Solution Manual For Chemistry: Principles and Reactions, 8th Edition

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|||||||ContentsPrefacevLecture ScheduleviiChapter 1Matter and Measurements1Chapter 2Atoms, Molecules, and Ions11Chapter 3Mass Relations in Chemistry; Stoichiometry21Chapter 4Reactions in Aqueous Solution37Chapter 5Gases49Chapter 6Electronic Structure and the Periodic Table61Chapter 7Covalent Bonding71Chapter 8Thermochemistry83Chapter 9Liquids and Solids95Chapter 10Solutions105Chapter 11Rate of Reaction119Chapter 12Gaseous Chemical Equilibrium133Chapter 13Acids and Bases147Chapter 14Equilibria in Acid-Base Solutions161Chapter 15Complex Ions and Precipitation Equilibra177Chapter 16Spontaneity of Reaction189Chapter 17Electrochemistry201Chapter 18Nuclear Reactions219Chapter 19Complex Ions229Chapter 20Chemistry of the Metals237Chapter 21Chemistry of the Nonmetals245Chapter 22Organic Chemistrry255Chapter 23Organic Polymers: Natural and Synthetic265iii

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|||||||PrefaceThis manual starts off with a section entitled “Lecture Schedule,” which you may find helpful in adapting thetext to your class schedule. Beyond that, the manual is organized by text chapters. For each chapter, weinclude three different features:1.“Lecture Notes,” which suggest the amount of time that we devote to each chapter and the topics weemphasize. Included are detailed lecture outlines (from our own lectures) that may serve as a guidefor your lectures. At a minimum, they indicate how we cover topics and how successive topics can beintegrated.2.A list of demonstrations illustrating topics in the chapter. These are taken from three sources:•The manualTested Demonstrations in Chemistry(1994), Volumes I and II, compiled and edited byby George Gilbert by arrangement with the Journal of Chemical Education. These are coded as“GILB” with the experiment number (e.g., M 12)•Chemical Demonstrations(1983-1992), Volumes 1-4, published by Bassam Shakashiri with manycollaborators and contributors.These are listed as “SHAK” followed by the volume and pagereference.•Demonstrations described in theJournal of Chemical Educationwith the journal reference.3.Answers and detailed solutions to:•The summary problem at the end of each chapter•Odd numbered text problems•Challenge problems at the end of each problem setNote that Appendix 6 has answers to all the even-numbered problems. Detailed solutions to many ofthese problems are in theStudent Solutions Manualavailable from Cengage Learning, Brooks/Cole.v

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|||||||Lecture ScheduleUnlike other general chemistry texts, ours can be covered in its entirety in a one-year course. A reasonableschedule appears below. Further comments on the time you should devote to each chapter are in the bodyof this manual. The underlying assumption is that you are teaching 14-week semesters with two 50-minutelectures per week.If three class periods are devoted to examinations each semester, that leaves 25 forcovering material. On that basis, you are able to complete Chapter 10 (Solutions) in the first semester. Theseond semester will then start with Chapter 11 (Rate of Reaction).FIRST SEMESTER SCHEDULEWeekLectureTopic11Chapter 1 (Matter and Measurements)2Chapter 123Chapter 2 (Atoms, Molecules, and Ions)4Chapter 235Chapter 3 (Mass Relations in Chemistry; Stoichiometry)6Chapter 347Chapter 38EXAM I59Chapter 4 (Reactions in Aqueous Solution)10Chapter 4611Chapter 412Chapter 5 (Gases)713Chapter 514Chapter 6 (Electronic Structure and the Periodic Table)815Chapter 616Chapter 6917EXAM II18Chapter 7 (Covalent Bonding)1019Chapter 720Chapter 71121Chapter 8 (Themochemistry)22Chapter 81223Chapter 9 (Liquids and Solids)24Chapter 91325Chapter 926EXAM III1427Chapter 10 (Solutions)28Chapter 10vii

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viiiLecture ScheduleSECOND SEMESTER SCHEDULEWeekLectureTopic11Chapter 11 (Rate of Reaction)2Chapter 1123Chapter 114Chapter 12 (Gaseous Chemical Equilibrium)35Chapter 126Chapter 13 (Acids and Bases)47Chapter 138Chapter 1359EXAM I10Chapter 14 (Equilibria in Acid-Base Solutions)611Chapter 1412Chapter 15 (Complex Ions)713Chapter 1514Chapter 16 (Precipitation Equilibria)815Chapter 17 (Spontaneity of Reaction)16Chapter 17917EXAM II18Chapter 18 (Electrochemistry)1019Chapter 1820Chapter 181121Chapter 19 (Nuclear Chemistry)22Chapter 20 (Chemistry of the Metals)1223Chapter 2024EXAM III1325Chapter 21 (Chemistry of the Nonmetals)26Chapter 211427Chapter 22 (Organic Chemistry)28Chapter 23 (Organic Polymers: Natural and Synthetic)

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Lecture ScheduleixIf you want to use lecture time for review, for going over assigned problems, or for doing a large number ofdemonstrations, you will have trouble keeping up with this schedule. As you’ve almost certainly learned bynow, the solution to this problem is not to talk faster. Judicious deletions work better. It’s been said, andwisely, that the secret of giving a good lecture is knowing what to leave out. Possible candidates include:•Introductory material on matter in Chapter 1 and atomic theory in Chapter 2. The chances are yourstudents have been exposed to this material more than once in high school and understood it reasonablywell the first time.•Boyle’s and Charles’s laws in Chapter 5. We start the chapter by writing the ideal gas law and go onfrom there.•The First Law discussion in Chapter 8. Quite frankly, this has very little to do with chemistry. Studentswill not be irreparably damaged if they are unaware of the distinction betweenHandE.•The discussion of colligative properties in Chapter 10 could be shortened. Raoult’s law could easily beomitted.•Reaction mechanisms in Chapter 11. Students have a lot of trouble with this. We are not sure it is worththe effort.•Polyprotic acids in Chapter 13.•The Second Law discussion in Chapter 17.Beyond these selective omissions, some instructors may want to delete one or another of the descriptivechapters at the end of the text (Chapters 20–22). If, in that way, you can squeeze out a couple of lectures,they can well be spent on Chapter 12 (three lectures instead of two) and Chapter 19 (two lectures insteadof one).Textbook authors sometimes tell you that chapters can be covered in almost any order, depending onyour preference. This isn’t really true for this textbook, or any other with structural integrity. It can be done,but only with very careful additions and deletions of material.Suppose, for example, you want to coverPrecipitation Equilibria (Chapter 16) immediately after Acid-Base Equilibria (Chapter 14).Keep in mindthat an understanding of formation constants (Complex Ions, Chapter 15) is assumed when methods ofdissolving precipitates are considered in Section 16.2 of Chapter 16.

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1||||||||MATTER ANDMEASUREMENTSLECTURE NOTESThis material ordinarily requires two lectures (100 minutes), allowing for a 10–15 minute introduction to thecourse in the first lecture. If you’re in a hurry, this can be cut to 112lectures by discussing only quantitativematerial (significant figures, unit conversions, density, solubility).A few points to keep in mind:•Virtually all of your students will be familiar with the metric system and prefixes.It may be worthdiscussing the rationale for SI, but you don’t have to dwell on it.•Students readily learn the rules of significant figures, but typically ignore them after Chapter 1. It mayhelp to emphasize that these are common-sense (albeit, approximate) rules for estimating experimentalerror.•Many (typically, the weaker) students resist using conversion factors, preferring instead a rote method.It may be useful to point out that conversion factors will be a recurring tool throughout the text, so arewell worth learning at this point.•Students often have trouble with solubility calculations. The approach in the text involves conversions(Example 1.8). The solubility is considered to be a conversion factor relating grams of solute to gramsof solvent.Lecture 1I. Types of SubstancesA.ElementsCannot be broken down into simpler substances.Examples:nitrogen, lead, sodium, arsenic.Symbols: N, Pb, Na, As.B.CompoundsContain two or more elements with fixed mass percents. Glucose: 40.00% C, 6.71% H, 53.29%O. Sodium chloride: 39.34% Na, 60.66% Cl.C.MixturesHomogeneous (solutions) vs. heterogeneous. Separation by filtration, distillation.II. Measured QuantitiesA.LengthBase unit is the meter. 1 km = 103m; 1 cm = 102m; 1 mm = 103m; 1 nm = 109m. Dimensionsof very tiny particles will be expressed in nanometers.1

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2Chapter 1B.Volume1 L = 103mL = 103cm3= 103m3. Buret, pipet, volumetric flask.C.Mass1 kg = 103g; 1 mg = 103g. Two different kinds of balances will be used in the lab. An analyticalbalance (±0.001 g) should be used only for accurate, quantitative work.D.TemperaturetF= 1.8 tC+ 32;TK= tC+ 273.15.Convert 68F toC and K:tC= (6832)/1.8 = 20CTK= 293Lecture 2III. Experimental Error; Significant FiguresSuppose an object is weighed on a crude balance to±0.1 g and the mass is found to be 23.6 g.This quantity contains three significant figures, that is, three experimentally significant digits. With ananalytical balance, the mass might be 23.582 g (five significant figures).A.Counting significant figures1.Volume of liquid = 24.0 mL; three significant figures.Zeroes at the end of the measuredquantity are significant when they follow nonzero digits.2.Volume = 0.0240 L; three significant figures (note that 0.0240 L = 24.0 mL). Zeroes at thebeginning of a measured quantity are not significant when they precede nonzero digits.B.Multiplication and divisionKeep only as many significant figures as there are in the least precise quantity. Density of a pieceof metal weighing 36.123 g with a volume of 13.4 mL = ?density = 36.123 g13.4 mL = 2.70 g/mLC.Addition and subtractionKeep only as many digits after the decimal point as there are in the least precise quantity.Add1.223 g of sugar to 154.5 g of coffee:total mass = 1.2 g + 154.5 g = 155.7 gNote that the rule for addition and subtraction does not apply to significant figures. The number ofsignificant figures may well decrease after subtraction.mass beaker + sample = 52.169 g(five significant figures)mass empty beaker = 52.120 g(five significant figures)mass sample =0.049 g(two significant figures)D.Exact numbersOneliter means 1.000000... L.

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Matter and Measurements3IV. Conversion FactorsA.One-step conversionA rainbow trout is measured to be 16.2 inches long. What is its length in centimeters?length in cm = 16.2 in⇥2.54 cm1 in= 41.1 cmNote the cancellation of units.To convert from centimeters to inches, use the conversion factor1 in = 2.54 cm. (Here, there areexactly2.54 cm in one inch.)B.Multiple conversion factorsA thrown baseball has speed 89.6 miles per hour. What is its speed in meters per second?1 mile = 1.609 km = 1.609⇥103m;1 h = 3600 sspeed =✓89.6 mileh◆⇥⇣1.609⇥103mmile⌘⇥✓1 h3600 s◆= 40.0 m/sV. Properties of SubstancesDistinguish between intensive and extensive, and between chemical and physical.A.DensityAn empty flask weighs 22.138 g. Pipet 5.00 mL of octane into the flask, producing a total mass of25.598 g. What volume is occupied by ten grams of octane?d= 3.460 g/5.00 mL = 0.692 g/mLV= 10.00 g⇥1 mL0.692 g = 14.5 mLNote that for density calculations, 1 mL = 1 cm3.B.SolubilityThis is often expressed as grams of solute per 100 g of solvent.Solubility of sugar at 20C = 210 g sugar/100 g water.A solution containing 210 g sugar/100 g water is saturated.A solution containing less than 210 g sugar/100 g water is unsaturated.A solution containing more than 210 g sugar/100 g water is supersaturated.1.How much water is required to dissolve 52 g of sugar at 20C?52 g sugar⇥100 g water210 g sugar = 25 g water

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4Chapter 12.A solution at 20C contains 25 g sugar and 125 g water. Is it unsaturated, saturated or super-saturated?mass sugar/100 g water =25 g sugar125 g water⇥100 g water = 20 g sugar(unsaturated)DEMONSTRATIONS1. Scientific method: GILB H 292. Decomposition of mercury(II) oxide: GILB A 83. Separation of a mixture: GILB A 144. Reaction of sodium with chlorine: GILB A 24, A 25; SHAK 1 61; J. Chem. Educ. 73 539 (1996)5. Chromatography: GILB Q 3, Q 136. Significant figures: J. Chem. Educ. 69 497 (1992)7. Density of liquids: GILB C 13; SHAK 3 2298. Supersaturation: GILB F 11; SHAK 1 27SUMMARY PROBLEM(a) K, Mn, O(b) density, melting point, solubility, color(c) mass = 2.703gcm3⇥48.7cm3= 132 g(d) 2.703gcm3⇥1 lb454 g⇥(2.54)3cm313in3⇥(12)3in31 ft= 169 lb/ft3(e)F = 95 (C) + 32 =95 (2.40⇥102) + 32 = 464F(2.40⇥102) + 273 = 513 K(f) 38.5 g H2O⇥6.38 g KMnO4100 g H2O= 2.46 g KMnO4(g) At 60C: 65.0 g H2O⇥25 g KMnO4100 g H2O= 16 g KMnO4The solution is unsaturated.At 20C: 65.0 g H2O⇥6.38 g KMnO4100 g H2O= 4.15 g KMnO4can be dissolved.The solution is supersaturated.

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Matter and Measurements5(h) 55.0 g H2O⇥25 g KMnO4100 g H2O= 14 g KMnO4can be dissolved at 60C.Yes, all the KMnO4added will dissolve.55.0 g H2O⇥6.38 g KMnO4100 g H2O= 3.51 g KMnO4can be dissolved at 20C.No, not all the KMnO4will dissolve.10.0 g - 3.51 g = 6.5 g KMnO4will remain undissolved.PROBLEMS1.(a) mixture(b) element(c) mixture(d) compound3.(a) solution(b) solution(c) heterogeneous mixture5.(a) distillation(b) filtration(c) gas chromatography7.(a) Ti(b) P(c) K(d) Mg9.(a) mercury(b) silicon(c) sodium(d) iodine11. (a) balance(b) thermometer(c) graduated cylinder13. tF= 1.8(52) + 32= 126F;tK= 52 + 273.15 = 325 K15. tC= (85.032)⇥59 = 29.4C ; solid17. (a) 3(b) ambiguous(c) 4(d) exact(e) 519. (a) 7.49 g(b) 298.69 cm(c) 1⇥101lb(d) 12.0 oz21. (a) 1.325⇥102cm(b) 8.83⇥104km(c) 6.432⇥109nm23. (c)25. 10,000: ambiguous1.71⇥105ft2: 3$22.00: exact20%: ambiguous27. (a) 80.0(b) 0.7615(c) 14.712(d) 0.03(e) 1.5⇥1022

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6Chapter 129. 4⇡(4.30 cm)33= 333 cm3;4⇡(4.33 cm)33= 3.40⇥102cm3; 7 cm331. (a) 303 m = 0.303 km < 303⇥103km(b) 500 g = 0.500 kg(c) 1.50 cm3= 1.50⇥1021nm3>1.50⇥103nm333. (a) 22.3 mL⇥1 L103mL = 2.23⇥102L(b) 22.3 cm3⇥1 in3(2.54 cm)3= 1.36 in3(c) 22.3 mL⇥1 L103mL⇥1.057 qt1 L= 0.0236 qt35. (a) 19.2 hands⇥13ft1 hand = 6.40 ft(b) 17.8 hands⇥13ft1 hand⇥12 in1 ft⇥1 m39.37 in = 1.81 m(c) 20.5 hands⇥13ft1 hand+ 3.0 ft = 9.8 ft37. 2.0 acre⇥4.356⇥104ft21 acre⇥(12)2in21 ft2⇥1 m2(39.37)2in2⇥1 hectare104m2= 0.81 hectare39. 5.0 mi1 mile⇥0.25 min1 Eng lap = 1.2 min0.50 km⇥1 mi1.609 km⇥5.0 min1 mi= 1.6 min41. (a) 3.0 qt plasma⇥1 L1.057 qt⇥1000 mL1 L⇥0.080 mL alcohol100 mL plasma= 2.3 mL(b) 3.0 qt plasma⇥1 L1.057 qt⇥1000 mL1 L⇥0.10 mL alcohol100 mL plasma = 2.8 mL(c) 2.8 mL2.3 mL = 0.5 mL43.235 kJ250 mL⇥103J1 kJ⇥1 cal4.18 J⇥1 kcal103cal⇥1000 mL1.057 qt⇥1 qt4 cups = 53.2 kcal/cup45.252 g0.750⇥225 mL = 1.49 g/mL

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Matter and Measurements747. Vmethanol= 43.7 g⇥1 mL0.791 g = 55.2 mL ,Vslug= 59.7 mL55.2 mL = 4.5 mLTherefore, dslug= 25.17 g4.5 mL = 5.6 g/mL49. (8.0⇥7.0⇥0.75) ft3⇥(12)3in31 ft3⇥(2.54)3cm31 in3⇥1.00 g1 cm3⇥1 kg1000 g = 1.2⇥103kg51. Volume of air = volume of roomV = 55 kg O2⇥1000 g1 kg⇥1 L1.31 g⇥100 L21 L= 2.0⇥105L53. At 30C: maximum amount of MgSO4that can be dissolved = 25.0 g H2O⇥38.9 g MgSO4100 g H2O= 9.72 gThe solution is unsaturated.(9.50 + 1.00 g) - 9.72 = 0.78 g will precipitate out55. (a) 46 g H2O⇥16 g NaHCO3100 g H2O= 7.4 g NaHCO3can be dissolved9.2 g in the mixture > 7.4 g, thus the solution is not homogeneous.9.2 g – 7.4 g = 1.8 g NaHCO3are undissolved.(b) 9.2 g NaHCO3⇥100 g H2O9.6 g NaHCO3= 96 g H2O needed to dissolve96 g46 g = 5.0⇥101g H2O needs to be added.57. 57.0 g25.0 g = 32.0 g of Pb(NO3)2dissolves in 64.0 g H2O at 10C. Solubility is32.0 g Pb(NO3)264.0 g H2O= 1.00 g PbNO3)22.00 g H2O= 50.0 g Pb(NO3)2100.0 g H2O59. (a) physical(b) physical(c) physical(d) chemical61. V = 35 ft⇥43 ft⇥28 in⇥1 ft12 in= 3.5⇥103ft33.5⇥103ft3⇥(12 in)3(1 ft)3⇥(2.54 cm)3(1 in)3= 9.9⇥107cm3mass = 9.9⇥107cm3⇥(0.35 g)(1 cm)3⇥1 lb453.6 g = 7.7⇥104lbs

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8Chapter 163. 108 carats⇥0.200 g1 carat⇥1 lb454 g = 0.0476 lb0.0476 lb⇥1 cm33.51 g⇥454 g1 lb⇥1 in3(2.54)3cm3= 0.376 in365. 153.2 g⇥1 cm34.55 g = 33.7 cm3= V;33.7 =⇡r2(7.75);r = 1.18 cm;d = 2.35 cm67. (a) Chemical properties show the behavior of the species in a reaction; physical propertiesare intrinsic qualities.(b) Distillation vaporizes the liquid; filtration removes the solid.(c) The solute is a component of the solution.69. The bottom layer is Hg; the middle layer is Pb; the top layer is ethyl alcohol.71. (a)⇡115 g; supersaturated(b)⇡30 g; unsaturated(c) Dissolve 30 g of compound in 100 g H2O.73. (a) See Figure in the answers to the problems in Appendix 5.(b)yx =100J0J78C(117.0C) = 0.512(c) 60J(d)J = 0.51(C) + 6074. 31.5 gal⇥4 qt1 gal⇥1 L1.057 qt⇥103m31 L⇥1 km3109m3= 1.2⇥1010km3area = 1.2⇥1010km3100 nm⇥1⇥1012nm1 km= 1.2 km275. V = 12.0 g⇥1 cm32.70 g =⇡(0.254 cm)2`;`= 21.9 cm76. 8.50⇥103L1 d⇥1 m3103L⇥7.0⇥106g Pb1 m3⇥0.75⇥0.50⇥365 d1 yr= 8.1⇥103g Pb

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Matter and Measurements978. mass of Hg in cylinder B = 145.20 gmass of Hg + metal in cylinder A = 92.60 gmass of metal = 145.20 g92.60 g = 52.60 gvolume of cylinder A = volume of cylinder B = volume of Hg in cylinder B= 145.2 g÷13.6 g/mL = 10.7 mLmass of Hg in cylinder A = 92.60 g – 52.60 g = 40.0 gvolume of Hg in cylinder A = 40.0 g÷13.6 g/mL = 2.94 mLvolume of metal = volume of cylinder A - volume of Hg in cylinder A = 10.7 mL2.94 mL = 7.76 mLdensity of metal = 52.60 g/7.76 mL = 6.78 g/mL

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2||||||||ATOMS, MOLECULESAND IONSLECTURE NOTESStudents find this material relatively easy to assimilate; it’s almost entirely qualitative. On the other hand,there’s a lot of memorizing (sorry, learning) to do. This chapter is coverable in two lectures.Some general observations:•Material in Sections 2.1–2.3 is generally well covered in high-school chemistry courses; no need todwell on it.•Students need to know the molecular formulas of the elements (Figure 2.13), the charges of ions withnoble-gas structures and the names and formulas of the common polyatomic ions (Table 2.2).Thecharges of transition-metal ions will be covered later, in Chapter 4.•Naming compounds requires students to distinguish between ionic and molecular substances. It helps topoint out that binary molecular compounds are composed of two nonmetals. Almost all ionic compoundscontain a metal cation combined with a nonmetal anion or negatively charged polyatomic ion. The flowcharts shown in Figures 2.18 and 2.19 should help visual learners.•The periodic table will be discussed in greater detail later in the text (Chapter 6).Lecture 1I. Atomic TheoryA.ElementsPostulates: Elements consist of tiny particles called atoms, which retain their identity in reactions.In a compound, atoms of two or more elements combine in a fixed ratio of small whole numbers(e.g., 1:1, 2:1, etc.).B.Componentsrelative massrelative chargelocationproton1+1nucleusneutron10nucleuselectron0.00051outsideC.Atomic numberIt is the number of protons in the nucleus or the number of electrons in a neutral atom.This ischaracteristic of a particular element: all H atoms have one proton, all He atoms have two protons,etc.11

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