E I G H T H E D I T I O N GENERAL CHEMISTRY...Instructor’s Resource Manual E I G H T H E D I T I O...

Instructor’s Resource Manual

E I G H T H E D I T I O N

GENERAL CHEMISTRY

Darrell D. EbbingWayne State University

Steven D. GammonUniversity of Idaho

HOUGHTON MIFFLIN COMPANY • BOSTON • NEW YORK

Vice President and Publisher: Charles HartfordExecutive Editor: Richard StrattonDevelopment Editor: Danielle RichardsonEditorial Associate: Rosemary MackSenior Project Editor: Nancy BlodgetSenior Manufacturing Coordinator: Priscilla BaileyExecutive Marketing Manager: Katherine GreigMarketing Associate: Alexandra Shaw

Copyright © 2005 by Houghton Mifflin Company. All rights reserved.

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Contents

Part I Introduction 1

Part II Chapter Essays 3

Part III Alternate Sequence of Text Coverage 5

Part IV Chapter Descriptions 8

Part V Operational Skills Masterlist 28

Part VI Correlation of Cumulative-Skills Problems with Text Sections 62

Part VII Alternate Examples for Lecture 66

Part VIII Brief Notes on Suggested Lecture Demonstrations 128

PART IIntroduction

General Chemistry, Eighth Edition, is designed to give the instructor the greatest flexibility increating a course for his or her students and to make the process of teaching with the text assmooth as possible. The careful, logical, and clear development of material in each chapter,with its appropriate division into parts, sections, and subsections, allows for flexible rear-rangement to meet individual syllabus configurations.

To smooth the process of teaching with the text, we have worked diligently in several areas.Each technical term is clearly defined at first mention, and each concept is carefully explainedand made as concrete as possible by using illustrations from everyday situations or by relatingthe concept clearly to its use in chemistry. Descriptive and applied chemistry is emphasizedearly on and throughout the book through the inclusion of interesting chemical facts in thetext, in problems, and in the boxed essays that occur within the chapters. We believe thisemphasis on descriptive chemistry is necessary to provide the motivation for learning chemi-cal concepts. We have also added Concept Checks and Conceptual Problems to aid the studentin learning the concepts. In these, we ask students questions that require them to think and tosolve problems by first asking, What are the chemical concepts that apply here? Thesequestions are phrased to force a thoughtful answer rather than allowing the student to lookfor a memorized algorithm. A Conceptual Guide is available that provides solutions to all ofthese Concept Checks and Conceptual Problems. By paying attention to these areas, we haveremoved a burden from the instructor, who can now concentrate his or her attention on themain requirements of teaching—motivating the students, emphasizing important points,discussing difficult concepts, drawing parallels, and so forth.

In the introduction to your course, it may be well to note for the students several featuresof the text that are specifically designed to help them in their study of chemistry. Indexes oftextbooks tend to be underutilized, but the one we have prepared for General Chemistry, EighthEdition, is especially thorough. When students want to find a topic they covered earlier butcan’t remember where in the text it is covered, they should be encouraged to consult the index.On the other hand, when they encounter a term whose definition escapes them, they shouldturn to the extensive glossary placed just before the index.

In order to understand where a chapter is going and how the material is developed, studentsshould examine the contents given at the start of each chapter. You can make use of thiscontents section as well. It will allow you to survey a chapter quickly to see how it correspondsto your course plan and to see what deletions or changes of order you might wish to make.You can refer the students to this contents section when you inform them of deletions orchanges in order or want to indicate parts of the chapter you intend to emphasize.

Note that important terms have been highlighted by black boldface type. To facilitate thestudent’s review, these terms are gathered together at the end of the chapter in the order inwhich they occur in the text. As the student goes through the list, he or she should recall the

Copyright © Houghton Mifflin Company. All rights reserved. 1

context in which the word occurred. All of these important terms are included in the glossaryat the end of the book.

Problem solving has received special attention in the text. Students should be aware thatkey statements or equations used for problem solving are highlighted. Also, page numbers oftables of data needed for problem solving are listed under Locations of Important Informationon the inside back cover of the book. The major problem-solving skills are explained inExamples, most of which include a Problem Strategy that underscores the thinking processinvolved in solving the problem. Some Examples include an Answer Check that employs a“check of reasonableness” of the answer, based on general knowledge of the problem. TheExamples are followed by Exercises for the student to work out. The answers to these Exercisesare given at the back of the book. Corresponding end-of-chapter Problems are noted at theends of the Exercises. Problems have been divided into categories: Conceptual Problems,Practice Problems, General Problems, and Cumulative-Skills problems (these have beenfollowed by Media Activities). Answers to odd-numbered Problems appear at the end of thebook. Complete solutions to Exercises, Review Questions, and Problems are available toinstructors in the Solutions Manual for General Chemistry. The Solutions Manual is alsoavailable for sale to students if the instructor approves. Alternatively, you may prefer thatyour students obtain the Student’s Solutions Manual, which contains solutions to only theodd-numbered Problems (along with complete solutions to Exercises and Review Questions).

In Part II of this manual, we describe and list the chapter essays. In Part III we describeseveral possible alternate sequences of text coverage, which can help you design your course.Part IV can also help you with this; for example, the chapter descriptions given there pointout alternate placements of chapters. Part IV also discusses the development of the chaptertext, gives special notes on the chapter, and offers suggestions on how to abbreviate thematerial if that seems appropriate. Part V gives an operational skills masterlist in whichoperational skills are correlated with Examples, Exercises, and Problems. You may find thisof use in making reading and problem assignments. Part VI lists sections of the text that covermaterial needed to solve each cumulative-skills problem. This list will help you avoidassigning problems that require text sections you have omitted. Part VII gives a selection ofAlternate Examples you can use in your lectures. Part VIII describes lecture demonstrationsyou may wish to try.

Copyright © Houghton Mifflin Company. All rights reserved.

2 PART I

PART IIChapter Essays

The eighth edition of General Chemistry includes two series of boxed essays whose purpose isto augment the main text. These essay series are titled A Chemist Looks At, and InstrumentalMethods.

The essays in the A Chemist Looks At series explore topics of general interest (such as humanvision) or subjects that are in the news (such as superconductors). Each essay applies theprinciples of chemistry described in the text, perhaps expanding on them.

The Instrumental Methods essays describe some of the most important instrumental methodsused by research chemists today, such as mass spectrometry and x-ray diffractometry. Thesedescriptions are purposely brief and are intended only to make students aware that chemiststoday routinely use sophisticated instruments in their work. Students are generally fascinatedto learn that modern chemistry relies so strongly on such instruments.

A Chemist Looks At

Essay Title Text Page Chapter Category

The Birth of the Post-it Note®

Thirty Seconds on the Island of StabilityNitric Oxide Gas and Biological SignalingCarbon Dioxide Gas and the Greenhouse EffectLucifers and Other MatchesZapping Hamburger with Gamma RaysLasers and Compact Disc PlayersLevitating Frogs and PeopleIonic Liquids and Green ChemistryChemical Bonds in NitroglycerinLeft-Handed and Right-Handed MoleculesHuman VisionStratospheric Ozone (An Absorber of Ultraviolet Rays)Removing Caffeine from CoffeeLiquid Crystal DisplaysWater (A Special Substance for Planet Earth)Hemoglobin Solubility and Sickle-Cell AnemiaThe World’s Smallest Test TubesSuperconductivtyBuckminsterfullerence—Third Form of CarbonSilica Aerogels, the Lightest “Solids”

5 56 188 213 235 271 276 311 335 344 384 409 410 433 453 465 486 514 537 542 550

1 2 5 5 6 7 7 8 9 91010101111111212131313

Everyday LifeFrontiersLife ScienceEnvironmentEveryday LifeEveryday LifeMaterialsFrontiersFrontiersEveryday LifeEveryday LifeLife ScienceEnvironmentEveryday LifeEveryday LifeEnvironmentLife ScienceFrontiersMaterialsMaterialsMaterials


Essay Title Text Page Chapter Category

Seeing Molecules ReactSlime Molds and Leopards’ SpotsTaking Your MedicineUnclogging the Sink and Other ChoresAcid RainLimestone CavesCoupling of ReactionsPositron Emission Tomography (PET)The Chernobyl Nuclear AccidentSalad Dressing and Chelate StabilityThe Cooperative Release of Oxygen from OxyhemoglobinDiscovery of NylonTobacco Mosaic Virus and Atomic Force Microscopy

602 629 667 682 700 748 784 883 892 973 99210401058

14151616171819212123232525

FrontiersEveryday LifeLife ScienceEveryday LifeEnvironmentEnvironmentLife ScienceLife ScienceEnvironmentEveryday LifeLife ScienceMaterialsLife Science

Instrumental Methods

Essay Title Text Page Chapter

Separation of Mixtures by ChromatographyMass Spectrometry and Molecular FormulaScanning Tunneling MicroscopyNuclear Magnetic Resonance (NMR)X Rays, Atomic Numbers, and Orbital Structure (Photoelectron Spectroscopy)Infrared Spectroscopy and Vibrations of Chemical BondsAutomated X-Ray Diffractometry

14 98282298

305363464

1 3 7 8

8 911


4 PART II

PART IIIAlternate Sequence of Text Coverage

For a two-semester course, the first semester might cover a selection of material from Chapters1 through 12, which treats basic chemistry, atomic and molecular structure, and states of matterand solutions. The second semester would then cover a selection of material from the last halfof the text, which treats kinetics, equilibrium (including thermodynamics and electrochemis-try), nuclear chemistry, and descriptive chemistry. You may want to look at Part IV of theInstructor’s Resource Manual for suggestions on ways to select or abbreviate the material fromthese chapters to fit your schedule.

A three-quarter course following the text sequence might begin with a selection fromChapters 1 through 8. Thus, the first term would cover basic chemistry and atomic structure.The second term would begin with chemical bonding (Chapters 9 and 10) and go through theintroductory chapters on chemical equilibrium (Chapters 15 and 16). The last term wouldcover aqueous equilibrium, thermodynamics, electrochemistry, nuclear chemistry, and aselection from the block of descriptive chemistry chapters (Chapters 13 through 25).

Alternate sequences of the text material can be easily designed. The figure given on thefollowing page may help you design your course by showing how the text chapters dependon previous ones. An arrow pointing to a box indicates that the preceding chapter is aprerequisite to the chapter given in the box. In addition, the chapter descriptions in Part IVgive suggestions on alternate placements of material and possible deletions of topics (seeunder Placement of the Chapter and Abbreviation of the Material).

One possible alternate lecture schedule follows. In this schedule, coverage of gases justprecedes discussion of liquids and solids, and thermodynamics is covered before equilibrium.

Alternate Two-Semester Sequence(Gases just before liquids; thermodynamics before equilibrium)

First SemesterChapters 1 through 4 Basic chemistryChapters 7 through 10 Atomic and molecular structureChapters 5, 11, 12 States of matter and solutionsChapter 13 Materials

Second SemesterChapter 14 Chemical kineticsChapter 6 ThermochemistryChapter 19 Thermodynamics (Sections 19.1 through 19.5)Chapter 15 Introduction to equilibrium (plus Sections 19.6 and 19.7)



Ch. 21 Nuclear Chemistry(Ch. 14 is useful but not required)

Ch. 20 Electrochemistry

Ch. 18 Solubility Equilibria

Ch. 15 Equilibrium

Ch. 16 Acids and Bases

Ch. 17 Acid–Base Equilibria

Ch. 7 Quantum Theory

Ch. 8 Electron Configurations

Ch. 9 Ionic and Covalent Bonding

Ch. 10 Molecular Geometry

Ch. 11 Liquids and Solids

Ch. 13 Materials

Ch. 14 Rates of Reaction

Ch. 19 Thermodynamics

Ch. 2 Atomic Theory

Ch. 1 Matter; Units

Ch. 3 Stoichiometry

Ch. 5 Gases

Ch. 6 Thermochemistry

Ch. 4 Reactions; Intro.

Ch. 22–25 Descriptive Chemistry

Ch. 12 Solutions

6 PART III

Chapters 16 through 18 Aqueous equilibriaChapter 20 ElectrochemistryChapters 21 through 25 Nuclear and descriptive chemistry (select chapters)

The following would be a similar three-quarter sequence:

Alternate Three-Quarter Sequence(Gases just before liquids; thermodynamics before equilibrium)

First TermChapters 1 through 4 Basic chemistryChapters 7 and 8 Atomic structureChapter 9 Chemical bonding

Second TermChapter 10 Molecular structureChapters 5, 11, 12 States of matter and solutionsChapter 14 Chemical kineticsChapter 6 ThermochemistryChapter 19 Thermodynamics (Sections 19.1 through 19.5)

Third TermChapter 15 Introduction to equilibrium (plus Sections 18.6 and 18.7)Chapters 16 through 18 Aqueous equilibriaChapter 20 ElectrochemistryChapters 13 and 21 Nuclear and descriptive chemistry (select chapters) through 25


Alternate Sequence of Text Coverage 7

PART IVChapter Descriptions

In this part of the Instructor’s Resource Manual, we look at each chapter of the text, describingthe logic of its present placement and possible alternate positions for it. We also describe thedevelopment of the chapter material, note any special points related to each chapter, andindicate possible ways to abbreviate the material if this is necessary and seems appropriate.This part of the manual should be especially useful in designing a syllabus for your course.As the lectures proceed, you may need to delete material to keep to your schedule; the sectionon abbreviation of the material for each chapter gives suggestions for ways to do this.

CHAPTER 1 Chemistry and Measurement

The chapter opens with a brief introduction to chemistry, followed by a discussion ofmeasurement and significant figures.

Placement of the Chapter

After the introductory material is presented, it is appropriate to discuss measurement becauseof its importance in problem solving.

Development of the Chapter

Chapter 1 is divided into two parts. The first part is a brief introduction to chemistry. Section1.1 describes the central role of chemistry in modern science and technology. Section 1.2describes the relationship between experiment and explanation, and Section 1.3 illustrates thismaterial with the law of conservation of mass. The last section of the first part (Section 1.4) isan introduction to the way the chemist describes matter.

The second part of the chapter concerns measurement. Section 1.5 discusses significantfigures and the limitations on experimental measurement. Section 1.6 describes SI units,including prefixes and base units. Section 1.7 discusses units such as volume and density thatare derived from the SI base units. Finally, Section 1.8 describes the conversion of units anddimensional analysis.

Special Notes

Students should understand the main features of the International System, including prefixes,base units, and derived units, but they also need to be familiar with traditional units, such asthe Angstrom and the liter. In any case, conversion of units is emphasized, so students can

8 Copyright © Houghton Mifflin Company. All rights reserved.

easily move from, say, picometers to angstroms. Use of the conversion-factor method appearsagain later in the text, particularly in Chapter 3 (stoichiometry).

Abbreviation of the Material

Most of the material in Chapter 1 is basic, and you will probably want to assign all of it asreading. However, students may be familiar with much of this material from a high schoolcourse; so after a brief introduction to chemistry, you might begin your lectures with signifi-cant figures and units, stressing unit conversions.

CHAPTER 2 Atoms, Molecules, and Ions

The chapter introduces basic concepts needed in the course: atomic theory, atomic structure,atomic weight, periodic table, molecular and ionic substances, formulas, organic compounds,naming of compounds, and chemical equations.


The early placement of this chapter is necessary because it introduces basic concepts neededfor subsequent work.


Atomic theory forms the thread of the chapter. Section 2.1 begins with atomic theory, andSections 2.2 and 2.3 discuss atomic and nuclear structure. Section 2.4 describes atomic weightsand how they are obtained. The periodic table is introduced in Section 2.5. Section 2.6, whichbegins the second part of the chapter, discusses molecular and ionic substances and how towrite chemical formulas. Section 2.7 gives a brief discussion of organic compounds. Thesecond part of the chapter ends with Section 2.8 on the naming of compounds. The final partof the chapter consists of Sections 2.9 and 2.10 on the writing and balancing of chemicalequations, respectively.

Special Notes

The periodic table is introduced in Section 2.5 but will be discussed again in Chapter 8 inconnection with electron configurations and periodicity of some atomic properties.


The chapter introduces basic concepts that students may have some familiarity with from aprevious course, so lecture time could be directed to the salient points. Sections 2.2 (on thestructure of the atom) and 2.3 (on nuclear structure and isotopes) and the portion of Section2.4 on mass spectrometry and atomic weights can be discussed later (just before Chapter 7,Quantum Theory of the Atom), except for a brief mention of atomic structure and isotopes.Nomenclature could be delayed until later, perhaps after Chapter 9 on bonding.


Chapter Descriptions 9

CHAPTER 3 Calculations with Chemical Formulas and Equations

This chapter uses the concepts of formula weight and the mole to obtain chemical formulasand to perform calculations with chemical equations.


This material follows Chapter 2 naturally, emphasizing the mole concept. However, you maywish to postpone it until you have covered the block of chapters on atomic and molecularstructure (Chapters 7 through 10).


This chapter consists of three parts. The first part, Sections 3.1 and 3.2, introduces the conceptsof formula weight and mole. The second part, Sections 3.3 through 3.5, describes how aformula is obtained from analytical data. The third part, Sections 3.6 through 3.8, uses thechemical equation to do mole–mass calculations.

Special Notes

The conversion-factor method (factor-label method) is used consistently to solve the problemsin this chapter. To illustrate how to obtain a formula, we begin with analytical data for aceticacid, the compound featured in the chapter opening, and calculate the mass percentages ofelements (Example 3.9) and then the molecular formula (Example 3.12).


If you are pressed for time, you might omit Section 3.4 on determining the percentage of carbonand hydrogen by combustion. Theoretical and percentage yields (last half of Section 3.8) mayalso be omitted.

CHAPTER 4 Chemical Reactions: An Introduction

This chapter introduces the basic concepts of reactions, particularly those concerned with ionicreactions in aqueous solution.


Early treatment of this material makes it possible to refer to various chemical reactions toillustrate the applications of principles. Moreover, this chapter can be useful in developinglaboratory work and as background for reading some of the essays. However, the subject canbe postponed, and even then some instructors may cover only parts of the chapter (say thesection on ionic equations).


10 PART IV


Because of the importance of ionic reactions in general chemistry, we begin by describing theionic theory of solutions (Section 4.1) and how ionic equations are used to represent ionicreactions (Section 4.2). The second part of the chapter discusses the three main types ofchemical reaction: precipitation reactions (Section 4.3), acid–base reactions (Section 4.4), andoxidation–reduction reactions (Sections 4.5 and 4.6). Section 4.6 treats only simple oxidation–reduction reactions; more complicated cases are discussed in Section 20.1. The third part,Sections 4.7 and 4.8, introduces the concept of molar concentration and then describescalculations pertaining to diluting a solution. The final part, Sections 4.9 and 4.10, looks atsome calculations in quantitative analysis.

Special Notes

Acids and bases will be discussed in detail in Chapter 16. Electrochemistry, Chapter 20, usesthe concepts of oxidation–reduction reactions, and Section 20.1 discusses the balancing ofmore complex oxidation–reduction reactions.


Sections 4.1 and 4.2 complement the treatment of chemical equations in Section 2.9, at the endof the previous chapter. The remainder of Chapter 4 can be treated to the extent appropriateto your course. Since acids and bases are discussed in detail in Chapter 16, you may wish togive only a brief treatment here. You can easily delay discussion of oxidation–reductionreactions to the second term, if you prefer.

The part of the chapter on solutions and molarity flows naturally from mole considerationsand from stoichiometry, and its inclusion here is useful in the laboratory. However, it may bepostponed until Chapter 12 on solutions, where other concentration units are discussed. Thesections on quantitative analysis could be omitted.

CHAPTER 5 The Gaseous State

This chapter treats the gas laws and the kinetic-molecular theory.


Opinion is divided on where this material is best placed. Stoichiometry and the measurementof gas volumes played key roles in the historical development of chemistry. Thus, there isprecedent for placing gases with stoichiometry. Moreover, the early discussion of gases allowsyou to use the gas laws in a range of laboratory experiments. In addition, the study of gasesgives an excellent opportunity to illustrate the interplay of experiment and theory. On theother hand, others prefer to present this chapter on gases immediately before Chapter 11 onliquids and solids, giving a unit on the states of matter.




The chapter stresses the role of experiment and theory. The first part begins with the meas-urement of pressure (Section 5.1) and moves to the empirical gas laws (Section 5.2), which canbe combined into the ideal gas law (Section 5.3). The ideal gas law is then applied tostoichiometry problems (Section 5.4). Section 5.5 on the law of partial pressures concludes thefirst part of the chapter. The second part deals with the kinetic theory of gases. Section 5.6states the postulates of the theory, relating them to experiment, and then gives a heuristicderivation of the ideal gas law. Section 5.7 on molecular speeds and diffusion and effusioncovers deductions from kinetic theory. Finally, Section 5.8 discusses deviations from idealityin the context of kinetic theory and introduces the van der Waals equation.

Special Notes

Boyle’s and Charles’s laws may be stated as proportionalities (V∝1/P, V∝T, V∝n). Perhapsbecause of the symmetry, Avogadro’s law is also sometimes stated this way. However, this isincorrect because the volume of any substance is proportional to moles. The essential contentof the law is that the molar volume is the same for all gases, which is the statement given onpage 187.

The “derivation” of the ideal gas law, given in Section 5.6, is purely heuristic, which seemsappropriate for general chemistry.


Most of the material in this chapter is basic. The text is sufficiently detailed and the conceptsare easily grasped, so a minimum of lecture time is needed. The last two sections of the chapter(Sections 5.7 and 5.8) can be covered to the extent that time allows.

CHAPTER 6 Thermochemistry

Heats of reaction and the concept of enthalpy are discussed. Concepts of entropy and freeenergy are deferred until Chapter 19 (the first law of thermodynamics is discussed explicitlyin that chapter).


In this position, the chapter follows soon after stoichiometry, so this aspect of thermochemistrycan be emphasized. The placement of this chapter also allows you to underscore the role ofenergy in chemistry before embarking on a discussion of chemical bonding. However, it ispossible to delay this chapter (for example, to precede Chapter 19), and the interveningchapters (Chapters 7–18) were written with this in mind. In these chapters, ∆H is brieflydefined as the heat of reaction. (The treatment of the Born–Haber cycle in Chapter 9 requiresa knowledge of Hess’s law.)


The first part of the chapter begins with a discussion of energy (Section 6.1) and then coversthe basic properties of heats of reaction. After some terms are defined (Section 6.2), the concept


12 PART IV

of enthalpy is discussed (Section 6.3), followed by a discussion of thermochemical equations(Section 6.4), the stoichiometry of heats of reaction (Section 6.5), and measurement of heats ofreaction (Section 6.6). In the second part of the chapter, heats of reactions are related to oneanother by Hess’s law (Section 6.7), and the concept of enthalpies of formation is discussed(Section 6.8). Thermochemistry is applied to fuels in the last section (6.9).

Special Notes

Enthalpy is introduced here as the heat of reaction at constant pressure, which is sufficient fordiscussing the elementary aspects of thermochemistry. However, a brief discussion at the endof Section 6.3 relates enthalpy to internal energy. Enthalpy is also defined precisely in Chapter19, where the first law of thermodynamics is discussed and where the distinction betweeninternal energy and enthalpy is stressed.


Sections 6.1 to 6.5 are basic; after covering those sections, you can abbreviate the material invarious ways. For example, Sections 6.6 and 6.9 might be omitted.

CHAPTER 7 Quantum Theory of the Atom

This chapter begins by presenting the properties of light as a prelude to describing Bohr’stheory of the hydrogen atom. The chapter ends with a discussion of quantum numbers andatomic orbitals. Electron configurations of atoms are dealt with in the next chapter.


This chapter could follow Chapter 2, which introduces atomic structure. However, theintervening chapters on chemical reactions and stoichiometry make possible a wealth oflaboratory experiments, whereas it is more difficult to come up with experiments for Chapters7 through 10.


The chapter consists of two parts. The first part introduces the concepts of light waves (Section7.1) and photons (Section 7.2), from which much of our information on atomic structure comes.The first part concludes with a look at the Bohr theory of the hydrogen atom (Section 7.3). Thesecond part of the chapter introduces quantum mechanics (Section 7.4) and quantum numbersand atomic orbitals (Section 7.5).

Special Notes

The section on the Bohr theory focuses on those aspects of the theory that carry over intomodern quantum theory: energy levels and transitions between levels. It does not emphasizeclassical orbits.




The most important section of the chapter is the concluding one on quantum numbers andatomic orbitals. You could simply discuss the basic concepts of light waves, photons, the Bohrtheory, and quantum mechanics without emphasizing calculations. Then you could go toSection 7.5, where you would concentrate on the quantum numbers.

CHAPTER 8 Electron Configurations and Periodicity

This chapter is a continuation of Chapter 7, which introduced the concepts of atomic orbitalsand quantum numbers. Here we look at the electron configurations of atoms and describe therelationship between these configurations and the periodic behavior of the elements.


The chapter is a continuation of Chapter 7 and should follow it.


In the first part of the chapter, we look at the electronic structure of atoms. Section 8.1 discusseselectron spin and the Pauli exclusion principle, Sections 8.2 and 8.3 describe the building-upprinciple for obtaining the ground-state electron configurations of atoms, and Section 8.4introduces Hund’s rule and orbital diagrams of atoms. In the second part of the chapter,Sections 8.5 and 8.6 describe the periodic table and its relationship to the electron configura-tions of atoms. Section 8.7 gives brief descriptions of the main-group elements.

Special Notes

Theoretical calculations show that the 3d subshell is just below the 4s subshell in energythroughout the transition elements, even though the 4s fills before the 3d. The explanation isthat the total energy of atoms, which is what determines the ground-state electron configura-tions, depends not only on the energy of the individual orbitals but also on the repulsions ofelectrons. You can still keep the discussion elementary by referring to the order of filling orbuilding-up order of the atomic orbitals, noting that this order is the same as the order ofenergy of the orbitals with some exceptions that occur when the sublevels are close in energy.

The building-up order reproduces most of the ground-state configurations correctly butotherwise has no fundamental significance. By ordering the subshells of a ground-stateconfiguration by the principal quantum number, you obtain the valence-shell configurationat the far right. Also, you place the most easily ionized electrons at the far right. For example,if you write the electron configuration for iron as 1s22s22p63s23p63d64s2, the configuration ofFe2+ is written by taking away the 4s electrons.

The chapter emphasizes how to obtain the electron configurations of atoms from theposition of the element in the periodic table. This helps students learn the relationship betweenthe configurations and the periodic table.


Most of the material given in this chapter is important for subsequent work. In particular,students should be able to obtain the electron configuration and orbital diagram for an atomand understand the relationship of these to the periodic table. Moreover, students need to


14 PART IV

understand the concepts of ionization energy and electron affinity to understand the discus-sion of the ionic bond in the next chapter. One could omit the last section, which brieflydescribes the main-group elements. Otherwise, lecture time can be saved by concentrating onthe salient points: electron configurations and orbital diagrams of the atoms, a brief descrip-tion of the periodic table, and a brief discussion of ionization energy and electron affinity.

CHAPTER 9 Ionic and Covalent Bonding

The chapter discusses the elementary aspects of chemical bonding, concluding with sectionson bond properties.


The chapter builds on the concepts of atomic structure introduced in Chapters 7 and 8. Thus,the concepts of ionization energy and electron affinity described toward the end of theprevious chapter flow smoothly into the subject of ionic bonding.


The chapter begins with ionic bonding, describing ionic bonds (Section 9.1) and the electronconfigurations of ions (Section 9.2). Section 9.3 discusses the concept of ionic radii. Then thechapter looks at covalent bonding in Sections 9.4 and 9.5. A general method of writing Lewiselectron-dot formulas is given in Section 9.6. The writing of Lewis formulas is elaborated onin Sections 9.7 through 9.9. Section 9.7 describes resonance and Section 9.8 discusses exceptionsto the octet rule. The final section on writing Lewis formulas (Section 9.9) introduces theconcept of formal charge and applies it to choosing the most appropriate Lewis formula. Thelast sections cover bond properties. Section 9.10 discusses bond length and bond order, andSection 9.11 discusses bond energy.

Special Notes

The electron-dot formula, as rough a description as it is, provides a lot of information in aclear and simple fashion. It is introduced first for atoms and is used to describe ions. Then itis used again when covalent bonding is described. The method given in Section 9.6 for writingelectron-dot formulas will work for most of the molecules encountered in general chemistry,including exceptions to the octet rule. Note that you first distribute electrons to the atomssurrounding the central atom (usually the most electropositive atom) to satisfy the octet rulefor them. However, the electrons on the central atom may or may not satisfy the octet rule.


The most important material is in Sections 9.1, 9.2, and 9.4 through 9.7. The remainder of thechapter can be covered to the extent that time permits.



CHAPTER 10 Molecular Geometry and Chemical Bonding Theory

This chapter covers more advanced bonding concepts than those presented in the previouschapter, including molecular geometry, hybrid orbitals, and molecular orbital theory.


The chapter follows Chapter 9 logically, but it can be postponed. For example, you can discussit just before Chapter 22 on the descriptive chemistry of the main-group elements.


The chapter begins with the VSEPR model (Section 10.1) because of its simplicity and itsreliance on electron-dot formulas, which were covered in the previous chapter. Section 10.2relates dipole moment and molecular geometry. The next two sections (10.3 and 10.4) discussthe valence bond description of the electronic structure of molecules. The last part of thechapter, Sections 10.5 through 10.7, describes molecular orbital theory.

Special Notes

In describing the order of filling of molecular orbitals, we give the order at the bottom of page404, but in the marginal note there we state that the order by energy of the σ2p orbital is belowthat of the π2p orbital in the case of O2 and of F2. However, this change in order has no essentialeffect on the electron configuration or on deductions of bond order or magnetic character.


Discussion of the VSEPR model can be abbreviated to cover up to four electron pairs, omittingfive and six pairs. In that case, you simply omit the last portion of Section 10.1. The remainderof the chapter can be condensed or omitted to suit your needs.

CHAPTER 11 States of Matter; Liquids and Solids

This chapter looks at the states of matter (particularly liquids and solids) and their transfor-mations from one state to another.


Chapters 11 and 12 deal with matter in bulk and require some knowledge of chemical bonding.


The chapter begins with a comparison of gases, liquids, and solids (Section 11.1). Then followtwo sections on changes of state: Section 11.2 on phase transitions and Section 11.3 on phasediagrams. Changes of state are discussed near the beginning of the chapter because they aresome of the most important properties of liquids and solids. The next section describes some


16 PART IV

additional properties of liquids (Section 11.4); these properties are then explained in terms ofintermolecular forces (Section 11.5). By describing the properties before intermolecular forces,we stress that experiment precedes explanation. The last part of the chapter describes the solidstate: types of solids (Section 11.6), crystalline solids (Sections 11.7 through 11.9), and deter-mining crystal structure by x-ray diffraction (Section 11.10).


You might choose to omit discussion of the Clausius–Clapeyron equation in Section 11.2 andphase diagrams in Section 11.3. In the part on the solid state, you might omit Section 11.8 andperhaps Sections 11.9 and 11.10.

CHAPTER 12 Solutions

This chapter looks at solution formation, colligative properties, concentration units, andcolloids.


The chapter continues the discussion of matter in bulk begun in Chapter 11.


The chapter begins by looking at the types of solutions (Section 12.1), then describes thesolution process (Section 12.2) and the effects of temperature and pressure on solubility(Section 12.3). In preparation for a discussion of colligative properties, Section 12.4 describesvarious units of concentration. Then Sections 12.5 through 12.8 look at colligative properties.The final section of the chapter describes colloids.

Special Notes

Some textbooks apply Le Chatelier’s principle and heats of solution to predict the temperaturedependence of the solubility of salts. The difficulty in doing this has been discussed in theliterature (G. M. Bodner, J. Chem. Educ. 1980, 57, 117; see also R. Treptow, J. Chem. Educ. 1984,61, 499). Essentially, the problem is that what is required is the differential heat of solution atsaturation, whereas what we usually have in mind is an integral heat of solution. Note, forexample, that NaOH is more soluble with increasing temperature even though the (integral)heat of solution is negative. It seems best simply to state that most ionic substances are moresoluble at higher temperature, noting a few exceptions, and leave it at that.


You could concentrate your attention on Sections 12.4 through 12.8, omitting or condensingthe material in the rest of the chapter.



CHAPTER 13 Materials of Technology

This chapter looks at some of the metallic and nonmetallic materials of modern technology.For example, the chapter briefly discusses nanotechnology, in which one studies materialswith a view toward developing useful applications.


The chapter follows the basic chapters (1 to 12) describing the structure of matter. So, itprovides an opportunity to apply the concepts just learned. Alternately, the chapter could betreated with the descriptive chapters at the end of the course.


The chapter is divided into two parts, a part on metals (Sections 13.1 to 13.3) and a part onnonmetallic materials (Sections 13.4 to 13.8). Section 13.1 describes the natural sources of themetallic elements, Section 21.2 discusses metallurgy, and Section 21.3 explains the bonding inmetals in terms of molecular orbital theory. Section 13.4 describes the different forms of carbon,diamond, graphite, and the fullerenes, materials that have important uses in modern technol-ogy. Section 13.5 discusses the theory of semiconductors, which are the basis of solid-stateelectronics devices. Section 13.6 describes some chemistry of silicon (the basic material insemiconductor devices), silica, and the silicates. The final two sections of the chapter, Sections13.7 and 13.8, cover ceramics and composites (a material constructed of two or more differentkinds of materials).


Various selections of material are possible. Some of the frontiers work has been in the area ofnonmetallic materials, so you might restrict yourself to a selection from Sections 13.4 to 13.8.

CHAPTER 14 Rates of Reaction

This chapter looks at some important questions concerning chemical reactions: how fast dochemical reactions occur, what factors affect this rate, and how do reactions occur at themolecular level?


Although not essential for the equilibrium chapters that follow, chemical kinetics can givesome insight into chemical equilibrium. An alternate position for this discussion would beafter Chapter 20 and before Chapter 21. Early drafts of the book placed the chapter here, sothere is no difficulty with this order.


The chapter has been written to stress the experimental basis of the subject. It begins with thedefinition of reaction rate (Section 14.1) and its experimental measurement (Section 14.2). The


18 PART IV

next sections describe the dependence of rate on concentrations of substances (Section 14.3)and how the concentrations vary with time (Section 14.4). The following two sections (14.5and 14.6) discuss how rates of reaction vary with temperature. The second part of the chapterlooks at reaction mechanisms, first discussing elementary reactions (Section 14.7) and thenshowing how the reaction mechanism is related to the rate law (Section 14.8). The last section(14.9) describes catalysis, explaining it in terms of the reaction mechanism.

Special Notes

Chemical kinetics can be a rather abstract subject. The treatment given here emphasizesexperimental results in order to reduce this abstract character. Reaction mechanisms arecarefully described as explanations of experimental observations, and the provisional statusof these explanations is emphasized.


If time is short, you can pick and choose topics. Sections 14.1 and 14.3 through 14.5 areimportant ones to cover. Then, after a brief introduction to mechanisms, Section 14.9 oncatalysis would complete the discussion of the factors affecting reaction rates.

CHAPTER 15 Chemical Equilibrium

This chapter begins a block of chapters dealing with chemical equilibrium. The method usedto solve equilibrium problems is described here, and that method is used uniformly through-out the chapters on equilibrium calculations. Gaseous reactions are used to illustrate theprinciples.


Although this placement of equilibrium is a typical one, an alternate sequence in whichthermodynamics precedes chemical equilibrium is described in Part III.


The chapter begins with three sections (15.1 through 15.3) describing chemical equilibriumand defining the equilibrium constant. The second part of the chapter (Sections 15.4 through15.6) discusses the use of the equilibrium constant. Thus, Section 15.5 introduces the conceptof reaction quotient, and Section 15.6 describes how to use the equilibrium constant tocalculate equilibrium concentrations. The last three sections of the chapter (Sections 15.7through 15.9) discuss how changing the conditions in a reaction affects the yield of product.

Special Notes

Section 15.1 introduces chemical equilibrium as a dynamic equilibrium. This discussion isself-contained and does not rely on Chapter 14. A brief subsection in Section 15.2 (Equilib-rium—A Kinetics Argument) does use the concepts of reaction rate from Chapter 14 to obtainthe equilibrium constant, and you may want to omit it if you have not covered Chapter 14.



Example 15.1 asks for the equilibrium amounts of substances given the starting amountsand the amount of one substance in the equilibrium mixture. The problem is essentially onein stoichiometry, but it is set up in the way equilibrium problems will be set up throughout(with a table of starting, change, and equilibrium amounts). Thus, the example forms a bridgefrom stoichiometry, which students are familiar with, to equilibrium problems. The firstexample of a calculation of equilibrium concentrations from the equilibrium constant is givenin Example 15.7. It is described as a three-step problem: set up a table with starting, change,and equilibrium concentrations; substitute the expressions for equilibrium concentrations intothe equilibrium equation; solve the equation. Note that the table is set up to parallel thechemical equation, with coefficients of x corresponding to coefficients in the chemical equa-tion.

Note that the common statement of Le Chatelier’s principle in terms of “stresses” is toovague and may give the wrong result unless the stresses are restricted to intensive variables.See R. S. Treptow, J. Chem. Educ. 1980, 57, 417–420; see also I. N. Levine, Physical Chemistry, 3rded.; Wiley: New York, 1988; p. 186.


Because of the importance of the topic, most of the chapter should be covered if possible. Eachsection gives added insight into the concept of equilibrium. However, a basic treatment wouldinclude the introduction to equilibrium and the equilibrium constant (Sections 15.1 through15.3), plus calculations with Kc (Section 15.6). Le Chatelier’s principle applied to adding orremoving substances (Section 15.7) gives students a qualitative way of looking at such thingsas the common-ion effect (treated later in Sections 17.5 and 18.2). The reaction quotient,introduced in Section 15.5, is used to discuss precipitation in Chapter 18, but the discussionthere (Section 18.3) will stand on its own.

CHAPTER 16 Acids and Bases

This chapter discusses the three main acid–base concepts and introduces the pH conceptneeded for the next chapter.


The chapter consists of three parts: acid–base concepts, acid and base strengths, and self-ioni-zation of water and pH. Section 16.1 reviews the Arrhenius concept, which was discussed inChapter 4. Section 16.2 describes the Brønsted–Lowry concept in more detail than was donein Chapter 4. Section 16.3 describes the Lewis concept of acids and bases. After describingthese acid–base concepts, Sections 16.4 and 16.5 discuss the relative strengths of acids andbases and the relationship of acid strength to molecular structure. The last sections (16.6through 16.8) discuss the concepts of self-ionization of water and of pH, which are central tounderstanding the following chapter.

Special Notes

Acids and bases were discussed in Chapter 4. Chapter 16 is a more complete discussion,including the pH concept.


20 PART IV


If you spent a fair bit of time on acids and bases in lecturing on Chapter 4, you could treat thefirst part of Chapter 16 briefly. Sections 16.3 through 16.5 are optional. The sections onself-ionization and pH are important for the next chapter.

CHAPTER 17 Acid–Base Equilibria

This chapter looks quantitatively at acid–base equilibria, including hydrolysis, common-ioneffect, and buffers.


The chapter could follow rather than precede Chapter 18 on solubility. In that case, Sections18.4 and 18.7 should be postponed until Ka is introduced.


The chapter is organized in two parts. The first part deals with solutions containing a weakacid or base, and the second part with solutions of a weak acid or base to which a commonion is added. Sections 17.1 through 17.4 treat acid–base equilibria in a unified way, progressingfrom monoprotic to polyprotic acids, then to bases and to salts. Once the method describedin Examples 17.2 and 17.3 for weak acids is mastered, the treatment of base and salt solutionsis seen to be essentially the same. Section 17.5 looks specifically at the common-ion effect;then Section 17.6 discusses buffers. The final section looks at pH changes during acid–basetitrations.

Special Notes

The unity of the subject of equilibrium calculations has been stressed; the general method wasdescribed in Example 15.7.

To obtain the 5% rule stated just before Example 17.3, start with the equilibrium equation(C is the acid concentration).

Ka = x2

C − x

Then,

x2 = Ka(C − x)

x = √KaC (1 − x/C)1/2 � √KaC [1 − x/(2C)]

The expression on the right was obtained by retaining only the first two terms of a powerseries expansion of [1 – x/(2C)]1/2. If you drop the term in x in this expression (equal toKaC[x/(2C)], the solution of the equation is equivalent to dropping x in the denominator of



the equilibrium equation, in which case x = (KaC)1/2. The fractional error in this approximationis

x2C

= √KaC

2C =

√Ka/C2

If the error is to be less than or equal to 5%,

√Ka/C2

× 100 ≤ 5

which is equivalent to the rule in the text.


The most important material in the chapter concerns acid–base ionization equilibria (Sections17.1 and 17.3). The remaining sections can be emphasized or not depending on the timeavailable.

CHAPTER 18 Solubility and Complex-Ion Equilibria

The chapter deals with solubility and precipitation, as well as complex-ion formation and itseffect on solubility.


This chapter can be treated before acid–base equilibria, except for Sections 18.4 and 18.7, whichshould be postponed until Ka is discussed.


Solubility equilibria are described in the first part of the chapter, starting with a discussion ofKsp (Section 18.1) and following with the common-ion effect (Section 18.2), precipitationcalculations (Section 18.3), and the effect of pH on solubility (Section 18.4). The next part looksat complex-ion equilibria, discussing the formation of complex ions (Section 18.5) and theeffect of complex-ion formation on solubility (Section 18.6). The final section of the chapterbriefly describes the qualitative analysis scheme for metal ions.

Special Notes

Both Chapters 17 and 18 are important to understanding the qualitative analysis of metal ions,but Chapter 18 is especially pertinent. Note the inclusion of fractional precipitation in Section18.3 and the separation of metal ions by sulfide precipitation in Section 18.4, both of whichare useful in discussing qualitative analysis.


22 PART IV


The most important topics involving Ksp are covered in Sections 18.1 through 18.3 (to the endof Example 18.7). The coverage of the rest of the chapter can be tailored to individualcircumstances. Thus, if the course is accompanied by extensive laboratory work in qualitativeanalysis, Section 18.7 could be omitted.

CHAPTER 19 Thermodynamics and Equilibrium

This chapter looks at the spontaneity of a chemical reaction in terms of the thermodynamicconcepts of entropy and free energy. The equilibrium constant is related to free energy in thelast two sections.


The chapter has been placed just before the chapter on electrochemistry and after the block ofchapters on chemical equilibrium, but it could be covered earlier. For example, it could easilyfollow Chapter 14, or it could even be covered before Chapter 14 if Sections 19.6 and 19.7(which refer to the equilibrium constant) are postponed until the equilibrium constant isdefined. Chapter 19 has been divided into three parts, with the last part treating free energyand equilibrium constants, to facilitate this rearrangement.


The chapter begins with a section that reviews thermochemistry and introduces the first lawof thermodynamics. It also gives the calculation of ∆H for the preparation of urea from NH3and CO2. Later, we calculate ∆S and ∆G for this reaction, which was introduced in the chapteropening, as an illustration of thermodynamic concepts. Entropy is discussed in Sections 19.2and 19.3, and free energy in the remainder of the chapter. Spontaneity of a reaction and freeenergy are related in Section 19.4, and further interpretation of free energy is given in Section19.5. Free energy is related to the equilibrium constant in Section 19.6, and then the change offree energy with temperature is discussed in Section 19.7.

Special Notes

The emphasis of the chapter is on using thermodynamics to obtain a criterion of spontaneityfor a chemical reaction. This provides the unifying thread of the chapter, giving a cleanpresentation without getting bogged down in detail. The criterion that the quantity ∆H – ∆Sis negative for a spontaneous change is developed immediately from the second law ofthermodynamics (at the end of Section 19.2). This sets the stage for the next two sections: first,how to find ∆S; then, defining free energy as a convenient way to express this criterion ofspontaneity in terms of a single quantity.




The basic sections of the chapter are 19.2 through 19.4. The remaining sections may be coveredto the extent desired. Section 19.5 simply amplifies the meaning of free energy, and Section19.6 links free energy to the equilibrium constant.

CHAPTER 20 Electrochemistry

This chapter looks at electrochemistry and what it has to say about the spontaneity of reactionand about the electrolytic decomposition of a substance.


The chapter has been placed after Chapter 19 to make use of the concepts of spontaneousreaction and free energy introduced there.


The chapter is divided into three parts. The first part (Section 20.1) treats the balancing ofoxidation–reduction reactions. The second part of the chapter deals with voltaic cells. Section20.2 describes the construction of a voltaic cell, and Section 20.3 introduces notation used torepresent such a cell. Sections 20.4 and 20.5 discuss the concept of cell emf, Section 20.6 relatesthe cell emf to the reaction equilibrium, and Section 20.7 looks at the dependence of cell emfon reactant concentrations. The final section (20.8) of this part describes some commercialvoltaic (galvanic) cells.

The third part of the chapter deals with electrolytic cells. Section 20.9 discusses theelectrolysis of molten salts; Section 20.10 describes the electrolysis of aqueous solutions. Thelast section (20.11) treats the stoichiometry of electrolysis.

Special Notes

Electrolytic cells have been treated at the end of the chapter to use the concept of electrodepotential in discussing the ease of electrolytic decomposition in Section 20.10. However,Section 20.11 on the stoichiometry of electrolysis could be moved to the beginning of thechapter.


The basic material is covered in Sections 20.2 through 20.5 and 20.11. Other sections may beomitted if you are pressed for time.

CHAPTER 21 Nuclear Chemistry

This chapter looks at nuclear reactions, the rate of radioactive decay, and the energy changethat results from nuclear fission and nuclear fusion.


24 PART IV


This chapter returns to rates of reaction (introduced in Chapter 14) after the block of equilib-rium chapters (Chapters 15–20). Chapter 14 provides background for discussing rate ofradioactive decay. However, the pertinent equations for rate of radioactive decay are given inthe text, so the chapter could be covered before chemical kinetics if desired. You could easilytreat the material on nuclear reactions in Sections 21.1 and 21.2 just after the discussion ofnuclear structure in Section 2.3 if that seems desirable.


The chapter is divided into two parts, the first one looking at various nuclear reactions andcharacteristics of radioactivity and the second looking at the energy of nuclear reactions. Thefirst part begins by describing the types of nuclear reactions: radioactivity (Section 21.1) andnuclear bombardment reactions (Section 21.2). The chapter then discusses ways to detect theradiations from radioactive decay (Section 21.3). The next section (21.4) looks at the rate ofradioactive decay and the half-life of a radioactive isotope. Section 21.5 discusses chemicaland medical applications of radioactive isotopes. The second part of the chapter begins byexplaining the concepts of mass–energy equivalence and nuclear binding energy (Section21.6). The chapter ends with a discussion of nuclear fission and fusion (Section 21.7).

Special Notes

We have avoided using the expression “conversion of mass to energy,” which is misleading.Consider a system consisting of an electron and a positron. The rest mass of this systemchanges from 2me to zero if the two particles collide and annihilate one another. But thesurroundings increase in mass by 2me by absorbing the two gamma-ray photons that areemitted. Therefore, the total mass remains constant. Similarly, the total energy remainsconstant. The equation E = mc2 simply states that if a system has a given quantity of mass, ithas a quantity of energy equal to mc2. See R. P. Bauman, J. Chem. Educ. 1966, 43, 366, and R. S.Treptow, J. Chem. Educ. 1986, 63, 103.


The basic material is in Sections 21.1, 21.2, 21.4, and 21.6. Other sections may be omitted.

CHAPTER 22 Chemistry of the Main-Group Elements

This chapter looks at the descriptive chemistry of the main-group or representative elements.


The principles introduced in the previous chapters are applied to the chemistry of the elementsin Chapters 22 through 25. Much of this material could be covered earlier, however.




The first section of the chapter (22.1) deals with the main-group elements in general beforegoing on to the sections describing the families of elements. The first part of the chapterdiscusses the main-group metals (Sections 22.2 through 22.4). The second part of the chapterdiscusses the main-group nonmetals (Sections 22.5 through 22.10).

Special Notes

The periodic table is used to correlate information about the main-group elements. Section22.1 provides the overview for this discussion. Within each group, we concentrate on one ortwo elements to focus the discussion. By discussing only one element in a section, you canpick and choose sections to meet your needs. You can include only a few elements or all ofthose described.


You can pick as many or as few elements as you wish from this chapter.

CHAPTER 23 The Transition Elements

This chapter looks at some descriptive chemistry of the transition elements and at the structureof coordination compounds.


This chapter continues the discussion of the descriptive chemistry of the elements, whichbegan in Chapter 22.


The chapter consists of two parts, the first looking at the properties of the transition elementsand the second looking at complexes and coordination compounds. Section 23.1 describes theperiodic trends seen in the properties of the transition elements, and Section 23.2 looks at somechemistry of chromium and copper. Section 23.3 begins the second part by discussing theformation and structure of complexes; basic terms are defined here. Section 23.4 introducesthe nomenclature of coordination compounds, and Section 23.5 discusses isomerism. The lasttwo sections look at the electronic structure of transition-metal complexes in terms of valencebond theory (Section 23.6) and crystal field theory (Section 23.7).


You can tailor the material to the needs of your particular course. If you wish to concentrateon coordination compounds, you can begin the chapter with Section 23.3. In looking at theelectronic structure of complexes, you could limit the treatment to octahedral complexes,omitting any discussion of tetrahedral and square planar complexes.


26 PART IV

CHAPTER 24 Organic Chemistry

This chapter gives a brief introduction to organic chemistry.


The chapter is part of the block of chapters covering the descriptive chemistry of the elements.Properties of various organic compounds were treated in earlier chapters, but this is the firstorganized treatment of the subject other than what was done in Section 2.7. It is requiredbackground for the next chapter on polymer molecules. Although it occurs near the end ofthe book, this discussion of organic chemistry is easily moved to an earlier position in thegeneral chemistry course.


The first section (24.1) discusses the bonding of carbon. After that the chapter is dividedlogically into the study of hydrocarbons and derivatives of hydrocarbons. In the first part,Sections 24.2 through 24.4 explores the structure of different series of hydrocarbons. Section24.5 then describes the naming of these hydrocarbons. In the second part of the chapter,Sections 24.6 and 24.7 discusses oxygen and nitrogen derivatives of the hydrocarbons.


One possibility for an abridged treatment would cover Sections 24.1, 24.2, 24.3, and 24.6.

CHAPTER 25 Polymer Materials: Synthetic and Biological

This chapter is a brief introduction into polymers.


The chapter relies on the discussion of organic chemistry in Chapter 24.


The chapter is divided into two parts, one on synthetic polymers and the other on biologicalpolymers, with an emphasis on proteins and their biosynthesis starting with the genetic code.In the first part of the chapter, Section 25.2 describes the synthesis of organic polymers. Section25.3 describes electrical conducting polymers. In the second part of the chapter, 25.3 describesproteins. Section 25.4 describes nucleic acids.


You could choose to cover only one part of the chapter, either the one on synthetic polymersor the one on biological polymers.



PART VOperational Skills Masterlist

Chapter 1 Chemistry and Measurement Examples Exercises Problems

1. Using the law of conservation ofmass Given the masses of all substancesin a chemical reaction except one, calcu-late the mass of this one substance.

1.1 1.1 1.31, 1.32,1.33, 1.34

2. Using significant figures in calcula-tions Given an arithmetic setup, reportthe answer to the correct number of sig-nificant figures and round it properly.

1.2 1.3 1.55, 1.56

3. Converting from one temperaturescale to another Given a temperaturereading on one scale, convert it to an-other scale—Celsius, Kelvin, or Fahren-heit.

1.3 1.5 1.63, 1.69,1.65, 1.66

4. Calculating the density of a sub-stance Given the mass and volume of asubstance, calculate the density.

1.4 1.6 1.67, 1.68,1.69, 1.70

5. Using the density to relate mass andvolume Given the mass and density of asubstance, calculate the volume; or giventhe volume and density, calculate themass.

1.5 1.7 1.71, 1.72,1.73, 1.74

6. Converting units Given an equationrelating one unit to another (or a series ofsuch equations), convert a measurementexpressed in one unit to a new unit.

1.6, 1.7, 1.8 1.8, 1.9,1.10

1.75, 1.76,1.77, 1.78,1.79, 1.80,1.81, 1.82,1.83, 1.84


Chapter 2 Atoms, Molecules, and Ions Examples Exercises Problems

1. Writing nuclide symbols Given thenumber of protons and neutrons in a nu-cleus, write its nuclide symbol.

2.1 2.1 2.41, 2.42

2. Determining atomic weight from iso-topic masses and fractional abun-dances Given the isotopic masses (inatomic mass units) and fractional isotopicabundances for a naturally occurring ele-ment, calculate its atomic weight.

2.2 2.2 2.45, 2.46,2.47, 2.48

3. Writing an ionic formula, given theions Given the formulas of a cation andan anion, write the formula of the ioniccompound of these ions.

2.3 2.4 2.69, 2.70

4. Writing a name of a compound fromits formula, or vice versa Given thename of a simple compound (ionic, bi-nary molecular, acid, or hydrate), writethe name, or vice versa.

2.4, 2.5,2.6, 2.7,2.10, 2.11

2.5, 2.6, 2.7,2.8, 2.11,2.12

2.71, 2.72,2.73, 2.74,2.77, 2.78,2.79, 2.80,2.85, 2.86,2.87, 2.88

5. Writing the name of a binary molecu-lar compound from its molecularmodel Given the molecular model of abinary compound, write the name.

2.8 2.9 2.81, 2.82

6. Writing the name and formula of ananion from the acid Given the nameand formula of an oxoacid, write thename and formula of the oxoanion; orfrom the name and formula of the oxoan-ion, write the formula and name of theoxoacid.

2.9 2.10 2.83, 2.84

7. Balancing simple equations Giventhe formulas of the reactants and prod-ucts in a chemical reaction, obtain thecoefficients of the balanced equation.

2.12 2.13 2.91, 2.92


Operational Skills Masterlist 29

Chapter 3 Calculations with ChemicalFormulas and Equations Examples Exercises Problems

Note: A table of atomic weights is necessary formost of these skills.

1. Calculating the formula weight froma formula or molecular model Giventhe formula of a compound and a table ofatomic weights, calculate the formulaweight.

3.1, 3.2 3.1, 3.2 3.21, 3.22,3.23, 3.24

2. Calculating the mass of an atom ormolecule Using the molar mass andAvogadro’s number, calculate the mass ofan atom or molecule in grams.

3.3 3.3 3.27, 3.28,3.29, 3.30

3. Converting moles of substance tograms, and vice versa Given the molesof a compound with a known formula,calculate the mass. Or, given the mass ofa compound with a known formula, cal-culate the moles.

3.4, 3.5 3.4, 3.5 3.31, 3.32,3.33, 3.34,3.35, 3.36

4. Calculating the number of moleculesin a given mass Given the mass of asample of a molecular substance and itsformula, calculate the number of mole-cules in the sample.

3.6 3.6 3.39, 3.40,3.41, 3.42

5. Calculating the percentage composi-tion from the formula Given the for-mula of a compound, calculate the masspercentages of the elements in it.

3.7 3.7 3.51, 3.52,3.53, 3.54

6. Calculating the mass of an elementin a given mass of compound Giventhe mass percentages of elements in agiven mass of a compound, calculate themass of any element.

3.8 3.8 3.55, 3.56

7. Calculating the percentages of C andH by combustion Given the masses ofCO2 and H2O obtained from the combus-tion of a known mass of a compound ofC, H, and O, compute the mass percent-ages of each element.

3.9 3.9 3.57, 3.58


30 PART V

8. Determining the empirical formulafrom percentage composition Given themasses of elements in a known mass ofcompound, or given its percentage com-position, obtain the empirical formula.

3.10, 3.11 3.10, 3.11 3.59, 3.60,3.61, 3.62,3.63, 3.64

9. Determining the molecular formulafrom percentage composition and mo-lecular weight Given the empirical for-mula and molecular weight of asubstance, obtain its molecular formula.

3.12 3.12 3.67, 3.68,3.69, 3.70

10. Relating quantities in a chemicalequation Given a chemical equationand the amount of one substance, calcu-late the amount of another substance in-volved in the reaction.

3.13, 3.14 3.14, 3.15,3.16

3.77, 3.78,3.79, 3.80,3.81, 3.82,3.83, 3.84

11. Calculating with a limiting reac-tant Given the amounts of reactants andthe chemical equation, find the limitingreactant; then calculate the amount of aproduct.

3.15, 3.16 3.17, 3.18 3.85, 3.86,3.87, 3.88

Examples Exercises Problems



Chapter 4 Chemical Reactions: An Introduction Examples Exercises Problems

1. Using the solubility rules Cover theformula of an ionic compound, predict itssolubility in water.

4.1 4.1 4.23, 4.24

2. Writing net ionic equations Given amolecular equation, write the correspond-ing net ionic equation.

4.2 4.2 4.27, 4.28

3. Deciding whether precipitation willoccur Using solubility rules, decidewhether two soluble ionic compoundswill react to form a precipitate. If theywill, write the net ionic equation.

4.3 4.3 4.31, 4.32,4.33, 4.34

4. Classifying acids and bases as strongor weak Given the formula of an acid orbase, classify it as strong or weak.

4.4 4.4 4.35, 4.36

5. Writing an equation for a neutraliza-tion Given an acid and a base, write themolecular equation and then the net ionicequation for the neutralization reaction.

4.5 4.5 4.37, 4.38,4.39, 4.40

6. Writing an equation for a reactionwith gas formation Given the reactionbetween a carbonate, sulfide, or sulfiteand an acid, write the molecular and netionic equations.

4.6 4.7 4.45, 4.46,4.47, 4.48

7. Assigning oxidation numbers Giventhe formula of a simple compound orion, obtain the oxidation numbers of theatoms, using the rules for assigning oxi-dation numbers.

4.7 4.8 4.49, 4.50,4.51, 4.52

8. Balancing equations by the half-reac-tion method Given the skeleton equa-tion for an oxidation–reduction equation,complete and balance it.

4.8 4.9 4.59, 4.60

9. Calculating molarity from mass andvolume Given the mass of the soluteand the volume of the solution, calculatethe molarity.

4.9 4.10 4.61, 4.62,4.63, 4.64


32 PART V

10. Using molarity as a conversion fac-tor Given the volume and molarity of asolution, calculate the amount of solute.Or, given the amount of solute and themolarity of a solution, calculate thevolume.

4.10 4.12 4.65, 4.66,4.67, 4.68,4.69, 4.70,4.71, 4.72

11. Diluting a solution Calculate thevolume of solution of known molarity re-quired to make a specified volume of so-lution with different molarity.

4.11 4.13 4.73, 4.74

12. Determining the amount of speciesby gravimetric analysis Given theamount of a precipitate in a gravimetricanalysis, calculate the amount of a re-lated species.

4.12 4.14 4.77, 4.78

13. Calculating the volume of reactantsolution needed Given the chemicalequation, calculate the volume of solu-tion of known molarity of one substancethat just reacts with a given volume of so-lution of another substance.

4.13 4.15 4.83, 4.84

14. Calculating the quantity of sub-stance in a titrated solution Calculatethe mass of one substance that reactswith a given volume of known molarityof solution of another substance.

4.14 4.16 4.85, 4.86




Chapter 5 The Gaseous State Examples Exercises Problems

1. Relating liquid height and pressureGiven the density of a liquid used in a ba-rometer or manometer and the height ofthe column of liquid, obtain the pressurereading in mmHg.

5.1 5.1 5.31, 5.32

2. Using the empirical gas laws Givenan initial volume occupied by a gas, cal-culate the final volume when the pres-sure changes at fixed temperature; whenthe temperature changes at fixed pres-sure; and when both pressure and tem-perature change.

5.2, 5.3, 5.4 5.2, 5.3, 5.4 5.33, 5.34,5.35, 5.36,5.39, 5.40,5.41, 5.42,5.45, 5.46

3. Deriving empirical gas laws from theideal gas law Starting from the ideal gaslaw, derive the relationship between anytwo variables.

5.5 5.5 5.49, 5.50

4. Using the ideal gas law Given anythree of the variables P, V, T, and n for agas, calculate the fourth one from theideal gas law.

5.6 5.6 5.51, 5.52,5.53, 5.54,5.55, 5.56

5. Relating gas density and molecularweight Given the molecular weight, cal-culate the density of a gas for a particulartemperature and pressure; or, given thegas density, calculate the molecularweight.

5.7, 5.8 5.7, 5.8 5.57, 5.58,5.59, 5.60,5.61, 5.62,5.63, 5.64

6. Solving stoichiometry problems in-volving gas volumes Given the volume(or mass) of one substance in a reaction,calculate the mass (or volume) of anotherproduced or used up.

5.9 5.9 5.67, 5.68,5.69, 5.70,5.71, 5.72

7. Calculating partial pressures andmole fractions of a gas in a mixtureGiven the masses of gases in a mixture,calculate the partial pressures and molefractions.

5.10 5.10 5.75, 5.76,5.77, 5.78

8. Calculating the amount of gas col-lected over water Given the volume, to-tal pressure, and temperature of gascollected over water, calculate the massof the dry gas.

5.11 5.11 5.81, 5.82


34 PART V

9. Calculating the rms speed of gasmolecules Given the molecular weightand temperature of a gas, calculate therms molecular speed.

5.12 5.12, 5.13 5.83, 5.84,5.85, 5.86,5.87, 5.88

10. Calculating the ratio of effusionrates of gases Given the molecularweights of two gases, calculate the ratioof rates of effusion; or, given the relativeeffusion rates of a known and an un-known gas, obtain the molecular weightof the unknown gas (as in Exercise 5.15).

5.13 5.14, 5.15 5.89, 5.90,5.91, 5.92,5.93, 5.94

11. Using the van der Waals equationGiven n, T, V, and the van der Waals con-stants a and b for a gas, calculate the pres-sure from the van der Waals equation.

5.14 5.16 5.95, 5.96




Chapter 6 Thermochemistry Examples Exercises Problems

1. Calculating kinetic energy Giventhe mass and speed of an object, calculatethe kinetic energy.

6.1 6.1 6.37, 6.38,6.39, 6.40

2. Writing thermochemical equationsGiven a chemical equation, states of sub-stances, and the quantity of heat ab-sorbed or evolved for molar amounts,write the thermochemical equation.

6.2 6.3 6.45, 6.46

3. Manipulating thermochemical equa-tions Given a thermochemical equation,write the thermochemical equation fordifferent multiples of the coefficients orfor the reverse reaction.

6.3 6.4 6.47, 6.48,6.49, 6.50

4. Calculating the heat of reaction fromthe stoichiometry Given the value of∆H for a chemical equation, calculate theheat of reaction for a given mass of reac-tant or product.

6.4 6.5 6.51, 6.52,6.53, 6.54

5. Relating heat and specific heatGiven any three of the quantities q, s, m,and ∆t, calculate the fourth one.

6.5 6.6 6.57, 6.58

6. Calculating ∆H from calorimetricdata Given the amounts of reactants andthe temperature change of a calorimeterof specified heat capacity, calculate theheat of reaction.

6.6 6.7 6.61, 6.62,6.63, 6.64

7. Applying Hess’s law Given a set ofreactions with enthalpy changes, calcu-late ∆H for a reaction obtained from theseother reactions by using Hess’s law.

6.7 6.8 6.65, 6.66,6.67, 6.68

8. Calculating the heat of phase tran-sition from standard enthalpies offormation Given a table of standardenthalpies of formation, calculate theheat of phase transition.

6.8 6.9 6.71, 6.72

9. Calculating the enthalpy of reactionfrom standard enthalpies of formationGiven a table of standard enthalpies offormation, calculate the enthalpy of reac-tion.

6.9 6.10, 6.11 6.73, 6.74,6.75, 6.76,6.77, 6.78


36 PART V

Chapter 7 Quantum Theory of the Atom Examples Exercises Problems

1. Relating wavelength and frequencyof light Given the frequency of light, cal-culate the wavelength, or vice versa.

7.1, 7.2 7.1, 7.2 7.29, 7.30,7.31, 7.32

2. Calculating the energy of a photonGiven the frequency or wavelength oflight, calculate the energy associated withone photon.

7.3 7.3 7.37, 7.38,7.39, 7.40

3. Determining the wavelength or fre-quency of a hydrogen atom transitionGiven the initial and final principal quan-tum numbers for an electron transition inthe hydrogen atom, calculate the fre-quency or wavelength of light emitted.You need the value of RH.

7.4 7.4 7.43, 7.44,7.45, 7.46

4. Applying the de Broglie relationGiven the mass and speed of a particle,calculate the wavelength of the associ-ated wave.

7.5 7.6 7.51, 7.52

5. Using the rules for quantum num-bers Given a set of quantum numbers n,l, ml, and ms, state whether that set is per-missible for an electron.

7.6 7.7 7.63, 7.64



Chapter 8 Electron Configurationsand Periodicity Examples Exercises Problems

1. Applying the Pauli exclusion princi-ple Given an orbital diagram or electronconfiguration, decide whether it is possi-ble or not, according to the Pauli exclu-sion principle.

8.1 8.1 8.35, 8.36,8.37, 8.38

2. Determining the configuration of anatom using the building-up principleGiven the atomic number of an atom,write the complete electron configurationfor the ground state, according to thebuilding-up principle.

8.2 8.2 8.41, 8.42,8.43, 8.44

3. Determining the configuration of anatom using the period and group num-bers Given the period and group for anelement, write the configuration of theouter electrons.

8.3 8.3, 8.4 8.45, 8.46,8.47, 8.48,8.49, 8.50

4. Applying Hund’s rule Given theelectron configuration for the groundstate of an atom, write the orbital dia-gram.

8.4 8.5 8.51, 8.52

5. Applying periodic trends Using theknown trends and referring to a periodictable, arrange a series of elements in or-der by atomic radius or ionization energy.

8.5, 8.6 8.6, 8.7 8.55, 8.56,8.57, 8.58


38 PART V

Chapter 9 Ionic and Covalent Bonding Examples Exercises Problems

1. Using Lewis symbols to representionic bond formation Given a metallicand a nonmetallic main-group element,use Lewis symbols to represent the trans-fer of electrons to form ions of noble-gasconfigurations.

9.1 9.1 9.31, 9.32

2. Writing electron configurations ofions Given an ion, write the electronconfiguration. For an ion of a main-groupelement, give the Lewis symbol.

9.2, 9.3 9.2, 9.3, 9.4 9.33, 9.34,9.35, 9.36,9.37, 9.38

3. Using periodic trends to obtain rela-tive ionic radii Given a series of ions, ar-range them in order of increasing ionicradius.

9.4 9.7 9.43, 9.44

4. Using electronegativities to obtainrelative bond polarities Given the elec-tronegativities of the atoms, arrange a se-ries of bonds in order by polarity.

9.5 9.8 9.51, 9.52

5. Writing Lewis formulas Given themolecular formula of a simple compoundor ion, write the Lewis electron-dot for-mula.

9.6, 9.7,9.8, 9.10

9.9, 9.10,9.11, 9.13

9.55, 9.56,9.57, 9.58,9.59, 9.60,9.65, 9.66

6. Writing resonance formulas Given asimple molecule with delocalized bond-ing, write the resonance description.

9.9 9.12 9.61, 9.62,9.63, 9.64

7. Using formal charges to determinethe best Lewis formula Given two ormore Lewis formulas, use formal chargesto determine which formula best de-scribes the electron distribution or givesthe most plausible molecular structure.

9.11 9.15 9.71, 9.72

8. Relating bond order and bondlength Know the relationship betweenbond order and bond length.

9.12 9.17 9.77, 9.78

9. Estimating ∆H from bond energiesGiven a table of bond energies, estimatethe heat of reaction.

9.13 9.18 9.79, 9.80



Chapter 10 Molecular Geometryand Chemical Bonding Theory Examples Exercises Problems

1. Predicting molecular geometriesGiven the formula of a simple molecule,predict its geometry, using the VSEPRmodel.

10.1, 10.2 10.1, 10.2 10.27, 10.28,10.29, 10.30,10.33, 10.34,10.35, 10.36

2. Relating dipole moment and molecu-lar geometry State what geometries of amolecule AXn are consistent with the in-formation that the molecule has anonzero dipole moment.

10.3 10.3, 10.4 10.37, 10.38,10.39, 10.40

3. Applying valence bond theoryGiven the formula of a simple molecule,describe its bonding, using valence bondtheory.

10.4, 10.5,10.6

10.5, 10.6,10.7

10.41, 10.42,10.43, 10.44,10.47, 10.48,10.49, 10.50,10.51, 10.52

4. Describing molecular orbital configu-rations Given the formula of a diatomicmolecule obtained from first- or second-period elements, deduce the molecularorbital configuration, the bond order, andwhether the molecular substance is dia-magnetic or paramagnetic.

10.7, 10.8 10.9, 10.10 10.55, 10.56,10.57, 10.58


40 PART V

Chapter 11 States of Matter; Liquids and Solids Examples Exercises Problems

1. Calculating the heat required for aphase change of a given mass of sub-stance Given the heat of fusion (or va-porization) of a substance, calculate theamount of heat required to melt (or va-porize) a given quantity of that substance.

11.1 11.1 11.37, 11.38,11.39, 11.40

2. Calculating vapor pressures andheats of vaporization Given the vaporpressure of a liquid at one temperatureand its heat of vaporization, calculate thevapor pressure at another temperature.Given the vapor pressures of a liquid attwo temperatures, calculate the heat ofvaporization.

11.2, 11.3 11.2, 11.3 11.43, 11.44,11.45, 11.46

3. Relating the conditions for the lique-faction of gases to the critical tempera-ture Given the critical temperature andpressure of a substance, describe the con-ditions necessary for liquefying the gase-ous substance.

11.4 11.4 11.51, 11.52

4. Identifying intermolecular forcesGiven the molecular structure, state thekinds of intermolecular forces expectedfor a substance.

11.5 11.5 11.57, 11.58

5. Determining relative vapor pressureon the basis of intermolecular attrac-tion Given two liquids, decide on the ba-sis of the intermolecular forces which hasthe higher vapor pressure at a given tem-perature or which has the lower boilingpoint.

11.6 11.6, 11.7 11.61, 11.62,11.63, 11.64

6. Identifying types of solids Fromwhat you know about the bonding in asolid, classify it as molecular, metallic,ionic, or covalent network.

11.7 11.8 11.67, 11.68,11.69, 11.70

7. Determining the relative meltingpoints based on types of solids Given alist of substances, arrange them in orderof increasing melting point from whatyou know of their structures.

11.8 11.9 11.71, 11.72



8. Determining the number of atomsper unit cell Given the description of aunit cell, find the number of atoms percell.

11.9 11.10 11.77, 11.78

9. Calculating atomic mass from unit-cell dimension and density Given theedge length of the unit cell, the crystalstructure, and the density of a metal, cal-culate the mass of a metal ion.

11.10 11.11 11.79, 11.80

10. Calculating unit-cell dimensionfrom unit-cell type and density Giventhe unit-cell structure, the density, andthe atomic weight for an element, calcu-late the edge length of the unit cell.

11.11 11.12 11.81, 11.82



42 PART V

Chapter 12 Solutions Examples Exercises Problems

1. Applying Henry’s law Given thesolubility of a gas at one pressure, find itssolubility at another pressure.

12.1 12.4 12.41, 12.42

2. Calculating solution concentrationGiven the mass percent of solute, statehow to prepare a given mass of solution.Given the masses of solute and solvent,find the molality and mole fractions.

12.2, 12.3,12.4

12.5, 12.6,12.7

12.43, 12.44,12.45, 12.46,12.47, 12.48,12.49, 12.50,12.51, 12.52

3. Converting concentration unitsGiven the molality of a solution, calculatethe mole fractions of solute and solvent;and given the mole fractions, calculatethe molality. Given the density, calculatethe molarity from the molality, and viceversa.

12.5, 12.6,12.7, 12.8

12.8, 12.9,12.10, 12.11

12.53, 12.54,12.55, 12.56,12.57, 12.58,12.59, 12.60

4. Calculating vapor-pressure lower-ing Given the mole fraction of solute ina solution of nonvolatile, undissociatedsolute and the vapor pressure of pure sol-vent, calculate the vapor-pressure lower-ing and vapor pressure of the solution.

12.9 12.12 12.61, 12.62

5. Calculating boiling-point elevationand freezing-point depression Giventhe molality of a solution of nonvolatile,undissociated solute, calculate the boil-ing-point elevation and freezing-point de-pression.

12.10 12.13 12.63, 12.64

6. Calculating molecular weightsGiven the masses of solvent and soluteand the molality of the solution, find themolecular weight of the solute. Given themasses of solvent and solute, the freezing-point depression, and Kf, find the molecu-lar weight of the solute.

12.11, 12.12 12.14, 12.15 12.67, 12.68,12.69, 12.70

7. Calculating osmotic pressure Giventhe molarity and the temperature of a so-lution, calculate its osmotic pressure.

12.13 12.16 12.71, 12.72



8. Determining colligative propertiesof ionic solutions Given the concentra-tion of ionic compound in a solution, cal-culate the magnitude of a colligativeproperty; if i is not given, assume thevalue based on the formula of the ioniccompound.

12.14 12.17 12.73, 12.74



44 PART V

Chapter 13 Materials of Technology Examples Exercises Problems

Note: The problem-solving skills used in this chapter are discussed in previous chapters.



Chapter 14 Rates of Reaction Examples Exercises Problems

1. Relating the different ways of ex-pressing reaction rates Given the bal-anced equation for a reaction, relate thedifferent possible ways of defining therate of the reaction.

14.1 14.1 14.33, 14.34

2. Calculating the average reactionrate Given the concentration of reactantor product at two different times, calcu-late the average rate of reaction over thattime interval.

14.2 14.2 14.37, 14.38

3. Determining the order of reactionfrom the rate law Given an empiricalrate law, obtain the orders with respect toeach reactant (and catalyst, if any) andthe overall order.

14.3 14.3 14.41, 14.42,14.43, 14.44

4. Determining the rate law from initialrates Given initial concentrations andinitial-rate data (in which the concentra-tions of all species are changed one at atime, holding the others constant), findthe rate law for the reaction.

14.4 14.4 14.45, 14.46,14.47, 14.48,14.49, 14.50

5. Using an integrated rate law Giventhe rate constant and initial reactant con-centration for a first-order, second-order,or zero-order reactions, calculate the reac-tant concentration after a definite time, orcalculate the time it takes for the concen-tration to decrease to a prescribed value.

14.5 14.5 14.51, 14.52,14.53, 14.54,14.55, 14.56

6. Relating the half-life of a reaction tothe rate constant Given the rate con-stant for a reaction, calculate the half-life.

14.6 14.6 14.57, 14.58

7. Using the Arrhenius equationGiven the values of the rate constant fortwo temperatures, find the activation en-ergy and calculate the rate constant at athird temperature.

14.7 14.7 14.71, 14.62,14.73, 14.74

8. Writing the overall chemical equa-tion from a mechanism Given a mecha-nism for a reaction, obtain the overallchemical equation.

14.8 14.8 14.77, 14.78


46 PART V

9. Determining the molecularity of anelementary reaction Given an elemen-tary reaction, state the molecularity.

14.9 14.9 14.79, 14.80

10. Writing the rate equation for an ele-mentary reaction Given an elementaryreaction, write the rate equation.

14.10 14.10 14.81, 14.82

11. Determining the rate law from amechanism Given a mechanism with aninitial slow step, obtain the rate law.Given a mechanism with an initial fast,equilibrium step, obtain the rate law.

14.11, 14.12 14.11, 14.12 14.83, 14.84,14.85, 14.86




Chapter 15 Chemical Equilibrium Examples Exercises Problems

1. Applying stoichiometry to an equi-librium mixture Given the startingamounts of reactants and the amount ofone substance at equilibrium, find theequilibrium composition.

15.1 15.1 15.23, 15.24,15.25, 15.26,15.27, 15.28

2. Writing equilibrium-constant expres-sions Given the chemical equation,write the equilibrium-constant expres-sion.

15.2, 15.4 15.2, 15.6 15.29, 15.30,15.49, 15.50

3. Obtaining an equilibrium constantfrom reaction composition Given theequilibrium composition, find Kc.

15.3 15.4 15.39, 15.40,15.41, 15.42

4. Using the reaction quotient Giventhe concentrations of substances in a reac-tion mixture, predict the direction of reac-tion.

15.5 15.8 15.55, 15.56

5. Obtaining one equilibrium concen-tration given the others Given Kc andall concentrations of substances but onein an equilibrium mixture, calculate theconcentration of this one substance.

15.6 15.9 15.59, 15.60

6. Solving equilibrium problemsGiven the starting composition and Kc ofa reaction mixture, calculate the equilib-rium composition.

15.7, 15.8 15.10, 15.11 15.61, 15.62,15.63, 15.64

7. Applying Le Chatelier’s principleGiven a reaction, use Le Chatelier’s prin-ciple to decide the effect of adding or re-moving a substance, changing thepressure, or changing the temperature.

15.9, 15.10,15.11

15.12, 15.13,15.14

15.67, 15.68,15.69, 15.70,15.71, 15.72


48 PART V

Chapter 16 Acids and Bases Examples Exercises Problems

1. Identifying acid and base speciesGiven a proton-transfer reaction, labelthe acids and bases, and name the conju-gate acid-base pairs.

16.1 16.1 16.29, 16.30

2. Identifying Lewis acid and base spe-cies Given a reaction involving the dona-tion of an electron pair, identify the Lewisacid and the Lewis base.

16.2 16.2 16.33, 16.34,16.35, 16.36

3. Deciding whether reactants or prod-ucts are favored in an acid–base reac-tion Given an acid–base reaction andthe relative strengths of acids (or bases),decide whether reactants or products arefavored.

16.3 16.3 16.39, 16.40,16.41, 16.42

4. Calculating concentrations of H3O+

and OH– in solutions of a strong acid orbase Given the concentration of a strongacid or base, calculate the hydronium-ionand hydroxide-ion concentrations.

16.4 16.5 16.47, 16.48

5. Calculating the pH from thehydronium-ion concentration, or viceversa Given the hydronium-ion concen-tration, calculate the pH; or given the pH,calculate the hydronium-ion concentra-tion.

16.5, 16.6 16.7, 16.8,16.9, 16.10

16.61, 16.62,16.67, 16.68,16.69, 16.70,16.71, 16.72



Chapter 17 Acid–Base Equilibria Examples Exercises Problems

1. Determining Ka (or Kb) from the solu-tion pH Given the molarity and pH of asolution of a weak acid, calculate Ka forthe acid. The Kb for a base can be deter-mined in a similar way (see Exercise 16.5).

17.1 17.1 17.29, 17.30

2. Calculating concentrations of speciesin a weak acid solution using Ka GivenKa, calculate the hydrogen-ion concentra-tion and pH of a solution of a weak acidof known molarity. Given Ka1, Ka2, andthe molarity of a diprotic acid solution,calculate the pH and the concentrationsof H+, HA–, and A2–.

17.2, 17.3,17.4

17.2, 17.3,17.4

17.31, 17.32,17.37, 17.38,17.41, 17.42

3. Calculating concentrations of speciesin a weak base solution using Kb GivenKb, calculate the hydrogen-ion concentra-tion and pH of a solution of a weak baseof known molarity.

17.5 17.6 17.47, 17.48

4. Predicting whether a salt solution isacidic, basic, or neutral Decide whetheran aqueous solution of a given salt isacidic, basic, or neutral.

17.6 17.7 17.53, 17.54

5. Obtaining Ka from Kb or Kb from KaCalculate Ka for a cation or Kb for an an-ion from the ionization constant of theconjugate base or acid.

17.7 17.8 17.57, 17.58

6. Calculating concentrations of speciesin a salt solution Given the concentra-tion of a solution of a salt in which oneion hydrolyzes, and given the ionizationconstant of the conjugate acid or base ofthis ion, calculate the H+ concentration.

17.8 17.9 17.59, 17.60,17.61, 17.62

7. Calculating the common-ion effecton acid ionization Given Ka and the con-centrations of weak acid and strong acidin a solution, calculate the degree of ioni-zation of the weak acid. Given Ka and theconcentrations of weak acid and its saltin a solution, calculate the pH.

17.9, 17.10 17.10, 17.11 17.63, 17.64,17.65, 17.66



50 PART V

8. Calculating the pH of a buffer fromgiven volumes of solution Given con-centrations and volumes of acid and con-jugate base from which a buffer isprepared, calculate the buffer pH.

17.11 17.12 17.69, 17.70

9. Calculating the pH of a buffer whena strong acid or strong base is addedCalculate the pH of a given volume ofbuffer solution (given the concentrationsof conjugate acid and base in the buffer)to which a specified amount of strongacid or strong base is added.

17.12 17.13 17.71, 17.72

10. Calculating the pH of a solution ofa strong acid and a strong base Calcu-late the pH during the titration of astrong acid by a strong base, given thevolumes and concentrations of the acidand base.

17.13 17.14 17.79, 17.80

11. Calculating the pH at the equiva-lence point in the titration of a weakacid by a strong base Calculate the pHat the equivalence point for the titrationof a weak acid by a strong base. Be ableto do the same type of calculation for thetitration of a weak base by a strong acid.

17.14 17.15 17.81, 17.82



Chapter 18 Solubility andComplex-Ion Equilibria Examples Exercises Problems

1. Writing solubility product expres-sions Write the solubility product ex-pression for a given ionic compound.

18.1 18.1 18.21, 18.22

2. Calculating Ksp from the solubility,or vice versa Given the solubility of aslightly soluble ionic compound, calcu-late Ksp. Given Ksp, calculate the solubilityof an ionic compound.

18.2, 18.3,18.4

18.2, 18.3,18.4

18.23, 18.24,18.25, 18.26,18.29, 18.30,18.31, 18.32

3. Calculating the solubility of aslightly soluble salt in a solution of acommon ion Given the solubility prod-uct constant, calculate the molar solubil-ity of a slightly soluble ionic compoundin a solution that contains a common ion.

18.5 18.5 18.33, 18.34,18.35, 18.36

4. Predicting whether precipitation willoccur Given the concentrations of ionsoriginally in solution, determine whethera precipitate is expected to form. Deter-mine whether a precipitate is expected toform when two solutions of known vol-ume and molarity are mixed. For bothproblems, you will need the solubilityproduct constant.

18.6, 18.7 18.6, 18.7 18.41, 18.42,18.43, 18.44

5. Determining the qualitative effect ofpH on solubility Decide whether thesolubility of a salt will be greatly in-creased by decreasing the pH.

18.8 18.8 18.53, 18.54

6. Calculating the concentration of ametal ion in equilibrium with a com-plex ion Calculate the concentration ofan aqueous metal ion in equilibrium withthe complex ion, given the original metal-ion and ligand concentrations. The forma-tion constant Kf of the complex ion isrequired.

18.9 18.9 18.57, 18.58

7. Predicting whether a precipitate willform in the presence of the complexion Predict whether an ionic compoundwill precipitate from a solution of knownconcentrations of cation, anion, and li-gand that complexes with the cation. Kfand Ksp are required.

18.10 18.10 18.59, 18.60


52 PART V

8. Calculating the solubility of aslightly soluble ionic compound in a so-lution of the complex ion Calculate themolar solubility of a slightly soluble ioniccompound in a solution of known con-centration of a ligand that complexeswith the cation. Ksp and Kf are required.

18.11 18.11 18.61, 18.62




Chapter 19 Thermodynamics and Equilibrium Examples Exercises Problems

1. Calculating the entropy change for aphase transition Given the heat ofphase transition and the temperature ofthe transition, calculate the entropychange of the system, ∆S.

19.1 19.3 19.29, 19.30

2. Predicting the sign of the entropychange of a reaction Predict the sign of∆S° for a reaction to which the ruleslisted in the text can be clearly applied.

19.2 19.4 19.33, 19.34

3. Calculating ∆S° for a reaction Giventhe standard entropies of reactants andproducts, calculate the change of entropy,∆S°, for the reaction.

19.3 19.5 19.35, 19.36

4. Calculating ∆G° from ∆H° and ∆S°Given enthalpies of formation and stand-ard entropies of reactants and products,calculate the standard free-energychange, ∆G°, for a reaction.

19.4 19.6 19.39, 19.40

5. Calculating ∆G° from standard freeenergies of formation Given the free en-ergies of formation of reactants and prod-ucts, calculate the standard free-energychange, ∆G°, for a reaction.

19.5 19.7 19.43, 19.44

6. Interpreting the sign of ∆G° Use thestandard free-energy change to deter-mine the spontaneity of a reaction.

19.6 19.8 19.47, 19.48

7. Writing the expression for a thermo-dynamic equilibrium constant For anybalanced chemical equation, write the ex-pression for the thermodynamic equilib-rium constant.

19.7 19.9 19.53, 19.54

8. Calculating K from the standard free-energy change Given the standard free-energy change for a reaction, calculatethe thermodynamic equilibrium constant.

19.8, 19.9 19.10, 19.11 19.55, 19.56,19.57, 19.58,19.59, 19.60

9. Calculating ∆G° and K at varioustemperatures Given ∆H° and ∆S° at25°C, calculate ∆G° and K for a reactionat a temperature other than 25°C.

19.10 19.12 19.61, 19.62


54 PART V

Chapter 20 Electrochemistry Examples Exercises Problems

1. Balancing equations in acid and ba-sic solutions by the half-reactionmethod Given the skeleton equation foran oxidation–reduction equation, com-plete and balance it.

20.1, 20.2 20.1, 20.2 20.29, 20.30,20.31, 20.32

2. Sketching and labeling a voltaic cellGiven a verbal description of a voltaiccell, sketch the cell, labeling the anodeand cathode, and give the directions ofelectron flow and ion migration.

20.3 20.3 20.37, 20.38

3. Writing the cell reaction from the cellnotation Given the notation for a voltaiccell, write the overall cell reaction. Alter-natively, given the cell reaction, write thecell notation.

20.4 20.5 20.47, 20.48

4. Calculating the quantity of workfrom a given amount of cell reactantGiven the emf and overall reaction for avoltaic cell, calculate the maximum workthat can be obtained from a givenamount of reactant.

20.5 20.6 20.51, 20.52,20.53, 20.54

5. Determining the relative strengths ofoxidizing and reducing agents Given atable of standard electrode potentials, listoxidizing or reducing agents by increas-ing strength.

20.6 20.7 20.55, 20.56,20.57, 20.58

6. Determining the direction of sponta-neity from electrode potentials Givenstandard electrode potentials, decide thedirection of spontaneity for an oxidation–reduction reaction under standard condi-tions.

20.7 20.8 20.59, 20.60

7. Calculating the emf from standardpotentials Given standard electrode po-tentials, calculate the standard emf of avoltaic cell.

20.8 20.9 20.63, 20.64

8. Calculating the free-energy changefrom electrode potentials Given stand-ard electrode potentials, calculate thestandard free-energy change for an oxida-tion–reduction reaction.

20.9 20.10 20.67, 20.68



9. Calculating the cell emf from free-energy change Given a table of stand-ard free energies of formation, calculatethe standard emf of a voltaic cell.

20.10 20.11 20.71, 20.72

10. Calculating the equilibrium con-stant from cell emf Given standard po-tentials (or standard emf), calculate theequilibrium constant of an oxidation–re-duction reaction.

20.11 20.12 20.75, 20.76

11. Calculating the cell emf for nonstan-dard conditions Given standard elec-trode potentials and the concentrations ofsubstances in a voltaic cell, calculate thecell emf.

20.12 20.13 20.79, 20.80

12. Predicting the half-reactions in anaqueous electrolysis Using values ofelectrode potentials, decide which elec-trode reactions actually occur in the elec-trolysis of an aqueous solution.

20.13 20.16 20.87, 20.88

13. Relating the amounts of charge andproduct in an electrolysis Given theamount of product obtained by electroly-sis, calculate the amount of charge thatflowed. Given the amount of charge thatflowed, calculate the amount of productobtained by electrolysis.

20.14, 20.15 20.17, 20.18 20.89, 20.90,20.91, 20.92



56 PART V

Chapter 21 Nuclear Chemistry Examples Exercises Problems

1. Writing a nuclear equation Given aword description of a radioactive decayprocess, write the nuclear equation.

21.1 21.1 21.29, 21.30,21.31, 21.32

2. Deducing a product or reactant in anuclear equation Given all but one ofthe reactants and products in a nuclear re-action, find that one nuclide.

21.2, 21.6 21.2, 21.6 21.33, 21.34,21.35, 21.36,21.49, 21.50,21.51, 21.52

3. Predicting the relative stabilities ofnuclides Given a number of nuclides,determine which are most likely to be ra-dioactive and which are most likely to bestable.

21.3 21.3 21.37, 21.38

4. Predicting the type of radioactive de-cay Predict the type of radioactive decaythat is most likely for given nuclides.

21.4 21.4 21.39, 21.40

5. Using the notation for a bombard-ment reaction Given an equation for anuclear bombardment reaction, write theabbreviated notation, or vice versa.

21.5 21.5 21.43, 21.44,21.45, 21.46

6. Calculating the decay constant fromthe activity Given the activity (disinte-grations per second) of a radioactive iso-tope, obtain the decay constant.

21.7 21.7 21.53, 21.54,21.55, 21.56

7. Relating the decay constant, half-life,and activity Given the decay constant ofa radioactive isotope, obtain the half-life,or vice versa. Given the decay constantand mass of a radioactive isotope, calcu-late the activity of the sample.

21.8, 21.9 21.8, 21.9 21.57, 21.58,21.59, 21.60,21.61, 21.62

8. Determining the fraction of nuclei re-maining after a specified time Giventhe half-life of a radioactive isotope, cal-culate the fraction remaining after a speci-fied time.

21.10 21.10 21.65, 21.66

9. Applying the carbon-14 datingmethod Given the disintegrations ofcarbon-14 nuclei per gram of carbon in adead organic object, calculate the age ofthe object—that is, the time since itsdeath.

21.11 21.11 21.71, 21.72



10. Calculating the energy change for anuclear reaction Given nuclear masses,calculate the energy change for a nuclearreaction. Obtain the answer in joules permole or MeV per particle.

21.12 21.12 21.77, 21.78



58 PART V

Chapter 22 Chemistry of the Main-Group Elements




Chapter 23 The Transition Elements Examples Exercises Problems

1. Writing the IUPAC name given thestructural formula of a coordinationcompound, and vice versa Given thestructural formulas of coordination com-pounds, write the IUPAC names; giventhe IUPAC names of complexes, write thestructural formulas.

23.1, 23.2 23.1, 23.2 23.43, 23.44,23.45, 23.46,23.47, 23.48

2. Deciding whether isomers are possi-ble Given the formula of a complex, de-cide whether geometric isomers arepossible and, if so, draw them. Given thestructural formula of a complex, decidewhether enantiomers (optical isomers)are possible and, if so, draw them.

23.3, 23.4 23.3, 23.4 23.49, 23.50,23.51, 23.52

3. Describing the bonding in a complexion Given a transition-metal complexion, describe the bonding types (high-spin and low-spin, if both exist), using va-lence bond theory for octahedral andfour-coordinate complexes. Give thenumber of unpaired electrons in the com-plex. Do the same using crystal fieldtheory.

23.5, 23.6 23.5, 23.6 23.53, 23.54,23.55, 23.56

4. Predicting the relative wavelengthsof absorption of complex ions Giventwo complexes that differ only in the li-gands, predict, on the basis of the spectro-chemical series, which complex absorbsat higher wavelength. Given the absorp-tion maxima, predict the colors of thecomplexes.

23.7 23.7 23.57, 23.58,23.59, 23.60


60 PART V

Chapter 24 Organic Chemistry Examples Exercises Problems

1. Writing a condensed structuralformula Given the structural formula ofa hydrocarbon, write the condensed struc-ture formula.

24.1 24.1 24.23, 24.24

2. Predicting cis–trans isomers Given acondensed structural formula of analkene, decide whether cis and trans iso-mers are possible, and, if so, draw thestructural formulas.

24.2 24.2 24.25, 24.26

3. Predicting the major product in anaddition reaction Predict the majorproduct in the addition of an unsymmet-rical reagent to an unsymmetrical alkene.

24.3 24.3 24.31, 24.32

4. Writing the IUPAC name of a hydro-carbon given the structural formula, andvice versa Given the structure of a hy-drocarbon, state the IUPAC name. Giventhe IUPAC name of a hydrocarbon, writethe structural formula.

24.4, 24.5,24.6

24.4, 24.5,24.6, 24.7,24.8, 24.9

24.33, 24.34,24.35, 24.36,24.37, 24.38,24.39, 24.40,24.41, 24.42,24.43, 24.44

Chapter 25 Polymer Materials: Syntheticand Biological Examples Exercises Problems




PART VICorrelation of Cumulative-Skills Problemswith Text Sections

Listed below are the cumulative-skills problems found at the end of the chapters and thesections of the text needed to solve each problem. You can use this list to be sure that a problemyou wish to assign does not require any sections you may have omitted from students’required reading.

Chapter 1

1.137 and 1.138: 1.3, 1.5, 1.71.139 and 1.140: 1.5, 1.71.141 and 1.142: 1.5, 1.7, 1.81.143 and 1.144: 1.3, 1.5, 1.71.145 and 1.146: 1.5, 1.7, 1.81.147 and 1.148: 1.5, 1.7, 1.8

Chapter 2

2.125 and 2.126: 1.7, 1.82.127 and 2.128: 1.2, 2.62.129 and 2.130: 1.7, 2.4

Chapter 3

3.109 and 3.110: 1.3, 1.8, 3.23.111 and 3.112: 1.8, 3.1, 3.23.113 and 3.114: 1.3, 3.2, 3.53.115 and 3.116: 1.8, 3.2, 3.3

Chapter 4

4.119 and 4.120: 2.6, 2.8, 4.24.121 and 4.122: 1.3, 1.8, 4.24.123 and 4.124: 1.3, 4.2, 4.3, 4.44.125 and 4.126: 2.6, 4.54.127 and 4.128: 1.7, 3.1, 3.2, 3.34.129 and 4.130: 3.7, 4.7, 4.10


4.131 and 4.132: 3.2, 3.74.133 and 4.134: 1.7, 2.10, 4.3, 4.94.135 and 4.136: 3.1, 3.3, 3.74.137 and 4.138: 1.7, 3.1, 4.7, 4.104.139 and 4.140: 3.3, 3.7, 4.4, 4.10

Chapter 5

5.125 and 5.126: 3.2, 3.3, 5.35.127 and 5.128: 3.2, 5.1, 5.3, 5.4, 5.55.129 and 5.130: 3.7, 3.8, 5.3, 5.45.131 and 5.132: 1.7, 3.2, 5.35.133 and 5.134: 1.7, 5.3, 5.7

Chapter 6

6.115 and 6.116: 6.1, 6.66.117 and 6.118: 2.10, 3.2, 3.3, 6.56.119 and 6.120: 3.2, 6.56.121 and 6.122: 3.2, 3.8, 6.56.123 and 6.124: 5.3, 6.66.125 and 6.126: 2.10, 3.2, 3.7, 6.8

Chapter 7

7.85 and 7.86: 1.8, 3.2, 7.27.87 and 7.88: 1.8, 6.6, 7.27.89 and 7.90: 1.8, 6.1, 7.27.91 and 7.92: 1.7, 6.1, 7.2, 7.4

Chapter 8

8.79 and 8.80: 2.9, 2.10, 5.3, 8.78.81 and 8.82: 2.4, 3.3, 8.78.83 and 8.84: 3.2, 8.68.85 and 8.86: 3.2, 7.3, 8.68.87 and 8.88: 6.7, 8.6

Chapter 9

9.105 and 9.106: 2.10, 4.4, 9.59.107 and 9.108: 2.8, 3.5, 9.4, 9.69.109 and 9.110: 3.5, 9.69.111 and 9.112: 5.3, 9.1, 9.4, 9.6, 9.89.113 and 9.114: 6.8, 9.119.115 and 9.116: 9.5, 9.119.117 and 9.118: 8.6, 9.5, 9.11


Correlation of Cumulative-Skills Problems with Text Sections 63

Chapter 10

10.77 and 10.78: 2.6, 3.3, 3.5, 9.6, 9.8, 10.1, 10.310.79 and 10.80: 2.10, 5.3, 10.310.81 and 10.82: 9.10, 10.1, 10.3, 10.410.83 and 10.84: 9.6, 9.7, 9.8, 9.11, 10.1, 10.3

Chapter 11

11.117 and 11.118: 3.2, 5.3, 5.5, 11.211.119 and 11.120: 1.7, 3.2, 6.2, 16.211.121 and 11.122: 3.2, 6.6, 11.211.123 and 11.124: 1.7, 3.2, 5.3, 11.2

Chapter 12

12.105 and 12.106: 3.2, 4.1, 4.7, 12.412.107 and 12.108: 6.7, 8.6, 12.212.109 and 12.110: 3.2, 12.412.111 and 12.112: 1.7, 3.2, 4.712.113 and 12.114: 12.4, 12.612.115 and 12.116: 3.2, 3.5, 12.4, 12.6

Chapter 14

14.123 and 14.124: 2.10, 5.3, 14.314.125 and 14.126: 2.10, 6.2, 6.8, 14.414.127 and 14.128: 5.3, 14.4

Chapter 15

15.107 and 15.108: 3.2, 4.7, 15.2, 15.615.109 and 15.110: 3.2, 5.3, 15.6

Chapter 16

16.103 and 16.104: 3.2, 3.7, 9.6, 16.416.105 and 16.106: 3.2, 5.3, 16.3

Chapter 17

17.125 and 17.126: 1.7, 3.3, 16.8, 17.117.127 and 17.128: 1.7, 12.4, 12.6, 17.117.129 and 17.130: 1.7, 4.4, 4.7, 16.8, 17.6


64 PART VI

Chapter 18

18.101 and 18.102: 17.6, 18.1, 18.3, 18.418.103 and 18.104: 15.2, 15.6, 17.1, 17.6, 18.1, 18.318.105 and 18.106: 3.2, 4.2, 4.3, 4.7, 18.1, 18.3

Chapter 19

19.99 and 19.100: 6.2, 15.2, 19.3, 19.4, 19.619.101 and 19.102: 6.2, 15.2, 19.3, 19.4, 19.619.103 and 19.104: 17.1, 19.4, 19.6, 19.7

Chapter 20

20.123 and 20.124: 6.8, 19.2, 19.4, 20.520.125 and 20.126: 16.8, 20.4, 20.620.127 and 20.128: 15.8, 17.6, 20.4, 20.620.129 and 20.130: 18.1, 20.4, 20.5, 20.6

Chapter 21

21.99 and 21.100: 2.6, 3.2, 14.4, 21.421.101 and 21.102: 5.3, 21.1, 21.621.103 and 21.104: 5.3, 6.2, 6.8, 21.1, 21.2, 21.6


Correlation of Cumulative-Skills Problems with Text Sections 65

PART VIIAlternate Examples for Lecture


Alternate Example 1.1 Using the Law of Conservation of Mass

Aluminum powder burns in oxygen to produce a substance called aluminum oxide. A sampleof 2.00 grams of aluminum is burned in oxygen and produces 3.78 grams of aluminum oxide.How many grams of oxygen were used in this reaction?

Answer: 1.78 grams

Alternate Example 1.2 Using Significant Figures in Calculations

Perform the following calculations, rounding the answers to the correct number of significantfigures.

(a) 5.8914

1.289 × 7.28 (b) 0.453 – 1.59 (c) 0.456 – 0.421 (d) 92.34 × (0.456 – 0.421)

Answers: (a) 0.628 (b) –1.14 (c) 0.035 (d) 3.2

Alternate Example 1.3 Converting from One Temperature Scale to Another

In winter, the average low temperature of interior Alaska is –30°F (two significant figures).What is this temperature in degrees Celsius? in kelvins?

Answer: –34°C; 239 K

Alternate Example 1.4 Calculating the Density of a Substance

Oil of wintergreen is a colorless liquid used as a flavoring. A 28.1-g sample of oil of wintergreenhas a volume of 23.7 mL. What is the density of oil of wintergreen?

Answer: 1.19 g/mL


Alternate Example 1.5 Using the Density to Relate Mass and Volume

A sample of gasoline has a density of 0.718 g/mL. What is the volume of 454 g of gasoline?

Answer: 632 mL

Alternate Example 1.6 Converting Units: Metric Unit to Metric Unit

A sample of sodium metal is burned in chlorine gas, producing 573 mg of sodium chloride.How many grams is this? How many kilograms?

Answer: 0.573 g; 5.73 × 10–4 kg

Alternate Example 1.7 Converting Units: Metric Volume to Metric Volume

An experiment calls for 54.3 mL of ethanol. What is this volume in cubic meters?

Answer: 5.43 × 10–5 m3

First Alternate Example 1.8 Converting Units: Any Unit to Another Unit

The Star of Asia sapphire in the Smithsonian Institution weighs 330 carats (three significantfigures). What is this weight in grams? One carat equals 200 mg (exact).

Answer: 66.0 g

Second Alternate Example 1.8 Converting Units: Any Unit to Another Unit

The dimensions of Noah’s ark were reported as 3.0 × 102 cubits by 5.0 × 101 cubits. Express thesize in units of feet and meters (1 cubit = 1.5 ft).

Answer: 4.5 × 102 ft by 75 ft; 1.4 × 102 m by 23 m


Alternate Example 2.1 Writing Nuclide Symbols

Write the nuclide symbol for the nucleus that has 19 protons and 20 neutrons.

Answer: 3919K


Alternate Examples for Lecture 67

Alternate Example 2.2 Determining Atomic Weight from Isotopic Masses andFractional Abundances

An element has four naturally occurring isotopes. The mass and percentage abundance ofeach isotope are as follows:

PercentageAbundance Mass (amu)

1.4823.622.652.3

203.973 205.9745206.9759207.9766

What is the atomic weight and the name of the element?

Answer: 207 amu; lead

Alternate Example 2.3 Writing an Ionic Formula, Given the Ions

(a) What is the formula of magnesium nitride, which is composed of the ions Mg2+ and N3–?(b) What is the formula of calcium phosphate, which is composed of the ions Ca2+ and PO4

3–?

Answers: (a) Mg3N2 (b) Ca3(PO4)2

Alternate Example 2.4 Naming an Ionic Compound from Its Formula

Name the following: (a) BaO, (b) Cr2(SO4)3.

Answers: (a) barium oxide (b) chromium(III) sulfate

Alternate Example 2.5 Writing the Formula from the Name of an Ionic Compound

Write the formulas for the following: (a) potassium carbonate, (b) manganese(II) sulfate,(c) selenium tetrafluoride.

Answers: (a) K2CO3 (b) MnSO4 (c) SeF4

Alternate Example 2.6 Naming a Binary Compound from Its Formula

Name the following compounds: (a) OF2, (b) S4N4, (c) BCl3.

Answers: (a) oxygen difluoride (b) tetrasulfur tetranitride (c) boron trichloride


68 PART VII

Alternate Example 2.7 Writing the Formula from the Name of a Binary Compound

Give the formula for each of the following: (a) carbon disulfide, (b) nitrogen tribromide,(c) dinitrogen tetrafluoride.

Answers: (a) CS2 (b) NBr3 (c) N2F4

Alternate Example 2.9 Writing the Name and Formula of an Anion from the Acid

Bromine has an oxoacid HBrO2, whose name is bromous acid (compare chlorous acid, HClO2).What is the name and formula of the corresponding anion?

Answer: bromite ion, BrO2–

Alternate Example 2.10 Naming a Hydrate from Its Formula

A compound whose common name is green vitriol has the chemical formula FeSO4⋅7H2O.What is the chemical name of this compound?

Answer: iron(II) sulfate heptahydrate

Alternate Example 2.11 Writing the Formula from the Name of a Hydrate

Calcium chloride hexahydrate is used to melt snow from roads. What is the formula of thiscompound?

Answer: CaCl2⋅6H2O

Alternate Example 2.12 Balancing Simple Equations

Balance the following equations.

(a) CS2 + O2 → CO2 + SO2(b) NH3 + O2 → NO + H2O(c) C2H5OH + O2 → CO2 + H2O

Answers:

(a) CS2 + 3O2 → CO2 + 2SO2 (b) 4NH3 + 5O2 → 4NO + 6H2O (c) C2H5OH + 3O2 → 2CO2 + 3H2O




Alternate Example 3.1 Calculating the Formula Weight from a Formula

Calculate the formula weight of the following compounds from their formulas (obtain theanswers to three significant figures): (a) calcium hydroxide, Ca(OH)2; (b) methylamine,CH3NH2.

Answers: (a) 74.1 amu (b) 31.1 amu

Alternate Example 3.3 Calculating the Mass of an Atom or Molecule

What is the mass of the nitric acid molecule, HNO3?

Answer: 1.05 × 10–22 g

Alternate Example 3.4 Converting Moles of Substance to Grams

A sample of nitric acid contains 0.253 mol HNO3. How many grams is this?

Answer: 15.9 g

First Alternate Example 3.5 Converting Grams of Substance to Moles

Calcite is a mineral composed of calcium carbonate, CaCO3. A sample of calcite composed ofpure calcium carbonate weighs 23.6 g. How many moles of calcium carbonate is this?

Answer: 0.236 mol CaCO3

Second Alternate Example 3.5 Converting Grams of Substance to Moles

The average daily requirement of the essential amino acid leucine, C6H14O2N, is 2.2 g for anadult. How many moles of leucine are required daily?

Answer: 0.017 mol

Alternate Example 3.6 Calculating the Number of Atoms in a Given Mass

The daily requirement of chromium in the human diet is 1.0 × 10–6 g. How many atoms ofchromium does this represent?

Answer: 1.2 × 1016 atoms


70 PART VII

Alternate Example 3.7 Calculating the Percentage Composition from the Formula

Lead(II) chromate, PbCrO4, is used as a paint pigment (chrome yellow). What is the percentagecomposition of lead(II) chromate?

Answer: 64.1% Pb, 16.1% Cr, and 19.8% O

Alternate Example 3.8 Calculating the Mass of an Element in a Given Mass ofCompound

The chemical name of table sugar is sucrose, C12H22O11. How many grams of carbon (C) arein 61.8 g sucrose?

Answer: 26.0 g

Alternate Example 3.9 Calculating the Percentages of C and H by Combustion

Benzene is a liquid compound composed of carbon and hydrogen; it is used in the preparationof polystyrene plastic. A sample of benzene weighing 342 mg is burned in oxygen and forms1156 mg of carbon dioxide. What is the percentage composition of benzene?

Answer: 92.3% C and 7.7% H

Alternate Example 3.10 Determining the Empirical Formula from PercentageComposition (Binary Compound)

In Alternate Example 3.8, we determined the percentage composition of benzene: 92.3% C and7.7% H. What is the empirical formula of benzene?

Answer: CH

Alternate Example 3.11 Determining the Empirical Formula from PercentageComposition (General)

Sodium pyrophosphate is used in detergent preparations. The mass percentages of theelements in this compound are 34.6% Na, 23.3% P, and 42.1% O. What is the empirical formulaof sodium pyrophosphate?

Answer: Na4P2O7

First Alternate Example 3.12 Determining the Molecular Formula from PercentageComposition and Molecular Weight

The percentage composition of benzene is 92.3% C and 7.7% H. In Alternate Example 3.9, wefound the empirical formula of benzene from these data to be CH. In a separate experiment,the molecular weight of benzene was determined to be 78.1 amu. What is the molecularformula of benzene?

Answer: C6H6



Second Alternate Example 3.12 Determining the Molecular Formula fromPercentage Composition and Molecular Weight

Hexamethylene is one of the materials used to produce a type of nylon. Elemental analysis ofthe substance gives 62.1% C, 13.8% H, and 24.1% N. Its molecular weight is 116 amu. What isits molecular formula?

Answer: C6H16N2

Alternate Example 3.13 Relating the Quantity of Reactant to Quantity of Product

Propane, C3H8, is normally a gas, but it is sold as a fuel compressed as a liquid in steelcylinders. The gas burns according to the equation

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

How many grams of CO2 are produced when 20.0 g of propane is burned?

Answer: 59.9 g CO2

Alternate Example 3.14 Relating the Quantities of Two Reactants (or Two Products)

How many grams of O2 are required to burn 20.0 g C3H8?

Answer: 72.6 g O2

Alternate Example 3.15 Calculating with a Limiting Reactant (Involving Moles)

Magnesium metal is used to prepare zirconium metal, which is used to make the containerfor nuclear fuel (the nuclear fuel rods).

ZrCl4(g) + 2Mg(l) → 2MgCl2(s) + Zr(s)

How many moles of zirconium metal can be produced from a reaction mixture containing0.20 mol ZrCl4 and 0.50 mol Mg?

Answer: 0.20 mol


72 PART VII

Alternate Example 3.16 Calculating with a Limiting Reactant (Involving Masses)

Urea, CH4N2O, is used as a nitrogen fertilizer. It is manufactured from ammonia and carbondioxide at high pressure and high temperature.

2NH3 + CO2 → CH4N2O + H2O

In a laboratory experiment, 10.0 g NH3 and 10.0 g CO2 were added to a reaction vessel. Whatis the maximum quantity (in grams) of urea that can be obtained? How many grams of theexcess reactant are left at the end of the reaction?

Answer: 13.6 g CH4N2O; 2.3 g NH3


Alternate Example 4.2 Writing Net Ionic Equations

Write a net ionic equation for the following molecular equations.

(a) H2SO4(aq) + Mg(OH)2(s) → MgSO4(aq) + 2H2O(l)(b) KCl(aq) + AgNO3(aq) → KNO3(aq) + AgCl(s)

Answers: (a) 2H+(aq) + Mg(OH)2(s) → Mg2+(aq) + 2H2O(l) (b) Cl–(aq) + Ag+(aq) →AgCl(s)

First Alternate Example 4.3 Deciding Whether Precipitation Occurs

For each of the following, decide whether precipitation will occur. If it does, write themolecular equation and the net ionic equation.

(a) KBr + MgSO4 →(b) NaOH + MgCl2 →

Answers: (a) no reaction (b) 2NaOH(aq) + MgCl2(aq) → 2NaCl(aq) + Mg(OH)2(s);2OH–(aq) + Mg2+(aq) → Mg(OH)2(s)

Second Alternate Example 4.3 Deciding Whether Precipitation Occurs

Decide whether a precipitation reaction will occur for the following. If precipitation doesoccur, write the molecular equation and the net ionic equation for the reaction.

K3PO4 + CaCl2 →

Answer: 2K3PO4(aq) + 3CaCl2(aq) → 6KCl(aq) + Ca3(PO4)2(s); 2PO43–(aq) + 3Ca2+(aq) →

Ca3(PO4)2(s)



Alternate Example 4.4 Classifying Acids and Bases as Strong or Weak

Classify each of the following as a strong or weak acid or base: (a) KOH, (b) H2S, (c) CH3NH2,(d) HClO4.

Answers: (a) strong base (b) weak acid (c) weak base (d) strong acid

Alternate Example 4.5 Writing an Equation for a Neutralization

Write the molecular equation and the net ionic equation for the neutralization of sulfurousacid, H2SO3, by potassium hydroxide, KOH.

Answer: H2SO3(aq) + 2KOH(aq) → K2SO3(aq) + 2H2O(l); H2SO3(aq) + 2OH–(aq) →SO3

2–(aq) + 2H2O(l)

Alternate Example 4.6 Writing an Equation for a Reaction with Gas Formation

Write the molecular equation and the net ionic equation for the reaction of copper(II) carbonatewith hydrochloric acid.

Answer: CuCO3(s) + 2HCl(aq) → CuCl2(aq) + H2O(l) + CO2(g); CuCO3(s) + 2H+(aq) →Cu2+(aq) + H2O(l) + CO2(g)

First Alternate Example 4.7 Assigning Oxidation Numbers

Potassium permanganate, KMnO4, is a purple-colored compound; potassium manganate,K2MnO4, is a green-colored compound. Obtain the oxidation numbers of the manganese atomin these compounds.

Answer: +7, +6

Second Alternate Example 4.7 Assigning Oxidation Numbers

What is the oxidation number of Cr in the dichromate ion, Cr2O72–?

Answer: +6

Alternate Example 4.9 Calculating Molarity from Mass and Volume

You place a 1.53-g sample of potassium dichromate, K2Cr2O7, into a 50.0-mL volumetric flaskand add water to bring the solution up to the mark on the neck of the flask. What is the molarityof K2Cr2O7 in the solution?

Answer: 0.104 M


74 PART VII

Alternate Example 4.10 Using Molarity as a Conversion Factor

A solution of sodium chloride used for intravenous transfusion (physiological saline solution)has a concentration of 0.154 M NaCl. How many moles of NaCl are contained in a 500-mLbottle of physiological saline solution? How many grams of NaCl are in the 500 mL of solution?

Answer: 0.0770 mol; 4.50 g

Alternate Example 4.11 Diluting a Solution

A saturated stock solution of NaCl is 6.00 M. How much of this stock solution is needed toprepare 1.00 L of physiological saline solution, which is 0.154 M NaCl?

Answer: 25.7 mL

Alternate Example 4.12 Determining the Amount of a Species byGravimetric Analysis

A soluble silver compound was analyzed for the percentage of silver by adding sodiumchloride solution to precipitate the silver ion as silver chloride. If 1.583 g of silver compoundgave 1.788 g of silver chloride, what is the mass percentage of silver in the compound?

Answer: 85.01%

Alternate Example 4.13 Calculating the Volume of Reactant Solution Needed

Zinc sulfide reacts with hydrochloric acid to produce hydrogen sulfide gas:

ZnS(s) + 2HCl(aq) → ZnCl2(aq) + H2S(g)

How many milliliters of 0.0512 M HCl are required to react with 0.392 g ZnS?

Answer: 157 mL

Alternate Example 4.14 Calculating the Quantity of Substance in a Titrated Solution

A dilute solution of hydrogen peroxide is sold in drugstores as a mild antiseptic. A typicalsolution was analyzed for the percentage of hydrogen peroxide by titrating it with potassiumpermanganate:

5H2O2(aq) + 2KMnO4(aq) + 6H+(aq) → 8H2O(l) + 5O2(g) + 2K+(aq) + 2Mn2+(aq)

What is the mass percentage of H2O2 in a solution if 57.5 g of solution required 38.9 mL of0.534 M KMnO4?

Answer: 3.07%




Alternate Example 5.2 Using Boyle’s Law

A volume of oxygen gas occupies 38.7 mL at 751 mmHg and 21°C. What is the volume if thepressure changes to 359 mmHg while the temperature remains constant?

Answer: 81.0 mL

Alternate Example 5.3 Using Charles’s Law

You prepared carbon dioxide by adding HCl(aq) to marble chips (CaCO3). According to yourcalculations, you should obtain 79.4 mL of CO2 at 0°C and 760 mmHg. How many millilitersof gas would you obtain at 27°C?

Answer: 87.2 mL

Alternate Example 5.4 Using the Combined Gas Law

Divers working from a North Sea drilling platform experience pressures of 5.0 × 101 atm at adepth of 5.0 × 102 m. If a balloon is inflated to a volume of 5.0 L (the volume of a lung) at thatdepth at a water temperature of 4.0°C, what would the volume of the balloon be on the surface(1.0 atm pressure) at a temperature of 11°C?

Answer: 2.6 × 102 L

Alternate Example 5.5 Deriving Empirical Gas Laws from the Ideal Gas Law

You put varying amounts of gas into a given container at a given temperature. Use the idealgas law to show that the amount (moles) of gas is proportional to pressure at constanttemperature and volume.

Answer: n ∝ P

Alternate Example 5.6 Using the Ideal Gas Law

A 50.0-L cylinder of nitrogen, N2, has a pressure of 17.1 atm at 23°C. What is the mass ofnitrogen in the cylinder?

Answer: 985 g N2

Alternate Example 5.7 Calculating Gas Density

What is the density of methane gas (natural gas), CH4, at 125°C and 3.50 atm?

Answer: 1.72 g/L


76 PART VII

Alternate Example 5.8 Determining the Molecular Weight of a Vapor

A 500.0-mL flask containing a sample of octane, a component of gasoline, is placed in a boilingwater bath in Denver, where the atmospheric pressure is 634 mmHg and water boils at 95.0°C.The mass of the vapor required to fill the flask is 1.57 g. What is the molecular weight of octane?The empirical formula of octane is C4H9. What is the molecular formula of octane?

Answer: 114 amu; C8H18

Alternate Example 5.9 Solving Stoichiometry Problems Involving Gas Volumes

When a 2.0-L bottle of concentrated HCl was spilled, 1.2 kg of CaCO3 was required toneutralize the spill. What volume of CO2 was released by the neutralization at 735 mmHg and20°C?

CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)

Answer: 3.0 × 102 L CO2

First Alternate Example 5.10 Calculating Partial Pressures of a Gas in a Mixture

A 100.0-mL sample of air exhaled from the lungs is analyzed and found to contain 0.0830 gN2, 0.0194 g O2, 0.00640 g CO2, and 0.00441 g water vapor at 35°C. What is the partial pressureof each component and the total pressure of the sample?

Answer: P(N2) = 0.749 atm; P(O2) = 0.153 atm; P(CO2) = 0.0368 atm; P(H2O) = 0.0619 atm;Ptotal = 1.00 atm

Second Alternate Example 5.10 Calculating Mole Fractions of a Gas in a Mixture

The partial pressure of air in the alveoli, the air sacs in the lungs, is as follows: nitrogen, 570.0mmHg; oxygen, 103.0 mmHg; carbon dioxide, 40.0 mmHg; and water vapor, 47.0 mmHg.What is the mole fraction of each component of alveolar air?

Answer: mol N2 = 0.7500; mol O2 = 0.1355; mol CO2 = 0.0526; mol H2O = 0.0618

Alternate Example 5.11 Calculating the Amount of Gas Collected over Water

You prepare nitrogen gas by heating ammonium nitrite:

NH4NO2(s) → N2(g) + 2H2O(l)

If you collected the nitrogen over water at 22°C and 727 mmHg, how many liters of gas wouldyou obtain from 5.68 g NH4NO2?

Answer: 2.31 L



Alternate Example 5.12 Calculating the rms Speed of Gas Molecules

What is the rms speed of carbon dioxide molecules in a container of gas at 23°C?

Answer: 4.10 × 102 m/s

Alternate Example 5.13 Calculating the Ratio of Effusion Rates of Gases

Both hydrogen and helium have been used as the buoyant gas in blimps. If a small leak wereto occur, which gas would effuse more rapidly and by what factor?

Answer: H2; 1.4 times as fast

Alternate Example 5.14 Using the van der Waals Equation

Use the van der Waals equation to calculate the pressure exerted by 2.00 mol CO2 that has avolume of 10.0 L at 25°C. Compare with the pressure obtained from the ideal gas law.

Answer: 4.79 atm (van der Waals equation); 4.89 atm (ideal gas law)


Alternate Example 6.1 Calculating Kinetic Energy

A person weighing 75.0 kg (165 lb) runs a course in 1.78 m/s (4.00 mph). What is this person’skinetic energy?

Answer: 119 J

Alternate Example 6.2 Writing Thermochemical Equations

Sulfur, S8, burns in air to produce sulfur dioxide. The reaction evolves 9.31 kJ of heat per gramof sulfur at constant pressure. Write the thermochemical equation for this reaction.

Answer: S8(s) + 8O2(g) → 8SO2(g); ∆H = –2.39 × 103 kJ

Alternate Example 6.3 Manipulating Thermochemical Equations

When sulfur burns in air, the following reaction occurs (see Alternate Example 6.2):

S8(s) + 8O2(g) → 8SO2(g); ∆H = –2.39 × 103 kJ

Write the thermochemical equation for the dissociation of one mole of sulfur dioxide into itselements.

Answer: SO2(g) → 18S8(s) + O2(g); ∆H = +296 kJ


78 PART VII

First Alternate Example 6.4 Calculating the Heat of Reaction from the Stoichiometry

You burn 15.0 g of sulfur in air. How much heat evolves from this amount of sulfur? Thethermochemical equation is

S8(s) + 8O2(g) → 8SO2(g); ∆H = –2.39 × 103 kJ

(This was obtained in Alternate Example 6.2.)

Answer: 1.40 × 102 kJ (q = –1.40 × 102 kJ)

Second Alternate Example 6.4 Calculating the Heat of Reactionfrom the Stoichiometry

The daily energy requirement for a 20-year-old male weighing 67 kg is 1.3 × 104 kJ. For a20-year-old female weighing 58 kg, the daily requirement is 8.8 × 103 kJ. If all this energy wereto be provided by the combustion of glucose, C6H12O6, how many grams of glucose wouldhave to be consumed by the male and the female each day?

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l); ∆H = –2.82 × 103 kJ

Answer: male, 830 g/day; female, 560 g/day

Alternate Example 6.5 Relating Heat and Specific Heat

A piece of zinc weighing 35.8 g was heated from 20.00°C to 28.00°C. How much heat wasrequired? The specific heat of zinc is 0.388 J/(g⋅°C).

Answer: 111 J

Alternate Example 6.6 Calculating ∆H from Calorimetric Data

Nitromethane, CH3NO2, an organic solvent, burns in oxygen to give the following reaction:

CH3NO2(l) + 34O2(g) → CO2(g) + 3

2H2O(l) + 1

2N2(g)

You place a 1.724-g sample of nitromethane in a calorimeter with oxygen. The nitromethaneis ignited and burns in oxygen. The temperature of the calorimeter increases from 22.23°C to28.81°C. In a separate experiment, you determine that the heat capacity of the calorimeter andits contents is 3.044 kJ/°C. What is the ∆H of reaction (expressed as a thermochemicalequation)?

Answer: CH3NO2(l) + 34O2(g) → CO2(g) + 3

2H2O(l) + 1

2N2(g); ∆H = –709 kJ



Alternate Example 6.7 Applying Hess’s Law

What is the enthalpy of reaction, ∆H, for the reaction of calcium metal with water?

Ca(s) + 2H2O(l) → Ca2+(aq) + 2OH–(aq) + H2(g)

This reaction occurs very slowly, so it is impractical to measure ∆H directly. However, ∆H ofthe following reactions can be measured:

H+(aq) + OH–(aq) → H2O(l); ∆H = –55.9 kJ

Ca(s) + 2H+(aq) → Ca2+(aq) + H2(g); ∆H = –543.0 kJ

Answer: –431.2 kJ

Alternate Example 6.8 Calculating the Heat of Phase Transition from StandardEnthalpies of Formation

What is the heat of vaporization of methanol, CH3OH, at 25°C and 1 atm? Use standardenthalpies of formation (Appendix C).

Answer: 37.4 kJ/mol

Alternate Example 6.9 Calculating the Enthalpy of Reaction from StandardEnthalpies of Formation

Methyl alcohol, CH3OH, is toxic because liver enzymes oxidize it to formaldehyde, HCHO,which can coagulate protein. Calculate ∆H° for the following reaction; standard enthalpies offormation are (in kJ/mol): CH3OH(aq), –245.9; HCHO(aq), –150.2; H2O(l), –285.8.

2CH3OH(aq) + O2(g) → 2HCHO(aq) + 2H2O(l)

Answer: –380.2 kJ


Alternate Example 7.1 Obtaining the Wavelength of Light from Its Frequency

What is the wavelength of blue light with a frequency of 6.4 × 1014/s?

Answer: 470 nm


80 PART VII

Alternate Example 7.2 Obtaining the Frequency of Light from Its Wavelength

What is the frequency of red light having a wavelength of 681 nm?

Answer: 4.41 × 1014/s

Alternate Example 7.3 Calculating the Energy of a Photon

The blue–green line of the hydrogen atom spectrum has a wavelength of 486 nm. What is theenergy of a photon of this light?

Answer: 4.09 × 10–19 J

Alternate Example 7.4 Determining the Wavelength or Frequency of a HydrogenAtom Transition

What is the wavelength of the light emitted when the electron in a hydrogen atom undergoesa transition from energy level n = 6 to level n = 3?

Answer: 1.09 × 103 nm (near infrared)

Alternate Example 7.5 Applying the de Broglie Relation

Compare the wavelengths of (a) an electron traveling at a speed of one-hundredth the speedof light with (b) that of a baseball of mass 0.145 kg having a speed of 26.8 m/s (60.0 mi/hr).

Answers: (a) 243 pm (b) 1.71 × 10–34 m (too small to measure)

Alternate Example 7.6 Using the Rules for Quantum Numbers

Which of the following are permissible as sets of quantum numbers for an atomic orbital?

(a) n = 4, l = 4, ml = 0, ms = 12

(b) n = 3, l = 2, ml = 1, ms = – 12

(c) n = 2, l = 0, ml = 0, ms = 32

(d) n = 5, l = 3, ml = –3, ms = 12

Answers: (a) impermissible (l equals n) (b) permissible (c) impermissible (32 is not al-

lowed for ms) (d) permissible




Alternate Example 8.1 Applying the Pauli Exclusion Principle

Which of the following electron configurations or orbital diagrams are allowed and which arenot allowed by the Pauli exclusion principle? If they are not allowed, explain why.

(a) 1s22s12p3

(b) 1s22s12p8

(c) 1s22s22p63s23p63d8 (d) 1s22s22p63s23p63d11

(e)

Answers: (a) allowed (b) not allowed; only six electrons can be put into a p subshell(c) allowed (d) not allowed; only ten electrons can be put into a d subshell (e) not allowed;two electrons in an orbital must have opposite spin

Alternate Example 8.2 Determining the Configuration of an Atom Using theBuilding-Up Principle

Write the complete electron configuration of the arsenic atom, As, using the building-upprinciple.

Answer: 1s22s22p63s23p6 3d104s24p3

Alternate Example 8.3 Determining the Configuration of an Atom Using the Periodand Group Numbers

What are the electron configurations for the valence electrons of arsenic and cadmium?

Answer: arsenic—4s24p3; cadmium—4d105s2

Alternate Example 8.4 Applying Hund’s Rule

Write an orbital diagram for the ground state of the nickel atom.

Answer:


82 PART VII

Alternate Example 8.5 Determining Relative Atomic Sizes from Periodic Trends

Refer to a periodic table and arrange the following in order of increasing atomic radius: Br,Se, Te.

Answer: Te, Se, Br

Alternate Example 8.6 Determining Relative Ionization Energies fromPeriodic Trends

Refer to the periodic table and arrange the following in order of increasing ionization energy:As, Br, Sb.

Answer: Sb, As, Br


Alternate Example 9.1 Using Lewis Symbols to Represent Ionic Bond Formation

Represent the transfer of electrons in forming calcium oxide, CaO, from atoms.

Answer:

Alternate Example 9.2 Writing the Electron Configuration and Lewis Symbol for aMain-Group Ion

Obtain the electron configuration and the Lewis symbol for the chloride ion, Cl–.

Answer:

Alternate Example 9.3 Writing Electron Configurations of Transition-Metal Ions

Obtain the electron configurations of Mn and Mn2+.

Answer: Mn—1s22s22p63s23p63d54s2; Mn2+—1s22s22p63s23p63d5

Alternate Example 9.4 Using Periodic Trends to Obtain Relative Ionic Radii

Using a periodic table only, arrange the following ions in order of increasing ionic radius: Br–,Se2–, Sr2+.

Answer: Sr2+, Br–, Se2–



Alternate Example 9.5 Using Electronegativities to Obtain Relative Bond Polarities

Using electronegativities, arrange the following bonds in order by increasing polarity: C—N,Na—F, O—H.

Answer: C—N, O—H, Na—F

Alternate Example 9.6 Writing Lewis Formulas (Single Bonds Only)

Write electron-dot formulas for the following: (a) OF2; (b) NF3; (c) hydroxylamine, NH2OH.

Answers:

Alternate Example 9.7 Writing Lewis Formulas (Including Multiple Bonds)

Write electron-dot formulas for the following: (a) CO2, (b) HCN.

Answers:

Alternate Example 9.8 Writing Lewis Formulas (Ionic Species)

Phosphorus pentachloride exists in the solid state as the ionic compound [PCl4+][PCl6

–]; itexists in the gas phase as the PCl5 molecule. Write the Lewis formula of the PCl4

+ ion.

Answer:

Alternate Example 9.9 Writing Resonance Formulas

Draw resonance formulas of the acetate ion, CH3COO–.

Answer:


84 PART VII

Alternate Example 9.10 Writing Lewis Formulas (Exceptions to the Octet Rule)

Obtain the Lewis formula of the IF5 molecule.

Answer:

Alternate Example 9.11 Using Formal Charges to Determine the Best Lewis Formula

Compare the formal charges for the following electron-dot formulas of CO2:

Which is the preferred electron-dot formula?

Answer:

Alternate Example 9.12 Relating Bond Order and Bond Length

Consider the propylene molecule:

One of the carbon–carbon bonds has a length of 150 pm; the other has a length of 134 pm.Identify each bond with a bond length.

Answer: The C—C bond length is 150 pm; the C——C bond length is 134 pm.



Alternate Example 9.13 Estimating ∆H from Bond Energies

Estimate the enthalpy change for the following reaction, using bond energies:

Answer: ∆H = –158 kJ


Alternate Example 10.1 Predicting Molecular Geometries (Two, Three, or FourElectron Pairs)

Use the VSEPR model to predict the geometries of the following molecules: (a) AsF3, (b) PH4+,

(c) BCl3.

Answers: (a) trigonal pyramidal (b) tetrahedral (c) trigonal planar

Alternate Example 10.2 Predicting Molecular Geometries (Five or Six Electron Pairs)

Using the VSEPR model, predict the geometry of the following species: (a) ICl3, (b) ICl4–.

Answers: (a) T-shaped (b) square planar

Alternate Example 10.3 Relating Dipole Moment and Molecular Geometry

Which of the following molecules would be expected to have a zero dipole moment on thebasis of their geometry?

(a) GeF4 (b) SF2 (c) XeF2 (d) AsF3

Answers: GeF4, XeF2

Alternate Example 10.4 Applying Valence Bond Theory (Two, Three, or FourElectron Pairs)

Use valence bond theory to describe the bonding about an N atom in N2F4.


86 PART VII

Answer: The orbital diagram of the ground-state N atom is

The hybridized atom is

Each nitrogen atom forms two N—F bonds, one N—N bond, and one lone pair. An N—F bondis formed by the overlap of an sp3 orbital on N with the singly occupied 2p orbital on the Fatom. One sp3 orbital on N is used for the lone pair.

Alternate Example 10.5 Applying Valence Bond Theory (Five or Six Electron Pairs)

Use valence bond theory to describe the bonding in the ClF2– ion.

Answer: The orbital diagram of the ground state of the Cl– ion is

The sp3d hybridized ion is

The equatorial sp3d hybrid orbitals are used for lone pairs; the axial hybrid orbitals are usedin forming Cl—F bonds. Each Cl—F bond is formed by overlapping an sp3d hybrid orbital onCl– with a singly occupied 2p orbital on F.

Alternate Example 10.6 Applying Valence Bond Theory (Multiple Bonding)

Describe the bonding about the C atom in formaldehyde, CH2O, using valence bond theory.

Answer: The C and O atoms are sp2 hybridized; each atom has an unhybridized 2p orbitalperpendicular to the plane of the hybrid orbitals on that atom. Each C—H bond is formed bythe overlap of the 1s orbital on the H atom with an sp2 hybrid orbital on C. The C——O bondconsists of a σ and a π orbital, each doubly occupied. The σ bond is formed by the overlap ofan sp2 hybrid orbital on the C atom with an sp2 hybrid orbital on the O atom. The π bond isformed by the overlap of the 2p orbital on C with the 2p orbital on O.



Alternate Example 10.7 Describing Molecular Orbital Configurations(Homonuclear Diatomic Molecules)

Give the orbital diagram and electron configuration of the F2 molecule. Is the molecularsubstance diamagnetic or paramagnetic? What is the order of the bond in F2?

Answer: The orbital diagram is

The configuration is

KK(σ2s)2(σ2s∗ )2(π2p)4(σ2p)2(π2p

∗ )4

The molecular substance is diamagnetic; the bond order is 1.

Alternate Example 10.8 Describing Molecular Orbital Configurations(Heteronuclear Diatomic Molecules)

A number of compounds of the nitrosonium ion, NO+, are known, including nitrosoniumhydrogen sulfate, (NO+)(HSO4

2–). Use the molecular orbitals similar to those of a homonucleardiatomic molecule and obtain the orbital diagram, electron configuration, bond order, andmagnetic characteristics of the NO+ ion. (Note that the stability of the positive ion results fromthe loss of an antibonding electron from NO.)

Answer: The orbital diagram is

The electron configuration is

KK(σ2s)2(σ2s∗ )2(π2p)4(σ2p)2

The bond order is 3; a substance containing the ion is diamagnetic (provided the anion isalso diamagnetic).

σ2s σ2s∗ π2p σ2p π2p

∗ σ2p∗

σ2s σ2s∗ π2p σ2p π2p

∗ σ2p∗


88 PART VII


First Alternate Example 11.1 Calculating the Heat Required for a Phase Change ofa Given Mass of Substance

The fuel requirements of some homes are supplied by propane gas, C3H8, contained as theliquid in steel cylinders. If a home uses 2.40 kg of propane in an average day, how much heatmust be absorbed by the propane cylinder each day to evaporate the liquid propane, formingthe gas that is subsequently burned? The heat of vaporization of propane is 16.9 kJ/mol.

Answer: 920 kJ

Second Alternate Example 11.1 Calculating the Heat Required for a PhaseChange of a Given Mass of Substance

A 25.0-g ice cube at 0°C is placed in a glass with 2.50 × 102 g of tea at 25.0°C. To whattemperature will the tea cool? Assume that no heat is lost to the surroundings and that thespecific heat of tea is 4.184 J/(g⋅°C). The heat of fusion of ice is 6.01 kJ/mol.

Answer: 15.5°C

Alternate Example 11.2 Calculating the Vapor Pressure at a Given Temperature

The vapor pressure of diethyl ether (commonly known simply as ether) is 439.8 mmHg at20.0°C. The heat of vaporization of ether is 28.2 kJ/mol. What is the vapor pressure at 34°C?

Answer: 746 mmHg (The normal b.p. is 34.0°C.)

Alternate Example 11.3 Calculating the Heat of Vaporization from Vapor Pressures

The vapor pressures of ethanol (“alcohol”) are 100 mmHg and 760 mmHg (three significantfigures for each) at 34.9°C and 78.4°C, respectively. What is the heat of vaporization of alcohol?

Answer: 42.0 kJ/mol

Alternate Example 11.4 Relating the Conditions for the Liquefaction of Gases to theCritical Temperature

Carbon dioxide is available in steel cylinders as the liquid at room temperature. Oxygen,however, is available in steel cylinders (at room temperature) as the compressed gas, not theliquid. Explain the difference. The critical temperatures of CO2 and O2 are 31°C and –119°C,respectively.

Answer: The carbon dioxide is below its critical temperature, so under sufficient pressureit liquefies. Oxygen, on the other hand, is above its critical temperature, so it cannot beliquefied no matter how great the pressure.



Alternate Example 11.5 Identifying Intermolecular Forces

Identify the intermolecular forces that you expect for each of the following substances: (a) O2,(b) H2O2, (c) CHBr3.

Answers: (a) London forces (b) dipole–dipole, hydrogen bonding, London forces(c) dipole–dipole, London forces

Alternate Example 11.6 Determining Relative Vapor Pressure on the Basis ofIntermolecular Attraction

Which substance in each of the following pairs has the higher vapor pressure? (a) BCl3 or PCl3,(b) H2O2 or H2S.

Answers: (a) BCl3 (b) H2S

Alternate Example 11.7 Identifying Types of Solids

Identify the type of solid that you would expect for each of the following substances: (a) NF3,(b) CaBr2, (c) Na, (d) Ge.

Answers: (a) molecular (b) ionic (c) metallic (d) covalent network

Alternate Example 11.8 Determining Relative Melting Points Based onTypes of Solids

For each of the following, identify the type of solid. Then arrange the substances in order byincreasing melting point.

CaO, CH3CH2OH, NaCl, CH3Cl

Answer: CH3Cl (molecular), CH3CH2OH (molecular), NaCl (ionic), CaO (ionic)

Alternate Example 11.9 Determining the Number of Atoms per Unit Cell

How many atoms are there in a body-centered cubic lattice of a potassium crystal, in whichthere are potassium atoms at each lattice point?

Answer: Two atoms

Alternate Example 11.10 Calculating Atomic Mass from Unit-Cell Dimensionand Density

Polonium crystallizes in a simple cubic lattice (one Po atom at each lattice point) with aunit-cell length of 336 pm. The density of polonium metal is 9.20 g/cm3. Calculate the atomicweight of polonium from these data. Avogadro’s number is 6.022 × 1023/mol.

Answer: 210 amu


90 PART VII

Alternate Example 11.11 Calculating Unit-Cell Dimension from Unit-Cell Type and Density

Sodium metal has a body-centered cubic lattice with one sodium atom at each lattice point.The density of sodium metal is 0.968 g/cm3. Calculate the length of an edge of the unit cell.The atomic weight of sodium is 22.99 amu.

Answer: 429 pm


Alternate Example 12.1 Applying Henry’s Law

Helium–oxygen mixtures are sometimes used as the breathing gas in deep-sea diving. At sealevel (where the pressure is 1.0 atm), the solubility of pure helium in blood is 0.94 g/100 mL.What is the solubility of pure helium at a depth of 1500 ft? Pressure increases by 1.0 atm foreach 33 ft of depth, so at 1500 ft the pressure is 46 atm. (For a helium–oxygen mixture, thesolubility of helium will depend on the initial partial pressure of helium in the mixture, whichwill be less than 1.0 atm.)

Answer: 43 g/100 mL

Alternate Example 12.2 Calculating Mass Percentage of Solute

An experiment calls for 36.0 g of a 5.00% aqueous solution of potassium bromide. Describehow you would make up such a solution.

Answer: Dissolve 1.8 g KBr in 34.2 g H2O.

Alternate Example 12.3 Calculating the Molality of Solute

Iodine dissolves in various organic solvents, such as methylene chloride, in which it forms anorange solution. What is the molality of I2 in a solution of 5.00 g of iodine, I2, in 30.0 g ofmethylene chloride, CH2Cl2?

Answer: 0.657 m

First Alternate Example 12.4 Calculating the Mole Fractions of Components

A solution of iodine in methylene chloride, CH2Cl2, contains 1.50 g I2 and 56.00 g CH2Cl2.What are the mole fractions of each component in the solution?

Answer: 8.89 × 10–3 mole fraction I2; 0.9911 mole fraction CH2Cl2



Second Alternate Example 12.4 Calculating the Mole Fractions of Components

A bottle of bourbon is labeled 94 proof, or 47% by volume alcohol in water. What is the molefraction of ethyl alcohol, C2H5OH, in the bourbon? The density of ethyl alcohol is 0.80 g/mL.

Answer: 0.22 mole fraction C2H5OH

Alternate Example 12.5 Converting Molality to Mole Fractions

A 3.6 m solution of calcium chloride is used in tractor tires to give them weight; the additionof CaCl2 prevents the water from freezing at temperatures above about –20°C. What are themole fractions of CaCl2 and water in such a solution?

Answer: 0.061 mole fraction CaCl2, 0.939 mole fraction H2O

Alternate Example 12.6 Converting Mole Fractions to Molality

A solution contains 8.89 × 10–3 mole fraction I2 dissolved in 0.9911 mole fraction CH2Cl2(methylene chloride). What is the molality of I2 in the solution?

Answer: 0.106 m

Alternate Example 12.7 Converting Molality to Molarity

Citric acid, HC6H7O7, is often used in fruit beverages to add tartness. An aqueous solution ofcitric acid is 2.331 m HC6H7O7. What is the molarity of citric acid in the solution? The densityof the solution is 1.1346 g/mL.

Answer: 1.772 M

Alternate Example 12.8 Converting Molarity to Molality

An aqueous solution of ethanol, C2H5OH, is 14.1 M C2H5OH. The density of the solution is0.853 g/cm3. What is the molality of ethanol in the solution?

Answer: 69.3 m

Alternate Example 12.9 Calculating Vapor-Pressure Lowering

Eugenol, C10H12O2, is the chief constituent of oil of clove. It is a pale yellow liquid thatdissolves in ethanol, C2H5OH; it has a boiling point of 255°C (thus, it has a relatively low vaporpressure at room temperature). What is the vapor-pressure lowering at 20.0°C of ethanolcontaining 8.56 g of eugenol in 50.0 g of ethanol? The vapor pressure of ethanol at 20.0°C is44.6 mmHg.

Answer: 2.04 mmHg


92 PART VII

Alternate Example 12.10 Calculating Boiling-Point Elevation andFreezing-Point Depression

A solution was made up of eugenol, C10H12O2, in diethyl ether (“ether”). If the solution was0.575 m eugenol in ether, what was the freezing point and the boiling point of the solution?The freezing point and the boiling point of pure ether are –116.3°C and 34.6°C, respectively;the freezing-point depression and boiling-point-elevation constants are 1.79°C/m and2.02°C/m, respectively.

Answer: –117.3°C; 35.8°C

Alternate Example 12.11 Calculating the Molecular Weight of a Solute fromMolality

Anethole is the chief constituent of oil of anise, a flavoring agent having a licorice-like flavor.A solution of 58.1 mg of anethole in 5.00 g of benzene is determined by freezing-pointdepression to have a molality of 0.0784 m. What is the molecular weight of anethole?

Answer: 148 amu

First Alternate Example 12.12 Calculating the Molecular Weight fromFreezing-Point Depression

In a freezing-point experiment, the molality of a solution of 58.1 mg of anethole in 5.00 g ofbenzene was determined to be 0.0784 m. What is the molecular weight of anethole?

Answer: 148 amu

Second Alternate Example 12.12 Calculating the Molecular Weight fromBoiling-Point Elevation

An 11.2-g sample of sulfur was dissolved in 40.0 g of carbon disulfide. The boiling-pointelevation of carbon disulfide was found to be 2.63°C. What is the molecular weight of thesulfur in solution? What is the formula of molecular sulfur?

Answer: 256 amu; S8

Alternative Example 12.13 Calculating Osmotic Pressure

Dextran, a polymer of glucose units, is produced by bacteria growing in sucrose solutions.Solutions of dextran in water have been used as a blood plasma substitute. What is the osmoticpressure (in mmHg) at 21°C of a solution containing 1.50 g of dextran dissolved in 100.0 mLof aqueous solution, if the average molecular weight of the dextran is 4.0 × 104 amu?

According to Example 5.1 on text page 182, 760.0 mmHg is equivalent to the pressureexerted by a column of water 10.334 m high. Thus, each 1.00 mmHg of pressure is equivalentto the pressure of a 1.36-cm column of water. If the density of this dextran solution is equal tothat of water, what height of solution would exert a pressure equal to its osmotic pressure?

Answer: 6.9 mmHg; 9.4 cm



Alternate Example 12.14 Determining Colligative Properties of Ionic Solutions

What is the osmotic pressure at 25.0°C of an isotonic saline solution (a solution having anosmotic pressure equal to that of blood) that contains 0.900 g NaCl in 100.0 mL of aqueoussolution? Assume that i has the ideal value (based on the formula).

Answer: 7.53 atm


Alternate Example 14.1 Relating the Different Ways of Expressing Reaction Rates

Peroxydisulfate ion oxidizes iodide ion to the triiodide ion, I3–. (The triiodide ion has a brown

color and is formed by the reaction of iodine with iodide ion.) The reaction is

S2O82–(aq) + 3I–(aq) → 2SO4

2–(aq) + I3–(aq)

How is the rate of reaction that is expressed as the rate of formation of I3– related to the rate

of reaction of I–?

Answer: ∆[I3 −]

∆t =

13

[I −]∆t

Alternate Example 14.2 Calculating the Average Reaction Rate

Calculate the average rate of formation of O2 in the following reaction during the time intervalfrom 1200 s to 1800 s using the data given in Figure 13.5 on text page 542.

2N2O5(g) → 4NO2(g) + O2(g)

The data are Time [O2]1200 0.00361800 0.0048

Answer: ∆[O2]

∆t = 2.0 × 10−6 M/s


94 PART VII

Alternate Example 14.3 Determining the Order of Reaction from the Rate Law

Hydrogen peroxide oxidizes iodide ion in acidic solution:

H2O2(aq) + 3I–(aq) + 2H+(aq) → I3–(aq) + 2H2O(l)

The rate law for this reaction is

Rate = k[H2O2][I–]

What is the order of reaction with respect to each reactant species? What is the overall order?

Answer: First order with respect to H2O2, first order with respect to I–, second order overall.

Alternate Example 14.4 Determining the Rate Law from Initial Rates

Iron(II) is oxidized to iron(III) by chlorine in an acidic solution:

2Fe2+(aq) + Cl2(aq) H+

→ 2Fe3+(aq) + 2Cl–(aq)

The following data were collected (the rates given are relative, not actual, rates):

Initial Concentrations (mol/L)

[Fe2+] [Cl2] [H+] RateExp. 1 0.0020 0.0020 1.0 1.0 × 10–5

Exp. 2 0.0040 0.0020 1.0 2.0 × 10–5

Exp. 3 0.0020 0.0040 1.0 2.0 × 10–5

Exp. 4 0.0040 0.0040 1.0 4.0 × 10–5

Exp. 5 0.0020 0.0020 0.5 2.0 × 10–5

Exp. 6 0.0020 0.0020 0.1 1.0 × 10–4

What is the reaction order with respect to Fe2+, Cl2, and H+? What is the rate law and therelative rate constant?

Answer: First order with respect to both Fe2+ and Cl2; –1 order with respect to H+.

Rate = k[Fe2+][Cl2]

[H+] ; k = 2.5.

Alternate Example 14.5 Using an Integrated Rate Law

Cyclopropane is used as an anesthetic. The isomerization of cyclopropane to propene is afirst-order reaction with a rate constant of 9.2/s at 1000°C. If an initial sample of cyclopropanehas a concentration of 6.00 M, what will the cyclopropane concentration be after 1.00 s?

Answer: 6.1 × 10–4 M



Alternate Example 14.6 Relating the Half-Life of a Reaction to the Rate Constant

Ammonium nitrite is unstable because ammonium ion reacts with nitrite ion to producenitrogen:

NH4+(aq) + NO2

–(aq) → N2(g) + 2H2O(l)

In a solution that is 10.0 M in NH4+, the reaction is first order in nitrite ion (for low

concentrations), and the rate constant at 25°C is 3.0 × 10–3/s. What is the half-life of thereaction?

Answer: 2.3 × 102 s

Alternate Example 14.7 Using the Arrhenius Equation

A convenient rule of thumb is that the rate of a reaction doubles for a 10°C change intemperature. What is the activation energy for a reaction whose rate doubles from 10.0°C to20.0°C? By what factor would the reaction rate increase if the temperature were increased from10.0°C to 25.0°C?

Answer: 47.8 kJ/mol; 2.78

Alternate Example 14.8 Writing the Overall Chemical Equation from a Mechanism

Chlorofluorocarbons, such as CCl2F2, decompose in the stratosphere from the irradiation withshort-wavelength ultraviolet light present at those altitudes. The decomposition yields chlo-rine atoms. These chlorine atoms catalyze the decomposition of ozone in the presence ofoxygen atoms (available in the stratosphere from the ultraviolet irradiation of O2) to giveoxygen molecules. The mechanism of the decomposition is

Cl(g) + O3(g) → ClO(g) + O2(g)

ClO(g) + O(g) → Cl(g) + O2(g)

What is the overall chemical equation for the decomposition of ozone?

Answer: Cl(g) + O3(g) → ClO(g) + O2(g)ClO(g) + O(g) → Cl(g) + O2(g)

O3(g) + O(g) → 2O2(g)

Note that Cl atoms are used up in the first step but regenerated in the second step, so Cl atomsdo not appear in the overall equation. In other words, Cl functions as a catalyst (see Section13.9).


96 PART VII

Alternate Example 14.9 Determining the Molecularity of an Elementary Reaction

What is the molecularity of each of the steps in the mechanism of ozone decompositiondescribed in Alternate Example 13.8?

Answer: Each step is bimolecular.

Alternate Example 14.10 Writing the Rate Equation for an Elementary Reaction

(a) Write the rate equation for the first step in the ozone decomposition mechanism describedin Alternate Example 13.8. (b) Write the rate equation for the following elementary reaction:NO(g) + NO(g) → N2O2(g).

Answers: (a) Rate = k[Cl][O3] (b) Rate = k[NO]2

Alternate Example 14.11 Determining the Rate Law from a Mechanism with anInitial Slow Step

The decomposition of hydrogen peroxide is catalyzed by iodide ion. The mechanism isthought to be

H2O2(aq) + I–(aq) → H2O(l) + IO–(aq)

IO–(aq) + H2O2(aq) → H2O(l) + O2(g) + I–(aq)

At 25°C, the first step is slow relative to the second step. What is the rate law predicted by thismechanism?

Answer: Rate = k[H2O2][I–]

Alternate Example 14.12 Determining the Rate Law from a Mechanism with anInitial Fast, Equilibrium Step

The mechanism for the decomposition of hydrogen peroxide in the presence of iodide ion isdescribed in Alternate Example 13.11. At 100°C, the first step is fast relative to the second step.What is the rate law predicted by this mechanism? Note that [H2O] can be taken as constant.

Answer: Rate = k[H2O2]2[I–]




Alternate Example 15.1 Applying Stoichiometry to an Equilibrium Mixture

When heated, phosphorus pentachloride, PCl5, forms PCl3 and Cl2 as follows:

PCl5(g) PCl3(g) + Cl2(g)

When 1.00 mol PCl5 in a 1.00-L container is allowed to come to equilibrium at a certaintemperature, the mixture is found to contain 0.135 mol PCl3. What is the molar compositionof the mixture; that is, how many moles of each substance are present?

Answer: 0.135 mol PCl3, 0.135 mol Cl2, and 0.865 mol PCl5

Alternate Example 15.2 Writing Equilibrium-Constant Expressions

Methanol, wood alcohol, is made commercially by hydrogenation of carbon monoxide atelevated temperature and pressure in the presence of a catalyst:

2H2(g) + CO(g) CH3OH(g)

What is the Kc expression for this reaction?

Answer: Kc = [CH3OH][H2]2[CO]

Alternate Example 15.3 Obtaining an Equilibrium Constant from ReactionComposition

Carbon dioxide decomposes at elevated temperatures to carbon monoxide and oxygen:

2CO2(g) 2CO(g) + O2(g)

At 3000 K, 2.00 mol CO2 is placed into a 1.00-L container and allowed to come to equilibrium.At equilibrium, 0.90 mol CO2 remains. What is the value for Kc at this temperature?

Answer: Kc = 0.82

Alternate Example 15.4 Writing Kc for a Reaction with Pure Solids or Liquids

When water, in the form of steam, is passed over hot coke (carbon), a mixture of hydrogenand carbon monoxide, called water gas, is formed. This mixture can be used as a fuel. Writethe equilibrium-constant (Kc) expression for this process.

H2O(g) + C(s) CO(g) + H2(g)

Answer: Kc = [CO][H2]

[H2O]


98 PART VII

Alternate Example 15.5 Using the Reaction Quotient

Nickel(II) oxide can be reduced to the metal by treatment with carbon monoxide.

CO(g) + NiO(s) CO2(g) + Ni(s)

If the partial pressure of CO is 100. mmHg and the total pressure of CO and CO2 does notexceed 1.0 atm, will reaction occur at 1500 K at equilibrium? (Kp = 700. at 1500 K.)

Answer: Reaction will occur; Q = 6.6, and more Ni(s) will form.

Alternate Example 15.6 Obtaining One Equilibrium Concentration Given the Others

Nitrogen and oxygen form nitric oxide.

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

If an equilibrium mixture at 25°C contains 0.040 mol/L of N2 and 0.010 mol/L of O2, what isthe concentration of NO in this mixture? The equilibrium constant at 25°C is 1 × 10–30.

Answer: 2 × 10–17 mol/L

Alternate Example 15.7 Solving an Equilibrium Problem (Involving a LinearEquation in x)

Hydrogen iodide decomposes to hydrogen gas and iodine gas.

2HI(g) H2(g) + I2(g)

At 800 K, the equilibrium constant, Kc, for this reaction is 0.016. If 0.50 mol HI is placed in a5.0-L flask, what will be the composition of the equilibrium mixture?

Answer: [HI] = 0.080 M; [H2] = [I2] = 0.010 M

Alternate Example 15.8 Solving an Equilibrium Problem (Involving a QuadraticEquation in x)

N2O4 decomposes to NO2; the equilibrium equation in the gaseous phase is

N2O4(g) 2NO2(g)

At 100°C, Kc = 0.36. If a 1.00-L flask initially contains 0.100 mol N2O4/L, what will be theconcentration of NO2 at equilibrium?

Answer: [NO2] = 0.12 mol/L



Alternate Example 15.9 Applying Le Chatelier’s Principle When aConcentration Is Altered

The Fischer–Tropsch process for the synthesis of gasoline consists of passing a mixture ofcarbon monoxide and hydrogen over an iron–cobalt catalyst. A typical reaction that occurs inthe process is as follows:

8CO(g) + 17H2(g) C8H18(g) + 8H2O(g)

Suppose the reaction mixture comes to equilibrium at 200°C, then is suddenly cooled to roomtemperature where octane (C8H18) liquefies. The remaining gases are then reheated to 200°C.What is the direction of the reaction as equilibrium is attained?

Answer: Left to right

Alternate Example 15.10 Applying Le Chatelier’s Principle When thePressure Is Altered

A typical reaction that occurs in the Fischer–Tropsch process is

8CO(g) + 17H2(g) C8H18(g) + 8H2O(g)

Would you expect more or less of the product octane, C8H18, at equilibrium as the pressureincreases?

Answer: More product at high pressure

Altered Example 15.11 Applying Le Chatelier’s Principle When theTemperature Is Altered

Calculate ∆H° for the chemical equation given in the previous alternate example, usingstandard heats of formation, ∆Hf°. From the result, predict whether more or less octane, C8H18,would be produced at 200°C than at 20°C. Values of ∆Hf° (in kJ/mol) are as follows: CO(g),–110; C8H18(g), –209; H2O(g), –242.

Answer: ∆H° = –1265 kJ. Higher temperature favors less product at equilibrium. (However,equilibrium is attained more quickly at higher temperature.)


100 PART VII


Alternate Example 16.1 Identifying Acid and Base Species

Identify the acid and base species in the following equations:

(a) CO32–(aq) + H2O(l) HCO3

–(aq) + OH–(aq)(b) C2H3O2

–(aq) + HNO2(aq) HC2H3O2(aq) + NO2–(aq)

Answers: (a) acid species—H2O, HCO3–; base species—CO3

2–, OH– (b) acid species—HNO2, HC2H3O2; base species—C2H3O2

–, NO2–

Alternate Example 16.2 Identifying Lewis Acid and Base Species

Calcium oxide reacts with sulfur dioxide to produce calcium sulfite. The reaction is useful inremoving sulfur dioxide from the gases produced in the combustion of sulfur-containingmaterials. We can represent this reaction as the reaction of the oxide ion with sulfur dioxide.

Label each species on the left as either Lewis acid or Lewis base.

Answer: O2–, Lewis base; SO2, Lewis acid

Alternate Example 16.3 Deciding Whether Reactants or Products Are Favored in anAcid–Base Reaction

Decide which species are favored at the completion of the following reaction:

HCN(aq) + HSO3–(aq) CN–(aq) + H2SO3(aq)

Answer: The reactants are favored.

Alternate Example 16.4 Calculating Concentrations of H3O+and OH– in Solutions ofa Strong Acid or Base

Calculate the concentrations of hydronium ion and hydroxide ion at 25°C in: (a) 0.10 M HCl,(b) 1.4 × 10–4 M Mg(OH)2, a strong base.

Answers: (a) [H3O+] = 0.10 M; [OH–] = 1.0 × 10–13 M(b) [H3O+] = 3.6 × 10–11 M; [OH–] = 2.8 × 10–4 M



Alternate Example 16.5 Calculating the pH from the Hydronium-Ion Concentration

Calculate the pH of typical adult blood, which has a hydronium-ion concentration of4.0 × 10–8 M.

Answer: pH = 7.40

Alternate Example 16.6 Calculating the Hydronium-Ion Concentration from the pH

The pH of natural rain is 5.60. Calculate its hydronium-ion concentration.

Answer: [H3O+] = 2.5 × 10–6 M


Alternate Example 17.1 Determining Ka from the Solution pH

Sore-throat medications sometimes contain the weak acid phenol, HC6H5O. A 0.10 M solutionof phenol has a pH of 5.43 at 25°C. What is the acid-ionization constant, Ka, for this acid at25°C? What is its degree of ionization?

Answer: Ka = 1.4 × 10–10; degree of ionization = 3.7 × 10–5

Alternate Example 17.2 Calculating Concentrations of Species in a Weak AcidSolution Using Ka (Approximation Method)

Para-hydroxybenzoic acid is used to make certain dyes. What are the concentrations of thisacid, of hydrogen ion, and of para-hydroxybenzoate anion in a 0.200 M aqueous solution at25°C? What is the pH of the solution and the degree of ionization of this acid? The Ka of thisacid is 2.6 × 10–5.

Answer: [p-hydroxybenzoic acid] = 0.198 M; [H+] = [anion] = 2.3 × 10–3 M; pH = 2.64; degreeof ionization = 0.012

Alternate Example 17.3 Calculating Concentrations of Species in a Weak AcidSolution Using Ka (Quadratic Formula)

What is the pH at 25°C of 400 mL of aqueous solution containing 0.400 mol of chloroaceticacid, a monoprotic acid? The Ka = 1.35 × 10–3.

Answer: [H+] = 3.6 × 10–2 M (using quadratic formula); pH = 1.44


102 PART VII

Alternate Example 17.4 Calculating Concentrations of Species in a Solution of aDiprotic Acid

Tartaric acid, H2C4H4O6, is a diprotic acid used in food products. What is the pH of a 0.10 Msolution and the concentration of the C4H4O6

2– ion? Ka1 = 9.2 × 10–4 and Ka2 = 4.3 × 10–5.

Answer: pH = 2.02; [C4H4O62–] = 4.3 × 10–5 (= Ka2)

Alternate Example 17.5 Calculating Concentrations of Species in a Weak BaseSolution Using Kb

Aniline, C6H5NH2, is used in the manufacturing of some perfumes. What is the pH of a 0.035M solution of aniline at 25°C ? The Kb = 4.2 × 10–10 at 25°C.

Answer: pH = 8.56

Alternate Example 17.6 Predicting Whether a Salt Solution Is Acidic, Basic,or Neutral

Ammonium nitrate, NH4NO3, is administered as an intravenous solution to patients whoseblood pH has deviated from the normal value of 7.40. Would this substance be used foracidosis (blood pH < 7.40) or alkalosis (blood pH > 7.40)?

Answer: NH4NO3 is a salt of a weak base and strong acid, so its solution would be acidic;it would be used for alkalosis.

Alternate Example 17.7 Obtaining Ka from Kb or Kb from Ka

Obtain the Kb for the F– ion, the ion added to public water supplies to protect teeth. For HF,Ka = 6.8 × 10–4.

Answer: Kb = 1.5 × 10–11

Alternate Example 17.8 Calculating Concentrations of Species in a Salt Solution

Household bleach is a 5% solution of sodium hypochlorite, NaClO. This corresponds to amolar concentration of about 0.70 M NaClO (Kb = 2.86 × 10–7). What are the OH– concentrationand the pH of such a solution?

Answer: [OH–] = 4.5 × 10–4; pH = 10.65



Alternate Example 17.9 Calculating the Common-Ion Effect on Acid Ionization(Effect of a Strong Acid)

Calculate the degree of ionization of benzoic acid, HC7H5O2, in a 0.15 M solution to whichsufficient HCl is added to make it also 0.010 M HCl. Compare the degree of ionization to thatof 0.15 M benzoic acid (no HCl). Ka = 6.3 × 10–5.

Answer: Degree of ionization = 0.0063; this is much smaller than the degree of ionizationof 0.15 M benzoic acid without HCl (0.020).

Alternate Example 17.10 Calculating the Common-Ion Effect on Acid Ionization(Effect of a Conjugate Base)

Calculate the pH of a 0.10 M solution of HF to which sufficient sodium fluoride is added tomake the concentration 0.20 M NaF. The Ka of HF = 6.8 × 10–4.

Answer: pH = 3.47

Alternate Example 17.11 Calculating the pH of a Buffer from Given Volumesof Solution

What is the pH of a buffer made by mixing 1.00 L of 0.020 M benzoic acid, HC7H5O2, with 3.00L of 0.060 M sodium benzoate, NaC7H5O2? The Ka for benzoic acid is 6.3 × 10–5.

Answer: pH = 5.15

Alternate Example 17.12 Calculating the pH of a Buffer When a Strong Acid orStrong Base Is Added

Calculate the pH change that will result from the addition of 5.0 mL of 0.10 M HCl to 50.0 mLof a buffer containing 0.10 M NH3 and 0.10 M NH4

+. How much would the pH of 50.0 mL ofwater change if the same amount of acid were added?

Answer: The pH of the buffer decreases by 0.09 pH units from 9.26 to 9.17. The pH of thewater decreases by 4.96 pH units from 7.00 to 2.04.

Alternate Example 17.13 Calculating the pH of a Solution of a Strong Acid and aStrong Base

Calculate the pOH and the pH of a solution in which 10.0 mL of 0.100 M HCl is added to 25.0mL of 0.100 M NaOH.

Answer: pOH = 1.368; pH = 12.632


104 PART VII

Alternate Example 17.14 Calculating the pH at the Equivalence Point in theTitration of a Weak Acid by a Strong Base

Calculate the [OH–] and the pH at the equivalence point for the titration of 500. mL of 0.10 Mpropionic acid with 0.050 M calcium hydroxide. (This can be used to prepare a preservativefor bread.) Ka = 1.3 × 10–5.

Answer: [OH–] = 6.2 × 10–6 M; pH = 8.79


Alternate Example 18.1 Writing Solubility Product Expressions

Write the solubility product expression for the following salts: (a) Hg2Cl2; (b) HgCl2.

Answers: (a) Hg2Cl2: Ksp = [Hg22+][Cl–]2

(b) HgCl2: Ksp = [Hg2+][Cl–]2

Alternate Example 18.2 Calculating Ksp from the Solubility (Simple Example)

Exactly 0.133 mg of AgBr will dissolve in 1.00 L of water. What is the value of Ksp for AgBr?

Answer: Ksp = 5.02 × 10–13

Alternate Example 18.3 Calculating Ksp from the Solubility(More Complicated Example)

An experimenter finds that the solubility of barium fluoride is 1.1 g in 1.00 L of water at 25°C.What is the value of Ksp for barium fluoride, BaF2, at this temperature?

Answer: Ksp = 1.0 × 10–6

Alternate Example 18.4 Calculating the Solubility from Ksp

Calomel, whose chemical name is mercury(I) chloride, Hg2Cl2, was once used in medicine (asa laxative and diuretic). It has a Ksp equal to 1.3 × 10–18. What is the solubility of Hg2Cl2 ingrams per liter?

Answer: 6.9 × 10–7 mol/L = 3.2 × 10–4 g/L



Alternate Example 18.5 Calculating the Solubility of a Slightly Soluble Salt in aSolution of a Common Ion

What is the molar solubility of silver chloride in 1.0 L of solution that contains 2.0 × 10–2 molof HCl?

Answer: 9.0 × 10–9 M

Alternate Example 18.6 Predicting Whether Precipitation Will Occur(Given the Ion Concentrations)

One form of kidney stones is calcium phosphate, Ca3(PO4)2, which has a Ksp of 1 × 10–26. If asample of urine contains 1.0 × 10–3 M Ca2+ and 1.0 × 10–8 M PO4

3– ion, calculate Qc and predictwhether Ca3(PO4)2 will precipitate.

Answer: Qc = 1.0 × 10–25. Precipitation will occur.

Alternate Example 18.7 Predicting Whether Precipitation Will Occur(Given Solution Volumes and Concentrations)

Exactly 0.400 L of 0.50 M Pb2+ and 1.60 L of 2.50 × 10–2 M Cl– are mixed together to form2.00 L of solution. Calculate Qc and predict whether PbCl2 will precipitate. The Ksp of PbCl2 is1.6 × 10–5.

Answer: Qc = 4.0 × 10–5. Precipitation will occur.

Alternate Example 18.8 Determining the Qualitative Effect of pH on Solubility

Consider the two slightly soluble salts barium fluoride and silver bromide. Which of thesewould have its solubility more affected by the addition of strong acid? Would the solubilityof that salt increase or decrease?

Answer: The barium fluoride is much more soluble in acidic solution, whereas the solubil-ity of the silver bromide is not affected.

Alternate Example 18.9 Calculating the Concentration of a Metal Ion inEquilibrium with a Complex Ion

What is the concentration of Ag+(aq) ion in 0.00010 M AgNO3 that is also 1.0 M CN–? The Kf

for Ag(CN)2– is 5.6 × 1018.

Answer: [Ag+] = 1.8 × 10–23 M


106 PART VII

Alternate Example 18.10 Predicting Whether a Precipitate Will Form in thePresence of the Complex Ion

Silver chloride usually does not precipitate in solutions of 1.00 M NH3 (see Example 17.10 ontext page 745). However, silver bromide has a smaller Ksp. Will silver bromide precipitate froma solution containing 0.010 M AgNO3, 0.010 M NaBr, and 1.00 M NH3? Calculate the Qc valueand compare it with silver bromide’s Ksp of 5.0 × 10–13.

Answer: Qc = 6.1 × 10–12; thus, AgBr will precipitate.

Alternate Example 18.11 Calculating the Solubility of a Slightly Soluble IonicCompound in a Solution of the Complex Ion

Calculate the molar solubility of AgBr in 1.0 M NH3 at 25°C.

Answer: [Ag(NH3)2+] = 2.9 × 10–3 M = molar solubility

CHAPTER 19 Thermodynamics and Equilibrium

Alternate Example 19.1 Calculating the Entropy Change for a Phase Transition

Acetone, CH3COCH3, is a volatile liquid solvent (it is used in nail polish, for example). Thestandard enthalpy of formation of the liquid at 25°C is –247.6 kJ/mol; the same quantity forthe vapor is –216.6 kJ/mol. What is the entropy change when 1.00 mol liquid acetone vaporizesat 25°C?

Answer: 104.0 J/(K⋅mol)

Alternate Example 19.2 Predicting the Sign of the Entropy Change of a Reaction

The opening to Chapter 6, on thermochemistry, describes the endothermic reaction of solidbarium hydroxide octahydrate and solid ammonium nitrate:

Ba(OH)2⋅8H2O(s) + 2NH4NO3(s) → 2NH3(g) + 10H2O(l) + Ba(NO3)2(aq)

Predict the sign of ∆S° for the reaction.

Answer: positive



Alternate Example 19.3 Calculating ∆S° for a Reaction

When wine is exposed to air in the presence of certain bacteria, the ethyl alcohol is oxidizedto acetic acid, giving vinegar. Calculate the standard entropy change at 25°C for the followingsimilar reaction:

CH3CH2OH(l) + O2(g) → CH3COOH(l) + H2O(l)

The standard entropies of the substances in J/(K⋅mol) at 25°C are CH3CH2OH(l), 161; O2(g),205; CH3COOH(l), 160; H2O(l), 69.9.

Answer: ∆S° = –136 J/K

Alternate Example 19.4 Calculating ∆G° from ∆H° and ∆S°

Using standard enthalpies of formation and the value of ∆S° obtained in Alternate Example18.3, calculate ∆G° at 25°C for the oxidation of ethyl alcohol to acetic acid. (See AlternateExample 18.3 for the equation.) The standard enthalpies of formation of the substances inkJ/mol at 25°C are CH3CH2OH(l), –277.6; CH3COOH(l), –487.0; H2O(l), –285.8.

Answer: ∆H° = –495.2 kJ, ∆G° = –454.7 kJ

Alternate Example 19.5 Calculating ∆G° from Standard Free Energies of Formation

Calculate the free-energy change, ∆G°, at 25°C for the oxidation of ethyl alcohol to acetic acidusing standard free energies of formation. (See Alternate Example 18.3 for the equation.) Thestandard free energies of formation of the substances in kJ/mol at 25°C are CH3CH2OH(l),–174.8; CH3COOH(l), –392.5; H2O(l), –237.2.

Answer: ∆G° = –454.9 kJ

Alternate Example 19.6 Interpreting the Sign of ∆G°

Consider the reaction discussed in Alternate Example 18.2:

Ba(OH)2⋅8H2O(s) + 2NH4NO3(s) → 2NH3(g) + 10H2O(l) + Ba(NO3)2(aq)

The standard enthalpy change at 25°C is 170.4 kJ; the standard entropy change at 25°C is 657J/K. Calculate ∆G° at 25°C for the reaction. Interpret the values of ∆H°, ∆S°, and ∆G°.

Answer: ∆G° = –25 kJ. The positive value of ∆H° indicates an endothermic reaction; thelarge positive ∆S° indicates the formation of considerable disorder (formation of gas, liquid,and solution from two crystalline solids); the negative ∆G° indicates a spontaneous reaction.Note that the negative ∆G° results from the fact that although ∆H° is positive, ∆S° is a largepositive number.


108 PART VII

Alternate Example 19.7 Writing the Expression for a Thermodynamic EquilibriumConstant

Write the expressions for the thermodynamic equilibrium constants for each of the followingreactions.

(a) N2O4(g) 2NO2(g)(b) Zn(s) + 2H+(aq) Zn2+(aq) + H2(g)

Answers: (a) K = p(NO2)2

p(N2O4) (b) K =

[Zn2+]p(H2)[H+]2

Alternate Example 19.8 Calculating K from the Standard Free-Energy Change(Molecular Equation)

Calculate the value of the thermodynamic equilibrium constant at 25°C for the reaction givenin Alternate Example 18.7(a):

N2O4(g) 2NO2(g)

The values of the standard free energy of formation of the substances in kJ/mol at 25°C areNO2, 51.30; N2O4, 97.82.

Answer: K = 0.145

Alternate Example 19.9 Calculating K from the Standard Free-Energy Change(Net Ionic Equation)

Calculate the value of the thermodynamic equilibrium constant at 25°C for the reaction givenin Alternate Example 18.7(b):

Zn(s) + 2H+(aq) Zn2+(aq) + H2(g)

The values of the standard free energy of formation of the substances in kJ/mol at 25°C areH+(aq), 0; Zn2+(aq), –147.2.

Answer: K = 6.1 × 1025



Alternate Example 19.10 Calculating ∆G° and K at Various Temperatures

Obtain the standard free-energy change and Kp at 35°C for the reaction whose free-energychange and equilibrium constant were obtained at 25°C in Alternate Example 18.8:

N2O4(g) 2NO2(g)

The standard enthalpies of formation of the substances in kJ/mol at 25°C are N2O4, 9.16; NO2,33.2. The standard entropies at 25°C, in J/(K⋅mol) are N2O4, 304.3; NO2, 239.9.

Answer: ∆G° at 35°C is 3.2 kJ; Ksp is 0.29.


Alternate Example 20.1 Balancing Equations by the Half-Reaction Method (AcidicSolution)

Nitrate ion in acid solution (nitric acid) is an oxidizing agent. When it reacts with zinc, themetal is oxidized to the zinc ion, Zn2+, and nitrate is reduced. Assume that nitrate is reducedto the ammonium ion, NH4

+. Write the balanced ionic equation for this reaction; use thehalf-reaction method.

Answer: 4Zn + NO3– + 10H+ → 4Zn2+ + NH4

+ + 3H2O

Alternate Example 20.2 Balancing Equations by the Half-Reaction Method (BasicSolution)

Lead(II) ion, Pb2+, yields the plumbite ion, Pb(OH)3–, in basic solution. In turn, this ion is

oxidized in basic hypochlorite solution, ClO–, to lead(IV) oxide, PbO2. Balance the equationfor this reaction, using the half-reaction method. The skeleton equation is

Pb(OH)3– + ClO– → PbO2 + Cl–

Answer: ClO– + Pb(OH)3– → Cl– + PbO2 + H2O + OH–


110 PART VII

Alternate Example 20.3 Sketching and Labeling a Voltaic Cell

You construct one half-cell of a voltaic cell by inserting a copper metal strip into a solution ofcopper(II) sulfate. You construct another half-cell by inserting an aluminum metal strip intoa solution of aluminum nitrate. You now connect the half-cells by a salt bridge. Whenconnected to an external circuit, the aluminum is oxidized. Sketch the resulting voltaic cell.Label the anode and cathode, showing the corresponding half-reactions. Indicate the directionof electron flow in the external circuit.

Answer: Aluminum is the anode; copper is the cathode. The electrons flow from the anodeto the cathode. The half-reactions are

Al(s) → Al3+(aq) + 3e–

Cu2+(aq) + 2e– → Cu(s)

Alternate Example 20.4 Writing the Cell Reaction from the Cell Notation

The cell notation for the voltaic cell in Alternate Example 19.3 is

Al(s)Al3+(aq) Cu2+(aq)Cu(s)

Write the cell reaction.

Answer: 2Al(s) + 3Cu2+(aq) → 2Al3+(aq) + 3Cu(s)

Alternate Example 20.5 Calculating the Quantity of Work from a Given Amount ofCell Reactant

The emf of a particular cell constructed as described in Alternate Example 19.3 is 0.500 V. Thecell reaction, given in Alternate Example 19.2, is

2Al(s) + 3Cu2+(aq) → 2Al3+(aq) + 3Cu(s)

Calculate the maximum electrical work of this cell obtained from 1.00 g of aluminum.

Answer: –5.36 kJ

Alternate Example 20.6 Determining the Relative Strengths of Oxidizing andReducing Agents

(a) Which is the stronger reducing agent under standard conditions, Sn2+ (to Sn4+) or Fe (toFe2+)? (b) Which is the stronger oxidizing agent under standard conditions, Cl2 or MnO4

–?

Answers: (a) Fe (b) MnO4–



Alternate Example 20.7 Determining the Direction of Spontaneity fromElectrode Potentials

Will dichromate ion oxidize manganese(II) ion to permanganate ion in acid solution understandard conditions?

Answer: no

Alternate Example 20.8 Calculating the emf from Standard Potentials

A fuel cell is simply a voltaic cell that uses a continuous supply of electrode materials toprovide a continuous supply of electrical energy. A fuel cell employed by NASA on spacecraftuses hydrogen and oxygen under basic conditions to produce electricity; the water alsoproduced can be used for drinking. The net reaction is

2H2(g) + O2(g) → 2H2O(l)

Calculate the standard emf of the oxygen–hydrogen fuel cell.

2H2O(l) + 2e– H2(g) + 2OH–(aq) E° = –0.83 V

O2(g) + 2H2O(l) + 4e– 4OH–(aq) E° = 0.40 V

Answer: 1.23 V

Alternate Example 20.9 Calculating the Free-Energy Change fromElectrode Potentials

Calculate the standard free-energy change for the net reaction used in the hydrogen–oxygenfuel cell described in Alternate Example 19.8.

2H2(g) + O2(g) → 2H2O(l)

The emf of the cell was calculated in that example. How does this compare with ∆Gf° of H2O(l)?

Answer: –475 kJ (for the formation of 2 mol H2O); this is 2∆Gf°

Alternate Example 20.10 Calculating the Cell emf from Free-Energy Change

A voltaic cell consists of one half-cell with Fe dipping into an aqueous solution of 1.0 M FeCl2and the other half-cell with Ag dipping into an aqueous solution of 1.0 M AgNO3. Obtain thestandard free-energy change for the cell reaction using standard free energies of formation.The standard free energies of formation of the ions in kJ/mol are Ag+(aq), 77; Fe2+(aq), –85.What is the emf of this cell?

Answer: –239 kJ, 1.24 V


112 PART VII

Alternate Example 20.11 Calculating the Equilibrium Constant from Cell emf

Calculate the equilibrium constant K at 25°C for the following reaction from the standard emf.

Pb2+(aq) + Fe(s) Pb(s) + Fe2+(aq)

Answer: Ecell° = 0.28 V, K = 2.9 × 109

Alternate Example 20.12 Calculating the Cell emf for Nonstandard Conditions

A pH meter is constructed using hydrogen gas bubbling over an inert platinum electrode (thehydrogen electrode) at a pressure of 1.2 atm. The other electrode is aluminum metal immersedin a 0.20 M Al3+ solution. What is the cell emf when the hydrogen electrode is immersed in asample of acid rain with a pH of 4.0 at 25°C? If the electrode is placed in a sample of shampoosolution and the emf is 1.17 V, what is the pH of the shampoo solution? The cell reaction is

2Al(s) + 6H+(aq) 2Al3+(aq) + 3H2(g)

Answer: emf = 1.43 V; pH = 8.4

Alternate Example 20.13 Predicting the Half-Reactions in an Aqueous Electrolysis

Describe what you expect to happen at the electrodes when an aqueous solution of sodiumiodide is electrolyzed.

Answer: H2O is reduced to H2 (in preference to the reduction of Na+ to Na); I– is oxidized(in preference to the oxidation of H2O to O2).

Alternate Example 20.14 Calculating the Amount of Charge from the Amount ofProduct in an Electrolysis

What electric charge is required to plate a piece of automobile molding with 1.00 g ofchromium metal using a chromium(III) ion solution? If the electrolysis current is 2.00 A, howlong does the plating take?

Answer: 5.57 × 103 C; 46.4 min

Alternate Example 20.15 Calculating the Amount of Product from the Amount ofCharge in an Electrolysis

A solution of nickel salt is electrolyzed to nickel metal by a current of 2.43 A. If this currentflows for 10.0 min, how many coulombs is this? How much nickel metal is deposited in theelectrolysis?

Answer: 1.46 × 103 C; 0.443 g




Alternate Example 21.1 Writing a Nuclear Equation

Radon-222 is a radioactive noble gas that is sometimes present as an air pollutant in homesbuilt over soil with high uranium content (uranium-238 decays to radium-226, which in turndecays to radon-222). A radon-222 nucleus decays to polonium-218 by emitting an alphaparticle. Write the nuclear equation for this decay process.

Answer: 22286Rn → 218

84Po + 42He

Alternate Example 21.2 Deducing a Product or Reactant in a Nuclear Equation

Iodine-131 is used in the diagnosis and treatment of thyroid cancer. This isotope decays bybeta emission. What is the product nucleus?

Answer: 13154Xe

Alternate Example 21.3 Predicting the Relative Stabilities of Nuclides

Predict which nucleus in each pair should be more stable and explain why: (a) astatine-210,lead-207; (b) molybdenum-91, molybdenum-92; (c) calcium-37, calcium-42.

Answers: (a) Pb-207. It has an atomic number less than 83, whereas At has an atomicnumber greater than 83. (b) Mo-92. It has a magic number of neutrons; Mo-91 does not. (c)Ca-42. It lies within the band of stability; Ca-37 lies below the band of stability.

Alternate Example 21.4 Predicting the Type of Radioactive Decay

Thallium-201 is a radioactive isotope used in the diagnosis of circulatory impairment andheart disease. How do you expect it to decay?

Answer: Positron emission or electron capture. (Electron capture is more likely becausethallium is a heavy element.)

Alternate Example 21.5 Using the Notation for a Bombardment Reaction

Sodium-22 is made by the bombardment of magnesium-24 (the most abundant isotope ofmagnesium) by deuterons. An alpha particle is the other product. Write the abbreviatednotation for the nuclear reaction.

Answer: 2412Mg(d, α)22

11


114 PART VII

Alternate Example 21.6 Determining the Product Nucleus in a NuclearBombardment Reaction

A neutron is produced when lithium-7 is bombarded with a proton. What product nucleus isobtained in this reaction?

Answer: Be-7

Alternate Example 21.7 Calculating the Decay Constant from the Activity

The thorium-234 isotope decays by emitting a beta particle. A 50.0-µg sample of thorium-234has an activity of 1.16 Ci. What is the decay constant for thorium-234?

Answer: 3.34 × 10–7/s

Alternate Example 21.8 Calculating the Half-Life from the Decay Constant

Thallium-201 is used in the diagnosis of heart disease. The isotope decays by electron capture;the decay constant is 2.63 × 10–6/s. What is the half-life of this isotope in days?

Answer: 3.05 days

Alternate Example 21.9 Calculating the Decay Constant and Activityfrom the Half-Life

Iodine-131 is used in the diagnosis and treatment of thyroid disorders. The half-life for thebeta decay of iodine-131 is 8.07 days. What is the decay constant (in /s)? What is the activityin curies of a 1.0-µg sample of iodine-131?

Answer: 9.94 × 10–7/s; 0.12 Ci

Alternate Example 21.10 Determining the Fraction of Nuclei Remaining After aSpecified Time

A 0.500-g sample of iodine-131 is obtained by a hospital. How much will remain after a periodof one week? The half-life is 8.07 days.

Answer: 54.8%, or 0.274 g

Alternate Example 21.11 Applying the Carbon-14 Dating Method

A sample of wheat recovered from a cave was analyzed and gave 12.8 disintegrations ofcarbon-14 per minute per gram of carbon. What is the age of the grain? Carbon from livingmaterials decays at a rate of 15.3 disintegrations per minute per gram of carbon. The half-lifeof carbon-14 is 5730 y.

Answer: 1.48 × 103 y



Alternate Example 21.12 Calculating the Energy Change for a Nuclear Reaction

Consider the following nuclear reaction in which a lithium-7 nucleus is bombarded with ahydrogen nucleus to produce two alpha particles:

73Li + 11H → 24

2He

What is the energy change of this reaction per gram of lithium? The nuclear masses are73Li, 7.01436 amu; 11H, 1.00728 amu; 42He, 4.00150 amu.

Answer: –2.387 × 1011 J/g


Alternate Example 23.1 Writing the IUPAC Name Given the Structural Formula of aCoordination Compound

Give the IUPAC name of the coordination compound [Cu(CN)4(H2O)2]Cl2.

Answer: tetracyanodiaquacopper(II) chloride

Alternate Example 23.2 Writing the Structural Formula Given the IUPAC Name of aCoordination Compound

What is the structural formula of hexaamminecobalt(II) tetrachloroaurate(III)?

Answer: [Co(NH3)6][AuCl4]2


116 PART VII

Alternate Example 23.3 Deciding Whether Geometric Isomers Are Possible

Sketch the geometric isomers of dichlorodiammineplatinum(II) and dichlorotetraammine-cobalt(III) ion.

Answer:

Alternate Example 23.4 Deciding Whether Optical Isomers Are Possible

Do any of the following have optical isomers? If so, describe the isomers. (a) trans-Co(NH3)2(en)2

3+ (b) cis-Co(NH3)2(en)23+

Answer: (b) only; the optical isomers are similar to those in Figure 23.16B in the text. (Note:Models are very useful to illustrate the optical isomers.)

Alternate Example 23.5 Describing the Bonding in an Octahedral Complex Ion(Crystal Field Theory)

Both Fe2+ and Co3+ have 3d6 configurations and form hexaammine complexes. However, theiron(II) complex is paramagnetic, and the cobalt(III) complex is diamagnetic. Using crystalfield theory, obtain the number of unpaired electrons in each complex; also note whether eachcomplex is high-spin or low-spin.

Answer: Fe(NH3)42+, four unpaired electrons, high-spin; Co(NH3)4

3+, no unpaired elec-trons, low-spin



Alternate Example 23.6 Describing the Bonding in a Four-Coordinate Complex Ion(Crystal Field Theory)

The qualitative analysis of nickel(II) ion is based on the reaction of the ion with the organiccompound dimethylglyoxime to form the chelate bis)dimethylglyoximato)nickel(II), a red-colored insoluble compound.

The lone pair on each nitrogen atom bonds to the nickel atom. The complex is diamagnetic.Describe the distribution of d electrons in the nickel(II) complex bis(dimethylglyoxi-mato)nickel(II). The complex has a square planar geometry.

Answer: The distribution of electrons is

Alternate Example 23.7 Predicting the Relative Wavelengths of Absorption ofComplex Ions

Earlier we described the square planar complex of nickel and dimethylglyoxime (AlternateExample 23.6), noting that it has a red color. The text describes the bonding in the square planarcomplex Ni(CN)4

2–; this complex ion has a yellow color. Imagine that the ligands in Ni(CN)42–

are exchanged for the dimethylglyoxime ligands. Is the color change from yellow to red in thedirection that you would expect? Note that the bonding of dimethylglyoxime to nickel isthrough lone pairs on nitrogen atoms (similar to bonding in ammonia).

Answer: The bonding changes from strong (with CN– ligands) to weak (with lone pairs onN atoms, as in NH3). The crystal field splitting becomes smaller, so the wavelength of thetransition (which gives rise to the color) becomes longer, which is in the direction expected.


118 PART VII


Alternate Example 24.2 Predicting cis–trans Isomers

Consider each of the alkenes given in Alternate Example 24.3. Can either of them exist ascis–trans isomers? If so, draw the structural formulas and label each as cis or trans.

Answers: (a) 3,4-dimethyl-3-hexene

(b) 3-ethyl-3-hexene; no geometric isomers

Alternate Example 24.3 Predicting the Major Product in an Addition Reaction

Water, HOH, can add across a double bond (in the presence of an acid). What would youexpect to be the major organic product when 1-pentene reacts with water in an additionreaction?

Answer:

Alternate Example 24.4 Writing the IUPAC Name of an Alkane Given the StructuralFormula

Give the IUPAC name for each of the following:

Answers: (a) 2,3-dimethyl-4-t-butylheptane (b) 2-methylhexane



Alternate Example 24.5 Writing the Structural Formula of an Alkane Given theIUPAC Name

Write the condensed structural formula of 2,3,5-trimethylhexane.

Answer:

Alternate Example 24.6 Writing the IUPAC Name of an Alkene Given the StructuralFormula

Name the alkenes whose structural formulas are given below.

Answers: (a) 3,4-dimethyl-3-hexene (b) 3-ethyl-3-hexene


120 PART VII

PART VIIIBrief Notes on Suggested Lecture Demonstrations

This part of the Instructor’s Resource Manual gives suggestions for lecture demonstrationscorrelated with appropriate sections of the text. These include demonstrations that aredescribed in the text, perhaps in a figure. Although the brief descriptions given here may besufficient (see “Caution” note below), the following references also give descriptions of lecturedemonstrations, and you may find them helpful:

“Test Demonstrations in Chemistry,” J. Chem. Educ., Easton, Pa., 1962.Hubert N. Alyea, “TOPS in General Chemistry,” J. Chem. Educ., Easton, Pa., 1967.Bassam Z. Shakhashiri, Chemical Demonstrations, Vol. 1 (1983), Vol. 2 (1985), Vol. 3 (1989), and

Vol. 4 (1992), University of Wisconsin Press, Madison.Lee R. Summerlin and James L. Ealy, Jr., Chemical Demonstrations, Vol. 1, American Chemical

Society, Washington, D.C., 1985.Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, Chemical Demonstrations, Vol. 2,

American Chemical Society, Washington, D.C., 1987.

CAUTION: These brief notes for suggested lecture demonstrations are intended for useby professional chemists who understand the reactions and procedures involved. Becausethese are brief notes, other references should be consulted for details in handling chemicalsand observing precautions to ensure the safety of students and demonstrator.


Section 1.3 Demonstration of the Conservation of Mass Using aMagnesium Flash Bulb

Weigh a flash bulb, flash it, and reweigh after cooling. Note constancy of mass. Use digitalbalance weighing to milligram accuracy. Have a student read the scale, putting the result onthe overhead or blackboard.

Section 1.4 Separation of a Mixture

Mix sodium chloride and silicon carbide or other water-insoluble material. Add water todissolve the NaCl and filter to remove silicon carbide.


Section 1.4 Separation of Ink by Paper Chromatography

Spot ink about 1–2 cm from the bottom of a strip of filter paper. Then dip the end in 50%ethanol to separate the dyes. When the dyes have separated, pass the sheet around forinspection.

Sections 1.6 and 1.8 Comparison of Metric and Common U.S. Units

Compare the following side by side: 1 qt water with 1 L water, 1 lb salt with 0.5 kg salt, 1 ozwater with 30 mL water, yardstick with meterstick.


Opening: Reaction of Sodium and Chlorine

Melt or ignite a small piece of sodium (about 1 g) in a deflagration spoon and lower into abottle of chlorine gas. Contrast the characteristics of the reactants and products.

Section 2.2 Cathode Rays

Show a demonstration-model cathode-ray tube (similar to the one shown in Figure 2.4 on textpage 46); show how the rays are bent by a magnetic field. (A similar demonstration of anelectric field requires that the electric plates be inside the tube; they cannot be outside, becauseof polarization of the glass.) Note that a paddle wheel is sometimes used to demonstrate themomentum of electrons, but the motion of the wheel is actually the result of convection fromresidual gas.

Section 2.6 Models of Molecules and NaCl Crystal

Compare various molecular models; discuss the structure of NaCl using a model of the crystal.Show a reaction in terms of molecular models.


Section 3.2 Mole of Substance

Show a mole of each of various substances (see Figure 3.2). Note the range of volumes. Later,when discussing gases, comment on the difference between gases, which have the same molarvolume, and other substances.


122 PART VIII


Section 4.1 Conductivity of Solutions

Test the electrical conductivity of various solutions using an apparatus like that described inthe text. The following are possibilities to test: distilled water, 0.1 M HCl, 0.1 M acetic acid,6 M acetic acid, glacial acetic acid, HCl in benzene, 0.1 M KClO3, molten KClO3, tap water.

A device that can be used to illustrate conductivity consists of two bulbs, one a neon glowbulb, the other a 7-watt night-light tungsten bulb wired through electrodes to the 110-volt line.Only the glow bulb will light for weak electrolytes, but both will light for strong electrolytes.

Section 4.3 Precipitation

Add a solution of NaI to a solution of Pb(NO3)2 to show the formation of a precipitate. Youcan use large volumes of solutions that will show up in a lecture room, or use small volumeswith a TOPS projector.

Section 4.5 Silver or Lead Tree (Displacement Reaction)

Bend copper wire in the shape of a tree with limbs and place in a solution of silver nitrate. Thewire grows crystals, giving the appearance of a tree with branches and leaves. Alternatively,use zinc wire placed in lead acetate solution.

Section 4.5 Oxidation of Sugar by Potassium Chlorate

Carefully mix equal volumes of potassium chlorate and sugar (do not grind!). Place in a pileon a square of asbestos and add a few drops of concentrated sulfuric acid. The mixture burstsinto violent flame.

Section 4.5 Oxidation of Glycerol by Potassium Permanganate

Place a pile of potassium permanganate crystals on a square of asbestos. Pour a small quantityof glycerol over the crystals. Within a few seconds, the mixture ignites.

Section 4.5 Hydrogen Peroxide—Oxidizing and Reducing Properties

Add 3% H2O2 to dilute KI to which some sulfuric acid has been added. Iodide ion is oxidizedin acid solution by hydrogen peroxide to iodine. The iodine forms a blue complex with starch.

Add 3% H2O2 to dilute KMnO4 to which some sulfuric acid has been added. The purplecolor of permanganate ion fades as the ion is reduced by hydrogen peroxide in acid solution.

Section 4.7 Ammonia Fountain

Fill a dry flask from a cylinder of ammonia, holding the flask upside-down when filling sothat the ammonia will displace air. Press the dropper to push some water into the flaskcontaining the ammonia to start the fountain. The demonstration is interesting and can beused to introduce some properties of an important reagent.


Brief Notes on Suggested Lecture Demonstrations 123


Section 5.1 Effect of Reducing the Pressure on a Column of Mercury

Insert a closed-end barometer tube into one hole of a two-holed stopper and a glass tubeconnected to rubber tubing into the other hole. Fill the barometer tube with mercury and placethe stopper and tube ends into a wide-mouth bottle containing mercury. Attach the rubbertube to an aspirator or vacuum pump. Comment on the change of mercury column height aspressure is reduced.

Section 5.2 Cartesian Diver

Fill a large bottle with water and place an inverted vial containing air into the water in theneck of the bottle. Fit a stopper into the neck. When you press the stopper, the air in the vialis compressed (water is not) and the vial sinks. Contrast compressibility of gases with therelative compressibility of liquids. See TOPS manual, page 74, for a description of a TOPSdemonstration.

Section 5.2 Effect of Temperature Change on Gas Volume

Fill a balloon with helium. Submerge the balloon in liquid nitrogen until it contracts to a smallvolume. When the balloon is thrown toward the audience, it will expand, hover for a periodof time, then rise. Pour the excess liquid nitrogen on the floor for a dramatic ending.

Section 5.2 Charles’s Law

Fit a flask with a one-hole stopper through which a short length of glass tubing passes. Heatthe flask over a water bath (note the bath temperature and the air temperature). Remove theflask from the water bath and quickly immerse upside-down in a large beaker of water atroom temperature (add food coloring to the water for visibility). As the air in the flask cools,water enters the flask. Adjust the pressure in the flask to atmospheric pressure by making thelevels of water inside and outside the flask equal. Remove the flask from the beaker withoutallowing more water to enter. Measure the volume of water in the flask, then measure thevolume of the flask. Confirm Charles’s law from the change in volume of air in the flask.

Section 5.2 Molar Volume of a Gas

Display a cube that has a volume of 22.4 L.


Section 6.2 An Endothermic Reaction

Place about 0.1 mol of barium hydroxide octahydrate crystals in a 250-mL Erlenmeyer flask.Also have ready 0.2 mol of ammonium nitrate (or ammonium thiocyanate) powder in a test


124 PART VIII

tube. Add water to a small, smooth board to form a puddle. Then add the ammonium salt tothe flask containing the barium hydroxide and stopper tightly to contain the ammonia fumes.A slurry will form soon after the solids are mixed. Place the flask in the puddle of water onthe board. Talk for about two minutes, giving the flask time to freeze to the board. Now movethe board from side to side to show that a solution has been formed; then carefully turn theboard upside-down to show that the flask has frozen to it. See Figure 6.1.

Section 6.2 An Exothermic Reaction (Rusting of Iron)

Wash steel wool in dilute hydrochloric acid or acetic acid. Rinse well in water. Wrap the steelwool around the bulb of a thermometer or around a projection thermocouple. The temperaturerises as the iron rusts.


Section 7.3 Atomic Line Spectra

Provide students with diffraction gratings in slide mounts (available from Edmund ScientificCompany). Have the students observe various atomic line spectra from discharge tubes (linesare to the right, as you look at the grating), as well as from a continuous source, such as atungsten bulb.


Section 8.7 Reactivity of Some Metallic Elements

Clean a strip of magnesium ribbon and a piece of aluminum wire with steel wool. Place eachin some water containing several drops of phenolphthalein. Also place small pieces of calcium,then sodium, and finally potassium into water containing phenolphthalein. Note the relativerates of reaction.


Section 9.2 Color of Ions in Aqueous Solution

Place aqueous solutions of ions in flat dishes, as in Figure 9.6. Project with overhead.




Section 10.1 Arrangements of Electron Pairs

Tie two to six balloons of about the same size to show the different arrangements. See Figure10.3.

Section 10.1 Molecular Shapes

Show lecture-size models of molecules with various geometries.

Section 10.6 Paramagnetism of Oxygen

To prepare liquid oxygen, place a tall test tube in a Dewar flask containing liquid nitrogen.Pass oxygen gas through a glass tube into the test tube. Blue liquid oxygen will form in thetest tube.

Pour liquid nitrogen over the poles of a strong magnet to cool it. Note that nitrogen doesnot stick to the magnet poles. Now pour liquid oxygen over the poles; note that the oxygendoes stick to the magnet poles. See Figure 10.29.


Section 11.2 Sublimation of Iodine

Place iodine crystals in a beaker and cover with watch glass or evaporating dish containingice. Heat with a low flame. Iodine will sublime and collect on the underside of the watch glassor dish. See Figure 11.3.


Section 12.2 Supersaturated Solutions

Fill a flask three-fourths full with sodium thiosulfate pentahydrate or sodium acetate trihy-drate crystals. Add just enough water to dampen the crystals. Cover and heat slowly until asolution forms without any solid (you may need to add some water). Allow it to cool slowlyto give the supersaturated solution. If you have difficulty keeping the solution from crystal-lizing prematurely, add a small quantity of water and reheat to give a solution. When ready,seed with a crystal of the solute. Crystallization is dramatic (see Figure 12.4). Have studentsnote how warm the resulting solution is. The solid mixture can be reheated to repeat thedemonstration.


126 PART VIII

Section 12.3 Effect of Temperature Change on Solubility

Potassium nitrate, ammonium nitrate, or boric acid may be used to demonstrate an increasein solubility with increasing temperature. Add more than the required amount of salt tosaturate the solution at room temperature. Then show that heating will dissolve more salt.

Ceric sulfate, calcium hydroxide, or calcium acetate may be used to demonstrate a decreasein solubility with temperature. Add more than the required amount to saturate a hot solution.Then show that more dissolves on cooling.

Section 12.7 Osmosis

Cover the wide end of a thistle-tube funnel with a semipermeable membrane. Add a sugarsyrup (to which a food dye has been added for visibility) to the tube end of the funnel. Immersethe wide end in distilled water. Note rise in height of liquid after a half-hour. See Figure 12.24.

Section 12.9 Colloids

The following are some colloids that can be used to demonstrate filterability and the Tyndalleffect:

1. Gelatin: Prepare a 2% solution by dissolving gelatin in boiling water.2. Colloidal sulfur: Saturate a half liter of cold water with SO2, then pass H2S through the

solution for several minutes. Colloidal sulfur forms.3. Colloidal arsenic sulfide: Add about 1 g As2O3 to a liter of water and bring to a boil.

Pass H2S into the hot solution for several minutes. Colloidal As2S3 forms.

Compare a true solution with a colloidal solution. Show that both pass through a filter, butonly the colloid gives the Tyndall effect. A laser pointer is a convenient light source for theTyndall effect.


Section 14.3 Iodine Clock Reaction

The essential reaction involves the oxidation of iodide ion by peroxydisulfate ion:

2I− + S2O82− → I2 + 2SO4

2−

This reaction is slow. The I2 reacts quickly with thiosulfate ion in the solution. When thiosulfateion is used up, the concentration of I2 increases and gives a blue color with starch indicator.

Prepare a solution containing 200 mL 0.2 M KI, 100 mL 0.005 M Na2S2O3, and 1 mL 1% starchindicator. Add to 100 mL (NH4)2S2O8. The concentrations may be varied to change the timebefore the blue color appears.




Section 15.7 Le Chatelier’s Principle: Iron Thiocyanate Complex

Mix small quantities of iron(III) nitrate and potassium thiocyanate in about a liter of water togive a dilute, orange-yellow solution of Fe(CNS)2+. Apportion the solution among threebeakers. Add Fe3+ to one beaker and CNS– to another to shift the equilibrium. Note the deepred color obtained in both cases, which results from the formation of more Fe(CNS)2+, aspredicted by Le Chatelier’s principle. See also Figure 15.8 for another demonstration.

Section 15.8 Effect of Changing the Temperature: NO2–N2O4 Equilibrium

Use three sealed tubes containing a small amount of nitrogen dioxide gas (tubes are commer-cially available). Place one in an ice bath and another in boiling water, leaving the third oneat room temperature. Colorless N2O4 is stabler at lower temperatures and reddish-brown NO2at higher temperatures. Note that conversion of the tetroxide to the dioxide is an endothermicprocess.

Section 15.9 Ostwald Process for Preparing Nitric Acid (Platinum Catalysis)

A small coil of platinum wire or platinum foil is affixed to a glass rod that passes through athree-hole rubber stopper. A glass tube also passes through the stopper and opens at its lowerend near the platinum. The other end is connected to a source of oxygen. When the rubber-stopper assembly is placed in a flask containing concentrated ammonia, the platinum wireshould be about a centimeter above the solution. Before putting the rubber stopper in the flask,heat the platinum. When the stopper assembly is placed in the flask, the platinum will continueto glow for several minutes from the exothermic reaction of NH3 and O2 to produce NO. Notethat the use of a copper wire instead of platinum yields N2 instead of NO.


Section 16.4 Relative Acid and Base Strengths

See Bassam Z. Shakhashiri, Chemical Demonstrations, Vol. 3 (Madison: University of Wiscon-sin Press, 1989), no. 8.25, pp. 158–161.

Section 16.8 Acid–Base Indicators

Add different indicators to solutions previously made up with various pH values. Colorchanges can be shown as acid and base are added. See Figures 16.10 and 16.11 in the text.


128 PART VIII


Section 17.4 Hydrolysis of Salts

Measure the pH of a number of salt solutions, using either a pH meter or acid–base indicators.Have a student read the pH meter.

Section 17.6 Effect of Adding Acid or Base to a Buffer

Prepare a buffer solution from equal volumes of 1 M acetic acid and 1 M sodium acetate (pHof buffer is 4.7). Add strong base (or strong acid) from a buret to a certain volume of waterand watch the change of pH. Repeat, but with the same volume of buffer. Compare the results.


Section 18.5 Amphoteric Hydroxides

Place a solution of zinc chloride in one beaker and a solution of sodium hydroxide in another.Pour some of the sodium hydroxide solution into the zinc chloride solution and note theformation of a white precipitate of zinc hydroxide. Then add more sodium hydroxide untilthe precipitate dissolves. See Figure 18.8.

Section 18.6 Solubility of Silver Salts and Complex-Ion Formation

Prepare 0.1 M solutions of the following compounds: AgNO3, Na2CO3, NaOH, NaCl, NaBr,Na2S2O3, NaI, NaCN, and Na2S. Also have available 6 M NH3. To the AgNO3 solution, addthe following solutions in order: Na2CO3 (gives pale yellow precipitate of Ag2CO3), NaOH(gives brown precipitate of Ag2O), NaCl (gives white precipitate of AgCl), NH3 (precipitatedissolves to give silver ammine complex ion), NaBr (gives pale yellow precipitate of AgBr),Na2S2O3 (dissolves precipitate to form thiosulfate complex ion), NaI (forms yellow precipitateof AgI), NaCN (precipitate dissolves to give dicyanoargentate ion), and Na2S (forms blackprecipitate of Ag2S).


Section 20.2 Voltaic Cells

Place a metal electrode in a beaker containing a solution of the metal ion. Couple this half-cellwith a similar half-cell of another metal using a salt bridge consisting of filter paper soakedin saturated KCl solution. Measure the emf of the voltaic cell with a voltmeter.



Section 20.8 Lead Storage Cell

Hang two strips of lead foil over the edge of a beaker filled with 0.1 M sulfuric acid. Attachclips to the lead electrodes and attach the wire leads to a 7.5-V battery or power supply. Notethe evolution of gas during the charging of the cell. Now attach the lead cell to a bell or 1.5-Vbulb to demonstrate that the cell has been charged. (Lead sulfate first forms on the lead strips;during charging at the positive plate, PbSO4 + 2H2O forms PbO2 + 4H+ + SO4

2−; at the negativeplate, PbSO4 gives Pb + SO4

2−.)

Section 20.10 Electrolysis of Water

Fill a Hoffman apparatus with 0.1 M sulfuric acid and operate at 22 V. Oxygen dissolves morereadily than hydrogen, so the volume ratio will not be exactly 2 to 1 unless that apparatus hasbeen operated previously to saturate the solution with the gases.

Section 20.11 Electrolysis of Copper Sulfate

Rinse several square centimeters of copper gauze (to be used as the cathode) in distilled water,dry, and weigh to nearest 0.01 g. Use a strip of copper as the anode. Immerse electrodes in asolution made from 1 L water, 200 g CuSO4, and 80 g concentrated H2SO4. Connect in seriesto resistance and ammeter; electrolyze for 30 min at 0.25 A. Rinse cathode and weigh. Comparewith calculated value.


Section 21.3 Cloud Chamber

Construct a cloud chamber from a screw-top jar. Glue pieces of felt on the bottom of the jarand the top of the lid. Saturate both pieces of felt with methanol and screw on the lid. Invertthe jar with the lid on a block of dry ice. Wait about 15 min. Shine a spotlight through the sideof the jar. Cloud tracks from cosmic rays or ambient radioactivity will be seen. A gamma-raysource will produce many tracks.

Section 21.3 Detection and Absorption of Beta Rays

Use uranyl nitrate as a beta-ray source and detect with a Geiger counter. Vary the position ofthe source and note decrease in counts with distance (inverse square law). Sheets of paper andmetal may be used to test for absorption of beta rays.


130 PART VIII

CHAPTER 22 Chemistry of the Main-Group Elements

Section 22.2 Properties of Sodium Metal

Slice off a piece of sodium metal to demonstrate the softness of the metal and its silveryappearance (see Figure 22.2). Show malleability by flattening the piece with the side of theknife. Put a piece of the metal in water to demonstrate its chemical reactivity. (It is advisableto use a plastic shield to protect the audience.)

Section 22.2 Solvay Process

Saturate concentrated ammonia solution with sodium chloride contained in a beaker. Addpieces of dry ice. Sodium hydrogen carbonate will precipitate. See Figure 22.8.

Section 22.3 Burning of Magnesium in Air, H2O, and CO2

Magnesium metal burns in air to give a mixture of the oxide and the nitride. In water vapor(steam), magnesium burns to give MgO and H2. In CO2, the metal gives MgO and C.Demonstrate the burning in air; then show that the metal continues to burn when placed inthe vapor over boiling water. Show that a match flame is extinguished when inserted in abeaker containing dry ice (or pour carbon dioxide gas over the flame), but magnesium willcontinue to burn.

Section 22.4 Reaction of Aluminum with Acid and Base

Demonstrate the reactions of aluminum metal with hydrochloric acid and with sodiumhydroxide solution. Note the evolution of gas (hydrogen) in both cases. You can place beakeron overhead to project.

Section 22.7 Burning of Phosphorus

Prepare a solution containing 1 g of white phosphorus in about 10 mL of carbon disulfide.(Use the solution carefully and do not store!) Place several drops of the solution on some paperon a square of asbestos. The paper will ignite as soon as the solvent evaporates.

Section 22.8 Preparation of Oxygen

Add water to sodium peroxide in a test tube or flask. Test the evolution of oxygen with asmoldering wood splint. As an alternative preparation, heat a mixture of potassium chloratewith a small amount of manganese dioxide. (CAUTION: Avoid organic contaminants!)

Section 22.8 Dehydrating Action of Concentrated Sulfuric Acid

Place a pile of sugar in a beaker, moisten slightly with water, and pour concentrated sulfuricacid on to the pile. A column of porous carbon forms. (A hood is advisable for this demon-stration. Otherwise, you might consider showing a video; see Video series A and C.)




Section 23.1 Oxidation States of Vanadium

Prepare a solution that is 1 M sodium vanadate (or vanadyl sulfate) in 1 M sulfuric acid. Pourthrough a Jones reductor to give vanadium(II) sulfate solution. Fill lower half of a cylinderwith this solution. Then carefully pour in 0.01 M potassium permanganate and let stand.Layers illustrating four oxidation states of vanadium will form: violet vanadous (+2), emeraldgreen vanadic (+3), deep blue vanadyl (+4), and yellow vanadate (+5). See Figure 23.2.


Section 24.3 Preparation of Acetylene

Drop several pieces of calcium carbide in a beaker containing water. Ignite with a match onthe end of a long stick. (Use a clear explosion shield!) The acetylene ignites with a pop. SeeFigure 24.10 for another demonstration.

Section 24.6 Silver-Mirror Test for Aldehydes

Dissolve about 10 g of silver nitrate in 0.5 L of water. Add several drops of sodium hydroxidesolution. Then add concentrated ammonia until the precipitate that first forms is just dis-solved. Do not add excess ammonia. Pour the solution into a very clean flask and add someformaldehyde solution or acetaldehyde. Note the formation of a silver mirror. Dilute thesolutions and dispose of them promptly after use.

CHAPTER 25 Polymer Materials: Synthetic and Biological

Section 25.1 Preparation of Nylon

Carefully pour a solution containing 60 g of hexamethylene diamine per liter of water over asolution containing 60 mL of sebacoyl chloride (or adipoyl chloride) per liter of hexane to formtwo layers. A film of nylon forms at the interface of the layers. Grab this film with forceps andpull upward to form a filament. Add a dye to one solution to increase visibility. Wash filamentwell with water to remove hydrochloric acid if it is to be passed around for inspection. SeeFigure 25.4.


132 PART VIII

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