Chemistry A Molecular Approach [4 ed.]

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Brief Contents 1 Units of Measurement for Physical and Chemical Change 2 Atoms and Elements 3 Molecules, Compounds, and Nomenclature 4 Chemical Reactions and Stoichiometry 5 Gases 6 Thermochemistry 7 The Quantum-Mechanical Model of the Atom 8 Periodic Properties of the Elements 9 Chemical Bonding I: Lewis Theory

1 29 57 101 149 199 244 302 336

10 Chemical Bonding II: Molecular Shapes, Valence Bond Theory, and Molecular Orbital Theory 381 11 Liquids, Solids, and Intermolecular Forces 436 12 Solutions 13 Chemical Kinetics 14 Chemical Equilibrium 15 Acids and Bases 16 Aqueous Ionic Equilibrium 17 Gibbs Energy and Thermodynamics 18 Electrochemistry 19 Radioactivity and Nuclear Chemistry 20 Organic Chemistry I: Structures 21 Organic Chemistry II: Reactions 22 Biochemistry 23 Chemistry of the Nonmetals 24 Metals and Metallurgy 25 Transition Metals and Coordination Compounds

494 542 595 639 694 749 799 846 882 924 970 1004 1038 1059

Appendix I: Common Mathematical Operations in Chemistry

A-1

Appendix II: Useful Data

A-7

Appendix III: Answers to Selected Exercises

A-17

Appendix IV: Answers to In-Chapter Practice Problems

A-60

Glossary

G-1

Essential Data and Equations

E-1

Credits

C-1

Index

I-1

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vii

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Contents Preface

xx

1

Units of Measurement for Physical and Chemical Change 1

1.1 Physical and Chemical Changes and Physical and Chemical Properties CONCEPTUAL CONNECTION 1.1: Chemical and Physical Changes

1.2 Energy: A Fundamental Part of Physical and Chemical Change 1.3 The Units of Measurement

2 3

3 4

12

18

19

23

42

44

47

47

CHAPTER IN REVIEW Key Terms  50  Key Concepts  51  Key Equations and Relationships  51  Key Skills 52

50

EXERCISES Review Questions  52  Problems by Topic  52 Cumulative Problems  55  Challenge Problems  56 Conceptual Problems  56

52

3

Molecules, Compounds, and Nomenclature

3.1 Hydrogen, Oxygen, and Water 3.2 Chemical Bonds

57 57 59

Ionic Bonds  59  Covalent Bonds   60

3.3 Representing Compounds: Chemical Formulas and Molecular Models

29 30 31 32

The Law of Conservation of Mass  32

CONCEPTUAL CONNECTION 2.1: The Law of

viii

41

Mass Spectrometry: Measuring the Mass of Atoms and Molecules  43

Ions and the Periodic Table  50

EXERCISES Review Questions  23  Problems by Topic  23  Cumulative Problems  26  Challenge Problems  27 Conceptual Problems  28

Conservation of Mass  The Law of Definite Proportions  33  The Law of Multiple Proportions  33  John Dalton and the Atomic Theory  34

2.5 Atomic Mass: The Average Mass of an Element’s Atoms

11

22

2.1 Imaging and Moving Individual Atoms 2.2 Early Ideas About the Building Blocks of Matter 2.3 Modern Atomic Theory and the Laws That Led to It

Come From?

CONCEPTUAL CONNECTION 2.3: The Mole 2.7 The Periodic Table of the Elements

10

CHAPTER IN REVIEW Key Terms  22  Key Concepts  22  Key Equations and Relationships 22  Key Skills 23

Atoms and Elements

41

The Mole: A Chemist’s “Dozen”  44  Converting Between Number of Moles and Number of Atoms  45  Converting Between Mass and Amount (Number of Moles)  45

General Problem-Solving Strategy  19  Order-ofMagnitude Estimations  20  Problems Involving an Equation 20

2

Isotopes, and Ions

CHEMISTRY IN YOUR DAY: Where Did Elements

2.6 Molar Mass: Counting Atoms by Weighing Them

Counting Significant Figures  13  Exact Numbers  14  Significant Figures in Calculations  15  Rules for Calculations  15  Rules for Rounding  16  Precision and Accuracy  17

THE NATURE OF SCIENCE: Integrity in Data Gathering 1.5 Solving Chemical Problems

34

The Discovery of the Electron  34  The Discovery of the Nucleus  36  Protons, the Atomic Number, and Neutrons  38  Isotopes: When the Number of Neutrons Varies  39  Ions: Losing and Gaining Electrons 40

CONCEPTUAL CONNECTION 2.2: The Nuclear Atom,

The Standard Units  4  The Metre: A Measure of Length  4  The Kilogram: A Measure of Mass  5  The Second: A Measure of Time  5  The Kelvin: A Measure of Temperature  5  SI Prefixes  6  Conversions Involving the SI Prefixes  7  Derived Units  8

CONCEPTUAL CONNECTION 1.2: Density CHEMISTRY AND MEDICINE: Bone Density 1.4 The Reliability of a Measurement

2.4 Atomic Structure

32

60

Types of Chemical Formulas  60

3.4 Formulas and Names

64

Ionic Compounds  64

CONCEPTUAL CONNECTION 3.1: Unambiguous Nomenclature Molecular Compounds 67  Naming Acids 69 CONCEPTUAL CONNECTION 3.2: Nomenclature 3.5 Organic Compounds

67 70

70

Naming Hydrocarbons  72  Cyclic Hydrocarbons  76  Aromatic Hydrocarbons  77

CONCEPTUAL CONNECTION 3.3: Organic Nomenclature 77 Functionalized Hydrocarbons  78

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Get complete eBook Order by email at [email protected] O N T E N T S       ix 3.6 Formula Mass and the Mole Concept for Compounds

81

Molar Mass of a Compound  81  Using Molar Mass to Count Molecules by Weighing  81

3.7 Composition of Compounds

83 86

87

CONCEPTUAL CONNECTION 3.4: Chemical Formula and Mass Percent Composition CHNS Analysis  90

89

CHAPTER IN REVIEW Key Terms  91  Key Concepts  92  Key Equations and Relationships 92  Key Skills 93

91

EXERCISES Review Questions  93  Problems by Topic  94 Cumulative Problems  99  Challenge Problems  100 Conceptual Problems  100

93

4

4.1 Chemistry of Cuisine 4.2 Writing and Balancing Chemical Equations

101 101 102

105

105

Electrolyte and Nonelectrolyte Solutions  106 The Solubility of Ionic Compounds  107

4.4 Precipitation Reactions 4.5 Acid–Base Reactions

109 111 115

Oxidation States  116  Identifying Redox Reactions 118

149

5.1 Breathing: Putting Pressure to Work 5.2 Pressure: The Result of Molecular Collisions

149 150

CHEMISTRY AND MEDICINE: Blood Pressure 5.3 The Gas Laws: Boyle’s Law, Charles’s Law, and Avogadro’s Law

153

154

Boyle’s Law: Volume and Pressure  154  Charles’s Law: Volume and Temperature  156  The Combined Gas Law  159  Avogadro’s Law: Volume and Amount (in Moles)  160

5.4 The Ideal Gas Law 5.5 Applications of the Ideal Gas Law: Molar Volume, Density, and Molar Mass of a Gas

161 163

CONCEPTUAL CONNECTION 5.1: Molar Volume

164

Density of a Gas  164  Molar Mass of a Gas  165

5.6 Mixtures of Gases and Partial Pressures CHEMISTRY IN THE ENVIRONMENT: Stack Sampling

167 168

Collecting Gases over Water  170

5.7 Gases in Chemical Reactions: Stoichiometry Revisited 172

CONCEPTUAL CONNECTION 4.2: Oxidation and Reduction

120

CHEMISTRY IN YOUR DAY: Bleached Blonde

120

Balancing Oxidation–Reduction Equations  121

4.7 Reaction Stoichiometry: How Much Is Produced?

5

141

Molar Volume at Standard Temperature and Pressure 163

Acid–Base Reactions Evolving a Gas  114

4.6 Oxidation–Reduction Reactions

EXERCISES Review Questions  141  Problems by Topic  141  Cumulative Problems  145  Challenge Problems  147 Conceptual Problems  148

The Barometer and Units of Pressure  151 The Manometer: Measuring Gas Pressures in the Laboratory 152

CONCEPTUAL CONNECTION 4.1: Balanced Chemical 4.3 Solutions and Solubility

135

CHAPTER IN REVIEW  139 Key Terms  139  Key Concepts  139  Key Equations and Relationships 140  Key Skills 140

Gases

How to Write Balanced Chemical Equations  104 Equations

CONCEPTUAL CONNECTION 4.6: Solutions Solution Stoichiometry  137

Calculating Molecular Formulas for Compounds  88

Chemical Reactions and Stoichiometry

133

Solution Concentration  133  Using Molarity in Calculations 134

Conversion Factors in Chemical Formulas  84

CHEMISTRY IN YOUR DAY: Drug Tablets and Capsules 3.8 Determining a Chemical Formula from Experimental Data

4.9 Solution Concentration and Solution Stoichiometry

125

Making Molecules: Mole-to-Mole Conversions  125

CHEMISTRY IN YOUR DAY: The Stoichiometry of Composting 125 Making Molecules: Mass-to-Mass Conversions  126 CONCEPTUAL CONNECTION 4.3: Stoichiometry 128 4.8 Limiting Reactant, Theoretical Yield, and Percent Yield 128 Limiting Reactant, Theoretical Yield, and Percent Yield from Initial Reactant Masses  129

Molar Volume and Stoichiometry  174

CONCEPTUAL CONNECTION 5.2: Pressure and Number of Moles

5.8 Kinetic Molecular Theory: A Model for Gases

175

175

Kinetic Molecular Theory and the Ideal Gas Law  177  Temperature and Molecular Velocities  178

CONCEPTUAL CONNECTION 5.3: Kinetic Molecular Theory

5.9 Mean Free Path, Diffusion, and Effusion of Gases 5.10 Real Gases: The Effects of Size and Intermolecular Forces

181

181 182

The Effect of the Finite Volume of Gas Particles  183  The Effect of Intermolecular Forces  184  Van der Waals Equation  185  Real Gases  186

CONCEPTUAL CONNECTION 4.4: Limiting Reactant and Theoretical Yield

129

CONCEPTUAL CONNECTION 4.5: Reactant in Excess

133

CONCEPTUAL CONNECTION 5.4: Real Gases CHEMISTRY IN YOUR DAY: Pressure in Outer Space

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187 187

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CHAPTER IN REVIEW 189 Key Terms  189  Key Concepts  189  Key Equations and Relationships 190  Key Skills 190 EXERCISES Review Questions  191  Problems by Topic  192 Cumulative Problems  196  Challenge Problems  198  Conceptual Problems  198

6

Thermochemistry

6.1 Chemical Hand Warmers 6.2 The Nature of Energy: Key Definitions

191

7

The Quantum-Mechanical Model of the Atom

199 199 200

Units of Energy  202

Internal Energy  203

CONCEPTUAL CONNECTION 6.1: System and CONCEPTUAL CONNECTION 6.2: Heat and Work 6.4 Quantifying Heat and Work

205 206

207

Heat 207

CONCEPTUAL CONNECTION 6.3: The Heat Capacity of Water

209 211

6.5 Measuring ¢rU for Chemical Reactions: Constant-Volume Calorimetry 213 6.6 Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure 216 CONCEPTUAL CONNECTION 6.5: The Difference Between ¢rH and ¢rU 217 Exothermic and Endothermic Processes: A Molecular View  219 219

Stoichiometry Involving ¢rH: Thermochemical Equations 219

6.7 Constant-Pressure Calorimetry: Measuring ¢rH 221 CONCEPTUAL CONNECTION 6.7: Constant-Pressure Versus Constant-Volume Calorimetry

223

6.8 Relationships Involving ¢rH 223 6.9 Determining Enthalpies of Reaction from Standard Enthalpies of Formation 226 Standard States and Standard Enthalpy Changes  226  Calculating the Standard Enthalpy Change for a Reaction  228

6.10 Energy Use and the Environment

245 245

The Wave Nature of Light  246  The Electromagnetic Spectrum  248  Interference and Diffraction  249 for Cancer The Particle Nature of Light  252

250

CONCEPTUAL CONNECTION 7.1 255 CONCEPTUAL CONNECTION 7.2: The Photoelectric Effect 256 7.3 Atomic Spectroscopy and the Bohr Model 256 CONCEPTUAL CONNECTION 7.3: Emission Spectra 260 CHEMISTRY IN YOUR DAY: Atomic Spectroscopy: A Bar Code for Atoms 

260

CONCEPTUAL CONNECTION 7.4 263 7.4 The Wave Nature of Matter: The de Broglie Wavelength, the Uncertainty Principle, and Indeterminacy 263 CONCEPTUAL CONNECTION 7.5: The de Broglie Wavelength of Macroscopic Objects The Uncertainty Principle  264  Indeterminacy and Probability Distribution Maps  266

7.5 Quantum Mechanics and Electrons in Atoms

231

Implications of Dependence on Fossil Fuels  231 CHAPTER IN REVIEW 234 Key Terms  234  Key Concepts  234  Key Equations and Relationships 235  Key Skills 236

264

267

Solutions to the Schrödinger Equation for the Particle in a Box: Quantum Numbers  268

CONCEPTUAL CONNECTION 7.6

270

Solutions to the Schrödinger Equation for the Hydrogen Atom  271

7.6 The Shapes of Atomic Orbitals

CONCEPTUAL CONNECTION 6.6: Exothermic and Endothermic Reactions

244

The de Broglie Wavelength  264

CONCEPTUAL CONNECTION 6.4: Thermal Energy Transfer Work: Pressure–Volume Work  211

7.1 Quantum Mechanics: The Theory That Explains the Behaviour of the Absolutely Small 7.2 The Nature of Light

236

CHEMISTRY AND MEDICINE: Radiation Treatment

6.3 The First Law of Thermodynamics: There Is No Free Lunch 202 CHEMISTRY IN YOUR DAY: Perpetual Motion Machines 203

Surroundings

EXERCISES Review Questions  236  Problems by Topic  237 Cumulative Problems  240  Challenge Problems  242 Conceptual Problems  243

273

s Orbitals (l = 0) 273  p Orbitals (l = 1)  276 d Orbitals (l = 2)  277  f Orbitals (l = 3)  277 The Phase of Orbitals  278

CONCEPTUAL CONNECTION 7.7: The Shapes of Atoms 279 The Hydrogen-Like Wave Functions  279

CONCEPTUAL CONNECTION 7.8: Radial Probabilities 281 7.7 Electron Configurations: How Electrons Occupy Orbitals 282 Electron Spin and the Pauli Exclusion Principle  282  Sublevel Energy Splitting in Multielectron Atoms  283

CONCEPTUAL CONNECTION 7.9: Penetration and Shielding 287 Electron Configurations for Multielectron Atoms  287 CONCEPTUAL CONNECTION 7.10: Electron Configurations and Quantum Numbers Electron Configurations for Transition Metals  290 Electron Configurations and Magnetic Properties of Ions 291

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289

Get complete eBook Order by email at [email protected] O N T E N T S      xi CHAPTER IN REVIEW  Key Terms  294  Key Concepts  294  Key Equations and Relationships  295  Key Skills  295

294

EXERCISES Review Questions  296  Problems by Topic  297 Cumulative Problems  299  Challenge Problems  300 Conceptual Problems  301

296

8

Periodic Properties of the Elements

302 303 303 304

Orbital Blocks in the Periodic Table  306  Writing an Electron Configuration for an Element from Its Position in the Periodic Table  307  The d-Block and f-Block Elements  307

8.4 The Explanatory Power of the Quantum-Mechanical Model 8.5 Periodic Trends in the Size of Atoms and Effective Nuclear Charge

CONCEPTUAL CONNECTION 8.1: Atomic Size 8.6 Ionic Radii CONCEPTUAL CONNECTION 8.2: Ions, Isotopes, and Atomic Size

8.7 Ionization Energy

Bonding Models and AIDS Drugs Types of Chemical Bonds Representing Valence Electrons with Dots Lewis Structures: An Introduction to Ionic and Covalent Bonding

337 337 339 340

CONCEPTUAL CONNECTION 9.1: Energy and the Octet Rule 343 Writing Lewis Structures for Polyatomic Ions  343  Ionic Bonding and Electron Transfer  344 9.5 The Ionic Bonding Model

345

The Born–Haber Cycle  346  Trends in Lattice Energies: Ion Size  347  Trends in Lattice Energies: Ion Charge  348  Ionic Bonding: Models and Reality  349

CONCEPTUAL CONNECTION 9.2: Melting 350

9.6 Covalent Bond Energies, Lengths, and Vibrations 308 309

350

Bond Energy  350  Using Average Bond Energies to Estimate Enthalpy Changes for Reactions  351

CONCEPTUAL CONNECTION 9.3: Bond Energies and ¢rH 352 Bond Lengths 353  Bond Vibrations 354

9.7 Electronegativity and Bond Polarity 315

316 318

318

Trends in First Ionization Energy  318  Exceptions to Trends in First Ionization Energy  320  Ionization Energies of Transition Metals  321  Trends in Second and Successive Ionization Energies 322

356

Electronegativity  357  Bond Polarity, Dipole Moment, and Percent Ionic Character  358

CONCEPTUAL CONNECTION 9.4: Percent Ionic Character 361 9.8 Resonance and Formal Charge 361 Resonance 361  Formal Charge 363

9.9 Exceptions to the Octet Rule: Drawing Lewis Structures for Odd-Electron Species and Incomplete Octets 366 Odd-Electron Species  366  Incomplete Octets  366

CHEMISTRY IN THE ENVIRONMENT: Free Radicals and the Atmospheric Vacuum Cleaner

CONCEPTUAL CONNECTION 8.3: Ionization 322

323

Electron Affinity 323  Metallic Character 323

8.9 Some Examples of Periodic Chemical Behaviour: The Alkali Metals, Alkaline Earth Metals, Halogens, and Noble Gases

9.1 9.2 9.3 9.4

Points of Ionic Solids

Effective Nuclear Charge  311  Slater’s Rules  313  Atomic Radii of d-Block Elements  314

8.8 Electron Affinities and Metallic Character



336

Drawing Lewis Structures for Molecular Compounds 340

8.1 Nerve Signal Transmission 8.2 The Development of the Periodic Table 8.3 Electron Configurations, Valence Electrons, and the Periodic Table

Energies and Chemical Bonding

9

Chemical Bonding I: Lewis Theory

367

9.10 Lewis Structures for Hypercoordinate Compounds CHEMISTRY IN THE ENVIRONMENT: The Lewis

368

Structure of Ozone

373

CHAPTER IN REVIEW  373 Key Terms  373  Key Concepts  374  Key Equations and Relationships 374  Key Skills 375

324

The Alkali Metals (Group 1)  325  The Alkaline Earth Metals (Group 2)  326  The Halogens (Group 17)  327  The Noble Gases (Group 18)  328

CHEMISTRY AND MEDICINE: Potassium Iodide in Radiation Emergencies

329

CHAPTER IN REVIEW  Key Terms  330  Key Concepts  330  Key Equations and Relationships  330  Key Skills  331

330

EXERCISES  Review Questions  331  Problems by Topic  332  Cumulative Problems  333  Challenge Problems  334 Conceptual Problems  335

331

EXERCISES  Review Questions  375  Problems by Topic  376 Cumulative Problems  378  Challenge Problems  379 Conceptual Problems  380

10

Chemical Bonding II: Molecular Shapes, Valence Bond Theory, and Molecular Orbital Theory

375

381

10.1 Artificial Sweeteners: Fooled by Molecular Shape 10.2 VSEPR Theory: The Five Basic Shapes Two Electron Groups: Linear Geometry  383  Three

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382 382

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CONCEPTUAL CONNECTION 10.1: Electron Groups and Molecular Geometry Four Electron Groups: Tetrahedral Geometry  384  Five Electron Groups: Trigonal Bipyramidal Geometry  385  Six Electron Groups: Octahedral Geometry  386 10.3 VSEPR Theory: The Effect of Lone Pairs

384

387

Four Electron Groups with Lone Pairs  387  Five Electron Groups with Lone Pairs  388  Six Electron Groups with Lone Pairs  389

CONCEPTUAL CONNECTION 10.2: Molecular Geometry and Electron Groups

10.4 VSEPR Theory: Predicting Molecular Geometries

391

391

Predicting the Shapes of Larger Molecules  393

10.5 Molecular Shape and Polarity CHEMISTRY IN YOUR DAY: How Soap Works 10.6 Valence Bond Theory: Orbital Overlap as a Chemical Bond CONCEPTUAL CONNECTION 10.3: What Is a

394 397

399

400

sp3 Hybridization  401  sp2 Hybridization and Double Bonds  402

CHEMISTRY IN YOUR DAY: The Chemistry of Vision CONCEPTUAL CONNECTION 10.4: Single and

407

Double Bonds sp Hybridization and Triple Bonds  407  Writing Hybridization and Bonding Schemes  409

407

10.8 Molecular Orbital Theory: Electron Delocalization

411

11

425

428

436

11.1 Climbing Geckos and Intermolecular Forces 437 11.2 Solids, Liquids, and Gases: A Molecular Comparison 437 Changes Between States  439

CONCEPTUAL CONNECTION 11.1: State Changes 11.3 Intermolecular Forces: The Forces That Hold Condensed States Together

CHEMISTRY IN YOUR DAY: Viscosity and Motor Oil 11.6 Vaporization and Vapour Pressure

453

455

456

The Process of Vaporization  456  The Energetics of Vaporization  457  Vapour Pressure and Dynamic Equilibrium 459 461

The Critical Point: The Transition to an Unusual State of Matter  465

11.7 Sublimation and Fusion

466

Sublimation 466  Fusion 467  Energetics of Melting and Freezing  467

11.8 Heating Curve for Water CONCEPTUAL CONNECTION 11.3: Cooling of Water with Ice

468 470

470

The Major Features of a Phase Diagram  470  Navigation Within a Phase Diagram  471  The Phase Diagrams of Other Substances  472

CONCEPTUAL CONNECTION 11.4: Phase Diagrams 11.10 Crystalline Solids: Determining Their Structure by X-Ray Crystallography 11.11 Crystalline Solids: Unit Cells and Basic Structures 11.12 Crystalline Solids: The Fundamental Types

427 CHAPTER IN REVIEW  Key Terms  427  Key Concepts  427  Key Equations and Relationships 428  Key Skills 428

Liquids, Solids, and Intermolecular Forces

453

473

473 475

Closest-Packed Structures  478

CONCEPTUAL CONNECTION 10.5: What Is a

EXERCISES  Review Questions  428  Problems by Topic  429  Cumulative Problems  432  Challenge Problems  434  Conceptual Problems  435

450

451

Surface Tension 453  Viscosity 454  Capillary Action  455

11.9 Phase Diagrams

Linear Combination of Atomic Orbitals (LCAO)  412  Period 2 Homonuclear Diatomic Molecules  415  Period 2 Heteronuclear Diatomic Molecules 421  Polyatomic Molecules 422  Larger Conjugated Pi Systems  423 Chemical Bond? Part II An Extension of MO Theory: Band Theory of Solids 425

CHEMISTRY AND MEDICINE: Hydrogen Bonding in DNA 11.4 Water: An Extraordinary Substance CHEMISTRY IN THE ENVIRONMENT: Water Pollution 11.5 Intermolecular Forces in Action: Surface Tension, Viscosity, and Capillary Action

CONCEPTUAL CONNECTION 11.2: Vapour Pressure 397

Chemical Bond? Part I

10.7 Valence Bond Theory: Hybridization of Atomic Orbitals

Ion-Induced Dipole Force  441  Dispersion Force 441  Dipole–Dipole Force 443  Hydrogen Bonding  446  Dipole-Induced Dipole Force 448  Ion–Dipole Force 448

440

440

481

Molecular Solids  481  Ionic Solids  481  Atomic Solids  483 CHAPTER IN REVIEW  485 Key Terms  485  Key Concepts  485  Key Equations and Relationships 486  Key Skills 486 EXERCISES  Review Questions  486  Problems by Topic  487  Cumulative Problems  491  Challenge Problems  493  Conceptual Problems  493

12

Solutions

486

494

12.1 Thirsty Solutions: Why You Shouldn’t Drink Seawater 494 12.2 Types of Solutions and Solubility 496 Nature’s Tendency Toward Mixing: Entropy  496  The Effect of Intermolecular Forces 497

CONCEPTUAL CONNECTION 12.1: Solubility CHEMISTRY IN YOUR DAY: Vitamin D in Foods

500

and Supplements

500

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12.3 Energetics of Solution Formation

501

Aqueous Solutions and Heats of Hydration  503

12.4 Solution Equilibrium and Factors Affecting Solubility 504 The Temperature Dependence of the Solubility of Solids  505  Factors Affecting the Solubility of Gases in Water  506

CONCEPTUAL CONNECTION 12.2: Solubility and Temperature

507

CONCEPTUAL CONNECTION 12.3: Henry’s Law 12.5 Expressing Solution Concentration CHEMISTRY IN THE ENVIRONMENT: Lake Nyos

508

509 509

Molarity  510  Molality  510  Parts by Mass and Parts by Volume  511  Mole Fraction and Mole Percent 513 513

12.6 Colligative Properties: Vapour Pressure Lowering, Freezing Point Depression, Boiling Point Elevation, and Osmotic Pressure

516

Vapour Pressure Lowering  516  Vapour Pressures of Solutions Containing a Volatile (Nonelectrolyte) Solute 519

CONCEPTUAL CONNECTION 12.4: Raoult’s Law

520

Freezing Point Depression and Boiling Point Elevation 522

CHEMISTRY IN THE ENVIRONMENT: Antifreeze in Frogs 524 CHEMISTRY IN YOUR DAY: Airplane De-icing 525 Osmotic Pressure  526

12.7 Colligative Properties of Strong Electrolyte Solutions 528 CONCEPTUAL CONNECTION 12.5: Colligative Properties 528 Strong Electrolytes and Vapour Pressure  529  Colligative Properties and Medical Solutions  530

12.8 Colloids

531

CHAPTER IN REVIEW  Key Terms  533  Key Concepts  533  Key Equations and Relationships  534  Key Skills  535

533

EXERCISES  Review Questions 535 Problems by Topic 536 Cumulative Problems  539  Challenge Problems  540 Conceptual Problems  541

535

542

13.1 Hibernating Frogs 13.2 The Rate of a Chemical Reaction

543 543

Measuring Reaction Rates  547

13.3 The Rate Law: The Effect of Concentration on Reaction Rate

567

567

Rate Laws for Elementary Steps  568  Rate-Determining Steps and Overall Reaction Rate Laws  568  The Steady-State Approximation 571

574

CHEMISTRY AND MEDICINE: Enzyme Catalysis and the Role of Chymotrypsin in Digestion

580

CHAPTER IN REVIEW  Key Terms  581  Key Concepts  581  Key Equations and Relationships  582  Key Skills  582

581

EXERCISES  Review Questions  583  Problems by Topic  583 Cumulative Problems  589  Challenge Problems  593 Conceptual Problems  594

583

14

Chemical Equilibrium

595

14.1 Hemoglobin and Equilibrium 14.2 The Concept of Dynamic Equilibrium 14.3 Le Châtelier’s Principle: How a System at Equilibrium Responds to Disturbances

596 596 598

The Effect of Changing the Amount of Reactant or Product on Equilibrium  598  The Effect of a Volume Change on Equilibrium  601  The Effect of Changing the Pressure by Adding an Inert Gas  602  The Effect of a Temperature Change on Equilibrium  602  The Effect of Adding a Catalyst on Equilibrium  605

14.4 The Expression for the Equilibrium Constant

605

CONCEPTUAL CONNECTION 14.1: Heterogeneous Equilibria, KP and Kc 611 14.5 The Equilibrium Constant (K  )

611

Units of Equilibrium Constants  611 The Significance of the Equilibrium Constant  611

CONCEPTUAL CONNECTION 14.2: Equilibrium Constants 612 548

Determining the Order of a Reaction  549  Reaction Order for Multiple Reactants  551

CONCEPTUAL CONNECTION 13.1: Rate and

Half-Life, Lifetime, and Decay Time  557

CONCEPTUAL CONNECTION 13.3: Collision Theory 13.6 Reaction Mechanisms

Relating KP and Kc  606  The Unitless Thermodynamic Equilibrium Constant  608  Heterogeneous Equilibria: Reactions Involving Solids and Liquids  609

Chemical Kinetics

13.4 The Integrated Rate Law: The Dependence of Concentration on Time

560

Arrhenius Plots: Experimental Measurements of the Frequency Factor and the Activation Energy  563  The Collision Model: A Closer Look at the Frequency Factor  565

Homogeneous and Heterogeneous Catalysis  575  Enzymes: Biological Catalysts  577

and Personal Care Products

Concentration

560

13.7 Catalysis

CHEMISTRY IN THE ENVIRONMENT: Pharmaceuticals

13

CONCEPTUAL CONNECTION 13.2: Kinetics Summary 13.5 The Effect of Temperature on Reaction Rate

552

552

Relationships Between the Equilibrium Constant and the Chemical Equation  612

CHEMISTRY AND MEDICINE: Life and Equilibrium 14.6 Calculating the Equilibrium Constant from Measured Quantities 14.7 The Reaction Quotient: Predicting the Direction of Change CONCEPTUAL CONNECTION 14.3: Q and K

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613

615 618 620

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14.8 Finding Equilibrium Concentrations

620

Finding Partial Pressures or Concentrations at Equilibrium Amounts When We Know the Equilibrium Constant and All But One of the Equilibrium Amounts of the Reactants and Products  620  Finding Equilibrium Concentrations When We Know the Equilibrium Constant and Initial Concentrations or Pressures   621  Simplifying Approximations in Working Equilibrium Problems  625

CONCEPTUAL CONNECTION 14.4: The x is small Approximation

629

CHAPTER IN REVIEW  629 Key Terms  629  Key Concepts  629  Key Equations and Relationships 630  Key Skills 630 EXERCISES  Review Questions  631  Problems by Topic  632 Cumulative Problems  636  Challenge Problems  637 Conceptual Problems  638

15

631

15.1 Heartburn 15.2 The Nature of Acids and Bases 15.3 Definitions of Acids and Bases

639 640 642

The Arrhenius Definition  642  The Brønsted–Lowry Definition 643

CONCEPTUAL CONNECTION 15.1: Conjugate 645

15.4 Acid Strength and the Acid Ionization Constant (Ka ) 645 Strong Acids  645  Weak Acids  646  The Acid Ionization Constant (Ka) 647

CONCEPTUAL CONNECTION 15.2: Conjugate Bases 15.5 Base Solutions

648

654

655

659

CONCEPTUAL CONNECTION 15.4: Percent Ionization

661

Mixtures of Acids  661

CONCEPTUAL CONNECTION 15.5: Strong and 663

665

Anions as Weak Bases  666  Cations as Weak Acids 669

Molecular Structure Amine Bases  683

683

CONCEPTUAL CONNECTION 15.8: Amine Base 683

683

CHAPTER IN REVIEW  Key Terms  685  Key Concepts  685  Key Equations and Relationships  686  Key Skills  686

685

EXERCISES  Review Questions  687  Problems by Topic  688 Cumulative Problems  691  Challenge Problems  693 Conceptual Problems  693

687

16

Aqueous Ionic Equilibrium

16.1 The Danger of Antifreeze 16.2 Buffers: Solutions That Resist pH Change

694 695 696

Calculating the pH of a Buffer Solution  697  The Henderson–Hasselbalch Equation  698 Calculating pH Changes in a Buffer Solution  701 Base to a Buffer 704 Buffers Containing a Base and Its Conjugate Acid  704

16.3 Buffer Effectiveness: Buffer Range and Buffer Capacity

706

Relative Amounts of Acid and Base  706  Absolute Concentrations of the Acid and Conjugate Base 706  Buffer Range 707

CHEMISTRY AND MEDICINE: Buffer Effectiveness in Human Blood Buffer Capacity  709

CONCEPTUAL CONNECTION 16.3: Buffer Capacity 16.4 Titrations and pH Curves

708 709

709

The Titration of a Strong Acid with a Strong Base  709 The Titration of a Weak Acid with a Strong Base  714  The Titration of a Weak Base with a Strong Acid  719  The Titration of a Polyprotic Acid  720

CONCEPTUAL CONNECTION 16.4: Acid–Base Titrations 720 Indicators: pH-Dependent Colours  720

CONCEPTUAL CONNECTION 15.6: Acidities of Metal Cations in Aqueous Solutions Classifying Salt Solutions as Acidic, Basic, or Neutral 670

Binary Acids 680  Oxyacids 681

CONCEPTUAL CONNECTION 15.7: Acid Strength and

650

CONCEPTUAL CONNECTION 15.3: The x is small Approximation Percent Ionization of a Weak Acid  659

15.8 The Acid–Base Properties of Ions and Salts

15.11 Strengths of Acids and Bases and Molecular Structure 680

CONCEPTUAL CONNECTION 16.2: Adding Acid or

Strong Acids  655  Weak Acids  655

Weak Acids Finding the [OH-] and pH of Basic Solutions  663

Molecules That Act as Lewis Acids  679  Cations That Act as Lewis Acids  680

CONCEPTUAL CONNECTION 16.1: pH of Buffer Solutions 701

The pH Scale: A Way to Quantify Acidity and Basicity  652  pOH and Other p Scales  653

CHEMISTRY AND MEDICINE: Ulcers 15.7 Finding [H3O+], [OH−], and pH of Acid or Base Solutions

677

678

648

Strong Bases  648  Weak Bases  648

15.6 Autoionization of Water and pH

CHEMISTRY IN YOUR DAY: Weak Acids in Wine 15.10 Lewis Acids and Bases

15.12 Ocean Acidification

639

673

Finding the pH of Polyprotic Acid Solutions  674  Finding the Concentration of the Anions for a Weak Diprotic Acid Solution  676

Strength and Molecular Structure

Acids and Bases

Acid–Base Pairs

15.9 Polyprotic Acids

16.5 Solubility Equilibria and the Solubility Product Constant 723 670

Ksp and Molar Solubility  723  Ksp and Relative Solubility  726  The Effect of a Common Ion on Solubility 726

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CHEMISTRY AND MEDICINE: Fluoride and Teeth

727

The Effect of an Uncommon Ion on Solubility (Salt Effect)  728  The Effect of pH on Solubility  728

16.6 Precipitation

729

Selective Precipitation  731

16.7 Qualitative Chemical Analysis

733

Group A: Insoluble Chlorides  734  Group B: AcidInsoluble Sulfides  734  Group C: Base-Insoluble Sulfides and Hydroxides  735  Group D: Insoluble Phosphates  735  Group E: Alkali Metals and NH4+ 735

17.10 Gibbs Energy Changes for Nonstandard States: The Relationship Between ¢rG° and ¢rG 781 The Gibbs Energy Change of a Reaction Under Nonstandard Conditions  782

CONCEPTUAL CONNECTION 17.3: Gibbs Energy Changes and Le Châtelier’s Principle

17.11 Gibbs Energy and Equilibrium: Relating ¢rG° to the Equilibrium Constant (K  )

784

785

The Temperature Dependence of the Equilibrium Constant 787

16.8 Complex–Ion Equilibria

735

The Effect of Complex–Ion Equilibria on Solubility 737

CONCEPTUAL CONNECTION 16.5: Solubility and Complex–ION Equilibria

738

CHAPTER IN REVIEW  738 Key Terms  738  Key Concepts  739  Key Equations and Relationships 739  Key Skills 739 EXERCISES  Review Questions  740  Problems by Topic  741 Cumulative Problems  746  Challenge Problems  747 Conceptual Problems  748

740

790 CHAPTER IN REVIEW  Key Terms  790  Key Concepts  790  Key Equations and Relationships 791  Key Skills 791 EXERCISES  Review Questions  792  Problems by Topic  792 Cumulative Problems  795  Challenge Problems  797 Conceptual Problems  798

18

Electrochemistry

18.1 Lightning and Batteries 18.2 Voltaic (or Galvanic) Cells: Generating Electricity from Spontaneous Chemical Reactions

792

799 800 800

Electrochemical Cell Notation  803

17

Gibbs Energy and Thermodynamics

18.3 Standard Electrode Potentials

749

17.1 Spontaneous and Nonspontaneous Processes 17.2 Entropy and the Second Law of Thermodynamics

750 751

Entropy  753  The Entropy Change Associated with a Change in State  758

17.3 Heat Transfer and Changes in the Entropy of the Surroundings

759

CONCEPTUAL CONNECTION 17.1: Entropy and Biological Systems

763

17.4 Entropy Changes for Phase Transitions 763 17.5 Entropy Changes in Chemical Reactions: Calculating ¢rS° 764 Standard Molar Entropies (S°) and the Third Law of Thermodynamics 764

17.6 Gibbs Energy

769

Predicting Whether a Metal Will Dissolve in Acid  812

CONCEPTUAL CONNECTION 18.2: Metals Dissolving

Calculating Gibbs Energy Changes with ¢rG° = ¢rH° - T ¢rS°  774 Calculating ¢rG° with Tabulated Values of Gibbs Energies of Formation 775  Calculating ¢rG° for a Stepwise Reaction from the Changes in Gibbs Energy for Each of the Steps  777

812

18.4 Cell Potential, Gibbs Energy, and the Equilibrium Constant 812 ° The Relationship Between ¢rG° and Ecell 813 CONCEPTUAL CONNECTION 18.3: Periodic Trends and the Direction of Spontaneity for Redox Reactions  814 The Relationship Between E°cell and K 815

18.5 Cell Potential and Concentration 816 CONCEPTUAL CONNECTION 18.4: Relating ∘ Q, K, Ecell, and Ecell 819 Concentration Cells  820

CHEMISTRY AND MEDICINE: Concentration Cells in Human Nerve Cells

18.6 Batteries: Using Chemistry to Generate Electricity

The Effect of ¢r H, ¢r S, and T on Spontaneity  771

CONCEPTUAL CONNECTION 17.2: ¢r H, ¢r S, and ¢r G 773 17.7 Gibbs Energy Changes in Chemical Reactions: Calculating ¢rG° 773

17.8 Making a Nonspontaneous Process Spontaneous 17.9 What Is Gibbs Energy?

CONCEPTUAL CONNECTION 18.1: Selective Oxidation 812

in Acids

The Temperature Dependence of ¢Ssurr 760  Quantifying Entropy Changes in the Surroundings  761

803

Predicting the Spontaneous Direction of an Oxidation–Reduction Reaction  809

823

823

Dry-Cell Batteries  823  Lead–Acid Storage Batteries  824  Other Rechargeable Batteries  824  Fuel Cells  826

CHEMISTRY IN YOUR DAY: Rechargeable Battery Recycling

18.7 Electrolysis: Driving Nonspontaneous Chemical Reactions with Electricity

827

827

Predicting the Products of Electrolysis  830  Stoichiometry of Electrolysis  833

778 780

18.8 Corrosion: Undesirable Redox Reactions Preventing Corrosion  837

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835

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CONCEPTUAL CONNECTION 18.5: Sacrificial Anodes

837

CHAPTER IN REVIEW  Key Terms  837  Key Concepts  837  Key Equations and Relationships  838  Key Skills  839 EXERCISES  Review Questions  839  Problems by Topic  840 Cumulative Problems  843  Challenge Problems  845 Conceptual Problems  845

19

Radioactivity and Nuclear Chemistry

876

837

CHAPTER IN REVIEW  Key Terms  876  Key Concepts  876  Key Equations and Relationships  877  Key Skills  877 EXERCISES  Review Questions  878  Problems by Topic  878 Cumulative Problems  880  Challenge Problems  881 Conceptual Problems  881

878

839

20 846

19.1 Medical Isotopes 19.2 The Discovery of Radioactivity 19.3 Types of Radioactivity

846 847 848

Alpha 1a2 Decay  849  Beta 1b2 Decay  850 Gamma 1g2 Ray Emission  851 Positron Emission  851  Electron Capture   851 852

19.4 The Valley of Stability: Predicting the Type of Radioactivity

853

Magic Numbers  855  Radioactive Decay Series 855

19.5 Measurements and Units of Radioactivity 19.6 The Kinetics of Radioactive Decay and Radiometric Dating

855 856 858

863

CHEMISTRY IN YOUR DAY: Uranium Isotopes and 19.8 Converting Mass to Energy in Nuclear Reactions and Nuclear Binding Energy CONCEPTUAL CONNECTION 19.3: Comparing the Energies of Nuclear Reactions to Chemical Reactions Nuclear Binding Energy  867

19.9 Nuclear Fusion: The Power of the Sun 19.10 Nuclear Transmutation and Transuranium Elements CONCEPTUAL CONNECTION 19.4: Nuclear Transformations

19.11 The Effects of Radiation on Life

866

866 867

869 870 872

872

Acute Radiation Damage  872  Increased Cancer Risk 872  Genetic Defects 872  Measuring Radiation Exposure  872

CONCEPTUAL CONNECTION 19.5: Radiation Exposure 873 19.12 Radioactivity in Medicine and Other Applications 874 Diagnosis in Medicine  874  Radiotherapy in Medicine 875  Other Applications 875

20.3 Hydrocarbons

884

885

Drawing Hydrocarbon Structures  885  Types of Hydrocarbons 888  Alkanes 888  Alkenes 889 Alkynes  889  Conjugated Alkenes and Aromatics  890

891

Halides 892  Amines 892  Alcohols 893 Ethers  894  Carbonyls: Aldehydes and Ketones  894 The Carboxylic Acid Family  895

20.5 Constitutional Isomerism 897 CONCEPTUAL CONNECTION 20.1: Constitutional Isomers 897 20.6 Stereoisomerism I: Conformational Isomerism 898 Conformational Isomerism: Rotation About Single Bonds  898  Ring Conformations of Cycloalkanes  900

CONCEPTUAL CONNECTION 20.2: Staggered 20.7 Stereoisomerism II: Configurational Isomerism

901

901

CONCEPTUAL CONNECTION 20.3: Chirality

Nuclear Power: Using Fission to Generate Electricity 864 The CANDU Reactor

Difference Between Organic and Inorganic

Cis–Trans Isomerism in Alkenes  902  Enantiomers: Chirality  905

Radiocarbon Dating: Using Radioactivity to Measure the Age of Fossils and Artifacts  860  Uranium/Lead Dating 861

19.7 The Discovery of Fission: The Atomic Bomb and Nuclear Power

883 883

Conformations in Rings

The Integrated Rate Law  857

CONCEPTUAL CONNECTION 19.2: Half-Life

20.1 Fragrances and Odours 20.2 Carbon: Why It Is Unique THE NATURE OF SCIENCE: Vitalism and the Perceived

20.4 Functional Groups

CONCEPTUAL CONNECTION 19.1: Alpha and Beta Decay

Organic Chemistry I: Structures 882

906

Absolute Configurations  907

CHEMISTRY AND MEDICINE: Anesthetics and Alcohol 20.8 Structure Determination

909

909

Using the Molecular Formula: The Index of Hydrogen Deficiency  909  Spectroscopic Methods for Structure Determination  912 CHAPTER IN REVIEW  Key Terms  915  Key Concepts  915  Key Equations and Relationships  916  Key Skills  917

915

EXERCISES  Review Questions  917  Problems by Topic  918 Cumulative Problems  920  Challenge Problems  921 Conceptual Problems  922

917

21

Organic Chemistry II: Reactions 924

21.1 Discovering New Drugs 21.2 Organic Acids and Bases The Range of Organic Acidities  925  Inductive Effects: Withdrawal of Electron Density  927 Resonance Effects: Charge Delocalization in the Conjugate Base  928  Acidic Hydrogen Atoms Bonded to Carbon  929  Mechanisms in Organic Chemistry  929  Acid and Base Reagents  930

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925 925

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21.3 Oxidation and Reduction

932

Redox Reactions  933 Reduction

935

CHEMISTRY IN YOUR DAY: Hydrogen and The Oil Sands

935

21.4 Nucleophilic Substitution Reactions at Saturated Carbon

936

CONCEPTUAL CONNECTION 22.1: Peptides 22.5 Protein Structure

985

985

Secondary Structure  988  Tertiary Structure  988 Quaternary Structure  988

Groups and Base Strength

940

21.5 Elimination Reactions

942

The E1 Mechanism  943

CONCEPTUAL CONNECTION 21.3: Can a Strong Acid React Like a Base? The E2 Mechanism  944  Elimination Versus Substitution 944

943

21.6 Electrophilic Additions to Alkenes

944

Hydrohalogenation 944

22.6 Nucleic Acids: Blueprints for Proteins

989

The Basic Structure of Nucleic Acids  990  The Genetic Code  992

CONCEPTUAL CONNECTION 22.2: The Genetic Code 22.7 DNA Replication, the Double Helix, and Protein Synthesis

993

993

DNA Replication and the Double Helix  993  Protein Synthesis  995 CHAPTER IN REVIEW  Key Terms 996  Key Concepts 996  Key Skills 997

CONCEPTUAL CONNECTION 21.4: H—X as an Electrophile Other Addition Reactions  945

945

CONCEPTUAL CONNECTION 21.5: Mechanism of Hydration 946 21.7 Nucleophilic Additions to Aldehydes and Ketones

946

Addition of Alcohols  946  The Grignard Reaction 947

21.8 Nucleophilic Substitutions of Acyl Compounds CONCEPTUAL CONNECTION 21.6: The Reaction

949

Between an Acyl Chloride and NH3 952

21.9 Electrophilic Aromatic Substitutions 953 CONCEPTUAL CONNECTION 21.7: Resonance Structures 953 21.10 Polymerization 954 Step-Growth Polymers  955  Addition Polymers  955 and Low-Density Polyethylene

956

CHAPTER IN REVIEW  Key Terms  957  Key Concepts  958  Key Equations and Relationships  959  Key Skills  960

957

EXERCISES  Review Questions  960  Problems by Topic  961 Cumulative Problems  967  Challenge Problems  968 Conceptual Problems  969

960

23

Chemistry of the Nonmetals 1004

23.1 Insulated Nanowires 1005 23.2 The Main-Group Elements: Bonding and Properties 1005 Atomic Size and Types of Bonds  1005

23.3 Silicates: The Most Abundant Matter in Earth’s Crust 1006 Quartz and Glass  1006  Aluminosilicates  1007  Individual Silicate Units, Silicate Chains, and Silicate Sheets 1007

970 971

Fatty Acids  971  Fats and Oils  973  Other Lipids  974

23.5 Carbon, Carbides, and Carbonates

975

1013

Carbon 1013 1013

CONCEPTUAL CONNECTION 23.2: Carbonate Solubility 1018 23.6 Nitrogen and Phosphorus: Essential Elements for Life 1018 Elemental Nitrogen and Phosphorus  1018  Nitrogen Compounds 1020  Phosphorus Compounds 1023

23.7 Oxygen

CHEMISTRY AND MEDICINE: Dietary Fat:

1010

Elemental Boron  1010  Boron–Halogen Compounds: Trihalides  1011  Boron–Oxygen Compounds  1012  Boron–Hydrogen Compounds: Boranes 1012

CONCEPTUAL CONNECTION 23.1: Phase Changes and Pressure Carbides  1016  Carbon Oxides  1017  Carbonates 1017

970

996

EXERCISES  997 Review Questions  997  Problems by Topic  998 Cumulative Problems  1001  Challenge Problems  1002 Conceptual Problems  1003

23.4 Boron and Its Remarkable Structures

CHEMISTRY IN YOUR DAY: High-, Medium-,

The Good, the Bad, and the Ugly

981

Amino Acids: The Building Blocks of Proteins  982  Peptide Bonding Between Amino Acids  984

CHEMISTRY AND MEDICINE: The Essential Amino Acids 987

CONCEPTUAL CONNECTION 21.2: Leaving

22.1 Diabetes and the Synthesis of Human Insulin 22.2 Lipids

22.4 Proteins and Amino Acids

Primary Structure  986

The SN1 Mechanism  936  The SN2 Mechanism  938  Factors Affecting Nucleophilic Substitution Reactions  939

Biochemistry

976

Simple Carbohydrates: Monosaccharides and Disaccharides 977  Complex Carbohydrates 980

CONCEPTUAL CONNECTION 21.1: Oxidation and

22

22.3 Carbohydrates

Elemental Oxygen  1025  Uses for Oxygen  1026  Oxides 1026  Ozone 1026

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23.8 Sulfur: A Dangerous but Useful Element

1027

Elemental Sulfur  1027  Hydrogen Sulfide and Metal Sulfides 1028  Sulfur Dioxide 1029  Sulfuric Acid 1029

23.9 Halogens: Reactive Elements with High Electronegativity

1033 CHAPTER IN REVIEW  Key Terms 1033  Key Concepts 1033  Key Skills 1034 1034 EXERCISES  Review Questions  1034  Problems by Topic  1034 Cumulative Problems  1036  Challenge Problems  1037 Conceptual Problems  1037

Metals and Metallurgy

24.1 Vanadium: A Problem and an Opportunity 24.2 The General Properties and Natural Distribution of Metals 24.3 Metallurgical Processes

1038 1039 1039 1041

Separation 1041  Pyrometallurgy 1042  Hydrometallurgy 1042  Electrometallurgy 1043 Powder Metallurgy  1044

24.4 Metal Structures and Alloys

1044

Alloys 1045

CONCEPTUAL CONNECTION 24.1: Interstitial Alloys 24.5 Sources, Properties, and Products of Some of the 3d Transition Metals

1050

1050

Titanium 1050  Chromium 1051  Manganese 1052 Cobalt 1052  Copper 1053  Nickel 1054 Zinc 1054 CHAPTER IN REVIEW  Key Terms 1055  Key Concepts 1055  Key Skills  1055

1055

1056 EXERCISES  Review Questions  1056  Problems by Topic  1056 Cumulative Problems  1057  Challenge Problems  1058 Conceptual Problems  1058

25

Transition Metals and Coordination Compounds

1059

25.1 The Colours of Rubies and Emeralds 25.2 Electron Configurations of Transition Metals

1060 1060

Electron Configurations  1060  Oxidation States 1061

25.3 Coordination Compounds

1067

Structural Isomerism  1067  Stereoisomerism 1067

25.5 Bonding in Coordination Compounds

1071

Ligand Field Theory  1071  Octahedral Complexes  1071  The Colour of Complex Ions and Ligand Field Strength  1072

1030

Elemental Fluorine and Hydrofluoric Acid  1031  Elemental Chlorine  1032  Halogen Oxides  1032

24

25.4 Structure and Isomerization

CONCEPTUAL CONNECTION 25.1: Weak- and Strong-Field Ligands 1074 Magnetic Properties  1074  Tetrahedral and Square Planar Complexes  1075 25.6 Applications of Coordination Compounds

1076

Chelating Agents  1076  Chemical Analysis 1076  Colouring Agents 1076  Biomolecules 1077 CHAPTER IN REVIEW  1079 Key Terms  1079  Key Concepts  1079  Key Equations and Relationships  1080  Key Skills  1080 EXERCISES  1080 Review Questions  1080  Problems by Topic  1081 Cumulative Problems  1082  Challenge Problems  1083 Conceptual Problems  1083

Appendix I: Common Mathematical Operations in Chemistry A  Scientific Notation B Logarithms C  Quadratic Equations D Graphs

A-1 A-1 A-3 A-4 A-5

Appendix II: Useful Data A-7 A  Atomic Colours A-7 B Standard Thermodynamic Quantities for Selected Substances at 25°C A-7 C  Aqueous Equilibrium Constants A-12 D  Standard Electrode Potentials at 25°C A-15 E Vapour Pressure of Water at Various TemperaturesA-16 Appendix III: Answers to Selected Exercises

A-17

Appendix IV: Answers to In-Chapter Practice Problems

A-60

Glossary

G-1

Essential Data and Equations 

E-1

Credits 

C-1

IndexI-1 1061

Naming Coordination Compounds  1065

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7

6

5

4

3

2

1

4 Be

9.012

12 Mg

24.31

20 Ca

40.08

38 Sr

87.62

56 Ba

137.33

88 Ra

[226.03]

3 Li

6.941

11 Na

22.99

19 K

39.10

37 Rb

85.47

55 Cs

132.91

87 Fr

[223.02]

58 Ce 140.12

90 Th 232.04

57 La 138.91

89 Ac [227.03]

231.04

91 Pa

140.91

59 Pr

[266.12]

106 Sg

183.84

74 W

95.96

42 Mo

52.00

6 24 Cr

238.03

92 U

144.24

60 Nd

[264.12]

107 Bh

186.21

75 Re

[98]

43 Tc

54.94

7 25 Mn

[237.05]

93 Np

[145]

61 Pm

[269.13]

108 Hs

190.23

76 Os

101.07

44 Ru

55.85

8 26 Fe

Transition metals

Metalloids

[244.06]

94 Pu

150.36

62 Sm

[268.14]

109 Mt

192.22

77 Ir

102.91

45 Rh

58.93

9 27 Co

[272]

64 Gd 157.25

96 Cm [247.07]

63 Eu 151.96

95 Am [243.06]

111 Rg

196.97

79 Au

107.87

47 Ag

63.55

11 29 Cu

[271]

110 Ds

195.08

78 Pt

106.42

46 Pd

58.69

10 28 Ni

Nonmetals

[247.07]

97 Bk

158.93

65 Tb

[277]

112 Cn

200.59

80 Hg

112.41

48 Cd

65.38

12 30 Zn

[251.08]

98 Cf

162.50

66 Dy

[286]

113 Nh

204.38

81 Tl

114.82

49 In

69.72

31 Ga

26.98

[252.08]

99 Es

164.93

67 Ho

[289]

114 Fl

207.2

82 Pb

118.71

50 Sn

72.64

32 Ge

14 Si

13 Al 28.09

12.01

10.81

5 B

14 6 C

13

Atomic masses in brackets are the masses of the longest-lived or most important isotope of radioactive elements.

Actinoid series

[262.11]

105 Db

180.95

73 Ta

92.91

[261.11]

104 Rf

178.49

72 Hf

91.22

40 Zr

39 Y

88.91

50.94

47.87

44.96

41 Nb

5 23 V

4 22 Ti

Metals

3 21 Sc

Lanthanoid series

2

1.008

1 1 H

Main groups

[257.10]

100 Fm

167.26

68 Er

[289]

115 Mc

208.98

83 Bi

121.76

51 Sb

74.92

33 As

30.97

15 P

14.01

15 7 N

[258.10]

101 Md

168.93

69 Tm

[292]

116 Lv

[208.98]

84 Po

127.60

52 Te

78.96

34 Se

32.07

16 S

16.00

8 O

16

Main groups

[259.10]

102 No

173.05

70 Yb

[294]

117 Ts

[209.99]

85 At

126.90

53 I

79.90

35 Br

35.45

17 Cl

19.00

9 F

17

[262.11]

103 Lr

174.97

71 Lu

[294]

118 Og

[222.02]

86 Rn

131.29

54 Xe

83.80

36 Kr

39.95

18 Ar

20.18

10 Ne

4.003

18 2 He

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Units of Measurement for Physical and Chemical Change

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1

Ancient science had sought knowledge of what things are, to be contemplated as an end in itself satisfying to the knower. In contrast, modern science seeks knowledge of how things work, to be used as a means for the relief and comfort of all humanity, knowers and non-knowers alike. —Leon R. Kass

O

NE OF THE MOST fundamental ideas in chemistry, perhaps in all of science, is that the properties of matter are determined by the properties of molecules and atoms. The properties of atoms and molecules determine

how matter behaves. The properties of water molecules determine how water behaves, the properties of sugar molecules determine how sugar behaves, and the properties of the molecules that compose our bodies determine how our bodies behave. Our understanding of matter at the molecular level gives us unprecedented control over that matter. In fact, our expanded understanding of the molecules that

1.1 Physical and Chemical Changes and Physical and Chemical Properties  2 1.2 Energy: A Fundamental Part of Physical and Chemical Change  3 1.3 The Units of Measurement  4 1.4 The Reliability of a Measurement  12 1.5 Solving Chemical Problems  19

compose living organisms has made possible the biological revolution of the last 50 years.

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1

2       Ch a pter

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1   Unit s of Me a s ur e me nt fo r Phy si c al an d C h em i c al C h an g e

1.1  Physical and Chemical Changes and Physical and Chemical Properties Every day, whether we know it or not, we witness changes in matter that are a result of the properties of the atoms and molecules composing that matter—ice melts, iron rusts, gasoline burns, fruit ripens, water evaporates. What happens to the molecules that compose the matter during such changes? The answer depends on the type of change. Physical changes are those that do not alter the chemical composition. For example, when water boils, it changes state from a liquid to a gas, but both are composed of water molecules (Figure 1.1 ▼). Chemical changes are those that alter the composition (the chemical structure) of matter. During a chemical change, atoms rearrange, transforming the original substance into a different one. For example, the rusting of iron is a chemical change (Figure 1.1). Iron atoms combine with oxygen molecules from air to form iron(III) oxide, the orange-coloured substance we call rust.

Iron atoms

Water molecules change from liquid to gaseous state: physical change.

H2O(g) Iron(III) oxide (rust)

H2O(l)

▲ FIGURE 1.1  Boiling Is a Physical Change, and Rusting Is a Chemical Change  When water boils, individual molecules leave the liquid as a gas, but the chemical identity is not altered—both the liquid and the vapour are composed of water molecules. When iron rusts, the iron atoms combine with oxygen to form a different chemical substance.

Physical and chemical changes are a manifestation of physical and chemical properties. A physical property is one that a substance displays without changing its composition, whereas a chemical property is one that a substance displays only by changing its composition via a chemical change. The smell of gasoline is a physical property—gasoline does not change its composition when it exhibits its odour. The combustibility of gasoline, in contrast, is a chemical property—gasoline does change its composition when it burns, turning into completely new substances (primarily carbon dioxide and water). Physical properties include odour, taste, colour, appearance, melting point, boiling point, and density. Chemical properties include corrosiveness, flammability, acidity, toxicity, and other such characteristics. If we want to understand the substances around us, we must understand the physical and chemical properties of the atoms and molecules that compose them—this is the central goal of chemistry. A good, simple definition of chemistry is: Chemistry—the science that seeks to understand the properties and behaviour of matter by studying the properties and behaviour of atoms and molecules.

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1.2  Energy: A Fundamental Part of Physical and Chemical Change       3 Get complete eBook Order by email at [email protected]

CONCEPTUAL CONNECTION 1.1 Chemical and Physical Changes The diagram to the left represents liquid water molecules in a pan. Which of the three diagrams on the right best represents the water molecules after they have been vaporized by the boiling of liquid water?

(a)

(b)

(c)

1.2  Energy: A Fundamental Part of Physical and Chemical Change Physical and chemical changes are usually accompanied by energy changes. For example, when water evaporates from your skin (a physical change), the water molecules absorb energy from your body, making you feel cooler. When you burn natural gas on the stove (a chemical change), energy is released, heating the food you are cooking. Understanding the physical and chemical changes of matter—that is, understanding chemistry—requires that we understand energy changes and energy flow. The total energy of an object is a sum of its kinetic energy, the energy associated with its motion, and its potential energy, the energy associated with its position or composition. An object held several metres from the ground has potential energy due to its position within Earth’s gravitational field (Figure 1.2 ▶). If you drop the object, it accelerates, and the potential energy is converted to kinetic energy. When the object hits the ground, its kinetic energy is converted primarily to thermal energy, the energy associated with the temperature of an object. Thermal energy is actually a type of kinetic energy because it arises from the motion of the individual atoms or molecules that make up an object. In other words, when the object hits the ground, its kinetic energy is essentially transferred to the atoms and molecules that compose the ground, raising the temperature of the ground ever so slightly. According to the law of conservation of energy, energy is neither created nor destroyed. The potential energy of the object becomes kinetic energy as it accelerates toward the ground. The kinetic energy then becomes thermal energy when the object hits the ground. The total amount of thermal energy that is released through the process is exactly equal to the initial potential energy of the object. Another principle to note is the tendency of systems with high potential energy to change in a way that lowers their potential energy. Objects lifted several metres from the ground, with high potential energy, tend to be unstable because they contain a significant amount of localized potential energy. Unless restrained, the object will naturally fall, lowering its potential energy. We can harness some of the raised object’s potential energy to do work. For example, we can attach the object to a rope that turns a paddle wheel or spins a drill as it falls. After it falls to the ground, the object contains less potential energy—it has become more stable. Some chemical substances are like the raised object just described. For example, the molecules that compose gasoline have a relatively high potential energy—energy is concentrated in them just as energy is concentrated in the raised object. The molecules in gasoline, therefore, tend to undergo chemical changes (specifically combustion) that will lower their potential energy. As the energy of the molecules is released, some of it can be harnessed to do work, such as moving a car down the street (Figure 1.3 ▼). The molecules that result from the chemical change have less potential energy than the original molecules in gasoline and are more stable.

10 kg

High potential energy

Kinetic energy

Low potential energy Heat

▲ FIGURE 1.2  Energy Conversions

When the object on the roof is released, gravitational potential energy is converted into kinetic energy as the object falls. The kinetic energy is converted mostly to thermal energy when the weight strikes the ground.

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▶ FIGURE 1.3  Using Chemical Energy

to Do Work  The compounds produced when gasoline burns have less chemical potential energy than the gasoline molecules.

Molecules in gasoline (high potential energy) Molecules in exhaust (low potential energy)

Some of released energy harnessed to do work

Car moves forward

Chemical potential energy, such as that contained in the molecules that compose gasoline, arises primarily from electrostatic forces between the electrically charged particles (protons and electrons) that compose atoms and molecules. We will learn more about these particles, as well as the properties of charge, in Chapter 2. For now know that molecules contain specific, sometimes complex, arrangements of these charged particles. Some of these arrangements—such as the one within the molecules that compose gasoline—have a much higher potential energy than others. When gasoline undergoes combustion, the arrangement of these particles changes, creating molecules with much lower potential energy and transferring a great deal of energy (mostly in the form of heat) to the surroundings.

1.3  The Units of Measurement On July 23, 1983, Air Canada Flight 143, a Boeing 767 jet, ran out of fuel at 7920 m altitude, halfway between its flight from Montreal, QC, to Edmonton, AB. Prior to take-off in Montreal, a fuel gauge malfunctioned, so the fuel level had to be measured manually. An investigation revealed that the ground crew had mistakenly used an incorrect conversion factor when determining how much fuel was needed to fill the plane. The jet required 22 300 kg of fuel for the flight. Unfortunately, the ground crew used the conversion factor of 1.77 lb L-1 to convert the number of kilograms to the number of litres required to fill the plane. The correct conversion factor is 0.803 kg L-1 (the density of jet fuel), so the plane had less than half the required amount of fuel for the flight. Fortunately, the pilots were able to glide the plane—with no engines running—to a safe landing at the former RCAF Station Gimli in Manitoba with no injuries to the 61 passengers or 5 crew members. As a result, Air Canada Flight 143 was nicknamed “The Gimli Glider.” In chemistry, as in aviation, units—the standard quantities used to specify measurement—are critical. If you get them wrong, the consequences can be disastrous. There are many measurement systems that you may come across, such as the Imperial system, used in the United States, and the metric system, used in most of the rest of the world. Scientists mainly use the International System of Units (SI), which is a subsystem of the metric system. We will focus on the SI system.

The Standard Units

TABLE 1.1  SI Base Units Quantity

Unit

Symbol

Length

metre

m

Mass

kilogram

kg

Time

second

s

Temperature

kelvin

K

Amount of substance

mole

mol

Electric current

ampere

A

Luminous intensity

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Table 1.1 shows the standard SI base units. For now, we will focus on the first four of these units: the metre, the standard unit of length; the kilogram, the standard unit of mass; the second, the standard unit of time; and the kelvin, the standard unit of temperature.

The Metre: A Measure of Length The metre (m) is defined as the distance light travels through a vacuum in precisely 1/299 792 458 s. A tall human is about 2 m tall email and the CN stands 553.3 m tall. by atTower [email protected]

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Scientists commonly deal with a wide range of lengths and distances. The separation between the sun and the closest star (Proxima Centauri) is about 3.8 * 1016 m, while many chemical bonds measure about 1.5 * 10-10 m.

The Kilogram: A Measure of Mass The kilogram (kg), defined as the mass of a metal cylinder kept at the International Bureau of Weights and Measures at Sèvres, France, is a measure of mass, a quantity different from weight. The mass of an object is a measure of the quantity of matter within it, while the weight of an object is a measure of the gravitational pull on its matter and has the unit N, newtons. If you weigh yourself on the moon, for example, its weaker gravity pulls on you with less force than does Earth’s gravity, resulting in a lower weight. A 75.0 kg person on Earth has a weight of 736 N. The same person would weigh just 122 N on the moon due to the moon’s smaller gravitational field. However, the person’s mass—the quantity of matter in his or her body—remains the same on every planet, 75.0 kg. This book has a mass of about 2.5 kg. A second common unit of mass is the gram (g). One gram is 1/1000 kg. A nickel has a mass of about 3.95 g.

553.3 m

2m

▲ The CN Tower is 553.3 metres tall. A hockey player is about 2 metres tall.

The Second: A Measure of Time The International Bureau of Weights and Measures originally defined the second (s) in terms of the day and the year, but a second is now defined more precisely as the duration of 9 192 631 770 periods of the radiation emitted from a certain transition in a cesium-133 atom. Scientists measure time on a large range of scales. The human heart beats about once every second, the age of the universe is estimated to be about 13.7 billion years, and some molecular bonds break or form in time periods as short as 1 * 10-15 s.

The Kelvin: A Measure of Temperature

100 °C

373 K

100 Celsius degrees

100 kelvins

0.00 °C

273 K

Water boils

Water freezes

The kelvin (K) is the SI unit of temperature. The temperature of a sample of matter is a measure of the average kinetic energy—the energy due to motion—of the atoms or molecules that compose the matter. The water molecules in a glass of hot water are, on average, moving faster than the molecules in a cold glass of water. Temperature is an indirect measure of this molecular motion. −273.15 °C 0K Absolute zero Temperature also determines the direction of thermal energy transfer, or what we commonly call heat. Thermal energy transfers from hot objects to cold ones. For example, when you touch another person’s warm hand (and yours is cold), thermal energy flows from their hand Celsius Kelvin to yours, making your hand feel warmer. However, if you touch an ice cube, thermal energy flows out of your hand to the ice, cooling your ▲ FIGURE 1.4  Comparison of the Celsius and Kelvin hand (and possibly melting some of the ice cube). Temperature Scales  The zero point of the Kelvin scale is In Figure 1.4 ▶, two common temperature scales are displayed. absolute zero (the lowest possible temperature), whereas the Most people (including scientists) use the Celsius (°C) scale. On this zero point of the Celsius scale is the freezing point of water. scale, pure water freezes at 0 °C and boils at 100 °C (at sea level). Room temperature is approximately 20 °C. The SI unit for temperature, as we have seen, is the kelvin. The Kelvin scale (some- Molecular motion does not completely stop times called the absolute temperature scale) avoids negative temperatures by assigning 0 K at absolute zero because of the uncertainty to the coldest temperature possible, absolute zero. Absolute zero (exactly -273.15 °C) is principle in quantum mechanics, which we will discuss in Chapter 7. the temperature at which translational motion of molecules stops. Lower temperatures do not exist. The size of the kelvin is identical to that of the Celsius degree—the only difference is the temperature that each designates as zero. You can convert between the two Note that we give Kelvin temperatures in kelvins (not “degrees kelvin”) or K (not °K). temperature scales with the following formula: TC T = + 273.15 K °C where T is the temperature in K and TC is the temperature in °C.

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Throughout this text, you will see examples worked out in formats that are designed to help you develop problem-solving skills. The most common format uses two columns to guide you through the worked example. The left column describes the thought processes and steps used in solving the problem, while the right column shows the implementation. The first example in this two-column format follows. EXAMPLE  1.1

CONVERTING BETWEEN TEMPERATURE SCALES

A sick child has a dangerous temperature of 40.00 °C. What is the child’s temperature in K? SOLUTION  Begin by finding the equation that relates the quantity that is given 1°C2 and the quantity you are trying to find (K). Since this equation gives the temperature in K directly, substitute in the correct value for the Celsius temperature and compute the answer.

TC T = + 273.15 K °C T K T K T K T

Tc + 273.15 °C 40.00 °C = + 273.15 °C =

= 313.15 = 313.15 K

FOR PRACTICE 1.1 Gallium is a solid metal at room temperature, but will melt to a liquid in your hand. The melting point of gallium is 303.0 K. What is this temperature in the Celsius scale?

SI Prefixes Scientific notation (see Appendix IA) allows us to express very large or very small quantities in a compact manner by using exponents. For example, the diameter of a proton can be written as 1.6 * 10-15 m. The International System of Units uses the prefix multipliers, shown in Table 1.2, with the standard units. These multipliers change the value of the unit

TABLE 1.2  Metric Prefixes Prefix

Symbol

Multiplier

exa

E

1 000 000 000 000 000 000

1018

peta

P

1 000 000 000 000 000

1015

tera

T

1 000 000 000 000

1012

giga

G

1 000 000 000

109

mega

M

1 000 000

106

kilo

k

1000

103

hecto

h

100

102

deca

da

10

101

deci

d

0.1

10-1

centi

c

0.01

10-2

milli

m

0.001

10-3

micro

m

0.000 001

10-6

nano

n

0.000 000 001

10-9

pico

p

0.000 000 000 001

10-12

femto

f

0.000 000 000 000 001

10-15

atto

a

0.000 000 000 000 000 001

10-18

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by powers of 10. For example, the kilometre has the prefix “kilo,” meaning 1000 times or 103 times. Therefore, 1 kilometre = 1000 metres = 103 metres Similarly, the millimetre has the prefix “milli,” meaning 0.001 times or 10-3 times. 1 millimetre = 0.001 metres = 10-3 metres

Conversions Involving the SI Prefixes When reporting the value of a measurement, we typically choose a prefix multiplier close to the size of the quantity being measured. For example, distances between cities are conveniently given in kilometres, but molecular dimensions such as bond length are typically reported in picometres. The distance between two singly bonded carbon atoms is 1.54 * 10-10 m, but this is sometimes more conveniently reported as 154 picometres (154 pm). Knowing how to work with units of measurement is central to solving chemical problems. Even though many different units are used to report measured values, the SI base units of length, mass, and time are metre, kilogram, and second respectively. You will soon see that it is necessary to be able to convert between different units, and that this is a skill you must master right away. We can use an example to illustrate how to convert between units using the metric prefixes. A hydrogen atom is 106 pm in diameter. To convert that diameter in pm to a value in m we must first find a conversion factor, a value that when multiplied by a measured quantity changes it to a different unit of measure. From Table 1.2, we find that 1 pm = 10-12 m Another way of expressing this is as a ratio. To do this, divide the side of the equality that has the unit you want to convert to, by the side of the equality you want to convert from. 10-12 m m = 10-12 pm 1 pm

or

10-12 m pm-1

Then, to convert the value from pm to m, we multiply by this factor. 106 pm * 10-12 m pm-1 = 106 * 10-12 m or 1.06 * 10-10 m

m and m pm-1 are both valid ways of pm expressing one unit divided by another, since 1 = pm-1. You will see both ways in pm science, and for the most part, the latter is the way you will see units expressed in this text.

Using units as a guide to solving problems is often called dimensional analysis. Units should always be included in calculations; they are multiplied, divided, and cancelled like any other algebraic quantity.

PROCEDURE FOR … Converting Between Different SI Prefixes

EXAMPLE  1.2 

EXAMPLE  1.3 

Converting Between Different SI Prefixes

Converting Between Different SI Prefixes

In 2018, half of the Nobel Prize in Physics was awarded to Donna Strickland and Gérard Mourou for their method of generating high-intensity, ultrashort laser pulses. The record for the shortest laser pulse is 43 as (Gaumnitz et al., Optics Express, 2017, 25, 27506).

If DNA from a single cell were stretched out, it would have a length of about 2 m and a diameter of about 2.4 nm. What is the diameter of the stretched DNA molecule in mm?

Convert this pulse duration to fs. Begin by writing down the important information that is provided in the question as well as the information you are asked to find.

GIVEN:  43 as

GIVEN:  2.4 nm

FIND:  time in fs

FIND:  diameter in mm

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(continued)

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PROCEDURE FOR …

EXAMPLE 1.2 

(continued )

Next, we need to derive a conversion factor to convert from the unit the quantity is expressed in to the desired unit. Table 1.2 has the SI prefixes. In each of these questions, we will require two conversion factors.

CHECK  You should always take a minute to make sure your answer makes sense.

(continued )

EXAMPLE 1.3 

From Table 1.2, we see that 1 as = 10-18 s. This allows us to first convert from as to s by dividing the side 10-18 s with s by the side with as, or 1 as 10-18 s as-1. Also from Table 1.2, we see that 1 fs = 10-15 s. This allows a conversion from s to fs by dividing the side with fs by 1 fs the side with s, -15 or 1015 fs s-1. 10 s

(continued )

From Table 1.2, we see that 1 nm = 10-9 m. This allows us to first convert from nm to m by dividing the side with m by the side with 10-9 m nm, or 10-9 m nm-1. 1 nm Also from Table 1.2, we see that 1 mm = 10-6 m. This allows a conversion from m to mm by dividing the side with mm by 1 mm the side with m, -6 or 106 mm m-1. 10 m

SOLVE:  43 as * 110-18 s as-12 * 11015 fs s-12 = 4.3 * 10-2 fs (or 43 * 10-3 fs)

SOLVE:  2.4 nm * 110-9 m nm-12 * 1106 mm m-12 = 2.4 * 10-3 mm

In this case, we wanted to convert from a smaller unit, as, to a larger unit, fs. It makes sense that the number has become smaller.

In this case, we wanted to convert from a smaller unit, nm, to a larger unit, mm. It makes sense that the number has become smaller.

FOR PRACTICE 1.2  The world’s longest carbon–carbon bond is 1.704 * 10-10 m (Schreiner et al., Nature, 2011, 477, 308). Convert this bond length to nm and to pm.

FOR PRACTICE 1.3  A chemist requires 150 mg of a certain chemical each time she conducts an experiment. She has 11.0 g of the chemical available. How many times can she repeat the experiment?

Derived Units Derived units are either combinations of different units or multiples of units of the same type. For example, the SI unit for speed is metres per second (m s -1). Notice that this unit is formed from two other SI base units—metres and seconds—put together. You are probably more familiar with speed in kilometres per hour—also an example of a derived unit. Other common derived units in chemistry are those for volume and density. We will look at each of these individually.

10 cm

Volume   The measure of space is referred to as volume. Any unit of length, when cubed (raised to the third power), becomes a unit of volume. Thus, the cubic metre (m3) and cubic centimetre (cm3) are units of volume. The cubic nature of volume is not always intuitive, and we have to think abstractly in order to think about volume. For example, consider the following question: How many small cubes measuring 1 cm on each side are required to construct a large cube measuring 10 cm (or 1 dm) on each side? The answer to this question, as you can see by carefully examining the unit cube in Figure 1.5 ◀, is 1000 small cubes. When you go from a linear, one-dimensional distance to three-dimensional volume, you must raise both the linear dimension and its unit to the third power. Thus, the volume of a cube is equal to the length of its edge cubed: volume of cube = 1edge length23

3 3 1 cm A cube with a 10 cm edge length has a volume of 110 cm2 (or 1000 cm ), and a cube

A 10 cm cube contains 1000 1 cm cubes.

▲ FIGURE 1.5  The Relationship

Between Length and Volume

with a 100 cm edge length has a volume of 1100 cm23 (or 1 * 106 cm3). To convert from m3 to mm3, we construct the conversion factor as follows: 1 m = 1000 mm

11 m23 = 11000 mm23 1 m3 = 109 mm3

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TABLE 1.3  Some Common Derived Units Quantity

Unit Name

Symbol

density

kilogram per cubic metre

kg m-3

gram per cubic centimetre

g cm-3

gram per millilitre

g mL-1

metre per second

m s-1

kilometre per hour

km h-1

cubic centimetre, millilitre

cm3 = mL

decimetre cubed, litre

dm3 = L

cubic metre, kilolitre

m3 = kL

speed

volume

Then, express this equality as a ratio by dividing the side of the equality with the units you want to convert to, by the side of the equality you want to convert from: 1 m3 = 109 mm3 109 mm3 m-3 This is the conversion factor to convert from m3 to mm3. You should also realize that to convert from mm3 to m3, the conversion factor is just the inverse: 10-9 m3 mm-3 Other common units of volume in chemistry are the litre (L), which is equivalent to 1 dm3; and the millilitre (mL), which is equivalent to 1 cm3. One m3 is equivalent to a kilolitre (kL). The common units of volume are listed in Table 1.3. Density   An old riddle asks, “Which weighs more, a tonne of bricks or a tonne of feathers?” The answer, of course, is neither—they both weigh the same (1 tonne). If you answered bricks, you confused weight with density. The density (d ) of a substance is the ratio of its mass (m) to its volume (V): density =

mass volume

or

d =

m V

Density is a characteristic physical property of materials and differs from one substance to another, as you can see in Table 1.4. The density of a substance also depends on its temperature. Density is an example of an intensive property, one that is independent of the amount of the substance. The density of aluminum, for example, is the same whether you have an ounce or a tonne. Intensive properties are often used to identify substances because these properties depend only on the type of substance, not on the amount of it. For example, from Table 1.4 you can see that pure gold has a density of 19.3 g cm-3. One way to determine whether a substance is pure gold is to measure its density and compare it to 19.3 g cm-3. Mass, in contrast, is an extensive property, one that depends on the amount of the substance. The units of density are those of mass divided by volume. Although the SI-derived unit for density is kg m-3, the density of liquids and solids is most often expressed in g cm-3 or g mL-1 (cm3 and mL are equivalent units; 1 cm3 = l mL). Aluminum is one of the least dense structural metals with a density of 2.70 g cm-3, while platinum is one of the densest metals with a density of 21.4 g cm-3. We calculate the density of a substance by dividing the mass of a given amount of the substance by its volume. For example, suppose a small nugget we suspect to be gold

The m in the equation for density is in italic type, meaning that it stands for mass rather than for metres. In general, the symbols for units such as metres (m), seconds (s), or kelvins (K ) appear in regular type, while those for variables such as mass (m ), volume (V  ), and time (t  ) appear in italics.

TABLE 1.4  Density of Some

Common Substances at 20 °C*

Substance

Density (g cm−3)

Charcoal (from oak)

0.57

Ethanol

0.789

Ice

0.917 (at 0 °C)

Water

1.00 (at 4 °C)

Sugar (sucrose)

1.58

Glass

2.6

Aluminum

2.70

Copper

8.96

Gold Iron

19.3 7.86

Lead

11.4

Mercury

13.55

Platinum

21.4

Titanium

4.51

* Unless otherwise indicated.

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has a mass of 22.5 g and a volume of 2.38 cm3. To find its density, we divide the mass by the volume: d =

22.5 g m = = 9.45 g cm-3 V 2.38 cm3

In this case, the density reveals that the nugget is not pure gold. EXAMPLE  1.4

CALCULATING DENSITY AND CONVERTING DENSITY UNITS

A student conducts an experiment to determine the density of air. He weighs an evacuated flask and one filled with air at ambient pressure, and determines the difference in mass (corresponding to the mass of air) to be 2.27 g. By displacement of water, he determines the volume of the flask to be 1.68 L. (a) Determine the density of air in g L-1. (b) Convert the density to units of g cm-3 in order to compare it to the density of a solid like aluminum. SOLUTION  Set up the problem by writing the important information that is given as well as the information that you are asked to find. Here, we are asked to find the density of air in the flask and then to convert the units of g L-1 to g cm-3 to compare to the density of aluminum metal. Next, write down the equation that defines density. (a) Solve the problem by substituting the correct values of mass and volume into the expression for density. (b) Next, determine a conversion factor to convert L to cm3 and apply this conversion factor to the quantity you need to convert. Since 1 cm3 = 1 mL, converting from L to mL is the same as converting from L to cm3.

GIVEN:  m = 2.27 g V = 1.68 L FIND:  (a) density in g L-1 (b) density in g cm-3 m EQUATION:  d = V SOLVE:  d =

2.27 g = 1.35 g L-1 1.68 L 1 L = 103 mL 1 L = 103 cm3

Thus, the conversion factor is 10 - 3L cm-3. 1.35 g L - 1 * 10-3 L cm-3 = 1.35 * 10-3 g cm-3 The density of aluminum is 2.70 g cm-3 and, as such, is 2000 times as dense as air.

CHECK  Ensure your answer makes sense. We know that air is not nearly as dense as a metal, so our answer, which says that air is 2000 times less dense than aluminum, makes sense. FOR PRACTICE 1.4 The mass of a helium-4 nucleus is 6.7 * 10-27 kg and the radius is 1.9 fm (femtometres). Calculate the density of a helium nucleus in kg fm-3 and in g cm-3. Assume the atom is a sphere (volume = 14>32pr3).

FOR MORE PRACTICE 1.4 A metal cube has an edge length of 11.4 mm and a mass of 6.67 g. Calculate the density of the metal and use Table 1.4 to determine the likely identity of the metal.

CONCEPTUAL CONNECTION 1.2 Density The density of copper decreases as temperature increases (as does the density of most substances). Which of the following will be true upon changing the temperature of a sample of copper from room temperature to 95 °C? (a) the copper sample will become lighter (b) the copper sample will become heavier (c) the copper sample will expand (d) the copper sample will contract

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The units of a quantity should always be written with the number, even when performing a calculation. Let’s return to the Gimli Glider example that we discussed at the beginning of the section. The correction factor that the ground crew were using to convert the mass of jet fuel to a volume was actually the density of the jet fuel, 1.77 lb L-1; however, it was expressed in incorrect units. Had the ground crew kept the units with their quantities, they would have immediately realized their mistake. V =

22 300 kg m = = 1.26 * 104 L kg lb - 1 d 1.77 lb L-1

The units of the answer are clearly not appropriate for the quantity they were trying to obtain, which should have the units of L, not L kg lb-1. The calculation with the density expressed in the correct units, 0.803 kg L-1, is: V =

22 300 kg m = = 2.78 * 104 L d 0.803 kg L-1

The volume of fuel calculated by the ground crew was less than half of the actual requirement. Derived Units with Special Names  Compiled in Table 1.5 are some common units in chemistry that are derived from the SI base units, which are also given special names. For example, the joule (J) is a derived unit and can be expressed in SI base units: J = kg m2 s-2 It is important to become familiar with how these derived units are expressed in the base units. For example, you know that energy or work (w) is a force (F) acting over a distance (d), or: w = F * d The unit of force is the newton (N) and the unit of distance is the metre (m), so the unit of work would be N # m. If we express the newton in its SI base units, work has the units: kg m s-2 # m = kg m2 s-2 = J

CHEMISTRY AND MEDICINE  Bone Density Osteoporosis—which means porous bone—is a condition in which bone density becomes too low. The healthy bones of a young adult have a density of about 1.0 g cm-3. Patients suffering from osteoporosis, however, can have bone densities as low as 0.22 g cm-3. These low densities mean the bones have deteriorated and weakened, resulting in increased

susceptibility to fractures, especially hip fractures. Patients suffering from osteoporosis can also experience height loss and disfiguration such as dowager’s hump, a condition in which the patient becomes hunched over due to compression of the vertebrae. Osteoporosis is most common in postmenopausal women, but it can also occur in people (including men) who have certain diseases, such as insulin-dependent diabetes, or who take certain medications, such as prednisone. Osteoporosis is usually diagnosed and monitored with hip X-rays. Low-density bones absorb fewer of the X-rays than do highdensity bones, producing characteristic differences in the X-ray image. Treatments for osteoporosis include additional calcium and vitamin D, drugs that prevent bone weakening, exercise and strength training, and, in extreme cases, hipreplacement surgery.

Question ▲ Magnified views of the bone matrix in a normal femur (left) and one weakened by osteoporosis (right).

Suppose you find a large animal bone in the woods, too large to fit in a beaker or flask. How might you approximate its density?

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TABLE 1.5  Common Derived Units with Special Names Quantity

Unit Name

Symbol

Expression in Terms of SI Base Units

energy, work, heat

joule

J

kg m2 s-2

pressure

pascal

Pa

kg m-1 s-2

bar

bar

( = 105 Pa = 105 kg m-1 s-2)

kilopascal

kPa

( = 103 Pa = 103 kg m-1 s-2)

force

newton

N

kg m s-2

electric charge

coulomb

C

As

voltage, electrochemical potential

volt

V

kg m2 s-2 C-1

frequency

hertz

Hz

s-1

1.4  The Reliability of a Measurement One thing we spend a lot of time doing is checking the temperature. Is it cold outside, just right, or hot? In fact, we keep records of temperatures and these records August Temperatures in St. John’s, NL date back to before the 1950s for many cities in the world. Table 1.6 lists the maxiand Windsor, ON mum and average temperatures experienced for the month of August over a oneMax (and Avg) Temp (°C) decade time period beginning in 2011 for both St. John’s, NL—a coastal city in the North Atlantic Ocean—and Windsor, ON, a city that is more centrally located Year St. John’s, NL Windsor, ON in North America. 2011 26.6 (14.7) 31.0 (22.5) The first thing you will notice is that St. John’s has a coastal climate. St. John’s 2012 28.2 (17.7) 32.2 (22.0) is considerably cooler than Windsor, mainly because the Atlantic Ocean, a vast 2013 25.3 (16.4) 31.0 (21.7) heat sink, keeps it cooler in the summer, much as it keeps St. John’s warmer than 2014 27.9 (16.0) 30.6 (20.8) Windsor in the winter. The second thing you might notice is the number of digits to which the measurements are reported. The number of digits in a reported mea2015 27.3 (17.3) 32.5 (21.0) surement indicates the certainty associated with that measurement. For example, 2016 25.7 (15.9) 33.8 (23.6) a less certain measurement of the maximum August temperatures for St. John’s 2017 26.3 (15.4) 30.3 (20.5) and Windsor in 2020 might be 29 and 34 °C. These temperatures are reported to 2018 29.3 (17.7) 32.4 (23.1) the nearest degree Celsius, while the data in Table 1.6 are reported to the nearest 2019 28.2 (16.6) 31.0 (22.0) 0.1 °C. Scientists agree on a standard way of reporting measured quantities in which 2020 28.6 (17.2) 33.6 (22.0) the number of reported digits reflects the certainty in the measurement: more digSource: https://www.weatherstats.ca/ its, more certainty; fewer digits, less certainty. Numbers are usually written so that the uncertainty is {1 in the last reported digit unless otherwise indicated. By reporting the temperature as 26.2 °C, scientists mean 26.2{ 0.1 °C. This means that the temperature is between 26.1 and 26.3 °C—it might be 26.3 °C, for example, but it could not be 27.0 °C. In contrast, if the reported value was 26 °C, this would mean 26{1 °C, or between 25 and 27 °C. In general, TABLE 1.6  Maximum and Average

scientific measurements are reported so that every digit is certain except the last, which is estimated. For example, consider the following reported number: 5.213 certain

estimated

The first three digits are certain; the last digit is estimated. The number of digits reported in a measurement depends on the measuring device. Consider weighing a pistachio nut on two different balances (Figure 1.6 ▶). The balance on the left has marks every 1 g, while the balance on the right has marks every 0.1 g. For the balance on the left, we mentally divide the space between the 1 and 2 g marks into ten equal spaces and estimate that the pointer is at about 1.2 g. We then write the measurement as 1.2 g, indicating that we are sure of the “1” but have estimated the “.2”.

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◀ FIGURE 1.6  Estimation in Weighing

(a) This scale has markings every 1 g, so we estimate to the tenths place by mentally dividing the space into ten equal spaces to estimate the last digit. This reading is 1.2 g. (b) Because this balance has markings every 0.1 g, we estimate to the hundredths place. This reading is 1.27 g.

(a)

(b)

Markings every 1 g Estimated reading 1.2 g

Markings every 0.1 g Estimated reading 1.27 g

The balance on the right, with marks every tenth of a gram, requires us to write the result with more digits. The pointer is between the 1.2 g mark and the 1.3 g mark. We again divide the space between the two marks into ten equal spaces and estimate the third digit. For the figure shown, we report 1.27 g.

EXAMPLE  1.5

REPORTING THE CORRECT NUMBER OF DIGITS

The graduated cylinder shown at right has markings every 0.1 mL. Report the volume (which is read at the bottom of the meniscus) to the correct number of digits. (Note: the meniscus is the crescent-shaped surface at the top of a column of liquid.) SOLUTION Since the bottom of the meniscus is between the 4.5 and 4.6 mL markings, mentally divide the space between the markings into ten equal spaces and estimate the next digit. In this case, you should report the result as 4.57 mL. What if you estimated a little differently and wrote 4.56 mL? In general, one unit difference in the last digit is acceptable because the last digit is estimated and different people might estimate it slightly differently. However, if you wrote 4.63 mL, you would have misreported the measurement.

Meniscus

FOR PRACTICE 1.5 Record the temperature on the thermometer shown at right to the correct number of digits.

Counting Significant Figures The certainty of a measurement—which depends on the instrument used to make the measurement—must be preserved, not only when recording the measurement, but also when performing calculations that use the measurement. We can accomplish the preservation of this certainty by using significant figures. In any reported measurement, the nonplaceholding digits—those that are not simply marking the decimal place—are called significant figures (or significant digits). The greater the number of significant figures, the greater the certainty of the measurement. For example, the number 23.5 has three significant figures, while the number 23.56 has four. To determine the number of significant figures in a number containing zeros, we must distinguish between zeros that are significant and those that simply mark the decimal place. For example,

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in the number 0.0008, the leading zeros mark the decimal place but do not add to the certainty of the measurement and are therefore not significant; this number has only one significant figure. In contrast, the trailing zeros in the number 0.000800 do add to the certainty of the measurement and are therefore counted as significant; this number has three significant figures. To determine the number of significant figures in a number, follow these rules (with examples shown on the right): Significant Figure Rules 1.  All nonzero digits are significant. 2. Zeros between nonzero digits are significant. 3. Zeros to the left of the first nonzero digit are not significant. They only serve to locate the decimal point. 4. Zeros at the end of a number are categorized as follows: a) Zeros after a decimal point are always significant. b) Zeros before a decimal point and after a nonzero number are always significant. c) Zeros before an implied decimal point are ambiguous and scientific notation should be used.

Examples 28.03 0.0540 408 7.0301 0.0032

0.0000 6

not significant

45.000

3.5600

140.00

2500.55

1200 ambiguous 1.2 * 103 2 significant figures 1.20 * 103 3 significant figures 1.200 * 103 4 significant figures

When reporting temperatures, the previous rules (specifically 4c) may not be obeyed. When a temperature such as 300 K is reported, according to rule 4c, the number of significant figures is ambiguous; there could be one or three significant figures. In chemistry, temperature is almost always measured to an uncertainty of at least {1 °C (or K). Therefore, if a temperature such as 100 °C, 300 K, or 420 K is reported, then there are three significant figures. Of course, if a temperature such as 300.0 K is given, then there are four significant figures.

Exact Numbers Exact numbers have no uncertainty, and thus do not limit the number of significant figures in any calculation. We can regard an exact number as having an unlimited number of significant figures. Exact numbers originate from three sources: ▶ ▶ From the accurate counting of discrete objects. For example, 3 atoms means

3.00000 . . . atoms. ▶ ▶ From defined quantities, such as the number of centimetres in 1 m. Because 100 cm

is defined as 1 m, 100 cm = 1 m means 100.000000 . . . cm = 1.000000 . . . m Another example is the conversion from °C to K, which is to add exactly 273.15 or 273.15000000. . . . ▶ ▶ From integral numbers that are part of an equation. For example, in the equation

diameter , the number 2 is exact and therefore has an unlimited number of 2 significant figures.

radius =

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EXAMPLE  1.6

DETERMINING THE NUMBER OF SIGNIFICANT FIGURES IN A NUMBER

How many significant figures are in each number? (a) (b) (c) (d)

(e) 0.00002 mm (f) 10 000 m (g) 350 K

0.04450 m 5.0003 km 100 cm m-1 1.000 * 105 s

(a) 0.04450 m

Four significant figures. The two 4s and the 5 are significant (rule 1). The last zero is after a decimal point and is therefore significant (rule 4a). The leading zeros only mark the decimal place and are therefore not significant (rule 3). Five significant figures. The 5 and 3 are significant (rule 1) as are the three interior zeros (rule 2). Unlimited significant figures. Defined quantities have an unlimited number of significant figures.

(b) 5.0003 km (c) 100 cm m-1 (d) 1.000 * 105 s (e) 0.00002 mm (f) 10 000 m

(g) 350 K

Four significant figures. The 1 is significant (rule 1). The trailing zeros are after a decimal point and therefore significant (rule 4a). One significant figure. The 2 is significant (rule 1). The leading zeros only mark the decimal place and are therefore not significant (rule 3). Ambiguous. The 1 is significant (rule 1) but the trailing zeros occur before an implied decimal point and are therefore ambiguous (rule 4). Without more information, we would assume one significant figure. It is better to write this than to indicate a specific number of significant figures (rule 4c). Three significant figures. Temperature can be assumed to be read to the nearest 1 °C at least.

FOR PRACTICE 1.6 How many significant figures are in each number? (e) 1.4500 km (a) 554 km (f) 21 000 m (b) 7 pennies 5 (g) 200 °C (c) 1.01 * 10 m (d) 0.00099 s

Significant Figures in Calculations When you use measured quantities in calculations, the results of the calculation must reflect the precision of the measured quantities. You should not lose or gain certainty during mathematical operations. Follow these rules when carrying significant figures through calculations.

Rules for Calculations 1. In multiplication or division, the result carries the same number of significant figures as the factor with the fewest significant figures. For example: 1.052

*

*

12.054

0.53

=

6.7208 = 6.7

(4 sig. figures)  (5 sig. figures)  (2 sig. figures)      (2 sig. figures) 2.0035 ,

3.20

=

0.626094 = 0.626

(5 sig. figures) (3 sig. figures)       (3 sig. figures) 2. In addition or subtraction, the result carries the same number of decimal places as the quantity with the fewest decimal places. Here are a few examples: 2.345 0.07 98.6122 5.9       2.9975 3.11 -0.221 5.4125 = 5.41 101.7222 = 101.72 5.679 = 5.7

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3. When computing the logarithm of a number, the result has the same number of significant figures after the decimal point (called the mantissa) as there are significant figures in the number whose logarithm is being calculated. The digits before the decimal place in the answer represent the order of magnitude. For example: In this text, log means take the base 10 logarithm (i.e., log10) and ln means take the natural logarithm.

log(45.250) = log(4.5250 * 101) = 1.65562 (5 sig. figures after the decimal) log(1106.26) = log(1.10626 * 103) = 3.043857 (6 sig. figures after the decimal) log(6.2 * 1010) = 10.79 (2 sig. figures after the decimal) The same is true for taking the natural logarithm: ln(6.89 * 105) = 13.443 (3 sig. figures after the decimal)

e is a mathematical constant. To 5 significant figures, e = 2.7183. Natural base logarithms and antilogarithms are used quite often in chemistry, so it is useful to become familiar with them.

4. When computing the antilogarithm (the inverse of the logarithm), the number of significant figures in the result is the same as the number of significant digits after the decimal (the mantissa). The number before the decimal place represents the order of magnitude. For example: 106.125 = 1.33 * 106 10-10.00 = 1.0 * 10-10

(3 sig. figures) (2 sig. figures)

e12.2 = 2 * 105

(1 sig. figure)

Rules for Rounding A common rule when rounding a large number of values to be averaged is the “round to even” rule in which 5.35 would round to 5.4 and 5.45 would round to 5.4 since the last significant figure before the 5 is an even number. We will use the simpler rule in this text since it is consistent with most calculators.

When rounding to the correct number of significant figures, round down if the last (or rightmost) digit dropped is four or less; round up if the last (or rightmost) digit dropped is five or more. To two significant figures: 5.37 rounds to 5.4 5.34 rounds to 5.3 5.35 rounds to 5.4 5.349 rounds to 5.3 Notice in the last example that only the rightmost digit being dropped determines in which direction to round—ignore all digits to the right of it. To avoid rounding errors in multistep calculations, round only the final answer—do not round intermediate steps. If you write down intermediate answers, keep track of significant figures by underlining the last significant digit. 6.78 * 5.903 * (5.489 - 5.01) = 6.78 * 5.903 * 0.479 = 19.1707 = 19 underline last significant digit Notice that for multiplication or division, the quantity with the fewest significant figures determines the number of significant figures in the answer, but for addition and subtraction, the quantity with the fewest decimal places determines the number of decimal places in the answer. In multiplication and division, we focus on significant figures, but in addition and subtraction, we focus on decimal places. When a problem involves addition or subtraction, the answer may have a different number of significant figures than the initial quantities. Keep this in mind when working problems that involve both addition or subtraction and multiplication or division. For example, 1.002 - 0.999 0.003 = 3.754 3.754 = 7.99 * 10-4 = 8 * 10-4 The answer has only one significant figure, even though the initial numbers had three or four.

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EXAMPLE  1.7

SIGNIFICANT FIGURES IN CALCULATIONS

Perform each calculation and report the answer to the correct number of significant figures. (a) 1.10 * 0.5120 * 4.0015 , 3.4555 (b) 0.355 + 105.1 - 100.5820 (c) 4.562 * 3.99870 , (452.6755 - 452.33) (d) log(16.45) + 2.34 (e) e8.125 * 2.551 * 10-3 (a) Round the intermediate result (in blue) to three significant figures to reflect the three significant figures in the least precisely known quantity (1.10).

1.10 * 0.5120 * 4.0015 , 3.4555 = 0.65219 = 0.652

(b) Round the intermediate answer (in blue) to one decimal place to reflect the quantity with the fewest decimal places (105.1). Notice that 105.1 is not the quantity with the fewest significant figures, but it has the fewest decimal places and therefore determines the number of decimal places in the answer.

0.355 105.1 -100.5820 4.8730 = 4.9

(c) Mark the intermediate result to two decimal places to reflect the number of decimal places in the quantity within the parentheses having the fewest number of decimal places (452.33). Round the final answer to two significant figures to reflect the two significant figures in the least precisely known quantity (0.3455).

4.562 * 3.99870 , (452.6755 - 452.33) = 4.562 * 3.99870 , 0.3455 = 52.79904 = 53

(d) Mark the intermediate result to the fourth digit after the decimal to reflect the number of significant figures in the result of the logarithm. Round the final answer to two decimal places to reflect the two decimal places in the least precisely known quantity (2.34).

log(16.45) + 2.34 = 1.21617 + 2.34 = 3.55617 = 3.56

(e) Mark the intermediate result to three significant figures to reflect the three significant figures in the antilogarithm. Round the final answer to three significant figures to reflect the three significant figures in the least precisely known quantity (the antilogarithm).

e8.125 = = =

* 2.551 * 10-3 3.3779 * 103 * 2.551 * 10-3 8.6169 8.62

FOR PRACTICE 1.7 Perform each calculation to the correct number of significant figures. (a) 3.10007 * 9.441 * 0.0301 , 2.31 (b) 0.881 + 132.1 - 12.02 (c) 2.5110 * 21.20 , (44.11 + 1.233) (d) ln(12.5) - 0.0284 (e) 108.1325 , 2.6 * 10-3

Precision and Accuracy Scientists often repeat measurements several times to increase confidence in the result. We can distinguish between two different kinds of certainty—called accuracy and precision—associated with such repeated measurements. Accuracy refers to how close the measured value is to the actual value. Precision refers to how close a series of measurements are to one another or how reproducible they are. A series of measurements can be precise (close to one another in value, and reproducible) but not accurate (not close to the true value). Consider the results of three students who repeatedly weighed a lead block known to have a true mass of 10.00 g (indicated by the solid horizontal blue line on the graphs on the next page).

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18       Cha pter Get 1   Unitcomplete s o f Me a s ur e meeBook nt fo r Phy siOrder c al an d C hby em i c al C h an g e at email Inaccurate, imprecise

Inaccurate, precise

11

11

Mass (g)

Average 9.79 g

10.49 10.31

10

9.79

True mass

9.92

9.5

9

Accurate, precise

11 Average 10.13 g

10.5

[email protected]

Average 10.01 g

10.5

10.5

10

10

9.82

9.78

9.75

9.80

2 3 Trial number

9

4

1

2 3 Trial number

Student A

Student B

Student A

Student B

Student C

Trial 1

10.49 g

9.78 g

10.03 g

Trial 2

9.79 g

9.82 g

9.99 g

Trial 3

9.92 g

9.75 g

10.03 g

Trial 4

10.31 g

9.80 g

9.98 g

Average

10.13 g

9.79 g

10.01 g

9.99

10.03

9.98

9.5

9.5

1

10.03

4

9

1

2 3 Trial number

4

Student C

▶ ▶ The results of student A are both inaccurate (not close to the

true value) and imprecise (not consistent with one another). The inconsistency is the result of random error, error that has equal probability of being too high or too low. Almost all measurements have some degree of random error. Random error can, with enough trials, average itself out. ▶ ▶ The results of student B are precise (close to one another in

value) but inaccurate. The inaccuracy is the result of systematic error, error that tends toward being either too high or too low. Systematic error does not average out with repeated trials. For instance, if a balance is not properly calibrated, it may systematically read too high or too low. ▶ ▶ The results of student C display little systematic error or random error—they are

both accurate and precise.

THE NATURE OF SCIENCE  Integrity in Data Gathering Most scientists spend many hours collecting data in the laboratory. Often, the data do not turn out exactly as the scientist had expected (or hoped). A scientist may then be tempted to “fudge” his or her results. For example, suppose you are expecting a particular set of measurements to follow a certain pattern. After working hard over several days or weeks to make the measurements, you notice that a few of them do not quite fit the pattern that you anticipated. You might find yourself wishing that you could simply change or omit the “faulty” measurements. Altering data in this way is considered highly unethical in the scientific community and, when discovered, the scientist is usually punished severely. In 2004, Dr. Hwang Woo Suk, a stem cell researcher at the Seoul National University in Korea, published a research paper in Science (a highly respected research journal) claiming that he and his colleagues had cloned human embryonic stem cells. As part of his evidence, he showed photographs of the cells. The paper was hailed as an incredible breakthrough, and Dr. Hwang travelled the world lecturing on his work. Time magazine even listed him among their “people that matter” for 2004. Several

months later, however, one of his co-workers revealed that the photographs were fraudulent. According to the co-worker, the photographs came from a computer data bank of stem cell photographs, not from a cloning experiment. A university panel investigated the results and confirmed that the photographs and other data had indeed been faked. Dr. Hwang was forced to resign his prestigious post at the university. Although not common, incidents like this do occur from time to time. They are damaging to a community that is largely built on trust. A scientist’s peers (other researchers in similar fields) review all published research papers, but usually they are judging whether the data support the conclusion—they assume that the experimental measurements are authentic. The pressure to succeed sometimes leads researchers to betray that trust. However, over time, the tendency of scientists to reproduce and build upon one another’s work results in the discovery of the fraudulent data. When that happens, the researchers at fault are usually banished from the community and their careers are ruined.

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1.5  Solving Chemical Problems Learning to solve problems is one of the most important skills you will acquire in this course. No one succeeds in chemistry—or in life, really—without the ability to solve problems. Although no simple formula applies to every chemistry problem, you can learn problem-solving strategies and begin to develop some chemical intuition. Many of the problems you will solve in this course can be thought of as unit conversion problems, where you are given one or more quantities and asked to convert them into different units. Other problems require that you use specific equations to get to the information you are trying to find. In the sections that follow, you will find strategies to help you solve both of these types of problems. Of course, many problems contain both conversions and equations, requiring the combination of these strategies, and some problems may require an altogether different approach.

General Problem-Solving Strategy In this text, we use a standard problem-solving procedure that can be adapted to many of the problems encountered in general chemistry and beyond. To solve any problem, you need to assess the information given in the problem and devise a way to get to the information asked for. In other words, you must: ▶ ▶ Identify the starting point (the given information). ▶ ▶ Identify the end point (what you must find). ▶ ▶ Devise a way to get from the starting point to the end point using what is given as

well as what you already know or can look up. (We call this the conceptual plan.) In graphic form, we can represent this progression as: Given S Conceptual Plan S Find One of the main difficulties students have when trying to solve problems in general chemistry is not knowing where to start. While no problem-solving procedure is applicable to all problems, the following four-step procedure can be helpful in working through many of the numerical problems you will encounter in this text. 1. Sort.  Begin by sorting the information in the problem. Given information is the basic data provided by the problem—often one or more numbers with their associated units. Find indicates what information you will need for your answer. 2. Strategize.  This is usually the hardest part of solving a problem. In this process, you must develop a conceptual plan—a series of steps that will get you from the given information to the information you are trying to find. You have already seen conceptual plans for simple unit conversion problems. Each arrow in a conceptual plan represents a computational step. On the left side of the arrow is the quantity you had before the step, on the right side of the arrow is the quantity you will have after the step, and below the arrow is the information you need to get from one to the other—the relationship between the quantities.   Often, such relationships will take the form of conversion factors or equations. These may be given in the problem, in which case you will have written them down under “Given” in step 1. Usually, however, you will need other information—which may include physical constants, formulas, or conversion factors—to help get you from what you are given to what you must find. This information comes from what you have learned or can look up in the chapter or in tables within the text. In some cases, you may get stuck at the strategize step. If you cannot figure out how to get from the given information to the information you are asked to find, you might try working backward. For example, you may want to look at the units of the quantity you are trying to find and try to find conversion factors to get to the units of the given quantity. You may even try a combination of strategies: work forward, backward, or some of both. If you persist, you will develop a strategy to solve the problem. 3. Solve.  This is the easiest part of solving a problem. Once you set up the problem properly and devise a conceptual plan, you simply follow the plan to solve the problem.

Most problems can be solved in more than one way. The solutions we derive in this text will tend to be the most straightforward but certainly not the only way to solve the problem.

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Carry out any mathematical operations (paying attention to the rules for significant figures in calculations) and cancel units as needed. 4. Check.  This is the step beginning students most often overlook. Experienced problem solvers always ask, Does this answer make sense? Are the units correct? Is the number of significant figures correct? When solving multistep problems, errors easily creep into the solution. You can catch most of these errors by simply checking the answer, and considering the physical meaning of your result. For example, suppose you are calculating the number of atoms in a gold coin and end up with the answer of 1.1 * 10 - 6 atoms. Could a gold coin really be composed of one-millionth of one atom? In the following pages, we apply this problem-solving procedure to unit conversion problems. The procedure is summarized in the left column, and two examples of applying the procedure are shown in the middle and right columns. This three-column format will be used in selected examples throughout this text. It allows you to see how you can apply a particular procedure to two different problems. Work through one problem first (from top to bottom) and then see how you can apply the same procedure to the other problem. Being able to see the commonalities and differences between problems is a key part of developing problem-solving skills.

Order-of-Magnitude Estimations Calculation plays a major role in chemical problem solving. But precise numerical calculation is not always necessary, or even possible. Sometimes data are only approximate, so there is no point in trying to determine an extremely precise answer. At other times, you simply don’t need a high degree of precision—a rough estimate or a simplified “back of the envelope” calculation is enough. Scientists often use these kinds of calculations to get an initial feel for a problem, or as a quick check to see whether a proposed solution is “in the right ballpark.” One way to make such estimates is to simplify the numbers so that they can be manipulated easily. The technique, known as order-of-magnitude estimation, is based on focusing only on the exponential part of numbers written in scientific notation, according to the following guidelines: ▶ ▶ If the decimal part of the number is less than 5, just drop it. Thus, 4.36 * 105

becomes 105 and 2.7 * 10 - 3 becomes 10-3.

▶ ▶ If the decimal part is 5 or more, round it up to 10 and rewrite the number as a power

of 10. Thus, 5.982 * 107 becomes 10 * 107 = 108, and 6.1101 * 10 - 3 becomes 10 * 10 - 3 = 10 - 2.

When you make these approximations, you are left with powers of 10, which are easily multiplied and divided—often in your head. It’s important to remember, however, that your answer is only as reliable as the numbers used to get it, so never assume that the results of an order-of-magnitude calculation are accurate to more than an order of magnitude. Suppose, for example, that you want to estimate the number of years it would take an immortal being to spend 1 mole (6.022 * 1023) of pennies at a rate of 100 million dollars per second. Since a year has 3.16 * 107 seconds, you can approximate the number of years as follows: 1024 pennies *

1 year 1 dollar 1 second * 8 * 7 ≈ 107 years 2 10 seconds 10 pennies 10 dollars

We have estimated that it will take about 10 million years to spend one mole of pennies at the rate of 100 million dollars per year—clearly impossible for a mere mortal. In our general problem-solving procedure, the last step is to check whether the results seem reasonable. Order-of-magnitude estimations can often help you catch the kinds of mistakes that may happen in a detailed calculation, such as entering an incorrect exponent or sign into your calculator, or multiplying when you should have divided.

Problems Involving an Equation Problems involving equations can be solved in much the same way as problems involving conversions. Usually, in problems involving equations, you must find one of the variables

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in the equation, given the others. The conceptual plan approach outlined previously can be used for problems involving equations. For example, suppose you are given the mass (m) and volume (V) of a sample and asked to calculate its density. The conceptual plan shows how the equation takes you from the given quantities to the find quantity. m,V

d d=

m V

Here, instead of a conversion factor under the arrow, this conceptual plan has an equation. The equation shows the relationship between the quantities on the left of the arrow and the quantities on the right. Note that at this point, the equation need not be solved for the quantity on the right (although in this particular case, it is). The procedure that follows, as well as the two examples, will guide you in developing a strategy to solve problems involving equations. We will use the three-column format here. Work through one problem from top to bottom and then see how you can apply the same general procedure to the second problem.

PROCEDURE FOR …

EXAMPLE   1.8

EXAMPLE   1.9

Solving Problems Involving Equations

Problems Using Equations Find the radius (r), in centimetres, of a spherical water droplet with a volume (V) of 0.058 cm3. For a sphere, V = (4>3)pr3.

Problems Using Equations Find the density, in g cm - 3, of a metal cylinder with a mass (m) of 8.3 g, a length (l ) of 1.94 cm, and a radius (r) of 0.55 cm. For a cylinder, V = pr2l.

SORT  Begin by sorting the information in the problem into given and find categories.

GIVEN:  V = 0.058 cm3 FIND:  r in cm

GIVEN: m = 8.3 g      l = 1.94 cm     r = 0.55 cm FIND:  d in g cm-3

STRATEGIZE Write a conceptual plan for the problem. Focus on the equation(s). The conceptual plan shows how the equation takes you from the given quantity (or quantities) to the find quantity. The conceptual plan may have several parts, involving other equations or required conversions. In these examples, you use the geometrical relationships given in the problem statements as well as the definition of density, d = m>V, which you learned in this chapter.

CONCEPTUAL PLAN

CONCEPTUAL PLAN

SOLVE Follow the conceptual plan. Solve the equation(s) for the find quantity (if it is not already in the proper form). Gather each of the quantities that must go into the equation in the correct units, as well as looking up any physical constants that are necessary. (Convert to the correct units if necessary.) Substitute the numerical values and their units into the equation(s) and compute the answer.

SOLUTION 4 V = pr3 3 3 3 r = V 4p

Round the answer to the correct number of significant figures.

V

r V=

4 3

V 2

V= r l

r3

RELATIONSHIPS USED V =

l,r

m,V

d d = m/V

4 3 pr 3

RELATIONSHIPS USED V = pr2l m d = V

r = a = a

3 Vb 4p

SOLUTION V = pr2l = p(0.55 cm)2(1.94 cm) 1>3

1>3 3 0.058 cm3 b 4p

= 0.24013 cm 0.24013 cm = 0.24 cm

= 1.8436 cm3 m d = V 8.3 g = = 4.50195 g cm-3 1.8436 cm3 4.50195 g cm-3 = 4.5 g cm-3

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(continued)

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PROCEDURE FOR …

EXAMPLE 1.8 

(continued )

CHECK  Check your answer. Are the units correct? Does the answer make sense?

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(continued )

EXAMPLE 1.9 

(continued )

The units (cm) are correct and the magnitude of the result makes sense.

The units (g cm-3) are correct. The magnitude of the answer seems correct for one of the lighter metals (see Table 1.4).

FOR PRACTICE 1.8 Find the radius (r) of an aluminum cylinder that is 2.00 cm long and has a mass of 12.4 g. For a cylinder, V = pr2l. Density of aluminum = 2.7 g cm-3.

FOR PRACTICE 1.9 Find the density, in g cm-3, of a metal cube with a mass of 50.3 g and an edge length (l ) of 2.65 cm. For a cube, V = l3.

CHAPTER IN REVIEW Key Terms Section 1.1

Section 1.3

physical change (2) chemical change (2) physical property (2) chemical property (2)

units (4) Imperial system (4) metric system (4) International System of Units (SI) (4) metre (m) (4) kilogram (kg) (5) mass (5) second (s) (5) kelvin (K) (5) temperature (5)

Section 1.2 kinetic energy (3) potential energy (3) thermal energy (3) law of conservation of energy (3)

Section 1.4

Celsius (°C) scale (5) prefix multipliers (6) conversion factor (7) dimensional analysis (7) derived unit (8) volume (8) litre (L) (9) millilitre (mL) (9) density (d) (9) intensive property (9) extensive property (9)

significant figures (significant digits) (13) exact numbers (14) accuracy (17) precision (17) random error (18) systematic error (18)

Key Concepts The Properties of Matter 1.1 The properties of matter can be divided into two kinds: physical and chemical. Matter displays its physical properties without changing its composition. Matter displays its chemical properties only through changing its composition. Changes in matter in which its composition does not change are called physical changes. Changes in matter in which its composition does change are called chemical changes.

Energy 1.2 In chemical and physical changes, matter often exchanges energy with its surroundings. In these exchanges, the total energy is always conserved; energy is neither created nor destroyed. Systems with

high potential energy tend to change in the direction of lower potential energy, releasing energy into the surroundings.

The Units of Measurement and Significant Figures (1.3, 1.4) Scientists use primarily SI units, which are based on the metric system. The SI base units include the metre (m) for length, the kilogram (kg) for mass, the second (s) for time, and the kelvin (K) for temperature. Derived units are those formed from a combination of base units. Common derived units include volume (cm3 or m3) and density (g cm - 3). Measured quantities are reported so that the number of digits reflects the uncertainty in the measurement. Significant figures are the nonplaceholding digits in a reported number.

Key Equations and Relationships Relationship between Kelvin (K) and Celsius (°C) temperature scales (1.3) TC T = + 273.15 K °C

Relationship between density (d  ), mass (m ), and volume (V  ) (1.3) d =

m V

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Get complete eBook Order by email at [email protected] Exercises      23

Key Skills Determining Physical and Chemical Changes and Properties (1.1) • Exercises 21–26

Converting Between Temperature Scales (1.3) • Example 1.1   • For Practice 1.1   • Exercises 27–30

Converting Between Different SI Prefixes (1.3) • Examples 1.2, 1.3  • For Practice 1.2  • For Practice 1.3  • Exercises 31–40

Calculating Density and Converting Density Units (1.3) • Example 1.4  • For Practice 1.4  • For More Practice 1.4  • Exercises 41–44

Reporting the Correct Number of Digits (1.4) • Example 1.5  • For Practice 1.5  • Exercises 49, 50

Working with Significant Figures (1.4) • Examples 1.6, 1.7  • For Practice 1.6, 1.7  • Exercises 51–66

Solving Problems Involving Equations (1.5) • Examples 1.8, 1.9  • For Practice 1.8, 1.9  • Exercises 45–48, 83–90

EXERCISES Review Questions 1. Explain the main goal of chemistry.

11. Explain the difference between density and mass.

2. How do solids, liquids, and gases differ?

12. Explain the difference between intensive and extensive properties.

3. What is the difference between a physical property and a chemical property?

13. What is the meaning of the number of digits reported in a measured quantity?

4. What is the difference between a physical change and a chemical change? Give some examples of each.

14. When multiplying or dividing measured quantities, what determines the number of significant figures in the result?

5. Explain the significance of the law of conservation of energy.

15. When adding or subtracting measured quantities, what determines the number of significant figures in the result?

6. What kind of energy is chemical energy? In what way is an elevated weight similar to a tank of gasoline? 7. What are the standard SI base units of length, mass, time, and temperature? 8. What are the two common temperature scales? Does the size of a unit differ between them?

16. When taking the logarithm of a number, what determines the number of decimal places in the result? 17. When taking the antilogarithm of a number, what determines the number of significant figures in the result? 18. Explain the difference between precision and accuracy.

9. What are prefix multipliers? Give some examples.

19. Explain the difference between random error and systematic error.

10. What is a derived unit? Give an example.

20. What is dimensional analysis?

Problems by Topic Note: Answers to all odd-numbered Problems, numbered in blue, can be found in Appendix III. Exercises in the Problems by Topic section are paired, with each odd-numbered problem followed by a similar even-numbered problem. Exercises in the Cumulative Problems section are also paired, but somewhat more loosely. (Challenge Problems and Conceptual Problems, because of their nature, are unpaired.)

Physical and Chemical Changes and Properties 21. Classify each property as physical or chemical. a. the tendency of ethanol to burn b. the shine of silver c. the odour of paint thinner d. the flammability of propane gas

22. Classify each property as physical or chemical. a. the boiling point of ethanol b. the temperature at which dry ice evaporates c. the tendency of iron to rust d. the colour of gold 23. . Classify each change as physical or chemical. a. Natural gas burns in a stove. b. The liquid propane in a gas grill evaporates because the valve was left open. c. The liquid propane in a gas grill burns in a flame. d. A bicycle frame rusts on repeated exposure to air and water.

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24       Cha pter Get 1   Unitcomplete s o f Me a s ur e meeBook nt fo r Phy siOrder c al an d C hby em i c al C h an g e at email 24. Classify each change as physical or chemical. a. Sugar burns when heated on a skillet. b. Sugar dissolves in water. c. A platinum ring becomes dull because of continued abrasion. d. A silver surface becomes tarnished after exposure to air for a long period of time. 25. . Based on the molecular diagram, classify each change as physical or chemical.

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Units of Measurement 27. . Convert each temperature. a. 0.00 °C to K (temperature at which water freezes) b. 77 K to °C (temperature of liquid nitrogen) c. 37.0 °C to K (body temperature) 28. Convert each temperature. 100.0 °C to K (temperature of boiling water at sea level) a. b. 2.735 K to °C (average temperature of the universe as measured from background blackbody radiation) c. 22 °C to K 29. . The coldest temperature ever recorded outside of the Antarctic was on Mount Logan in Yukon on May 26, 1991, when the temperature dipped to -77.5 °C. Convert this temperature to K. 30. The warmest temperature ever measured in the world was 56.7 °C on July 10, 1913, at Furnace Creek Ranch, Death Valley, California. Convert this temperature to K.

(a)

31. . Use the prefix multipliers to express each measurement without any exponents. a. 1.2 * 10-9 m b. 22 * 10-15 s c. 1.5 * 109 g d. 3.5 * 106 L

(b)

32. Use prefix multipliers to express each measurement without any exponents. a. 38.8 * 105 g b. 55.2 * 10-10 s 11 c. 23.4 * 10 m d. 87.9 * 10-7 L

(c) 26. Based on the molecular diagram, classify each change as physical or chemical.

33. . Use scientific notation to express each quantity with only the base units (no prefix multipliers). a. 4.5 ns b. 18 fs c. 128 pm d. 35 mm 34. Use scientific notation to express each quantity with only the base units (no prefix multipliers). a. 35 mL b. 225 Mm c. 133 Tg d. 1.5 cg 35. . Complete the table: a. 1245 kg 1.245 * 106 g  1.245 * 109 mg b. 515 km _______dm   _______ cm c. 122.355 s _______ms   _______ ks d. 3.345 kJ _______J    _______ mJ 36. Complete the table: a. 355 km s-1 _______ cm s-1 _______ m ms-1 b. 1228 g L-1 _______ g mL-1 _______ kg ML-1 c. 556 mK s-1 _______ K s-1 _______ mK ms-1 d. 2.554 mg mL-1 _______ g L-1 _______ mg mL-1

(b)

(a)

37. . Express the quantity 254 998 m in each unit. a. km b. Mm c. mm d. cm 38. Express the quantity 556.2 * 10-12 s in each unit. a. ms b. ns c. ps d. fs 39. . How many 1 cm squares would it take to construct a square that is 1 m on each side? 40. How many 1 cm cubes would it take to construct a cube that is 4 cm on its edge?

Density (c)

41. . A Canadian penny was found to have a mass of 2.35 g and a volume of 0.302 cm3. Is the penny made of pure copper? Explain. 42. A titanium bicycle frame displaces 0.314 L of water and has a mass of 1.41 kg. What is the density of the titanium in g cm-3?

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Get complete eBook Order by email at [email protected] Exercises      25 43. . Glycerol is a viscous liquid often used in cosmetics and soaps. A 3.25 L sample of pure glycerol has a mass of 4.10 * 103 g. What is the density of glycerol in g cm-3? 44. A supposedly gold nugget is tested to determine its density. It is found to displace 19.3 mL of water and has a mass of 371 grams. Could the nugget be made of gold? 45. . Ethylene glycol (antifreeze) has a density of 1.11 g cm-3. a. What is the mass, in g, of 417 mL of this liquid? b. What is the volume, in L, of 4.1 kg of this liquid? 46. Acetone (nail polish remover) has a density of 0.7857 g cm-3. a. What is the mass, in g, of 28.56 mL of acetone? b. What is the volume, in mL, of 6.54 g of acetone? 47. . A small airplane takes on 245 L of fuel. If the density of the fuel is 0.803 g mL-1, what mass of fuel has the airplane taken on? 48. Human fat has a density of 0.918 g cm-3. How much volume (in cm3) is gained by a person who gains 5.00 kg of pure fat?

The Reliability of a Measurement and Significant Figures 49. . Read each measurement to the correct number of significant figures. Note: Laboratory glassware should always be read from the bottom of the meniscus.

(a)

(b)

(c)

50. Read each measurement to the correct number of significant figures. Note: Laboratory glassware should always be read from the bottom of the meniscus. Digital balances normally display mass to the correct number of significant figures for that particular balance.

53. How many significant figures are in each number? a. 0.000312 m b. 312 000 s c. 3.12 * 105 km d. 13 127 s e. 2000 54. How many significant figures are in each number? a. 0.1111 s b. 0.007 m c. 108 700 km d. 1.563300 * 1011 m e. 30 800 55. . Indicate the number of significant figures in each number. If the number is an exact number, indicate an unlimited number of significant figures. a. p = 3.14 b. 1 m3 = 1000 dm3 c. 5683.91 km2 (land area of Prince Edward Island) d. 3.0 * 108 m s-1 (speed of light in a vacuum) 56. Indicate the number of significant figures in each number. If the number is an exact number, indicate an unlimited number of significant figures. a. 12 = 1 dozen b. 11.4 g cm-3 (density of lead) c. 2.42 * 1019 km (distance to Andromeda galaxy) d. 1640 km3 (volume of Lake Ontario) 57. . Round each number to four significant figures. a. 156.852 b. 156.842 c. 156.849 d. 156.899 58. Round each number to three significant figures. a. 79 845.82 b. 1.549837 * 107 c. 2.3499999995 d. 0.000045389

Significant Figures in Calculations 59. . Calculate to the correct number of significant figures. a. 9.15 , 4.970 b. 1.54 * 0.03060 * 0.69 c. 27.5 * 1.82 , 100.04 d. (2.290 * 106) , (6.7 * 104) 60. Calculate to the correct number of significant figures. a. 89.3 * 77.0 * 0.08 b. (5.01 * 105) , (7.8 * 102) c. 4.005 * 74 * 0.007 d. 453 , 2.031

(a)

(b)

(c)

61. . Calculate to the correct number of significant figures. a. 43.7 - 2.341 b. 17.6 + 2.838 + 2.3 + 110.77 c. 19.6 + 58.33 - 4.974 d. 5.99 - 5.572

51. . For each number, underline the zeros that are significant and draw a line through the zeros that are not: a. 1 050 501 km b. 0.0020 m c. 0.000000000000002 s d. 0.001090 cm

62. Calculate to the correct number of significant figures. a. 0.004 + 0.09879 b. 1239.3 + 9.73 + 3.42 c. 2.4 - 1.777 d. 532 + 7.3 - 48.523

52. For each number, underline the zeros that are significant and draw a line through the zeros that are not: a. 180 701 mi b. 0.001040 m c. 0.005710 km d. 90 201 m

63. . Acceleration is a derived unit and can be expressed in SI base units as m s-2. Suppose a 1 kg mass is accelerated through 1 m at 1 m s-2. Express the product of these three numbers in SI base units and a derived unit with a special name.

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26       Cha pter Get 1   Unitcomplete s o f Me a s ur e meeBook nt fo r Phy siOrder c al an d C hby em i c al C h an g e at email 64. Divide a force of 1 N (1 newton) by an area of 1 m2 and express the quotient in SI base units and a derived unit with a special name. Hint: A newton is a derived unit with a special name, so you must first express it in SI base units. 65. . Calculate to the correct number of significant figures. a. (24.6681 * 2.38) + 332.58 b. (85.3 - 21.489) , 0.0059 c. (512 , 986.7) + 5.44 d. [(28.7 * 105) , 48.533] + 144.99 66. Calculate to the correct number of significant figures. a. [(1.7 * 106) , (2.63 * 105)] + 7.33 b. (568.99 - 232.1) , 5.3 c. (9443 + 45 - 9.9) * 8.1 * 106 d. (3.14 * 2.4367) - 2.34

Unit Conversions 67. . Convert: a. 3.25 kg to g b. 250 ms to s c. 0.345 L to cm3 d. 257 dm to km 68. Convert: 1405 mg to mg b. 62 500 fs to ns a. c. 1056 cm2 to m2 d. 25.0 mL to L

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70. A cyclist rides at an average speed of 8.1 m s-1. If she wants to bike 212 km, how many hours must she ride? 71. . A house has an area of 195 m2. Convert its area to: a. km2 b. dm2 c. cm2 72. A bedroom has a volume of 115 m3. Convert its volume to: a. km3 b. dm3 c. cm3 73. . The average area of land burned by forest fires every year in Canada is 2.5 million hm2 (square hectometres, or hectares). What is this value in m2 and km2? 74. Besides Vatican City, the smallest country in the world is Monaco, which has a land area of 1.8 km2. What is the area of Monaco in hm2 and m2? 75. . An acetaminophen suspension for infants contains 80 mg/ 0.80 mL suspension. The recommended dosage is 15 mg/kg body weight. How many mL of this suspension should be given to an infant weighing 6.5 kg? 76. An ibuprofen suspension for infants contains 100 mg/5.0 mL suspension. The recommended dose is 10 mg/kg body weight. How many mL of this suspension should be given to an infant weighing 8.2 kg?

Solving Chemical Problems 69. . A runner wants to run 10.0 km. She knows that her running pace is 3.5 m s-1. How many minutes must she run?

Cumulative Problems 77. . There are exactly 60 seconds in a minute, there are exactly 60 minutes in an hour, there are exactly 24 hours in a mean solar day, and there are 365.24 solar days in a solar year. Find the number of seconds in a solar year. Be sure to give your answer with the correct number of significant figures.

83. . A thief wants to use a can of sand to replace a solid gold cylinder that sits on a weight-sensitive, alarmed pedestal. The gold cylinder has a length of 22 cm and a radius of 3.8 cm. Given that the density of gold and sand are 19.3 g cm-3 and 3.00 g cm-3, respectively, what volume of sand must the thief use?

78. Use scientific notation to indicate the number of significant figures in each statement: a. Fifty million Frenchmen can’t be wrong. b. “For every ten jokes, thou hast got an hundred enemies” (Laurence Sterne, 1713–1768). c. The diameter of a Ca atom is 1.8 one hundred millionths of a centimetre. d. Sixty thousand dollars is a lot of money to pay for a car. e. The density of platinum is 21.4 g cm-3.

84. The proton has a radius of approximately 1.0 * 10-13 cm and a mass of 1.7 * 10-24 g. Determine the density of a proton. For a sphere, V = (4>3)pr3.

79. . Classify each property as intensive or extensive. a. volume b. boiling point c. temperature d. electrical conductivity e. energy 80. A temperature measurement of 25 °C has three significant figures, while a temperature measurement of - 196 °C has only two significant figures. Explain. 81. . Do each calculation without using your calculator and give the answers to the correct number of significant figures. a. 1.76 * 10-3>8.0 * 102 b. 1.87 * 10-2 + 2 * 10-4 - 3.0 * 10-3 c. 3(1.36 * 105)(0.000322)>0.0824(129.2)

82. Using the value of the euro at C$1.37 (Canadian dollars) and the price of 1 litre of gasoline in France at 1.67 euros, calculate the price of gasoline in C$ in France.

85. . The density of titanium is 4.51 g cm-3. What is the volume (in litres) of 3.5 kg of titanium? 86. The density of iron is 7.86 g cm-3. What is its density in kilograms per cubic metre (kg m-3)? 87. . A steel cylinder has a length of 5.49 cm, a radius of 0.56 cm, and a mass of 41 g. What is the density of the steel in g cm-3? 88. A solid aluminum sphere has a mass of 85 g. Use the density of aluminum to find the radius of the sphere in decimetres. 89. . A backyard swimming pool holds 185 cubic metres of water. What is the mass of the water in kilograms? 90. An iceberg has a volume of 8921 L. What is the mass of the ice (in kg) composing the iceberg? 91. . The Toyota Prius, a hybrid electric vehicle, has a combined consumption rating of 3.8 L/100 km. How many kilometres can the Prius travel on 15 litres of gasoline? 92. The Ford Mustang (GT Coupe) has a consumption rating of 7.6 L/100 km on the highway or 12 L/100 km in the city. How far could you travel on a full tank of gas (61.0 L) on the highway and in the city?

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Get complete eBook Order by email at [email protected] Exercises      27 93. The single proton that forms the nucleus of the hydrogen atom has a radius of approximately 1.0 * 10-13 cm. The hydrogen atom itself has a radius of approximately 52.9 pm. What fraction of the space within the atom is occupied by the nucleus? 94. A sample of gaseous neon atoms at atmospheric pressure and 0 °C contains 2.69 * 1022 atoms per litre. The atomic radius of neon is 69 pm. What fraction of the space is occupied by the atoms themselves? What does this reveal about the separation between atoms in the gaseous phase? 95. . The length of a hydrogen molecule is 212 pm. Find the length in kilometres of a row of 6.02 * 1023 hydrogen molecules. The diameter of a ping pong ball is 4.0 cm. Find the length in kilometres of a row of 6.02 * 1023 ping pong balls. 96. The world’s record in the 100 m dash is 9.58 s. Find the speed in km hr-1 of the runner who set the record. 97. . Table salt contains 39.33 g of sodium per 100 g of salt. Health Canada recommends that adults consume less than 2.40 g of sodium per day. A particular snack mix contains 1.25 g of salt per 100 g of the mix. What mass of the snack mix can you consume per day and still be within the Health Canada limit?

98. Lead metal can be extracted from a mineral called galena, which contains 86.6% lead by mass. A particular ore contains 68.5% galena by mass. If the lead can be extracted with 92.5% efficiency, what mass of ore is required to make a lead sphere with a 5.00 cm radius? 99. . Liquid nitrogen has a density of 0.808 g mL-1 and boils at 77 K. Researchers often purchase liquid nitrogen in insulated 175 L tanks. The liquid vaporizes quickly to gaseous nitrogen (which has a density of 1.15 g L-1 at room temperature and atmospheric pressure) when the liquid is removed from the tank. Suppose that all 175 L of liquid nitrogen in a tank accidentally vaporized in a lab that measured 10.00 m * 10.00 m * 2.50 m. What is the maximum fraction of the air in the room that could be displaced by the gaseous nitrogen? 100. Mercury is often used as an expansion medium in a thermometer. The mercury sits in a bulb on the bottom of the thermometer and rises up a thin capillary as the temperature rises. Suppose a mercury thermometer contains 3.380 g of mercury and has a capillary that is 0.200 mm in diameter. How far does the mercury rise in the capillary when the temperature changes from 0.0 °C to 25.0 °C? The density of mercury at these temperatures is 13.596 g cm-3 and 13.534 g cm-3, respectively.

Challenge Problems 101. A force of 2.31 * 104 N is applied to a diver’s face mask that has an area of 125 cm2. Find the pressure in bar on the face mask. 102. The SI unit of force is the newton, derived from the base units by using the definition of force, F = ma. The dyne is a non-SI unit of force in which mass is measured in grams and time is measured in seconds. The relationship between the two units is 1 dyne = 10-5 N. Find the unit of length used to define the dyne. 103. Kinetic energy can be defined as ½ mv2 or as 3/2 PV. Show that the derived SI units of each of these terms are those of energy. (Pressure is force/area and force is mass * acceleration.) 104. In 1999, scientists discovered a new class of black holes with masses 100 to 10 000 times the mass of our sun, but occupying less space than our moon. Suppose that one of these black holes has a mass of 1 * 103 suns and a radius equal to one-half the radius of our moon. What is the density of the black hole in g cm-3? The radius of our sun is 7.0 * 105 km, and it has an average density of 1.4 * 103 kg m-3. The diameter of the moon is 3.48 * 103 km. 105. Nanotechnology, the field of trying to build ultrasmall structures one atom at a time, has progressed in recent years. One potential application of nanotechnology is the construction of artificial cells. The simplest cells would probably mimic red blood cells and be used as oxygen transports within the body. For example, nanocontainers, perhaps constructed of carbon, could be pumped full of oxygen and injected into a person’s

bloodstream. If the person needed additional oxygen—due to a heart attack perhaps, or for the purpose of space travel—these containers could slowly release oxygen into the blood, allowing tissues that would otherwise die to remain alive. Suppose that the nanocontainers were cubic and had an edge length of 25 nm. a. What is the volume of one nanocontainer? (Ignore the thickness of the nanocontainer’s wall.) b. Suppose that each nanocontainer could contain pure oxygen pressurized to a density of 85 g L-1. How many grams of oxygen could be contained by each nanocontainer? c. Normal air contains about 0.28 g of oxygen per litre. An average human inhales about 0.50 L of air per breath and takes about 20 breaths per minute. How many grams of oxygen does a human inhale per hour? (Assume two significant figures.) d. What is the minimum number of nanocontainers that a person would need in their bloodstream to provide one hour’s worth of oxygen? e. What is the minimum volume occupied by the number of nanocontainers computed in part (d)? Is such a volume feasible, given that total blood volume in an adult is about 5 litres? 106. A box contains a mixture of small copper spheres and small lead spheres. The total volume of both metals, measured by the displacement of water, is 427 cm3, and the total mass is 4.36 kg. What percentage of the spheres are copper?

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Conceptual Problems 107. A volatile liquid (one that easily evaporates) is put into a jar and the jar is then sealed. Does the mass of the sealed jar and its contents change upon vaporization of the liquid? 108. The diagram represents solid carbon dioxide, also known as dry ice. Which of the diagrams below best represents the dry ice after it has sublimed into a gas?

(a)

(b)

(c)

109. A cube has an edge length of 7 cm. If it is divided up into 1 cm cubes, how many 1 cm cubes would there be? 110. Substance A has a density of 1.7 g cm-3. Substance B has a density of 1.7 kg m-3. Without doing any calculations, determine which substance is more dense. 111. . For each box, examine the blocks attached to the balances. Based on their positions and sizes, determine which block is more dense (the dark block or the lighter-coloured block), or if the relative densities cannot be determined. (Think carefully about the information being shown.)

(a)

(b)

(c)

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