massively by people pu�ing plastic on their face and

throwing it away.

Workers exposed to polypropylene plastic fibres known as ‘flock’

have developed ‘flock worker’s lung’ from inhaling small pieces of

the flock fibres which can damage lung tissue, reduce breathing

capacity and exacerbate other respiratory problems. Now …

commonly used surgical masks have three layers of melt-blown

textiles made of … polypropylene. We have billions of people

pu�ing these microplastics against their mouth, nose and face for

hours at a time day a�er day in the form of masks. How does

anyone think that will work out? I mean – what could possibly go

wrong? We posted a number of scientific studies on this at

davidicke.com, but when I went back to them as I was writing this

book the links to the science research website where they were

hosted were dead. Anything that challenges the official narrative in

any way is either censored or vilified. The official narrative is so

unsupportable by the evidence that only deleting the truth can

protect it. A study by Chinese scientists still survived – with the

usual twist which it why it was still active, I guess. Yes, they found

that virtually all the masks they tested increased the daily intake of

microplastic fibres, but people should still wear them because the

danger from the ‘virus’ was worse said the crazy ‘team’ from the

Institute of Hydrobiology in Wuhan. Scientists first discovered

microplastics in lung tissue of some patients who died of lung cancer

in the 1990s. Subsequent studies have confirmed the potential health

damage with the plastic degrading slowly and remaining in the

lungs to accumulate in volume. Wuhan researchers used a machine

simulating human breathing to establish that masks shed up to

nearly 4,000 microplastic fibres in a month with reused masks

producing more. Scientists said some masks are laced with toxic

chemicals and a variety of compounds seriously restricted for both

health and environmental reasons. They include cobalt (used in blue

dye) and formaldehyde known to cause watery eyes, burning

sensations in the eyes, nose, and throat, plus coughing, wheezing

and nausea. No – that must be ‘Covid-19’.

Mask ‘worms’

There is another and potentially even more sinister content of masks.

Mostly new masks of different makes filmed under a microscope

around the world have been found to contain strange black fibres or

‘worms’ that appear to move or ‘crawl’ by themselves and react to

heat and water. The nearest I have seen to them are the self-

replicating fibres that are pulled out through the skin of those

suffering from Morgellons disease which has been connected to the

phenomena of ‘chemtrails’ which I will bring into the story later on.

Morgellons fibres continue to grow outside the body and have a

form of artificial intelligence. Black ‘worm’ fibres in masks have that

kind of feel to them and there is a nanotechnology technique called

‘worm micelles’ which carry and release drugs or anything else you

want to deliver to the body. For sure the suppression of humanity by

mind altering drugs is the Cult agenda big time and the more

excuses they can find to gain access to the body the more

opportunities there are to make that happen whether through

‘vaccines’ or masks pushed against the mouth and nose for hours on

end.

So let us summarise the pros and cons of masks:

Against masks: Breathing in your own carbon dioxide; depriving the

body and brain of sufficient oxygen; build-up of toxins in the mask

that can be breathed into the lungs and cause rashes on the face and

‘mask-mouth’; breathing microplastic fibres and toxic chemicals into

the lungs; dehumanisation and deleting individualisation by literally

making people faceless; destroying human emotional interaction

through facial expression and deleting parental connection with

their babies which look for guidance to their facial expression.

For masks: They don’t protect you from a ‘virus’ that doesn’t exist

and even if it did ‘viral’ particles are so minute they are smaller than

the holes in the mask.

Governments, police, supermarkets, businesses, transport

companies, and all the rest who seek to impose masks have done no

risk assessment on their consequences for health and psychology

and are now open to group lawsuits when the impact becomes clear

with a cumulative epidemic of respiratory and other disease.

Authorities will try to exploit these effects and hide the real cause by

dubbing them ‘Covid-19’. Can you imagine se�ing out to force the

population to wear health-destroying masks without doing any

assessment of the risks? It is criminal and it is evil, but then how

many people targeted in this way, who see their children told to

wear them all day at school, have asked for a risk assessment?

Billions can’t be imposed upon by the few unless the billions allow it.

Oh, yes, with just a tinge of irony, 85 percent of all masks made

worldwide come from China.

Wash your hands in toxic shite

‘Covid’ rules include the use of toxic sanitisers and again the health

consequences of constantly applying toxins to be absorbed through

the skin is obvious to any level of Renegade Mind. America’s Food

and Drug Administration (FDA) said that sanitisers are drugs and

issued a warning about 75 dangerous brands which contain

methanol used in antifreeze and can cause death, kidney damage

and blindness. The FDA circulated the following warning even for

those brands that it claims to be safe:

Store hand sanitizer out of the reach of pets and children, and children should use it only with adult supervision. Do not drink hand sanitizer. This is particularly important for young children, especially toddlers, who may be attracted by the pleasant smell or brightly colored bottles of hand sanitizer.

Drinking even a small amount of hand sanitizer can cause alcohol poisoning in children. (However, there is no need to be concerned if your children eat with or lick their hands after using hand sanitizer.) During this coronavirus pandemic, poison control centers have had an increase in calls about accidental ingestion of hand sanitizer, so it is important that adults monitor young children’s use.

Do not allow pets to swallow hand sanitizer. If you think your pet has eaten something potentially dangerous, call your veterinarian or a pet poison control center right away. Hand sanitizer is flammable and should be stored away from heat and flames. When using hand sanitizer, rub your hands until they feel completely dry before performing activities that may involve heat, sparks, static electricity, or open flames.

There you go, perfectly safe, then, and that’s without even a mention

of the toxins absorbed through the skin. Come on kids – sanitise

your hands everywhere you go. It will save you from the ‘virus’. Put

all these elements together of the ‘Covid’ normal and see how much

health and psychology is being cumulatively damaged, even

devastated, to ‘protect your health’. Makes sense, right? They are

only imposing these things because they care, right? Right?

Submitting to insanity

Psychological reframing of the population goes very deep and is

done in many less obvious ways. I hear people say how

contradictory and crazy ‘Covid’ rules are and how they are ever

changing. This is explained away by dismissing those involved as

idiots. It is a big mistake. The Cult is delighted if its cold calculation

is perceived as incompetence and idiocy when it is anything but. Oh,

yes, there are idiots within the system – lots of them – but they are

administering the Cult agenda, mostly unknowingly. They are not

deciding and dictating it. The bulwark against tyranny is self-

respect, always has been, always will be. It is self-respect that has

broken every tyranny in history. By its very nature self-respect will

not bow to oppression and its perpetrators. There is so li�le self-

respect that it’s always the few that overturn dictators. Many may

eventually follow, but the few with the iron spines (self-respect) kick

it off and generate the momentum. The Cult targets self-respect in

the knowledge that once this has gone only submission remains.

Crazy, contradictory, ever-changing ‘Covid’ rules are systematically

applied by psychologists to delete self-respect. They want you to see

that the rules make no sense. It is one thing to decide to do

something when you have made the choice based on evidence and

logic. You still retain your self-respect. It is quite another when you

can see what you are being told to do is insane, ridiculous and

makes no sense, and yet you still do it. Your self-respect is

extinguished and this has been happening as ever more obviously

stupid and nonsensical things have been demanded and the great

majority have complied even when they can see they are stupid and

nonsensical.

People walk around in face-nappies knowing they are damaging

their health and make no difference to a ‘virus’. They do it in fear of

not doing it. I know it’s da�, but I’ll do it anyway. When that

happens something dies inside of you and submissive reframing has

begun. Next there’s a need to hide from yourself that you have

conceded your self-respect and you convince yourself that you have

not really submi�ed to fear and intimidation. You begin to believe

that you are complying with craziness because it’s the right thing to

do. When first you concede your self-respect of 2+2 = 4 to 2+2 = 5 you

know you are compromising your self-respect. Gradually to avoid

facing that fact you begin to believe that 2+2=5. You have been

reframed and I have been watching this process happening in the

human psyche on an industrial scale. The Cult is working to break

your spirit and one of its major tools in that war is humiliation. I

read how former American soldier Bradley Manning (later Chelsea

Manning a�er a sex-change) was treated a�er being jailed for

supplying WikiLeaks with documents exposing the enormity of

government and elite mendacity. Manning was isolated in solitary

confinement for eight months, put under 24-hour surveillance,

forced to hand over clothing before going to bed, and stand naked

for every roll call. This is systematic humiliation. The introduction of

anal swab ‘Covid’ tests in China has been done for the same reason

to delete self-respect and induce compliant submission. Anal swabs

are mandatory for incoming passengers in parts of China and

American diplomats have said they were forced to undergo the

indignity which would have been calculated humiliation by the

Cult-owned Chinese government that has America in its sights.

Government-people: An abusive relationship

Spirit-breaking psychological techniques include giving people hope

and apparent respite from tyranny only to take it away again. This

happened in the UK during Christmas, 2020, when the psycho-

psychologists and their political lackeys announced an easing of

restrictions over the holiday only to reimpose them almost

immediately on the basis of yet another lie. There is a big

psychological difference between ge�ing used to oppression and

being given hope of relief only to have that dashed. Psychologists

know this and we have seen the technique used repeatedly. Then

there is traumatising people before you introduce more extreme

regulations that require compliance. A perfect case was the

announcement by the dark and sinister Whi�y and Vallance in the

UK that ‘new data’ predicted that 4,000 could die every day over the

winter of 2020/2021 if we did not lockdown again. I think they call it

lying and a�er traumatising people with that claim out came

Jackboot Johnson the next day with new curbs on human freedom.

Psychologists know that a frightened and traumatised mind

becomes suggestable to submission and behaviour reframing.

Underpinning all this has been to make people fearful and

suspicious of each other and see themselves as a potential danger to

others. In league with deleted self-respect you have the perfect

psychological recipe for self-loathing. The relationship between

authority and public is now demonstrably the same as that of

subservience to an abusive partner. These are signs of an abusive

relationship explained by psychologist Leslie Becker-Phelps:

Psychological and emotional abuse: Undermining a partner’s self-worth with verbal a�acks, name-calling, and beli�ling.

Humiliating the partner in public, unjustly accusing them of having

an affair, or interrogating them about their every behavior. Keeping

partner confused or off balance by saying they were just kidding or

blaming the partner for ‘making’ them act this way … Feigning in

public that they care while turning against them in private. This

leads to victims frequently feeling confused, incompetent, unworthy,

hopeless, and chronically self-doubting. [Apply these techniques to

how governments have treated the population since New Year, 2020,

and the parallels are obvious.]

Physical abuse: The abuser might physically harm their partner in a range of ways, such as grabbing, hi�ing, punching, or shoving

them. They might throw objects at them or harm them with a

weapon. [Observe the physical harm imposed by masks, lockdown,

and so on.]

Threats and intimidation: One way abusers keep their partners in line is by instilling fear. They might be verbally threatening, or give

threatening looks or gestures. Abusers o�en make it known that

they are tracking their partner’s every move. They might destroy

their partner’s possessions, threaten to harm them, or threaten to

harm their family members. Not surprisingly, victims of this abuse

o�en feel anxiety, fear, and panic. [No words necessary.]

Isolation: Abusers o�en limit their partner’s activities, forbidding them to talk or interact with friends or family. They might limit

access to a car or even turn off their phone. All of this might be done

by physically holding them against their will, but is o�en

accomplished through psychological abuse and intimidation. The

more isolated a person feels, the fewer resources they have to help

gain perspective on their situation and to escape from it. [No words

necessary.]

Economic abuse: Abusers o�en make their partners beholden to them for money by controlling access to funds of any kind. They

might prevent their partner from ge�ing a job or withhold access to

money they earn from a job. This creates financial dependency that

makes leaving the relationship very difficult. [See destruction of

livelihoods and the proposed meagre ‘guaranteed income’ so long as

you do whatever you are told.]

Using children: An abuser might disparage their partner’s parenting skills, tell their children lies about their partner, threaten

to take custody of their children, or threaten to harm their children.

These tactics instil fear and o�en elicit compliance. [See reframed

social service mafia and how children are being mercilessly abused

by the state over ‘Covid’ while their parents look on too frightened

to do anything.]

A further recurring trait in an abusive relationship is the abused

blaming themselves for their abuse and making excuses for the

abuser. We have the public blaming each other for lockdown abuse

by government and many making excuses for the government while

a�acking those who challenge the government. How o�en we have

heard authorities say that rules are being imposed or reimposed only

because people have refused to ‘behave’ and follow the rules. We

don’t want to do it – it’s you.

Renegade Minds are an antidote to all of these things. They will

never concede their self-respect no ma�er what the circumstances.

Even when apparent humiliation is heaped upon them they laugh in

its face and reflect back the humiliation on the abuser where it

belongs. Renegade Minds will never wear masks they know are only

imposed to humiliate, suppress and damage both physically and

psychologically. Consequences will take care of themselves and they

will never break their spirit or cause them to concede to tyranny. UK

newspaper columnist Peter Hitchens was one of the few in the

mainstream media to speak out against lockdowns and forced

vaccinations. He then announced he had taken the jab. He wanted to

see family members abroad and he believed vaccine passports were

inevitable even though they had not yet been introduced. Hitchens

has a questioning and critical mind, but not a Renegade one. If he

had no amount of pressure would have made him concede. Hitchens

excused his action by saying that the ba�le has been lost. Renegade

Minds never accept defeat when freedom is at stake and even if they

are the last one standing the self-respect of not submi�ing to tyranny

is more important than any outcome or any consequence.

That’s why Renegade Minds are the only minds that ever changed

anything worth changing.

‘R

CHAPTER EIGHT

‘Reframing’ insanity

Insanity is relative. It depends on who has who locked in what cage

Ray Bradbury

eframing’ a mind means simply to change its perception and

behaviour. This can be done subconsciously to such an extent

that subjects have no idea they have been ‘reframed’ while to any

observer changes in behaviour and a�itudes are obvious.

Human society is being reframed on a ginormous scale since the

start of 2020 and here we have the reason why psychologists rather

than doctors have been calling the shots. Ask most people who have

succumbed to ‘Covid’ reframing if they have changed and most will

say ‘no’; but they have and fundamentally. The Cult’s long-game has

been preparing for these times since way back and crucial to that has

been to prepare both population and officialdom mentally and

emotionally. To use the mind-control parlance they had to reframe

the population with a mentality that would submit to fascism and

reframe those in government and law enforcement to impose

fascism or at least go along with it. The result has been the fact-

deleted mindlessness of ‘Wokeness’ and officialdom that has either

enthusiastically or unquestioningly imposed global tyranny

demanded by reframed politicians on behalf of psychopathic and

deeply evil cultists. ‘Cognitive reframing’ identifies and challenges

the way someone sees the world in the form of situations,

experiences and emotions and then restructures those perceptions to

view the same set of circumstances in a different way. This can have

benefits if the a�itudes are personally destructive while on the other

side it has the potential for individual and collective mind control

which the subject has no idea has even happened.

Cognitive therapy was developed in the 1960s by Aaron T. Beck

who was born in Rhode Island in 1921 as the son of Jewish

immigrants from the Ukraine. He became interested in the

techniques as a treatment for depression. Beck’s daughter Judith S.

Beck is prominent in the same field and they founded the Beck

Institute for Cognitive Behavior Therapy in Philadelphia in 1994.

Cognitive reframing, however, began to be used worldwide by those

with a very dark agenda. The Cult reframes politicians to change

their a�itudes and actions until they are completely at odds with

what they once appeared to stand for. The same has been happening

to government administrators at all levels, law enforcement, military

and the human population. Cultists love mind control for two main

reasons: It allows them to control what people think, do and say to

secure agenda advancement and, by definition, it calms their

legendary insecurity and fear of the unexpected. I have studied mind

control since the time I travelled America in 1996. I may have been

talking to next to no one in terms of an audience in those years, but

my goodness did I gather a phenomenal amount of information and

knowledge about so many things including the techniques of mind

control. I have described this in detail in other books going back to

The Biggest Secret in 1998. I met a very large number of people

recovering from MKUltra and its offshoots and successors and I

began to see how these same techniques were being used on the

population in general. This was never more obvious than since the

‘Covid’ hoax began.

Reframing the enforcers

I have observed over the last two decades and more the very clear

transformation in the dynamic between the police, officialdom and

the public. I tracked this in the books as the relationship mutated

from one of serving the public to seeing them as almost the enemy

and certainly a lower caste. There has always been a class divide

based on income and always been some psychopathic, corrupt, and

big-I-am police officers. This was different. Wholesale change was

unfolding in the collective dynamic; it was less about money and far

more about position and perceived power. An us-and-them was

emerging. Noses were li�ed skyward by government administration

and law enforcement and their a�itude to the public they were

supposed to be serving changed to one of increasing contempt,

superiority and control. The transformation was so clear and

widespread that it had to be planned. Collective a�itudes and

dynamics do not change naturally and organically that quickly on

that scale. I then came across an organisation in Britain called

Common Purpose created in the late 1980s by Julia Middleton who

would work in the office of Deputy Prime Minister John Presco�

during the long and disastrous premiership of war criminal Tony

Blair. When Blair speaks the Cult is speaking and the man should

have been in jail a long time ago. Common Purpose proclaims itself

to be one of the biggest ‘leadership development’ organisations in

the world while functioning as a charity with all the financial benefits

which come from that. It hosts ‘leadership development’ courses and

programmes all over the world and claims to have ‘brought

together’ what it calls ‘leaders’ from more than 100 countries on six

continents. The modus operandi of Common Purpose can be

compared with the work of the UK government’s reframing network

that includes the Behavioural Insights Team ‘nudge unit’ and

‘Covid’ reframing specialists at SPI-B. WikiLeaks described

Common Purpose long ago as ‘a hidden virus in our government

and schools’ which is unknown to the general public: ‘It recruits and

trains “leaders” to be loyal to the directives of Common Purpose and

the EU, instead of to their own departments, which they then

undermine or subvert, the NHS [National Health Service] being an

example.’ This is a vital point to understand the ‘Covid’ hoax. The

NHS, and its equivalent around the world, has been u�erly reframed

in terms of administrators and much of the medical personnel with

the transformation underpinned by recruitment policies. The

outcome has been the criminal and psychopathic behaviour of the

NHS over ‘Covid’ and we have seen the same in every other major

country. WikiLeaks said Common Purpose trainees are ‘learning to

rule without regard to democracy’ and to usher in a police state

(current events explained). Common Purpose operated like a ‘glue’

and had members in the NHS, BBC, police, legal profession, church,

many of Britain’s 7,000 quangos, local councils, the Civil Service,

government ministries and Parliament, and controlled many RDA’s

(Regional Development Agencies). Here we have one answer for

how and why British institutions and their like in other countries

have changed so negatively in relation to the public. This further

explains how and why the beyond-disgraceful reframed BBC has

become a propaganda arm of ‘Covid’ fascism. They are all part of a

network pursuing the same goal.

By 2019 Common Purpose was quoting a figure of 85,000 ‘leaders’

that had a�ended its programmes. These ‘students’ of all ages are

known as Common Purpose ‘graduates’ and they consist of

government, state and local government officials and administrators,

police chiefs and officers, and a whole range of others operating

within the national, local and global establishment. Cressida Dick,

Commissioner of the London Metropolitan Police, is the Common

Purpose graduate who was the ‘Gold Commander’ that oversaw

what can only be described as the murder of Brazilian electrician

Jean Charles de Menezes in 2005. He was held down by

psychopathic police and shot seven times in the head by a

psychopathic lunatic a�er being mistaken for a terrorist when he

was just a bloke going about his day. Dick authorised officers to

pursue and keep surveillance on de Menezes and ordered that he be

stopped from entering the underground train system. Police

psychopaths took her at her word clearly. She was ‘disciplined’ for

this outrage by being promoted – eventually to the top of the ‘Met’

police where she has been a disaster. Many Chief Constables

controlling the police in different parts of the UK are and have been

Common Purpose graduates. I have heard the ‘graduate’ network

described as a sort of Mafia or secret society operating within the

fabric of government at all levels pursuing a collective policy

ingrained at Common Purpose training events. Founder Julia

Middleton herself has said:

Locally and internationally, Common Purpose graduates will be ‘lighting small fires’ to create change in their organisations and communities … The Common Purpose effect is best illustrated by the many stories of small changes brought about by leaders, who themselves have changed.

A Common Purpose mission statement declared:

Common Purpose aims to improve the way society works by expanding the vision, decision- making ability and influence of all kinds of leaders. The organisation runs a variety of educational programmes for leaders of all ages, backgrounds and sectors, in order to provide them with the inspirational, information and opportunities they need to change the world.

Yes, but into what? Since 2020 the answer has become clear.

NLP and the Delphi technique

Common Purpose would seem to be a perfect name or would

common programming be be�er? One of the foundation methods of

reaching ‘consensus’ (group think) is by se�ing the agenda theme

and then encouraging, cajoling or pressuring everyone to agree a

‘consensus’ in line with the core theme promoted by Common

Purpose. The methodology involves the ‘Delphi technique’, or an

adaption of it, in which opinions are expressed that are summarised

by a ‘facilitator or change agent’ at each stage. Participants are

‘encouraged’ to modify their views in the light of what others have

said. Stage by stage the former individual opinions are merged into

group consensus which just happens to be what Common Purpose

wants them to believe. A key part of this is to marginalise anyone

refusing to concede to group think and turn the group against them

to apply pressure to conform. We are seeing this very technique used

on the general population to make ‘Covid’ group-thinkers hostile to

those who have seen through the bullshit. People can be reframed by

using perception manipulation methods such as Neuro-Linguistic

Programming (NLP) in which you change perception with the use of

carefully constructed language. An NLP website described the

technique this way:

… A method of influencing brain behaviour (the ‘neuro’ part of the phrase) through the use of language (the ‘linguistic’ part) and other types of communication to enable a person to ‘recode’ the way the brain responds to stimuli (that’s the ‘programming’) and manifest new and better behaviours. Neuro-Linguistic Programming often incorporates hypnosis and self- hypnosis to help achieve the change (or ‘programming’) that is wanted.

British alternative media operation UKColumn has done very

detailed research into Common Purpose over a long period. I quoted

co-founder and former naval officer Brian Gerrish in my book

Remember Who You Are, published in 2011, as saying the following

years before current times:

It is interesting that many of the mothers who have had children taken by the State speak of the Social Services people being icily cool, emotionless and, as two ladies said in slightly different words, ‘… like little robots’. We know that NLP is cumulative, so people can be given small imperceptible doses of NLP in a course here, another in a few months, next year etc. In this way, major changes are accrued in their personality, but the day by day change is almost unnoticeable.

In these and other ways ‘graduates’ have had their perceptions

uniformly reframed and they return to their roles in the institutions

of government, law enforcement, legal profession, military,

‘education’, the UK National Health Service and the whole swathe of

the establishment structure to pursue a common agenda preparing

for the ‘post-industrial’, ‘post-democratic’ society. I say ‘preparing’

but we are now there. ‘Post-industrial’ is code for the Great Reset

and ‘post-democratic’ is ‘Covid’ fascism. UKColumn has spoken to

partners of those who have a�ended Common Purpose ‘training’.

They have described how personalities and a�itudes of ‘graduates’

changed very noticeably for the worse by the time they had

completed the course. They had been ‘reframed’ and told they are

the ‘leaders’ – the special ones – who know be�er than the

population. There has also been the very demonstrable recruitment

of psychopaths and narcissists into government administration at all

levels and law enforcement. If you want psychopathy hire

psychopaths and you get a simple cause and effect. If you want

administrators, police officers and ‘leaders’ to perceive the public as

lesser beings who don’t ma�er then employ narcissists. These

personalities are identified using ‘psychometrics’ that identifies

knowledge, abilities, a�itudes and personality traits, mostly through

carefully-designed questionnaires and tests. As this policy has

passed through the decades we have had power-crazy, power-

trippers appointed into law enforcement, security and government

administration in preparation for current times and the dynamic

between public and law enforcement/officialdom has been

transformed. UKColumn’s Brian Gerrish said of the narcissistic

personality:

Their love of themselves and power automatically means that they will crush others who get in their way. I received a major piece of the puzzle when a friend pointed out that when they made public officials re-apply for their own jobs several years ago they were also required to do psychometric tests. This was undoubtedly the start of the screening process to get ‘their’ sort of people in post.

How obvious that has been since 2020 although it was clear what

was happening long before if people paid a�ention to the changing

public-establishment dynamic.

Change agents

At the centre of events in ‘Covid’ Britain is the National Health

Service (NHS) which has behaved disgracefully in slavishly

following the Cult agenda. The NHS management structure is awash

with Common Purpose graduates or ‘change agents’ working to a

common cause. Helen Bevan, a Chief of Service Transformation at

the NHS Institute for Innovation and Improvement, co-authored a

document called ‘Towards a million change agents, a review of the

social movements literature: implications for large scale change in

the NHS‘. The document compared a project management approach

to that of change and social movements where ‘people change

themselves and each other – peer to peer’. Two definitions given for

a ‘social movement’ were:

A group of people who consciously attempt to build a radically new social

order; involves people of a broad range of social backgrounds; and deploys

politically confrontational and socially disruptive tactics – Cyrus

Zirakzadeh 1997

Collective challenges, based on common purposes and social solidarities, in

sustained interaction with elites, opponents, and authorities – Sidney

Tarrow 1994

Helen Bevan wrote another NHS document in which she defined

‘framing’ as ‘the process by which leaders construct, articulate and

put across their message in a powerful and compelling way in order

to win people to their cause and call them to action’. I think I could

come up with another definition that would be rather more accurate.

The National Health Service and institutions of Britain and the wider

world have been taken over by reframed ‘change agents’ and that

includes everything from the United Nations to national

governments, local councils and social services which have been

kidnapping children from loving parents on an extraordinary and

gathering scale on the road to the end of parenthood altogether.

Children from loving homes are stolen and kidnapped by the state

and put into the ‘care’ (inversion) of the local authority through

council homes, foster parents and forced adoption. At the same time

children are allowed to be abused without response while many are

under council ‘care’. UKColumn highlighted the Common Purpose

connection between South Yorkshire Police and Rotherham council

officers in the case of the scandal in that area of the sexual

exploitation of children to which the authorities turned not one blind

eye, but both:

We were alarmed to discover that the Chief Executive, the Strategic Director of Children and Young People’s Services, the Manager for the Local Strategic Partnership, the Community Cohesion Manager, the Cabinet Member for Cohesion, the Chief Constable and his predecessor had all attended Leadership training courses provided by the pseudo-charity Common Purpose.

Once ‘change agents’ have secured positions of hire and fire within

any organisation things start to move very quickly. Personnel are

then hired and fired on the basis of whether they will work towards

the agenda the change agent represents. If they do they are rapidly

promoted even though they may be incompetent. Those more

qualified and skilled who are pre-Common Purpose ‘old school’ see

their careers stall and even disappear. This has been happening for

decades in every institution of state, police, ‘health’ and social

services and all of them have been transformed as a result in their

a�itudes to their jobs and the public. Medical professions, including

nursing, which were once vocations for the caring now employ

many cold, callous and couldn’t give a shit personality types. The

UKColumn investigation concluded:

By blurring the boundaries between people, professions, public and private sectors, responsibility and accountability, Common Purpose encourages ‘graduates’ to believe that as new selected leaders, they can work together, outside of the established political and social structures, to achieve a paradigm shift or CHANGE – so called ‘Leading Beyond Authority’. In doing so, the allegiance of the individual becomes ‘reframed’ on CP colleagues and their NETWORK.

Reframing the Face-Nappies

Nowhere has this process been more obvious than in the police

where recruitment of psychopaths and development of

unquestioning mind-controlled group-thinkers have transformed

law enforcement into a politically-correct ‘Woke’ joke and a travesty

of what should be public service. Today they wear their face-nappies

like good li�le gofers and enforce ‘Covid’ rules which are fascism

under another name. Alongside the specifically-recruited

psychopaths we have so�ware minds incapable of free thought.

Brian Gerrish again:

An example is the policeman who would not get on a bike for a press photo because he had not done the cycling proficiency course. Normal people say this is political correctness gone mad. Nothing could be further from the truth. The policeman has been reframed, and in his reality it is perfect common sense not to get on the bike ‘because he hasn’t done the cycling course’.

Another example of this is where the police would not rescue a boy from a pond until they had taken advice from above on the ‘risk assessment’. A normal person would have arrived, perhaps thought of the risk for a moment, and dived in. To the police now ‘reframed’, they followed ‘normal’ procedure.

There are shocking cases of reframed ambulance crews doing the

same. Sheer unthinking stupidity of London Face-Nappies headed

by Common Purpose graduate Cressida Dick can be seen in their

behaviour at a vigil in March, 2021, for a murdered woman, Sarah

Everard. A police officer had been charged with the crime. Anyone

with a brain would have le� the vigil alone in the circumstances.

Instead they ‘manhandled’ women to stop them breaking ‘Covid

rules’ to betray classic reframing. Minds in the thrall of perception

control have no capacity for seeing a situation on its merits and

acting accordingly. ‘Rules is rules’ is their only mind-set. My father

used to say that rules and regulations are for the guidance of the

intelligent and the blind obedience of the idiot. Most of the

intelligent, decent, coppers have gone leaving only the other kind

and a few old school for whom the job must be a daily nightmare.

The combination of psychopaths and rule-book so�ware minds has

been clearly on public display in the ‘Covid’ era with automaton

robots in uniform imposing fascistic ‘Covid’ regulations on the

population without any personal initiative or judging situations on

their merits. There are thousands of examples around the world, but

I’ll make my point with the infamous Derbyshire police in the

English East Midlands – the ones who think pouring dye into beauty

spots and using drones to track people walking in the countryside

away from anyone is called ‘policing’. To them there are rules

decreed by the government which they have to enforce and in their

bewildered state a group gathering in a closed space and someone

walking alone in the countryside are the same thing. It is beyond

idiocy and enters the realm of clinical insanity.

Police officers in Derbyshire said they were ‘horrified’ – horrified –

to find 15 to 20 ‘irresponsible’ kids playing a football match at a

closed leisure centre ‘in breach of coronavirus restrictions’. When

they saw the police the kids ran away leaving their belongings

behind and the reframed men and women of Derbyshire police were

seeking to establish their identities with a view to fining their

parents. The most natural thing for youngsters to do – kicking a ball

about – is turned into a criminal activity and enforced by the

moronic so�ware programs of Derbyshire police. You find the same

mentality in every country. These barely conscious ‘horrified’ officers

said they had to take action because ‘we need to ensure these rules

are being followed’ and ‘it is of the utmost importance that you

ensure your children are following the rules and regulations for

Covid-19’. Had any of them done ten seconds of research to see if

this parroting of their masters’ script could be supported by any

evidence? Nope. Reframed people don’t think – others think for

them and that’s the whole idea of reframing. I have seen police

officers one a�er the other repeating without question word for

word what officialdom tells them just as I have seen great swathes of

the public doing the same. Ask either for ‘their’ opinion and out

spews what they have been told to think by the official narrative.

Police and public may seem to be in different groups, but their

mentality is the same. Most people do whatever they are told in fear

not doing so or because they believe what officialdom tells them;

almost the entirety of the police do what they are told for the same

reason. Ultimately it’s the tiny inner core of the global Cult that’s

telling both what to do.

So Derbyshire police were ‘horrified’. Oh, really? Why did they

think those kids were playing football? It was to relieve the

psychological consequences of lockdown and being denied human

contact with their friends and interaction, touch and discourse vital

to human psychological health. Being denied this month a�er month

has dismantled the psyche of many children and young people as

depression and suicide have exploded. Were Derbyshire police

horrified by that? Are you kidding? Reframed people don’t have those

mental and emotional processes that can see how the impact on the

psychological health of youngsters is far more dangerous than any

‘virus’ even if you take the mendacious official figures to be true. The

reframed are told (programmed) how to act and so they do. The

Derbyshire Chief Constable in the first period of lockdown when the

black dye and drones nonsense was going on was Peter Goodman.

He was the man who severed the connection between his force and

the Derbyshire Constabulary Male Voice Choir when he decided that

it was not inclusive enough to allow women to join. The fact it was a

male voice choir making a particular sound produced by male voices

seemed to elude a guy who terrifyingly ran policing in Derbyshire.

He retired weeks a�er his force was condemned as disgraceful by

former Supreme Court Justice Jonathan Sumption for their

behaviour over extreme lockdown impositions. Goodman was

replaced by his deputy Rachel Swann who was in charge when her

officers were ‘horrified’. The police statement over the boys

commi�ing the hanging-offence of playing football included the line

about the youngsters being ‘irresponsible in the times we are all

living through’ missing the point that the real relevance of the ‘times

we are all living through’ is the imposition of fascism enforced by

psychopaths and reframed minds of police officers playing such a

vital part in establishing the fascist tyranny that their own children

and grandchildren will have to live in their entire lives. As a

definition of insanity that is hard to beat although it might be run

close by imposing masks on people that can have a serious effect on

their health while wearing a face nappy all day themselves. Once

again public and police do it for the same reason – the authorities tell

them to and who are they to have the self-respect to say no?

Wokers in uniform

How reframed do you have to be to arrest a six-year-old and take him

to court for picking a flower while waiting for a bus? Brain dead police

and officialdom did just that in North Carolina where criminal

proceedings happen regularly for children under nine. A�orney

Julie Boyer gave the six-year-old crayons and a colouring book

during the ‘flower’ hearing while the ‘adults’ decided his fate.

County Chief District Court Judge Jay Corpening asked: ‘Should a

child that believes in Santa Claus, the Easter Bunny and the tooth

fairy be making life-altering decisions?’ Well, of course not, but

common sense has no meaning when you have a common purpose

and a reframed mind. Treating children in this way, and police

operating in American schools, is all part of the psychological

preparation for children to accept a police state as normal all their

adult lives. The same goes for all the cameras and biometric tracking

technology in schools. Police training is focused on reframing them

as snowflake Wokers and this is happening in the military. Pentagon

top brass said that ‘training sessions on extremism’ were needed for

troops who asked why they were so focused on the Capitol Building

riot when Black Lives Ma�er riots were ignored. What’s the

difference between them some apparently and rightly asked.

Actually, there is a difference. Five people died in the Capitol riot,

only one through violence, and that was a police officer shooting an

unarmed protestor. BLM riots killed at least 25 people and cost

billions. Asking the question prompted the psychopaths and

reframed minds that run the Pentagon to say that more ‘education’

(programming) was needed. Troop training is all based on

psychological programming to make them fodder for the Cult –

‘Military men are just dumb, stupid animals to be used as pawns in

foreign policy’ as Cult-to-his-DNA former Secretary of State Henry

Kissinger famously said. Governments see the police in similar terms

and it’s time for those among them who can see this to defend the

people and stop being enforcers of the Cult agenda upon the people.

The US military, like the country itself, is being targeted for

destruction through a long list of Woke impositions. Cult-owned

gaga ‘President’ Biden signed an executive order when he took office

to allow taxpayer money to pay for transgender surgery for active

military personnel and veterans. Are you a man soldier? No, I’m a

LGBTQIA+ with a hint of Skoliosexual and Spectrasexual. Oh, good

man. Bad choice of words you bigot. The Pentagon announced in

March, 2021, the appointment of the first ‘diversity and inclusion

officer’ for US Special Forces. Richard Torres-Estrada arrived with

the publication of a ‘D&I Strategic Plan which will guide the

enterprise-wide effort to institutionalize and sustain D&I’. If you

think a Special Forces ‘Strategic Plan’ should have something to do

with defending America you haven’t been paying a�ention.

Defending Woke is now the military’s new role. Torres-Estrada has

posted images comparing Donald Trump with Adolf Hitler and we

can expect no bias from him as a representative of the supposedly

non-political Pentagon. Cable news host Tucker Carlson said: ‘The

Pentagon is now the Yale faculty lounge but with cruise missiles.’

Meanwhile Secretary of Defense Lloyd Austin, a board member of

weapons-maker Raytheon with stock and compensation interests in

October, 2020, worth $1.4 million, said he was purging the military

of the ‘enemy within’ – anyone who isn’t Woke and supports Donald

Trump. Austin refers to his targets as ‘racist extremists’ while in true

Woke fashion being himself a racist extremist. Pentagon documents

pledge to ‘eradicate, eliminate and conquer all forms of racism,

sexism and homophobia’. The definitions of these are decided by

‘diversity and inclusion commi�ees’ peopled by those who see

racism, sexism and homophobia in every situation and opinion.

Woke (the Cult) is dismantling the US military and purging

testosterone as China expands its military and gives its troops

‘masculinity training’. How do we think that is going to end when

this is all Cult coordinated? The US military, like the British military,

is controlled by Woke and spineless top brass who just go along with

it out of personal career interests.

‘Woke’ means fast asleep

Mind control and perception manipulation techniques used on

individuals to create group-think have been unleashed on the global

population in general. As a result many have no capacity to see the

obvious fascist agenda being installed all around them or what

‘Covid’ is really all about. Their brains are firewalled like a computer

system not to process certain concepts, thoughts and realisations that

are bad for the Cult. The young are most targeted as the adults they

will be when the whole fascist global state is planned to be fully

implemented. They need to be prepared for total compliance to

eliminate all pushback from entire generations. The Cult has been

pouring billions into taking complete control of ‘education’ from

schools to universities via its operatives and corporations and not

least Bill Gates as always. The plan has been to transform ‘education’

institutions into programming centres for the mentality of ‘Woke’.

James McConnell, professor of psychology at the University of

Michigan, wrote in Psychology Today in 1970:

The day has come when we can combine sensory deprivation with drugs, hypnosis, and astute manipulation of reward and punishment, to gain almost absolute control over an individual’s behaviour. It should then be possible to achieve a very rapid and highly effective type of brainwashing that would allow us to make dramatic changes in a person’s behaviour and personality ...

… We should reshape society so that we all would be trained from birth to want to do what society wants us to do. We have the techniques to do it... no-one owns his own personality you acquired, and there’s no reason to believe you should have the right to refuse to acquire a new personality if your old one is anti-social.

This was the potential for mass brainwashing in 1970 and the

mentality there displayed captures the arrogant psychopathy that

drives it forward. I emphasise that not all young people have

succumbed to Woke programming and those that haven’t are

incredibly impressive people given that today’s young are the most

perceptually-targeted generations in history with all the technology

now involved. Vast swathes of the young generations, however, have

fallen into the spell – and that’s what it is – of Woke. The Woke

mentality and perceptual program is founded on inversion and you

will appreciate later why that is so significant. Everything with Woke

is inverted and the opposite of what it is claimed to be. Woke was a

term used in African-American culture from the 1900s and referred

to an awareness of social and racial justice. This is not the meaning

of the modern version or ‘New Woke’ as I call it in The Answer. Oh,

no, Woke today means something very different no ma�er how

much Wokers may seek to hide that and insist Old Woke and New

•

•

•

•

•

Woke are the same. See if you find any ‘awareness of social justice’

here in the modern variety:

Woke demands ‘inclusivity’ while excluding anyone with a

different opinion and calls for mass censorship to silence other

views.

Woke claims to stand against oppression when imposing

oppression is the foundation of all that it does. It is the driver of

political correctness which is nothing more than a Cult invention

to manipulate the population to silence itself.

Woke believes itself to be ‘liberal’ while pursuing a global society

that can only be described as fascist (see ‘anti-fascist’ fascist

Antifa).

Woke calls for ‘social justice’ while spreading injustice wherever it

goes against the common ‘enemy’ which can be easily identified

as a differing view.

Woke is supposed to be a metaphor for ‘awake’ when it is solid-

gold asleep and deep in a Cult-induced coma that meets the

criteria for ‘off with the fairies’.

I state these points as obvious facts if people only care to look. I

don’t do this with a sense of condemnation. We need to appreciate

that the onslaught of perceptual programming on the young has

been incessant and merciless. I can understand why so many have

been reframed, or, given their youth, framed from the start to see the

world as the Cult demands. The Cult has had access to their minds

day a�er day in its ‘education’ system for their entire formative

years. Perception is formed from information received and the Cult-

created system is a life-long download of information delivered to

elicit a particular perception, thus behaviour. The more this has

expanded into still new extremes in recent decades and ever-

increasing censorship has deleted other opinions and information

why wouldn’t that lead to a perceptual reframing on a mass scale? I

have described already cradle-to-grave programming and in more

recent times the targeting of young minds from birth to adulthood

has entered the stratosphere. This has taken the form of skewing

what is ‘taught’ to fit the Cult agenda and the omnipresent

techniques of group-think to isolate non-believers and pressure them

into line. There has always been a tendency to follow the herd, but

we really are in a new world now in relation to that. We have parents

who can see the ‘Covid’ hoax told by their children not to stop them

wearing masks at school, being ‘Covid’ tested or having the ‘vaccine’

in fear of the peer-pressure consequences of being different. What is

‘peer-pressure’ if not pressure to conform to group-think? Renegade

Minds never group-think and always retain a set of perceptions that

are unique to them. Group-think is always underpinned by

consequences for not group-thinking. Abuse now aimed at those

refusing DNA-manipulating ‘Covid vaccines’ are a potent example

of this. The biggest pressure to conform comes from the very group

which is itself being manipulated. ‘I am programmed to be part of a

hive mind and so you must be.’

Woke control structures in ‘education’ now apply to every

mainstream organisation. Those at the top of the ‘education’

hierarchy (the Cult) decide the policy. This is imposed on

governments through the Cult network; governments impose it on

schools, colleges and universities; their leadership impose the policy

on teachers and academics and they impose it on children and

students. At any level where there is resistance, perhaps from a

teacher or university lecturer, they are targeted by the authorities

and o�en fired. Students themselves regularly demand the dismissal

of academics (increasingly few) at odds with the narrative that the

students have been programmed to believe in. It is quite a thought

that students who are being targeted by the Cult become so

consumed by programmed group-think that they launch protests

and demand the removal of those who are trying to push back

against those targeting the students. Such is the scale of perceptual

inversion. We see this with ‘Covid’ programming as the Cult

imposes the rules via psycho-psychologists and governments on

shops, transport companies and businesses which impose them on

their staff who impose them on their customers who pressure

Pushbackers to conform to the will of the Cult which is in the

process of destroying them and their families. Scan all aspects of

society and you will see the same sequence every time.

Fact free Woke and hijacking the ‘left’

There is no more potent example of this than ‘Woke’, a mentality

only made possible by the deletion of factual evidence by an

‘education’ system seeking to produce an ever more uniform society.

Why would you bother with facts when you don’t know any?

Deletion of credible history both in volume and type is highly

relevant. Orwell said: ‘Who controls the past controls the future:

who controls the present controls the past.’ They who control the

perception of the past control the perception of the future and they

who control the present control the perception of the past through

the writing and deleting of history. Why would you oppose the

imposition of Marxism in the name of Wokeism when you don’t

know that Marxism cost at least 100 million lives in the 20th century

alone? Watch videos and read reports in which Woker generations

are asked basic historical questions – it’s mind-blowing. A survey of

2,000 people found that six percent of millennials (born

approximately early1980s to early 2000s) believed the Second World

War (1939-1945) broke out with the assassination of President

Kennedy (in 1963) and one in ten thought Margaret Thatcher was

British Prime Minister at the time. She was in office between 1979

and 1990. We are in a post-fact society. Provable facts are no defence

against the fascism of political correctness or Silicon Valley

censorship. Facts don’t ma�er anymore as we have witnessed with

the ‘Covid’ hoax. Sacrificing uniqueness to the Woke group-think

religion is all you are required to do and that means thinking for

yourself is the biggest Woke no, no. All religions are an expression of

group-think and censorship and Woke is just another religion with

an orthodoxy defended by group-think and censorship. Burned at

the stake becomes burned on Twi�er which leads back eventually to

burned at the stake as Woke humanity regresses to ages past.

The biggest Woke inversion of all is its creators and funders. I

grew up in a traditional le� of centre political household on a

council estate in Leicester in the 1950s and 60s – you know, the le�

that challenged the power of wealth-hoarding elites and threats to

freedom of speech and opinion. In those days students went on

marches defending freedom of speech while today’s Wokers march

for its deletion. What on earth could have happened? Those very

elites (collectively the Cult) that we opposed in my youth and early

life have funded into existence the antithesis of that former le� and

hijacked the ‘brand’ while inverting everything it ever stood for. We

have a mentality that calls itself ‘liberal’ and ‘progressive’ while

acting like fascists. Cult billionaires and their corporations have

funded themselves into control of ‘education’ to ensure that Woke

programming is unceasing throughout the formative years of

children and young people and that non-Wokers are isolated (that

word again) whether they be students, teachers or college professors.

The Cult has funded into existence the now colossal global network

of Woke organisations that have spawned and promoted all the

‘causes’ on the Cult wish-list for global transformation and turned

Wokers into demanders of them. Does anyone really think it’s a

coincidence that the Cult agenda for humanity is a carbon (sorry)

copy of the societal transformations desired by Woke?? These are

only some of them:

Political correctness: The means by which the Cult deletes all public debates that it knows it cannot win if we had the free-flow of

information and evidence.

Human-caused ‘climate change’: The means by which the Cult seeks to transform society into a globally-controlled dictatorship

imposing its will over the fine detail of everyone’s lives ‘to save the

planet’ which doesn’t actually need saving.

Transgender obsession: Preparing collective perception to accept the ‘new human’ which would not have genders because it would be

created technologically and not through procreation. I’ll have much

more on this in Human 2.0.

Race obsession: The means by which the Cult seeks to divide and rule the population by triggering racial division through the

perception that society is more racist than ever when the opposite is

the case. Is it perfect in that regard? No. But to compare today with

the racism of apartheid and segregation brought to an end by the

civil rights movement in the 1960s is to insult the memory of that

movement and inspirations like Martin Luther King. Why is the

‘anti-racism’ industry (which it is) so dominated by privileged white

people?

White supremacy: This is a label used by privileged white people to demonise poor and deprived white people pushing back on tyranny

to marginalise and destroy them. White people are being especially

targeted as the dominant race by number within Western society

which the Cult seeks to transform in its image. If you want to change

a society you must weaken and undermine its biggest group and

once you have done that by using the other groups you next turn on

them to do the same … ‘Then they came for the Jews and I was not a

Jew so I did nothing.’

Mass migration: The mass movement of people from the Middle East, Africa and Asia into Europe, from the south into the United

States and from Asia into Australia are another way the Cult seeks to

dilute the racial, cultural and political influence of white people on

Western society. White people ask why their governments appear to

be working against them while being politically and culturally

biased towards incoming cultures. Well, here’s your answer. In the

same way sexually ‘straight’ people, men and women, ask why the

authorities are biased against them in favour of other sexualities. The

answer is the same – that’s the way the Cult wants it to be for very

sinister motives.

These are all central parts of the Cult agenda and central parts of the

Woke agenda and Woke was created and continues to be funded to

an immense degree by Cult billionaires and corporations. If anyone

begins to say ‘coincidence’ the syllables should stick in their throat.

Billionaire ‘social justice warriors’

Joe Biden is a 100 percent-owned asset of the Cult and the Wokers’

man in the White House whenever he can remember his name and

for however long he lasts with his rapidly diminishing cognitive

function. Even walking up the steps of an aircra� without falling on

his arse would appear to be a challenge. He’s not an empty-shell

puppet or anything. From the minute Biden took office (or the Cult

did) he began his executive orders promoting the Woke wish-list.

You will see the Woke agenda imposed ever more severely because

it’s really the Cult agenda. Woke organisations and activist networks

spawned by the Cult are funded to the extreme so long as they

promote what the Cult wants to happen. Woke is funded to promote

‘social justice’ by billionaires who become billionaires by destroying

social justice. The social justice mantra is only a cover for

dismantling social justice and funded by billionaires that couldn’t

give a damn about social justice. Everything makes sense when you

see that. One of Woke’s premier funders is Cult billionaire financier

George Soros who said: ‘I am basically there to make money, I

cannot and do not look at the social consequences of what I do.’ This

is the same Soros who has given more than $32 billion to his Open

Society Foundations global Woke network and funded Black Lives

Ma�er, mass immigration into Europe and the United States,

transgender activism, climate change activism, political correctness

and groups targeting ‘white supremacy’ in the form of privileged

white thugs that dominate Antifa. What a scam it all is and when

you are dealing with the unquestioning fact-free zone of Woke

scamming them is child’s play. All you need to pull it off in all these

organisations are a few in-the-know agents of the Cult and an army

of naïve, reframed, uninformed, narcissistic, know-nothings

convinced of their own self-righteousness, self-purity and virtue.

Soros and fellow billionaires and billionaire corporations have

poured hundreds of millions into Black Lives Ma�er and connected

groups and promoted them to a global audience. None of this is

motivated by caring about black people. These are the billionaires

that have controlled and exploited a system that leaves millions of

black people in abject poverty and deprivation which they do

absolutely nothing to address. The same Cult networks funding

BLM were behind the slave trade! Black Lives Ma�er hijacked a

phrase that few would challenge and they have turned this laudable

concept into a political weapon to divide society. You know that

BLM is a fraud when it claims that All Lives Ma�er, the most

inclusive statement of all, is ‘racist’. BLM and its Cult masters don’t

want to end racism. To them it’s a means to an end to control all of

humanity never mind the colour, creed, culture or background.

What has destroying the nuclear family got to do with ending

racism? Nothing – but that is one of the goals of BLM and also

happens to be a goal of the Cult as I have been exposing in my books

for decades. Stealing children from loving parents and giving

schools ever more power to override parents is part of that same

agenda. BLM is a Marxist organisation and why would that not be

the case when the Cult created Marxism and BLM? Patrisse Cullors, a

BLM co-founder, said in a 2015 video that she and her fellow

organisers, including co-founder Alicia Garza, are ‘trained Marxists’.

The lady known a�er marriage as Patrisse Khan-Cullors bought a

$1.4 million home in 2021 in one of the whitest areas of California

with a black population of just 1.6 per cent and has so far bought four

high-end homes for a total of $3.2 million. How very Marxist. There

must be a bit of spare in the BLM coffers, however, when Cult

corporations and billionaires have handed over the best part of $100

million. Many black people can see that Black Lives Ma�er is not

working for them, but against them, and this is still more

confirmation. Black journalist Jason Whitlock, who had his account

suspended by Twi�er for simply linking to the story about the

‘Marxist’s’ home buying spree, said that BLM leaders are ‘making

millions of dollars off the backs of these dead black men who they

wouldn’t spit on if they were on fire and alive’.

Black Lies Matter

Cult assets and agencies came together to promote BLM in the wake

of the death of career criminal George Floyd who had been jailed a

number of times including for forcing his way into the home of a

black woman with others in a raid in which a gun was pointed at her

stomach. Floyd was filmed being held in a Minneapolis street in 2020

with the knee of a police officer on his neck and he subsequently

died. It was an appalling thing for the officer to do, but the same

technique has been used by police on peaceful protestors of

lockdown without any outcry from the Woke brigade. As

unquestioning supporters of the Cult agenda Wokers have

supported lockdown and all the ‘Covid’ claptrap while a�acking

anyone standing up to the tyranny imposed in its name. Court

documents would later include details of an autopsy on Floyd by

County Medical Examiner Dr Andrew Baker who concluded that

Floyd had taken a fatal level of the drug fentanyl. None of this

ma�ered to fact-free, question-free, Woke. Floyd’s death was

followed by worldwide protests against police brutality amid calls to

defund the police. Throwing babies out with the bathwater is a

Woke speciality. In the wake of the murder of British woman Sarah

Everard a Green Party member of the House of Lords, Baroness

Jones of Moulescoomb (Nincompoopia would have been be�er),

called for a 6pm curfew for all men. This would be in breach of the

Geneva Conventions on war crimes which ban collective

punishment, but that would never have crossed the black and white

Woke mind of Baroness Nincompoopia who would have been far

too convinced of her own self-righteousness to compute such details.

Many American cities did defund the police in the face of Floyd riots

and a�er $15 million was deleted from the police budget in

Washington DC under useless Woke mayor Muriel Bowser car-

jacking alone rose by 300 percent and within six months the US

capital recorded its highest murder rate in 15 years. The same

happened in Chicago and other cities in line with the Cult/Soros

plan to bring fear to streets and neighbourhoods by reducing the

police, releasing violent criminals and not prosecuting crime. This is

the mob-rule agenda that I have warned in the books was coming for

so long. Shootings in the area of Minneapolis where Floyd was

arrested increased by 2,500 percent compared with the year before.

Defunding the police over George Floyd has led to a big increase in

dead people with many of them black. Police protection for

politicians making these decisions stayed the same or increased as

you would expect from professional hypocrites. The Cult doesn’t

actually want to abolish the police. It wants to abolish local control

over the police and hand it to federal government as the

psychopaths advance the Hunger Games Society. Many George

Floyd protests turned into violent riots with black stores and

businesses destroyed by fire and looting across America fuelled by

Black Lives Ma�er. Woke doesn’t do irony. If you want civil rights

you must loot the liquor store and the supermarket and make off

with a smart TV. It’s the only way.

It’s not a race war – it’s a class war

Black people are patronised by privileged blacks and whites alike

and told they are victims of white supremacy. I find it extraordinary

to watch privileged blacks supporting the very system and bloodline

networks behind the slave trade and parroting the same Cult-serving

manipulative crap of their privileged white, o�en billionaire,

associates. It is indeed not a race war but a class war and colour is

just a diversion. Black Senator Cory Booker and black

Congresswoman Maxine Waters, more residents of Nincompoopia,

personify this. Once you tell people they are victims of someone else

you devalue both their own responsibility for their plight and the

power they have to impact on their reality and experience. Instead

we have: ‘You are only in your situation because of whitey – turn on

them and everything will change.’ It won’t change. Nothing changes

in our lives unless we change it. Crucial to that is never seeing

yourself as a victim and always as the creator of your reality. Life is a

simple sequence of choice and consequence. Make different choices

and you create different consequences. You have to make those

choices – not Black Lives Ma�er, the Woke Mafia and anyone else

that seeks to dictate your life. Who are they these Wokers, an

emotional and psychological road traffic accident, to tell you what to

do? Personal empowerment is the last thing the Cult and its Black

Lives Ma�er want black people or anyone else to have. They claim to

be defending the underdog while creating and perpetuating the

underdog. The Cult’s worst nightmare is human unity and if they

are going to keep blacks, whites and every other race under

economic servitude and control then the focus must be diverted

from what they have in common to what they can be manipulated to

believe divides them. Blacks have to be told that their poverty and

plight is the fault of the white bloke living on the street in the same

poverty and with the same plight they are experiencing. The

difference is that your plight black people is due to him, a white

supremacist with ‘white privilege’ living on the street. Don’t unite as

one human family against your mutual oppressors and suppressors

– fight the oppressor with the white face who is as financially

deprived as you are. The Cult knows that as its ‘Covid’ agenda

moves into still new levels of extremism people are going to respond

and it has been spreading the seeds of disunity everywhere to stop a

united response to the evil that targets all of us.

Racist a�acks on ‘whiteness’ are ge�ing ever more outrageous and

especially through the American Democratic Party which has an

appalling history for anti-black racism. Barack Obama, Joe Biden,

Hillary Clinton and Nancy Pelosi all eulogised about Senator Robert

Byrd at his funeral in 2010 a�er a nearly 60-year career in Congress.

Byrd was a brutal Ku Klux Klan racist and a violent abuser of Cathy

O’Brien in MKUltra. He said he would never fight in the military

‘with a negro by my side’ and ‘rather I should die a thousand times,

and see Old Glory trampled in the dirt never to rise again, than to

see this beloved land of ours become degraded by race mongrels, a

throwback to the blackest specimen from the wilds’. Biden called

Byrd a ‘very close friend and mentor’. These ‘Woke’ hypocrites are

not anti-racist they are anti-poor and anti-people not of their

perceived class. Here is an illustration of the scale of anti-white

racism to which we have now descended. Seriously Woke and

moronic New York Times contributor Damon Young described

whiteness as a ‘virus’ that ‘like other viruses will not die until there

are no bodies le� for it to infect’. He went on: ‘… the only way to

stop it is to locate it, isolate it, extract it, and kill it.’ Young can say

that as a black man with no consequences when a white man saying

the same in reverse would be facing a jail sentence. That’s racism. We

had super-Woke numbskull senators Tammy Duckworth and Mazie

Hirono saying they would object to future Biden Cabinet

appointments if he did not nominate more Asian Americans and

Pacific Islanders. Never mind the ability of the candidate what do

they look like? Duckworth said: ‘I will vote for racial minorities and I

will vote for LGBTQ, but anyone else I’m not voting for.’ Appointing

people on the grounds of race is illegal, but that was not a problem

for this ludicrous pair. They were on-message and that’s a free pass

in any situation.

Critical race racism

White children are told at school they are intrinsically racist as they

are taught the divisive ‘critical race theory’. This claims that the law

and legal institutions are inherently racist and that race is a socially

constructed concept used by white people to further their economic

and political interests at the expense of people of colour. White is a

‘virus’ as we’ve seen. Racial inequality results from ‘social,

economic, and legal differences that white people create between

races to maintain white interests which leads to poverty and

criminality in minority communities‘. I must tell that to the white

guy sleeping on the street. The principal of East Side Community

School in New York sent white parents a manifesto that called on

them to become ‘white traitors’ and advocate for full ‘white

abolition’. These people are teaching your kids when they urgently

need a psychiatrist. The ‘school’ included a chart with ‘eight white

identities’ that ranged from ‘white supremacist’ to ‘white abolition’

and defined the behaviour white people must follow to end ‘the

regime of whiteness’. Woke blacks and their privileged white

associates are acting exactly like the slave owners of old and Ku Klux

Klan racists like Robert Byrd. They are too full of their own self-

purity to see that, but it’s true. Racism is not a body type; it’s a state

of mind that can manifest through any colour, creed or culture.

Another racial fraud is ‘equity’. Not equality of treatment and

opportunity – equity. It’s a term spun as equality when it means

something very different. Equality in its true sense is a raising up

while ‘equity’ is a race to the bo�om. Everyone in the same level of

poverty is ‘equity’. Keep everyone down – that’s equity. The Cult

doesn’t want anyone in the human family to be empowered and

BLM leaders, like all these ‘anti-racist’ organisations, continue their

privileged, pampered existence by perpetuating the perception of

gathering racism. When is the last time you heard an ‘anti-racist’ or

‘anti-Semitism’ organisation say that acts of racism and

discrimination have fallen? It’s not in the interests of their fund-

raising and power to influence and the same goes for the

professional soccer anti-racism operation, Kick It Out. Two things

confirmed that the Black Lives Ma�er riots in the summer of 2020

were Cult creations. One was that while anti-lockdown protests were

condemned in this same period for ‘transmi�ing ‘Covid’ the

authorities supported mass gatherings of Black Lives Ma�er

supporters. I even saw self-deluding people claiming to be doctors

say the two types of protest were not the same. No – the non-existent

‘Covid’ was in favour of lockdowns and a�acked those that

protested against them while ‘Covid’ supported Black Lives Ma�er

and kept well away from its protests. The whole thing was a joke

and as lockdown protestors were arrested, o�en brutally, by

reframed Face-Nappies we had the grotesque sight of police officers

taking the knee to Black Lives Ma�er, a Cult-funded Marxist

organisation that supports violent riots and wants to destroy the

nuclear family and white people.

He’s not white? Shucks!

Woke obsession with race was on display again when ten people

were shot dead in Boulder, Colorado, in March, 2021. Cult-owned

Woke TV channels like CNN said the shooter appeared to be a white

man and Wokers were on Twi�er condemning ‘violent white men’

with the usual mantras. Then the shooter’s name was released as

Ahmad Al Aliwi Alissa, an anti-Trump Arab-American, and the sigh

of disappointment could be heard five miles away. Never mind that

ten people were dead and what that meant for their families. Race

baiting was all that ma�ered to these sick Cult-serving people like

Barack Obama who exploited the deaths to further divide America

on racial grounds which is his job for the Cult. This is the man that

‘racist’ white Americans made the first black president of the United

States and then gave him a second term. Not-very-bright Obama has

become filthy rich on the back of that and today appears to have a

big influence on the Biden administration. Even so he’s still a

downtrodden black man and a victim of white supremacy. This

disingenuous fraud reveals the contempt he has for black people

when he puts on a Deep South Alabama accent whenever he talks to

them, no, at them.

Another BLM red flag was how the now fully-Woke (fully-Cult)

and fully-virtue-signalled professional soccer authorities had their

teams taking the knee before every match in support of Marxist

Black Lives Ma�er. Soccer authorities and clubs displayed ‘Black

Lives Ma�er’ on the players’ shirts and flashed the name on

electronic billboards around the pitch. Any fans that condemned

what is a Freemasonic taking-the-knee ritual were widely

condemned as you would expect from the Woke virtue-signallers of

professional sport and the now fully-Woke media. We have reverse

racism in which you are banned from criticising any race or culture

except for white people for whom anything goes – say what you like,

no problem. What has this got to do with racial harmony and

equality? We’ve had black supremacists from Black Lives Ma�er

telling white people to fall to their knees in the street and apologise

for their white supremacy. Black supremacists acting like white

supremacist slave owners of the past couldn’t breach their self-

obsessed, race-obsessed sense of self-purity. Joe Biden appointed a

race-obsessed black supremacist Kristen Clarke to head the Justice

Department Civil Rights Division. Clarke claimed that blacks are

endowed with ‘greater mental, physical and spiritual abilities’ than

whites. If anyone reversed that statement they would be vilified.

Clarke is on-message so no problem. She’s never seen a black-white

situation in which the black figure is anything but a virtuous victim

and she heads the Civil Rights Division which should treat everyone

the same or it isn’t civil rights. Another perception of the Renegade

Mind: If something or someone is part of the Cult agenda they will

be supported by Woke governments and media no ma�er what. If

they’re not, they will be condemned and censored. It really is that

simple and so racist Clarke prospers despite (make that because of)

her racism.

The end of culture

Biden’s administration is full of such racial, cultural and economic

bias as the Cult requires the human family to be divided into

warring factions. We are now seeing racially-segregated graduations

and everything, but everything, is defined through the lens of

perceived ‘racism. We have ‘racist’ mathematics, ‘racist’ food and

even ‘racist’ plants. World famous Kew Gardens in London said it

was changing labels on plants and flowers to tell its pre-‘Covid’

more than two million visitors a year how racist they are. Kew

director Richard Deverell said this was part of an effort to ‘move

quickly to decolonise collections’ a�er they were approached by one

Ajay Chhabra ‘an actor with an insight into how sugar cane was

linked to slavery’. They are plants you idiots. ‘Decolonisation’ in the

Woke manual really means colonisation of society with its mentality

and by extension colonisation by the Cult. We are witnessing a new

Chinese-style ‘Cultural Revolution’ so essential to the success of all

Marxist takeovers. Our cultural past and traditions have to be swept

away to allow a new culture to be built-back-be�er. Woke targeting

of long-standing Western cultural pillars including historical

monuments and cancelling of historical figures is what happened in

the Mao revolution in China which ‘purged remnants of capitalist

and traditional elements from Chinese society‘ and installed Maoism

as the dominant ideology‘. For China see the Western world today

and for ‘dominant ideology’ see Woke. Be�er still see Marxism or

Maoism. The ‘Covid’ hoax has specifically sought to destroy the arts

and all elements of Western culture from people meeting in a pub or

restaurant to closing theatres, music venues, sports stadiums, places

of worship and even banning singing. Destruction of Western society

is also why criticism of any religion is banned except for Christianity

which again is the dominant religion as white is the numerically-

dominant race. Christianity may be fading rapidly, but its history

and traditions are weaved through the fabric of Western society.

Delete the pillars and other structures will follow until the whole

thing collapses. I am not a Christian defending that religion when I

say that. I have no religion. It’s just a fact. To this end Christianity

has itself been turned Woke to usher its own downfall and its ranks

are awash with ‘change agents’ – knowing and unknowing – at

every level including Pope Francis (definitely knowing) and the

clueless Archbishop of Canterbury Justin Welby (possibly not, but

who can be sure?). Woke seeks to coordinate a�acks on Western

culture, traditions, and ways of life through ‘intersectionality’

defined as ‘the complex, cumulative way in which the effects of

multiple forms of discrimination (such as racism, sexism, and

classism) combine, overlap, or intersect especially in the experiences

of marginalised individuals or groups’. Wade through the Orwellian

Woke-speak and this means coordinating disparate groups in a

common cause to overthrow freedom and liberal values.

The entire structure of public institutions has been infested with

Woke – government at all levels, political parties, police, military,

schools, universities, advertising, media and trade unions. This

abomination has been achieved through the Cult web by appointing

Wokers to positions of power and ba�ering non-Wokers into line

through intimidation, isolation and threats to their job. Many have

been fired in the wake of the empathy-deleted, vicious hostility of

‘social justice’ Wokers and the desire of gutless, spineless employers

to virtue-signal their Wokeness. Corporations are filled with Wokers

today, most notably those in Silicon Valley. Ironically at the top they

are not Woke at all. They are only exploiting the mentality their Cult

masters have created and funded to censor and enslave while the

Wokers cheer them on until it’s their turn. Thus the Woke ‘liberal

le�’ is an inversion of the traditional liberal le�. Campaigning for

justice on the grounds of power and wealth distribution has been

replaced by campaigning for identity politics. The genuine

traditional le� would never have taken money from today’s

billionaire abusers of fairness and justice and nor would the

billionaires have wanted to fund that genuine le�. It would not have

been in their interests to do so. The division of opinion in those days

was between the haves and have nots. This all changed with Cult

manipulated and funded identity politics. The division of opinion

today is between Wokers and non-Wokers and not income brackets.

Cult corporations and their billionaires may have taken wealth

disparity to cataclysmic levels of injustice, but as long as they speak

the language of Woke, hand out the dosh to the Woke network and

censor the enemy they are ‘one of us’. Billionaires who don’t give a

damn about injustice are laughing at them till their bellies hurt.

Wokers are not even close to self-aware enough to see that. The

transformed ‘le�’ dynamic means that Wokers who drone on about

‘social justice’ are funded by billionaires that have destroyed social

justice the world over. It’s why they are billionaires.

The climate con

Nothing encapsulates what I have said more comprehensively than

the hoax of human-caused global warming. I have detailed in my

books over the years how Cult operatives and organisations were the

pump-primers from the start of the climate con. A purpose-built

vehicle for this is the Club of Rome established by the Cult in 1968

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with the Rockefellers and Rothschilds centrally involved all along.

Their gofer frontman Maurice Strong, a Canadian oil millionaire,

hosted the Earth Summit in Rio de Janeiro, Brazil, in 1992 where the

global ‘green movement’ really expanded in earnest under the

guiding hand of the Cult. The Earth Summit established Agenda 21

through the Cult-created-and-owned United Nations to use the

illusion of human-caused climate change to justify the

transformation of global society to save the world from climate

disaster. It is a No-Problem-Reaction-Solution sold through

governments, media, schools and universities as whole generations

have been terrified into believing that the world was going to end in

their lifetimes unless what old people had inflicted upon them was

stopped by a complete restructuring of how everything is done.

Chill, kids, it’s all a hoax. Such restructuring is precisely what the

Cult agenda demands (purely by coincidence of course). Today this

has been given the codename of the Great Reset which is only an

updated term for Agenda 21 and its associated Agenda 2030. The

la�er, too, is administered through the UN and was voted into being

by the General Assembly in 2015. Both 21 and 2030 seek centralised

control of all resources and food right down to the raindrops falling

on your own land. These are some of the demands of Agenda 21

established in 1992. See if you recognise this society emerging today:

End national sovereignty

State planning and management of all land resources, ecosystems,

deserts, forests, mountains, oceans and fresh water; agriculture;

rural development; biotechnology; and ensuring ‘equity’

The state to ‘define the role’ of business and financial resources

Abolition of private property

‘Restructuring’ the family unit (see BLM)

Children raised by the state

People told what their job will be

Major restrictions on movement

Creation of ‘human se�lement zones’

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Mass rese�lement as people are forced to vacate land where they

live

Dumbing down education

Mass global depopulation in pursuit of all the above

The United Nations was created as a Trojan horse for world

government. With the climate con of critical importance to

promoting that outcome you would expect the UN to be involved.

Oh, it’s involved all right. The UN is promoting Agenda 21 and

Agenda 2030 justified by ‘climate change’ while also driving the

climate hoax through its Intergovernmental Panel on Climate

Change (IPCC), one of the world’s most corrupt organisations. The

IPCC has been lying ferociously and constantly since the day it

opened its doors with the global media hanging unquestioningly on

its every mendacious word. The Green movement is entirely Woke

and has long lost its original environmental focus since it was co-

opted by the Cult. An obsession with ‘global warming’ has deleted

its values and scrambled its head. I experienced a small example of

what I mean on a beautiful country walk that I have enjoyed several

times a week for many years. The path merged into the fields and

forests and you felt at one with the natural world. Then a ‘Green’

organisation, the Hampshire and Isle of Wight Wildlife Trust, took

over part of the land and proceeded to cut down a large number of

trees, including mature ones, to install a horrible big, bright steel

‘this-is-ours-stay-out’ fence that destroyed the whole atmosphere of

this beautiful place. No one with a feel for nature would do that. Day

a�er day I walked to the sound of chainsaws and a magnificent

mature weeping willow tree that I so admired was cut down at the

base of the trunk. When I challenged a Woke young girl in a green

shirt (of course) about this vandalism she replied: ‘It’s a weeping

willow – it will grow back.’ This is what people are paying for when

they donate to the Hampshire and Isle of Wight Wildlife Trust and

many other ‘green’ organisations today. It is not the environmental

movement that I knew and instead has become a support-system –

as with Extinction Rebellion – for a very dark agenda.

Private jets for climate justice

The Cult-owned, Gates-funded, World Economic Forum and its

founder Klaus Schwab were behind the emergence of Greta

Thunberg to harness the young behind the climate agenda and she

was invited to speak to the world at … the UN. Schwab published a

book, Covid-19: The Great Reset in 2020 in which he used the ‘Covid’

hoax and the climate hoax to lay out a new society straight out of

Agenda 21 and Agenda 2030. Bill Gates followed in early 2021 when

he took time out from destroying the world to produce a book in his

name about the way to save it. Gates flies across the world in private

jets and admi�ed that ‘I probably have one of the highest

greenhouse gas footprints of anyone on the planet … my personal

flying alone is gigantic.’ He has also bid for the planet’s biggest

private jet operator. Other climate change saviours who fly in private

jets include John Kerry, the US Special Presidential Envoy for

Climate, and actor Leonardo DiCaprio, a ‘UN Messenger of Peace

with special focus on climate change’. These people are so full of

bullshit they could corner the market in manure. We mustn’t be

sceptical, though, because the Gates book, How to Avoid a Climate

Disaster: The Solutions We Have and the Breakthroughs We Need, is a

genuine a�empt to protect the world and not an obvious pile of

excrement a�ributed to a mega-psychopath aimed at selling his

masters’ plans for humanity. The Gates book and the other shite-pile

by Klaus Schwab could have been wri�en by the same person and

may well have been. Both use ‘climate change’ and ‘Covid’ as the

excuses for their new society and by coincidence the Cult’s World

Economic Forum and Bill and Melinda Gates Foundation promote

the climate hoax and hosted Event 201 which pre-empted with a

‘simulation’ the very ‘coronavirus’ hoax that would be simulated for

real on humanity within weeks. The British ‘royal’ family is

promoting the ‘Reset’ as you would expect through Prince ‘climate

change caused the war in Syria’ Charles and his hapless son Prince

William who said that we must ‘reset our relationship with nature

and our trajectory as a species’ to avoid a climate disaster. Amazing

how many promotors of the ‘Covid’ and ‘climate change’ control

systems are connected to Gates and the World Economic Forum. A

‘study’ in early 2021 claimed that carbon dioxide emissions must fall

by the equivalent of a global lockdown roughly every two years for

the next decade to save the planet. The ‘study’ appeared in the same

period that the Schwab mob claimed in a video that lockdowns

destroying the lives of billions are good because they make the earth

‘quieter’ with less ‘ambient noise’. They took down the video amid a

public backlash for such arrogant, empathy-deleted stupidity You

see, however, where they are going with this. Corinne Le Quéré, a

professor at the Tyndall Centre for Climate Change Research,

University of East Anglia, was lead author of the climate lockdown

study, and she writes for … the World Economic Forum. Gates calls

in ‘his’ book for changing ‘every aspect of the economy’ (long-time

Cult agenda) and for humans to eat synthetic ‘meat’ (predicted in

my books) while cows and other farm animals are eliminated.

Australian TV host and commentator Alan Jones described what

carbon emission targets would mean for farm animals in Australia

alone if emissions were reduced as demanded by 35 percent by 2030

and zero by 2050:

Well, let’s take agriculture, the total emissions from agriculture are about 75 million tonnes of carbon dioxide, equivalent. Now reduce that by 35 percent and you have to come down to 50 million tonnes, I’ve done the maths. So if you take for example 1.5 million cows, you’re going to have to reduce the herd by 525,000 [by] 2030, nine years, that’s 58,000 cows a year. The beef herd’s 30 million, reduce that by 35 percent, that’s 10.5 million, which means 1.2 million cattle have to go every year between now and 2030. This is insanity!

There are 75 million sheep. Reduce that by 35 percent, that’s 26 million sheep, that’s almost 3 million a year. So under the Paris Agreement over 30 million beasts. dairy cows, cattle, pigs and sheep would go. More than 8,000 every minute of every hour for the next decade, do these people know what they’re talking about?

Clearly they don’t at the level of campaigners, politicians and

administrators. The Cult does know; that’s the outcome it wants. We

are faced with not just a war on humanity. Animals and the natural

world are being targeted and I have been saying since the ‘Covid’

hoax began that the plan eventually was to claim that the ‘deadly

virus’ is able to jump from animals, including farm animals and

domestic pets, to humans. Just before this book went into production

came this story: ‘Russia registers world’s first Covid-19 vaccine for

cats & dogs as makers of Sputnik V warn pets & farm animals could

spread virus’. The report said ‘top scientists warned that the deadly

pathogen could soon begin spreading through homes and farms’

and ‘the next stage is the infection of farm and domestic animals’.

Know the outcome and you’ll see the journey. Think what that

would mean for animals and keep your eye on a term called

zoonosis or zoonotic diseases which transmit between animals and

humans. The Cult wants to break the connection between animals

and people as it does between people and people. Farm animals fit

with the Cult agenda to transform food from natural to synthetic.

The gas of life is killing us

There can be few greater examples of Cult inversion than the

condemnation of carbon dioxide as a dangerous pollutant when it is

the gas of life. Without it the natural world would be dead and so we

would all be dead. We breathe in oxygen and breathe out carbon

dioxide while plants produce oxygen and absorb carbon dioxide. It

is a perfect symbiotic relationship that the Cult wants to dismantle

for reasons I will come to in the final two chapters. Gates, Schwab,

other Cult operatives and mindless repeaters, want the world to be

‘carbon neutral’ by at least 2050 and the earlier the be�er. ‘Zero

carbon’ is the cry echoed by lunatics calling for ‘Zero Covid’ when

we already have it. These carbon emission targets will

deindustrialise the world in accordance with Cult plans – the post-

industrial, post-democratic society – and with so-called renewables

like solar and wind not coming even close to meeting human energy

needs blackouts and cold are inevitable. Texans got the picture in the

winter of 2021 when a snow storm stopped wind turbines and solar

panels from working and the lights went down along with water

which relies on electricity for its supply system. Gates wants

everything to be powered by electricity to ensure that his masters

have the kill switch to stop all human activity, movement, cooking,

water and warmth any time they like. The climate lie is so

stupendously inverted that it claims we must urgently reduce

carbon dioxide when we don’t have enough.

Co2 in the atmosphere is a li�le above 400 parts per million when

the optimum for plant growth is 2,000 ppm and when it falls

anywhere near 150 ppm the natural world starts to die and so do we.

It fell to as low as 280 ppm in an 1880 measurement in Hawaii and

rose to 413 ppm in 2019 with industrialisation which is why the

planet has become greener in the industrial period. How insane then

that psychopathic madman Gates is not satisfied only with blocking

the rise of Co2. He’s funding technology to suck it out of the

atmosphere. The reason why will become clear. The industrial era is

not destroying the world through Co2 and has instead turned

around a potentially disastrous ongoing fall in Co2. Greenpeace co-

founder and scientist Patrick Moore walked away from Greenpeace

in 1986 and has exposed the green movement for fear-mongering

and lies. He said that 500 million years ago there was 17 times more

Co2 in the atmosphere than we have today and levels have been

falling for hundreds of millions of years. In the last 150 million years

Co2 levels in Earth’s atmosphere had reduced by 90 percent. Moore

said that by the time humanity began to unlock carbon dioxide from

fossil fuels we were at ‘38 seconds to midnight’ and in that sense:

‘Humans are [the Earth’s] salvation.’ Moore made the point that only

half the Co2 emi�ed by fossil fuels stays in the atmosphere and we

should remember that all pollution pouring from chimneys that we

are told is carbon dioxide is in fact nothing of the kind. It’s pollution.

Carbon dioxide is an invisible gas.

William Happer, Professor of Physics at Princeton University and

long-time government adviser on climate, has emphasised the Co2

deficiency for maximum growth and food production. Greenhouse

growers don’t add carbon dioxide for a bit of fun. He said that most

of the warming in the last 100 years, a�er the earth emerged from

the super-cold period of the ‘Li�le Ice Age’ into a natural warming

cycle, was over by 1940. Happer said that a peak year for warming in

1988 can be explained by a ‘monster El Nino’ which is a natural and

cyclical warming of the Pacific that has nothing to do with ‘climate

change’. He said the effect of Co2 could be compared to painting a

wall with red paint in that once two or three coats have been applied

it didn’t ma�er how much more you slapped on because the wall

will not get much redder. Almost all the effect of the rise in Co2 has

already happened, he said, and the volume in the atmosphere would

now have to double to increase temperature by a single degree.

Climate hoaxers know this and they have invented the most

ridiculously complicated series of ‘feedback’ loops to try to

overcome this rather devastating fact. You hear puppet Greta going

on cluelessly about feedback loops and this is why.

The Sun affects temperature? No you climate denier

Some other nonsense to contemplate: Climate graphs show that rises

in temperature do not follow rises in Co2 – it’s the other way round

with a lag between the two of some 800 years. If we go back 800

years from present time we hit the Medieval Warm Period when

temperatures were higher than now without any industrialisation

and this was followed by the Li�le Ice Age when temperatures

plummeted. The world was still emerging from these centuries of

serious cold when many climate records began which makes the

ever-repeated line of the ‘ho�est year since records began’

meaningless when you are not comparing like with like. The coldest

period of the Li�le Ice Age corresponded with the lowest period of

sunspot activity when the Sun was at its least active. Proper

scientists will not be at all surprised by this when it confirms the

obvious fact that earth temperature is affected by the scale of Sun

activity and the energetic power that it subsequently emits; but

when is the last time you heard a climate hoaxer talking about the

Sun as a source of earth temperature?? Everything has to be focussed

on Co2 which makes up just 0.117 percent of so-called greenhouse

gases and only a fraction of even that is generated by human activity.

The rest is natural. More than 90 percent of those greenhouse gases

are water vapour and clouds (Fig 9). Ban moisture I say. Have you

noticed that the climate hoaxers no longer use the polar bear as their

promotion image? That’s because far from becoming extinct polar

bear communities are stable or thriving. Joe Bastardi, American

meteorologist, weather forecaster and outspoken critic of the climate

lie, documents in his book The Climate Chronicles how weather

pa�erns and events claimed to be evidence of climate change have

been happening since long before industrialisation: ‘What happened

before naturally is happening again, as is to be expected given the

cyclical nature of the climate due to the design of the planet.’ If you

read the detailed background to the climate hoax in my other books

you will shake your head and wonder how anyone could believe the

crap which has spawned a multi-trillion dollar industry based on

absolute garbage (see HIV causes AIDs and Sars-Cov-2 causes

‘Covid-19’). Climate and ‘Covid’ have much in common given they

have the same source. They both have the contradictory everything

factor in which everything is explained by reference to them. It’s hot

– ‘it’s climate change’. It’s cold – ‘it’s climate change’. I got a sniffle –

‘it’s Covid’. I haven’t got a sniffle – ‘it’s Covid’. Not having a sniffle

has to be a symptom of ‘Covid’. Everything is and not having a

sniffle is especially dangerous if you are a slow walker. For sheer

audacity I offer you a Cambridge University ‘study’ that actually

linked ‘Covid’ to ‘climate change’. It had to happen eventually. They

concluded that climate change played a role in ‘Covid-19’ spreading

from animals to humans because … wait for it … I kid you not … the

two groups were forced closer together as populations grow. Er, that’s it.

The whole foundation on which this depended was that ‘Bats are the

likely zoonotic origin of SARS-CoV-1 and SARS-CoV-2’. Well, they

are not. They are nothing to do with it. Apart from bats not being the

origin and therefore ‘climate change’ effects on bats being irrelevant

I am in awe of their academic insight. Where would we be without

them? Not where we are that’s for sure.

Figure 9: The idea that the gas of life is disastrously changing the climate is an insult to brain cell activity.

One other point about the weather is that climate modification is

now well advanced and not every major weather event is natural –

or earthquake come to that. I cover this subject at some length in

other books. China is openly planning a rapid expansion of its

weather modification programme which includes changing the

climate in an area more than one and a half times the size of India.

China used weather manipulation to ensure clear skies during the

2008 Olympics in Beijing. I have quoted from US military documents

detailing how to employ weather manipulation as a weapon of war

and they did that in the 1960s and 70s during the conflict in Vietnam

with Operation Popeye manipulating monsoon rains for military

purposes. Why would there be international treaties on weather

modification if it wasn’t possible? Of course it is. Weather is

energetic information and it can be changed.

How was the climate hoax pulled off? See ‘Covid’

If you can get billions to believe in a ‘virus’ that doesn’t exist you can

get them to believe in human-caused climate change that doesn’t

exist. Both are being used by the Cult to transform global society in

the way it has long planned. Both hoaxes have been achieved in

pre�y much the same way. First you declare a lie is a fact. There’s a

‘virus’ you call SARS-Cov-2 or humans are warming the planet with

their behaviour. Next this becomes, via Cult networks, the

foundation of government, academic and science policy and belief.

Those who parrot the mantra are given big grants to produce

research that confirms the narrative is true and ever more

‘symptoms’ are added to make the ‘virus’/’climate change’ sound

even more scary. Scientists and researchers who challenge the

narrative have their grants withdrawn and their careers destroyed.

The media promote the lie as the unquestionable truth and censor

those with an alternative view or evidence. A great percentage of the

population believe what they are told as the lie becomes an

everybody-knows-that and the believing-masses turn on those with

a mind of their own. The technique has been used endlessly

throughout human history. Wokers are the biggest promotors of the

climate lie and ‘Covid’ fascism because their minds are owned by the

Cult; their sense of self-righteous self-purity knows no bounds; and

they exist in a bubble of reality in which facts are irrelevant and only

get in the way of looking without seeing.

Running through all of this like veins in a blue cheese is control of

information, which means control of perception, which means

control of behaviour, which collectively means control of human

society. The Cult owns the global media and Silicon Valley fascists

for the simple reason that it has to. Without control of information it

can’t control perception and through that human society. Examine

every facet of the Cult agenda and you will see that anything

supporting its introduction is never censored while anything

pushing back is always censored. I say again: Psychopaths that know

why they are doing this must go before Nuremberg trials and those

that follow their orders must trot along behind them into the same

dock. ‘I was just following orders’ didn’t work the first time and it

must not work now. Nuremberg trials must be held all over the

world before public juries for politicians, government officials,

police, compliant doctors, scientists and virologists, and all Cult

operatives such as Gates, Tedros, Fauci, Vallance, Whi�y, Ferguson,

Zuckerberg, Wojcicki, Brin, Page, Dorsey, the whole damn lot of

them – including, no especially, the psychopath psychologists.

Without them and the brainless, gutless excuses for journalists that

have repeated their lies, none of this could be happening. Nobody

can be allowed to escape justice for the psychological and economic

Armageddon they are all responsible for visiting upon the human

race.

As for the compliant, unquestioning, swathes of humanity, and the

self-obsessed, all-knowing ignorance of the Wokers … don’t start me.

God help their kids. God help their grandkids. God help them.

I

CHAPTER NINE

We must have it? So what is it?

Well I won’t back down. No, I won’t back down. You can stand me

up at the Gates of Hell. But I won’t back down

Tom Petty

will now focus on the genetically-manipulating ‘Covid vaccines’

which do not meet this official definition of a vaccine by the US

Centers for Disease Control (CDC): ‘A product that stimulates a

person’s immune system to produce immunity to a specific disease,

protecting the person from that disease.’ On that basis ‘Covid

vaccines’ are not a vaccine in that the makers don’t even claim they

stop infection or transmission.

They are instead part of a multi-levelled conspiracy to change the

nature of the human body and what it means to be ‘human’ and to

depopulate an enormous swathe of humanity. What I shall call

Human 1.0 is on the cusp of becoming Human 2.0 and for very

sinister reasons. Before I get to the ‘Covid vaccine’ in detail here’s

some background to vaccines in general. Government regulators do

not test vaccines – the makers do – and the makers control which

data is revealed and which isn’t. Children in America are given 50

vaccine doses by age six and 69 by age 19 and the effect of the whole

combined schedule has never been tested. Autoimmune diseases

when the immune system a�acks its own body have soared in the

mass vaccine era and so has disease in general in children and the

young. Why wouldn’t this be the case when vaccines target the

immune system? The US government gave Big Pharma drug

companies immunity from prosecution for vaccine death and injury

in the 1986 National Childhood Vaccine Injury Act (NCVIA) and

since then the government (taxpayer) has been funding

compensation for the consequences of Big Pharma vaccines. The

criminal and satanic drug giants can’t lose and the vaccine schedule

has increased dramatically since 1986 for this reason. There is no

incentive to make vaccines safe and a big incentive to make money

by introducing ever more. Even against a ridiculously high bar to

prove vaccine liability, and with the government controlling the

hearing in which it is being challenged for compensation, the vaccine

court has so far paid out more than $4 billion. These are the vaccines

we are told are safe and psychopaths like Zuckerberg censor posts

saying otherwise. The immunity law was even justified by a ruling

that vaccines by their nature were ‘unavoidably unsafe’.

Check out the ingredients of vaccines and you will be shocked if

you are new to this. They put that in children’s bodies?? What?? Try

aluminium, a brain toxin connected to dementia, aborted foetal

tissue and formaldehyde which is used to embalm corpses. World-

renowned aluminium expert Christopher Exley had his research into

the health effect of aluminium in vaccines shut down by Keele

University in the UK when it began taking funding from the Bill and

Melinda Gates Foundation. Research when diseases ‘eradicated’ by

vaccines began to decline and you will find the fall began long before

the vaccine was introduced. Sometimes the fall even plateaued a�er

the vaccine. Diseases like scarlet fever for which there was no

vaccine declined in the same way because of environmental and

other factors. A perfect case in point is the polio vaccine. Polio began

when lead arsenate was first sprayed as an insecticide and residues

remained in food products. Spraying started in 1892 and the first US

polio epidemic came in Vermont in 1894. The simple answer was to

stop spraying, but Rockefeller-created Big Pharma had a be�er idea.

Polio was decreed to be caused by the poliovirus which ‘spreads from

person to person and can infect a person’s spinal cord’. Lead

arsenate was replaced by the lethal DDT which had the same effect

of causing paralysis by damaging the brain and central nervous

system. Polio plummeted when DDT was reduced and then banned,

but the vaccine is still given the credit for something it didn’t do.

Today by far the biggest cause of polio is the vaccines promoted by

Bill Gates. Vaccine justice campaigner Robert Kennedy Jr, son of

assassinated (by the Cult) US A�orney General Robert Kennedy,

wrote:

In 2017, the World Health Organization (WHO) reluctantly admitted that the global explosion in polio is predominantly vaccine strain. The most frightening epidemics in Congo, Afghanistan, and the Philippines, are all linked to vaccines. In fact, by 2018, 70% of global polio cases were vaccine strain.

Vaccines make fortunes for Cult-owned Gates and Big Pharma

while undermining the health and immune systems of the

population. We had a glimpse of the mentality behind the Big

Pharma cartel with a report on WION (World is One News), an

international English language TV station based in India, which

exposed the extraordinary behaviour of US drug company Pfizer

over its ‘Covid vaccine’. The WION report told how Pfizer had made

fantastic demands of Argentina, Brazil and other countries in return

for its ‘vaccine’. These included immunity from prosecution, even

for Pfizer negligence, government insurance to protect Pfizer from

law suits and handing over as collateral sovereign assets of the

country to include Argentina’s bank reserves, military bases and

embassy buildings. Pfizer demanded the same of Brazil in the form

of waiving sovereignty of its assets abroad; exempting Pfizer from

Brazilian laws; and giving Pfizer immunity from all civil liability.

This is a ‘vaccine’ developed with government funding. Big Pharma

is evil incarnate as a creation of the Cult and all must be handed

tickets to Nuremberg.

Phantom ‘vaccine’ for a phantom ‘disease’

I’ll expose the ‘Covid vaccine’ fraud and then go on to the wider

background of why the Cult has set out to ‘vaccinate’ every man,

woman and child on the planet for an alleged ‘new disease’ with a

survival rate of 99.77 percent (or more) even by the grotesquely-

manipulated figures of the World Health Organization and Johns

Hopkins University. The ‘infection’ to ‘death’ ratio is 0.23 to 0.15

percent according to Stanford epidemiologist Dr John Ioannidis and

while estimates vary the danger remains tiny. I say that if the truth

be told the fake infection to fake death ratio is zero. Never mind all

the evidence I have presented here and in The Answer that there is no

‘virus’ let us just focus for a moment on that death-rate figure of say

0.23 percent. The figure includes all those worldwide who have

tested positive with a test not testing for the ‘virus’ and then died

within 28 days or even longer of any other cause – any other cause.

Now subtract all those illusory ‘Covid’ deaths on the global data

sheets from the 0.23 percent. What do you think you would be le�

with? Zero. A vaccination has never been successfully developed for

a so-called coronavirus. They have all failed at the animal testing

stage when they caused hypersensitivity to what they were claiming

to protect against and made the impact of a disease far worse. Cult-

owned vaccine corporations got around that problem this time by

bypassing animal trials, going straight to humans and making the

length of the ‘trials’ before the public rollout as short as they could

get away with. Normally it takes five to ten years or more to develop

vaccines that still cause demonstrable harm to many people and

that’s without including the long-term effects that are never officially

connected to the vaccination. ‘Covid’ non-vaccines have been

officially produced and approved in a ma�er of months from a

standing start and part of the reason is that (a) they were developed

before the ‘Covid’ hoax began and (b) they are based on computer

programs and not natural sources. Official non-trials were so short

that government agencies gave emergency, not full, approval. ‘Trials’

were not even completed and full approval cannot be secured until

they are. Public ‘Covid vaccination’ is actually a continuation of the

trial. Drug company ‘trials’ are not scheduled to end until 2023 by

which time a lot of people are going to be dead. Data on which

government agencies gave this emergency approval was supplied by

the Big Pharma corporations themselves in the form of

Pfizer/BioNTech, AstraZeneca, Moderna, Johnson & Johnson, and

others, and this is the case with all vaccines. By its very nature

emergency approval means drug companies do not have to prove that

the ‘vaccine’ is ‘safe and effective’. How could they with trials way

short of complete? Government regulators only have to believe that

they could be safe and effective. It is criminal manipulation to get

products in circulation with no testing worth the name. Agencies

giving that approval are infested with Big Pharma-connected place-

people and they act in the interests of Big Pharma (the Cult) and not

the public about whom they do not give a damn.

More human lab rats

‘Covid vaccines’ produced in record time by Pfizer/BioNTech and

Moderna employ a technique never approved before for use on humans.

They are known as mRNA ‘vaccines’ and inject a synthetic version of

‘viral’ mRNA or ‘messenger RNA’. The key is in the term

‘messenger’. The body works, or doesn’t, on the basis of information

messaging. Communications are constantly passing between and

within the genetic system and the brain. Change those messages and

you change the state of the body and even its very nature and you

can change psychology and behaviour by the way the brain

processes information. I think you are going to see significant

changes in personality and perception of many people who have had

the ‘Covid vaccine’ synthetic potions. Insider Aldous Huxley

predicted the following in 1961 and mRNA ‘vaccines’ can be

included in the term ‘pharmacological methods’:

There will be, in the next generation or so, a pharmacological method of making people love their servitude, and producing dictatorship without tears, so to speak, producing a kind of painless concentration camp for entire societies, so that people will in fact have their own liberties taken away from them, but rather enjoy it, because they will be distracted from any desire to rebel by propaganda or brainwashing, or brainwashing enhanced by pharmacological methods. And this seems to be the final revolution.

Apologists claim that mRNA synthetic ‘vaccines’ don’t change the

DNA genetic blueprint because RNA does not affect DNA only the

other way round. This is so disingenuous. A process called ‘reverse

transcription’ can convert RNA into DNA and be integrated into

DNA in the cell nucleus. This was highlighted in December, 2020, by

scientists at Harvard and Massachuse�s Institute of Technology

(MIT). Geneticists report that more than 40 percent of mammalian

genomes results from reverse transcription. On the most basic level

if messaging changes then that sequence must lead to changes in

DNA which is receiving and transmi�ing those communications.

How can introducing synthetic material into cells not change the

cells where DNA is located? The process is known as transfection

which is defined as ‘a technique to insert foreign nucleic acid (DNA

or RNA) into a cell, typically with the intention of altering the

properties of the cell’. Researchers at the Sloan Ke�ering Institute in

New York found that changes in messenger RNA can deactivate

tumour-suppressing proteins and thereby promote cancer. This is

what happens when you mess with messaging. ‘Covid vaccine’

maker Moderna was founded in 2010 by Canadian stem cell

biologist Derrick J. Rossi a�er his breakthrough discovery in the field

of transforming and reprogramming stem cells. These are neutral

cells that can be programmed to become any cell including sperm

cells. Moderna was therefore founded on the principle of genetic

manipulation and has never produced any vaccine or drug before its

genetically-manipulating synthetic ‘Covid’ shite. Look at the name –

Mode-RNA or Modify-RNA. Another important point is that the US

Supreme Court has ruled that genetically-modified DNA, or

complementary DNA (cDNA) synthesized in the laboratory from

messenger RNA, can be patented and owned. These psychopaths are

doing this to the human body.

Cells replicate synthetic mRNA in the ‘Covid vaccines’ and in

theory the body is tricked into making antigens which trigger

antibodies to target the ‘virus spike proteins’ which as Dr Tom

Cowan said have never been seen. Cut the crap and these ‘vaccines’

deliver self-replicating synthetic material to the cells with the effect of

changing human DNA. The more of them you have the more that

process is compounded while synthetic material is all the time self-

replicating. ‘Vaccine’-maker Moderna describes mRNA as ‘like

so�ware for the cell’ and so they are messing with the body’s

so�ware. What happens when you change the so�ware in a

computer? Everything changes. For this reason the Cult is preparing

a production line of mRNA ‘Covid vaccines’ and a long list of

excuses to use them as with all the ‘variants’ of a ‘virus’ never shown

to exist. The plan is further to transfer the mRNA technique to other

vaccines mostly given to children and young people. The cumulative

consequences will be a transformation of human DNA through a

constant infusion of synthetic genetic material which will kill many

and change the rest. Now consider that governments that have given

emergency approval for a vaccine that’s not a vaccine; never been

approved for humans before; had no testing worth the name; and

the makers have been given immunity from prosecution for any

deaths or adverse effects suffered by the public. The UK government

awarded permanent legal indemnity to itself and its employees for

harm done when a patient is being treated for ‘Covid-19’ or

‘suspected Covid-19’. That is quite a thought when these are possible

‘side-effects’ from the ‘vaccine’ (they are not ‘side’, they are effects)

listed by the US Food and Drug Administration:

Guillain-Barre syndrome; acute disseminated encephalomyelitis;

transverse myelitis; encephalitis; myelitis; encephalomyelitis;

meningoencephalitis; meningitis; encephalopathy; convulsions;

seizures; stroke; narcolepsy; cataplexy; anaphylaxis; acute

myocardial infarction (heart a�ack); myocarditis; pericarditis;

autoimmune disease; death; implications for pregnancy, and birth

outcomes; other acute demyelinating diseases; non anaphylactic

allergy reactions; thrombocytopenia ; disseminated intravascular

coagulation; venous thromboembolism; arthritis; arthralgia; joint

pain; Kawasaki disease; multisystem inflammatory syndrome in

children; vaccine enhanced disease. The la�er is the way the

‘vaccine’ has the potential to make diseases far worse than they

would otherwise be.

UK doctor and freedom campaigner Vernon Coleman described

the conditions in this list as ‘all unpleasant, most of them very

serious, and you can’t get more serious than death’. The thought that

anyone at all has had the ‘vaccine’ in these circumstances is

testament to the potential that humanity has for clueless,

unquestioning, stupidity and for many that programmed stupidity

has already been terminal.

An insider speaks

Dr Michael Yeadon is a former Vice President, head of research and

Chief Scientific Adviser at vaccine giant Pfizer. Yeadon worked on

the inside of Big Pharma, but that did not stop him becoming a vocal

critic of ‘Covid vaccines’ and their potential for multiple harms,

including infertility in women. By the spring of 2021 he went much

further and even used the no, no, term ‘conspiracy’. When you begin

to see what is going on it is impossible not to do so. Yeadon spoke

out in an interview with freedom campaigner James Delingpole and

I mentioned earlier how he said that no one had samples of ‘the

virus’. He explained that the mRNA technique originated in the anti-

cancer field and ways to turn on and off certain genes which could

be advantageous if you wanted to stop cancer growing out of

control. ‘That’s the origin of them. They are a very unusual

application, really.’ Yeadon said that treating a cancer patient with

an aggressive procedure might be understandable if the alternative

was dying, but it was quite another thing to use the same technique

as a public health measure. Most people involved wouldn’t catch the

infectious agent you were vaccinating against and if they did they

probably wouldn’t die:

If you are really using it as a public health measure you really want to as close as you can get to zero sides-effects … I find it odd that they chose techniques that were really cutting their teeth in the field of oncology and I’m worried that in using gene-based vaccines that have to be injected in the body and spread around the body, get taken up into some cells, and the regulators haven’t quite told us which cells they get taken up into … you are going to be generating a wide range of responses … with multiple steps each of which could go well or badly.

I doubt the Cult intends it to go well. Yeadon said that you can put

any gene you like into the body through the ‘vaccine’. ‘You can

certainly give them a gene that would do them some harm if you

wanted.’ I was intrigued when he said that when used in the cancer

field the technique could turn genes on and off. I explore this process

in The Answer and with different genes having different functions

you could create mayhem – physically and psychologically – if you

turned the wrong ones on and the right ones off. I read reports of an

experiment by researchers at the University of Washington’s school

of computer science and engineering in which they encoded DNA to

infect computers. The body is itself a biological computer and if

human DNA can inflict damage on a computer why can’t the

computer via synthetic material mess with the human body? It can.

The Washington research team said it was possible to insert

malicious malware into ‘physical DNA strands’ and corrupt the

computer system of a gene sequencing machine as it ‘reads gene

le�ers and stores them as binary digits 0 and 1’. They concluded that

hackers could one day use blood or spit samples to access computer

systems and obtain sensitive data from police forensics labs or infect

genome files. It is at this level of digital interaction that synthetic

‘vaccines’ need to be seen to get the full picture and that will become

very clear later on. Michael Yeadon said it made no sense to give the

‘vaccine’ to younger people who were in no danger from the ‘virus’.

What was the benefit? It was all downside with potential effects:

The fact that my government in what I thought was a civilised, rational country, is raining [the ‘vaccine’] on people in their 30s and 40s, even my children in their 20s, they’re getting letters and phone calls, I know this is not right and any of you doctors who are vaccinating you know it’s not right, too. They are not at risk. They are not at risk from the disease, so you are now hoping that the side-effects are so rare that you get away with it. You don’t give new technology … that you don’t understand to 100 percent of the population.

Blood clot problems with the AstraZeneca ‘vaccine’ have been

affecting younger people to emphasise the downside risks with no

benefit. AstraZeneca’s version, produced with Oxford University,

does not use mRNA, but still gets its toxic cocktail inside cells where

it targets DNA. The Johnson & Johnson ‘vaccine’ which uses a

similar technique has also produced blood clot effects to such an

extent that the United States paused its use at one point. They are all

‘gene therapy’ (cell modification) procedures and not ‘vaccines’. The

truth is that once the content of these injections enter cells we have

no idea what the effect will be. People can speculate and some can

give very educated opinions and that’s good. In the end, though,

only the makers know what their potions are designed to do and

even they won’t know every last consequence. Michael Yeadon was

scathing about doctors doing what they knew to be wrong.

‘Everyone’s mute’, he said. Doctors in the NHS must know this was

not right, coming into work and injecting people. ‘I don’t know how

they sleep at night. I know I couldn’t do it. I know that if I were in

that position I’d have to quit.’ He said he knew enough about

toxicology to know this was not a good risk-benefit. Yeadon had

spoken to seven or eight university professors and all except two

would not speak out publicly. Their universities had a policy that no

one said anything that countered the government and its medical

advisors. They were afraid of losing their government grants. This is

how intimidation has been used to silence the truth at every level of

the system. I say silence, but these people could still speak out if they

made that choice. Yeadon called them ‘moral cowards’ – ‘This is

about your children and grandchildren’s lives and you have just

buggered off and le� it.’

‘Variant’ nonsense

Some of his most powerful comments related to the alleged

‘variants’ being used to instil more fear, justify more lockdowns, and

introduce more ‘vaccines’. He said government claims about

‘variants’ were nonsense. He had checked the alleged variant ‘codes’

and they were 99.7 percent identical to the ‘original’. This was the

human identity difference equivalent to pu�ing a baseball cap on

and off or wearing it the other way round. A 0.3 percent difference

would make it impossible for that ‘variant’ to escape immunity from

the ‘original’. This made no sense of having new ‘vaccines’ for

‘variants’. He said there would have to be at least a 30 percent

difference for that to be justified and even then he believed the

immune system would still recognise what it was. Gates-funded

‘variant modeller’ and ‘vaccine’-pusher John Edmunds might care to

comment. Yeadon said drug companies were making new versions

of the ‘vaccine’ as a ‘top up’ for ‘variants’. Worse than that, he said,

the ‘regulators’ around the world like the MHRA in the UK had got

together and agreed that because ‘vaccines’ for ‘variants’ were so

similar to the first ‘vaccines’ they did not have to do safety studies. How

transparently sinister that is. This is when Yeadon said: ‘There is a

conspiracy here.’ There was no need for another vaccine for

‘variants’ and yet we were told that there was and the country had

shut its borders because of them. ‘They are going into hundreds of

millions of arms without passing ‘go’ or any regulator. Why did they

do that? Why did they pick this method of making the vaccine?’

The reason had to be something bigger than that it seemed and

‘it’s not protection against the virus’. It’s was a far bigger project that

meant politicians and advisers were willing to do things and not do

things that knowingly resulted in avoidable deaths – ‘that’s already

happened when you think about lockdown and deprivation of

health care for a year.’ He spoke of people prepared to do something

that results in the avoidable death of their fellow human beings and

it not bother them. This is the penny-drop I have been working to

get across for more than 30 years – the level of pure evil we are

dealing with. Yeadon said his friends and associates could not

believe there could be that much evil, but he reminded them of

Stalin, Pol Pot and Hitler and of what Stalin had said: ‘One death is a

tragedy. A million? A statistic.’ He could not think of a benign

explanation for why you need top-up vaccines ‘which I’m sure you

don’t’ and for the regulators ‘to just get out of the way and wave

them through’. Why would the regulators do that when they were

still wrestling with the dangers of the ‘parent’ vaccine? He was

clearly shocked by what he had seen since the ‘Covid’ hoax began

and now he was thinking the previously unthinkable:

If you wanted to depopulate a significant proportion of the world and to do it in a way that doesn’t involve destruction of the environment with nuclear weapons, poisoning everyone with anthrax or something like that, and you wanted plausible deniability while you had a multi-year infectious disease crisis, I actually don’t think you could come up with a better plan of work than seems to be in front of me. I can’t say that’s what they are going to do, but I can’t think of a benign explanation why they are doing it.

He said he never thought that they would get rid of 99 percent of

humans, but now he wondered. ‘If you wanted to that this would be

a hell of a way to do it – it would be unstoppable folks.’ Yeadon had

concluded that those who submi�ed to the ‘vaccine’ would be

allowed to have some kind of normal life (but for how long?) while

screws were tightened to coerce and mandate the last few percent. ‘I

think they’ll put the rest of them in a prison camp. I wish I was

wrong, but I don’t think I am.’ Other points he made included: There

were no coronavirus vaccines then suddenly they all come along at

the same time; we have no idea of the long term affect with trials so

short; coercing or forcing people to have medical procedures is

against the Nuremberg Code instigated when the Nazis did just that;

people should at least delay having the ‘vaccine’; a quick Internet

search confirms that masks don’t reduce respiratory viral

transmission and ‘the government knows that’; they have smashed

civil society and they know that, too; two dozen peer-reviewed

studies show no connection between lockdown and reducing deaths;

he knew from personal friends the elite were still flying around and

going on holiday while the public were locked down; the elite were

not having the ‘vaccines’. He was also asked if ‘vaccines’ could be

made to target difference races. He said he didn’t know, but the

document by the Project for the New American Century in

September, 2000, said developing ‘advanced forms of biological

warfare that can target specific genotypes may transform biological

warfare from the realm of terror to a politically useful tool.’ Oh,

they’re evil all right. Of that we can be absolutely sure.

Another cull of old people

We have seen from the CDC definition that the mRNA ‘Covid

vaccine’ is not a vaccine and nor are the others that claim to reduce

‘severity of symptoms’ in some people, but not protect from infection

or transmission. What about all the lies about returning to ‘normal’ if

people were ‘vaccinated’? If they are not claimed to stop infection

and transmission of the alleged ‘virus’, how does anything change?

This was all lies to manipulate people to take the jabs and we are

seeing that now with masks and distancing still required for the

‘vaccinated’. How did they think that elderly people with fragile

health and immune responses were going to be affected by infusing

their cells with synthetic material and other toxic substances? They

knew that in the short and long term it would be devastating and

fatal as the culling of the old that began with the first lockdowns was

continued with the ‘vaccine’. Death rates in care homes soared

immediately residents began to be ‘vaccinated’ – infused with

synthetic material. Brave and commi�ed whistleblower nurses put

their careers at risk by exposing this truth while the rest kept their

heads down and their mouths shut to put their careers before those

they are supposed to care for. A long-time American Certified

Nursing Assistant who gave his name as James posted a video in

which he described emotionally what happened in his care home

when vaccination began. He said that during 2020 very few residents

were sick with ‘Covid’ and no one died during the entire year; but

shortly a�er the Pfizer mRNA injections 14 people died within two

weeks and many others were near death. ‘They’re dropping like

flies’, he said. Residents who walked on their own before the shot

could no longer and they had lost their ability to conduct an

intelligent conversation. The home’s management said the sudden

deaths were caused by a ‘super-spreader’ of ‘Covid-19’. Then how

come, James asked, that residents who refused to take the injections

were not sick? It was a case of inject the elderly with mRNA

synthetic potions and blame their illness and death that followed on

the ‘virus’. James described what was happening in care homes as

‘the greatest crime of genocide this country has ever seen’.

Remember the NHS staff nurse from earlier who used the same

word ‘genocide’ for what was happening with the ‘vaccines’ and

that it was an ‘act of human annihilation’. A UK care home

whistleblower told a similar story to James about the effect of the

‘vaccine’ in deaths and ‘outbreaks’ of illness dubbed ‘Covid’ a�er

ge�ing the jab. She told how her care home management and staff

had zealously imposed government regulations and no one was

allowed to even question the official narrative let alone speak out

against it. She said the NHS was even worse. Again we see the

results of reframing. A worker at a local care home where I live said

they had not had a single case of ‘Covid’ there for almost a year and

when the residents were ‘vaccinated’ they had 19 positive cases in

two weeks with eight dying.

It’s not the ‘vaccine’ – honest

The obvious cause and effect was being ignored by the media and

most of the public. Australia’s health minister Greg Hunt (a former

head of strategy at the World Economic Forum) was admi�ed to

hospital a�er he had the ‘vaccine’. He was suffering according to

reports from the skin infection ‘cellulitis’ and it must have been a

severe case to have warranted days in hospital. Immediately the

authorities said this was nothing to do with the ‘vaccine’ when an

effect of some vaccines is a ‘cellulitis-like reaction’. We had families

of perfectly healthy old people who died a�er the ‘vaccine’ saying

that if only they had been given the ‘vaccine’ earlier they would still

be alive. As a numbskull rating that is off the chart. A father of four

‘died of Covid’ at aged 48 when he was taken ill two days a�er

having the ‘vaccine’. The man, a health administrator, had been

‘shielding during the pandemic’ and had ‘not really le� the house’

until he went for the ‘vaccine’. Having the ‘vaccine’ and then falling

ill and dying does not seem to have qualified as a possible cause and

effect and ‘Covid-19’ went on his death certificate. His family said

they had no idea how he ‘caught the virus’. A family member said:

‘Tragically, it could be that going for a vaccination ultimately led to

him catching Covid …The sad truth is that they are never going to

know where it came from.’ The family warned people to remember

that the virus still existed and was ‘very real’. So was their stupidity.

Nurses and doctors who had the first round of the ‘vaccine’ were

collapsing, dying and ending up in a hospital bed while they or their

grieving relatives were saying they’d still have the ‘vaccine’ again

despite what happened. I kid you not. You mean if your husband

returned from the dead he’d have the same ‘vaccine’ again that killed

him??

Doctors at the VCU Medical Center in Richmond, Virginia, said

the Johnson & Johnson ‘vaccine’ was to blame for a man’s skin

peeling off. Patient Richard Terrell said: ‘It all just happened so fast.

My skin peeled off. It’s still coming off on my hands now.’ He said it

was stinging, burning and itching and when he bent his arms and

legs it was very painful with ‘the skin swollen and rubbing against

itself’. Pfizer/BioNTech and Moderna vaccines use mRNA to change

the cell while the Johnson & Johnson version uses DNA in a process

similar to AstraZeneca’s technique. Johnson & Johnson and

AstraZeneca have both had their ‘vaccines’ paused by many

countries a�er causing serious blood problems. Terrell’s doctor Fnu

Nutan said he could have died if he hadn’t got medical a�ention. It

sounds terrible so what did Nutan and Terrell say about the ‘vaccine’

now? Oh, they still recommend that people have it. A nurse in a

hospital bed 40 minutes a�er the vaccination and unable to swallow

due to throat swelling was told by a doctor that he lost mobility in

his arm for 36 hours following the vaccination. What did he say to

the ailing nurse? ‘Good for you for ge�ing the vaccination.’ We are

dealing with a serious form of cognitive dissonance madness in both

public and medical staff. There is a remarkable correlation between

those having the ‘vaccine’ and trumpeting the fact and suffering bad

happenings shortly a�erwards. Witold Rogiewicz, a Polish doctor,

made a video of his ‘vaccination’ and ridiculed those who were

questioning its safety and the intentions of Bill Gates: ‘Vaccinate

yourself to protect yourself, your loved ones, friends and also

patients. And to mention quickly I have info for anti-vaxxers and

anti-Coviders if you want to contact Bill Gates you can do this

through me.’ He further ridiculed the dangers of 5G. Days later he

was dead, but naturally the vaccination wasn’t mentioned in the

verdict of ‘heart a�ack’.

Lies, lies and more lies

So many members of the human race have slipped into extreme

states of insanity and unfortunately they include reframed doctors

and nursing staff. Having a ‘vaccine’ and dying within minutes or

hours is not considered a valid connection while death from any

cause within 28 days or longer of a positive test with a test not

testing for the ‘virus’ means ‘Covid-19’ goes on the death certificate.

How could that ‘vaccine’-death connection not have been made

except by calculated deceit? US figures in the initial rollout period to

February 12th, 2020, revealed that a third of the deaths reported to

the CDC a�er ‘Covid vaccines’ happened within 48 hours. Five men

in the UK suffered an ‘extremely rare’ blood clot problem a�er

having the AstraZeneca ‘vaccine’, but no causal link was established

said the Gates-funded Medicines and Healthcare products

Regulatory Agency (MHRA) which had given the ‘vaccine’

emergency approval to be used. Former Pfizer executive Dr Michael

Yeadon explained in his interview how the procedures could cause

blood coagulation and clots. People who should have been at no risk

were dying from blood clots in the brain and he said he had heard

from medical doctor friends that people were suffering from skin

bleeding and massive headaches. The AstraZeneca ‘shot’ was

stopped by some 20 countries over the blood clo�ing issue and still

the corrupt MHRA, the European Medicines Agency (EMA) and the

World Health Organization said that it should continue to be given

even though the EMA admi�ed that it ‘still cannot rule out

definitively’ a link between blood clo�ing and the ‘vaccine’. Later

Marco Cavaleri, head of EMA vaccine strategy, said there was indeed

a clear link between the ‘vaccine’ and thrombosis, but they didn’t

know why. So much for the trials showing the ‘vaccine’ is safe. Blood

clots were affecting younger people who would be under virtually

no danger from ‘Covid’ even if it existed which makes it all the more

stupid and sinister.

The British government responded to public alarm by wheeling

out June Raine, the terrifyingly weak infant school headmistress

sound-alike who heads the UK MHRA drug ‘regulator’. The idea

that she would stand up to Big Pharma and government pressure is

laughable and she told us that all was well in the same way that she

did when allowing untested, never-used-on-humans-before,

genetically-manipulating ‘vaccines’ to be exposed to the public in the

first place. Mass lying is the new normal of the ‘Covid’ era. The

MHRA later said 30 cases of rare blood clots had by then been

connected with the AstraZeneca ‘vaccine’ (that means a lot more in

reality) while stressing that the benefits of the jab in preventing

‘Covid-19’ outweighed any risks. A more ridiculous and

disingenuous statement with callous disregard for human health it is

hard to contemplate. Immediately a�er the mendacious ‘all-clears’

two hospital workers in Denmark experienced blood clots and

cerebral haemorrhaging following the AstraZeneca jab and one died.

Top Norwegian health official Pål Andre Holme said the ‘vaccine’

was the only common factor: ‘There is nothing in the patient history

of these individuals that can give such a powerful immune response

… I am confident that the antibodies that we have found are the

cause, and I see no other explanation than it being the vaccine which

triggers it.’ Strokes, a clot or bleed in the brain, were clearly

associated with the ‘vaccine’ from word of mouth and whistleblower

reports. Similar consequences followed with all these ‘vaccines’ that

we were told were so safe and as the numbers grew by the day it

was clear we were witnessing human carnage.

Learning the hard way

A woman interviewed by UKColumn told how her husband

suffered dramatic health effects a�er the vaccine when he’d been in

good health all his life. He went from being a li�le unwell to losing

all feeling in his legs and experiencing ‘excruciating pain’.

Misdiagnosis followed twice at Accident and Emergency (an

‘allergy’ and ‘sciatica’) before he was admi�ed to a neurology ward

where doctors said his serious condition had been caused by the

‘vaccine’. Another seven ‘vaccinated’ people were apparently being

treated on the same ward for similar symptoms. The woman said he

had the ‘vaccine’ because they believed media claims that it was safe.

‘I didn’t think the government would give out a vaccine that does

this to somebody; I believed they would be bringing out a

vaccination that would be safe.’ What a tragic way to learn that

lesson. Another woman posted that her husband was transporting

stroke patients to hospital on almost every shi� and when he asked

them if they had been ‘vaccinated’ for ‘Covid’ they all replied ‘yes’.

One had a ‘massive brain bleed’ the day a�er his second dose. She

said her husband reported the ‘just been vaccinated’ information

every time to doctors in A and E only for them to ignore it, make no

notes and appear annoyed that it was even mentioned. This

particular report cannot be verified, but it expresses a common

theme that confirms the monumental underreporting of ‘vaccine’

consequences. Interestingly as the ‘vaccines’ and their brain blood

clot/stroke consequences began to emerge the UK National Health

Service began a publicity campaign telling the public what to do in

the event of a stroke. A Sco�ish NHS staff nurse who quit in disgust

in March, 2021, said:

I have seen traumatic injuries from the vaccine, they’re not getting reported to the yellow card [adverse reaction] scheme, they’re treating the symptoms, not asking why, why it’s happening. It’s just treating the symptoms and when you speak about it you’re dismissed like you’re crazy, I’m not crazy, I’m not crazy because every other colleague I’ve spoken to is terrified to speak out, they’ve had enough.

Videos appeared on the Internet of people uncontrollably shaking

a�er the ‘vaccine’ with no control over muscles, limbs and even their

face. A Sco�ish mother broke out in a severe rash all over her body

almost immediately a�er she was given the AstraZeneca ‘vaccine’.

The pictures were horrific. Leigh King, a 41-year-old hairdresser

from Lanarkshire said: ‘Never in my life was I prepared for what I

was about to experience … My skin was so sore and constantly hot

… I have never felt pain like this …’ But don’t you worry, the

‘vaccine’ is perfectly safe. Then there has been the effect on medical

staff who have been pressured to have the ‘vaccine’ by psychopathic

‘health’ authorities and government. A London hospital consultant

who gave the name K. Polyakova wrote this to the British Medical

Journal or BMJ:

I am currently struggling with … the failure to report the reality of the morbidity caused by our current vaccination program within the health service and staff population. The levels of sickness after vaccination is unprecedented and staff are getting very sick and some with neurological symptoms which is having a huge impact on the health service function. Even the young and healthy are off for days, some for weeks, and some requiring medical treatment. Whole teams are being taken out as they went to get vaccinated together.

Mandatory vaccination in this instance is stupid, unethical and irresponsible when it comes to protecting our staff and public health. We are in the voluntary phase of vaccination, and encouraging staff to take an unlicensed product that is impacting on their immediate health … it is clearly stated that these vaccine products do not offer immunity or stop transmission. In which case why are we doing it?

Not to protect health that’s for sure. Medical workers are lauded by

governments for agenda reasons when they couldn’t give a toss

about them any more than they can for the population in general.

Schools across America faced the same situation as they closed due

to the high number of teachers and other staff with bad reactions to

the Pfizer/BioNTech, Moderna, and Johnson & Johnson ‘Covid

vaccines’ all of which were linked to death and serious adverse

effects. The BMJ took down the consultant’s comments pre�y

quickly on the grounds that they were being used to spread

‘disinformation’. They were exposing the truth about the ‘vaccine’

was the real reason. The cover-up is breathtaking.

Hiding the evidence

The scale of the ‘vaccine’ death cover-up worldwide can be

confirmed by comparing official figures with the personal experience

of the public. I heard of many people in my community who died

immediately or soon a�er the vaccine that would never appear in the

media or even likely on the official totals of ‘vaccine’ fatalities and

adverse reactions when only about ten percent are estimated to be

reported and I have seen some estimates as low as one percent in a

Harvard study. In the UK alone by April 29th, 2021, some 757,654

adverse reactions had been officially reported from the

Pfizer/BioNTech, Oxford/AstraZeneca and Moderna ‘vaccines’ with

more than a thousand deaths linked to jabs and that means an

estimated ten times this number in reality from a ten percent

reporting rate percentage. That’s seven million adverse reactions and

10,000 potential deaths and a one percent reporting rate would be

ten times those figures. In 1976 the US government pulled the swine

flu vaccine a�er 53 deaths. The UK data included a combined 10,000

eye disorders from the ‘Covid vaccines’ with more than 750 suffering

visual impairment or blindness and again multiply by the estimated

reporting percentages. As ‘Covid cases’ officially fell hospitals

virtually empty during the ‘Covid crisis’ began to fill up with a

range of other problems in the wake of the ‘vaccine’ rollout. The

numbers across America have also been catastrophic. Deaths linked

to all types of vaccine increased by 6,000 percent in the first quarter of

2021 compared with 2020. A 39-year-old woman from Ogden, Utah,

died four days a�er receiving a second dose of Moderna’s ‘Covid

vaccine’ when her liver, heart and kidneys all failed despite the fact

that she had no known medical issues or conditions. Her family

sought an autopsy, but Dr Erik Christensen, Utah’s chief medical

examiner, said proving vaccine injury as a cause of death almost

never happened. He could think of only one instance where an

autopsy would name a vaccine as the official cause of death and that

would be anaphylaxis where someone received a vaccine and died

almost instantaneously. ‘Short of that, it would be difficult for us to

definitively say this is the vaccine,’ Christensen said. If that is true

this must be added to the estimated ten percent (or far less)

reporting rate of vaccine deaths and serious reactions and the

conclusion can only be that vaccine deaths and serious reactions –

including these ‘Covid’ potions’ – are phenomenally understated in

official figures. The same story can be found everywhere. Endless

accounts of deaths and serious reactions among the public, medical

and care home staff while official figures did not even begin to

reflect this.

Professional script-reader Dr David Williams, a ‘top public-health

official’ in Ontario, Canada, insulted our intelligence by claiming

only four serious adverse reactions and no deaths from the more

than 380,000 vaccine doses then given. This bore no resemblance to

what people knew had happened in their owns circles and we had

Dirk Huyer in charge of ge�ing millions vaccinated in Ontario while

at the same time he was Chief Coroner for the province investigating

causes of death including possible death from the vaccine. An aide

said he had stepped back from investigating deaths, but evidence

indicated otherwise. Rosemary Frei, who secured a Master of Science

degree in molecular biology at the Faculty of Medicine at Canada’s

University of Calgary before turning to investigative journalism, was

one who could see that official figures for ‘vaccine’ deaths and

reactions made no sense. She said that doctors seldom reported

adverse events and when people got really sick or died a�er ge�ing

a vaccination they would a�ribute that to anything except the

vaccines. It had been that way for years and anyone who wondered

aloud whether the ‘Covid vaccines’ or other shots cause harm is

immediately branded as ‘anti-vax’ and ‘anti-science’. This was

‘career-threatening’ for health professionals. Then there was the

huge pressure to support the push to ‘vaccinate’ billions in the

quickest time possible. Frei said:

So that’s where we’re at today. More than half a million vaccine doses have been given to people in Ontario alone. The rush is on to vaccinate all 15 million of us in the province by September. And the mainstream media are screaming for this to be sped up even more. That all adds up to only a very slim likelihood that we’re going to be told the truth by officials about how many people are getting sick or dying from the vaccines.

What is true of Ontario is true of everywhere.

They KNEW – and still did it

The authorities knew what was going to happen with multiple

deaths and adverse reactions. The UK government’s Gates-funded

and Big Pharma-dominated Medicines and Healthcare products

Regulatory Agency (MHRA) hired a company to employ AI in

compiling the projected reactions to the ‘vaccine’ that would

otherwise be uncountable. The request for applications said: ‘The

MHRA urgently seeks an Artificial Intelligence (AI) so�ware tool to

process the expected high volume of Covid-19 vaccine Adverse Drug

Reaction …’ This was from the agency, headed by the disingenuous

June Raine, that gave the ‘vaccines’ emergency approval and the

company was hired before the first shot was given. ‘We are going to

kill and maim you – is that okay?’ ‘Oh, yes, perfectly fine – I’m very

grateful, thank you, doctor.’ The range of ‘Covid vaccine’ adverse

reactions goes on for page a�er page in the MHRA criminally

underreported ‘Yellow Card’ system and includes affects to eyes,

ears, skin, digestion, blood and so on. Raine’s MHRA amazingly

claimed that the ‘overall safety experience … is so far as expected

from the clinical trials’. The death, serious adverse effects, deafness

and blindness were expected? When did they ever mention that? If

these human tragedies were expected then those that gave approval

for the use of these ‘vaccines’ must be guilty of crimes against

humanity including murder – a definition of which is ‘killing a

person with malice aforethought or with recklessness manifesting

extreme indifference to the value of human life.’ People involved at

the MHRA, the CDC in America and their equivalent around the

world must go before Nuremberg trials to answer for their callous

inhumanity. We are only talking here about the immediate effects of

the ‘vaccine’. The longer-term impact of the DNA synthetic

manipulation is the main reason they are so hysterically desperate to

inoculate the entire global population in the shortest possible time.

Africa and the developing world are a major focus for the ‘vaccine’

depopulation agenda and a mass vaccination sales-pitch is

underway thanks to caring people like the Rockefellers and other

Cult assets. The Rockefeller Foundation, which pre-empted the

‘Covid pandemic’ in a document published in 2010 that ‘predicted’

what happened a decade later, announced an initial $34.95 million

grant in February, 2021, ‘to ensure more equitable access to Covid-19

testing and vaccines’ among other things in Africa in collaboration

with ‘24 organizations, businesses, and government agencies’. The

pan-Africa initiative would focus on 10 countries: Burkina Faso,

Ethiopia, Ghana, Kenya, Nigeria, Rwanda, South Africa, Tanzania,

Uganda, and Zambia’. Rajiv Shah, President of the Rockefeller

Foundation and former administrator of CIA-controlled USAID, said

that if Africa was not mass-vaccinated (to change the DNA of its

people) it was a ‘threat to all of humanity’ and not fair on Africans.

When someone from the Rockefeller Foundation says they want to

do something to help poor and deprived people and countries it is

time for a belly-laugh. They are doing this out of the goodness of

their ‘heart’ because ‘vaccinating’ the entire global population is

what the ‘Covid’ hoax set out to achieve. Official ‘decolonisation’ of

Africa by the Cult was merely a prelude to financial colonisation on

the road to a return to physical colonisation. The ‘vaccine’ is vital to

that and the sudden and convenient death of the ‘Covid’ sceptic

president of Tanzania can be seen in its true light. A lot of people in

Africa are aware that this is another form of colonisation and

exploitation and they need to stand their ground.

The ‘vaccine is working’ scam

A potential problem for the Cult was that the ‘vaccine’ is meant to

change human DNA and body messaging and not to protect anyone

from a ‘virus’ never shown to exist. The vaccine couldn’t work

because it was not designed to work and how could they make it

appear to be working so that more people would have it? This was

overcome by lowering the amplification rate of the PCR test to

produce fewer ‘cases’ and therefore fewer ‘deaths’. Some of us had

been pointing out since March, 2020, that the amplification rate of

the test not testing for the ‘virus’ had been made artificially high to

generate positive tests which they could call ‘cases’ to justify

lockdowns. The World Health Organization recommended an

absurdly high 45 amplification cycles to ensure the high positives

required by the Cult and then remained silent on the issue until

January 20th, 2021 – Biden’s Inauguration Day. This was when the

‘vaccinations’ were seriously underway and on that day the WHO

recommended a�er discussions with America’s CDC that

laboratories lowered their testing amplification. Dr David Samadi, a

certified urologist and health writer, said the WHO was encouraging

all labs to reduce their cycle count for PCR tests. He said the current

cycle was much too high and was ‘resulting in any particle being

declared a positive case’. Even one mainstream news report I saw

said this meant the number of ‘Covid’ infections may have been

‘dramatically inflated’. Oh, just a li�le bit. The CDC in America

issued new guidance to laboratories in April, 2021, to use 28 cycles

but only for ‘vaccinated’ people. The timing of the CDC/WHO

interventions were cynically designed to make it appear the

‘vaccines’ were responsible for falling cases and deaths when the real

reason can be seen in the following examples. New York’s state lab,

the Wadsworth Center, identified 872 positive tests in July, 2020,

based on a threshold of 40 cycles. When the figure was lowered to 35

cycles 43 percent of the 872 were no longer ‘positives’. At 30 cycles

the figure was 63 percent. A Massachuse�s lab found that between

85 to 90 percent of people who tested positive in July with a cycle

threshold of 40 would be negative at 30 cycles, Ashish Jha, MD,

director of the Harvard Global Health Institute, said: ‘I’m really

shocked that it could be that high … Boy, does it really change the

way we need to be thinking about testing.’ I’m shocked that I could

see the obvious in the spring of 2020, with no medical background,

and most medical professionals still haven’t worked it out. No, that’s

not shocking – it’s terrifying.

Three weeks a�er the WHO directive to lower PCR cycles the

London Daily Mail ran this headline: ‘Why ARE Covid cases

plummeting? New infections have fallen 45% in the US and 30%

globally in the past 3 weeks but experts say vaccine is NOT the main

driver because only 8% of Americans and 13% of people worldwide

have received their first dose.’ They acknowledged that the drop

could not be a�ributed to the ‘vaccine’, but soon this morphed

throughout the media into the ‘vaccine’ has caused cases and deaths

to fall when it was the PCR threshold. In December, 2020, there was

chaos at English Channel ports with truck drivers needing negative

‘Covid’ tests before they could board a ferry home for Christmas.

The government wanted to remove the backlog as fast as possible

and they brought in troops to do the ‘testing’. Out of 1,600 drivers

just 36 tested positive and the rest were given the all clear to cross

the Channel. I guess the authorities thought that 36 was the least

they could get away with without the unquestioning catching on.

The amplification trick which most people believed in the absence of

information in the mainstream applied more pressure on those

refusing the ‘vaccine’ to succumb when it ‘obviously worked’. The

truth was the exact opposite with deaths in care homes soaring with

the ‘vaccine’ and in Israel the term used was ‘skyrocket’. A re-

analysis of published data from the Israeli Health Ministry led by Dr

Hervé Seligmann at the Medicine Emerging Infectious and Tropical

Diseases at Aix-Marseille University found that Pfizer’s ‘Covid

vaccine’ killed ‘about 40 times more [elderly] people than the disease

itself would have killed’ during a five-week vaccination period and

260 times more younger people than would have died from the

‘virus’ even according to the manipulated ‘virus’ figures. Dr

Seligmann and his co-study author, Haim Yativ, declared a�er

reviewing the Israeli ‘vaccine’ death data: ‘This is a new Holocaust.’

Then, in mid-April, 2021, a�er vast numbers of people worldwide

had been ‘vaccinated’, the story changed with clear coordination.

The UK government began to prepare the ground for more future

lockdowns when Nuremberg-destined Boris Johnson told yet

another whopper. He said that cases had fallen because of lockdowns

not ‘vaccines’. Lockdowns are irrelevant when there is no ‘virus’ and

the test and fraudulent death certificates are deciding the number of

‘cases’ and ‘deaths’. Study a�er study has shown that lockdowns

don’t work and instead kill and psychologically destroy people.

Meanwhile in the United States Anthony Fauci and Rochelle

Walensky, the ultra-Zionist head of the CDC, peddled the same line.

More lockdown was the answer and not the ‘vaccine’, a line repeated

on cue by the moron that is Canadian Prime Minister Justin Trudeau.

Why all the hysteria to get everyone ‘vaccinated’ if lockdowns and

not ‘vaccines’ made the difference? None of it makes sense on the

face of it. Oh, but it does. The Cult wants lockdowns and the

‘vaccine’ and if the ‘vaccine’ is allowed to be seen as the total answer

lockdowns would no longer be justified when there are still

livelihoods to destroy. ‘Variants’ and renewed upward manipulation

of PCR amplification are planned to instigate never-ending

lockdown and more ‘vaccines’.

You must have it – we’re desperate Israel, where the Jewish and Arab population are ruled by the

Sabbatian Cult, was the front-runner in imposing the DNA-

manipulating ‘vaccine’ on its people to such an extent that Jewish

refusers began to liken what was happening to the early years of

Nazi Germany. This would seem to be a fantastic claim. Why would

a government of Jewish people be acting like the Nazis did? If you

realise that the Sabbatian Cult was behind the Nazis and that

Sabbatians hate Jews the pieces start to fit and the question of why a

‘Jewish’ government would treat Jews with such callous disregard

for their lives and freedom finds an answer. Those controlling the

government of Israel aren’t Jewish – they’re Sabbatian. Israeli lawyer

Tamir Turgal was one who made the Nazi comparison in comments

to German lawyer Reiner Fuellmich who is leading a class action

lawsuit against the psychopaths for crimes against humanity. Turgal

described how the Israeli government was vaccinating children and

pregnant women on the basis that there was no evidence that this

was dangerous when they had no evidence that it wasn’t dangerous

either. They just had no evidence. This was medical experimentation

and Turgal said this breached the Nuremberg Code about medical

experimentation and procedures requiring informed consent and

choice. Think about that. A Nuremberg Code developed because of

Nazi experimentation on Jews and others in concentration camps by

people like the evil-beyond-belief Josef Mengele is being breached by

the Israeli government; but when you know that it’s a Sabbatian

government along with its intelligence and military agencies like

Mossad, Shin Bet and the Israeli Defense Forces, and that Sabbatians

were the force behind the Nazis, the kaleidoscope comes into focus.

What have we come to when Israeli Jews are suing their government

for violating the Nuremberg Code by essentially making Israelis

subject to a medical experiment using the controversial ‘vaccines’?

It’s a shocker that this has to be done in the light of what happened

in Nazi Germany. The Anshe Ha-Emet, or ‘People of the Truth’,

made up of Israeli doctors, lawyers, campaigners and public, have

launched a lawsuit with the International Criminal Court. It says:

When the heads of the Ministry of Health as well as the prime minister presented the vaccine in Israel and began the vaccination of Israeli residents, the vaccinated were not advised, that, in practice, they are taking part in a medical experiment and that their consent is required for this under the Nuremberg Code.

The irony is unbelievable, but easily explained in one word:

Sabbatians. The foundation of Israeli ‘Covid’ apartheid is the ‘green

pass’ or ‘green passport’ which allows Jews and Arabs who have

had the DNA-manipulating ‘vaccine’ to go about their lives – to

work, fly, travel in general, go to shopping malls, bars, restaurants,

hotels, concerts, gyms, swimming pools, theatres and sports venues,

while non-’vaccinated’ are banned from all those places and

activities. Israelis have likened the ‘green pass’ to the yellow stars

that Jews in Nazi Germany were forced to wear – the same as the

yellow stickers that a branch of UK supermarket chain Morrisons

told exempt mask-wears they had to display when shopping. How

very sensitive. The Israeli system is blatant South African-style

apartheid on the basis of compliance or non-compliance to fascism

rather than colour of the skin. How appropriate that the Sabbatian

Israeli government was so close to the pre-Mandela apartheid

regime in Pretoria. The Sabbatian-instigated ‘vaccine passport’ in

Israel is planned for everywhere. Sabbatians struck a deal with

Pfizer that allowed them to lead the way in the percentage of a

national population infused with synthetic material and the result

was catastrophic. Israeli freedom activist Shai Dannon told me how

chairs were appearing on beaches that said ‘vaccinated only’. Health

Minister Yuli Edelstein said that anyone unwilling or unable to get

the jabs that ‘confer immunity’ will be ‘le� behind’. The man’s a liar.

Not even the makers claim the ‘vaccines’ confer immunity. When

you see those figures of ‘vaccine’ deaths these psychopaths were

saying that you must take the chance the ‘vaccine’ will kill you or

maim you while knowing it will change your DNA or lockdown for

you will be permanent. That’s fascism. The Israeli parliament passed

a law to allow personal information of the non-vaccinated to be

shared with local and national authorities for three months. This was

claimed by its supporters to be a way to ‘encourage’ people to be

vaccinated. Hadas Ziv from Physicians for Human Rights described

this as a ‘draconian law which crushed medical ethics and the

patient rights’. But that’s the idea, the Sabbatians would reply.

Your papers, please

Sabbatian Israel was leading what has been planned all along to be a

global ‘vaccine pass’ called a ‘green passport’ without which you

would remain in permanent lockdown restriction and unable to do

anything. This is how badly – desperately – the Cult is to get everyone

‘vaccinated’. The term and colour ‘green’ was not by chance and

related to the psychology of fusing the perception of the green

climate hoax with the ‘Covid’ hoax and how the ‘solution’ to both is

the same Great Reset. Lying politicians, health officials and

psychologists denied there were any plans for mandatory

vaccinations or restrictions based on vaccinations, but they knew

that was exactly what was meant to happen with governments of all

countries reaching agreements to enforce a global system. ‘Free’

Denmark and ‘free’ Sweden unveiled digital vaccine certification.

Cyprus, Czech Republic, Estonia, Greece, Hungary, Iceland, Italy,

Poland, Portugal, Slovakia, and Spain have all commi�ed to a

vaccine passport system and the rest including the whole of the EU

would follow. The satanic UK government will certainly go this way

despite mendacious denials and at the time of writing it is trying to

manipulate the public into having the ‘vaccine’ so they could go

abroad on a summer holiday. How would that work without

something to prove you had the synthetic toxicity injected into you?

Documents show that the EU’s European Commission was moving

towards ‘vaccine certificates’ in 2018 and 2019 before the ‘Covid’

hoax began. They knew what was coming. Abracadabra – Ursula

von der Leyen, the German President of the Commission,

announced in March, 2021, an EU ‘Digital Green Certificate’ – green

again – to track the public’s ‘Covid status’. The passport sting is

worldwide and the Far East followed the same pa�ern with South

Korea ruling that only those with ‘vaccination’ passports – again the

green pass – would be able to ‘return to their daily lives’.

Bill Gates has been preparing for this ‘passport’ with other Cult

operatives for years and beyond the paper version is a Gates-funded

‘digital ta�oo’ to identify who has been vaccinated and who hasn’t.

The ‘ta�oo’ is reported to include a substance which is externally

readable to confirm who has been vaccinated. This is a bio-luminous

light-generating enzyme (think fireflies) called … Luciferase. Yes,

named a�er the Cult ‘god’ Lucifer the ‘light bringer’ of whom more

to come. Gates said he funded the readable ta�oo to ensure children

in the developing world were vaccinated and no one was missed out.

He cares so much about poor kids as we know. This was just the

cover story to develop a vaccine tagging system for everyone on the

planet. Gates has been funding the ID2020 ‘alliance’ to do just that in

league with other lovely people at Microso�, GAVI, the Rockefeller

Foundation, Accenture and IDEO.org. He said in interviews in

March, 2020, before any ‘vaccine’ publicly existed, that the world

must have a globalised digital certificate to track the ‘virus’ and who

had been vaccinated. Gates knew from the start that the mRNA

vaccines were coming and when they would come and that the plan

was to tag the ‘vaccinated’ to marginalise the intelligent and stop

them doing anything including travel. Evil just doesn’t suffice. Gates

was exposed for offering a $10 million bribe to the Nigerian House

of Representatives to invoke compulsory ‘Covid’ vaccination of all

Nigerians. Sara Cunial, a member of the Italian Parliament, called

Gates a ‘vaccine criminal’. She urged the Italian President to hand

him over to the International Criminal Court for crimes against

humanity and condemned his plans to ‘chip the human race’

through ID2020.

You know it’s a long-planned agenda when war criminal and Cult

gofer Tony Blair is on the case. With the scale of arrogance only

someone as dark as Blair can muster he said: ‘Vaccination in the end

is going to be your route to liberty.’ Blair is a disgusting piece of

work and he confirms that again. The media has given a lot of

coverage to a bloke called Charlie Mullins, founder of London’s

biggest independent plumbing company, Pimlico Plumbers, who has

said he won’t employ anyone who has not been vaccinated or have

them go to any home where people are not vaccinated. He said that

if he had his way no one would be allowed to walk the streets if they

have not been vaccinated. Gates was cheering at the time while I was

alerting the white coats. The plan is that people will qualify for

‘passports’ for having the first two doses and then to keep it they

will have to have all the follow ups and new ones for invented

‘variants’ until human genetics is transformed and many are dead

who can’t adjust to the changes. Hollywood celebrities – the usual

propaganda stunt – are promoting something called the WELL

Health-Safety Rating to verify that a building or space has ‘taken the

necessary steps to prioritize the health and safety of their staff,

visitors and other stakeholders’. They included Lady Gaga, Jennifer

Lopez, Michael B. Jordan, Robert DeNiro, Venus Williams, Wolfgang

Puck, Deepak Chopra and 17th Surgeon General Richard Carmona.

Yawn. WELL Health-Safety has big connections with China. Parent

company Delos is headed by former Goldman Sachs partner Paul

Scialla. This is another example – and we will see so many others –

of using the excuse of ‘health’ to dictate the lives and activities of the

population. I guess one confirmation of the ‘safety’ of buildings is

that only ‘vaccinated’ people can go in, right?

Electronic concentration camps

I wrote decades ago about the plans to restrict travel and here we are

for those who refuse to bow to tyranny. This can be achieved in one

go with air travel if the aviation industry makes a blanket decree.

The ‘vaccine’ and guaranteed income are designed to be part of a

global version of China’s social credit system which tracks behaviour

24/7 and awards or deletes ‘credits’ based on whether your

behaviour is supported by the state or not. I mean your entire

lifestyle – what you do, eat, say, everything. Once your credit score

falls below a certain level consequences kick in. In China tens of

millions have been denied travel by air and train because of this. All

the locations and activities denied to refusers by the ‘vaccine’

passports will be included in one big mass ban on doing almost

anything for those that don’t bow their head to government. It’s

beyond fascist and a new term is required to describe its extremes – I

guess fascist technocracy will have to do. The way the Chinese

system of technological – technocratic – control is sweeping the West

can be seen in the Los Angeles school system and is planned to be

expanded worldwide. Every child is required to have a ‘Covid’-

tracking app scanned daily before they can enter the classroom. The

so-called Daily Pass tracking system is produced by Gates’ Microso�

which I’m sure will shock you rigid. The pass will be scanned using

a barcode (one step from an inside-the-body barcode) and the

information will include health checks, ‘Covid’ tests and

vaccinations. Entry codes are for one specific building only and

access will only be allowed if a student or teacher has a negative test

with a test not testing for the ‘virus’, has no symptoms of anything

alleged to be related to ‘Covid’ (symptoms from a range of other

illness), and has a temperature under 100 degrees. No barcode, no

entry, is planned to be the case for everywhere and not only schools.

Kids are being psychologically prepared to accept this as ‘normal’

their whole life which is why what they can impose in schools is so

important to the Cult and its gofers. Long-time American freedom

campaigner John Whitehead of the Rutherford Institute was not

exaggerating when he said: ‘Databit by databit, we are building our

own electronic concentration camps.’ Canada under its Cult gofer

prime minister Justin Trudeau has taken a major step towards the

real thing with people interned against their will if they test positive

with a test not testing for the ‘virus’ when they arrive at a Canadian

airport. They are jailed in internment hotels o�en without food or

water for long periods and with many doors failing to lock there

have been sexual assaults. The interned are being charged

sometimes $2,000 for the privilege of being abused in this way.

Trudeau is fully on board with the Cult and says the ‘Covid

pandemic’ has provided an opportunity for a global ‘reset’ to

permanently change Western civilisation. His number two, Deputy

Prime Minister Chrystia Freeland, is a trustee of the World Economic

Forum and a Rhodes Scholar. The Trudeau family have long been

servants of the Cult. See The Biggest Secret and Cathy O’Brien’s book

Trance-Formation of America for the horrific background to Trudeau’s

father Pierre Trudeau another Canadian prime minister. Hide your

fascism behind the façade of a heart-on-the-sleeve liberal. It’s a well-

honed Cult technique.

What can the ‘vaccine’ really do? We have a ‘virus’ never shown to exist and ‘variants’ of the ‘virus’

that have also never been shown to exist except, like the ‘original’, as

computer-generated fictions. Even if you believe there’s a ‘virus’ the

‘case’ to ‘death’ rate is in the region of 0.23 to 0.15 percent and those

‘deaths’ are concentrated among the very old around the same

average age that people die anyway. In response to this lack of threat

(in truth none) psychopaths and idiots, knowingly and unknowingly

answering to Gates and the Cult, are seeking to ‘vaccinate’ every

man, woman and child on Planet Earth. Clearly the ‘vaccine’ is not

about ‘Covid’ – none of this ever has been. So what is it all about

really? Why the desperation to infuse genetically-manipulating

synthetic material into everyone through mRNA fraudulent

‘vaccines’ with the intent of doing this over and over with the

excuses of ‘variants’ and other ‘virus’ inventions? Dr Sherri

Tenpenny, an osteopathic medical doctor in the United States, has

made herself an expert on vaccines and their effects as a vehement

campaigner against their use. Tenpenny was board certified in

emergency medicine, the director of a level two trauma centre for 12

years, and moved to Cleveland in 1996 to start an integrative

medicine practice which has treated patients from all 50 states and

some 17 other countries. Weaning people off pharmaceutical drugs is

a speciality.

She became interested in the consequences of vaccines a�er

a�ending a meeting at the National Vaccine Information Center in

Washington DC in 2000 where she ‘sat through four days of listening

to medical doctors and scientists and lawyers and parents of vaccine

injured kids’ and asked: ‘What’s going on?’ She had never been

vaccinated and never got ill while her father was given a list of

vaccines to be in the military and was ‘sick his entire life’. The

experience added to her questions and she began to examine vaccine

documents from the Centers for Disease Control (CDC). A�er

reading the first one, the 1998 version of The General Recommendations

of Vaccination, she thought: ‘This is it?’ The document was poorly

wri�en and bad science and Tenpenny began 20 years of research

into vaccines that continues to this day. She began her research into

‘Covid vaccines’ in March, 2020, and she describes them as ‘deadly’.

For many, as we have seen, they already have been. Tenpenny said

that in the first 30 days of the ‘vaccine’ rollout in the United States

there had been more than 40,000 adverse events reported to the

vaccine adverse event database. A document had been delivered to

her the day before that was 172 pages long. ‘We have over 40,000

adverse events; we have over 3,100 cases of [potentially deadly]

anaphylactic shock; we have over 5,000 neurological reactions.’

Effects ranged from headaches to numbness, dizziness and vertigo,

to losing feeling in hands or feet and paraesthesia which is when

limbs ‘fall asleep’ and people have the sensation of insects crawling

underneath their skin. All this happened in the first 30 days and

remember that only about ten percent (or far less) of adverse reactions

and vaccine-related deaths are estimated to be officially reported.

Tenpenny said:

So can you think of one single product in any industry, any industry, for as long as products have been made on the planet that within 30 days we have 40,000 people complaining of side effects that not only is still on the market but … we’ve got paid actors telling us how great

they are for getting their vaccine. We’re offering people $500 if they will just get their vaccine and we’ve got nurses and doctors going; ‘I got the vaccine, I got the vaccine’.

Tenpenny said they were not going to be ‘happy dancing folks’

when they began to suffer Bell’s palsy (facial paralysis),

neuropathies, cardiac arrhythmias and autoimmune reactions that

kill through a blood disorder. ‘They’re not going to be so happy,

happy then, but we’re never going to see pictures of those people’

she said. Tenpenny described the ‘vaccine’ as ‘a well-designed killing

tool’.

No off-switch

Bad as the initial consequences had been Tenpenny said it would be

maybe 14 months before we began to see the ‘full ravage’ of what is

going to happen to the ‘Covid vaccinated’ with full-out

consequences taking anything between two years and 20 years to

show. You can understand why when you consider that variations of

the ‘Covid vaccine’ use mRNA (messenger RNA) to in theory

activate the immune system to produce protective antibodies

without using the actual ‘virus’. How can they when it’s a computer

program and they’ve never isolated what they claim is the ‘real

thing’? Instead they use synthetic mRNA. They are inoculating

synthetic material into the body which through a technique known

as the Trojan horse is absorbed into cells to change the nature of

DNA. Human DNA is changed by an infusion of messenger RNA

and with each new ‘vaccine’ of this type it is changed even more. Say

so and you are banned by Cult Internet platforms. The contempt the

contemptuous Mark Zuckerberg has for the truth and human health

can be seen in an internal Facebook video leaked to the Project

Veritas investigative team in which he said of the ‘Covid vaccines’:

‘… I share some caution on this because we just don’t know the long

term side-effects of basically modifying people’s DNA and RNA.’ At

the same time this disgusting man’s Facebook was censoring and

banning anyone saying exactly the same. He must go before a

Nuremberg trial for crimes against humanity when he knows that he

is censoring legitimate concerns and denying the right of informed

consent on behalf of the Cult that owns him. People have been killed

and damaged by the very ‘vaccination’ technique he cast doubt on

himself when they may not have had the ‘vaccine’ with access to

information that he denied them. The plan is to have at least annual

‘Covid vaccinations’, add others to deal with invented ‘variants’, and

change all other vaccines into the mRNA system. Pfizer executives

told shareholders at a virtual Barclays Global Healthcare Conference

in March, 2021, that the public may need a third dose of ‘Covid

vaccine’, plus regular yearly boosters and the company planned to

hike prices to milk the profits in a ‘significant opportunity for our

vaccine’. These are the professional liars, cheats and opportunists

who are telling you their ‘vaccine’ is safe. Given this volume of

mRNA planned to be infused into the human body and its ability to

then replicate we will have a transformation of human genetics from

biological to synthetic biological – exactly the long-time Cult plan for

reasons we’ll see – and many will die. Sherri Tenpenny said of this

replication:

It’s like having an on-button but no off-button and that whole mechanism … they actually give it a name and they call it the Trojan horse mechanism, because it allows that [synthetic] virus and that piece of that [synthetic] virus to get inside of your cells, start to replicate and even get inserted into other parts of your DNA as a Trojan-horse.

Ask the overwhelming majority of people who have the ‘vaccine’

what they know about the contents and what they do and they

would reply: ‘The government says it will stop me ge�ing the virus.’

Governments give that false impression on purpose to increase take-

up. You can read Sherri Tenpenny’s detailed analysis of the health

consequences in her blog at Vaxxter.com, but in summary these are

some of them. She highlights the statement by Bill Gates about how

human beings can become their own ‘vaccine manufacturing

machine’. The man is insane. [‘Vaccine’-generated] ‘antibodies’ carry

synthetic messenger RNA into the cells and the damage starts,

Tenpenny contends, and she says that lungs can be adversely

affected through varying degrees of pus and bleeding which

obviously affects breathing and would be dubbed ‘Covid-19’. Even

more sinister was the impact of ‘antibodies’ on macrophages, a white

blood cell of the immune system. They consist of Type 1 and Type 2

which have very different functions. She said Type 1 are ‘hyper-

vigilant’ white blood cells which ‘gobble up’ bacteria etc. However,

in doing so, this could cause inflammation and in extreme

circumstances be fatal. She says these affects are mitigated by Type 2

macrophages which kick in to calm down the system and stop it

going rogue. They clear up dead tissue debris and reduce

inflammation that the Type 1 ‘fire crews’ have caused. Type 1 kills

the infection and Type 2 heals the damage, she says. This is her

punchline with regard to ‘Covid vaccinations’: She says that mRNA

‘antibodies’ block Type 2 macrophages by a�aching to them and

deactivating them. This meant that when the Type 1 response was

triggered by infection there was nothing to stop that ge�ing out of

hand by calming everything down. There’s an on-switch, but no off-

switch, she says. What follows can be ‘over and out, see you when I

see you’.

Genetic suicide

Tenpenny also highlights the potential for autoimmune disease – the

body a�acking itself – which has been associated with vaccines since

they first appeared. Infusing a synthetic foreign substance into cells

could cause the immune system to react in a panic believing that the

body is being overwhelmed by an invader (it is) and the

consequences can again be fatal. There is an autoimmune response

known as a ‘cytokine storm’ which I have likened to a homeowner

panicked by an intruder and picking up a gun to shoot randomly in

all directions before turning the fire on himself. The immune system

unleashes a storm of inflammatory response called cytokines to a

threat and the body commits hara-kiri. The lesson is that you mess

with the body’s immune response at your peril and these ‘vaccines’

seriously – fundamentally – mess with immune response. Tenpenny

refers to a consequence called anaphylactic shock which is a severe

and highly dangerous allergic reaction when the immune system

floods the body with chemicals. She gives the example of having a

bee sting which primes the immune system and makes it sensitive to

those chemicals. When people are stung again maybe years later the

immune response can be so powerful that it leads to anaphylactic

shock. Tenpenny relates this ‘shock’ with regard to the ‘Covid

vaccine’ to something called polyethylene glycol or PEG. Enormous

numbers of people have become sensitive to this over decades of use

in a whole range of products and processes including food, drink,

skin creams and ‘medicine’. Studies have claimed that some 72

percent of people have antibodies triggered by PEG compared with

two percent in the 1960s and allergic hypersensitive reactions to this

become a gathering cause for concern. Tenpenny points out that the

‘mRNA vaccine’ is coated in a ‘bubble’ of polyethylene glycol which

has the potential to cause anaphylactic shock through immune

sensitivity. Many reports have appeared of people reacting this way

a�er having the ‘Covid vaccine’. What do we think is going to

happen as humanity has more and more of these ‘vaccines’?

Tenpenny said: ‘All these pictures we have seen with people with

these rashes … these weepy rashes, big reactions on their arms and

things like that – it’s an acute allergic reaction most likely to the

polyethylene glycol that you’ve been previously primed and

sensitised to.’

Those who have not studied the conspiracy and its perpetrators at

length might think that making the population sensitive to PEG and

then pu�ing it in these ‘vaccines’ is just a coincidence. It is not. It is

instead testament to how carefully and coldly-planned current

events have been and the scale of the conspiracy we are dealing

with. Tenpenny further explains that the ‘vaccine’ mRNA procedure

can breach the blood-brain barrier which protects the brain from

toxins and other crap that will cause malfunction. In this case they

could make two proteins corrupt brain function to cause

Amyotrophic lateral sclerosis (ALS) , a progressive nervous system

disease leading to loss of muscle control, and frontal lobe

degeneration – Alzheimer’s and dementia. Immunologist J. Bart

Classon published a paper connecting mRNA ‘vaccines’ to prion

disease which can lead to Alzheimer’s and other forms of

neurogenerative disease while others have pointed out the potential

to affect the placenta in ways that make women infertile. This will

become highly significant in the next chapter when I will discuss

other aspects of this non-vaccine that relate to its nanotechnology

and transmission from the injected to the uninjected.

Qualified in idiocy

Tenpenny describes how research has confirmed that these ‘vaccine’-

generated antibodies can interact with a range of other tissues in the

body and a�ack many other organs including the lungs. ‘This means

that if you have a hundred people standing in front of you that all

got this shot they could have a hundred different symptoms.’

Anyone really think that Cult gofers like the Queen, Tony Blair,

Christopher Whi�y, Anthony Fauci, and all the other psychopaths

have really had this ‘vaccine’ in the pictures we’ve seen? Not a

bloody chance. Why don’t doctors all tell us about all these dangers

and consequences of the ‘Covid vaccine’? Why instead do they

encourage and pressure patients to have the shot? Don’t let’s think

for a moment that doctors and medical staff can’t be stupid, lazy, and

psychopathic and that’s without the financial incentives to give the

jab. Tenpenny again:

Some people are going to die from the vaccine directly but a large number of people are going to start to get horribly sick and get all kinds of autoimmune diseases 42 days to maybe a year out. What are they going to do, these stupid doctors who say; ‘Good for you for getting that vaccine.’ What are they going to say; ‘Oh, it must be a mutant, we need to give an extra dose of that vaccine.’

Because now the vaccine, instead of one dose or two doses we need three or four because the stupid physicians aren’t taking the time to learn anything about it. If I can learn this sitting in my living room reading a 19 page paper and several others so can they. There’s nothing special about me, I just take the time to do it.

Remember how Sara Kayat, the NHS and TV doctor, said that the

‘Covid vaccine’ would ‘100 percent prevent hospitalisation and

death’. Doctors can be idiots like every other profession and they

should not be worshipped as infallible. They are not and far from it.

Behind many medical and scientific ‘experts’ lies an uninformed prat

trying to hide themselves from you although in the ‘Covid’ era many

have failed to do so as with UK narrative-repeating ‘TV doctor’

Hilary Jones. Pushing back against the minority of proper doctors

and scientists speaking out against the ‘vaccine’ has been the entire

edifice of the Cult global state in the form of governments, medical

systems, corporations, mainstream media, Silicon Valley, and an

army of compliant doctors, medical staff and scientists willing to say

anything for money and to enhance their careers by promoting the

party line. If you do that you are an ‘expert’ and if you won’t you are

an ‘anti-vaxxer’ and ‘Covidiot’. The pressure to be ‘vaccinated’ is

incessant. We have even had reports claiming that the ‘vaccine’ can

help cure cancer and Alzheimer’s and make the lame walk. I am

waiting for the announcement that it can bring you coffee in the

morning and cook your tea. Just as the symptoms of ‘Covid’ seem to

increase by the week so have the miracles of the ‘vaccine’. American

supermarket giant Kroger Co. offered nearly 500,000 employees in

35 states a $100 bonus for having the ‘vaccine’ while donut chain

Krispy Kreme promised ‘vaccinated’ customers a free glazed donut

every day for the rest of 2021. Have your DNA changed and you will

get a doughnut although we might not have to give you them for

long. Such offers and incentives confirm the desperation.

Perhaps the worse vaccine-stunt of them all was UK ‘Health’

Secretary Ma�-the-prat Hancock on live TV a�er watching a clip of

someone being ‘vaccinated’ when the roll-out began. Hancock faked

tears so badly it was embarrassing. Brain-of-Britain Piers Morgan,

the lockdown-supporting, ‘vaccine’ supporting, ‘vaccine’ passport-

supporting, TV host played along with Hancock – ‘You’re quite

emotional about that’ he said in response to acting so atrocious it

would have been called out at a school nativity which will

presumably today include Mary and Jesus in masks, wise men

keeping their camels six feet apart, and shepherds under tent arrest.

System-serving Morgan tweeted this: ‘Love the idea of covid vaccine

passports for everywhere: flights, restaurants, clubs, football, gyms,

shops etc. It’s time covid-denying, anti-vaxxer loonies had their

bullsh*t bluff called & bar themselves from going anywhere that

responsible citizens go.’ If only I could aspire to his genius. To think

that Morgan, who specialises in shouting over anyone he disagrees

with, was lauded as a free speech hero when he lost his job a�er

storming off the set of his live show like a child throwing his dolly

out of the pram. If he is a free speech hero we are in real trouble. I

have no idea what ‘bullsh*t’ means, by the way, the * throws me

completely.

The Cult is desperate to infuse its synthetic DNA-changing

concoction into everyone and has been using every lie, trick and

intimidation to do so. The question of ‘Why?’ we shall now address.

I

CHAPTER TEN

Human 2.0

I believe that at the end of the century the use of words and general

educated opinion will have altered so much that one will be able to

speak of machines thinking without expecting to be contradicted –

Alan Turing (1912-1954), the ‘Father of artificial intelligence‘

have been exposing for decades the plan to transform the human

body from a biological to a synthetic-biological state. The new

human that I will call Human 2.0 is planned to be connected to

artificial intelligence and a global AI ‘Smart Grid’ that would operate

as one global system in which AI would control everything from

your fridge to your heating system to your car to your mind.

Humans would no longer be ‘human’, but post-human and sub-

human, with their thinking and emotional processes replaced by AI.

What I said sounded crazy and beyond science fiction and I could

understand that. To any balanced, rational, mind it is crazy. Today,

however, that world is becoming reality and it puts the ‘Covid

vaccine’ into its true context. Ray Kurzweil is the ultra-Zionist

‘computer scientist, inventor and futurist’ and co-founder of the

Singularity University. Singularity refers to the merging of humans

with machines or ‘transhumanism’. Kurzweil has said humanity

would be connected to the cyber ‘cloud’ in the period of the ever-

recurring year of 2030:

Our thinking … will be a hybrid of biological and non-biological thinking … humans will be able to extend their limitations and ‘think in the cloud’ … We’re going to put gateways to the

cloud in our brains ... We’re going to gradually merge and enhance ourselves ... In my view, that’s the nature of being human – we transcend our limitations. As the technology becomes vastly superior to what we are then the small proportion that is still human gets smaller and smaller and smaller until it’s just utterly negligible.

They are trying to sell this end-of-humanity-as-we-know-it as the

next stage of ‘evolution’ when we become super-human and ‘like the

gods’. They are lying to you. Shocked, eh? The population, and again

especially the young, have been manipulated into addiction to

technologies designed to enslave them for life. First they induced an

addiction to smartphones (holdables); next they moved to

technology on the body (wearables); and then began the invasion of

the body (implantables). I warned way back about the plan for

microchipped people and we are now entering that era. We should

not be diverted into thinking that this refers only to chips we can see.

Most important are the nanochips known as smart dust, neural dust

and nanobots which are far too small to be seen by the human eye.

Nanotechnology is everywhere, increasingly in food products, and

released into the atmosphere by the geoengineering of the skies

funded by Bill Gates to ‘shut out the Sun’ and ‘save the planet from

global warming’. Gates has been funding a project to spray millions

of tonnes of chalk (calcium carbonate) into the stratosphere over

Sweden to ‘dim the Sun’ and cool the Earth. Scientists warned the

move could be disastrous for weather systems in ways no one can

predict and opposition led to the Swedish space agency announcing

that the ‘experiment’ would not be happening as planned in the

summer of 2021; but it shows where the Cult is going with dimming

the impact of the Sun and there’s an associated plan to change the

planet’s atmosphere. Who gives psychopath Gates the right to

dictate to the entire human race and dismantle planetary systems?

The world will not be safe while this man is at large.

The global warming hoax has made the Sun, like the gas of life,

something to fear when both are essential to good health and human

survival (more inversion). The body transforms sunlight into vital

vitamin D through a process involving … cholesterol. This is the

cholesterol we are also told to fear. We are urged to take Big Pharma

statin drugs to reduce cholesterol and it’s all systematic. Reducing

cholesterol means reducing vitamin D uptake with all the multiple

health problems that will cause. At least if you take statins long term

it saves the government from having to pay you a pension. The

delivery system to block sunlight is widely referred to as chemtrails

although these have a much deeper agenda, too. They appear at first

to be contrails or condensation trails streaming from aircra� into

cold air at high altitudes. Contrails disperse very quickly while

chemtrails do not and spread out across the sky before eventually

their content falls to earth. Many times I have watched aircra� cross-

cross a clear blue sky releasing chemtrails until it looks like a cloudy

day. Chemtrails contain many things harmful to humans and the

natural world including toxic heavy metals, aluminium (see

Alzheimer’s) and nanotechnology. Ray Kurzweil reveals the reason

without actually saying so: ‘Nanobots will infuse all the ma�er

around us with information. Rocks, trees, everything will become

these intelligent creatures.’ How do you deliver that? From the sky.

Self-replicating nanobots would connect everything to the Smart

Grid. The phenomenon of Morgellons disease began in the chemtrail

era and the correlation has led to it being dubbed the ‘chemtrail

disease’. Self-replicating fibres appear in the body that can be pulled

out through the skin. Morgellons fibres continue to grow outside the

body and have a form of artificial intelligence. I cover this at greater

length in Phantom Self.

‘Vaccine’ operating system

‘Covid vaccines’ with their self-replicating synthetic material are also

designed to make the connection between humanity and Kurzweil’s

‘cloud’. American doctor and dedicated campaigner for truth, Carrie

Madej, an Internal Medicine Specialist in Georgia with more than 20

years medical experience, has highlighted the nanotechnology aspect

of the fake ‘vaccines’. She explains how one of the components in at

least the Moderna and Pfizer synthetic potions are ‘lipid

nanoparticles’ which are ‘like li�le tiny computer bits’ – a ‘sci-fi

substance’ known as nanobots and hydrogel which can be ‘triggered

at any moment to deliver its payload’ and act as ‘biosensors’. The

synthetic substance had ‘the ability to accumulate data from your

body like your breathing, your respiration, thoughts and emotions,

all kind of things’ and each syringe could carry a million nanobots:

This substance because it’s like little bits of computers in your body, crazy, but it’s true, it can do that, [and] obviously has the ability to act through Wi-Fi. It can receive and transmit energy, messages, frequencies or impulses. That issue has never been addressed by these companies. What does that do to the human?

Just imagine getting this substance in you and it can react to things all around you, the 5G, your smart device, your phones, what is happening with that? What if something is triggering it, too, like an impulse, a frequency? We have something completely foreign in the human body.

Madej said her research revealed that electromagnetic (EMF)

frequencies emi�ed by phones and other devices had increased

dramatically in the same period of the ‘vaccine’ rollout and she was

seeing more people with radiation problems as 5G and other

electromagnetic technology was expanded and introduced to schools

and hospitals. She said she was ‘floored with the EMF coming off’

the devices she checked. All this makes total sense and syncs with

my own work of decades when you think that Moderna refers in

documents to its mRNA ‘vaccine’ as an ‘operating system’:

Recognizing the broad potential of mRNA science, we set out to create an mRNA technology platform that functions very much like an operating system on a computer. It is designed so that it can plug and play interchangeably with different programs. In our case, the ‘program’ or ‘app’ is our mRNA drug – the unique mRNA sequence that codes for a protein …

… Our MRNA Medicines – ‘The ‘Software Of Life’: When we have a concept for a new mRNA medicine and begin research, fundamental components are already in place. Generally, the only thing that changes from one potential mRNA medicine to another is the coding region – the actual genetic code that instructs ribosomes to make protein. Utilizing these instruction sets gives our investigational mRNA medicines a software-like quality. We also have the ability to combine different mRNA sequences encoding for different proteins in a single mRNA investigational medicine.

Who needs a real ‘virus’ when you can create a computer version to

justify infusing your operating system into the entire human race on

the road to making living, breathing people into cyborgs? What is

missed with the ‘vaccines’ is the digital connection between synthetic

material and the body that I highlighted earlier with the study that

hacked a computer with human DNA. On one level the body is

digital, based on mathematical codes, and I’ll have more about that

in the next chapter. Those who ridiculously claim that mRNA

‘vaccines’ are not designed to change human genetics should explain

the words of Dr Tal Zaks, chief medical officer at Moderna, in a 2017

TED talk. He said that over the last 30 years ‘we’ve been living this

phenomenal digital scientific revolution, and I’m here today to tell

you, that we are actually hacking the software of life, and that it’s

changing the way we think about prevention and treatment of

disease’:

In every cell there’s this thing called messenger RNA, or mRNA for short, that transmits the critical information from the DNA in our genes to the protein, which is really the stuff we’re all made out of. This is the critical information that determines what the cell will do. So we think about it as an operating system. So if you could change that, if you could introduce a line of code, or change a line of code, it turns out, that has profound implications for everything, from the flu to cancer.

Zaks should more accurately have said that this has profound

implications for the human genetic code and the nature of DNA.

Communications within the body go both ways and not only one.

But, hey, no, the ‘Covid vaccine’ will not affect your genetics. Cult

fact-checkers say so even though the man who helped to develop the

mRNA technique says that it does. Zaks said in 2017:

If you think about what it is we’re trying to do. We’ve taken information and our understanding of that information and how that information is transmitted in a cell, and we’ve taken our understanding of medicine and how to make drugs, and we’re fusing the two. We think of it as information therapy.

I have been writing for decades that the body is an information

field communicating with itself and the wider world. This is why

radiation which is information can change the information field of

body and mind through phenomena like 5G and change their nature

and function. ‘Information therapy’ means to change the body’s

information field and change the way it operates. DNA is a receiver-

transmi�er of information and can be mutated by information like

mRNA synthetic messaging. Technology to do this has been ready

and waiting in the underground bases and other secret projects to be

rolled out when the ‘Covid’ hoax was played. ‘Trials’ of such short

and irrelevant duration were only for public consumption. When

they say the ‘vaccine’ is ‘experimental’ that is not true. It may appear

to be ‘experimental’ to those who don’t know what’s going on, but

the trials have already been done to ensure the Cult gets the result it

desires. Zaks said that it took decades to sequence the human

genome, completed in 2003, but now they could do it in a week. By

‘they’ he means scientists operating in the public domain. In the

secret projects they were sequencing the genome in a week long

before even 2003.

Deluge of mRNA

Highly significantly the Moderna document says the guiding

premise is that if using mRNA as a medicine works for one disease

then it should work for many diseases. They were leveraging the

flexibility afforded by their platform and the fundamental role

mRNA plays in protein synthesis to pursue mRNA medicines for a

broad spectrum of diseases. Moderna is confirming what I was

saying through 2020 that multiple ‘vaccines’ were planned for

‘Covid’ (and later invented ‘variants’) and that previous vaccines

would be converted to the mRNA system to infuse the body with

massive amounts of genetically-manipulating synthetic material to

secure a transformation to a synthetic-biological state. The ‘vaccines’

are designed to kill stunning numbers as part of the long-exposed

Cult depopulation agenda and transform the rest. Given this is the

goal you can appreciate why there is such hysterical demand for

every human to be ‘vaccinated’ for an alleged ‘disease’ that has an

estimated ‘infection’ to ‘death’ ratio of 0.23-0.15 percent. As I write

•

•

•

children are being given the ‘vaccine’ in trials (their parents are a

disgrace) and ever-younger people are being offered the vaccine for

a ‘virus’ that even if you believe it exists has virtually zero chance of

harming them. Horrific effects of the ‘trials’ on a 12-year-old girl

were revealed by a family member to be serious brain and gastric

problems that included a bowel obstruction and the inability to

swallow liquids or solids. She was unable to eat or drink without

throwing up, had extreme pain in her back, neck and abdomen, and

was paralysed from the waist down which stopped her urinating

unaided. When the girl was first taken to hospital doctors said it was

all in her mind. She was signed up for the ‘trial’ by her parents for

whom no words suffice. None of this ‘Covid vaccine’ insanity makes

any sense unless you see what the ‘vaccine’ really is – a body-

changer. Synthetic biology or ‘SynBio’ is a fast-emerging and

expanding scientific discipline which includes everything from

genetic and molecular engineering to electrical and computer

engineering. Synthetic biology is defined in these ways:

A multidisciplinary area of research that seeks to create new

biological parts, devices, and systems, or to redesign systems that

are already found in nature.

The use of a mixture of physical engineering and genetic

engineering to create new (and therefore synthetic) life forms.

An emerging field of research that aims to combine the

knowledge and methods of biology, engineering and related

disciplines in the design of chemically-synthesized DNA to create

organisms with novel or enhanced characteristics and traits

(synthetic organisms including humans).

We now have synthetic blood, skin, organs and limbs being

developed along with synthetic body parts produced by 3D printers.

These are all elements of the synthetic human programme and this

comment by Kurzweil’s co-founder of the Singularity University,

Peter Diamandis, can be seen in a whole new light with the ‘Covid’

hoax and the sanctions against those that refuse the ‘vaccine’:

Anybody who is going to be resisting the progress forward [to transhumanism] is going to be resisting evolution and, fundamentally, they will die out. It’s not a matter of whether it’s good or bad. It’s going to happen.

‘Resisting evolution’? What absolute bollocks. The arrogance of these

people is without limit. His ‘it’s going to happen’ mantra is another

way of saying ‘resistance is futile’ to break the spirit of those pushing

back and we must not fall for it. Ge�ing this genetically-

transforming ‘vaccine’ into everyone is crucial to the Cult plan for

total control and the desperation to achieve that is clear for anyone

to see. Vaccine passports are a major factor in this and they, too, are a

form of resistance is futile. It’s NOT. The paper funded by the

Rockefeller Foundation for the 2013 ‘health conference’ in China

said:

We will interact more with artificial intelligence. The use of robotics, bio-engineering to augment human functioning is already well underway and will advance. Re-engineering of humans into potentially separate and unequal forms through genetic engineering or mixed human-robots raises debates on ethics and equality.

A new demography is projected to emerge after 2030 [that year again] of technologies (robotics, genetic engineering, nanotechnology) producing robots, engineered organisms, ‘nanobots’ and artificial intelligence (AI) that can self-replicate. Debates will grow on the implications of an impending reality of human designed life.

What is happening today is so long planned. The world army

enforcing the will of the world government is intended to be a robot

army, not a human one. Today’s military and its technologically

‘enhanced’ troops, pilotless planes and driverless vehicles are just

stepping stones to that end. Human soldiers are used as Cult fodder

and its time they woke up to that and worked for the freedom of the

population instead of their own destruction and their family’s

destruction – the same with the police. Join us and let’s sort this out.

The phenomenon of enforce my own destruction is widespread in

the ‘Covid’ era with Woker ‘luvvies’ in the acting and entertainment

industries supporting ‘Covid’ rules which have destroyed their

profession and the same with those among the public who put signs

on the doors of their businesses ‘closed due to Covid – stay safe’

when many will never reopen. It’s a form of masochism and most

certainly insanity.

Transgender = transhumanism

When something explodes out of nowhere and is suddenly

everywhere it is always the Cult agenda and so it is with the tidal

wave of claims and demands that have infiltrated every aspect of

society under the heading of ‘transgenderism’. The term ‘trans’ is so

‘in’ and this is the dictionary definition:

A prefix meaning ‘across’, ’through’, occurring … in loanwords from Latin, used in particular for denoting movement or conveyance from place to place (transfer; transmit; transplant) or complete change (transform; transmute), or to form adjectives meaning ’crossing’, ‘on the other side of’, or ‘going beyond’ the place named (transmontane; transnational; trans- Siberian).

Transgender means to go beyond gender and transhuman means

to go beyond human. Both are aspects of the Cult plan to transform

the human body to a synthetic state with no gender. Human 2.0 is not

designed to procreate and would be produced technologically with

no need for parents. The new human would mean the end of parents

and so men, and increasingly women, are being targeted for the

deletion of their rights and status. Parental rights are disappearing at

an ever-quickening speed for the same reason. The new human

would have no need for men or women when there is no procreation

and no gender. Perhaps the transgender movement that appears to

be in a permanent state of frenzy might now contemplate on how it

is being used. This was never about transgender rights which are

only the interim excuse for confusing gender, particularly in the

young, on the road to fusing gender. Transgender activism is not an

end; it is a means to an end. We see again the technique of creative

destruction in which you destroy the status quo to ‘build back be�er’

in the form that you want. The gender status quo had to be

destroyed by persuading the Cult-created Woke mentality to believe

that you can have 100 genders or more. A programme for 9 to 12

year olds produced by the Cult-owned BBC promoted the 100

genders narrative. The very idea may be the most monumental

nonsense, but it is not what is true that counts, only what you can

make people believe is true. Once the gender of 2 + 2 = 4 has been

dismantled through indoctrination, intimidation and 2 + 2 = 5 then

the new no-gender normal can take its place with Human 2.0.

Aldous Huxley revealed the plan in his prophetic Brave New World in

1932:

Natural reproduction has been done away with and children are created, decanted’, and raised in ‘hatcheries and conditioning centres’. From birth, people are genetically designed to fit into one of five castes, which are further split into ‘Plus’ and ‘Minus’ members and designed to fulfil predetermined positions within the social and economic strata of the World State.

How could Huxley know this in 1932? For the same reason George

Orwell knew about the Big Brother state in 1948, Cult insiders I have

quoted knew about it in 1969, and I have known about it since the

early 1990s. If you are connected to the Cult or you work your balls

off to uncover the plan you can predict the future. The process is

simple. If there is a plan for the world and nothing intervenes to stop

it then it will happen. Thus if you communicate the plan ahead of

time you are perceived to have predicted the future, but you haven’t.

You have revealed the plan which without intervention will become

the human future. The whole reason I have done what I have is to

alert enough people to inspire an intervention and maybe at last that

time has come with the Cult and its intentions now so obvious to

anyone with a brain in working order.

The future is here

Technological wombs that Huxley described to replace parent

procreation are already being developed and they are only the

projects we know about in the public arena. Israeli scientists told The

Times of Israel in March, 2021, that they have grown 250-cell embryos

into mouse foetuses with fully formed organs using artificial wombs

in a development they say could pave the way for gestating humans

outside the womb. Professor Jacob Hanna of the Weizmann Institute

of Science said:

We took mouse embryos from the mother at day five of development, when they are just of 250 cells, and had them in the incubator from day five until day 11, by which point they had grown all their organs.

By day 11 they make their own blood and have a beating heart, a fully developed brain. Anybody would look at them and say, ‘this is clearly a mouse foetus with all the characteristics of a mouse.’ It’s gone from being a ball of cells to being an advanced foetus.

A special liquid is used to nourish embryo cells in a laboratory

dish and they float on the liquid to duplicate the first stage of

embryonic development. The incubator creates all the right

conditions for its development, Hanna said. The liquid gives the

embryo ‘all the nutrients, hormones and sugars they need’ along

with a custom-made electronic incubator which controls gas

concentration, pressure and temperature. The cu�ing-edge in the

underground bases and other secret locations will be light years

ahead of that, however, and this was reported by the London

Guardian in 2017:

We are approaching a biotechnological breakthrough. Ectogenesis, the invention of a complete external womb, could completely change the nature of human reproduction. In April this year, researchers at the Children’s Hospital of Philadelphia announced their development of an artificial womb.

The article was headed ‘Artificial wombs could soon be a reality.

What will this mean for women?’ What would it mean for children is

an even bigger question. No mother to bond with only a machine in

preparation for a life of soulless interaction and control in a world

governed by machines (see the Matrix movies). Now observe the

calculated manipulations of the ‘Covid’ hoax as human interaction

and warmth has been curtailed by distancing, isolation and fear with

people communicating via machines on a scale never seen before.

These are all dots in the same picture as are all the personal

assistants, gadgets and children’s toys through which kids and

adults communicate with AI as if it is human. The AI ‘voice’ on Sat-

Nav should be included. All these things are psychological

preparation for the Cult endgame. Before you can make a physical

connection with AI you have to make a psychological connection

and that is what people are being conditioned to do with this ever

gathering human-AI interaction. Movies and TV programmes

depicting the transhuman, robot dystopia relate to a phenomenon

known as ‘pre-emptive programming’ in which the world that is

planned is portrayed everywhere in movies, TV and advertising.

This is conditioning the conscious and subconscious mind to become

familiar with the planned reality to dilute resistance when it

happens for real. What would have been a shock such is the change

is made less so. We have young children put on the road to

transgender transition surgery with puberty blocking drugs at an

age when they could never be able to make those life-changing

decisions.

Rachel Levine, a professor of paediatrics and psychiatry who

believes in treating children this way, became America’s highest-

ranked openly-transgender official when she was confirmed as US

Assistant Secretary at the Department of Health and Human

Services a�er being nominated by Joe Biden (the Cult). Activists and

governments press for laws to deny parents a say in their children’s

transition process so the kids can be isolated and manipulated into

agreeing to irreversible medical procedures. A Canadian father

Robert Hoogland was denied bail by the Vancouver Supreme Court

in 2021 and remained in jail for breaching a court order that he stay

silent over his young teenage daughter, a minor, who was being

offered life-changing hormone therapy without parental consent. At

the age of 12 the girl’s ‘school counsellor’ said she may be

transgender, referred her to a doctor and told the school to treat her

like a boy. This is another example of state-serving schools imposing

ever more control over children’s lives while parents have ever less.

Contemptible and extreme child abuse is happening all over the

world as the Cult gender-fusion operation goes into warp-speed.

Why the war on men – and now women?

The question about what artificial wombs mean for women should

rightly be asked. The answer can be seen in the deletion of women’s

rights involving sport, changing rooms, toilets and status in favour

of people in male bodies claiming to identify as women. I can

identify as a mountain climber, but it doesn’t mean I can climb a

mountain any more than a biological man can be a biological

woman. To believe so is a triumph of belief over factual reality which

is the very perceptual basis of everything Woke. Women’s sport is

being destroyed by allowing those with male bodies who say they

identify as female to ‘compete’ with girls and women. Male body

‘women’ dominate ‘women’s’ competition with their greater muscle

mass, bone density, strength and speed. With that disadvantage

sport for women loses all meaning. To put this in perspective nearly

300 American high school boys can run faster than the quickest

woman sprinter in the world. Women are seeing their previously

protected spaces invaded by male bodies simply because they claim

to identify as women. That’s all they need to do to access all women’s

spaces and activities under the Biden ‘Equality Act’ that destroys

equality for women with the usual Orwellian Woke inversion. Male

sex offenders have already commi�ed rapes in women’s prisons a�er

claiming to identify as women to get them transferred. Does this not

ma�er to the Woke ‘equality’ hypocrites? Not in the least. What

ma�ers to Cult manipulators and funders behind transgender

activists is to advance gender fusion on the way to the no-gender

‘human’. When you are seeking to impose transparent nonsense like

this, or the ‘Covid’ hoax, the only way the nonsense can prevail is

through censorship and intimidation of dissenters, deletion of

factual information, and programming of the unquestioning,

bewildered and naive. You don’t have to scan the world for long to

see that all these things are happening.

Many women’s rights organisations have realised that rights and

status which took such a long time to secure are being eroded and

that it is systematic. Kara Dansky of the global Women’s Human

Rights Campaign said that Biden’s transgender executive order

immediately he took office, subsequent orders, and Equality Act

legislation that followed ‘seek to erase women and girls in the law as

a category’. Exactly. I said during the long ago-started war on men

(in which many women play a crucial part) that this was going to

turn into a war on them. The Cult is phasing out both male and

female genders. To get away with that they are brought into conflict

so they are busy fighting each other while the Cult completes the job

with no unity of response. Unity, people, unity. We need unity

everywhere. Transgender is the only show in town as the big step

towards the no-gender human. It’s not about rights for transgender

people and never has been. Woke political correctness is deleting

words relating to genders to the same end. Wokers believe this is to

be ‘inclusive’ when the opposite is true. They are deleting words

describing gender because gender itself is being deleted by Human

2.0. Terms like ‘man’, ‘woman’, ‘mother’ and ‘father’ are being

deleted in the universities and other institutions to be replaced by

the no-gender, not trans-gender, ‘individuals’ and ‘guardians’.

Women’s rights campaigner Maria Keffler of Partners for Ethical

Care said: ‘Children are being taught from kindergarten upward that

some boys have a vagina, some girls have a penis, and that kids can

be any gender they want to be.’ Do we really believe that suddenly

countries all over the world at the same time had the idea of having

drag queens go into schools or read transgender stories to very

young children in the local library? It’s coldly-calculated confusion

of gender on the way to the fusion of gender. Suzanne Vierling, a

psychologist from Southern California, made another important

point:

Yesterday’s slave woman who endured gynecological medical experiments is today’s girl- child being butchered in a booming gender-transitioning sector. Ovaries removed, pushing her into menopause and osteoporosis, uncharted territory, and parents’ rights and authority decimated.

The erosion of parental rights is a common theme in line with the

Cult plans to erase the very concept of parents and ‘ovaries removed,

pushing her into menopause’ means what? Those born female lose

the ability to have children – another way to discontinue humanity

as we know it.

Eliminating Human 1.0 (before our very eyes)

To pave the way for Human 2.0 you must phase out Human 1.0. This

is happening through plummeting sperm counts and making

women infertile through an onslaught of chemicals, radiation

(including smartphones in pockets of men) and mRNA ‘vaccines’.

Common agriculture pesticides are also having a devastating impact

on human fertility. I have been tracking collapsing sperm counts in

the books for a long time and in 2021 came a book by fertility

scientist and reproductive epidemiologist Shanna Swan, Count

Down: How Our Modern World Is Threatening Sperm Counts, Altering

Male and Female Reproductive Development and Imperiling the Future of

the Human Race. She reports how the global fertility rate dropped by

half between 1960 and 2016 with America’s birth rate 16 percent

below where it needs to be to sustain the population. Women are

experiencing declining egg quality, more miscarriages, and more

couples suffer from infertility. Other findings were an increase in

erectile dysfunction, infant boys developing more genital

abnormalities, male problems with conception, and plunging levels

of the male hormone testosterone which would explain why so

many men have lost their backbone and masculinity. This has been

very evident during the ‘Covid’ hoax when women have been

prominent among the Pushbackers and big strapping blokes have

bowed their heads, covered their faces with a nappy and quietly

submi�ed. Mind control expert Cathy O’Brien also points to how

global education introduced the concept of ‘we’re all winners’ in

sport and classrooms: ‘Competition was defused, and it in turn

defused a sense of fighting back.’ This is another version of the

‘equity’ doctrine in which you drive down rather than raise up.

What a contrast in Cult-controlled China with its global ambitions

where the government published plans in January, 2021, to ‘cultivate

masculinity’ in boys from kindergarten through to high school in the

face of a ‘masculinity crisis’. A government adviser said boys would

be soon become ‘delicate, timid and effeminate’ unless action was

taken. Don’t expect any similar policy in the targeted West. A 2006

study showed that a 65-year-old man in 2002 had testosterone levels

15 percent lower than a 65-year-old man in 1987 while a 2020 study

found a similar story with young adults and adolescents. Men are

ge�ing prescriptions for testosterone replacement therapy which

causes an even greater drop in sperm count with up to 99 percent

seeing sperm counts drop to zero during the treatment. More sperm

is defective and malfunctioning with some having two heads or not

pursuing an egg.

A class of synthetic chemicals known as phthalates are being

blamed for the decline. These are found everywhere in plastics,

shampoos, cosmetics, furniture, flame retardants, personal care

products, pesticides, canned foods and even receipts. Why till

receipts? Everyone touches them. Let no one delude themselves that

all this is not systematic to advance the long-time agenda for human

body transformation. Phthalates mimic hormones and disrupt the

hormone balance causing testosterone to fall and genital birth

defects in male infants. Animals and fish have been affected in the

same way due to phthalates and other toxins in rivers. When fish

turn gay or change sex through chemicals in rivers and streams it is

a pointer to why there has been such an increase in gay people and

the sexually confused. It doesn’t ma�er to me what sexuality people

choose to be, but if it’s being affected by chemical pollution and

consumption then we need to know. Does anyone really think that

this is not connected to the transgender agenda, the war on men and

the condemnation of male ‘toxic masculinity’? You watch this being

followed by ‘toxic femininity’. It’s already happening. When

breastfeeding becomes ‘chest-feeding’, pregnant women become

pregnant people along with all the other Woke claptrap you know

that the world is going insane and there’s a Cult scam in progress.

Transgender activists are promoting the Cult agenda while Cult

billionaires support and fund the insanity as they laugh themselves

to sleep at the sheer stupidity for which humans must be infamous

in galaxies far, far away.

‘Covid vaccines’ and female infertility

We can now see why the ‘vaccine’ has been connected to potential

infertility in women. Dr Michael Yeadon, former Vice President and

Chief Scientific Advisor at Pfizer, and Dr Wolfgang Wodarg in

Germany, filed a petition with the European Medicines Agency in

December, 2020, urging them to stop trials for the Pfizer/BioNTech

shot and all other mRNA trials until further studies had been done.

They were particularly concerned about possible effects on fertility

with ‘vaccine’-produced antibodies a�acking the protein Syncytin-1

which is responsible for developing the placenta. The result would

be infertility ‘of indefinite duration’ in women who have the

‘vaccine’ with the placenta failing to form. Section 10.4.2 of the

Pfizer/BioNTech trial protocol says that pregnant women or those

who might become so should not have mRNA shots. Section 10.4

warns men taking mRNA shots to ‘be abstinent from heterosexual

intercourse’ and not to donate sperm. The UK government said that

it did not know if the mRNA procedure had an effect on fertility. Did

not know? These people have to go to jail. UK government advice did

not recommend at the start that pregnant women had the shot and

said they should avoid pregnancy for at least two months a�er

‘vaccination’. The ‘advice’ was later updated to pregnant women

should only have the ‘vaccine’ if the benefits outweighed the risks to

mother and foetus. What the hell is that supposed to mean? Then

‘spontaneous abortions’ began to appear and rapidly increase on the

adverse reaction reporting schemes which include only a fraction of

adverse reactions. Thousands and ever-growing numbers of

‘vaccinated’ women are describing changes to their menstrual cycle

with heavier blood flow, irregular periods and menstruating again

a�er going through the menopause – all links to reproduction

effects. Women are passing blood clots and the lining of their uterus

while men report erectile dysfunction and blood effects. Most

significantly of all unvaccinated women began to report similar

menstrual changes a�er interaction with ‘vaccinated’ people and men

and children were also affected with bleeding noses, blood clots and

other conditions. ‘Shedding’ is when vaccinated people can emit the

content of a vaccine to affect the unvaccinated, but this is different.

‘Vaccinated’ people were not shedding a ‘live virus’ allegedly in

‘vaccines’ as before because the fake ‘Covid vaccines’ involve

synthetic material and other toxicity. Doctors exposing what is

happening prefer the term ‘transmission’ to shedding. Somehow

those that have had the shots are transmi�ing effects to those that

haven’t. Dr Carrie Madej said the nano-content of the ‘vaccines’ can

‘act like an antenna’ to others around them which fits perfectly with

my own conclusions. This ‘vaccine’ transmission phenomenon was

becoming known as the book went into production and I deal with

this further in the Postscript.

Vaccine effects on sterility are well known. The World Health

Organization was accused in 2014 of sterilising millions of women in

Kenya with the evidence confirmed by the content of the vaccines

involved. The same WHO behind the ‘Covid’ hoax admi�ed its

involvement for more than ten years with the vaccine programme.

Other countries made similar claims. Charges were lodged by

Tanzania, Nicaragua, Mexico, and the Philippines. The Gardasil

vaccine claimed to protect against a genital ‘virus’ known as HPV

has also been linked to infertility. Big Pharma and the WHO (same

thing) are criminal and satanic entities. Then there’s the Bill Gates

Foundation which is connected through funding and shared

interests with 20 pharmaceutical giants and laboratories. He stands

accused of directing the policy of United Nations Children’s Fund

(UNICEF), vaccine alliance GAVI, and other groupings, to advance

the vaccine agenda and silence opposition at great cost to women

and children. At the same time Gates wants to reduce the global

population. Coincidence?

Great Reset = Smart Grid = new human

The Cult agenda I have been exposing for 30 years is now being

openly promoted by Cult assets like Gates and Klaus Schwab of the

World Economic Forum under code-terms like the ‘Great Reset’,

‘Build Back Be�er’ and ‘a rare but narrow window of opportunity to

reflect, reimagine, and reset our world’. What provided this ‘rare but

narrow window of opportunity’? The ‘Covid’ hoax did. Who created

that? They did. My books from not that long ago warned about the

planned ‘Internet of Things’ (IoT) and its implications for human

freedom. This was the plan to connect all technology to the Internet

and artificial intelligence and today we are way down that road with

an estimated 36 billion devices connected to the World Wide Web

and that figure is projected to be 76 billion by 2025. I further warned

that the Cult planned to go beyond that to the Internet of Everything

when the human brain was connected via AI to the Internet and

Kurzweil’s ‘cloud’. Now we have Cult operatives like Schwab calling

for precisely that under the term ‘Internet of Bodies’, a fusion of the

physical, digital and biological into one centrally-controlled Smart

Grid system which the Cult refers to as the ‘Fourth Industrial

Revolution’. They talk about the ‘biological’, but they really mean

the synthetic-biological which is required to fully integrate the

human body and brain into the Smart Grid and artificial intelligence

planned to replace the human mind. We have everything being

synthetically manipulated including the natural world through

GMO and smart dust, the food we eat and the human body itself

with synthetic ‘vaccines’. I said in The Answer that we would see the

Cult push for synthetic meat to replace animals and in February,

2021, the so predictable psychopath Bill Gates called for the

introduction of synthetic meat to save us all from ‘climate change’.

The climate hoax just keeps on giving like the ‘Covid’ hoax. The war

on meat by vegan activists is a carbon (oops, sorry) copy of the

manipulation of transgender activists. They have no idea (except

their inner core) that they are being used to promote and impose the

agenda of the Cult or that they are only the vehicle and not the reason.

This is not to say those who choose not to eat meat shouldn’t be

respected and supported in that right, but there are ulterior motives

•

•

•

for those in power. A Forbes article in December, 2019, highlighted

the plan so beloved of Schwab and the Cult under the heading:

‘What Is The Internet of Bodies? And How Is It Changing Our

World?’ The article said the human body is the latest data platform

(remember ‘our vaccine is an operating system’). Forbes described

the plan very accurately and the words could have come straight out

of my books from long before:

The Internet of Bodies (IoB) is an extension of the IoT and basically connects the human body to a network through devices that are ingested, implanted, or connected to the body in some way. Once connected, data can be exchanged, and the body and device can be remotely monitored and controlled.

They were really describing a human hive mind with human

perception centrally-dictated via an AI connection as well as

allowing people to be ‘remotely monitored and controlled’.

Everything from a fridge to a human mind could be directed from a

central point by these insane psychopaths and ‘Covid vaccines’ are

crucial to this. Forbes explained the process I mentioned earlier of

holdable and wearable technology followed by implantable. The

article said there were three generations of the Internet of Bodies that

include:

Body external: These are wearable devices such as Apple Watches

or Fitbits that can monitor our health.

Body internal: These include pacemakers, cochlear implants, and

digital pills that go inside our bodies to monitor or control various

aspects of health.

Body embedded: The third generation of the Internet of Bodies is

embedded technology where technology and the human body are

melded together and have a real-time connection to a remote

machine.

Forbes noted the development of the Brain Computer Interface (BCI)

which merges the brain with an external device for monitoring and

controlling in real-time. ‘The ultimate goal is to help restore function

to individuals with disabilities by using brain signals rather than

conventional neuromuscular pathways.’ Oh, do fuck off. The goal of

brain interface technology is controlling human thought and

emotion from the central point in a hive mind serving its masters

wishes. Many people are now agreeing to be chipped to open doors

without a key. You can recognise them because they’ll be wearing a

mask, social distancing and lining up for the ‘vaccine’. The Cult

plans a Great Reset money system a�er they have completed the

demolition of the global economy in which ‘money’ will be

exchanged through communication with body operating systems.

Rand Corporation, a Cult-owned think tank, said of the Internet of

Bodies or IoB:

Internet of Bodies technologies fall under the broader IoT umbrella. But as the name suggests, IoB devices introduce an even more intimate interplay between humans and gadgets. IoB devices monitor the human body, collect health metrics and other personal information, and transmit those data over the Internet. Many devices, such as fitness trackers, are already in use … IoB devices … and those in development can track, record, and store users’ whereabouts, bodily functions, and what they see, hear, and even think.

Schwab’s World Economic Forum, a long-winded way of saying

‘fascism’ or ‘the Cult’, has gone full-on with the Internet of Bodies in

the ‘Covid’ era. ‘We’re entering the era of the Internet of Bodies’, it

declared, ‘collecting our physical data via a range of devices that can

be implanted, swallowed or worn’. The result would be a huge

amount of health-related data that could improve human wellbeing

around the world, and prove crucial in fighting the ‘Covid-19

pandemic’. Does anyone think these clowns care about ‘human

wellbeing’ a�er the death and devastation their pandemic hoax has

purposely caused? Schwab and co say we should move forward with

the Internet of Bodies because ‘Keeping track of symptoms could

help us stop the spread of infection, and quickly detect new cases’.

How wonderful, but keeping track’ is all they are really bothered

about. Researchers were investigating if data gathered from

smartwatches and similar devices could be used as viral infection

alerts by tracking the user’s heart rate and breathing. Schwab said in

his 2018 book Shaping the Future of the Fourth Industrial Revolution:

The lines between technologies and beings are becoming blurred and not just by the ability to create lifelike robots or synthetics. Instead it is about the ability of new technologies to literally become part of us. Technologies already influence how we understand ourselves, how we think about each other, and how we determine our realities. As the technologies … give us deeper access to parts of ourselves, we may begin to integrate digital technologies into our bodies.

You can see what the game is. Twenty-four hour control and people

– if you could still call them that – would never know when

something would go ping and take them out of circulation. It’s the

most obvious rush to a global fascist dictatorship and the complete

submission of humanity and yet still so many are locked away in

their Cult-induced perceptual coma and can’t see it.

Smart Grid control centres

The human body is being transformed by the ‘vaccines’ and in other

ways into a synthetic cyborg that can be a�ached to the global Smart

Grid which would be controlled from a central point and other sub-

locations of Grid manipulation. Where are these planned to be? Well,

China for a start which is one of the Cult’s biggest centres of

operation. The technological control system and technocratic rule

was incubated here to be unleashed across the world a�er the

‘Covid’ hoax came out of China in 2020. Another Smart Grid location

that will surprise people new to this is Israel. I have exposed in The

Trigger how Sabbatian technocrats, intelligence and military

operatives were behind the horrors of 9/11 and not 1̀9 Arab hijackers’

who somehow manifested the ability to pilot big passenger airliners

when instructors at puddle-jumping flying schools described some

of them as a joke. The 9/11 a�acks were made possible through

control of civilian and military air computer systems and those of the

White House, Pentagon and connected agencies. See The Trigger – it

will blow your mind. The controlling and coordinating force were

the Sabbatian networks in Israel and the United States which by then

had infiltrated the entire US government, military and intelligence

system. The real name of the American Deep State is ‘Sabbatian

State’. Israel is a tiny country of only nine million people, but it is

one of the global centres of cyber operations and fast catching Silicon

Valley in importance to the Cult. Israel is known as the ‘start-up

nation’ for all the cyber companies spawned there with the

Sabbatian specialisation of ‘cyber security’ that I mentioned earlier

which gives those companies access to computer systems of their

clients in real time through ‘backdoors’ wri�en into the coding when

security so�ware is downloaded. The Sabbatian centre of cyber

operations outside Silicon Valley is the Israeli military Cyber

Intelligence Unit, the biggest infrastructure project in Israel’s history,

headquartered in the desert-city of Beersheba and involving some

20,000 ‘cyber soldiers’. Here are located a literal army of Internet

trolls scanning social media, forums and comment lists for anyone

challenging the Cult agenda. The UK military has something similar

with its 77th Brigade and associated operations. The Beersheba

complex includes research and development centres for other Cult

operations such as Intel, Microso�, IBM, Google, Apple, Hewle�-

Packard, Cisco Systems, Facebook and Motorola. Techcrunch.com

ran an article about the Beersheba global Internet technology centre

headlined ‘Israel’s desert city of Beersheba is turning into a cybertech

oasis’:

The military’s massive relocation of its prestigious technology units, the presence of multinational and local companies, a close proximity to Ben Gurion University and generous government subsidies are turning Beersheba into a major global cybertech hub. Beersheba has all of the ingredients of a vibrant security technology ecosystem, including Ben Gurion University with its graduate program in cybersecurity and Cyber Security Research Center, and the presence of companies such as EMC, Deutsche Telekom, PayPal, Oracle, IBM, and Lockheed Martin. It’s also the future home of the INCB (Israeli National Cyber Bureau); offers a special income tax incentive for cyber security companies, and was the site for the relocation of the army’s intelligence corps units.

Sabbatians have taken over the cyber world through the following

process: They scan the schools for likely cyber talent and develop

them at Ben Gurion University and their period of conscription in

the Israeli Defense Forces when they are stationed at the Beersheba

complex. When the cyber talented officially leave the army they are

funded to start cyber companies with technology developed by

themselves or given to them by the state. Much of this is stolen

through backdoors of computer systems around the world with

America top of the list. Others are sent off to Silicon Valley to start

companies or join the major ones and so we have many major

positions filled by apparently ‘Jewish’ but really Sabbatian

operatives. Google, YouTube and Facebook are all run by ‘Jewish’

CEOs while Twi�er is all but run by ultra-Zionist hedge-fund shark

Paul Singer. At the centre of the Sabbatian global cyber web is the

Israeli army’s Unit 8200 which specialises in hacking into computer

systems of other countries, inserting viruses, gathering information,

instigating malfunction, and even taking control of them from a

distance. A long list of Sabbatians involved with 9/11, Silicon Valley

and Israeli cyber security companies are operatives of Unit 8200.

This is not about Israel. It’s about the Cult. Israel is planned to be a

Smart Grid hub as with China and what is happening at Beersheba is

not for the benefit of Jewish people who are treated disgustingly by

the Sabbatian elite that control the country. A glance at the

Nuremberg Codes will tell you that.

The story is much bigger than ‘Covid’, important as that is to

where we are being taken. Now, though, it’s time to really strap in.

There’s more … much more …

I

CHAPTER ELEVEN

Who controls the Cult?

Awake, arise or be forever fall’n

John Milton, Paradise Lost

have exposed this far the level of the Cult conspiracy that operates

in the world of the seen and within the global secret society and

satanic network which operates in the shadows one step back from

the seen. The story, however, goes much deeper than that.

The ‘Covid’ hoax is major part of the Cult agenda, but only part,

and to grasp the biggest picture we have to expand our a�ention

beyond the realm of human sight and into the infinity of possibility

that we cannot see. It is from here, ultimately, that humanity is being

manipulated into a state of total control by the force which dictates

the actions of the Cult. How much of reality can we see? Next to

damn all is the answer. We may appear to see all there is to see in the

‘space’ our eyes survey and observe, but li�le could be further from

the truth. The human ‘world’ is only a tiny band of frequency that

the body’s visual and perceptual systems can decode into perception

of a ‘world’. According to mainstream science the electromagnetic

spectrum is 0.005 percent of what exists in the Universe (Fig 10). The

maximum estimate I have seen is 0.5 percent and either way it’s

miniscule. I say it is far, far, smaller even than 0.005 percent when

you compare reality we see with the totality of reality that we don’t.

Now get this if you are new to such information: Visible light, the

only band of frequency that we can see, is a fraction of the 0.005

percent (Fig 11 overleaf). Take this further and realise that our

universe is one of infinite universes and that universes are only a

fragment of overall reality – infinite reality. Then compare that with

the almost infinitesimal frequency band of visible light or human

sight. You see that humans are as near blind as it is possible to be

without actually being so. Artist and filmmaker, Sergio Toporek,

said:

Figure 10: Humans can perceive such a tiny band of visual reality it’s laughable.

Figure 11: We can see a smear of the 0.005 percent electromagnetic spectrum, but we still know it all. Yep, makes sense.

Consider that you can see less than 1% of the electromagnetic spectrum and hear less than 1% of the acoustic spectrum. 90% of the cells in your body carry their own microbial DNA and are not ‘you’. The atoms in your body are 99.9999999999999999% empty space and none of them are the ones you were born with ... Human beings have 46 chromosomes, two less than a potato.

The existence of the rainbow depends on the conical photoreceptors in your eyes; to animals without cones, the rainbow does not exist. So you don’t just look at a rainbow, you create it. This is pretty amazing, especially considering that all the beautiful colours you see represent less than 1% of the electromagnetic spectrum.

Suddenly the ‘world’ of humans looks a very different place. Take

into account, too, that Planet Earth when compared with the

projected size of this single universe is the equivalent of a billionth of

a pinhead. Imagine the ratio that would be when compared to

infinite reality. To think that Christianity once insisted that Earth and

humanity were the centre of everything. This background is vital if

we are going to appreciate the nature of ‘human’ and how we can be

manipulated by an unseen force. To human visual reality virtually

everything is unseen and yet the prevailing perception within the

institutions and so much of the public is that if we can’t see it, touch

it, hear it, taste it and smell it then it cannot exist. Such perception is

indoctrinated and encouraged by the Cult and its agents because it

isolates believers in the strictly limited, village-idiot, realm of the five

senses where perceptions can be firewalled and information

controlled. Most of those perpetuating the ‘this-world-is-all-there-is’

insanity are themselves indoctrinated into believing the same

delusion. While major players and influencers know that official

reality is laughable most of those in science, academia and medicine

really believe the nonsense they peddle and teach succeeding

generations. Those who challenge the orthodoxy are dismissed as

nu�ers and freaks to protect the manufactured illusion from

exposure. Observe the dynamic of the ‘Covid’ hoax and you will see

how that takes the same form. The inner-circle psychopaths knows

it’s a gigantic scam, but almost the entirety of those imposing their

fascist rules believe that ‘Covid’ is all that they’re told it is.

Stolen identity

Ask people who they are and they will give you their name, place of

birth, location, job, family background and life story. Yet that is not

who they are – it is what they are experiencing. The difference is

absolutely crucial. The true ‘I’, the eternal, infinite ‘I’, is consciousness,

a state of being aware. Forget ‘form’. That is a vehicle for a brief

experience. Consciousness does not come from the brain, but through

the brain and even that is more symbolic than literal. We are

awareness, pure awareness, and this is what withdraws from the

body at what we call ‘death’ to continue our eternal beingness,

isness, in other realms of reality within the limitlessness of infinity or

the Biblical ‘many mansions in my father’s house’. Labels of a

human life, man, woman, transgender, black, white, brown,

nationality, circumstances and income are not who we are. They are

what we are – awareness – is experiencing in a brief connection with a

band of frequency we call ‘human’. The labels are not the self; they

are, to use the title of one of my books, a Phantom Self. I am not

David Icke born in Leicester, England, on April 29th, 1952. I am the

consciousness having that experience. The Cult and its non-human

masters seek to convince us through the institutions of ‘education’,

science, medicine, media and government that what we are

experiencing is who we are. It’s so easy to control and direct

perception locked away in the bewildered illusions of the five senses

with no expanded radar. Try, by contrast, doing the same with a

humanity aware of its true self and its true power to consciously

create its reality and experience. How is it possible to do this? We do

it all day every day. If you perceive yourself as ‘li�le me’ with no

power to impact upon your life and the world then your life

experience will reflect that. You will hand the power you don’t think

you have to authority in all its forms which will use it to control your

experience. This, in turn, will appear to confirm your perception of

‘li�le me’ in a self-fulfilling feedback loop. But that is what ‘li�le me’

really is – a perception. We are all ‘big-me’, infinite me, and the Cult

has to make us forget that if its will is to prevail. We are therefore

manipulated and pressured into self-identifying with human labels

and not the consciousness/awareness experiencing those human

labels.

The phenomenon of identity politics is a Cult-instigated

manipulation technique to sub-divide previous labels into even

smaller ones. A United States university employs this list of le�ers to

describe student identity: LGBTTQQFAGPBDSM or lesbian, gay,

bisexual, transgender, transsexual, queer, questioning, flexual,

asexual, gender-fuck, polyamorous, bondage/discipline,

dominance/submission and sadism/masochism. I’m sure other lists

are even longer by now as people feel the need to self-identity the ‘I’

with the minutiae of race and sexual preference. Wokers

programmed by the Cult for generations believe this is about

‘inclusivity’ when it’s really the Cult locking them away into smaller

and smaller versions of Phantom Self while firewalling them from

the influence of their true self, the infinite, eternal ‘I’. You may notice

that my philosophy which contends that we are all unique points of

a�ention/awareness within the same infinite whole or Oneness is the

ultimate non-racism. The very sense of Oneness makes the

judgement of people by their body-type, colour or sexuality u�erly

ridiculous and confirms that racism has no understanding of reality

(including anti-white racism). Yet despite my perception of life Cult

agents and fast-asleep Wokers label me racist to discredit my

information while they are themselves phenomenally racist and

sexist. All they see is race and sexuality and they judge people as

good or bad, demons or untouchables, by their race and sexuality.

All they see is Phantom Self and perceive themselves in terms of

Phantom Self. They are pawns and puppets of the Cult agenda to

focus a�ention and self-identity in the five senses and play those

identities against each other to divide and rule. Columbia University

has introduced segregated graduations in another version of social

distancing designed to drive people apart and teach them that

different racial and cultural groups have nothing in common with

each other. The last thing the Cult wants is unity. Again the pump-

primers of this will be Cult operatives in the knowledge of what they

are doing, but the rest are just the Phantom Self blind leading the

Phantom Self blind. We do have something in common – we are all

the same consciousness having different temporary experiences.

What is this ‘human’?

Yes, what is ‘human’? That is what we are supposed to be, right? I

mean ‘human’? True, but ‘human’ is the experience not the ‘I’. Break

it down to basics and ‘human’ is the way that information is

processed. If we are to experience and interact with this band of

frequency we call the ‘world’ we must have a vehicle that operates

within that band of frequency. Our consciousness in its prime form

cannot do that; it is way beyond the frequency of the human realm.

My consciousness or awareness could not tap these keys and pick up

the cup in front of me in the same way that radio station A cannot

interact with radio station B when they are on different frequencies.

The human body is the means through which we have that

interaction. I have long described the body as a biological computer

which processes information in a way that allows consciousness to

experience this reality. The body is a receiver, transmi�er and

processor of information in a particular way that we call human. We

visually perceive only the world of the five senses in a wakened state

– that is the limit of the body’s visual decoding system. In truth it’s

not even visual in the way we experience ‘visual reality’ as I will

come to in a moment. We are ‘human’ because the body processes

the information sources of human into a reality and behaviour

system that we perceive as human. Why does an elephant act like an

elephant and not like a human or a duck? The elephant’s biological

computer is a different information field and processes information

according to that program into a visual and behaviour type we call

an elephant. The same applies to everything in our reality. These

body information fields are perpetuated through procreation (like

making a copy of a so�ware program). The Cult wants to break that

cycle and intervene technologically to transform the human

information field into one that will change what we call humanity. If

it can change the human information field it will change the way

that field processes information and change humanity both

‘physically’ and psychologically. Hence the messenger (information)

RNA ‘vaccines’ and so much more that is targeting human genetics

by changing the body’s information – messaging – construct through

food, drink, radiation, toxicity and other means.

Reality that we experience is nothing like reality as it really is in

the same way that the reality people experience in virtual reality

games is not the reality they are really living in. The game is only a

decoded source of information that appears to be a reality. Our

world is also an information construct – a simulation (more later). In

its base form our reality is a wavefield of information much the same

in theme as Wi-Fi. The five senses decode wavefield information into

electrical information which they communicate to the brain to

decode into holographic (illusory ‘physical’) information. Different

parts of the brain specialise in decoding different senses and the

information is fused into a reality that appears to be outside of us

but is really inside the brain and the genetic structure in general (Fig

12 overleaf). DNA is a receiver-transmi�er of information and a vital

part of this decoding process and the body’s connection to other

realities. Change DNA and you change the way we decode and

connect with reality – see ‘Covid vaccines’. Think of computers

decoding Wi-Fi. You have information encoded in a radiation field

and the computer decodes that information into a very different

form on the screen. You can’t see the Wi-Fi until its information is

made manifest on the screen and the information on the screen is

inside the computer and not outside. I have just described how we

decode the ‘human world’. All five senses decode the waveform ‘Wi-

Fi’ field into electrical signals and the brain (computer) constructs

reality inside the brain and not outside – ‘You don’t just look at a

rainbow, you create it’. Sound is a simple example. We don’t hear

sound until the brain decodes it. Waveform sound waves are picked

up by the hearing sense and communicated to the brain in an

electrical form to be decoded into the sounds that we hear.

Everything we hear is inside the brain along with everything we see,

feel, smell and taste. Words and language are waveform fields

generated by our vocal chords which pass through this process until

they are decoded by the brain into words that we hear. Different

languages are different frequency fields or sound waves generated

by vocal chords. Late British philosopher Alan Wa�s said:

Figure 12: The brain receives information from the five senses and constructs from that our perceived reality.

[Without the brain] the world is devoid of light, heat, weight, solidity, motion, space, time or any other imaginable feature. All these phenomena are interactions, or transactions, of vibrations with a certain arrangement of neurons.

That’s exactly what they are and scientist Robert Lanza describes in

his book, Biocentrism, how we decode electromagnetic waves and

energy into visual and ‘physical’ experience. He uses the example of

a flame emi�ing photons, electromagnetic energy, each pulsing

electrically and magnetically:

… these … invisible electromagnetic waves strike a human retina, and if (and only if) the waves happen to measure between 400 and 700 nano meters in length from crest to crest, then their energy is just right to deliver a stimulus to the 8 million cone-shaped cells in the retina.

Each in turn send an electrical pulse to a neighbour neuron, and on up the line this goes, at 250 mph, until it reaches the … occipital lobe of the brain, in the back of the head. There, a cascading complex of neurons fire from the incoming stimuli, and we subjectively perceive this experience as a yellow brightness occurring in a place we have been conditioned to call the ‘external world’.

You hear what you decode

If a tree falls or a building collapses they make no noise unless

someone is there to decode the energetic waves generated by the

disturbance into what we call sound. Does a falling tree make a

noise? Only if you hear it – decode it. Everything in our reality is a

frequency field of information operating within the overall ‘Wi-Fi’

field that I call The Field. A vibrational disturbance is generated in

The Field by the fields of the falling tree or building. These

disturbance waves are what we decode into the sound of them

falling. If no one is there to do that then neither will make any noise.

Reality is created by the observer – decoder – and the perceptions of

the observer affect the decoding process. For this reason different

people – different perceptions – will perceive the same reality or

situation in a different way. What one may perceive as a nightmare

another will see as an opportunity. The question of why the Cult is

so focused on controlling human perception now answers itself. All

experienced reality is the act of decoding and we don’t experience

Wi-Fi until it is decoded on the computer screen. The sight and

sound of an Internet video is encoded in the Wi-Fi all around us, but

we don’t see or hear it until the computer decodes that information.

Taste, smell and touch are all phenomena of the brain as a result of

the same process. We don’t taste, smell or feel anything except in the

brain and there are pain relief techniques that seek to block the

signal from the site of discomfort to the brain because if the brain

doesn’t decode that signal we don’t feel pain. Pain is in the brain and

only appears to be at the point of impact thanks to the feedback loop

between them. We don’t see anything until electrical information

from the sight senses is decoded in an area at the back of the brain. If

that area is damaged we can go blind when our eyes are perfectly

okay. So why do we go blind if we damage an eye? We damage the

information processing between the waveform visual information

and the visual decoding area of the brain. If information doesn’t

reach the brain in a form it can decode then we can’t see the visual

reality that it represents. What’s more the brain is decoding only a

fraction of the information it receives and the rest is absorbed by the

sub-conscious mind. This explanation is from the science magazine,

Wonderpedia:

Every second, 11 million sensations crackle along these [brain] pathways ... The brain is confronted with an alarming array of images, sounds and smells which it rigorously filters down until it is left with a manageable list of around 40. Thus 40 sensations per second make up what we perceive as reality.

The ‘world’ is not what people are told to believe that is it and the

inner circles of the Cult know that.

Illusory ‘physical’ reality

We can only see a smear of 0.005 percent of the Universe which is

only one of a vast array of universes – ‘mansions’ – within infinite

reality. Even then the brain decodes only 40 pieces of information

(‘sensations’) from a potential 11 million that we receive every

second. Two points strike you from this immediately: The sheer

breathtaking stupidity of believing we know anything so rigidly that

there’s nothing more to know; and the potential for these processes

to be manipulated by a malevolent force to control the reality of the

population. One thing I can say for sure with no risk of contradiction

is that when you can perceive an almost indescribable fraction of

infinite reality there is always more to know as in tidal waves of it.

Ancient Greek philosopher Socrates was so right when he said that

wisdom is to know how li�le we know. How obviously true that is

when you think that we are experiencing a physical world of solidity

that is neither physical nor solid and a world of apartness when

everything is connected. Cult-controlled ‘science’ dismisses the so-

called ‘paranormal’ and all phenomena related to that when the

‘para’-normal is perfectly normal and explains the alleged ‘great

mysteries’ which dumbfound scientific minds. There is a reason for

this. A ‘scientific mind’ in terms of the mainstream is a material

mind, a five-sense mind imprisoned in see it, touch it, hear it, smell it

and taste it. Phenomena and happenings that can’t be explained that

way leave the ‘scientific mind’ bewildered and the rule is that if they

can’t account for why something is happening then it can’t, by

definition, be happening. I beg to differ. Telepathy is thought waves

passing through The Field (think wave disturbance again) to be

decoded by someone able to connect with that wavelength

(information). For example: You can pick up the thought waves of a

friend at any distance and at the very least that will bring them to

mind. A few minutes later the friend calls you. ‘My god’, you say,

‘that’s incredible – I was just thinking of you.’ Ah, but they were

thinking of you before they made the call and that’s what you

decoded. Native peoples not entrapped in five-sense reality do this

so well it became known as the ‘bush telegraph’. Those known as

psychics and mediums (genuine ones) are doing the same only

across dimensions of reality. ‘Mind over ma�er’ comes from the fact

that ma�er and mind are the same. The state of one influences the

state of the other. Indeed one and the other are illusions. They are

aspects of the same field. Paranormal phenomena are all explainable

so why are they still considered ‘mysteries’ or not happening? Once

you go down this road of understanding you begin to expand

awareness beyond the five senses and that’s the nightmare for the

Cult.

Figure 13: Holograms are not solid, but the best ones appear to be.

Figure 14: How holograms are created by capturing a waveform version of the subject image.

Holographic ‘solidity’

Our reality is not solid, it is holographic. We are now well aware of

holograms which are widely used today. Two-dimensional

information is decoded into a three-dimensional reality that is not

solid although can very much appear to be (Fig 13). Holograms are

created with a laser divided into two parts. One goes directly onto a

holographic photographic print (‘reference beam’) and the other

takes a waveform image of the subject (‘working beam’) before being

directed onto the print where it ‘collides’ with the other half of the

laser (Fig 14). This creates a waveform interference pa�ern which

contains the wavefield information of whatever is being

photographed (Fig 15 overleaf). The process can be likened to

dropping pebbles in a pond. Waves generated by each one spread

out across the water to collide with the others and create a wave

representation of where the stones fell and at what speed, weight

and distance. A waveform interference pa�ern of a hologram is akin

to the waveform information in The Field which the five senses

decode into electrical signals to be decoded by the brain into a

holographic illusory ‘physical’ reality. In the same way when a laser

(think human a�ention) is directed at the waveform interference

pa�ern a three-dimensional version of the subject is projected into

apparently ‘solid’ reality (Fig 16). An amazing trait of holograms

reveals more ‘paranormal mysteries’. Information of the whole

hologram is encoded in waveform in every part of the interference

pa�ern by the way they are created. This means that every part of a

hologram is a smaller version of the whole. Cut the interference

wave-pa�ern into four and you won’t get four parts of the image.

You get quarter-sized versions of the whole image. The body is a

hologram and the same applies. Here we have the basis of

acupuncture, reflexology and other forms of healing which identify

representations of the whole body in all of the parts, hands, feet,

ears, everywhere. Skilled palm readers can do what they do because

the information of whole body is encoded in the hand. The concept

of as above, so below, comes from this.

Figure 15: A waveform interference pattern that holds the information that transforms into a hologram.

Figure 16: Holographic people including ‘Elvis’ holographically inserted to sing a duet with Celine Dion.

The question will be asked of why, if solidity is illusory, we can’t

just walk through walls and each other. The resistance is not solid

against solid; it is electromagnetic field against electromagnetic field

and we decode this into the experience of solid against solid. We

should also not underestimate the power of belief to dictate reality.

What you believe is impossible will be. Your belief impacts on your

decoding processes and they won’t decode what you think is

impossible. What we believe we perceive and what we perceive we

experience. ‘Can’t dos’ and ‘impossibles’ are like a firewall in a

computer system that won’t put on the screen what the firewall

blocks. How vital that is to understanding how human experience

has been hijacked. I explain in The Answer, Everything You Need To

Know But Have Never Been Told and other books a long list of

‘mysteries’ and ‘paranormal’ phenomena that are not mysterious

and perfectly normal once you realise what reality is and how it

works. ‘Ghosts’ can be seen to pass through ‘solid’ walls because the

walls are not solid and the ghost is a discarnate entity operating on a

frequency so different to that of the wall that it’s like two radio

stations sharing the same space while never interfering with each

other. I have seen ghosts do this myself. The apartness of people and

objects is also an illusion. Everything is connected by the Field like

all sea life is connected by the sea. It’s just that within the limits of

our visual reality we only ‘see’ holographic information and not the

field of information that connects everything and from which the

holographic world is made manifest. If you can only see holographic

‘objects’ and not the field that connects them they will appear to you

as unconnected to each other in the same way that we see the

computer while not seeing the Wi-Fi.

What you don’t know can hurt you

Okay, we return to those ‘two worlds’ of human society and the Cult

with its global network of interconnecting secret societies and

satanic groups which manipulate through governments,

corporations, media, religions, etc. The fundamental difference

between them is knowledge. The idea has been to keep humanity

ignorant of the plan for its total enslavement underpinned by a

crucial ignorance of reality – who we are and where we are – and

how we interact with it. ‘Human’ should be the interaction between

our expanded eternal consciousness and the five-sense body

experience. We are meant to be in this world in terms of the five

senses but not of this world in relation to our greater consciousness

and perspective. In that state we experience the small picture of the

five senses within the wider context of the big picture of awareness

beyond the five senses. Put another way the five senses see the dots

and expanded awareness connects them into pictures and pa�erns

that give context to the apparently random and unconnected.

Without the context of expanded awareness the five senses see only

apartness and randomness with apparently no meaning. The Cult

and its other-dimensional controllers seek to intervene in the

frequency realm where five-sense reality is supposed to connect with

expanded reality and to keep the two apart (more on this in the final

chapter). When that happens five-sense mental and emotional

processes are no longer influenced by expanded awareness, or the

True ‘I’, and instead are driven by the isolated perceptions of the

body’s decoding systems. They are in the world and of it. Here we

have the human plight and why humanity with its potential for

infinite awareness can be so easily manipulatable and descend into

such extremes of stupidity.

Once the Cult isolates five-sense mind from expanded awareness

it can then program the mind with perceptions and beliefs by

controlling information that the mind receives through the

‘education’ system of the formative years and the media perceptual

bombardment and censorship of an entire lifetime. Limit perception

and a sense of the possible through limiting knowledge by limiting

and skewing information while censoring and discrediting that

which could set people free. As the title of another of my books says

… And The Truth Shall Set You Free. For this reason the last thing the

Cult wants in circulation is the truth about anything – especially the

reality of the eternal ‘I’ – and that’s why it is desperate to control

information. The Cult knows that information becomes perception

which becomes behaviour which, collectively, becomes human

society. Cult-controlled and funded mainstream ‘science’ denies the

existence of an eternal ‘I’ and seeks to dismiss and trash all evidence

to the contrary. Cult-controlled mainstream religion has a version of

‘God’ that is li�le more than a system of control and dictatorship

that employs threats of damnation in an a�erlife to control

perceptions and behaviour in the here and now through fear and

guilt. Neither is true and it’s the ‘neither’ that the Cult wishes to

suppress. This ‘neither’ is that everything is an expression, a point of

a�ention, within an infinite state of consciousness which is the real

meaning of the term ‘God’.

Perceptual obsession with the ‘physical body’ and five-senses

means that ‘God’ becomes personified as a bearded bloke si�ing

among the clouds or a raging bully who loves us if we do what ‘he’

wants and condemns us to the fires of hell if we don’t. These are no

more than a ‘spiritual’ fairy tales to control and dictate events and

behaviour through fear of this ‘God’ which has bizarrely made ‘God-

fearing’ in religious circles a state to be desired. I would suggest that

fearing anything is not to be encouraged and celebrated, but rather

deleted. You can see why ‘God fearing’ is so beneficial to the Cult

and its religions when they decide what ‘God’ wants and what ‘God’

demands (the Cult demands) that everyone do. As the great

American comedian Bill Hicks said satirising a Christian zealot: ‘I

think what God meant to say.’ How much of this infinite awareness

(‘God’) that we access is decided by how far we choose to expand

our perceptions, self-identity and sense of the possible. The scale of

self-identity reflects itself in the scale of awareness that we can

connect with and are influenced by – how much knowing and

insight we have instead of programmed perception. You cannot

expand your awareness into the infinity of possibility when you

believe that you are li�le me Peter the postman or Mary in marketing

and nothing more. I’ll deal with this in the concluding chapter

because it’s crucial to how we turnaround current events.

Where the Cult came from

When I realised in the early 1990s there was a Cult network behind

global events I asked the obvious question: When did it start? I took

it back to ancient Rome and Egypt and on to Babylon and Sumer in

Mesopotamia, the ‘Land Between Two Rivers’, in what we now call

Iraq. The two rivers are the Tigris and Euphrates and this region is of

immense historical and other importance to the Cult, as is the land

called Israel only 550 miles away by air. There is much more going

with deep esoteric meaning across this whole region. It’s not only

about ‘wars for oil’. Priceless artefacts from Mesopotamia were

stolen or destroyed a�er the American and British invasion of Iraq in

2003 justified by the lies of Boy Bush and Tony Blair (their Cult

masters) about non-existent ‘weapons of mass destruction’.

Mesopotamia was the location of Sumer (about 5,400BC to 1,750BC),

and Babylon (about 2,350BC to 539BC). Sabbatians may have become

immensely influential in the Cult in modern times but they are part

of a network that goes back into the mists of history. Sumer is said by

historians to be the ‘cradle of civilisation’. I disagree. I say it was the

re-start of what we call human civilisation a�er cataclysmic events

symbolised in part as the ‘Great Flood’ destroyed the world that

existed before. These fantastic upheavals that I have been describing

in detail in the books since the early1990s appear in accounts and

legends of ancient cultures across the world and they are supported

by geological and biological evidence. Stone tablets found in Iraq

detailing the Sumer period say the cataclysms were caused by non-

human ‘gods’ they call the Anunnaki. These are described in terms

of extraterrestrial visitations in which knowledge supplied by the

Anunnaki is said to have been the source of at least one of the

world’s oldest writing systems and developments in astronomy,

mathematics and architecture that were way ahead of their time. I

have covered this subject at length in The Biggest Secret and Children

of the Matrix and the same basic ‘Anunnaki’ story can be found in

Zulu accounts in South Africa where the late and very great Zulu

high shaman Credo Mutwa told me that the Sumerian Anunnaki

were known by Zulus as the Chitauri or ‘children of the serpent’. See

my six-hour video interview with Credo on this subject entitled The

Reptilian Agenda recorded at his then home near Johannesburg in

1999 which you can watch on the Ickonic media platform.

The Cult emerged out of Sumer, Babylon and Egypt (and

elsewhere) and established the Roman Empire before expanding

with the Romans into northern Europe from where many empires

were savagely imposed in the form of Cult-controlled societies all

over the world. Mass death and destruction was their calling card.

The Cult established its centre of operations in Europe and European

Empires were Cult empires which allowed it to expand into a global

force. Spanish and Portuguese colonialists headed for Central and

South America while the British and French targeted North America.

Africa was colonised by Britain, France, Belgium, the Netherlands,

Portugal, Spain, Italy, and Germany. Some like Britain and France

moved in on the Middle East. The British Empire was by far the

biggest for a simple reason. By now Britain was the headquarters of

the Cult from which it expanded to form Canada, the United States,

Australia and New Zealand. The Sun never set on the British Empire

such was the scale of its occupation. London remains a global centre

for the Cult along with Rome and the Vatican although others have

emerged in Israel and China. It is no accident that the ‘virus’ is

alleged to have come out of China while Italy was chosen as the

means to terrify the Western population into compliance with

‘Covid’ fascism. Nor that Israel has led the world in ‘Covid’ fascism

and mass ‘vaccination’.

You would think that I would mention the United States here, but

while it has been an important means of imposing the Cult’s will it is

less significant than would appear and is currently in the process of

having what power it does have deleted. The Cult in Europe has

mostly loaded the guns for the US to fire. America has been

controlled from Europe from the start through Cult operatives in

Britain and Europe. The American Revolution was an illusion to

make it appear that America was governing itself while very

different forces were pulling the strings in the form of Cult families

such as the Rothschilds through the Rockefellers and other

subordinates. The Rockefellers are extremely close to Bill Gates and

established both scalpel and drug ‘medicine’ and the World Health

Organization. They play a major role in the development and

circulation of vaccines through the Rockefeller Foundation on which

Bill Gates said his Foundation is based. Why wouldn’t this be the

case when the Rockefellers and Gates are on the same team? Cult

infiltration of human society goes way back into what we call history

and has been constantly expanding and centralising power with the

goal of establishing a global structure to dictate everything. Look

how this has been advanced in great leaps with the ‘Covid’ hoax.

The non-human dimension

I researched and observed the comings and goings of Cult operatives

through the centuries and even thousands of years as they were

born, worked to promote the agenda within the secret society and

satanic networks, and then died for others to replace them. Clearly

there had to be a coordinating force that spanned this entire period

while operatives who would not have seen the end goal in their

lifetimes came and went advancing the plan over millennia. I went

in search of that coordinating force with the usual support from the

extraordinary synchronicity of my life which has been an almost

daily experience since 1990. I saw common themes in religious texts

and ancient cultures about a non-human force manipulating human

society from the hidden. Christianity calls this force Satan, the Devil

and demons; Islam refers to the Jinn or Djinn; Zulus have their

Chitauri (spelt in other ways in different parts of Africa); and the

Gnostic people in Egypt in the period around and before 400AD

referred to this phenomena as the ‘Archons’, a word meaning rulers

in Greek. Central American cultures speak of the ‘Predators’ among

other names and the same theme is everywhere. I will use ‘Archons’

as a collective name for all of them. When you see how their nature

and behaviour is described all these different sources are clearly

talking about the same force. Gnostics described the Archons in

terms of ‘luminous fire’ while Islam relates the Jinn to ‘smokeless

fire’. Some refer to beings in form that could occasionally be seen,

but the most common of common theme is that they operate from

unseen realms which means almost all existence to the visual

processes of humans. I had concluded that this was indeed the

foundation of human control and that the Cult was operating within

the human frequency band on behalf of this hidden force when I

came across the writings of Gnostics which supported my

conclusions in the most extraordinary way.

A sealed earthen jar was found in 1945 near the town of Nag

Hammadi about 75-80 miles north of Luxor on the banks of the River

Nile in Egypt. Inside was a treasure trove of manuscripts and texts

le� by the Gnostic people some 1,600 years earlier. They included 13

leather-bound papyrus codices (manuscripts) and more than 50 texts

wri�en in Coptic Egyptian estimated to have been hidden in the jar

in the period of 400AD although the source of the information goes

back much further. Gnostics oversaw the Great or Royal Library of

Alexandria, the fantastic depository of ancient texts detailing

advanced knowledge and accounts of human history. The Library

was dismantled and destroyed in stages over a long period with the

death-blow delivered by the Cult-established Roman Church in the

period around 415AD. The Church of Rome was the Church of

Babylon relocated as I said earlier. Gnostics were not a race. They

were a way of perceiving reality. Whenever they established

themselves and their information circulated the terrorists of the

Church of Rome would target them for destruction. This happened

with the Great Library and with the Gnostic Cathars who were

burned to death by the psychopaths a�er a long period of

oppression at the siege of the Castle of Monségur in southern France

in 1244. The Church has always been terrified of Gnostic information

which demolishes the official Christian narrative although there is

much in the Bible that supports the Gnostic view if you read it in

another way. To anyone studying the texts of what became known as

the Nag Hammadi Library it is clear that great swathes of Christian

and Biblical belief has its origin with Gnostics sources going back to

Sumer. Gnostic themes have been twisted to manipulate the

perceived reality of Bible believers. Biblical texts have been in the

open for centuries where they could be changed while Gnostic

documents found at Nag Hammadi were sealed away and

untouched for 1,600 years. What you see is what they wrote.

Use your pneuma not your nous

Gnosticism and Gnostic come from ‘gnosis’ which means

knowledge, or rather secret knowledge, in the sense of spiritual

awareness – knowledge about reality and life itself. The desperation

of the Cult’s Church of Rome to destroy the Gnostics can be

understood when the knowledge they were circulating was the last

thing the Cult wanted the population to know. Sixteen hundred

years later the same Cult is working hard to undermine and silence

me for the same reason. The dynamic between knowledge and

ignorance is a constant. ‘Time’ appears to move on, but essential

themes remain the same. We are told to ‘use your nous’, a Gnostic

word for head/brain/intelligence. They said, however, that spiritual

awakening or ‘salvation’ could only be secured by expanding

awareness beyond what they called nous and into pneuma or Infinite

Self. Obviously as I read these texts the parallels with what I have

been saying since 1990 were fascinating to me. There is a universal

truth that spans human history and in that case why wouldn’t we be

talking the same language 16 centuries apart? When you free

yourself from the perception program of the five senses and explore

expanded realms of consciousness you are going to connect with the

same information no ma�er what the perceived ‘era’ within a

manufactured timeline of a single and tiny range of manipulated

frequency. Humans working with ‘smart’ technology or knocking

rocks together in caves is only a timeline appearing to operate within

the human frequency band. Expanded awareness and the

knowledge it holds have always been there whether the era be Stone

Age or computer age. We can only access that knowledge by

opening ourselves to its frequency which the five-sense prison cell is

designed to stop us doing. Gates, Fauci, Whi�y, Vallance,

Zuckerberg, Brin, Page, Wojcicki, Bezos, and all the others behind

the ‘Covid’ hoax clearly have a long wait before their range of

frequency can make that connection given that an open heart is

crucial to that as we shall see. Instead of accessing knowledge

directly through expanded awareness it is given to Cult operatives

by the secret society networks of the Cult where it has been passed

on over thousands of years outside the public arena. Expanded

realms of consciousness is where great artists, composers and

writers find their inspiration and where truth awaits anyone open

enough to connect with it. We need to go there fast.

Archon hijack

A fi�h of the Nag Hammadi texts describe the existence and

manipulation of the Archons led by a ‘Chief Archon’ they call

‘Yaldabaoth’, or the ‘Demiurge’, and this is the Christian ‘Devil’,

‘Satan’, ‘Lucifer’, and his demons. Archons in Biblical symbolism are

the ‘fallen ones’ which are also referred to as fallen angels a�er the

angels expelled from heaven according to the Abrahamic religions of

Judaism, Christianity and Islam. These angels are claimed to tempt

humans to ‘sin’ ongoing and you will see how accurate that

symbolism is during the rest of the book. The theme of ‘original sin’

is related to the ‘Fall’ when Adam and Eve were ‘tempted by the

serpent’ and fell from a state of innocence and ‘obedience’

(connection) with God into a state of disobedience (disconnection).

The Fall is said to have brought sin into the world and corrupted

everything including human nature. Yaldabaoth, the ‘Lord Archon’,

is described by Gnostics as a ‘counterfeit spirit’, ‘The Blind One’,

‘The Blind God’, and ‘The Foolish One’. The Jewish name for

Yaldabaoth in Talmudic writings is Samael which translates as

‘Poison of God’, or ‘Blindness of God’. You see the parallels.

Yaldabaoth in Islamic belief is the Muslim Jinn devil known as

Shaytan – Shaytan is Satan as the same themes are found all over the

world in every religion and culture. The ‘Lord God’ of the Old

Testament is the ‘Lord Archon’ of Gnostic manuscripts and that’s

why he’s such a bloodthirsty bastard. Satan is known by Christians

as ‘the Demon of Demons’ and Gnostics called Yaldabaoth the

‘Archon of Archons’. Both are known as ‘The Deceiver’. We are

talking about the same ‘bloke’ for sure and these common themes

using different names, storylines and symbolism tell a common tale

of the human plight.

Archons are referred to in Nag Hammadi documents as mind

parasites, inverters, guards, gatekeepers, detainers, judges, pitiless

ones and deceivers. The ‘Covid’ hoax alone is a glaring example of

all these things. The Biblical ‘God’ is so different in the Old and New

Testaments because they are not describing the same phenomenon.

The vindictive, angry, hate-filled, ‘God’ of the Old Testament, known

as Yahweh, is Yaldabaoth who is depicted in Cult-dictated popular

culture as the ‘Dark Lord’, ‘Lord of Time’, Lord (Darth) Vader and

Dormammu, the evil ruler of the ‘Dark Dimension’ trying to take

over the ‘Earth Dimension’ in the Marvel comic movie, Dr Strange.

Yaldabaoth is both the Old Testament ‘god’ and the Biblical ‘Satan’.

Gnostics referred to Yaldabaoth as the ‘Great Architect of the

Universe’and the Cult-controlled Freemason network calls their god

‘the ‘Great Architect of the Universe’ (also Grand Architect). The

‘Great Architect’ Yaldabaoth is symbolised by the Cult as the all-

seeing eye at the top of the pyramid on the Great Seal of the United

States and the dollar bill. Archon is encoded in arch-itect as it is in

arch-angels and arch-bishops. All religions have the theme of a force

for good and force for evil in some sort of spiritual war and there is a

reason for that – the theme is true. The Cult and its non-human

masters are quite happy for this to circulate. They present

themselves as the force for good fighting evil when they are really

the force of evil (absence of love). The whole foundation of Cult

modus operandi is inversion. They promote themselves as a force for

good and anyone challenging them in pursuit of peace, love,

fairness, truth and justice is condemned as a satanic force for evil.

This has been the game plan throughout history whether the Church

of Rome inquisitions of non-believers or ‘conspiracy theorists’ and

‘anti-vaxxers’ of today. The technique is the same whatever the

timeline era.

Yaldabaoth is revolting (true)

Yaldabaoth and the Archons are said to have revolted against God

with Yaldabaoth claiming to be God – the All That Is. The Old

Testament ‘God’ (Yaldabaoth) demanded to be worshipped as such: ‘

I am the LORD, and there is none else, there is no God beside me’

(Isaiah 45:5). I have quoted in other books a man who said he was

the unofficial son of the late Baron Philippe de Rothschild of the

Mouton-Rothschild wine producing estates in France who died in

1988 and he told me about the Rothschild ‘revolt from God’. The

man said he was given the name Phillip Eugene de Rothschild and

we shared long correspondence many years ago while he was living

under another identity. He said that he was conceived through

‘occult incest’ which (within the Cult) was ‘normal and to be

admired’. ‘Phillip’ told me about his experience a�ending satanic

rituals with rich and famous people whom he names and you can

see them and the wider background to Cult Satanism in my other

books starting with The Biggest Secret. Cult rituals are interactions

with Archontic ‘gods’. ‘Phillip’ described Baron Philippe de

Rothschild as ‘a master Satanist and hater of God’ and he used the

same term ‘revolt from God’ associated with

Yaldabaoth/Satan/Lucifer/the Devil in describing the Sabbatian

Rothschild dynasty. ‘I played a key role in my family’s revolt from

God’, he said. That role was to infiltrate in classic Sabbatian style the

Christian Church, but eventually he escaped the mind-prison to live

another life. The Cult has been targeting religion in a plan to make

worship of the Archons the global one-world religion. Infiltration of

Satanism into modern ‘culture’, especially among the young,

through music videos, stage shows and other means, is all part of

this.

Nag Hammadi texts describe Yaldabaoth and the Archons in their

prime form as energy – consciousness – and say they can take form if

they choose in the same way that consciousness takes form as a

human. Yaldabaoth is called ‘formless’ and represents a deeply

inverted, distorted and chaotic state of consciousness which seeks to

a�ached to humans and turn them into a likeness of itself in an

a�empt at assimilation. For that to happen it has to manipulate

humans into low frequency mental and emotional states that match

its own. Archons can certainly appear in human form and this is the

origin of the psychopathic personality. The energetic distortion

Gnostics called Yaldabaoth is psychopathy. When psychopathic

Archons take human form that human will be a psychopath as an

expression of Yaldabaoth consciousness. Cult psychopaths are

Archons in human form. The principle is the same as that portrayed

in the 2009 Avatar movie when the American military travelled to a

fictional Earth-like moon called Pandora in the Alpha Centauri star

system to infiltrate a society of blue people, or Na’vi, by hiding

within bodies that looked like the Na’vi. Archons posing as humans

have a particular hybrid information field, part human, part Archon,

(the ancient ‘demigods’) which processes information in a way that

manifests behaviour to match their psychopathic evil, lack of

empathy and compassion, and stops them being influenced by the

empathy, compassion and love that a fully-human information field

is capable of expressing. Cult bloodlines interbreed, be they royalty

or dark suits, for this reason and you have their obsession with

incest. Interbreeding with full-blown humans would dilute the

Archontic energy field that guarantees psychopathy in its

representatives in the human realm.

Gnostic writings say the main non-human forms that Archons

take are serpentine (what I have called for decades ‘reptilian’ amid

unbounded ridicule from the Archontically-programmed) and what

Gnostics describe as ‘an unborn baby or foetus with grey skin and

dark, unmoving eyes’. This is an excellent representation of the ET

‘Greys’ of UFO folklore which large numbers of people claim to have

seen and been abducted by – Zulu shaman Credo Mutwa among

them. I agree with those that believe in extraterrestrial or

interdimensional visitations today and for thousands of years past.

No wonder with their advanced knowledge and technological

capability they were perceived and worshipped as gods for

technological and other ‘miracles’ they appeared to perform.

Imagine someone arriving in a culture disconnected from the

modern world with a smartphone and computer. They would be

seen as a ‘god’ capable of ‘miracles’. The Renegade Mind, however,

wants to know the source of everything and not only the way that

source manifests as human or non-human. In the same way that a

Renegade Mind seeks the original source material for the ‘Covid

virus’ to see if what is claimed is true. The original source of

Archons in form is consciousness – the distorted state of

consciousness known to Gnostics as Yaldabaoth.

‘Revolt from God’ is energetic disconnection

Where I am going next will make a lot of sense of religious texts and

ancient legends relating to ‘Satan’, Lucifer’ and the ‘gods’. Gnostic

descriptions sync perfectly with the themes of my own research over

the years in how they describe a consciousness distortion seeking to

impose itself on human consciousness. I’ve referred to the core of

infinite awareness in previous books as Infinite Awareness in

Awareness of Itself. By that I mean a level of awareness that knows

that it is all awareness and is aware of all awareness. From here

comes the frequency of love in its true sense and balance which is

what love is on one level – the balance of all forces into a single

whole called Oneness and Isness. The more we disconnect from this

state of love that many call ‘God’ the constituent parts of that

Oneness start to unravel and express themselves as a part and not a

whole. They become individualised as intellect, mind, selfishness,

hatred, envy, desire for power over others, and such like. This is not

a problem in the greater scheme in that ‘God’, the All That Is, can

experience all these possibilities through different expressions of

itself including humans. What we as expressions of the whole

experience the All That Is experiences. We are the All That Is

experiencing itself. As we withdraw from that state of Oneness we

disconnect from its influence and things can get very unpleasant and

very stupid. Archontic consciousness is at the extreme end of that. It

has so disconnected from the influence of Oneness that it has become

an inversion of unity and love, an inversion of everything, an

inversion of life itself. Evil is appropriately live wri�en backwards.

Archontic consciousness is obsessed with death, an inversion of life,

and so its manifestations in Satanism are obsessed with death. They

use inverted symbols in their rituals such as the inverted pentagram

and cross. Sabbatians as Archontic consciousness incarnate invert

Judaism and every other religion and culture they infiltrate. They

seek disunity and chaos and they fear unity and harmony as they

fear love like garlic to a vampire. As a result the Cult, Archons

incarnate, act with such evil, psychopathy and lack of empathy and

compassion disconnected as they are from the source of love. How

could Bill Gates and the rest of the Archontic psychopaths do what

they have to human society in the ‘Covid’ era with all the death,

suffering and destruction involved and have no emotional

consequence for the impact on others? Now you know. Why have

Zuckerberg, Brin, Page, Wojcicki and company callously censored

information warning about the dangers of the ‘vaccine’ while

thousands have been dying and having severe, sometimes life-

changing reactions? Now you know. Why have Tedros, Fauci,

Whi�y, Vallance and their like around the world been using case and

death figures they’re aware are fraudulent to justify lockdowns and

all the deaths and destroyed lives that have come from that? Now

you know. Why did Christian Drosten produce and promote a

‘testing’ protocol that he knew couldn’t test for infectious disease

which led to a global human catastrophe. Now you know. The

Archontic mind doesn’t give a shit (Fig 17). I personally think that

Gates and major Cult insiders are a form of AI cyborg that the

Archons want humans to become.

Figure 17: Artist Neil Hague’s version of the ‘Covid’ hierarchy.

Human batteries

A state of such inversion does have its consequences, however. The

level of disconnection from the Source of All means that you

withdraw from that source of energetic sustenance and creativity.

This means that you have to find your own supply of energetic

power and it has – us. When the Morpheus character in the first

Matrix movie held up a ba�ery he spoke a profound truth when he

said: ‘The Matrix is a computer-generated dream world built to keep

us under control in order to change the human being into one of

these.’ The statement was true in all respects. We do live in a

technologically-generated virtual reality simulation (more very

shortly) and we have been manipulated to be an energy source for

Archontic consciousness. The Disney-Pixar animated movie

Monsters, Inc. in 2001 symbolised the dynamic when monsters in

their world had no energy source and they would enter the human

world to terrify children in their beds, catch the child’s scream, terror

(low-vibrational frequencies), and take that energy back to power

the monster world. The lead character you might remember was a

single giant eye and the symbolism of the Cult’s all-seeing eye was

obvious. Every thought and emotion is broadcast as a frequency

unique to that thought and emotion. Feelings of love and joy,

empathy and compassion, are high, quick, frequencies while fear,

depression, anxiety, suffering and hate are low, slow, dense

frequencies. Which kind do you think Archontic consciousness can

connect with and absorb? In such a low and dense frequency state

there’s no way it can connect with the energy of love and joy.

Archons can only feed off energy compatible with their own

frequency and they and their Cult agents want to delete the human

world of love and joy and manipulate the transmission of low

vibrational frequencies through low-vibrational human mental and

emotional states. We are their energy source. Wars are energetic

banquets to the Archons – a world war even more so – and think

how much low-frequency mental and emotional energy has been

generated from the consequences for humanity of the ‘Covid’ hoax

orchestrated by Archons incarnate like Gates.

The ancient practice of human sacrifice ‘to the gods’, continued in

secret today by the Cult, is based on the same principle. ‘The gods’

are Archontic consciousness in different forms and the sacrifice is

induced into a state of intense terror to generate the energy the

Archontic frequency can absorb. Incarnate Archons in the ritual

drink the blood which contains an adrenaline they crave which

floods into the bloodstream when people are terrorised. Most of the

sacrifices, ancient and modern, are children and the theme of

‘sacrificing young virgins to the gods’ is just code for children. They

have a particular pre-puberty energy that Archons want more than

anything and the energy of the young in general is their target. The

California Department of Education wants students to chant the

names of Aztec gods (Archontic gods) once worshipped in human

sacrifice rituals in a curriculum designed to encourage them to

‘challenge racist, bigoted, discriminatory, imperialist/colonial

beliefs’, join ‘social movements that struggle for social justice’, and

‘build new possibilities for a post-racist, post-systemic racism

society’. It’s the usual Woke crap that inverts racism and calls it anti-

racism. In this case solidarity with ‘indigenous tribes’ is being used

as an excuse to chant the names of ‘gods’ to which people were

sacrificed (and still are in secret). What an example of Woke’s

inability to see beyond black and white, us and them, They condemn

the colonisation of these tribal cultures by Europeans (quite right),

but those cultures sacrificing people including children to their

‘gods’, and mass murdering untold numbers as the Aztecs did, is

just fine. One chant is to the Aztec god Tezcatlipoca who had a man

sacrificed to him in the 5th month of the Aztec calendar. His heart

was cut out and he was eaten. Oh, that’s okay then. Come on

children … a�er three … Other sacrificial ‘gods’ for the young to

chant their allegiance include Quetzalcoatl, Huitzilopochtli and Xipe

Totec. The curriculum says that ‘chants, affirmations, and energizers

can be used to bring the class together, build unity around ethnic

studies principles and values, and to reinvigorate the class following

a lesson that may be emotionally taxing or even when student

engagement may appear to be low’. Well, that’s the cover story,

anyway. Chanting and mantras are the repetition of a particular

frequency generated from the vocal cords and chanting the names of

these Archontic ‘gods’ tunes you into their frequency. That is the last

thing you want when it allows for energetic synchronisation,

a�achment and perceptual influence. Initiates chant the names of

their ‘Gods’ in their rituals for this very reason.

Vampires of the Woke

Paedophilia is another way that Archons absorb the energy of

children. Paedophiles possessed by Archontic consciousness are

used as the conduit during sexual abuse for discarnate Archons to

vampire the energy of the young they desire so much. Stupendous

numbers of children disappear every year never to be seen again

although you would never know from the media. Imagine how

much low-vibrational energy has been generated by children during

the ‘Covid’ hoax when so many have become depressed and

psychologically destroyed to the point of killing themselves.

Shocking numbers of children are now taken by the state from

loving parents to be handed to others. I can tell you from long

experience of researching this since 1996 that many end up with

paedophiles and assets of the Cult through corrupt and Cult-owned

social services which in the reframing era has hired many

psychopaths and emotionless automatons to do the job. Children are

even stolen to order using spurious reasons to take them by the

corrupt and secret (because they’re corrupt) ‘family courts’. I have

wri�en in detail in other books, starting with The Biggest Secret in

1997, about the ubiquitous connections between the political,

corporate, government, intelligence and military elites (Cult

operatives) and Satanism and paedophilia. If you go deep enough

both networks have an interlocking leadership. The Woke mentality

has been developed by the Cult for many reasons: To promote

almost every aspect of its agenda; to hijack the traditional political

le� and turn it fascist; to divide and rule; and to target agenda

pushbackers. But there are other reasons which relate to what I am

describing here. How many happy and joyful Wokers do you ever

see especially at the extreme end? They are a mental and

psychological mess consumed by emotional stress and constantly

emotionally cocked for the next explosion of indignation at someone

referring to a female as a female. They are walking, talking, ba�eries

as Morpheus might say emi�ing frequencies which both enslave

them in low-vibrational bubbles of perceptual limitation and feed

the Archons. Add to this the hatred claimed to be love; fascism

claimed to ‘anti-fascism’, racism claimed to be ‘anti-racism’;

exclusion claimed to inclusion; and the abuse-filled Internet trolling.

You have a purpose-built Archontic energy system with not a wind

turbine in sight and all founded on Archontic inversion. We have

whole generations now manipulated to serve the Archons with their

actions and energy. They will be doing so their entire adult lives

unless they snap out of their Archon-induced trance. Is it really a

surprise that Cult billionaires and corporations put so much money

their way? Where is the energy of joy and laughter, including

laughing at yourself which is confirmation of your own emotional

security? Mark Twain said: ‘The human race has one really effective

weapon, and that is laughter.‘ We must use it all the time. Woke has

destroyed comedy because it has no humour, no joy, sense of irony,

or self-deprecation. Its energy is dense and intense. Mmmmm, lunch

says the Archontic frequency. Rudolf Steiner (1861-1925) was the

Austrian philosopher and famous esoteric thinker who established

Waldorf education or Steiner schools to treat children like unique

expressions of consciousness and not minds to be programmed with

the perceptions determined by authority. I’d been writing about this

energy vampiring for decades when I was sent in 2016 a quote by

Steiner. He was spot on:

There are beings in the spiritual realms for whom anxiety and fear emanating from human beings offer welcome food. When humans have no anxiety and fear, then these creatures starve. If fear and anxiety radiates from people and they break out in panic, then these creatures find welcome nutrition and they become more and more powerful. These beings are hostile towards humanity. Everything that feeds on negative feelings, on anxiety, fear and superstition, despair or doubt, are in reality hostile forces in super-sensible worlds, launching cruel attacks on human beings, while they are being fed ... These are exactly the feelings that belong to contemporary culture and materialism; because it estranges people from the spiritual world, it is especially suited to evoke hopelessness and fear of the unknown in people, thereby calling up the above mentioned hostile forces against them.

Pause for a moment from this perspective and reflect on what has

happened in the world since the start of 2020. Not only will pennies

drop, but billion dollar bills. We see the same theme from Don Juan

Matus, a Yaqui Indian shaman in Mexico and the information source

for Peruvian-born writer, Carlos Castaneda, who wrote a series of

books from the 1960s to 1990s. Don Juan described the force

manipulating human society and his name for the Archons was the

predator:

We have a predator that came from the depths of the cosmos and took over the rule of our lives. Human beings are its prisoners. The predator is our lord and master. It has rendered us docile, helpless. If we want to protest, it suppresses our protest. If we want to act independently, it demands that we don’t do so ... indeed we are held prisoner!

They took us over because we are food to them, and they squeeze us mercilessly because we are their sustenance. Just as we rear chickens in coops, the predators rear us in human coops, humaneros. Therefore, their food is always available to them.

Different cultures, different eras, same recurring theme.

The ‘ennoia’ dilemma

Nag Hammadi Gnostic manuscripts say that Archon consciousness

has no ‘ennoia’. This is directly translated as ‘intentionality’, but I’ll

use the term ‘creative imagination’. The All That Is in awareness of

itself is the source of all creativity – all possibility – and the more

disconnected you are from that source the more you are

subsequently denied ‘creative imagination’. Given that Archon

consciousness is almost entirely disconnected it severely lacks

creativity and has to rely on far more mechanical processes of

thought and exploit the creative potential of those that do have

‘ennoia’. You can see cases of this throughout human society. Archon

consciousness almost entirely dominates the global banking system

and if we study how that system works you will appreciate what I

mean. Banks manifest ‘money’ out of nothing by issuing lines of

‘credit’ which is ‘money’ that has never, does not, and will never

exist except in theory. It’s a confidence trick. If you think ‘credit’

figures-on-a-screen ‘money’ is worth anything you accept it as

payment. If you don’t then the whole system collapses through lack

of confidence in the value of that ‘money’. Archontic bankers with

no ‘ennoia’ are ‘lending’ ‘money’ that doesn’t exist to humans that do

have creativity – those that have the inspired ideas and create

businesses and products. Archon banking feeds off human creativity

which it controls through ‘money’ creation and debt. Humans have

the creativity and Archons exploit that for their own benefit and

control while having none themselves. Archon Internet platforms

like Facebook claim joint copyright of everything that creative users

post and while Archontic minds like Zuckerberg may officially head

that company it will be human creatives on the staff that provide the

creative inspiration. When you have limitless ‘money’ you can then

buy other companies established by creative humans. Witness the

acquisition record of Facebook, Google and their like. Survey the

Archon-controlled music industry and you see non-creative dark

suit executives making their fortune from the human creativity of

their artists. The cases are endless. Research the history of people

like Gates and Zuckerberg and how their empires were built on

exploiting the creativity of others. Archon minds cannot create out of

nothing, but they are skilled (because they have to be) in what

Gnostic texts call ‘countermimicry’. They can imitate, but not

innovate. Sabbatians trawl the creativity of others through

backdoors they install in computer systems through their

cybersecurity systems. Archon-controlled China is globally infamous

for stealing intellectual property and I remember how Hong Kong,

now part of China, became notorious for making counterfeit copies

of the creativity of others – ‘countermimicry’. With the now

pervasive and all-seeing surveillance systems able to infiltrate any

computer you can appreciate the potential for Archons to vampire

the creativity of humans. Author John Lamb Lash wrote in his book

about the Nag Hammadi texts, Not In His Image:

Although they cannot originate anything, because they lack the divine factor of ennoia (intentionality), Archons can imitate with a vengeance. Their expertise is simulation (HAL, virtual reality). The Demiurge [Yaldabaoth] fashions a heaven world copied from the fractal patterns [of the original] ... His construction is celestial kitsch, like the fake Italianate villa of a Mafia don complete with militant angels to guard every portal.

This brings us to something that I have been speaking about since

the turn of the millennium. Our reality is a simulation; a virtual

reality that we think is real. No, I’m not kidding.

Human reality? Well, virtually

I had pondered for years about whether our reality is ‘real’ or some

kind of construct. I remembered being immensely affected on a visit

as a small child in the late 1950s to the then newly-opened

Planetarium on the Marylebone Road in London which is now

closed and part of the adjacent Madame Tussauds wax museum. It

was in the middle of the day, but when the lights went out there was

the night sky projected in the Planetarium’s domed ceiling and it

appeared to be so real. The experience never le� me and I didn’t

know why until around the turn of the millennium when I became

certain that our ‘night sky’ and entire reality is a projection, a virtual

reality, akin to the illusory world portrayed in the Matrix movies. I

looked at the sky one day in this period and it appeared to me like

the domed roof of the Planetarium. The release of the first Matrix

movie in 1999 also provided a synchronistic and perfect visual

representation of where my mind had been going for a long time. I

hadn’t come across the Gnostic Nag Hammadi texts then. When I

did years later the correlation was once again astounding. As I read

Gnostic accounts from 1,600 years and more earlier it was clear that

they were describing the same simulation phenomenon. They tell

how the Yaldabaoth ‘Demiurge’ and Archons created a ‘bad copy’ of

original reality to rule over all that were captured by its illusions and

the body was a prison to trap consciousness in the ‘bad copy’ fake

reality. Read how Gnostics describe the ‘bad copy’ and update that

to current times and they are referring to what we would call today a

virtual reality simulation.

Author John Lamb Lash said ‘the Demiurge fashions a heaven

world copied from the fractal pa�erns’ of the original through

expertise in ‘HAL’ or virtual reality simulation. Fractal pa�erns are

part of the energetic information construct of our reality, a sort of

blueprint. If these pa�erns were copied in computer terms it would

indeed give you a copy of a ‘natural’ reality in a non-natural

frequency and digital form. The principle is the same as making a

copy of a website. The original website still exists, but now you can

change the copy version to make it whatever you like and it can

become very different to the original website. Archons have done

this with our reality, a synthetic copy of prime reality that still exists

beyond the frequency walls of the simulation. Trapped within the

illusions of this synthetic Matrix, however, were and are human

consciousness and other expressions of prime reality and this is why

the Archons via the Cult are seeking to make the human body

synthetic and give us synthetic AI minds to complete the job of

turning the entire reality synthetic including what we perceive to be

the natural world. To quote Kurzweil: ‘Nanobots will infuse all the

ma�er around us with information. Rocks, trees, everything will

become these intelligent creatures.’ Yes, synthetic ‘creatures’ just as

‘Covid’ and other genetically-manipulating ‘vaccines’ are designed

to make the human body synthetic. From this perspective it is

obvious why Archons and their Cult are so desperate to infuse

synthetic material into every human with their ‘Covid’ scam.

Let there be (electromagnetic) light

Yaldabaoth, the force that created the simulation, or Matrix, makes

sense of the Gnostic reference to ‘The Great Architect’ and its use by

Cult Freemasonry as the name of its deity. The designer of the Matrix

in the movies is called ‘The Architect’ and that trilogy is jam-packed

with symbolism relating to these subjects. I have contended for years

that the angry Old Testament God (Yaldabaoth) is the ‘God’ being

symbolically ‘quoted’ in the opening of Genesis as ‘creating the

world’. This is not the creation of prime reality – it’s the creation of

the simulation. The Genesis ‘God’ says: ‘Let there be Light: and there

was light.’ But what is this ‘Light’? I have said for decades that the

speed of light (186,000 miles per second) is not the fastest speed

possible as claimed by mainstream science and is in fact the

frequency walls or outer limits of the Matrix. You can’t have a fastest

or slowest anything within all possibility when everything is

possible. The human body is encoded to operate within the speed of

light or within the simulation and thus we see only the tiny frequency

band of visible light. Near-death experiencers who perceive reality

outside the body during temporary ‘death’ describe a very different

form of light and this is supported by the Nag Hammadi texts.

Prime reality beyond the simulation (‘Upper Aeons’ to the Gnostics)

is described as a realm of incredible beauty, bliss, love and harmony

– a realm of ‘watery light’ that is so powerful ‘there are no shadows’.

Our false reality of Archon control, which Gnostics call the ‘Lower

Aeons’, is depicted as a realm with a different kind of ‘light’ and

described in terms of chaos, ‘Hell’, ‘the Abyss’ and ‘Outer Darkness’,

where trapped souls are tormented and manipulated by demons

(relate that to the ‘Covid’ hoax alone). The watery light theme can be

found in near-death accounts and it is not the same as simulation

‘light’ which is electromagnetic or radiation light within the speed of

light – the ‘Lower Aeons’. Simulation ‘light’ is the ‘luminous fire’

associated by Gnostics with the Archons. The Bible refers to

Yaldabaoth as ‘that old serpent, called the Devil, and Satan, which

deceiveth the whole world’ (Revelation 12:9). I think that making a

simulated copy of prime reality (‘countermimicry’) and changing it

dramatically while all the time manipulating humanity to believe it

to be real could probably meet the criteria of deceiving the whole

world. Then we come to the Cult god Lucifer – the Light Bringer.

Lucifer is symbolic of Yaldabaoth, the bringer of radiation light that

forms the bad copy simulation within the speed of light. ‘He’ is

symbolised by the lighted torch held by the Statue of Liberty and in

the name ‘Illuminati’. Sabbatian-Frankism declares that Lucifer is the

true god and Lucifer is the real god of Freemasonry honoured as

their ‘Great or Grand Architect of the Universe’ (simulation).

I would emphasise, too, the way Archontic technologically-

generated luminous fire of radiation has deluged our environment

since I was a kid in the 1950s and changed the nature of The Field

with which we constantly interact. Through that interaction

technological radiation is changing us. The Smart Grid is designed to

operate with immense levels of communication power with 5G

expanding across the world and 6G, 7G, in the process of

development. Radiation is the simulation and the Archontic

manipulation system. Why wouldn’t the Archon Cult wish to

unleash radiation upon us to an ever-greater extreme to form

Kurzweil’s ‘cloud’? The plan for a synthetic human is related to the

need to cope with levels of radiation beyond even anything we’ve

seen so far. Biological humans would not survive the scale of

radiation they have in their script. The Smart Grid is a technological

sub-reality within the technological simulation to further disconnect

five-sense perception from expanded consciousness. It’s a

technological prison of the mind.

Infusing the ‘spirit of darkness’

A recurring theme in religion and native cultures is the

manipulation of human genetics by a non-human force and most

famously recorded as the biblical ‘sons of god’ (the gods plural in the

original) who interbred with the daughters of men. The Nag

Hammadi Apocryphon of John tells the same story this way:

He [Yaldabaoth] sent his angels [Archons/demons] to the daughters of men, that they might take some of them for themselves and raise offspring for their enjoyment. And at first they did not succeed. When they had no success, they gathered together again and they made a plan together ... And the angels changed themselves in their likeness into the likeness of their mates, filling them with the spirit of darkness, which they had mixed for them, and with evil ... And they took women and begot children out of the darkness according to the likeness of their spirit.

Possession when a discarnate entity takes over a human body is an

age-old theme and continues today. It’s very real and I’ve seen it.

Satanic and secret society rituals can create an energetic environment

in which entities can a�ach to initiates and I’ve heard many stories

of how people have changed their personality a�er being initiated

even into lower levels of the Freemasons. I have been inside three

Freemasonic temples, one at a public open day and two by just

walking in when there was no one around to stop me. They were in

Ryde, the town where I live, Birmingham, England, when I was with

a group, and Boston, Massachuse�s. They all felt the same

energetically – dark, dense, low-vibrational and sinister. Demonic

a�achment can happen while the initiate has no idea what is going

on. To them it’s just a ritual to get in the Masons and do a bit of good

business. In the far more extreme rituals of Satanism human

possession is even more powerful and they are designed to make

possession possible. The hierarchy of the Cult is dictated by the

power and perceived status of the possessing Archon. In this way

the Archon hierarchy becomes the Cult hierarchy. Once the entity

has a�ached it can influence perception and behaviour and if it

a�aches to the extreme then so much of its energy (information)

infuses into the body information field that the hologram starts to

reflect the nature of the possessing entity. This is the Exorcist movie

type of possession when facial features change and it’s known as

shapeshi�ing. Islam’s Jinn are said to be invisible tricksters who

change shape, ‘whisper’, confuse and take human form. These are all

traits of the Archons and other versions of the same phenomenon.

Extreme possession could certainty infuse the ‘spirit of darkness’

into a partner during sex as the Nag Hammadi texts appear to

describe. Such an infusion can change genetics which is also

energetic information. Human genetics is information and the ‘spirit

of darkness’ is information. Mix one with the other and change must

happen. Islam has the concept of a ‘Jinn baby’ through possession of

the mother and by Jinn taking human form. There are many ways

that human genetics can be changed and remember that Archons

have been aware all along of advanced techniques to do this. What is

being done in human society today – and far more – was known

about by Archons at the time of the ‘fallen ones’ and their other

versions described in religions and cultures.

Archons and their human-world Cult are obsessed with genetics

as we see today and they know this dictates how information is

processed into perceived reality during a human life. They needed to

produce a human form that would decode the simulation and this is

symbolically known as ‘Adam and Eve’ who le� the ‘garden’ (prime

reality) and ‘fell’ into Matrix reality. The simulation is not a

‘physical’ construct (there is no ‘physical’); it is a source of

information. Think Wi-Fi again. The simulation is an energetic field

encoded with information and body-brain systems are designed to

decode that information encoded in wave or frequency form which

is transmi�ed to the brain as electrical signals. These are decoded by

the brain to construct our sense of reality – an illusory ‘physical’

world that only exists in the brain or the mind. Virtual reality games

mimic this process using the same sensory decoding system.

Information is fed to the senses to decode a virtual reality that can

appear so real, but isn’t (Figs 18 and 19). Some scientists believe –

and I agree with them – that what we perceive as ‘physical’ reality

only exists when we are looking or observing. The act of perception

or focus triggers the decoding systems which turn waveform

information into holographic reality. When we are not observing

something our reality reverts from a holographic state to a waveform

state. This relates to the same principle as a falling tree not making a

noise unless someone is there to hear it or decode it. The concept

makes sense from the simulation perspective. A computer is not

decoding all the information in a Wi-Fi field all the time and only

decodes or brings into reality on the screen that part of Wi-Fi that it’s

decoding – focusing upon – at that moment.

Figure 18: Virtual reality technology ‘hacks’ into the body’s five-sense decoding system.

Figure 19: The result can be experienced as very ‘real’.

Interestingly, Professor Donald Hoffman at the Department of

Cognitive Sciences at the University of California, Irvine, says that

our experienced reality is like a computer interface that shows us

only the level with which we interact while hiding all that exists

beyond it: ‘Evolution shaped us with a user interface that hides the

truth. Nothing that we see is the truth – the very language of space

and time and objects is the wrong language to describe reality.’ He is

correct in what he says on so many levels. Space and time are not a

universal reality. They are a phenomenon of decoded simulation

reality as part of the process of enslaving our sense of reality. Near-

death experiencers report again and again how space and time did

not exist as we perceive them once they were free of the body – body

decoding systems. You can appreciate from this why Archons and

their Cult are so desperate to entrap human a�ention in the five

senses where we are in the Matrix and of the Matrix. Opening your

mind to expanded states of awareness takes you beyond the

information confines of the simulation and you become aware of

knowledge and insights denied to you before. This is what we call

‘awakening’ – awakening from the Matrix – and in the final chapter I

will relate this to current events.

Where are the ‘aliens’?

A simulation would explain the so-called ‘Fermi Paradox’ named

a�er Italian physicist Enrico Fermi (1901-1954) who created the first

nuclear reactor. He considered the question of why there is such a

lack of extraterrestrial activity when there are so many stars and

planets in an apparently vast universe; but what if the night sky that

we see, or think we do, is a simulated projection as I say? If you

control the simulation and your aim is to hold humanity fast in

essential ignorance would you want other forms of life including

advanced life coming and going sharing information with

humanity? Or would you want them to believe they were isolated

and apparently alone? Themes of human isolation and apartness are

common whether they be the perception of a lifeless universe or the

fascist isolation laws of the ‘Covid’ era. Paradoxically the very

existence of a simulation means that we are not alone when some

force had to construct it. My view is that experiences that people

have reported all over the world for centuries with Reptilians and

Grey entities are Archon phenomena as Nag Hammadi texts

describe; and that benevolent ‘alien’ interactions are non-human

groups that come in and out of the simulation by overcoming

Archon a�empts to keep them out. It should be highlighted, too, that

Reptilians and Greys are obsessed with genetics and technology as

related by cultural accounts and those who say they have been

abducted by them. Technology is their way of overcoming some of

the limitations in their creative potential and our technology-driven

and controlled human society of today is archetypical Archon-

Reptilian-Grey modus operandi. Technocracy is really Archontocracy.

The Universe does not have to be as big as it appears with a

simulation. There is no space or distance only information decoded

into holographic reality. What we call ‘space’ is only the absence of

holographic ‘objects’ and that ‘space’ is The Field of energetic

information which connects everything into a single whole. The

same applies with the artificially-generated information field of the

simulation. The Universe is not big or small as a physical reality. It is

decoded information, that’s all, and its perceived size is decided by

the way the simulation is encoded to make it appear. The entire

night sky as we perceive it only exists in our brain and so where are

those ‘millions of light years’? The ‘stars’ on the ceiling of the

Planetarium looked a vast distance away.

There’s another point to mention about ‘aliens’. I have been

highlighting since the 1990s the plan to stage a fake ‘alien invasion’

to justify the centralisation of global power and a world military.

Nazi scientist Werner von Braun, who was taken to America by

Operation Paperclip a�er World War Two to help found NASA, told

his American assistant Dr Carol Rosin about the Cult agenda when

he knew he was dying in 1977. Rosin said that he told her about a

sequence that would lead to total human control by a one-world

government. This included threats from terrorism, rogue nations,

meteors and asteroids before finally an ‘alien invasion’. All of these

things, von Braun said, would be bogus and what I would refer to as

a No-Problem-Reaction-Solution. Keep this in mind when ‘the aliens

are coming’ is the new mantra. The aliens are not coming – they are

already here and they have infiltrated human society while looking

human. French-Canadian investigative journalist Serge Monast said

in 1994 that he had uncovered a NASA/military operation called

Project Blue Beam which fits with what Werner von Braun predicted.

Monast died of a ‘heart a�ack’ in 1996 the day a�er he was arrested

and spent a night in prison. He was 51. He said Blue Beam was a

plan to stage an alien invasion that would include religious figures

beamed holographically into the sky as part of a global manipulation

to usher in a ‘new age’ of worshipping what I would say is the Cult

‘god’ Yaldabaoth in a one-world religion. Fake holographic asteroids

are also said to be part of the plan which again syncs with von

Braun. How could you stage an illusory threat from asteroids unless

they were holographic inserts? This is pre�y straightforward given

the advanced technology outside the public arena and the fact that

our ‘physical’ reality is holographic anyway. Information fields

would be projected and we would decode them into the illusion of a

‘physical’ asteroid. If they can sell a global ‘pandemic’ with a ‘virus’

that doesn’t exist what will humans not believe if government and

media tell them?

All this is particularly relevant as I write with the Pentagon

planning to release in June, 2021, information about ‘UFO sightings’.

I have been following the UFO story since the early 1990s and the

common theme throughout has been government and military

denials and cover up. More recently, however, the Pentagon has

suddenly become more talkative and apparently open with Air

Force pilot radar images released of unexplained cra� moving and

changing direction at speeds well beyond anything believed possible

with human technology. Then, in March, 2021, former Director of

National Intelligence John Ratcliffe said a Pentagon report months

later in June would reveal a great deal of information about UFO

sightings unknown to the public. He said the report would have

‘massive implications’. The order to do this was included bizarrely

in a $2.3 trillion ‘coronavirus’ relief and government funding bill

passed by the Trump administration at the end of 2020. I would add

some serious notes of caution here. I have been pointing out since

the 1990s that the US military and intelligence networks have long

had cra� – ‘flying saucers’ or anti-gravity cra� – which any observer

would take to be extraterrestrial in origin. Keeping this knowledge

from the public allows cra� flown by humans to be perceived as alien

visitations. I am not saying that ‘aliens’ do not exist. I would be the

last one to say that, but we have to be streetwise here. President

Ronald Reagan told the UN General Assembly in 1987: ‘I

occasionally think how quickly our differences worldwide would

vanish if we were facing an alien threat from outside this world.’

That’s the idea. Unite against a common ‘enemy’ with a common

purpose behind your ‘saviour force’ (the Cult) as this age-old

technique of mass manipulation goes global.

Science moves this way …

I could find only one other person who was discussing the

simulation hypothesis publicly when I concluded it was real. This

was Nick Bostrom, a Swedish-born philosopher at the University of

Oxford, who has explored for many years the possibility that human

reality is a computer simulation although his version and mine are

not the same. Today the simulation and holographic reality

hypothesis have increasingly entered the scientific mainstream. Well,

the more open-minded mainstream, that is. Here are a few of the

ever-gathering examples. American nuclear physicist Silas Beane led

a team of physicists at the University of Bonn in Germany pursuing

the question of whether we live in a simulation. They concluded that

we probably do and it was likely based on a la�ice of cubes. They

found that cosmic rays align with that specific pa�ern. The team

highlighted the Greisen–Zatsepin–Kuzmin (GZK) limit which refers

to cosmic ray particle interaction with cosmic background radiation

that creates an apparent boundary for cosmic ray particles. They say

in a paper entitled ‘Constraints on the Universe as a Numerical

Simulation’ that this ‘pa�ern of constraint’ is exactly what you

would find with a computer simulation. They also made the point

that a simulation would create its own ‘laws of physics’ that would

limit possibility. I’ve been making the same point for decades that

the perceived laws of physics relate only to this reality, or what I

would later call the simulation. When designers write codes to create

computer and virtual reality games they are the equivalent of the

laws of physics for that game. Players interact within the limitations

laid out by the coding. In the same way those who wrote the codes

for the simulation decided the laws of physics that would apply.

These can be overridden by expanded states of consciousness, but

not by those enslaved in only five-sense awareness where simulation

codes rule. Overriding the codes is what people call ‘miracles’. They

are not. They are bypassing the encoded limits of the simulation. A

population caught in simulation perception would have no idea that

this was their plight. As the Bonn paper said: ‘Like a prisoner in a

pitch-black cell we would not be able to see the “walls” of our

prison,’ That’s true if people remain mesmerised by the five senses.

Open to expanded awareness and those walls become very clear. The

main one is the speed of light.

American theoretical physicist James Gates is another who has

explored the simulation question and found considerable evidence

to support the idea. Gates was Professor of Physics at the University

of Maryland, Director of The Center for String and Particle Theory,

and on Barack Obama’s Council of Advisors on Science and

Technology. He and his team found computer codes of digital data

embedded in the fabric of our reality. They relate to on-off electrical

charges of 1 and 0 in the binary system used by computers. ‘We have

no idea what they are doing there’, Gates said. They found within

the energetic fabric mathematical sequences known as error-

correcting codes or block codes that ‘reboot’ data to its original state

or ‘default se�ings’ when something knocks it out of sync. Gates was

asked if he had found a set of equations embedded in our reality

indistinguishable from those that drive search engines and browsers

and he said: ‘That is correct.’ Rich Terrile, director of the Centre for

Evolutionary Computation and Automated Design at NASA’s Jet

Propulsion Laboratory, has said publicly that he believes the

Universe is a digital hologram that must have been created by a form

of intelligence. I agree with that in every way. Waveform information

is delivered electrically by the senses to the brain which constructs a

digital holographic reality that we call the ‘world’. This digital level

of reality can be read by the esoteric art of numerology. Digital

holograms are at the cu�ing edge of holographics today. We have

digital technology everywhere designed to access and manipulate

our digital level of perceived reality. Synthetic mRNA in ‘Covid

vaccines’ has a digital component to manipulate the body’s digital

‘operating system’.

Reality is numbers

How many know that our reality can be broken down to numbers

and codes that are the same as computer games? Max Tegmark, a

physicist at the Massachuse�s Institute of Technology (MIT), is the

author of Our Mathematical Universe in which he lays out how reality

can be entirely described by numbers and maths in the way that a

video game is encoded with the ‘physics’ of computer games. Our

world and computer virtual reality are essentially the same.

Tegmark imagines the perceptions of characters in an advanced

computer game when the graphics are so good they don’t know they

are in a game. They think they can bump into real objects

(electromagnetic resistance in our reality), fall in love and feel

emotions like excitement. When they began to study the apparently

‘physical world’ of the video game they would realise that

everything was made of pixels (which have been found in our

energetic reality as must be the case when on one level our world is

digital). What computer game characters thought was physical

‘stuff’, Tegmark said, could actually be broken down into numbers:

And we’re exactly in this situation in our world. We look around and it doesn’t seem that mathematical at all, but everything we see is made out of elementary particles like quarks and electrons. And what properties does an electron have? Does it have a smell or a colour or a texture? No! ... We physicists have come up with geeky names for [Electron] properties, like

electric charge, or spin, or lepton number, but the electron doesn’t care what we call it, the properties are just numbers.

This is the illusory reality Gnostics were describing. This is the

simulation. The A, C, G, and T codes of DNA have a binary value –

A and C = 0 while G and T = 1. This has to be when the simulation is

digital and the body must be digital to interact with it. Recurring

mathematical sequences are encoded throughout reality and the

body. They include the Fibonacci sequence in which the two

previous numbers are added to get the next one, as in ... 1, 1, 2, 3, 5,

8, 13, 21, 34, 55, etc. The sequence is encoded in the human face and

body, proportions of animals, DNA, seed heads, pine cones, trees,

shells, spiral galaxies, hurricanes and the number of petals in a

flower. The list goes on and on. There are fractal pa�erns – a ‘never-

ending pa�ern that is infinitely complex and self-similar across all

scales in the as above, so below, principle of holograms. These and

other famous recurring geometrical and mathematical sequences

such as Phi, Pi, Golden Mean, Golden Ratio and Golden Section are

computer codes of the simulation. I had to laugh and give my head a

shake the day I finished this book and it went into the production

stage. I was sent an article in Scientific American published in April,

2021, with the headline ‘Confirmed! We Live in a Simulation’. Two

decades a�er I first said our reality is a simulation and the speed of

light is it’s outer limit the article suggested that we do live in a

simulation and that the speed of light is its outer limit. I le� school at

15 and never passed a major exam in my life while the writer was up

to his eyes in qualifications. As I will explain in the final chapter

knowing is far be�er than thinking and they come from very different

sources. The article rightly connected the speed of light to the

processing speed of the ‘Matrix’ and said what has been in my books

all this time … ‘If we are in a simulation, as it appears, then space is

an abstract property wri�en in code. It is not real’. No it’s not and if

we live in a simulation something created it and it wasn’t us. ‘That

David Icke says we are manipulated by aliens’ – he’s crackers.’

Wow …

The reality that humanity thinks is so real is an illusion. Politicians,

governments, scientists, doctors, academics, law enforcement,

media, school and university curriculums, on and on, are all

founded on a world that does not exist except as a simulated prison

cell. Is it such a stretch to accept that ‘Covid’ doesn’t exist when our

entire ‘physical’ reality doesn’t exist? Revealed here is the

knowledge kept under raps in the Cult networks of

compartmentalised secrecy to control humanity’s sense of reality by

inducing the population to believe in a reality that’s not real. If it

wasn’t so tragic in its experiential consequences the whole thing

would be hysterically funny. None of this is new to Renegade Minds.

Ancient Greek philosopher Plato (about 428 to about 347BC) was a

major influence on Gnostic belief and he described the human plight

thousands of years ago with his Allegory of the Cave. He told the

symbolic story of prisoners living in a cave who had never been

outside. They were chained and could only see one wall of the cave

while behind them was a fire that they could not see. Figures walked

past the fire casting shadows on the prisoners’ wall and those

moving shadows became their sense of reality. Some prisoners began

to study the shadows and were considered experts on them (today’s

academics and scientists), but what they studied was only an illusion

(today’s academics and scientists). A prisoner escaped from the cave

and saw reality as it really is. When he returned to report this

revelation they didn’t believe him, called him mad and threatened to

kill him if he tried to set them free. Plato’s tale is not only a brilliant

analogy of the human plight and our illusory reality. It describes,

too, the dynamics of the ‘Covid’ hoax. I have only skimmed the

surface of these subjects here. The aim of this book is to crisply

connect all essential dots to put what is happening today into its true

context. All subject areas and their connections in this chapter are

covered in great evidential detail in Everything You Need To Know,

But Have Never Been Told and The Answer.

They say that bewildered people ‘can’t see the forest for the trees’.

Humanity, however, can’t see the forest for the twigs. The five senses

see only twigs while Renegade Minds can see the forest and it’s the

forest where the answers lie with the connections that reveals.

Breaking free of perceptual programming so the forest can be seen is

the way we turn all this around. Not breaking free is how humanity

got into this mess. The situation may seem hopeless, but I promise

you it’s not. We are a perceptual heartbeat from paradise if only we

knew.

R

CHAPTER TWELVE

Escaping Wetiko

Life is simply a vacation from the infinite

Dean Cavanagh

enegade Minds weave the web of life and events and see

common themes in the apparently random. They are always

there if you look for them and their pursuit is aided by incredible

synchronicity that comes when your mind is open rather than

mesmerised by what it thinks it can see.

Infinite awareness is infinite possibility and the more of infinite

possibility that we access the more becomes infinitely possible. That

may be stating the apparently obvious, but it is a devastatingly-

powerful fact that can set us free. We are a point of a�ention within

an infinity of consciousness. The question is how much of that

infinity do we choose to access? How much knowledge, insight,

awareness, wisdom, do we want to connect with and explore? If

your focus is only in the five senses you will be influenced by a

fraction of infinite awareness. I mean a range so tiny that it gives

new meaning to infinitesimal. Limitation of self-identity and a sense

of the possible limit accordingly your range of consciousness. We are

what we think we are. Life is what we think it is. The dream is the

dreamer and the dreamer is the dream. Buddhist philosophy puts it

this way: ‘As a thing is viewed, so it appears.’ Most humans live in

the realm of touch, taste, see, hear, and smell and that’s the limit of

their sense of the possible and sense of self. Many will follow a

religion and speak of a God in his heaven, but their lives are still

dominated by the five senses in their perceptions and actions. The

five senses become the arbiter of everything. When that happens all

except a smear of infinity is sealed away from influence by the rigid,

unyielding, reality bubbles that are the five-sense human or

Phantom Self. Archon Cult methodology is to isolate consciousness

within five-sense reality – the simulation – and then program that

consciousness with a sense of self and the world through a deluge of

life-long information designed to instil the desired perception that

allows global control. Efforts to do this have increased dramatically

with identity politics as identity bubbles are squeezed into the

minutiae of five-sense detail which disconnect people even more

profoundly from the infinite ‘I’.

Five-sense focus and self-identity are like a firewall that limits

access to the infinite realms. You only perceive one radio or

television station and no other. We’ll take that literally for a moment.

Imagine a vast array of stations giving different information and

angles on reality, but you only ever listen to one. Here we have the

human plight in which the population is overwhelmingly confined

to CultFM. This relates only to the frequency range of CultFM and

limits perception and insight to that band – limits possibility to that

band. It means you are connecting with an almost imperceptibly

minuscule range of possibility and creative potential within the

infinite Field. It’s a world where everything seems apart from

everything else and where synchronicity is rare. Synchronicity is

defined in the dictionary as ‘the happening by chance of two or more

related or similar events at the same time‘. Use of ‘by chance’ betrays

a complete misunderstanding of reality. Synchronicity is not ‘by

chance’. As people open their minds, or ‘awaken’ to use the term,

they notice more and more coincidences in their lives, bits of ‘luck’,

apparently miraculous happenings that put them in the right place

at the right time with the right people. Days become peppered with

‘fancy meeting you here’ and ‘what are the chances of that?’ My

entire life has been lived like this and ever more so since my own

colossal awakening in 1990 and 91 which transformed my sense of

reality. Synchronicity is not ‘by chance’; it is by accessing expanded

realms of possibility which allow expanded potential for

manifestation. People broadcasting the same vibe from the same

openness of mind tend to be drawn ‘by chance’ to each other

through what I call frequency magnetism and it’s not only people. In

the last more than 30 years incredible synchronicity has also led me

through the Cult maze to information in so many forms and to

crucial personal experiences. These ‘coincidences’ have allowed me

to put the puzzle pieces together across an enormous array of

subjects and situations. Those who have breached the bubble of five-

sense reality will know exactly what I mean and this escape from the

perceptual prison cell is open to everyone whenever they make that

choice. This may appear super-human when compared with the

limitations of ‘human’, but it’s really our natural state. ‘Human’ as

currently experienced is consciousness in an unnatural state of

induced separation from the infinity of the whole. I’ll come to how

this transformation into unity can be made when I have described in

more detail the force that holds humanity in servitude by denying

this access to infinite self.

The Wetiko factor

I have been talking and writing for decades about the way five-sense

mind is systematically barricaded from expanded awareness. I have

used the analogy of a computer (five-sense mind) and someone at

the keyboard (expanded awareness). Interaction between the

computer and the operator is symbolic of the interaction between

five-sense mind and expanded awareness. The computer directly

experiences the Internet and the operator experiences the Internet

via the computer which is how it’s supposed to be – the two working

as one. Archons seek to control that point where the operator

connects with the computer to stop that interaction (Fig 20). Now the

operator is banging the keyboard and clicking the mouse, but the

computer is not responding and this happens when the computer is

taken over – possessed – by an appropriately-named computer ‘virus’.

The operator has lost all influence over the computer which goes its

own way making decisions under the control of the ‘virus’. I have

just described the dynamic through which the force known to

Gnostics as Yaldabaoth and Archons disconnects five-sense mind

from expanded awareness to imprison humanity in perceptual

servitude.

Figure 20: The mind ‘virus’ I have been writing about for decades seeks to isolate five-sense mind (the computer) from the true ‘I’. (Image by Neil Hague).

About a year ago I came across a Native American concept of

Wetiko which describes precisely the same phenomenon. Wetiko is

the spelling used by the Cree and there are other versions including

wintiko and windigo used by other tribal groups. They spell the

name with lower case, but I see Wetiko as a proper noun as with

Archons and prefer a capital. I first saw an article about Wetiko by

writer and researcher Paul Levy which so synced with what I had

been writing about the computer/operator disconnection and later

the Archons. I then read his book, the fascinating Dispelling Wetiko,

Breaking the Spell of Evil. The parallels between what I had concluded

long before and the Native American concept of Wetiko were so

clear and obvious that it was almost funny. For Wetiko see the

Gnostic Archons for sure and the Jinn, the Predators, and every

other name for a force of evil, inversion and chaos. Wetiko is the

Native American name for the force that divides the computer from

the operator (Fig 21). Indigenous author Jack D. Forbes, a founder of

the Native American movement in the 1960s, wrote another book

about Wetiko entitled Columbus And Other Cannibals – The Wetiko

Disease of Exploitation, Imperialism, and Terrorism which I also read.

Forbes says that Wetiko refers to an evil person or spirit ‘who

terrorizes other creatures by means of terrible acts, including

cannibalism’. Zulu shaman Credo Mutwa told me that African

accounts tell how cannibalism was brought into the world by the

Chitauri ‘gods’ – another manifestation of Wetiko. The distinction

between ‘evil person or spirit’ relates to Archons/Wetiko possessing

a human or acting as pure consciousness. Wetiko is said to be a

sickness of the soul or spirit and a state of being that takes but gives

nothing back – the Cult and its operatives perfectly described. Black

Hawk, a Native American war leader defending their lands from

confiscation, said European invaders had ‘poisoned hearts’ – Wetiko

hearts – and that this would spread to native societies. Mention of

the heart is very significant as we shall shortly see. Forbes writes:

‘Tragically, the history of the world for the past 2,000 years is, in

great part, the story of the epidemiology of the wetiko disease.’ Yes,

and much longer. Forbes is correct when he says: ‘The wetikos

destroyed Egypt and Babylon and Athens and Rome and

Tenochtitlan [capital of the Aztec empire] and perhaps now they will

destroy the entire earth.’ Evil, he said, is the number one export of a

Wetiko culture – see its globalisation with ‘Covid’. Constant war,

mass murder, suffering of all kinds, child abuse, Satanism, torture

and human sacrifice are all expressions of Wetiko and the Wetiko

possessed. The world is Wetiko made manifest, but it doesn’t have to

be. There is a way out of this even now.

Figure 21: The mind ‘virus’ is known to Native Americans as ‘Wetiko’. (Image by Neil Hague).

Cult of Wetiko

Wetiko is the Yaldabaoth frequency distortion that seeks to a�ach to

human consciousness and absorb it into its own. Once this

connection is made Wetiko can drive the perceptions of the target

which they believe to be coming from their own mind. All the

horrors of history and today from mass killers to Satanists,

paedophiles like Jeffrey Epstein and other psychopaths, are the

embodiment of Wetiko and express its state of being in all its

grotesqueness. The Cult is Wetiko incarnate, Yaldabaoth incarnate,

and it seeks to facilitate Wetiko assimilation of humanity in totality

into its distortion by manipulating the population into low

frequency states that match its own. Paul Levy writes:

‘Holographically enforced within the psyche of every human being

the wetiko virus pervades and underlies the entire field of

consciousness, and can therefore potentially manifest through any

one of us at any moment if we are not mindful.’ The ‘Covid’ hoax

has achieved this with many people, but others have not fallen into

Wetiko’s frequency lair. Players in the ‘Covid’ human catastrophe

including Gates, Schwab, Tedros, Fauci, Whi�y, Vallance, Johnson,

Hancock, Ferguson, Drosten, and all the rest, including the

psychopath psychologists, are expressions of Wetiko. This is why

they have no compassion or empathy and no emotional consequence

for what they do that would make them stop doing it. Observe all

the people who support the psychopaths in authority against the

Pushbackers despite the damaging impact the psychopaths have on

their own lives and their family’s lives. You are again looking at

Wetiko possession which prevents them seeing through the lies to

the obvious scam going on. Why can’t they see it? Wetiko won’t let

them see it. The perceptual divide that has now become a chasm is

between the Wetikoed and the non-Wetikoed.

Paul Levy describes Wetiko in the same way that I have long

described the Archontic force. They are the same distorted

consciousness operating across dimensions of reality: ‘… the subtle

body of wetiko is not located in the third dimension of space and

time, literally existing in another dimension … it is able to affect

ordinary lives by mysteriously interpenetrating into our three-

dimensional world.’ Wetiko does this through its incarnate

representatives in the Cult and by weaving itself into The Field

which on our level of reality is the electromagnetic information field

of the simulation or Matrix. More than that, the simulation is Wetiko

/ Yaldabaoth. Caleb Scharf, Director of Astrobiology at Columbia

University, has speculated that ‘alien life’ could be so advanced that

it has transcribed itself into the quantum realm to become what we

call physics. He said intelligence indistinguishable from the fabric of

the Universe would solve many of its greatest mysteries:

Perhaps hyper-advanced life isn’t just external. Perhaps it’s already all around. It is embedded in what we perceive to be physics itself, from the root behaviour of particles and fields to the phenomena of complexity and emergence ... In other words, life might not just be in the equations. It might BE the equations [My emphasis].

Scharf said it is possible that ‘we don’t recognise advanced life

because it forms an integral and unsuspicious part of what we’ve

considered to be the natural world’. I agree. Wetiko/Yaldabaoth is the

simulation. We are literally in the body of the beast. But that doesn’t

mean it has to control us. We all have the power to overcome Wetiko

influence and the Cult knows that. I doubt it sleeps too well because

it knows that.

Which Field?

This, I suggest, is how it all works. There are two Fields. One is the

fierce electromagnetic light of the Matrix within the speed of light;

the other is the ‘watery light’ of The Field beyond the walls of the

Matrix that connects with the Great Infinity. Five-sense mind and the

decoding systems of the body a�ach us to the Field of Matrix light.

They have to or we could not experience this reality. Five-sense mind

sees only the Matrix Field of information while our expanded

consciousness is part of the Infinity Field. When we open our minds,

and most importantly our hearts, to the Infinity Field we have a

mission control which gives us an expanded perspective, a road

map, to understand the nature of the five-sense world. If we are

isolated only in five-sense mind there is no mission control. We’re on

our own trying to understand a world that’s constantly feeding us

information to ensure we do not understand. People in this state can

feel ‘lost’ and bewildered with no direction or radar. You can see

ever more clearly those who are influenced by the Fields of Big

Infinity or li�le five-sense mind simply by their views and behaviour

with regard to the ‘Covid’ hoax. We have had this division

throughout known human history with the mass of the people on

one side and individuals who could see and intuit beyond the walls

of the simulation – Plato’s prisoner who broke out of the cave and

saw reality for what it is. Such people have always been targeted by

Wetiko/Archon-possessed authority, burned at the stake or

demonised as mad, bad and dangerous. The Cult today and its

global network of ‘anti-hate’, ‘anti-fascist’ Woke groups are all

expressions of Wetiko a�acking those exposing the conspiracy,

‘Covid’ lies and the ‘vaccine’ agenda.

Woke as a whole is Wetiko which explains its black and white

mentality and how at one it is with the Wetiko-possessed Cult. Paul

Levy said: ‘To be in this paradigm is to still be under the thrall of a

two-valued logic – where things are either true or false – of a

wetikoized mind.’ Wetiko consciousness is in a permanent rage,

therefore so is Woke, and then there is Woke inversion and

contradiction. ‘Anti-fascists’ act like fascists because fascists and ‘anti-

fascists’ are both Wetiko at work. Political parties act the same while

claiming to be different for the same reason. Secret society and

satanic rituals are a�aching initiates to Wetiko and the cold, ruthless,

psychopathic mentality that secures the positions of power all over

the world is Wetiko. Reframing ‘training programmes’ have the

same cumulative effect of a�aching Wetiko and we have their

graduates described as automatons and robots with a cold,

psychopathic, uncaring demeanour. They are all traits of Wetiko

possession and look how many times they have been described in

this book and elsewhere with regard to personnel behind ‘Covid’

including the police and medical profession. Climbing the greasy

pole in any profession in a Wetiko society requires traits of Wetiko to

get there and that is particularly true of politics which is not about

fair competition and pre-eminence of ideas. It is founded on how

many backs you can stab and arses you can lick. This culminated in

the global ‘Covid’ coordination between the Wetiko possessed who

pulled it off in all the different countries without a trace of empathy

and compassion for their impact on humans. Our sight sense can see

only holographic form and not the Field which connects holographic

form. Therefore we perceive ‘physical’ objects with ‘space’ in

between. In fact that ‘space’ is energy/consciousness operating on

multiple frequencies. One of them is Wetiko and that connects the

Cult psychopaths, those who submit to the psychopaths, and those

who serve the psychopaths in the media operations of the world.

Wetiko is Gates. Wetiko is the mask-wearing submissive. Wetiko is

the fake journalist and ‘fact-checker’. The Wetiko Field is

coordinating the whole thing. Psychopaths, gofers, media

operatives, ‘anti-hate’ hate groups, ‘fact-checkers’ and submissive

people work as one unit even without human coordination because they

are a�ached to the same Field which is organising it all (Fig 22). Paul

Levy is here describing how Wetiko-possessed people are drawn

together and refuse to let any information breach their rigid

perceptions. He was writing long before ‘Covid’, but I think you will

recognise followers of the ‘Covid’ religion oh just a little bit:

People who are channelling the vibratory frequency of wetiko align with each other through psychic resonance to reinforce their unspoken shared agreement so as to uphold their deranged view of reality. Once an unconscious content takes possession of certain individuals, it irresistibly draws them together by mutual attraction and knits them into groups tied together by their shared madness that can easily swell into an avalanche of insanity.

A psychic epidemic is a closed system, which is to say that it is insular and not open to any new information or informing influences from the outside world which contradict its fixed, limited, and limiting perspective.

There we have the Woke mind and the ‘Covid’ mind. Compatible

resonance draws the awakening together, too, which is clearly

happening today.

Figure 22: The Wetiko Field from which the Cult pyramid and its personnel are made manifest. (Image by Neil Hague).

Spiritual servitude

Wetiko doesn’t care about humans. It’s not human; it just possesses

humans for its own ends and the effect (depending on the scale of

possession) can be anything from extreme psychopathy to

unquestioning obedience. Wetiko’s worst nightmare is for human

consciousness to expand beyond the simulation. Everything is

focussed on stopping that happening through control of

information, thus perception, thus frequency. The ‘education

system’, media, science, medicine, academia, are all geared to

maintaining humanity in five-sense servitude as is the constant

stimulation of low-vibrational mental and emotional states (see

‘Covid’). Wetiko seeks to dominate those subconscious spaces

between five-sense perception and expanded consciousness where

the computer meets the operator. From these subconscious hiding

places Wetiko speaks to us to trigger urges and desires that we take

to be our own and manipulate us into anything from low-vibrational

to psychopathic states. Remember how Islam describes the Jinn as

invisible tricksters that ‘whisper’ and confuse. Wetiko is the origin of

the ‘trickster god’ theme that you find in cultures all over the world.

Jinn, like the Archons, are Wetiko which is terrified of humans

awakening and reconnecting with our true self for then its energy

source has gone. With that the feedback loop breaks between Wetiko

and human perception that provides the energetic momentum on

which its very existence depends as a force of evil. Humans are both

its target and its source of survival, but only if we are operating in

low-vibrational states of fear, hate, depression and the background

anxiety that most people suffer. We are Wetiko’s target because we

are its key to survival. It needs us, not the other way round. Paul

Levy writes:

A vampire has no intrinsic, independent, substantial existence in its own right; it only exists in relation to us. The pathogenic, vampiric mind-parasite called wetiko is nothing in itself – not being able to exist from its own side – yet it has a ‘virtual reality’ such that it can potentially destroy our species …

…The fact that a vampire is not reflected by a mirror can also mean that what we need to see is that there’s nothing, no-thing to see, other than ourselves. The fact that wetiko is the expression of something inside of us means that the cure for wetiko is with us as well. The critical issue is finding this cure within us and then putting it into effect.

Evil begets evil because if evil does not constantly expand and

find new sources of energetic sustenance its evil, its distortion, dies

with the assimilation into balance and harmony. Love is the garlic to

Wetiko’s vampire. Evil, the absence of love, cannot exist in the

presence of love. I think I see a way out of here. I have emphasised

so many times over the decades that the Archons/Wetiko and their

Cult are not all powerful. They are not. I don’t care how it looks even

now they are not. I have not called them li�le boys in short trousers

for effect. I have said it because it is true. Wetiko’s insatiable desire

for power over others is not a sign of its omnipotence, but its

insecurity. Paul Levy writes: ‘Due to the primal fear which

ultimately drives it and which it is driven to cultivate, wetiko’s body

politic has an intrinsic and insistent need for centralising power and

control so as to create imagined safety for itself.’ Yeeeeeees! Exactly!

Why does Wetiko want humans in an ongoing state of fear? Wetiko

itself is fear and it is petrified of love. As evil is an absence of love, so

love is an absence of fear. Love conquers all and especially Wetiko

which is fear. Wetiko brought fear into the world when it wasn’t here

before. Fear was the ‘fall’, the fall into low-frequency ignorance and

illusion – fear is False Emotion Appearing Real. The simulation is

driven and energised by fear because Wetiko/Yaldabaoth (fear) are

the simulation. Fear is the absence of love and Wetiko is the absence

of love.

Wetiko today

We can now view current events from this level of perspective. The

‘Covid’ hoax has generated momentous amounts of ongoing fear,

anxiety, depression and despair which have empowered Wetiko. No

wonder people like Gates have been the instigators when they are

Wetiko incarnate and exhibit every trait of Wetiko in the extreme.

See how cold and unemotional these people are like Gates and his

cronies, how dead of eye they are. That’s Wetiko. Sabbatians are

Wetiko and everything they control including the World Health

Organization, Big Pharma and the ‘vaccine’ makers, national ‘health’

hierarchies, corporate media, Silicon Valley, the banking system, and

the United Nations with its planned transformation into world

government. All are controlled and possessed by the Wetiko

distortion into distorting human society in its image. We are with

this knowledge at the gateway to understanding the world.

Divisions of race, culture, creed and sexuality are diversions to hide

the real division between those possessed and influenced by Wetiko

and those that are not. The ‘Covid’ hoax has brought both clearly

into view. Human behaviour is not about race. Tyrants and

dictatorships come in all colours and creeds. What unites the US

president bombing the innocent and an African tribe commi�ing

genocide against another as in Rwanda? What unites them? Wetiko.

All wars are Wetiko, all genocide is Wetiko, all hunger over centuries

in a world of plenty is Wetiko. Children going to bed hungry,

including in the West, is Wetiko. Cult-generated Woke racial

divisions that focus on the body are designed to obscure the reality

that divisions in behaviour are manifestations of mind, not body.

Obsession with body identity and group judgement is a means to

divert a�ention from the real source of behaviour – mind and

perception. Conflict sown by the Woke both within themselves and

with their target groups are Wetiko providing lunch for itself

through still more agents of the division, chaos, and fear on which it

feeds. The Cult is seeking to assimilate the entirety of humanity and

all children and young people into the Wetiko frequency by

manipulating them into states of fear and despair. Witness all the

suicide and psychological unravelling since the spring of 2020.

Wetiko psychopaths want to impose a state of unquestioning

obedience to authority which is no more than a conduit for Wetiko to

enforce its will and assimilate humanity into itself. It needs us to

believe that resistance is futile when it fears resistance and even

more so the game-changing non-cooperation with its impositions. It

can use violent resistance for its benefit. Violent impositions and

violent resistance are both Wetiko. The Power of Love with its Power

of No will sweep Wetiko from our world. Wetiko and its Cult know

that. They just don’t want us to know.

AI Wetiko

This brings me to AI or artificial intelligence and something else

Wetikos don’t want us to know. What is AI really? I know about

computer code algorithms and AI that learns from data input. These,

however, are more diversions, the expeditionary force, for the real AI

that they want to connect to the human brain as promoted by Silicon

Valley Wetikos like Kurzweil. What is this AI? It is the frequency of

Wetiko, the frequency of the Archons. The connection of AI to the

human brain is the connection of the Wetiko frequency to create a

Wetiko hive mind and complete the job of assimilation. The hive

mind is planned to be controlled from Israel and China which are

both 100 percent owned by Wetiko Sabbatians. The assimilation

process has been going on minute by minute in the ‘smart’ era which

fused with the ‘Covid’ era. We are told that social media is

scrambling the minds of the young and changing their personality.

This is true, but what is social media? Look more deeply at how it

works, how it creates divisions and conflict, the hostility and cruelty,

the targeting of people until they are destroyed. That’s Wetiko. Social

media is manipulated to tune people to the Wetiko frequency with

all the emotional exploitation tricks employed by platforms like

Facebook and its Wetiko front man, Zuckerberg. Facebook’s

Instagram announced a new platform for children to overcome a

legal bar on them using the main site. This is more Wetiko

exploitation and manipulation of kids. Amnesty International

likened the plan to foxes offering to guard the henhouse and said it

was incompatible with human rights. Since when did Wetiko or

Zuckerberg (I repeat myself) care about that? Would Brin and Page

at Google, Wojcicki at YouTube, Bezos at Amazon and whoever the

hell runs Twi�er act as they do if they were not channelling Wetiko?

Would those who are developing technologies for no other reason

than human control? How about those designing and selling

technologies to kill people and Big Pharma drug and ‘vaccine’

producers who know they will end or devastate lives? Quite a

thought for these people to consider is that if you are Wetiko in a

human life you are Wetiko on the ‘other side’ unless your frequency

changes and that can only change by a change of perception which

becomes a change of behaviour. Where Gates is going does not bear

thinking about although perhaps that’s exactly where he wants to go.

Either way, that’s where he’s going. His frequency will make it so.

The frequency lair

I have been saying for a long time that a big part of the addiction to

smartphones and devices is that a frequency is coming off them that

entraps the mind. People spend ages on their phones and sometimes

even a minute or so a�er they put them down they pick them up

again and it all repeats. ‘Covid’ lockdowns will have increased this

addiction a million times for obvious reasons. Addictions to alcohol

overindulgence and drugs are another way that Wetiko entraps

consciousness to a�ach to its own. Both are symptoms of low-

vibrational psychological distress which alcoholism and drug

addiction further compound. Do we think it’s really a coincidence

that access to them is made so easy while potions that can take

people into realms beyond the simulation are banned and illegal? I

have explored smartphone addiction in other books, the scale is

mind-blowing, and that level of addiction does not come without

help. Tech companies that make these phones are Wetiko and they

will have no qualms about destroying the minds of children. We are

seeing again with these companies the Wetiko perceptual

combination of psychopathic enforcers and weak and meek

unquestioning compliance by the rank and file.

The global Smart Grid is the Wetiko Grid and it is crucial to

complete the Cult endgame. The simulation is radiation and we are

being deluged with technological radiation on a devastating scale.

Wetiko frauds like Elon Musk serve Cult interests while occasionally

criticising them to maintain his street-cred. 5G and other forms of

Wi-Fi are being directed at the earth from space on a volume and

scale that goes on increasing by the day. Elon Musk’s (officially)

SpaceX Starlink project is in the process of pu�ing tens of thousands

of satellites in low orbit to cover every inch of the planet with 5G

and other Wi-Fi to create Kurzweil’s global ‘cloud’ to which the

human mind is planned to be a�ached very soon. SpaceX has

approval to operate 12,000 satellites with more than 1,300 launched

at the time of writing and applications filed for 30,000 more. Other

operators in the Wi-Fi, 5G, low-orbit satellite market include

OneWeb (UK), Telesat (Canada), and AST & Science (US). Musk tells

us that AI could be the end of humanity and then launches a

company called Neuralink to connect the human brain to computers.

Musk’s (in theory) Tesla company is building electric cars and the

driverless vehicles of the smart control grid. As frauds and

bullshi�ers go Elon Musk in my opinion is Major League.

5G and technological radiation in general are destructive to

human health, genetics and psychology and increasing the strength

of artificial radiation underpins the five-sense perceptual bubbles

which are themselves expressions of radiation or electromagnetism.

Freedom activist John Whitehead was so right with his ‘databit by

databit, we are building our own electronic concentration camps’.

The Smart Grid and 5G is a means to control the human mind and

infuse perceptual information into The Field to influence anyone in

sync with its frequency. You can change perception and behaviour

en masse if you can manipulate the population into those levels of

frequency and this is happening all around us today. The arrogance

of Musk and his fellow Cult operatives knows no bounds in the way

that we see with Gates. Musk’s satellites are so many in number

already they are changing the night sky when viewed from Earth.

The astronomy community has complained about this and they have

seen nothing yet. Some consequences of Musk’s Wetiko hubris

include: Radiation; visible pollution of the night sky; interference

with astronomy and meteorology; ground and water pollution from

intensive use of increasingly many spaceports; accumulating space

debris; continual deorbiting and burning up of aging satellites,

polluting the atmosphere with toxic dust and smoke; and ever-

increasing likelihood of collisions. A collective public open le�er of

complaint to Musk said:

We are writing to you … because SpaceX is in process of surrounding the Earth with a network of thousands of satellites whose very purpose is to irradiate every square inch of the

Earth. SpaceX, like everyone else, is treating the radiation as if it were not there. As if the mitochondria in our cells do not depend on electrons moving undisturbed from the food we digest to the oxygen we breathe.

As if our nervous systems and our hearts are not subject to radio frequency interference like any piece of electronic equipment. As if the cancer, diabetes, and heart disease that now afflict a majority of the Earth’s population are not metabolic diseases that result from interference with our cellular machinery. As if insects everywhere, and the birds and animals that eat them, are not starving to death as a result.

People like Musk and Gates believe in their limitless Wetiko

arrogance that they can do whatever they like to the world because

they own it. Consequences for humanity are irrelevant. It’s

absolutely time that we stopped taking this shit from these self-

styled masters of the Earth when you consider where this is going.

Why is the Cult so anti-human?

I hear this question o�en: Why would they do this when it will affect

them, too? Ah, but will it? Who is this them? Forget their bodies.

They are just vehicles for Wetiko consciousness. When you break it

all down to the foundations we are looking at a state of severely

distorted consciousness targeting another state of consciousness for

assimilation. The rest is detail. The simulation is the fly-trap in

which unique sensations of the five senses create a cycle of addiction

called reincarnation. Renegade Minds see that everything which

happens in our reality is a smaller version of the whole picture in

line with the holographic principle. Addiction to the radiation of

smart technology is a smaller version of addiction to the whole

simulation. Connecting the body/brain to AI is taking that addiction

on a giant step further to total ongoing control by assimilating

human incarnate consciousness into Wetiko. I have watched during

the ‘Covid’ hoax how many are becoming ever more profoundly

a�ached to Wetiko’s perceptual calling cards of aggressive response

to any other point of view (‘There is no other god but me’),

psychopathic lack of compassion and empathy, and servile

submission to the narrative and will of authority. Wetiko is the

psychopaths and subservience to psychopaths. The Cult of Wetiko is

so anti-human because it is not human. It embarked on a mission to

destroy human by targeting everything that it means to be human

and to survive as human. ‘Covid’ is not the end, just a means to an

end. The Cult with its Wetiko consciousness is seeking to change

Earth systems, including the atmosphere, to suit them, not humans.

The gathering bombardment of 5G alone from ground and space is

dramatically changing The Field with which the five senses interact.

There is so much more to come if we sit on our hands and hope it

will all go away. It is not meant to go away. It is meant to get ever

more extreme and we need to face that while we still can – just.

Carbon dioxide is the gas of life. Without that human is over.

Kaput, gone, history. No natural world, no human. The Cult has

created a cock and bull story about carbon dioxide and climate

change to justify its reduction to the point where Gates and the

ignoramus Biden ‘climate chief’ John Kerry want to suck it out of the

atmosphere. Kerry wants to do this because his master Gates does.

Wetikos have made the gas of life a demon with the usual support

from the Wokers of Extinction Rebellion and similar organisations

and the bewildered puppet-child that is Greta Thunberg who was

put on the world stage by Klaus Schwab and the World Economic

Forum. The name Extinction Rebellion is both ironic and as always

Wetiko inversion. The gas that we need to survive must be reduced

to save us from extinction. The most basic need of human is oxygen

and we now have billions walking around in face nappies depriving

body and brain of this essential requirement of human existence.

More than that 5G at 60 gigahertz interacts with the oxygen

molecule to reduce the amount of oxygen the body can absorb into

the bloodstream. The obvious knock-on consequences of that for

respiratory and cognitive problems and life itself need no further

explanation. Psychopaths like Musk are assembling a global system

of satellites to deluge the human atmosphere with this insanity. The

man should be in jail. Here we have two most basic of human needs,

oxygen and carbon dioxide, being dismantled.

Two others, water and food, are ge�ing similar treatment with the

United Nations Agendas 21 and 2030 – the Great Reset – planning to

centrally control all water and food supplies. People will not even

own rain water that falls on their land. Food is affected at the most

basic level by reducing carbon dioxide. We have genetic modification

or GMO infiltrating the food chain on a mass scale, pesticides and

herbicides polluting the air and destroying the soil. Freshwater fish

that provide livelihoods for 60 million people and feed hundreds of

millions worldwide are being ‘pushed to the brink’ according the

conservationists while climate change is the only focus. Now we

have Gates and Schwab wanting to dispense with current food

sources all together and replace them with a synthetic version which

the Wetiko Cult would control in terms of production and who eats

and who doesn’t. We have been on the Totalitarian Tiptoe to this for

more than 60 years as food has become ever more processed and full

of chemical shite to the point today when it’s not natural food at all.

As Dr Tom Cowan says: ‘If it has a label don’t eat it.’ Bill Gates is

now the biggest owner of farmland in the United States and he does

nothing without an ulterior motive involving the Cult. Klaus Schwab

wrote: ‘To feed the world in the next 50 years we will need to

produce as much food as was produced in the last 10,000 years …

food security will only be achieved, however, if regulations on

genetically modified foods are adapted to reflect the reality that gene

editing offers a precise, efficient and safe method of improving

crops.’ Liar. People and the world are being targeted with

aluminium through vaccines, chemtrails, food, drink cans, and

endless other sources when aluminium has been linked to many

health issues including dementia which is increasing year a�er year.

Insects, bees and wildlife essential to the food chain are being

deleted by pesticides, herbicides and radiation which 5G is

dramatically increasing with 6G and 7G to come. The pollinating bee

population is being devastated while wildlife including birds,

dolphins and whales are having their natural radar blocked by the

effects of ever-increasing radiation. In the summer windscreens used

to be spla�ered with insects so numerous were they. It doesn’t

happen now. Where have they gone?

Synthetic everything

The Cult is introducing genetically-modified versions of trees, plants

and insects including a Gates-funded project to unleash hundreds of

millions of genetically-modified, lab-altered and patented male

mosquitoes to mate with wild mosquitoes and induce genetic flaws

that cause them to die out. Clinically-insane Gates-funded Japanese

researchers have developed mosquitos that spread vaccine and are

dubbed ‘flying vaccinators’. Gates is funding the modification of

weather pa�erns in part to sell the myth that this is caused by carbon

dioxide and he’s funding geoengineering of the skies to change the

atmosphere. Some of this came to light with the Gates-backed plan

to release tonnes of chalk into the atmosphere to ‘deflect the Sun and

cool the planet’. Funny how they do this while the heating effect of

the Sun is not factored into climate projections focussed on carbon

dioxide. The reason is that they want to reduce carbon dioxide (so

don’t mention the Sun), but at the same time they do want to reduce

the impact of the Sun which is so essential to human life and health.

I have mentioned the sun-cholesterol-vitamin D connection as they

demonise the Sun with warnings about skin cancer (caused by the

chemicals in sun cream they tell you to splash on). They come from

the other end of the process with statin drugs to reduce cholesterol

that turns sunlight into vitamin D. A lack of vitamin D leads to a

long list of health effects and how vitamin D levels must have fallen

with people confined to their homes over ‘Covid’. Gates is funding

other forms of geoengineering and most importantly chemtrails

which are dropping heavy metals, aluminium and self-replicating

nanotechnology onto the Earth which is killing the natural world.

See Everything You Need To Know, But Have Never Been Told for the

detailed background to this.

Every human system is being targeted for deletion by a force that’s

not human. The Wetiko Cult has embarked on the process of

transforming the human body from biological to synthetic biological

as I have explained. Biological is being replaced by the artificial and

synthetic – Archontic ‘countermimicry’ – right across human society.

The plan eventually is to dispense with the human body altogether

and absorb human consciousness – which it wouldn’t really be by

then – into cyberspace (the simulation which is Wetiko/Yaldabaoth).

Preparations for that are already happening if people would care to

look. The alternative media rightly warns about globalism and ‘the

globalists’, but this is far bigger than that and represents the end of

the human race as we know it. The ‘bad copy’ of prime reality that

Gnostics describe was a bad copy of harmony, wonder and beauty to

start with before Wetiko/Yaldabaoth set out to change the simulated

‘copy’ into something very different. The process was slow to start

with. Entrapped humans in the simulation timeline were not

technologically aware and they had to be brought up to intellectual

speed while being suppressed spiritually to the point where they

could build their own prison while having no idea they were doing

so. We have now reached that stage where technological intellect has

the potential to destroy us and that’s why events are moving so fast.

Central American shaman Don Juan Matus said:

Think for a moment, and tell me how you would explain the contradictions between the intelligence of man the engineer and the stupidity of his systems of belief, or the stupidity of his contradictory behaviour. Sorcerers believe that the predators have given us our systems of beliefs, our ideas of good and evil; our social mores. They are the ones who set up our dreams of success or failure. They have given us covetousness, greed, and cowardice. It is the predator who makes us complacent, routinary, and egomaniacal.

In order to keep us obedient and meek and weak, the predators engaged themselves in a stupendous manoeuvre – stupendous, of course, from the point of view of a fighting strategist; a horrendous manoeuvre from the point of those who suffer it. They gave us their mind. The predators’ mind is baroque, contradictory, morose, filled with the fear of being discovered any minute now.

For ‘predators’ see Wetiko, Archons, Yaldabaoth, Jinn, and all the

other versions of the same phenomenon in cultures and religions all

over the world. The theme is always the same because it’s true and

it’s real. We have reached the point where we have to deal with it.

The question is – how?

Don’t fight – walk away

I thought I’d use a controversial subheading to get things moving in

terms of our response to global fascism. What do you mean ‘don’t

fight’? What do you mean ‘walk away’? We’ve got to fight. We can’t

walk away. Well, it depends what we mean by fight and walk away.

If fighting means physical combat we are playing Wetiko’s game and

falling for its trap. It wants us to get angry, aggressive, and direct

hate and hostility at the enemy we think we must fight. Every war,

every ba�le, every conflict, has been fought with Wetiko leading

both sides. It’s what it does. Wetiko wants a fight, anywhere, any

place. Just hit me, son, so I can hit you back. Wetiko hits Wetiko and

Wetiko hits Wetiko in return. I am very forthright as you can see in

exposing Wetikos of the Cult, but I don’t hate them. I refuse to hate

them. It’s what they want. What you hate you become. What you

fight you become. Wokers, ‘anti-haters’ and ‘anti-fascists’ prove this

every time they reach for their keyboards or don their balaclavas. By

walk away I mean to disengage from Wetiko which includes ceasing

to cooperate with its tyranny. Paul Levy says of Wetiko:

The way to ‘defeat’ evil is not to try to destroy it (for then, in playing evil’s game, we have already lost), but rather, to find the invulnerable place within ourselves where evil is unable to vanquish us – this is to truly ‘win’ our battle with evil.

Wetiko is everywhere in human society and it’s been on steroids

since the ‘Covid’ hoax. Every shouting match over wearing masks

has Wetiko wearing a mask and Wetiko not wearing one. It’s an

electrical circuit of push and resist, push and resist, with Wetiko

pushing and resisting. Each polarity is Wetiko empowering itself.

Dictionary definitions of ‘resist’ include ‘opposing, refusing to accept

or comply with’ and the word to focus on is ‘opposing’. What form

does this take – se�ing police cars alight or ‘refusing to accept or

comply with’? The former is Wetiko opposing Wetiko while the

other points the way forward. This is the difference between those

aggressively demanding that government fascism must be obeyed

who stand in stark contrast to the great majority of Pushbackers. We

saw this clearly with a march by thousands of Pushbackers against

lockdown in London followed days later by a Woker-hijacked

protest in Bristol in which police cars were set on fire. Masks were

virtually absent in London and widespread in Bristol. Wetiko wants

lockdown on every level of society and infuses its aggression to

police it through its unknowing stooges. Lockdown protesters are

the ones with the smiling faces and the hugs, The two blatantly

obvious states of being – ge�ing more obvious by the day – are the

result of Wokers and their like becoming ever more influenced by

the simulation Field of Wetiko and Pushbackers ever more

influenced by The Field of a far higher vibration beyond the

simulation. Wetiko can’t invade the heart which is where most

lockdown opponents are coming from. It’s the heart that allows them

to see through the lies to the truth in ways I will be highlighting.

Renegade Minds know that calmness is the place from which

wisdom comes. You won’t find wisdom in a hissing fit and wisdom

is what we need in abundance right now. Calmness is not weakness

– you don’t have to scream at the top of your voice to be strong.

Calmness is indeed a sign of strength. ‘No’ means I’m not doing it.

NOOOO!!! doesn’t mean you’re not doing it even more. Volume

does not advance ‘No – I’m not doing it’. You are just not doing it.

Wetiko possessed and influenced don’t know how to deal with that.

Wetiko wants a fight and we should not give it one. What it needs

more than anything is our cooperation and we should not give that

either. Mass rallies and marches are great in that they are a visual

representation of feeling, but if it ends there they are irrelevant. You

demand that Wetikos act differently? Well, they’re not going to are

they? They are Wetikos. We don’t need to waste our time demanding

that something doesn’t happen when that will make no difference.

We need to delete the means that allows it to happen. This, invariably,

is our cooperation. You can demand a child stop firing a peashooter

at the dog or you can refuse to buy the peashooter. If you provide

the means you are cooperating with the dog being smacked on the

nose with a pea. How can the authorities enforce mask-wearing if

millions in a country refuse? What if the 74 million Pushbackers that

voted for Trump in 2020 refused to wear masks, close their

businesses or stay in their homes. It would be unenforceable. The

few control the many through the compliance of the many and that’s

always been the dynamic be it ‘Covid’ regulations or the Roman

Empire. I know people can find it intimidating to say no to authority

or stand out in a crowd for being the only one with a face on display;

but it has to be done or it’s over. I hope I’ve made clear in this book

that where this is going will be far more intimidating than standing

up now and saying ‘No’ – I will not cooperate with my own

enslavement and that of my children. There might be consequences

for some initially, although not so if enough do the same. The

question that must be addressed is what is going to happen if we

don’t? It is time to be strong and unyieldingly so. No means no. Not

here and there, but everywhere and always. I have refused to wear a

mask and obey all the other nonsense. I will not comply with

tyranny. I repeat: Fascism is not imposed by fascists – there are never

enough of them. Fascism is imposed by the population acquiescing

to fascism. I will not do it. I will die first, or my body will. Living

meekly under fascism is a form of death anyway, the death of the

spirit that Martin Luther King described.

Making things happen

We must not despair. This is not over till it’s over and it’s far from

that. The ‘fat lady’ must refuse to sing. The longer the ‘Covid’ hoax

has dragged on and impacted on more lives we have seen an

awakening of phenomenal numbers of people worldwide to the

realisation that what they have believed all their lives is not how the

world really is. Research published by the system-serving University

of Bristol and King’s College London in February, 2021, concluded:

‘One in every 11 people in Britain say they trust David Icke’s take on

the coronavirus pandemic.’ It will be more by now and we have

gathering numbers to build on. We must urgently progress from

seeing the scam to ceasing to cooperate with it. Prominent German

lawyer Reiner Fuellmich, also licenced to practice law in America, is

doing a magnificent job taking the legal route to bring the

psychopaths to justice through a second Nuremberg tribunal for

crimes against humanity. Fuellmich has an impressive record of

beating the elite in court and he formed the German Corona

Investigative Commi�ee to pursue civil charges against the main

perpetrators with a view to triggering criminal charges. Most

importantly he has grasped the foundation of the hoax – the PCR

test not testing for the ‘virus’ – and Christian Drosten is therefore on

his charge sheet along with Gates frontman Tedros at the World

Health Organization. Major players must be not be allowed to inflict

their horrors on the human race without being brought to book. A

life sentence must follow for Bill Gates and the rest of them. A group

of researchers has also indicted the government of Norway for

crimes against humanity with copies sent to the police and the

International Criminal Court. The lawsuit cites participation in an

internationally-planned false pandemic and violation of

international law and human rights, the European Commission’s

definition of human rights by coercive rules, Nuremberg and Hague

rules on fundamental human rights, and the Norwegian

constitution. We must take the initiative from hereon and not just

complain, protest and react.

There are practical ways to support vital mass non-cooperation.

Organising in numbers is one. Lockdown marches in London in the

spring in 2021 were mass non-cooperation that the authorities could

not stop. There were too many people. Hundreds of thousands

walked the London streets in the centre of the road for mile a�er

mile while the Face-Nappies could only look on. They were

determined, but calm, and just did it with no histrionics and lots of

smiles. The police were impotent. Others are organising group

shopping without masks for mutual support and imagine if that was

happening all over. Policing it would be impossible. If the store

refuses to serve people in these circumstances they would be faced

with a long line of trolleys full of goods standing on their own and

everything would have to be returned to the shelves. How would

they cope with that if it kept happening? I am talking here about

moving on from complaining to being pro-active; from watching

things happen to making things happen. I include in this our

relationship with the police. The behaviour of many Face-Nappies

•

•

•

•

has been disgraceful and anyone who thinks they would never find

concentration camp guards in the ‘enlightened’ modern era have

had that myth busted big-time. The period and se�ing may change –

Wetikos never do. I watched film footage from a London march in

which a police thug viciously kicked a protestor on the floor who

had done nothing. His fellow Face-Nappies stood in a ring

protecting him. What he did was a criminal assault and with a

crowd far outnumbering the police this can no longer be allowed to

happen unchallenged. I get it when people chant ‘shame on you’ in

these circumstances, but that is no longer enough. They have no

shame those who do this. Crowds needs to start making a citizen’s

arrest of the police who commit criminal offences and brutally a�ack

innocent people and defenceless women. A citizen’s arrest can be

made under section 24A of the UK Police and Criminal Evidence

(PACE) Act of 1984 and you will find something similar in other

countries. I prefer to call it a Common Law arrest rather than

citizen’s for reasons I will come to shortly. Anyone can arrest a

person commi�ing an indictable offence or if they have reasonable

grounds to suspect they are commi�ing an indictable offence. On

both counts the a�ack by the police thug would have fallen into this

category. A citizen’s arrest can be made to stop someone:

Causing physical injury to himself or any other person

Suffering physical injury

Causing loss of or damage to property

Making off before a constable can assume responsibility for him

A citizen’s arrest may also be made to prevent a breach of the

peace under Common Law and if they believe a breach of the peace

will happen or anything related to harm likely to be done or already

done in their presence. This is the way to go I think – the Common

Law version. If police know that the crowd and members of the

public will no longer be standing and watching while they commit

their thuggery and crimes they will think twice about acting like

Brownshirts and Blackshirts.

Common Law – common sense

Mention of Common Law is very important. Most people think the

law is the law as in one law. This is not the case. There are two

bodies of law, Common Law and Statute Law, and they are not the

same. Common Law is founded on the simple premise of do no

harm. It does not recognise victimless crimes in which no harm is

done while Statute Law does. There is a Statute Law against almost

everything. So what is Statute Law? Amazingly it’s the law of the sea

that was brought ashore by the Cult to override the law of the land

which is Common Law. They had no right to do this and as always

they did it anyway. They had to. They could not impose their will on

the people through Common Law which only applies to do no harm.

How could you stitch up the fine detail of people’s lives with that?

Instead they took the law of the sea, or Admiralty Law, and applied

it to the population. Statute Law refers to all the laws spewing out of

governments and their agencies including all the fascist laws and

regulations relating to ‘Covid’. The key point to make is that Statute

Law is contract law. It only applies between contracting corporations.

Most police officers don’t even know this. They have to be kept in

the dark, too. Long ago when merchants and their sailing ships

began to trade with different countries a contractual law was

developed called Admiralty Law and other names. Again it only

applied to contracts agreed between corporate entities. If there is no

agreed contract the law of the sea had no jurisdiction and that still

applies to its new alias of Statute Law. The problem for the Cult when

the law of the sea was brought ashore was an obvious one. People

were not corporations and neither were government entities. To

overcome the la�er they made governments and all associated

organisations corporations. All the institutions are private

corporations and I mean governments and their agencies, local

councils, police, courts, military, US states, the whole lot. Go to the

Dun and Bradstreet corporate listings website for confirmation that

they are all corporations. You are arrested by a private corporation

called the police by someone who is really a private security guard

and they take you to court which is another private corporation.

Neither have jurisdiction over you unless you consent and contract

with them. This is why you hear the mantra about law enforcement

policing by consent of the people. In truth the people ‘consent’ only

in theory through monumental trickery.

Okay, the Cult overcame the corporate law problem by making

governments and institutions corporate entities; but what about

people? They are not corporations are they? Ah ... well in a sense,

and only a sense, they are. Not people exactly – the illusion of

people. The Cult creates a corporation in the name of everyone at the

time that their birth certificate is issued. Note birth/ berth certificate

and when you go to court under the law of the sea on land you stand

in a dock. These are throwbacks to the origin. My Common Law

name is David Vaughan Icke. The name of the corporation created

by the government when I was born is called Mr David Vaughan

Icke usually wri�en in capitals as MR DAVID VAUGHAN ICKE.

That is not me, the living, breathing man. It is a fictitious corporate

entity. The trick is to make you think that David Vaughan Icke and

MR DAVID VAUGHAN ICKE are the same thing. They are not. When

police charge you and take you to court they are prosecuting the

corporate entity and not the living, breathing, man or woman. They

have to trick you into identifying as the corporate entity and

contracting with them. Otherwise they have no jurisdiction. They do

this through a language known as legalese. Lawful and legal are not

the same either. Lawful relates to Common Law and legal relates to

Statute Law. Legalese is the language of Statue Law which uses

terms that mean one thing to the public and another in legalese.

Notice that when a police officer tells someone why they are being

charged he or she will say at the end: ‘Do you understand?’ To the

public that means ‘Do you comprehend?’ In legalese it means ‘Do

you stand under me?’ Do you stand under my authority? If you say

yes to the question you are unknowingly agreeing to give them

jurisdiction over you in a contract between two corporate entities.

This is a confidence trick in every way. Contracts have to be agreed

between informed parties and if you don’t know that David

Vaughan Icke is agreeing to be the corporation MR DAVID

VAUGHAN ICKE you cannot knowingly agree to contract. They are

deceiving you and another way they do this is to ask for proof of

identity. You usually show them a driving licence or other document

on which your corporate name is wri�en. In doing so you are

accepting that you are that corporate entity when you are not.

Referring to yourself as a ‘person’ or ‘citizen’ is also identifying with

your corporate fiction which is why I made the Common Law point

about the citizen’s arrest. If you are approached by a police officer

you identify yourself immediately as a living, breathing, man or

woman and say ‘I do not consent, I do not contract with you and I do

not understand’ or stand under their authority. I have a Common

Law birth certificate as a living man and these are available at no

charge from commonlawcourt.com. Businesses registered under the

Statute Law system means that its laws apply. There are, however,

ways to run a business under Common Law. Remember all ‘Covid’

laws and regulations are Statute Law – the law of contracts and you

do not have to contract. This doesn’t mean that you can kill someone

and get away with it. Common Law says do no harm and that

applies to physical harm, financial harm etc. Police are employees of

private corporations and there needs to be a new system of non-

corporate Common Law constables operating outside the Statute

Law system. If you go to davidicke.com and put Common Law into

the search engine you will find videos that explain Common Law in

much greater detail. It is definitely a road we should walk.

With all my heart

I have heard people say that we are in a spiritual war. I don’t like the

term ‘war’ with its Wetiko dynamic, but I know what they mean.

Sweep aside all the bodily forms and we are in a situation in which

two states of consciousness are seeking very different realities.

Wetiko wants upheaval, chaos, fear, suffering, conflict and control.

The other wants love, peace, harmony, fairness and freedom. That’s

where we are. We should not fall for the idea that Wetiko is all-

powerful and there’s nothing we can do. Wetiko is not all-powerful.

It’s a joke, pathetic. It doesn’t have to be, but it has made that choice

for now. A handful of times over the years when I have felt the

presence of its frequency I have allowed it to a�ach briefly so I could

consciously observe its nature. The experience is not pleasant, the

energy is heavy and dark, but the ease with which you can kick it

back out the door shows that its real power is in persuading us that

it has power. It’s all a con. Wetiko is a con. It’s a trickster and not a

power that can control us if we unleash our own. The con is founded

on manipulating humanity to give its power to Wetiko which

recycles it back to present the illusion that it has power when its

power is ours that we gave away. This happens on an energetic level

and plays out in the world of the seen as humanity giving its power

to Wetiko authority which uses that power to control the population

when the power is only the power the population has handed over.

How could it be any other way for billions to be controlled by a

relative few? I have had experiences with people possessed by

Wetiko and again you can kick its arse if you do it with an open

heart. Oh yes – the heart which can transform the world of perceived

‘ma�er’.

We are receiver-transmi�ers and processors of information, but

what information and where from? Information is processed into

perception in three main areas – the brain, the heart and the belly.

These relate to thinking, knowing, and emotion. Wetiko wants us to

be head and belly people which means we think within the confines

of the Matrix simulation and low-vibrational emotional reaction

scrambles balance and perception. A few minutes on social media

and you see how emotion is the dominant force. Woke is all emotion

and is therefore thought-free and fact-free. Our heart is something

different. It knows while the head thinks and has to try to work it out

because it doesn’t know. The human energy field has seven prime

vortexes which connect us with wider reality (Fig 23). Chakra means

‘wheels of light’ in the Sanskrit language of ancient India. The main

ones are: The crown chakra on top of the head; brow (or ‘third eye’)

chakra in the centre of the forehead; throat chakra; heart chakra in

the centre of the chest; solar plexus chakra below the sternum; sacral

chakra beneath the navel; and base chakra at the bo�om of the spine.

Each one has a particular function or functions. We feel anxiety and

nervousness in the belly where the sacral chakra is located and this

processes emotion that can affect the colon to give people ‘the shits’

or make them ‘shit scared’ when they are nervous. Chakras all play

an important role, but the Mr and Mrs Big is the heart chakra which

sits at the centre of the seven, above the chakras that connect us to

the ‘physical’ and below those that connect with higher realms (or at

least should). Here in the heart chakra we feel love, empathy and

compassion – ‘My heart goes out to you’. Those with closed hearts

become literally ‘heart-less’ in their a�itudes and behaviour (see Bill

Gates). Native Americans portrayed Wetiko with what Paul Levy

calls a ‘frigid, icy heart, devoid of mercy’ (see Bill Gates).

Figure 23: The chakra system which interpenetrates the human energy field. The heart chakra is the governor – or should be.

Wetiko trembles at the thought of heart energy which it cannot

infiltrate. The frequency is too high. What it seeks to do instead is

close the heart chakra vortex to block its perceptual and energetic

influence. Psychopaths have ‘hearts of stone’ and emotionally-

damaged people have ‘heartache’ and ‘broken hearts’. The

astonishing amount of heart disease is related to heart chakra

disruption with its fundamental connection to the ‘physical’ heart.

Dr Tom Cowan has wri�en an outstanding book challenging the

belief that the heart is a pump and making the connection between

the ‘physical’ and spiritual heart. Rudolph Steiner who was way

ahead of his time said the same about the fallacy that the heart is a

pump. What? The heart is not a pump? That’s crazy, right?

Everybody knows that. Read Cowan’s Human Heart, Cosmic Heart

and you will realise that the very idea of the heart as a pump is

ridiculous when you see the evidence. How does blood in the feet so

far from the heart get pumped horizontally up the body by the

heart?? Cowan explains in the book the real reason why blood

moves as it does. Our ‘physical’ heart is used to symbolise love when

the source is really the heart vortex or spiritual heart which is our

most powerful energetic connection to ‘out there’ expanded

consciousness. That’s why we feel knowing – intuitive knowing – in

the centre of the chest. Knowing doesn’t come from a process of

thoughts leading to a conclusion. It is there in an instant all in one

go. Our heart knows because of its connection to levels of awareness

that do know. This is the meaning and source of intuition – intuitive

knowing.

For the last more than 30 years of uncovering the global game and

the nature of reality my heart has been my constant antenna for

truth and accuracy. An American intelligence insider once said that I

had quoted a disinformer in one of my books and yet I had only

quoted the part that was true. He asked: ‘How do you do that?’ By

using my heart antenna was the answer and anyone can do it. Heart-

centred is how we are meant to be. With a closed heart chakra we

withdraw into a closed mind and the bubble of five-sense reality. If

you take a moment to focus your a�ention on the centre of your

chest, picture a spinning wheel of light and see it opening and

expanding. You will feel it happening, too, and perceptions of the

heart like joy and love as the heart impacts on the mind as they

interact. The more the chakra opens the more you will feel

expressions of heart consciousness and as the process continues, and

becomes part of you, insights and knowings will follow. An open

heart is connected to that level of awareness that knows all is One.

You will see from its perspective that the fault-lines that divide us

are only illusions to control us. An open heart does not process the

illusions of race, creed and sexuality except as brief experiences for a

consciousness that is all. Our heart does not see division, only unity

(Figs 24 and 25). There’s something else, too. Our hearts love to

laugh. Mark Twain’s quote that says ‘The human race has one really

effective weapon, and that is laughter’ is really a reference to the

heart which loves to laugh with the joy of knowing the true nature of

infinite reality and that all the madness of human society is an

illusion of the mind. Twain also said: ‘Against the assault of laughter

nothing can stand.’ This is so true of Wetiko and the Cult. Their

insecurity demands that they be taken seriously and their power and

authority acknowledged and feared. We should do nothing of the

sort. We should not get aggressive or fearful which their insecurity

so desires. We should laugh in their face. Even in their no-face as

police come over in their face-nappies and expect to be taken

seriously. They don’t take themselves seriously looking like that so

why should we? Laugh in the face of intimidation. Laugh in the face

of tyranny. You will see by its reaction that you have pressed all of its

bu�ons. Wetiko does not know what to do in the face of laughter or

when its targets refuse to concede their joy to fear. We have seen

many examples during the ‘Covid’ hoax when people have

expressed their energetic power and the string puppets of Wetiko

retreat with their tail limp between their knees. Laugh – the world is

bloody mad a�er all and if it’s a choice between laughter and tears I

know which way I’m going.

Figure 24: Head consciousness without the heart sees division and everything apart from everything else.

Figure 25: Heart consciousness sees everything as One.

‘Vaccines’ and the soul

The foundation of Wetiko/Archon control of humans is the

separation of incarnate five-sense mind from the infinite ‘I’ and

closing the heart chakra where the True ‘I’ lives during a human life.

The goal has been to achieve complete separation in both cases. I was

interested therefore to read an account by a French energetic healer

of what she said she experienced with a patient who had been given

the ‘Covid’ vaccine. Genuine energy healers can sense information

and consciousness fields at different levels of being which are

referred to as ‘subtle bodies’. She described treating the patient who

later returned a�er having, without the healer’s knowledge, two

doses of the ‘Covid vaccine’. The healer said:

I noticed immediately the change, very heavy energy emanating from [the] subtle bodies. The scariest thing was when I was working on the heart chakra, I connected with her soul: it was detached from the physical body, it had no contact and it was, as if it was floating in a state of total confusion: a damage to the consciousness that loses contact with the physical body, i.e. with our biological machine, there is no longer any communication between them.

I continued the treatment by sending light to the heart chakra, the soul of the person, but it seemed that the soul could no longer receive any light, frequency or energy. It was a very powerful experience for me. Then I understood that this substance is indeed used to detach consciousness so that this consciousness can no longer interact through this body that it possesses in life, where there is no longer any contact, no frequency, no light, no more energetic balance or mind.

This would create a human that is rudderless and at the extreme

almost zombie-like operating with a fractional state of consciousness

at the mercy of Wetiko. I was especially intrigued by what the healer

said in the light of the prediction by the highly-informed Rudolf

Steiner more than a hundred years ago. He said:

In the future, we will eliminate the soul with medicine. Under the pretext of a ‘healthy point of view’, there will be a vaccine by which the human body will be treated as soon as possible directly at birth, so that the human being cannot develop the thought of the existence of soul and Spirit. To materialistic doctors will be entrusted the task of removing the soul of humanity.

As today, people are vaccinated against this disease or that disease, so in the future, children will be vaccinated with a substance that can be produced precisely in such a way that people, thanks to this vaccination, will be immune to being subjected to the ‘madness’ of spiritual life. He would be extremely smart, but he would not develop a conscience, and that is the true goal of some materialistic circles.

Steiner said the vaccine would detach the physical body from the

etheric body (subtle bodies) and ‘once the etheric body is detached

the relationship between the universe and the etheric body would

become extremely unstable, and man would become an automaton’.

He said ‘the physical body of man must be polished on this Earth by

spiritual will – so the vaccine becomes a kind of arymanique

(Wetiko) force’ and ‘man can no longer get rid of a given

materialistic feeling’. Humans would then, he said, become

‘materialistic of constitution and can no longer rise to the spiritual’. I

have been writing for years about DNA being a receiver-transmi�er

of information that connects us to other levels of reality and these

‘vaccines’ changing DNA can be likened to changing an antenna and

what it can transmit and receive. Such a disconnection would clearly

lead to changes in personality and perception. Steiner further

predicted the arrival of AI. Big Pharma ‘Covid vaccine’ makers,

expressions of Wetiko, are testing their DNA-manipulating evil on

children as I write with a view to giving the ‘vaccine’ to babies. If it’s

a soul-body disconnector – and I say that it is or can be – every child

would be disconnected from ‘soul’ at birth and the ‘vaccine’ would

create a closed system in which spiritual guidance from the greater

self would play no part. This has been the ambition of Wetiko all

along. A Pentagon video from 2005 was leaked of a presentation

explaining the development of vaccines to change behaviour by their

effect on the brain. Those that believe this is not happening with the

‘Covid’ genetically-modifying procedure masquerading as a

‘vaccine’ should make an urgent appointment with Naivety

Anonymous. Klaus Schwab wrote in 2018:

Neurotechnologies enable us to better influence consciousness and thought and to understand many activities of the brain. They include decoding what we are thinking in fine levels of detail through new chemicals and interventions that can influence our brains to correct for errors or enhance functionality.

The plan is clear and only the heart can stop it. With every heart that

opens, every mind that awakens, Wetiko is weakened. Heart and

love are far more powerful than head and hate and so nothing like a

majority is needed to turn this around.

Beyond the Phantom

Our heart is the prime target of Wetiko and so it must be the answer

to Wetiko. We are our heart which is part of one heart, the infinite

heart. Our heart is where the true self lives in a human life behind

firewalls of five-sense illusion when an imposter takes its place –

Phantom Self; but our heart waits patiently to be set free any time we

choose to see beyond the Phantom, beyond Wetiko. A Wetikoed

Phantom Self can wreak mass death and destruction while the love

of forever is locked away in its heart. The time is here to unleash its

power and let it sweep away the fear and despair that is Wetiko.

Heart consciousness does not seek manipulated, censored,

advantage for its belief or religion, its activism and desires. As an

expression of the One it treats all as One with the same rights to

freedom and opinion. Our heart demands fairness for itself no more

than for others. From this unity of heart we can come together in

mutual support and transform this Wetikoed world into what reality

is meant to be – a place of love, joy, happiness, fairness, justice and

freedom. Wetiko has another agenda and that’s why the world is as

it is, but enough of this nonsense. Wetiko can’t stay where hearts are

open and it works so hard to keep them closed. Fear is its currency

and its food source and love in its true sense has no fear. Why would

love have fear when it knows it is All That Is, Has Been, And Ever Can

Be on an eternal exploration of all possibility? Love in this true sense

is not the physical a�raction that passes for love. This can be an

expression of it, yes, but Infinite Love, a love without condition, goes

far deeper to the core of all being. It is the core of all being. Infinite

realty was born from love beyond the illusions of the simulation.

Love infinitely expressed is the knowing that all is One and the

swi�ly-passing experience of separation is a temporary

hallucination. You cannot disconnect from Oneness; you can only

perceive that you have and withdraw from its influence. This is the

most important of all perception trickery by the mind parasite that is

Wetiko and the foundation of all its potential for manipulation.

If we open our hearts, open the sluice gates of the mind, and

redefine self-identity amazing things start to happen. Consciousness

expands or contracts in accordance with self-identity. When true self

is recognised as infinite awareness and label self – Phantom Self – is

seen as only a series of brief experiences life is transformed.

Consciousness expands to the extent that self-identity expands and

everything changes. You see unity, not division, the picture, not the

pixels. From this we can play the long game. No more is an

experience something in and of itself, but a fleeting moment in the

eternity of forever. Suddenly people in uniform and dark suits are no

longer intimidating. Doing what your heart knows to be right is no

longer intimidating and consequences for those actions take on the

same nature of a brief experience that passes in the blink of an

infinite eye. Intimidation is all in the mind. Beyond the mind there is

no intimidation.

An open heart does not consider consequences for what it knows

to be right. To do so would be to consider not doing what it knows to

be right and for a heart in its power that is never an option. The

Renegade Mind is really the Renegade Heart. Consideration of

consequences will always provide a getaway car for the mind and

the heart doesn’t want one. What is right in the light of what we face

today is to stop cooperating with Wetiko in all its forms and to do it

without fear or compromise. You cannot compromise with tyranny

when tyranny always demands more until it has everything. Life is

your perception and you are your destiny. Change your perception

and you change your life. Change collective perception and we

change the world.

Come on people … One human family, One heart, One goal …

FREEEEEEDOM!

We must se�le for nothing less.

T

Postscript

he big scare story as the book goes to press is the ‘Indian’

variant and the world is being deluged with propaganda about

the ‘Covid catastrophe’ in India which mirrors in its lies and

misrepresentations what happened in Italy before the first lockdown

in 2020.

The New York Post published a picture of someone who had

‘collapsed in the street from Covid’ in India in April, 2021, which

was actually taken during a gas leak in May, 2020. Same old, same

old. Media articles in mid-February were asking why India had been

so untouched by ‘Covid’ and then as their vaccine rollout gathered

pace the alleged ‘cases’ began to rapidly increase. Indian ‘Covid

vaccine’ maker Bharat Biotech was funded into existence by the Bill

and Melinda Gates Foundation (the pair announced their divorce in

May, 2021, which is a pity because they so deserve each other). The

Indian ‘Covid crisis’ was ramped up by the media to terrify the

world and prepare people for submission to still more restrictions.

The scam that worked the first time was being repeated only with far

more people seeing through the deceit. Davidicke.com and

Ickonic.com have sought to tell the true story of what is happening

by talking to people living through the Indian nightmare which has

nothing to do with ‘Covid’. We posted a le�er from ‘Alisha’ in Pune

who told a very different story to government and media mendacity.

She said scenes of dying people and overwhelmed hospitals were

designed to hide what was really happening – genocide and

starvation. Alisha said that millions had already died of starvation

during the ongoing lockdowns while government and media were

lying and making it look like the ‘virus’:

Restaurants, shops, gyms, theatres, basically everything is shut. The cities are ghost towns. Even so-called ‘essential’ businesses are only open till 11am in the morning. You basically have just an hour to buy food and then your time is up.

Inter-state travel and even inter-district travel is banned. The cops wait at all major crossroads to question why you are traveling outdoors or to fine you if you are not wearing a mask.

The medical community here is also complicit in genocide, lying about hospitals being full and turning away people with genuine illnesses, who need immediate care. They have even created a shortage of oxygen cylinders.

This is the classic Cult modus operandi played out in every country.

Alisha said that people who would not have a PCR test not testing

for the ‘virus’ were being denied hospital treatment. She said the

people hit hardest were migrant workers and those in rural areas.

Most businesses employed migrant workers and with everything

closed there were no jobs, no income and no food. As a result

millions were dying of starvation or malnutrition. All this was

happening under Prime Minister Narendra Modi, a 100-percent

asset of the Cult, and it emphasises yet again the scale of pure anti-

human evil we are dealing with. Australia banned its people from

returning home from India with penalties for trying to do so of up to

five years in jail and a fine of £37,000. The manufactured ‘Covid’

crisis in India was being prepared to justify further fascism in the

West. Obvious connections could be seen between the Indian

‘vaccine’ programme and increased ‘cases’ and this became a

common theme. The Seychelles, the most per capita ‘Covid

vaccinated’ population in the world, went back into lockdown a�er a

‘surge of cases’.

Long ago the truly evil Monsanto agricultural biotechnology

corporation with its big connections to Bill Gates devastated Indian

farming with genetically-modified crops. Human rights activist

Gurcharan Singh highlighted the efforts by the Indian government

to complete the job by destroying the food supply to hundreds of

millions with ‘Covid’ lockdowns. He said that 415 million people at

the bo�om of the disgusting caste system (still going whatever they

say) were below the poverty line and struggled to feed themselves

every year. Now the government was imposing lockdown at just the

time to destroy the harvest. This deliberate policy was leading to

mass starvation. People may reel back at the suggestion that a

government would do that, but Wetiko-controlled ‘leaders’ are

capable of any level of evil. In fact what is described in India is in the

process of being instigated worldwide. The food chain and food

supply are being targeted at every level to cause world hunger and

thus control. Bill Gates is not the biggest owner of farmland in

America for no reason and destroying access to food aids both the

depopulation agenda and the plan for synthetic ‘food’ already being

funded into existence by Gates. Add to this the coming hyper-

inflation from the suicidal creation of fake ‘money’ in response to

‘Covid’ and the breakdown of container shipping systems and you

have a cocktail that can only lead one way and is meant to. The Cult

plan is to crash the entire system to ‘build back be�er’ with the Great

Reset.

‘Vaccine’ transmission

Reports from all over the world continue to emerge of women

suffering menstrual and fertility problems a�er having the fake

‘vaccine’ and of the non-’vaccinated’ having similar problems when

interacting with the ‘vaccinated’. There are far too many for

‘coincidence’ to be credible. We’ve had menopausal women ge�ing

periods, others having periods stop or not stopping for weeks,

passing clots, sometimes the lining of the uterus, breast

irregularities, and miscarriages (which increased by 400 percent in

parts of the United States). Non-‘vaccinated’ men and children have

suffered blood clots and nose bleeding a�er interaction with the

‘vaccinated’. Babies have died from the effects of breast milk from a

‘vaccinated’ mother. Awake doctors – the small minority –

speculated on the cause of non-’vaccinated’ suffering the same

effects as the ‘vaccinated’. Was it nanotechnology in the synthetic

substance transmi�ing frequencies or was it a straight chemical

bioweapon that was being transmi�ed between people? I am not

saying that some kind of chemical transmission is not one possible

answer, but the foundation of all that the Cult does is frequency and

this is fertile ground for understanding how transmission can

happen. American doctor Carrie Madej, an internal medicine

physician and osteopath, has been practicing for the last 20 years,

teaching medical students, and she says a�ending different meetings

where the agenda for humanity was discussed. Madej, who operates

out of Georgia, did not dismiss other possible forms of transmission,

but she focused on frequency in search of an explanation for

transmission. She said the Moderna and Pfizer ‘vaccines’ contained

nano-lipid particles as a key component. This was a brand new

technology never before used on humanity. ‘They’re using a

nanotechnology which is pre�y much li�le tiny computer bits …

nanobots or hydrogel.’ Inside the ‘vaccines’ was ‘this sci-fi kind of

substance’ which suppressed immune checkpoints to get into the

cell. I referred to this earlier as the ‘Trojan horse’ technique that

tricks the cell into opening a gateway for the self-replicating

synthetic material and while the immune system is artificially

suppressed the body has no defences. Madej said the substance

served many purposes including an on-demand ability to ‘deliver

the payload’ and using the nano ‘computer bits’ as biosensors in the

body. ‘It actually has the ability to accumulate data from your body,

like your breathing, your respiration, thoughts, emotions, all kinds

of things.’

She said the technology obviously has the ability to operate

through Wi-Fi and transmit and receive energy, messages,

frequencies or impulses. ‘Just imagine you’re ge�ing this new

substance in you and it can react to things all around you, the 5G,

your smart device, your phones.’ We had something completely

foreign in the human body that had never been launched large scale

at a time when we were seeing 5G going into schools and hospitals

(plus the Musk satellites) and she believed the ‘vaccine’ transmission

had something to do with this: ‘… if these people have this inside of

them … it can act like an antenna and actually transmit it outwardly

as well.’ The synthetic substance produced its own voltage and so it

could have that kind of effect. This fits with my own contention that

the nano receiver-transmi�ers are designed to connect people to the

Smart Grid and break the receiver-transmi�er connection to

expanded consciousness. That would explain the French energy

healer’s experience of the disconnection of body from ‘soul’ with

those who have had the ‘vaccine’. The nanobots, self-replicating

inside the body, would also transmit the synthetic frequency which

could be picked up through close interaction by those who have not

been ‘vaccinated’. Madej speculated that perhaps it was 5G and

increased levels of other radiation that was causing the symptoms

directly although interestingly she said that non-‘vaccinated’

patients had shown improvement when they were away from the

‘vaccinated’ person they had interacted with. It must be remembered

that you can control frequency and energy with your mind and you

can consciously create energetic barriers or bubbles with the mind to

stop damaging frequencies from penetrating your field. American

paediatrician Dr Larry Palevsky said the ‘vaccine’ was not a ‘vaccine’

and was never designed to protect from a ‘viral’ infection. He called

it ‘a massive, brilliant propaganda of genocide’ because they didn’t

have to inject everyone to get the result they wanted. He said the

content of the jabs was able to infuse any material into the brain,

heart, lungs, kidneys, liver, sperm and female productive system.

‘This is genocide; this is a weapon of mass destruction.’ At the same

time American colleges were banning students from a�ending if

they didn’t have this life-changing and potentially life-ending

‘vaccine’. Class action lawsuits must follow when the consequences

of this college fascism come to light. As the book was going to press

came reports about fertility effects on sperm in ‘vaccinated’ men

which would absolutely fit with what I have been saying and

hospitals continued to fill with ‘vaccine’ reactions. Another question

is what about transmission via blood transfusions? The NHS has

extended blood donation restrictions from seven days a�er a ‘Covid

vaccination’ to 28 days a�er even a sore arm reaction.

I said in the spring of 2020 that the then touted ‘Covid vaccine’

would be ongoing each year like the flu jab. A year later Pfizer CEO,

the appalling Albert Bourla, said people would ‘likely’ need a

‘booster dose’ of the ‘vaccine’ within 12 months of ge�ing ‘fully

vaccinated’ and then a yearly shot. ‘Variants will play a key role’, he

said confirming the point. Johnson & Johnson CEO Alex Gorsky also

took time out from his ‘vaccine’ disaster to say that people may need

to be vaccinated against ‘Covid-19’ each year. UK Health Secretary,

the psychopath Ma� Hancock, said additional ‘boosters’ would be

available in the autumn of 2021. This is the trap of the ‘vaccine

passport’. The public will have to accept every last ‘vaccine’ they

introduce, including for the fake ‘variants’, or it would cease to be

valid. The only other way in some cases would be continuous testing

with a test not testing for the ‘virus’ and what is on the swabs

constantly pushed up your noise towards the brain every time?

‘Vaccines’ changing behaviour

I mentioned in the body of the book how I believed we would see

gathering behaviour changes in the ‘vaccinated’ and I am already

hearing such comments from the non-‘vaccinated’ describing

behaviour changes in friends, loved ones and work colleagues. This

will only increase as the self-replicating synthetic material and

nanoparticles expand in body and brain. An article in the Guardian in

2016 detailed research at the University of Virginia in Charlo�esville

which developed a new method for controlling brain circuits

associated with complex animal behaviour. The method, dubbed

‘magnetogenetics’, involves genetically-engineering a protein called

ferritin, which stores and releases iron, to create a magnetised

substance – ‘Magneto’ – that can activate specific groups of nerve

cells from a distance. This is claimed to be an advance on other

methods of brain activity manipulation known as optogenetics and

chemogenetics (the Cult has been developing methods of brain

control for a long time). The ferritin technique is said to be non-

invasive and able to activate neurons ‘rapidly and reversibly’. In

other words, human thought and perception. The article said that

earlier studies revealed how nerve cell proteins ‘activated by heat

and mechanical pressure can be genetically engineered so that they

become sensitive to radio waves and magnetic fields, by a�aching

them to an iron-storing protein called ferritin, or to inorganic

paramagnetic particles’. Sensitive to radio waves and magnetic

fields? You mean like 5G, 6G and 7G? This is the human-AI Smart

Grid hive mind we are talking about. The Guardian article said:

… the researchers injected Magneto into the striatum of freely behaving mice, a deep brain structure containing dopamine-producing neurons that are involved in reward and motivation, and then placed the animals into an apparatus split into magnetised and non-magnetised sections.

Mice expressing Magneto spent far more time in the magnetised areas than mice that did not, because activation of the protein caused the striatal neurons expressing it to release dopamine, so that the mice found being in those areas rewarding. This shows that Magneto can remotely control the firing of neurons deep within the brain, and also control complex behaviours.

Make no mistake this basic methodology will be part of the ‘Covid

vaccine’ cocktail and using magnetics to change brain function

through electromagnetic field frequency activation. The Pentagon is

developing a ‘Covid vaccine’ using ferritin. Magnetics would explain

changes in behaviour and why videos are appearing across the

Internet as I write showing how magnets stick to the skin at the

point of the ‘vaccine’ shot. Once people take these ‘vaccines’

anything becomes possible in terms of brain function and illness

which will be blamed on ‘Covid-19’ and ‘variants’. Magnetic field

manipulation would further explain why the non-‘vaccinated’ are

reporting the same symptoms as the ‘vaccinated’ they interact with

and why those symptoms are reported to decrease when not in their

company. Interestingly ‘Magneto’, a ‘mutant’, is a character in the

Marvel Comic X-Men stories with the ability to manipulate magnetic

fields and he believes that mutants should fight back against their

human oppressors by any means necessary. The character was born

Erik Lehnsherr to a Jewish family in Germany.

Cult-controlled courts

The European Court of Human Rights opened the door for

mandatory ‘Covid-19 vaccines’ across the continent when it ruled in

a Czech Republic dispute over childhood immunisation that legally

enforced vaccination could be ‘necessary in a democratic society’.

The 17 judges decided that compulsory vaccinations did not breach

human rights law. On the face of it the judgement was so inverted

you gasp for air. If not having a vaccine infused into your body is not

a human right then what is? Ah, but they said human rights law

which has been specifically wri�en to delete all human rights at the

behest of the state (the Cult). Article 8 of the European Convention

on Human Rights relates to the right to a private life. The crucial

word here is ‘except’:

There shall be no interference by a public authority with the exercise of this right EXCEPT such as is in accordance with the law and is necessary in a democratic society in the interests of national security, public safety or the economic wellbeing of the country, for the prevention of disorder or crime, for the protection of health or morals, or for the protection of the rights and freedoms of others [My emphasis].

No interference except in accordance with the law means there are no

‘human rights’ except what EU governments decide you can have at

their behest. ‘As is necessary in a democratic society’ explains that

reference in the judgement and ‘in the interests of national security,

public safety or the economic well-being of the country, for the

prevention of disorder or crime, for the protection of health or

morals, or for the protection of the rights and freedoms of others’

gives the EU a coach and horses to ride through ‘human rights’ and

sca�er them in all directions. The judiciary is not a check and

balance on government extremism; it is a vehicle to enforce it. This

judgement was almost laughably predictable when the last thing the

Cult wanted was a decision that went against mandatory

vaccination. Judges rule over and over again to benefit the system of

which they are a part. Vaccination disputes that come before them

are invariably delivered in favour of doctors and authorities

representing the view of the state which owns the judiciary. Oh, yes,

and we have even had calls to stop pu�ing ‘Covid-19’ on death

certificates within 28 days of a ‘positive test’ because it is claimed the

practice makes the ‘vaccine’ appear not to work. They are laughing

at you.

The scale of madness, inhumanity and things to come was

highlighted when those not ‘vaccinated’ for ‘Covid’ were refused

evacuation from the Caribbean island of St Vincent during massive

volcanic eruptions. Cruise ships taking residents to the safety of

another island allowed only the ‘vaccinated’ to board and the rest

were le� to their fate. Even in life and death situations like this we

see ‘Covid’ stripping people of their most basic human instincts and

the insanity is even more extreme when you think that fake

‘vaccine’-makers are not even claiming their body-manipulating

concoctions stop ‘infection’ and ‘transmission’ of a ‘virus’ that

doesn’t exist. St Vincent Prime Minister Ralph Gonsalves said: ‘The

chief medical officer will be identifying the persons already

vaccinated so that we can get them on the ship.’ Note again the

power of the chief medical officer who, like Whi�y in the UK, will be

answering to the World Health Organization. This is the Cult

network structure that has overridden politicians who ‘follow the

science’ which means doing what WHO-controlled ‘medical officers’

and ‘science advisers’ tell them. Gonsalves even said that residents

who were ‘vaccinated’ a�er the order so they could board the ships

would still be refused entry due to possible side effects such as

‘wooziness in the head’. The good news is that if they were woozy

enough in the head they could qualify to be prime minister of St

Vincent.

Microchipping freedom

The European judgement will be used at some point to justify moves

to enforce the ‘Covid’ DNA-manipulating procedure. Sandra Ro,

CEO of the Global Blockchain Business Council, told a World

Economic Forum event that she hoped ‘vaccine passports’ would

help to ‘drive forced consent and standardisation’ of global digital

identity schemes: ‘I’m hoping with the desire and global demand for

some sort of vaccine passport – so that people can get travelling and

working again – [it] will drive forced consent, standardisation, and

frankly, cooperation across the world.’ The lady is either not very

bright, or thoroughly mendacious, to use the term ‘forced consent’.

You do not ‘consent’ if you are forced – you submit. She was

describing what the plan has been all along and that’s to enforce a

digital identity on every human without which they could not

function. ‘Vaccine passports’ are opening the door and are far from

the end goal. A digital identity would allow you to be tracked in

everything you do in cyberspace and this is the same technique used

by Cult-owned China to enforce its social credit system of total

control. The ultimate ‘passport’ is planned to be a microchip as my

books have warned for nearly 30 years. Those nice people at the

Pentagon working for the Cult-controlled Defense Advanced

Research Projects Agency (DARPA) claimed in April, 2021, they

have developed a microchip inserted under the skin to detect

‘asymptomatic Covid-19 infection’ before it becomes an outbreak

and a ‘revolutionary filter’ that can remove the ‘virus’ from the

blood when a�ached to a dialysis machine. The only problems with

this are that the ‘virus’ does not exist and people transmi�ing the

‘virus’ with no symptoms is brain-numbing bullshit. This is, of

course, not a ruse to get people to be microchipped for very different

reasons. DARPA also said it was producing a one-stop ‘vaccine’ for

the ‘virus’ and all ‘variants’. One of the most sinister organisations

on Planet Earth is doing this? Be�er have it then. These people are

insane because Wetiko that possesses them is insane.

Researchers from the Salk Institute in California announced they

have created an embryo that is part human and part monkey. My

books going back to the 1990s have exposed experiments in top

secret underground facilities in the United States where humans are

being crossed with animal and non-human ‘extraterrestrial’ species.

They are now easing that long-developed capability into the public

arena and there is much more to come given we are dealing with

psychiatric basket cases. Talking of which – Elon Musk’s scientists at

Neuralink trained a monkey to play Pong and other puzzles on a

computer screen using a joystick and when the monkey made the

correct move a metal tube squirted banana smoothie into his mouth

which is the basic technique for training humans into unquestioning

compliance. Two Neuralink chips were in the monkey’s skull and

more than 2,000 wires ‘fanned out’ into its brain. Eventually the

monkey played a video game purely with its brain waves.

Psychopathic narcissist Musk said the ‘breakthrough’ was a step

towards pu�ing Neuralink chips into human skulls and merging

minds with artificial intelligence. Exactly. This man is so dark and

Cult to his DNA.

World Economic Fascism (WEF)

The World Economic Forum is telling you the plan by the statements

made at its many and various events. Cult-owned fascist YouTube

CEO Susan Wojcicki spoke at the 2021 WEF Global Technology

Governance Summit (see the name) in which 40 governments and

150 companies met to ensure ‘the responsible design and

deployment of emerging technologies’. Orwellian translation:

‘Ensuring the design and deployment of long-planned technologies

will advance the Cult agenda for control and censorship.’ Freedom-

destroyer and Nuremberg-bound Wojcicki expressed support for

tech platforms like hers to censor content that is ‘technically legal but

could be harmful’. Who decides what is ‘harmful’? She does and

they do. ‘Harmful’ will be whatever the Cult doesn’t want people to

see and we have legislation proposed by the UK government that

would censor content on the basis of ‘harm’ no ma�er if the

information is fair, legal and provably true. Make that especially if it

is fair, legal and provably true. Wojcicki called for a global coalition

to be formed to enforce content moderation standards through

automated censorship. This is a woman and mega-censor so self-

deluded that she shamelessly accepted a ‘free expression’ award –

Wojcicki – in an event sponsored by her own YouTube. They have no

shame and no self-awareness.

You know that ‘Covid’ is a scam and Wojcicki a Cult operative

when YouTube is censoring medical and scientific opinion purely on

the grounds of whether it supports or opposes the Cult ‘Covid’

narrative. Florida governor Ron DeSantis compiled an expert panel

with four professors of medicine from Harvard, Oxford, and

Stanford Universities who spoke against forcing children and

vaccinated people to wear masks. They also said there was no proof

that lockdowns reduced spread or death rates of ‘Covid-19’. Cult-

gofer Wojcicki and her YouTube deleted the panel video ‘because it

included content that contradicts the consensus of local and global

health authorities regarding the efficacy of masks to prevent the

spread of Covid-19’. This ‘consensus’ refers to what the Cult tells the

World Health Organization to say and the WHO tells ‘local health

authorities’ to do. Wojcicki knows this, of course. The panellists

pointed out that censorship of scientific debate was responsible for

deaths from many causes, but Wojcicki couldn’t care less. She would

not dare go against what she is told and as a disgrace to humanity

she wouldn’t want to anyway. The UK government is seeking to pass

a fascist ‘Online Safety Bill’ to specifically target with massive fines

and other means non-censored video and social media platforms to

make them censor ‘lawful but harmful’ content like the Cult-owned

Facebook, Twi�er, Google and YouTube. What is ‘lawful but

harmful’ would be decided by the fascist Blair-created Ofcom.

Another WEF obsession is a cyber-a�ack on the financial system

and this is clearly what the Cult has planned to take down the bank

accounts of everyone – except theirs. Those that think they have

enough money for the Cult agenda not to ma�er to them have got a

big lesson coming if they continue to ignore what is staring them in

the face. The World Economic Forum, funded by Gates and fronted

by Klaus Schwab, announced it would be running a ‘simulation’

with the Russian government and global banks of just such an a�ack

called Cyber Polygon 2021. What they simulate – as with the ‘Covid’

Event 201 – they plan to instigate. The WEF is involved in a project

with the Cult-owned Carnegie Endowment for International Peace

called the WEF-Carnegie Cyber Policy Initiative which seeks to

merge Wall Street banks, ‘regulators’ (I love it) and intelligence

agencies to ‘prevent’ (arrange and allow) a cyber-a�ack that would

bring down the global financial system as long planned by those that

control the WEF and the Carnegie operation. The Carnegie

Endowment for International Peace sent an instruction to First World

War US President Woodrow Wilson not to let the war end before

society had been irreversibly transformed.

The Wuhan lab diversion

As I close, the Cult-controlled authorities and lapdog media are

systematically pushing ‘the virus was released from the Wuhan lab’

narrative. There are two versions – it happened by accident and it

happened on purpose. Both are nonsense. The perceived existence of

the never-shown-to-exist ‘virus’ is vital to sell the impression that

there is actually an infective agent to deal with and to allow the

endless potential for terrifying the population with ‘variants’ of a

‘virus’ that does not exist. The authorities at the time of writing are

going with the ‘by accident’ while the alternative media is

promoting the ‘on purpose’. Cable news host Tucker Carlson who

has questioned aspects of lockdown and ‘vaccine’ compulsion has

bought the Wuhan lab story. ‘Everyone now agrees’ he said. Well, I

don’t and many others don’t and the question is why does the system

and its media suddenly ‘agree’? When the media moves as one unit

with a narrative it is always a lie – witness the hour by hour

mendacity of the ‘Covid’ era. Why would this Cult-owned

combination which has unleashed lies like machine gun fire

suddenly ‘agree’ to tell the truth??

Much of the alternative media is buying the lie because it fits the

conspiracy narrative, but it’s the wrong conspiracy. The real

conspiracy is that there is no virus and that is what the Cult is

desperate to hide. The idea that the ‘virus’ was released by accident

is ludicrous when the whole ‘Covid’ hoax was clearly long-planned

and waiting to be played out as it was so fast in accordance with the

Rockefeller document and Event 201. So they prepared everything in

detail over decades and then sat around strumming their fingers

waiting for an ‘accidental’ release from a bio-lab? What?? It’s crazy.

Then there’s the ‘on purpose’ claim. You want to circulate a ‘deadly

virus’ and hide the fact that you’ve done so and you release it down

the street from the highest-level bio-lab in China? I repeat – What??

You would release it far from that lab to stop any association being

made. But, no, we’ll do it in a place where the connection was certain

to be made. Why would you need to scam ‘cases’ and ‘deaths’ and

pay hospitals to diagnose ‘Covid-19’ if you had a real ‘virus’? What

are sections of the alternative media doing believing this crap?

Where were all the mass deaths in Wuhan from a ‘deadly pathogen’

when the recovery to normal life a�er the initial propaganda was

dramatic in speed? Why isn’t the ‘deadly pathogen’ now circulating

all over China with bodies in the street? Once again we have the

technique of tell them what they want to hear and they will likely

believe it. The alternative media has its ‘conspiracy’ and with

Carlson it fits with his ‘China is the danger’ narrative over years.

China is a danger as a global Cult operations centre, but not for this

reason. The Wuhan lab story also has the potential to instigate

conflict with China when at some stage the plan is to trigger a

Problem-Reaction-Solution confrontation with the West. Question

everything – everything – and especially when the media agrees on a

common party line.

Third wave … fourth wave … fifth wave …

As the book went into production the world was being set up for

more lockdowns and a ‘third wave’ supported by invented ‘variants’

that were increasing all the time and will continue to do so in public

statements and computer programs, but not in reality. India became

the new Italy in the ‘Covid’ propaganda campaign and we were told

to be frightened of the new ‘Indian strain’. Somehow I couldn’t find

it within myself to do so. A document produced for the UK

government entitled ‘Summary of further modelling of easing of

restrictions – Roadmap Step 2’ declared that a third wave was

inevitable (of course when it’s in the script) and it would be the fault

of children and those who refuse the health-destroying fake ‘Covid

vaccine’. One of the computer models involved came from the Cult-

owned Imperial College and the other from Warwick University

which I wouldn’t trust to tell me the date in a calendar factory. The

document states that both models presumed extremely high uptake

of the ‘Covid vaccines’ and didn’t allow for ‘variants’. The document

states: ‘The resurgence is a result of some people (mostly children)

being ineligible for vaccination; others choosing not to receive the

vaccine; and others being vaccinated but not perfectly protected.’

The mendacity takes the breath away. Okay, blame those with a

brain who won’t take the DNA-modifying shots and put more

pressure on children to have it as ‘trials’ were underway involving

children as young as six months with parents who give insanity a

bad name. Massive pressure is being put on the young to have the

fake ‘vaccine’ and child age consent limits have been systematically

lowered around the world to stop parents intervening. Most

extraordinary about the document was its claim that the ‘third wave’

would be driven by ‘the resurgence in both hospitalisations and

deaths … dominated by those that have received two doses of the vaccine,

comprising around 60-70% of the wave respectively’. The predicted

peak of the ‘third wave’ suggested 300 deaths per day with 250 of

them fully ‘vaccinated’ people. How many more lies do acquiescers

need to be told before they see the obvious? Those who took the jab

to ‘protect themselves’ are projected to be those who mostly get sick

and die? So what’s in the ‘vaccine’? The document went on:

It is possible that a summer of low prevalence could be followed by substantial increases in incidence over the following autumn and winter. Low prevalence in late summer should not be taken as an indication that SARS-CoV-2 has retreated or that the population has high enough levels of immunity to prevent another wave.

They are telling you the script and while many British people

believed ‘Covid’ restrictions would end in the summer of 2021 the

government was preparing for them to be ongoing. Authorities were

awarding contracts for ‘Covid marshals’ to police the restrictions

with contracts starting in July, 2021, and going through to January

31st, 2022, and the government was advertising for ‘Media Buying

Services’ to secure media propaganda slots worth a potential £320

million for ‘Covid-19 campaigns’ with a contract not ending until

March, 2022. The recipient – via a list of other front companies – was

reported to be American media marketing giant Omnicom Group

Inc. While money is no object for ‘Covid’ the UK waiting list for all

other treatment – including life-threatening conditions – passed 4.5

million. Meantime the Cult is seeking to control all official ‘inquiries’

to block revelations about what has really been happening and why.

It must not be allowed to – we need Nuremberg jury trials in every

country. The cover-up doesn’t get more obvious than appointing

ultra-Zionist professor Philip Zelikow to oversee two dozen US

virologists, public health officials, clinicians, former government

officials and four American ‘charitable foundations’ to ‘learn the

lessons’ of the ‘Covid’ debacle. The personnel will be those that

created and perpetuated the ‘Covid’ lies while Zelikow is the former

executive director of the 9/11 Commission who ensured that the

truth about those a�acks never came out and produced a report that

must be among the most mendacious and manipulative documents

ever wri�en – see The Trigger for the detailed exposure of the almost

unimaginable 9/11 story in which Sabbatians can be found at every

level.

Passive no more

People are increasingly challenging the authorities with amazing

numbers of people taking to the streets in London well beyond the

ability of the Face-Nappies to stop them. Instead the Nappies choose

situations away from the mass crowds to target, intimidate, and seek

to promote the impression of ‘violent protestors’. One such incident

happened in London’s Hyde Park. Hundreds of thousands walking

through the streets in protest against ‘Covid’ fascism were ignored

by the Cult-owned BBC and most of the rest of the mainstream

media, but they delighted in reporting how police were injured in

‘clashes with protestors’. The truth was that a group of people

gathered in Hyde Park at the end of one march when most had gone

home and they were peacefully having a good time with music and

chat. Face-Nappies who couldn’t deal with the full-march crowd

then waded in with their batons and got more than they bargained

for. Instead of just standing for this criminal brutality the crowd

used their numerical superiority to push the Face-Nappies out of the

park. Eventually the Nappies turned and ran. Unfortunately two or

three idiots in the crowd threw drink cans striking two officers

which gave the media and the government the image they wanted to

discredit the 99.9999 percent who were peaceful. The idiots walked

straight into the trap and we must always be aware of potential

agent provocateurs used by the authorities to discredit their targets.

This response from the crowd – the can people apart – must be a

turning point when the public no longer stand by while the innocent

are arrested and brutally a�acked by the Face-Nappies. That doesn’t

mean to be violent, that’s the last thing we need. We’ll leave the

violence to the Face-Nappies and government. But it does mean that

when the Face-Nappies use violence against peaceful people the

numerical superiority is employed to stop them and make citizen’s

arrests or Common Law arrests for a breach of the peace. The time

for being passive in the face of fascism is over.

We are the many, they are the few, and we need to make that count

before there is no freedom le� and our children and grandchildren

face an ongoing fascist nightmare.

COME ON PEOPLE – IT’S TIME.

One final thought …

The power of love

A force from above

Cleaning my soul

Flame on burn desire

Love with tongues of fire

Purge the soul

Make love your goal

I’ll protect you from the hooded claw

Keep the vampires from your door

When the chips are down I’ll be around

With my undying, death-defying

Love for you

Envy will hurt itself

Let yourself be beautiful

Sparkling love, flowers

And pearls and pre�y girls

Love is like an energy

Rushin’ rushin’ inside of me

This time we go sublime

Lovers entwine, divine, divine,

Love is danger, love is pleasure

Love is pure – the only treasure

I’m so in love with you

Purge the soul

Make love your goal

The power of love

A force from above

Cleaning my soul

The power of love

A force from above

A sky-scraping dove

Flame on burn desire

Love with tongues of fire

Purge the soul

Make love your goal

Frankie Goes To Hollywood

T

•

•

•

•

Appendix

Cowan-Kaufman-Morell Statement on Virus Isolation

(SOVI)

Isolation: The action of isolating; the fact or condition of being

isolated or standing alone; separation from other things or persons;

solitariness

Oxford English Dictionary

he controversy over whether the SARS-CoV-2 virus has ever

been isolated or purified continues. However, using the above

definition, common sense, the laws of logic and the dictates of

science, any unbiased person must come to the conclusion that the

SARS-CoV-2 virus has never been isolated or purified. As a result, no

confirmation of the virus’ existence can be found. The logical,

common sense, and scientific consequences of this fact are:

the structure and composition of something not shown to exist

can’t be known, including the presence, structure, and function of

any hypothetical spike or other proteins;

the genetic sequence of something that has never been found can’t

be known;

“variants” of something that hasn’t been shown to exist can’t be

known;

it’s impossible to demonstrate that SARS-CoV-2 causes a disease

called Covid-19.

1

2

In as concise terms as possible, here’s the proper way to isolate,

characterize and demonstrate a new virus. First, one takes samples

(blood, sputum, secretions) from many people (e.g. 500) with

symptoms which are unique and specific enough to characterize an

illness. Without mixing these samples with ANY tissue or products

that also contain genetic material, the virologist macerates, filters

and ultracentrifuges i.e. purifies the specimen. This common virology

technique, done for decades to isolate bacteriophages 1 and so-called

giant viruses in every virology lab, then allows the virologist to

demonstrate with electron microscopy thousands of identically sized

and shaped particles. These particles are the isolated and purified

virus.

These identical particles are then checked for uniformity by

physical and/or microscopic techniques. Once the purity is

determined, the particles may be further characterized. This would

include examining the structure, morphology, and chemical

composition of the particles. Next, their genetic makeup is

characterized by extracting the genetic material directly from the

purified particles and using genetic-sequencing techniques, such as

Sanger sequencing, that have also been around for decades. Then

one does an analysis to confirm that these uniform particles are

exogenous (outside) in origin as a virus is conceptualized to be, and

not the normal breakdown products of dead and dying tissues. 2 (As

of May 2020, we know that virologists have no way to determine

whether the particles they’re seeing are viruses or just normal break-

down products of dead and dying tissues.) 3

Isolation, characterization and analysis of bacteriophages from the haloalkaline lake Elmenteita, KenyaJuliah Khayeli Akhwale et al, PLOS One, Published: April 25, 2019. https://journals.plos.org/plosone/article?id=10.1371/journal.pone.0215734 – accessed 2/15/21

“Extracellular Vesicles Derived From Apoptotic Cells: An Essential Link Between Death and Regeneration,” Maojiao Li1 et al, Frontiers in Cell and Developmental Biology, 2020 October 2. https://www.frontiersin.org/articles/10.3389/fcell.2020.573511/full – accessed 2/15/21

3 “The Role of Extraellular Vesicles as Allies of HIV, HCV and SARS Viruses,” Flavia Giannessi, et al, Viruses, 2020 May

If we have come this far then we have fully isolated, characterized,

and genetically sequenced an exogenous virus particle. However, we

still have to show it is causally related to a disease. This is carried

out by exposing a group of healthy subjects (animals are usually

used) to this isolated, purified virus in the manner in which the

disease is thought to be transmi�ed. If the animals get sick with the

same disease, as confirmed by clinical and autopsy findings, one has

now shown that the virus actually causes a disease. This

demonstrates infectivity and transmission of an infectious agent.

None of these steps has even been a�empted with the SARS-CoV-2

virus, nor have all these steps been successfully performed for any

so-called pathogenic virus. Our research indicates that a single study

showing these steps does not exist in the medical literature.

Instead, since 1954, virologists have taken unpurified samples

from a relatively few people, o�en less than ten, with a similar

disease. They then minimally process this sample and inoculate this

unpurified sample onto tissue culture containing usually four to six

other types of material – all of which contain identical genetic

material as to what is called a “virus.” The tissue culture is starved

and poisoned and naturally disintegrates into many types of

particles, some of which contain genetic material. Against all

common sense, logic, use of the English language and scientific

integrity, this process is called “virus isolation.” This brew

containing fragments of genetic material from many sources is then

subjected to genetic analysis, which then creates in a computer-

simulation process the alleged sequence of the alleged virus, a so

called in silico genome. At no time is an actual virus confirmed by

electron microscopy. At no time is a genome extracted and

sequenced from an actual virus. This is scientific fraud.

The observation that the unpurified specimen — inoculated onto

tissue culture along with toxic antibiotics, bovine fetal tissue,

amniotic fluid and other tissues — destroys the kidney tissue onto

which it is inoculated is given as evidence of the virus’ existence and

pathogenicity. This is scientific fraud.

From now on, when anyone gives you a paper that suggests the

SARS-CoV-2 virus has been isolated, please check the methods

sections. If the researchers used Vero cells or any other culture

method, you know that their process was not isolation. You will hear

the following excuses for why actual isolation isn’t done:

1. There were not enough virus particles found in samples from patients to analyze.

2. Viruses are intracellular parasites; they can’t be found outside the cell in this manner.

If No. 1 is correct, and we can’t find the virus in the sputum of sick

people, then on what evidence do we think the virus is dangerous or

even lethal? If No. 2 is correct, then how is the virus spread from

person to person? We are told it emerges from the cell to infect

others. Then why isn’t it possible to find it?

Finally, questioning these virology techniques and conclusions is

not some distraction or divisive issue. Shining the light on this truth

is essential to stop this terrible fraud that humanity is confronting.

For, as we now know, if the virus has never been isolated, sequenced

or shown to cause illness, if the virus is imaginary, then why are we

wearing masks, social distancing and pu�ing the whole world into

prison?

Finally, if pathogenic viruses don’t exist, then what is going into

those injectable devices erroneously called “vaccines,” and what is

their purpose? This scientific question is the most urgent and

relevant one of our time.

We are correct. The SARS-CoV2 virus does not exist.

Sally Fallon Morell, MA

Dr. Thomas Cowan, MD

Dr. Andrew Kaufman, MD

Bibliography

Alinsky, Saul: Rules for Radicals (Vintage, 1989)

Antelman, Rabbi Marvin: To Eliminate the Opiate (Zahavia, 1974)

Bastardi, Joe: The Climate Chronicles (Relentless Thunder Press, 2018)

Cowan, Tom: Human Heart, Cosmic Heart (Chelsea Green Publishing, 2016)

Cowan, Tom, and Fallon Morell, Sally: The Contagion Myth (Skyhorse Publishing, 2020)

Forbes, Jack D: Columbus And Other Cannibals – The Wetiko Disease of Exploitation,

Imperialism, and Terrorism (Seven Stories Press, 2008 – originally published in 1979)

Gates, Bill: How to Avoid a Climate Disaster: The Solutions We Have and the Breakthroughs We

Need (Allen Lane, 2021)

Huxley, Aldous: Brave New World (Cha�o & Windus, 1932)

Köhnlein, Dr Claus, and Engelbrecht, Torsten: Virus Mania (emu-Vertag, Lahnstein, 2020)

Lanza, Robert, and Berman, Bob: Biocentrism (BenBella Books, 2010)

Lash, John Lamb: Not In His Image (Chelsea Green Publishing, 2006)

Lester, Dawn, and Parker, David: What Really Makes You Ill – Why everything you thought you

knew about disease is wrong (Independently Published, 2019)

Levy, Paul: Dispelling Wetiko, Breaking the Spell of Evil (North Atlantic Books, 2013)

Marx, Karl: A World Without Jews (Philosophical Library, first edition, 1959)

Mullis, Kary: Dancing Naked in the Mine Field (Bloomsbury, 1999)

O’Brien, Cathy: Trance-Formation of America (Reality Marketing, 1995)

Scholem, Gershon: The Messianic Idea in Judaism (Schocken Books, 1994)

Schwab, Klaus, and Davis, Nicholas: Shaping the Future of the Fourth Industrial Revolution: A

guide to building a better world (Penguin Books, 2018)

Schwab, Klaus: The Great Reset (Agentur Schweiz, 2020)

Sunstein, Cass and Thaler, Richard: Nudge: Improving Decisions About Health, Wealth, and

Happiness (Penguin, 2009)

Swan, Shanna: Count Down: How Our Modern World Is Threatening Sperm Counts, Altering

Male and Female Reproductive Development and Imperiling the Future of the Human Race

(Scribner, 2021)

Tegmark, Max: Our Mathematical Universe: My Quest for the Ultimate Nature of Reality (Penguin, 2015)

Velikovsky, Immanuel: Worlds in Collision (Paradigma, 2009)

Wilton, Robert: The Last Days of the Romanovs (Blurb, 2018, first published 1920)

Index

A

abusive relationships blaming themselves, abused as ref1

children ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10

conspiracy theories ref1

domestic abuse ref1, ref2

economic abuse and dependency ref1

isolation ref1

physical abuse ref1

psychological abuse ref1

signs of abuse ref1

addiction alcoholism ref1

frequencies ref1

substance abuse ref1, ref2

technology ref1, ref2, ref3

Adelson, Sheldon ref1, ref2, ref3

Agenda 21/Agenda 2030 (UN) ref1, ref2, ref3, ref4

AIDs/HIV ref1

causal link between HIV and AIDs ref1, ref2

retroviruses ref1

testing ref1, ref2

trial-run for Covid-19, as ref1, ref2

aliens/extraterrestrials ref1, ref2

aluminium ref1

Amazon ref1, ref2, ref3

amplification cycles ref1, ref2

anaphylactic shock ref1, ref2, ref3, ref4

animals ref1, ref2, ref3

antibodies ref1, ref2, ref3, ref4, ref5

Antifa ref1, ref2, ref3, ref4

antigens ref1, ref2

anti-Semitism ref1, ref2, ref3

Archons ref1, ref2

consciousness ref1, ref2, ref3

energy ref1, ref2, ref3

ennoia ref1

genetic manipulation ref1, ref2

inversion ref1, ref2, ref3

lockdowns ref1

money ref1

radiation ref1

religion ref1, ref2

technology ref1, ref2, ref3

Wetiko factor ref1, ref2, ref3, ref4

artificial intelligence (AI) ref1

army made up of robots ref1, ref2

Human 2.0 ref1, ref2

Internet ref1

MHRA ref1

Morgellons fibres ref1, ref2

Smart Grid ref1

Wetiko factor ref1

asymptomatic, Covid-19 as ref1, ref2, ref3

aviation industry ref1

B

banking, finance and money ref1, ref2, ref3

2008 crisis ref1, ref2

boom and bust ref1

cashless digital money systems ref1

central banks ref1

credit ref1

digital currency ref1

fractional reserve lending ref1

Great Reset ref1

guaranteed income ref1, ref2, ref3

Human 2.0 ref1

incomes, destruction of ref1, ref2

interest ref1

one per cent ref1, ref2

scams ref1

BBC ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8

Becker-Phelps, Leslie ref1

Behavioural Insights Team (BIT) (Nudge Unit) ref1, ref2, ref3

behavioural scientists and psychologists, advice from ref1, ref2

Bezos, Jeff ref1, ref2, ref3, ref4

Biden, Hunter ref1

Biden, Joe ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10, ref11, ref12, ref13, ref14, ref15, ref16, ref17

Big Pharma cholesterol ref1

health professionals ref1, ref2

immunity from prosecution in US ref1

vaccines ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8

Wetiko factor ref1, ref2

WHO ref1, ref2, ref3

Bill and Melinda Gates Foundation ref1, ref2, ref3, ref4, ref5, ref6, ref7

billionaires ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9 ref10, ref11

bird flu (H5N1) ref1

Black Lives Matter (BLM) ref1, ref2, ref3, ref4, ref5

Blair, Tony ref1, ref2, ref3, ref4, ref5, ref6, ref7

Brin, Sergei ref1, ref2, ref3, ref4, ref5, ref6, ref7

British Empire ref1

Bush, George HW ref1, ref2

Bush, George W ref1, ref2, ref3, ref4

Byrd, Robert ref1

C

Canada Global Cult ref1

hate speech ref1

internment ref1

masks ref1

old people ref1

SARS-COV-2 ref1

satellites ref1

vaccines ref1

wearable technology ref1

Capitol Hill riot ref1, ref2

agents provocateur ref1

Antifa ref1

Black Lives Ma�er (BLM) ref1, ref2

QAnon ref1

security precautions, lack of ref1, ref2, ref3

carbon dioxide ref1, ref2

care homes, deaths in ref1, ref2

cashless digital money systems ref1

censorship ref1, ref2, ref3, ref4, ref5

fact-checkers ref1

masks ref1

media ref1, ref2

private messages ref1

social media ref1, ref2, ref3, ref4, ref5, ref6

transgender persons ref1

vaccines ref1, ref2, ref3

Wokeness ref1

Centers for Disease Control (CDC) (United States) ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10, ref11, ref12, ref13

centralisation ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8

chakras ref1

change agents ref1, ref2, ref3

chemtrails ref1, ref2, ref3

chief medical officers and scientific advisers ref1, ref2, ref3, ref4, ref5, ref6

children see also young people

abuse ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10

care, taken into ref1, ref2, ref3

education ref1, ref2, ref3, ref4

energy ref1

family courts ref1

hand sanitisers ref1

human sacrifice ref1

lockdowns ref1, ref2, ref3

masks ref1, ref2, ref3, ref4, ref5

mental health ref1

old people ref1

parents, replacement of ref1, ref2

Psyop (psychological operation), Covid as a ref1, ref2

reframing ref1

smartphone addiction ref1

social distancing and isolation ref1

social media ref1

transgender persons ref1, ref2

United States ref1

vaccines ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10

Wetiko factor ref1

China ref1, ref2, ref3, ref4

anal swab tests ref1

Chinese Revolution ref1, ref2, ref3

digital currency ref1

Global Cult ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9

guaranteed income ref1

Imperial College ref1

Israel ref1

lockdown ref1, ref2

masculinity crisis ref1

masks ref1

media ref1

origins of virus in China ref1, ref2, ref3, ref4, ref5

pollution causing respiratory diseases ref1

Sabbatians ref1, ref2

Smart Grid ref1, ref2

social credit system ref1

testing ref1, ref2

United States ref1, ref2

vaccines ref1, ref2

Wetiko factor ref1

wet market conspiracy ref1

Wuhan ref1, ref2, ref3, ref4, ref5, ref6, ref7

cholesterol ref1, ref2

Christianity ref1, ref2, ref3, ref4, ref5

criticism ref1

cross, inversion of the ref1

Nag Hammadi texts ref1, ref2, ref3

Roman Catholic Church ref1, ref2

Sabbatians ref1, ref2

Satan ref1, ref2, ref3, ref4

Wokeness ref1

class ref1, ref2

climate change hoax ref1, ref2, ref3, ref4, ref5

Agenda 21/Agenda 2030 ref1, ref2, ref3

carbon dioxide ref1, ref2

Club of Rome ref1, ref2, ref3, ref4, ref5

fear ref1

funding ref1

Global Cult ref1

green new deals ref1

green parties ref1

inversion ref1

perception, control of ref1

PICC ref1

reframing ref1

temperature, increases in ref1

United Nations ref1, ref2

Wikipedia ref1

Wokeness ref1, ref2

Clinton, Bill ref1, ref2, ref3, ref4, ref5, ref6

Clinton, Hillary ref1, ref2, ref3

the cloud ref1, ref2, ref3, ref4, ref5, ref6, ref7

Club of Rome and climate change hoax ref1, ref2, ref3, ref4, ref5

cognitive therapy ref1

Cohn, Roy ref1

Common Law ref1

Admiralty Law ref1

arrests ref1, ref2

contractual law, Statute Law as ref1

corporate entities, people as ref1

legalese ref1

sea, law of the ref1

Statute Law ref1

Common Purpose leadership programme ref1, ref2

communism ref1, ref2

co-morbidities ref1

computer-generated virus,

Covid-19 as ref1, ref2, ref3

computer models ref1, ref2, ref3, ref4, ref5

connections ref1, ref2, ref3, ref4

consciousness ref1, ref2, ref3, ref4

Archons ref1, ref2, ref3

expanded ref1, ref2, ref3, ref4, ref5, ref6, ref7

experience ref1

heart ref1

infinity ref1, ref2

religion ref1, ref2

self-identity ref1

simulation thesis ref1

vaccines ref1

Wetiko factor ref1, ref2

conspiracy theorists ref1, ref2, ref3, ref4, ref5

contradictory rules ref1

contrails ref1

Corman-Drosten test ref1, ref2, ref3, ref4

countermimicry ref1, ref2, ref3

Covid-19 vaccines see vaccines

Covidiots ref1, ref2

Cowan, Tom ref1, ref2, ref3, ref4

crimes against humanity ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8

cyber-operations ref1

cyberwarfare ref1

D

DARPA (Defense Advanced Research Projects Agency) ref1

deaths care homes ref1

certificates ref1, ref2, ref3, ref4

mortality rate ref1

post-mortems/autopsies ref1

recording ref1, ref2, ref3, ref4, ref5, ref6, ref7

vaccines ref1, ref2, ref3, ref4, ref5

deceit pyramid of deceit ref1, ref2

sequence of deceit ref1

decoding ref1, ref2, ref3

dehumanisation ref1, ref2, ref3

Delphi technique ref1

democracy ref1

dependency ref1, ref2, ref3, ref4, ref5

Descartes, René ref1

DNA numbers ref1

vaccines ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10

DNR (do not resuscitate)

orders ref1

domestic abuse ref1, ref2

downgrading of Covid-19 ref1

Drosten, Christian ref1, ref2, ref3, ref4, ref5, ref6, ref7

Duesberg, Peter ref1, ref2

E

economic abuse ref1

Edmunds, John ref1, ref2

education ref1, ref2, ref3, ref4

electromagnetic spectrum ref1, ref2

Enders, John ref1

energy Archons ref1, ref2, ref3

children and young people ref1

consciousness ref1

decoding ref1

frequencies ref1, ref2, ref3, ref4

heart ref1

human energy field ref1

source, humans as an energy ref1, ref2

vaccines ref1

viruses ref1

ennoia ref1

Epstein, Jeffrey ref1, ref2

eternal ‘I’ ref1, ref2

ethylene oxide ref1

European Union ref1, ref2, ref3, ref4

Event ref1 and Bill Gates ref2

exosomes, Covid-19 as natural defence mechanism called ref1

experience ref1, ref2

Extinction Rebellion ref1, ref2

F

Facebook addiction ref1, 448–50

Facebook

Archons ref1

censorship ref1, ref2, ref3

hate speech ref1

monopoly, as ref1

private messages, censorship of ref1

Sabbatians ref1

United States election fraud ref1

vaccines ref1

Wetiko factor ref1

fact-checkers ref1

Fauci, Anthony ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10, ref11, ref12

fear ref1, ref2, ref3, ref4

climate change ref1

computer models ref1

conspiracy theories ref1

empty hospitals ref1

Italy ref1, ref2, ref3

lockdowns ref1, ref2, ref3, ref4

masks ref1, ref2

media ref1, ref2

medical staff ref1

Psyop (psychological operation), Covid as a ref1

Wetiko factor ref1, ref2

female infertility ref1

Fermi Paradox ref1

Ferguson, Neil ref1, ref2, ref3, ref4, ref5, ref6, ref7

fertility, decline in ref1

The Field ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8

finance see banking, finance and money

five-senses ref1, ref2

Archons ref1, ref2, ref3

censorship ref1

consciousness, expansion of ref1, ref2, ref3, ref4, ref5, ref6

decoding ref1

education ref1, ref2

the Field ref1, ref2

God, personification of ref1

infinity ref1, ref2

media ref1

paranormal ref1

perceptual programming ref1, ref2

Phantom Self ref1

pneuma not nous, using ref1

reincarnation ref1

self-identity ref1

Wetiko factor ref1, ref2, ref3, ref4, ref5, ref6

5G ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8

Floyd, George and protests, killing of ref1

flu, re-labelling of ref1, ref2, ref3

food and water, control of ref1, ref2

Freemasons ref1, ref2, ref3, ref4, ref5, ref6

Frei, Rosemary ref1

frequencies addictions ref1

Archons ref1, ref2, ref3

awareness ref1

chanting and mantras ref1

consciousness ref1

decoding ref1, ref2

education ref1

electromagnetic (EMF) frequencies ref1

energy ref1, ref2, ref3, ref4

fear ref1

the Field ref1, ref2 5G ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10

five-senses ref1, ref2

ghosts ref1

Gnostics ref1

hive-minds ref1

human, meaning of ref1

light ref1, ref2

love ref1, ref2

magnetism ref1

perception ref1

reality ref1, ref2, ref3

simulation ref1

terror ref1

vaccines ref1

Wetiko ref1, ref2, ref3

Fuellmich, Reiner ref1, ref2, ref3

furlough/rescue payments ref1

G

Gallo, Robert ref1, ref2, ref3

Gates, Bill Archons ref1, ref2, ref3

climate change ref1, ref2, ref3, ref4

Daily Pass tracking system ref1

Epstein ref1

fascism ref1

five senses ref1

GAVI ref1

Great Reset ref1

GSK ref1

Imperial College ref1, ref2

Johns Hopkins University ref1, ref2, ref3

lockdowns ref1, ref2

masks ref1

Nuremberg trial, proposal for ref1, ref2

Rockefellers ref1, ref2

social distancing and isolation ref1

Sun, dimming the ref1

synthetic meat ref1, ref2

vaccines ref1, ref2, ref3, ref4, ref5, ref6, ref7

Wellcome Trust ref1

Wetiko factor ref1, ref2, ref3

WHO ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10

Wokeness ref1

World Economic Forum ref1, ref2, ref3, ref4

Gates, Melinda ref1, ref2, ref3

GAVI vaccine alliance ref1

genetics, manipulation of ref1, ref2, ref3

Germany ref1, ref2, ref3, ref4, ref5, ref6 see also Nazi Germany

Global Cult ref1, ref2, ref3, ref4, ref5

anti-human, why Global Cult is ref1

Black Lives Ma�er (BLM) ref1, ref2, ref3, ref4

China ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9

climate change hoax ref1

contradictory rules ref1

Covid-19 ref1, ref2, ref3

fascism ref1

geographical origins ref1

immigration ref1

Internet ref1

mainstream media ref1, ref2

masks ref1, ref2

monarchy ref1

non-human dimension ref1

perception ref1

political parties ref1, ref2

pyramidal hierarchy ref1, ref2, ref3

reframing ref1

Sabbantian-Frankism ref1, ref2

science, manipulation of ref1

spider and the web ref1

transgender persons ref1

vaccines ref1

who controls the Cult ref1

Wokeness ref1, ref2, ref3, ref4

globalisation ref1, ref2

Gnostics ref1, ref2, ref3, ref4, ref5

Google ref1, ref2, ref3, ref4

government behavioural scientists and psychologists, advice from ref1, ref2

definition ref1

Joint Biosecurity Centre (JBC) ref1

people, abusive relationship with ref1

Great Reset ref1, ref2, ref3, ref4, ref5, ref6

fascism ref1, ref2, ref3

financial system ref1

Human 2.0 ref1

water and food, control of ref1

green parties ref1

Griesz-Brisson, Margarite ref1

guaranteed income ref1, ref2, ref3

H

Hancock, Matt ref1, ref2, ref3, ref4, ref5

hand sanitisers ref1

heart ref1, ref2

hive-minds/groupthink ref1, ref2, ref3

holographs ref1, ref2, ref3, ref4

hospitals, empty ref1

human, meaning of ref1

Human 2.0 ref1

addiction to technology ref1

artificial intelligence (AI) ref1, ref2

elimination of Human 1.0 ref1

fertility, decline in ref1

Great Reset ref1

implantables ref1

money ref1

mRNA ref1

nanotechnology ref1

parents, replacement of ref1, ref2

Smart Grid, connection to ref1, ref2

synthetic biology ref1, ref2, ref3, ref4

testosterone levels, decrease in ref1

transgender = transhumanism ref1, ref2, ref3

vaccines ref1, ref2, ref3, ref4

human sacrifice ref1, ref2, ref3

Hunger Games Society ref1, ref2, ref3, ref4, ref5, ref6, ref7

Huxley, Aldous ref1, ref2, ref3

I

identity politics ref1, ref2, ref3

Illuminati ref1, ref2

illusory physical reality ref1

immigration ref1, ref2, ref3, ref4

Imperial College ref1, ref2, ref3, ref4, ref5, ref6

implantables ref1, ref2

incomes, destruction of ref1, ref2

Infinite Awareness ref1, ref2, ref3, ref4

Internet ref1, ref2 see also social media

artificial intelligence (AI) ref1

independent journalism, lack of ref1

Internet of Bodies (IoB) ref1

Internet of Everything (IoE) ref1, ref2

Internet of Things (IoT) ref1, ref2

lockdowns ref1

Psyop (psychological operation), Covid as a ref1

trolls ref1

intersectionality ref1

inversion Archons ref1, ref2, ref3

climate change hoax ref1

energy ref1

Judaism ref1, ref2, ref3

symbolism ref1

Wetiko factor ref1

Wokeness ref1, ref2, ref3

Islam Archons ref1

crypto-Jews ref1

Islamic State ref1, ref2

Jinn and Djinn ref1, ref2, ref3

O�oman Empire ref1

Wahhabism ref1

isolation see social distancing and isolation

Israel China ref1

Cyber Intelligence Unit Beersheba complex ref1

expansion of illegal se�lements ref1

formation ref1

Global Cult ref1

Judaism ref1, ref2, ref3, ref4, ref5

medical experiments, consent for ref1

Mossad ref1, ref2, ref3, ref4

Palestine-Israel conflict ref1, ref2, ref3

parents, replacement of ref1

Sabbatians ref1, ref2, ref3, ref4, ref5

September 11, 2001, terrorist a�acks on United States ref1

Silicon Valley ref1

Smart Grid ref1, ref2

United States ref1, ref2

vaccines ref1

Wetiko factor ref1

Italy fear ref1, ref2, ref3

Lombardy ref1, ref2, ref3

vaccines ref1

J

Johns Hopkins University ref1, ref2, ref3, ref4, ref5, ref6, ref7

Johnson, Boris ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8

Joint Biosecurity Centre (JBC) ref1

Judaism anti-Semitism ref1, ref2, ref3

Archons ref1, ref2

crypto-Jews ref1

inversion ref1, ref2, ref3

Israel ref1, ref2, ref3, ref4, ref5

Labour Party ref1

Nazi Germany ref1, ref2, ref3, ref4

Sabbatians ref1, ref2, ref3, ref4, ref5

Silicon Valley ref1

Torah ref1

United States ref1, ref2

Zionists ref1, ref2, ref3

K

Kaufman, Andrew ref1, ref2, ref3, ref4

knowledge ref1, ref2, ref3, ref4, ref5, ref6

Koch’s postulates ref1

Kurzweil, Ray ref1, ref2, ref3, ref4, ref5, ref6, ref7

Kushner, Jared ref1, ref2

L

Labour Party ref1, ref2

Lanka, Stefan ref1, ref2

Lateral Flow Device (LFD) ref1

Levy, Paul ref1, ref2, ref3

Life Program ref1

lockdowns ref1, ref2, ref3

amplification tampering ref1

Archons ref1

Behavioural Insights Team ref1

Black Lives Ma�er (BLM) ref1

care homes, deaths in ref1

children

abuse ref1, ref2

mental health ref1

China ref1, ref2

computer models ref1

consequences ref1, ref2

dependency ref1, ref2, ref3

domestic abuse ref1

fall in cases ref1

fear ref1, ref2, ref3, ref4

guaranteed income ref1

Hunger Games Society ref1, ref2, ref3

interaction, destroying ref1

Internet ref1, ref2

overdoses ref1

perception ref1

police-military state ref1, ref2

protests ref1, ref2, ref3, ref4, ref5

psychopathic personality ref1, ref2, ref3

reporting/snitching, encouragement of ref1, ref2

testing ref1

vaccines ref1

Wetiko factor ref1

WHO ref1

love ref1, ref2, ref3

Lucifer ref1, ref2, ref3

M

Madej, Carrie ref1, ref2

Magufuli, John ref1, ref2

mainstream media ref1

BBC ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8

censorship ref1, ref2

China ref1

climate change hoax ref1

fear ref1, ref2

Global Cult ref1, ref2

independent journalism, lack of ref1

Ofcom ref1, ref2, ref3

perception ref1, ref2

Psyop (psychological operation), Covid as a ref1

Sabbatians ref1, ref2

social disapproval ref1

social distancing and isolation ref1

United States ref1, ref2

vaccines ref1, ref2, ref3, ref4, ref5

Mao Zedong ref1, ref2, ref3

Marx and Marxism ref1, ref2, ref3, ref4, ref5, ref6

masculinity ref1

masks/face coverings ref1, ref2, ref3

censorship ref1

children ref1, ref2, ref3, ref4, ref5

China, made in ref1

dehumanisation ref1, ref2, ref3

fear ref1, ref2

flu ref1

health professionals ref1, ref2, ref3, ref4

isolation ref1

laughter ref1

mass non-cooperation ref1

microplastics, risk of ref1

mind control ref1

multiple masks ref1

oxygen deficiency ref1, ref2, ref3

police ref1, ref2, ref3, ref4, ref5

pollution, as cause of plastic ref1

Psyop (psychological operation), Covid as a ref1

reframing ref1, ref2

risk assessments, lack of ref1, ref2

self-respect ref1

surgeons ref1

United States ref1

vaccines ref1, ref2, ref3, ref4, ref5

Wetiko factor ref1

‘worms’ ref1

The Matrix movies ref1, ref2, ref3

measles ref1, ref2

media see mainstream media

Medicines and Healthcare products Regulatory Agency (MHRA)

ref1, ref2, ref3, ref4

Mesopotamia ref1

messaging ref1

military-police state ref1, ref2, ref3

mind control ref1, ref2, ref3, ref4, ref5, ref6 see also MKUltra

MKUltra ref1, ref2, ref3

monarchy ref1

money see banking, finance and money

Montagnier, Luc ref1, ref2, ref3

Mooney, Bel ref1

Morgellons disease ref1, ref2

mortality rate ref1

Mullis, Kary ref1, ref2, ref3

Musk, Elon ref1

N

Nag Hammadi texts ref1, ref2, ref3

nanotechnology ref1, ref2, ref3

narcissism ref1

Nazi Germany ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8

near-death experiences ref1, ref2

Neocons ref1, ref2, ref3

Neuro-Linguistic Programming (NLP) and the Delphi technique

ref1

NHS (National Health Service) amplification cycles ref1

Common Purpose ref1, ref2

mind control ref1

NHS England ref1

saving the NHS ref1, ref2

vaccines ref1, ref2, ref3, ref4, ref5

whistle-blowers ref1, ref2, ref3

No-Problem-Reaction-Solution ref1, ref2, ref3, ref4

non-human dimension of Global Cult ref1

nous ref1

numbers, reality as ref1

Nuremberg Codes ref1, ref2, ref3

Nuremberg-like tribunal, proposal for ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10, ref11, ref12

O

Obama, Barack ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10

O’Brien, Cathy ref1, ref2, ref3, ref4

Ochel, Evita ref1

Ofcom ref1, ref2, ref3

old people ref1, ref2, ref3, ref4, ref5

Oneness ref1, ref2, ref3

Open Society Foundations (Soros) ref1, ref2, ref3

oxygen 406, 528–34

P

paedophilia ref1, ref2

Page, Larry ref1, ref2, ref3, ref4, ref5, ref6, ref7

Palestine-Israel conflict ref1, ref2, ref3

pandemic, definition of ref1

pandemic and health crisis scenarios/simulations ref1, ref2, ref3, ref4

paranormal ref1

PCR tests ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8

Pearl Harbor attacks, prior knowledge of ref1

Pelosi, Nancy ref1, ref2, ref3

perception ref1, ref2, ref3, ref4

climate change hoax ref1

control ref1, ref2, ref3

decoding ref1, ref2

enslavement ref1

externally-delivered perceptions ref1

five senses ref1

human labels ref1

media ref1, ref2

political parties ref1, ref2

Psyop (psychological operation), Covid as a ref1

sale of perception ref1

self-identity ref1, ref2

Wokeness ref1

Phantom Self ref1, ref2, ref3

pharmaceutical industry see Big Pharma

phthalates ref1

Plato’s Allegory of the Cave ref1, ref2

pneuma ref1

police Black Lives Ma�er (BLM) ref1

brutality ref1

citizen’s arrests ref1, ref2

common law arrests ref1, ref2

Common Purpose ref1

defunding ref1

lockdowns ref1, ref2

masks ref1, ref2, ref3, ref4

police-military state ref1, ref2, ref3

psychopathic personality ref1, ref2, ref3, ref4

reframing ref1

United States ref1, ref2, ref3, ref4

Wokeness ref1

polio ref1

political correctness ref1, ref2, ref3, ref4

political parties ref1, ref2, ref3, ref4

political puppets ref1

pollution ref1, ref2, ref3

post-mortems/autopsies ref1

Postage Stamp Consensus ref1, ref2

pre-emptive programming ref1

Problem-Reaction-Solution ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8

Project for the New American Century ref1, ref2, ref3, ref4

psychopathic personality ref1

Archons ref1

heart energy ref1

lockdowns ref1, ref2, ref3

police ref1, ref2, ref3, ref4

recruitment ref1, ref2

vaccines ref1

wealth ref1

Wetiko ref1, ref2

Psyop (psychological operation), Covid as a ref1, ref2, ref3, ref4, ref5

Pushbackers ref1, ref2, ref3, ref4

pyramid structure ref1, ref2, ref3, ref4

Q

QAnon Psyop ref1, ref2, ref3

R

racism see also Black Lives

Ma�er (BLM)

anti-racism industry ref1

class ref1

critical race theory ref1

culture ref1

intersectionality ref1

reverse racism ref1

white privilege ref1, ref2

white supremacy ref1, ref2, ref3, ref4, ref5

Wokeness ref1, ref2, ref3

radiation ref1, ref2

randomness, illusion of ref1, ref2, ref3

reality ref1, ref2, ref3

reframing ref1, ref2

change agents ref1, ref2

children ref1

climate change ref1

Common Purpose leadership programme ref1, ref2

contradictory rules ref1

enforcers ref1

masks ref1, ref2

NLP and the Delphi technique ref1

police ref1

Wetiko factor ref1

Wokeness ref1, ref2

religion see also particular religions

alien invasions ref1

Archons ref1, ref2

consciousness ref1, ref2

control, system of ref1, ref2, ref3

criticism, prohibition on ref1

five senses ref1

good and evil, war between ref1

hidden non-human forces ref1, ref2

Sabbatians ref1

save me syndrome ref1

Wetiko ref1

Wokeness ref1

repetition and mind control ref1, ref2, ref3

reporting/snitching, encouragement of ref1, ref2

Reptilians/Grey entities ref1

rewiring the mind ref1

Rivers, Thomas Milton ref1, ref2

Rockefeller family ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9

Rockefeller Foundation documents ref1, ref2, ref3, ref4

Roman Empire ref1

Rothschild family ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9

RT-PCR tests ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8

Russia collusion inquiry in US ref1

Russian Revolution ref1, ref2

Sabbatians ref1

S

Sabbantian-Frankism ref1, ref2

anti-Semitism ref1, ref2

banking and finance ref1, ref2, ref3

China ref1, ref2

Israel ref1, ref2, ref3, ref4, ref5

Judaism ref1, ref2, ref3, ref4, ref5

Lucifer ref1

media ref1, ref2

Nazis ref1, ref2

QAnon ref1

Rothschilds ref1, ref2, ref3, ref4, ref5, ref6

Russia ref1

Saudi Arabia ref1

Silicon Valley ref1

Sumer ref1

United States ref1, ref2, ref3

Wetiko factor ref1

Wokeness ref1, ref2, ref3

SAGE (Scientific Advisory Group for Emergencies) ref1, ref2, ref3, ref4

SARS-1 ref1

SARs-CoV-2 ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8

Satan/Satanism ref1, ref2, ref3, ref4, ref5, ref6, ref7

satellites in low-orbit ref1

Saudi Arabia ref1

Save Me Syndrome ref1

scapegoating ref1

Schwab, Klaus ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10, ref11, ref12

science, manipulation of ref1

self-identity ref1, ref2, ref3, ref4

self-respect, attacks on ref1

September 11, 2001, terrorist attacks on United States ref1, ref2, ref3, ref4

77th Brigade of UK military ref1, ref2, ref3

Silicon Valley/tech giants ref1, ref2, ref3, ref4, ref5, ref6 see also

Facebook

Israel ref1

Sabbatians ref1

technocracy ref1

Wetiko factor ref1

Wokeness ref1

simulation hypothesis ref1, ref2, ref3, ref4, ref5

Smart Grid ref1, ref2, ref3

artificial intelligence (AI) ref1

China ref1, ref2

control centres ref1

the Field ref1

Great Reset ref1

Human 2.0 ref1, ref2

Israel ref1, ref2

vaccines ref1

Wetiko factor ref1

social disapproval ref1

social distancing and isolation ref1, ref2, ref3

abusive relationships ref1, ref2

children ref1

flats and apartments ref1

heart issues ref1

hugs ref1

Internet ref1

masks ref1

media ref1

older people ref1, ref2

one-metre (three feet) rule ref1

rewiring the mind ref1

simulation, universe as a ref1

SPI-B ref1

substance abuse ref1

suicide and self-harm ref1, ref2, ref3, ref4, ref5

technology ref1

torture, as ref1, ref2

two-metre (six feet) rule ref1

women ref1

social justice ref1, ref2, ref3, ref4

social media see also Facebook bans on alternative views ref1

censorship ref1, ref2, ref3, ref4, ref5, ref6

children ref1

emotion ref1

perception ref1

private messages ref1

Twi�er ref1, ref2, ref3, ref4, ref5, ref6, ref7

Wetiko factor ref1

YouTube ref1, ref2, ref3, ref4, ref5

Soros, George ref1, ref2, ref3, ref4, ref5, ref6

Spain ref1

SPI-B (Scientific Pandemic Insights Group on Behaviours) ref1, ref2, ref3, ref4

spider and the web ref1, ref2, ref3, ref4

Starmer, Keir ref1

Statute Law ref1

Steiner, Rudolf ref1, ref2, ref3

Stockholm syndrome ref1

streptomycin ref1

suicide and self-harm ref1, ref2, ref3, ref4, ref5

Sumer ref1, ref2

Sunstein, Cass ref1, ref2, ref3

swine flu (H1N1) ref1, ref2, ref3

synchronicity ref1

synthetic biology ref1, ref2, ref3, ref4

synthetic meat ref1, ref2

T

technology see also artificial intelligence (AI); Internet;

social media addiction ref1, ref2, ref3, ref4

Archons ref1, ref2

the cloud ref1, ref2, ref3, ref4, ref5, ref6, ref7

cyber-operations ref1

cyberwarfare ref1

radiation ref1, ref2

social distancing and isolation ref1

technocracy ref1

Tedros Adhanom Ghebreyesus ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10, ref11, ref12, ref13

telepathy ref1

Tenpenny, Sherri ref1

Tesla, Nikola ref1

testosterone levels, decrease in ref1

testing for Covid-19 ref1, ref2

anal swab tests ref1

cancer ref1

China ref1, ref2, ref3

Corman-Drosten test ref1, ref2, ref3, ref4

death certificates ref1, ref2

fraudulent testing ref1

genetic material, amplification of ref1

Lateral Flow Device (LFD) ref1

PCR tests ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8

vaccines ref1, ref2, ref3

Thunberg, Greta ref1, ref2, ref3

Totalitarian Tiptoe ref1, ref2, ref3, ref4

transgender persons activism ref1

artificial wombs ref1

censorship ref1

child abuse ref1, ref2

Human 2.0 ref1, ref2, ref3

Wokeness ref1, ref2, ref3, ref4, ref5

women, deletion of rights and status of ref1, ref2

young persons ref1

travel restrictions ref1

Trudeau, Justin ref1, ref2, ref3

Trump, Donald ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10, ref11

Twitter ref1, ref2, ref3, ref4, ref5, ref6, ref7

U

UKColumn ref1, ref2

United Nations (UN) ref1, ref2, ref3, ref4, ref5 see also Agenda

21/Agenda 2030 (UN)

United States ref1, ref2

American Revolution ref1

borders ref1, ref2

Capitol Hill riot ref1, ref2

children ref1

China ref1, ref2

CIA ref1, ref2

Daily Pass tracking system ref1

demographics by immigration, changes in ref1

Democrats ref1, ref2, ref3, ref4, ref5, ref6, ref7

election fraud ref1

far-right domestic terrorists, pushbackers as ref1

Federal Reserve ref1

flu/respiratory diseases statistics ref1

Global Cult ref1, ref2

hand sanitisers, FDA warnings on ref1

immigration, effects of illegal ref1

impeachment ref1

Israel ref1, ref2

Judaism ref1, ref2, ref3

lockdown ref1

masks ref1

mass media ref1, ref2

nursing homes ref1

Pentagon ref1, ref2, ref3, ref4

police ref1, ref2, ref3, ref4

pushbackers ref1

Republicans ref1, ref2

borders ref1, ref2

Democrats ref1, ref2, ref3, ref4, ref5

Russia, inquiry into collusion with ref1

Sabbatians ref1, ref2, ref3

September 11, 2001, terrorist a�acks ref1, ref2, ref3, ref4

UFO sightings, release of information on ref1

vaccines ref1

white supremacy ref1, ref2, ref3, ref4

Woke Democrats ref1, ref2

V

vaccines ref1, ref2, ref3

adverse reactions ref1, ref2, ref3, ref4, ref5

Africa ref1

anaphylactic shock ref1, ref2, ref3, ref4

animals ref1, ref2

anti-vax movement ref1, ref2, ref3, ref4, ref5

AstraZeneca/Oxford ref1, ref2, ref3, ref4

autoimmune diseases, rise in ref1, ref2

Big Pharma ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8

bioweapon, as real ref1, ref2

black and ethnic minority communities ref1

blood clots ref1, ref2

Brain Computer Interface (BCI) ref1

care homes, deaths in ref1

censorship ref1, ref2, ref3

chief medical officers and scientific advisers, financial interests of

ref1, ref2

children ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10

China ref1, ref2

clinical trials ref1, ref2, ref3, ref4, ref5, ref6

compensation ref1

compulsory vaccinations ref1, ref2, ref3

computer programs ref1

consciousness ref1

cover-ups ref1

creation before Covid ref1

cytokine storm ref1

deaths and illnesses caused by vaccines ref1, ref2, ref3, ref4, ref5

definition ref1

developing countries ref1

digital ta�oos ref1

DNA-manipulation ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9,

ref10

emergency approval ref1, ref2, ref3, ref4, ref5

female infertility ref1

funding ref1

genetic suicide ref1

Global Cult ref1

heart chakras ref1

hesitancy ref1

Human 2.0 ref1, ref2, ref3, ref4

immunity from prosecution ref1, ref2, ref3

implantable technology ref1

Israel ref1

Johnson & Johnson ref1, ref2, ref3, ref4

lockdowns ref1

long-term effects ref1

mainstream media ref1, ref2, ref3, ref4, ref5

masks ref1, ref2, ref3, ref4, ref5

Medicines and Healthcare products Regulatory Agency (MHRA)

ref1, ref2

messaging ref1

Moderna ref1, ref2, ref3, ref4, ref5, ref6

mRNA vaccines ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9

nanotechnology ref1, ref2

NHS ref1, ref2, ref3, ref4, ref5

older people ref1, ref2

operating system ref1

passports ref1, ref2, ref3, ref4

Pfizer/BioNTech ref1, ref2, ref3, ref4, ref5, ref6, ref7

polyethylene glycol ref1

pregnant women ref1

psychopathic personality ref1

races, targeting different ref1

reverse transcription ref1

Smart Grid ref1

social distancing ref1

social media ref1

sterility ref1

synthetic material, introduction of ref1

tests ref1, ref2, ref3

travel restrictions ref1

variants ref1, ref2

viruses, existence of ref1

whistle-blowing ref1

WHO ref1, ref2, ref3, ref4

Wokeness ref1

working, vaccine as ref1

young people ref1

Vallance, Patrick ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9

variants ref1, ref2, ref3

vegans ref1

ventilators ref1, ref2

virology ref1, ref2

virtual reality ref1, ref2, ref3

viruses, existence of ref1

visual reality ref1, ref2

vitamin D ref1, ref2

von Braun, Wernher ref1, ref2

W

war-zone hospital myths ref1

waveforms ref1, ref2

wealth ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9 ref10, ref11

wet market conspiracy ref1

Wetiko factor ref1

alcoholism and drug addiction ref1

anti-human, why Global Cult is ref1

Archons ref1, ref2, ref3, ref4

artificial intelligence (AI) ref1

Big Pharma ref1, ref2

children ref1

China ref1

consciousness ref1, ref2

education ref1

Facebook ref1

fear ref1, ref2

frequency ref1, ref2

Gates ref1, ref2

Global Cult ref1, ref2

heart ref1, ref2

lockdowns ref1

masks ref1

Native American concept ref1

psychopathic personality ref1, ref2

reframing/retraining programmes ref1

religion ref1

Silicon Valley ref1

Smart Grid ref1

smartphone addiction ref1, ref2

social media ref1

war ref1, ref2

WHO ref1

Wokeness ref1, ref2, ref3

Yaldabaoth ref1, ref2, ref3, ref4

whistle-blowing ref1, ref2, ref3, ref4, ref5, ref6, ref7

white privilege ref1, ref2

white supremacy ref1, ref2, ref3, ref4, ref5

Whitty, Christopher ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10

‘who benefits’ ref1

Wi-Fi ref1, ref2, ref3, ref4

Wikipedia ref1, ref2

Wojcicki, Susan ref1, ref2, ref3, ref4, ref5, ref6, ref7

Wokeness Antifa ref1, ref2, ref3, ref4

anti-Semitism ref1

billionaire social justice warriors ref1, ref2, ref3

Capitol Hill riot ref1, ref2

censorship ref1

Christianity ref1

climate change hoax ref1, ref2

culture ref1

education, control of ref1

emotion ref1

facts ref1

fascism ref1, ref2, ref3

Global Cult ref1, ref2, ref3, ref4

group-think ref1

immigration ref1

indigenous people, solidarity with ref1

inversion ref1, ref2, ref3

le�, hijacking the ref1, ref2

Marxism ref1, ref2, ref3

mind control ref1

New Woke ref1

Old Woke ref1

Oneness ref1

perceptual programming ref1

Phantom Self ref1

police ref1

defunding the ref1

reframing ref1

public institutions ref1

Pushbackers ref1, ref2, ref3

racism ref1, ref2, ref3

reframing ref1, ref2

religion, as ref1

Sabbatians ref1, ref2, ref3

Silicon Valley ref1

social justice ref1, ref2, ref3, ref4

transgender ref1, ref2, ref3, ref4, ref5

United States ref1, ref2

vaccines ref1

Wetiko factor ref1, ref2, ref3

young people ref1, ref2, ref3

women, deletion of rights and status of ref1, ref2

World Economic Forum (WEF) ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9

World Health Organization (WHO) ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9

AIDs/HIV ref1

amplification cycles ref1

Big Pharma ref1, ref2, ref3

cooperation in health emergencies ref1

creation ref1, ref2

fatality rate ref1

funding ref1, ref2, ref3

Gates ref1

Internet ref1

lockdown ref1

vaccines ref1, ref2, ref3, ref4

Wetiko factor ref1

world number 1 (masses) ref1, ref2

world number 2 ref1

Wuhan ref1, ref2, ref3, ref4, ref5, ref6, ref7 ref8

Y

Yaldabaoth ref1, ref2, ref3, ref4, ref5, ref6

Yeadon, Michael ref1, ref2, ref3, ref4

young people see also children addiction to technology ref1

Human 2.0 ref1

vaccines ref1, ref2

Wokeness ref1, ref2, ref3

YouTube ref1, ref2, ref3, ref4, ref5

WHO 548

Z

Zaks, Tal ref1

Zionism ref1, ref2, ref3

Zuckerberg, Mark ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10, ref11, ref12

Zulus ref1

Before you go …

For more detail, background and evidence about the subjects in

Perceptions of a Renegade Mind – and so much more – see my

others books including And The Truth Shall Set You Free; The

Biggest Secret; Children of the Matrix; The David Icke Guide to the

Global Conspiracy; Tales from the Time Loop; The Perception

Deception; Remember Who You Are; Human Race Get Off Your

Knees; Phantom Self; Everything You Need To Know But Have Never

Been Told, The Trigger and The Answer.

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  • Cover
  • Half Title Page
  • Tilte Page
  • Copyright Page
  • Brief Contents
  • Contents
  • 1 Chemistry in Our Lives
    • CAREER Forensic Scientist
    • 1.1 Chemistry and Chemicals
    • 1.2 Scientific Method: Thinking Like a Scientist
      • Chemistry Link to Health Early Chemist: Paracelsus
    • 1.3 Studying and Learning Chemistry
    • 1.4 Key Math Skills for Chemistry
    • 1.5 Writing Numbers in Scientific Notation
      • UPDATE Forensic Evidence Helps Solve the Crime
    • Concept Map
    • Chapter Review
    • Key Terms
    • Key Math Skills
    • Understanding the Concepts
    • Additional Practice Problems
    • Challenge Problems
    • Answers to Engage Questions
    • Answers to Selected Problems
  • 2 Chemistry and Measurements
    • CAREER Registered Nurse
    • 2.1 Units of Measurement
    • 2.2 Measured Numbers and Significant Figures
    • 2.3 Significant Figures in Calculations
    • 2.4 Prefixes and Equalities
    • 2.5 Writing Conversion Factors
    • 2.6 Problem Solving Using Unit Conversion
      • Chemistry Link to Health Toxicology and Risk–Benefit Assessment
    • 2.7 Density
      • Chemistry Link to Health Bone Density
    • UPDATE Greg’s Visit with His Doctor
    • Concept Map
    • Chapter Review
    • Key Terms
    • Key Math Skills
    • Core Chemistry Skills
    • Understanding the Concepts
    • Additional Practice Problems
    • Challenge Problems
    • Answers to Engage Questions
    • Answers to Selected Problems
  • 3 Matter and Energy
    • CAREER Dietitian
    • 3.1 Classification of Matter
      • Chemistry Link to Health Breathing Mixtures
    • 3.2 States and Properties of Matter
    • 3.3 Temperature
      • Chemistry Link to Health Variation in Body Temperature
    • 3.4 Energy
    • 3.5 Specific Heat
    • 3.6 Energy and Nutrition
      • Chemistry Link to Health Losing and Gaining Weight
    • UPDATE A Diet and Exercise Program
    • Concept Map
    • Chapter Review
    • Key Terms
    • Core Chemistry Skills
    • Understanding the Concepts
    • Additional Practice Problems
    • Challenge Problems
    • Answers to Engage Questions
    • Answers to Selected Problems
    • Combining Ideas from Chapters 1 to 3
  • 4 Atoms and Elements
    • CAREER Farmer
    • 4.1 Elements and Symbols
    • 4.2 The Periodic Table
      • Chemistry Link to Health Elements Essential to Health
    • 4.3 The Atom
    • 4.4 Atomic Number and Mass Number
    • 4.5 Isotopes and Atomic Mass
    • UPDATE Improving Crop Production
    • Concept Map
    • Chapter Review
    • Key Terms
    • Core Chemistry Skills
    • Understanding the Concepts
    • Additional Practice Problems
    • Challenge Problems
    • Answers to Engage Questions
    • Answers to Selected Problems
  • 5 Electronic Structure of Atoms and Periodic Trends
    • CAREER Materials Engineer
    • 5.1 Electromagnetic Radiation
      • Chemistry Link to Health Biological Reactions to UV Light
    • 5.2 Atomic Spectra and Energy Levels
    • 5.3 Sublevels and Orbitals
    • 5.4 Orbital Diagrams and Electron Configurations
    • 5.5 Electron Configurations and the Periodic Table
    • 5.6 Trends in Periodic Properties
    • UPDATE Developing New Materials for Computer Chips
    • Concept Map
    • Chapter Review
    • Key Terms
    • Core Chemistry Skills
    • Understanding The Concepts
    • Additional Practice Problems
    • Challenge Problems
    • Answers to Engage Questions
    • Answers to Selected Problems
  • 6 Ionic and Molecular Compounds
    • CAREER Pharmacist
    • 6.1 Ions: Transfer of Electrons
      • Chemistry Link to Health Some Important Ions in the Body
    • 6.2 Ionic Compounds
    • 6.3 Naming and Writing Ionic Formulas
    • 6.4 Polyatomic Ions
    • 6.5 Molecular Compounds: Sharing Electrons
    • UPDATE Compounds at the Pharmacy
    • Concept Map
    • Chapter Review
    • Key Terms
    • Core Chemistry Skills
    • Understanding the Concepts
    • Additional Practice Problems
    • Challenge Problems
    • Answers to Engage Questions
    • Answers to Selected Problems
  • 7 Chemical Quantities
    • CAREER Veterinarian
    • 7.1 The Mole
    • 7.2 Molar Mass
    • 7.3 Calculations Using Molar Mass
    • 7.4 Mass Percent Composition
      • Chemistry Link to the Environment Fertilizers
    • 7.5 Empirical Formulas
    • 7.6 Molecular Formulas
    • UPDATE Prescriptions for Max
    • Concept Map
    • Chapter Review
    • Key Terms
    • Core Chemistry Skills
    • Understanding the Concepts
    • Additional Practice Problems
    • Challenge Problems
    • Answers to Engage Questions
    • Answers to Selected Problems
    • Combining Ideas from Chapters 4 to 7
  • 8 Chemical Reactions
    • CAREER Exercise Physiologist
    • 8.1 Equations for Chemical Reactions
    • 8.2 Balancing a Chemical Equation
    • 8.3 Types of Chemical Reactions
      • Chemistry Link to Health Incomplete Combustion: Toxicity of Carbon Monoxide
    • 8.4 Oxidation–Reduction Reactions
    • UPDATE Improving Natalie’s Overall Fitness
    • Concept Map
    • Chapter Review
    • Key Terms
    • Core Chemistry Skills
    • Understanding the Concepts
    • Additional Practice Problems
    • Challenge Problems
    • Answers to Engage Questions
    • Answers to Selected Problems
  • 9 Chemical Quantities in Reactions
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    • 9.1 Conservation of Mass
    • 9.2 Mole Relationships in Chemical Equations
    • 9.3 Mass Calculations for Chemical Reactions
    • 9.4 Limiting Reactants
    • 9.5 Percent Yield
    • 9.6 Energy in Chemical Reactions
      • Chemistry Link to Health Cold Packs andHot Packs
    • UPDATE Testing Water Samples for Insecticides
    • Concept Map
    • Chapter Review
    • Key Terms
    • Core Chemistry Skills
    • Understanding the Concepts
    • Additional Practice Problems
    • Challenge Problems
    • Answers to Engage Questions
    • Answers to Selected Problems
  • 10 Bonding and Properties of Solids and Liquids
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    • 10.1 Lewis Structures for Molecules and Polyatomic Ions
    • 10.2 Resonance Structures
    • 10.3 Shapes of Molecules and Polyatomic Ions (VSEPR Theory)
    • 10.4 Electronegativity and Bond Polarity
    • 10.5 Polarity of Molecules
    • 10.6 Intermolecular Forces Between Atoms or Molecules
    • 10.7 Changes of State
      • Chemistry Link to Health Steam Burns
    • UPDATE Histologist Stains Tissue with Dye
    • Concept Map
    • Chapter Review
    • Key Terms
    • Core Chemistry Skills
    • Understanding the Concepts
    • Additional Practice Problems
    • Challenge Problems
    • Answers to Engage Questions
    • Answers to Selected Problems
    • Combining Ideas from Chapters 8 to 10
  • 11 Gases
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    • 11.1 Properties of Gases
      • Chemistry Link to Health Measuring Blood Pressure
    • 11.2 Pressure and Volume (Boyle’s Law)
      • Chemistry Link to Health Pressure–Volume Relationship in Breathing
    • 11.3 Temperature and Volume (Charles’s Law)
    • 11.4 Temperature and Pressure (Gay‐Lussac’s Law)
    • 11.5 The Combined Gas Law
    • 11.6 Volume and Moles (Avogadro’s Law)
    • 11.7 The Ideal Gas Law
    • 11.8 Gas Laws and Chemical Reactions
    • 11.9 Partial Pressures (Dalton’s Law)
      • Chemistry Link to Health Blood Gases
      • Chemistry Link to Health Hyperbaric Chambers
    • UPDATE Exercise‐Induced Asthma
    • Concept Map
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    • Key Terms
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    • Understanding the Concepts
    • Additional Practice Problems
    • Challenge Problems
    • Answers to Engage Questions
    • Answers to Selected Problems
  • 12 Solutions
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    • 12.1 Solutions
      • Chemistry Link to Health Water in the Body
    • 12.2 Electrolytes and Nonelectrolytes
      • Chemistry Link to Health Electrolytes in Body Fluids
    • 12.3 Solubility
      • Chemistry Link to Health Gout and Kidney Stones: Saturation in Body Fluids
    • 12.4 Solution Concentrations
    • 12.5 Dilution of Solutions
    • 12.6 Chemical Reactions in Solution
    • 12.7 Molality and Freezing Point Lowering/Boiling Point Elevation
    • 12.8 Properties of Solutions: Osmosis
      • Chemistry Link to Health Hemodialysis and the Artificial Kidney
    • UPDATE Using Dialysis for Renal Failure
    • Concept Map
    • Chapter Review
    • Key Terms
    • Core Chemistry Skills
    • Understanding the Concepts
    • Additional Practice Problems
    • Challenge Problems
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    • Answers to Selected Problems
  • 13 Reaction Rates and Chemical Equilibrium
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    • 13.1 Rates of Reactions
    • 13.2 Chemical Equilibrium
    • 13.3 Equilibrium Constants
    • 13.4 Using Equilibrium Constants
    • 13.5 Changing Equilibrium Conditions: Le ChÂtelier’s Principle
      • Chemistry Link to Health Oxygen–Hemoglobin Equilibrium and Hypoxia
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    • 13.6 Equilibrium in Saturated Solutions
    • UPDATE Equilibrium of in the Ocean
    • Concept Map
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    • Understanding the Concepts
    • Additional Practice Problems
    • Challenge Problems
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    • Answers to Selected Problems
  • 14 Acids and Bases
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    • 14.1 Acids and Bases
    • 14.2 BrØnsted–Lowry Acids and Bases
    • 14.3 Strengths of Acids and Bases
    • 14.4 Dissociation of Weak Acids and Bases
    • 14.5 Dissociation of Water
    • 14.6 The pH Scale
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    • 14.7 Reactions of Acids and Bases
      • Chemistry Link to Health Antacids
    • 14.8 Acid–Base Titration
    • 14.9 Buffers
      • Chemistry Link to Health Buffers in the Blood Plasma
    • UPDATE Acid Reflux Disease
    • Concept Map
    • Chapter Review
    • Key Terms
    • Key Math Skills
    • Core Chemistry Skills
    • Understanding the Concepts
    • Additional Practice Problems
    • Challenge Problems
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    • Combining Ideas from Chapters 11 to 14
  • 15 Oxidation and Reduction
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    • 15.1 Oxidation and Reduction
    • 15.2 Balancing Oxidation–Reduction Equations Using Half‐Reactions
    • 15.3 Electrical Energy from Oxidation–‐Reduction Reactions
      • Chemistry Link to the Environment Corrosion: Oxidation of Metals
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    • 15.4 Oxidation–Reduction Reactions That Require Electrical Energy
    • UPDATE Whitening Kimberly’s Teeth
    • Concept Map
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    • Answers to Selected Problems
  • 16 Nuclear Chemistry
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    • 16.1 Natural Radioactivity
    • 16.2 Nuclear Reactions
      • Chemistry Link to Health Radon in Our Homes
    • 16.3 Radiation Measurement
      • Chemistry Link to Health Radiation and Food
    • 16.4 Half‐Life of a Radioisotope
      • Chemistry Link to the Environment Dating Ancient Objects
    • 16.5 Medical Applications Using Radioactivity
      • Chemistry Link to Health Brachytherapy
    • 16.6 Nuclear Fission and Fusion
    • UPDATE Cardiac Imaging Using a Radioisotope
    • Concept Map
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    • Combining Ideas from Chapters 15 and 16
  • 17 Organic Chemistry
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    • 17.1 Alkanes
    • 17.2 Alkenes, Alkynes, and Polymers
      • Chemistry Link to Health Hydrogenation of Unsaturated Fats
    • 17.3 Aromatic Compounds
      • Chemistry Link to Health Some Common Aromatic Compounds
      • Chemistry Link to Health Polycyclic Aromatic Hydrocarbons (PAHs)
    • 17.4 Alcohols and Ethers
      • Chemistry Link to Health Some Important Alcohols, Phenols, and Ethers
    • 17.5 Aldehydes and Ketones
      • Chemistry Link to Health Some Important Aldehydes and Ketones
    • 17.6 Carboxylic Acids and Esters
      • Chemistry Link to Health Carboxylic Acids in Metabolism
    • 17.7 Amines and Amides
      • Chemistry Link to the Environment Alkaloids: Amines in Plants
    • UPDATE Diane’s Treatment in the Burn Unit
    • Concept Map
    • Chapter Review
    • Summary of Naming
    • Summary of Reactions
    • Key Terms
    • Core Chemistry Skills
    • Understanding the Concepts
    • Additional Practice Problems
    • Challenge Problems
    • Answers to Engage Questions
    • Answers to Selected Problems
  • 18 Biochemistry
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    • 18.1 Carbohydrates
      • Chemistry Link to Health Hyperglycemia and Hypoglycemia
    • 18.2 Disaccharides and Polysaccharides
      • Chemistry Link to Health How Sweet is My Sweetener?
    • 18.3 Lipids
    • 18.4 Amino Acids and Proteins
      • Chemistry Link to Health Essential Amino Acids and Complete Proteins
    • 18.5 Protein Structure
    • 18.6 Proteins as Enzymes
    • 18.7 Nucleic Acids
    • 18.8 Protein Synthesis
      • UPDATE Kate’s Program for Type 2 Diabetes
    • Concept Map
    • Chapter Review
    • Key Terms
    • Core Chemistry Skills
    • Understanding the Concepts
    • Additional Practice Problems
    • Challenge Problems
    • Answers to Engage Questions
    • Answers to Selected Problems
    • Combining Ideas from Chapters 17 and 18
  • Credits
  • Glossary/Index