General chemistry 9th ebbing, gammon(solution manual)

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Complete Solutions Manual GENERAL CHEMISTRY NINTH EDITION Ebbing/Gammon David Bookin Mt San Jacinto College Darrell D Ebbing Wayne State University, Emeritus Steven D Gammon Western Washington University HOUGHTON MIFFLIN COMPANY BOSTON NEW YORK Vice President and Executive Publisher: George Hoffman Vice President and Publisher: Charles Hartford Senior Marketing Manager: Laura McGinn Development Editor: Kate Heinle Assistant Editor: Amy Galvin Editorial Assistant: Chip Cheek Marketing Assistant: Kris Bishop Copyright © 2009 by Houghton Mifflin Company All rights reserved Houghton Mifflin Company hereby grants you permission to reproduce the Houghton Mifflin material contained in this work in classroom quantities, solely for use with the accompanying Houghton Mifflin textbook All reproductions must include the Houghton Mifflin copyright notice, and no fee may be collected except to cover cost of duplication If you wish to make any other use of this material, including reproducing or transmitting the material or portions thereof in any form or by any electronic or mechanical means including any information storage or retrieval system, you must obtain prior written permission from Houghton Mifflin Company, unless such use is expressly permitted by federal copyright law If you wish to reproduce material acknowledging a rights holder other than Houghton Mifflin Company, you must obtain permission from the rights holder Address inquiries to College Permissions, Houghton Mifflin Company, 222 Berkeley Street, Boston, MA 02116-3764 ISBN 13: 978-0-618-95211-3 ISBN 10: 0-618-95211-X Contents PREFACE XI CHAPTER – CHEMISTRY AND MEASUREMENT Solutions To Exercises Answers To Concept Checks Answers To Self-Assessment And Review Questions Answers To Concept Explorations Answers To Conceptual Problems Solutions To Practice Problems 11 Solutions To General Problems 20 Solutions To Strategy Problems 29 Solutions To Cumulative-Skills Problems 30 CHAPTER – ATOMS, MOLECULES, AND IONS 36 Solutions To Exercises 36 Answers To Concept Checks 38 Answers To Self-Assessment And Review Questions 39 Answers To Concept Explorations 42 Answers To Conceptual Problems 43 Solutions To Practice Problems 45 Solutions To General Problems 56 Solutions To Strategy Problems 63 Solutions To Cumulative-Skills Problems 64 CHAPTER – CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS 66 Solutions To Exercises 66 Answers To Concept Checks 71 Answers To Self-Assessment And Review Questions 73 Answers To Concept Explorations 75 Answers To Conceptual Problems 77 Solutions To Practice Problems 79 Solutions To General Problems 103 Solutions To Strategy Problems 109 Solutions To Cumulative-Skills Problems 113 Copyright © Houghton Mifflin Company All rights reserved iv Contents CHAPTER – CHEMICAL REACTIONS 116 Solutions To Exercises 116 Answers To Concept Checks 120 Answers To Self-Assessment And Review Questions 122 Answers To Concept Explorations 124 Answers To Conceptual Problems 127 Solutions To Practice Problems 129 Solutions To General Problems 145 Solutions To Strategy Problems 155 Solutions To Cumulative-Skills Problems 157 CHAPTER – THE GASEOUS STATE 164 Solutions To Exercises 164 Answers To Concept Checks 170 Answers To Self-Assessment And Review Questions 172 Answers To Concept Explorations 176 Answers To Conceptual Problems 177 Solutions To Practice Problems 179 Solutions To General Problems 192 Solutions To Strategy Problems 200 Solutions To Cumulative-Skills Problems 202 CHAPTER – THERMOCHEMISTRY 207 Solutions To Exercises 207 Answers To Concept Checks 209 Answers To Self-Assessment And Review Questions 210 Answers To Concept Explorations 213 Answers To Conceptual Problems 214 Solutions To Practice Problems 216 Solutions To General Problems 225 Solutions To Strategy Problems 233 Solutions To Cumulative-Skills Problems 235 CHAPTER – QUANTUM THEORY OF THE ATOM 242 Solutions To Exercises 242 Answers To Concept Checks 243 Answers To Self-Assessment And Review Questions 244 Answers To Concept Explorations 246 Copyright © Houghton Mifflin Company All rights reserved Contents v Answers To Conceptual Problems 248 Solutions To Practice Problems 250 Solutions To General Problems 257 Solutions To Strategy Problems 262 Solutions To Cumulative-Skills Problems 264 CHAPTER – ELECTRON CONFIGURATIONS AND PERIODICITY 268 Solutions To Exercises 268 Answers To Concept Checks 269 Answers To Self-Assessment And Review Questions 270 Answers To Concept Explorations 273 Answers To Conceptual Problems 274 Solutions To Practice Problems 276 Solutions To General Problems 279 Solutions To Strategy Problems 282 Solutions To Cumulative-Skills Problems 283 CHAPTER – IONIC AND COVALENT BONDING 286 Solutions To Exercises 286 Answers To Concept Checks 289 Answers To Self-Assessment And Review Questions 291 Answers To Concept Explorations 294 Answers To Conceptual Problems 295 Solutions To Practice Problems 298 Solutions To General Problems 317 Solutions To Strategy Problems 328 Solutions To Cumulative-Skills Problems 330 CHAPTER 10 – MOLECULAR GEOMETRY AND CHEMICAL BONDING THEORY 336 Solutions To Exercises 336 Answers To Concept Checks 340 Answers To Self-Assessment And Review Questions 341 Answers To Concept Explorations 344 Answers To Conceptual Problems 345 Solutions To Practice Problems 348 Solutions To General Problems 361 Solutions To Strategy Problems 368 Solutions To Cumulative-Skills Problems 370 Copyright © Houghton Mifflin Company All rights reserved vi Contents CHAPTER 11 – STATES OF MATTER; LIQUIDS AND SOLIDS 375 Solutions To Exercises 375 Answers To Concept Checks 377 Answers To Self-Assessment And Review Questions 379 Answers To Concept Explorations 381 Answers To Conceptual Problems 384 Solutions To Practice Problems 387 Solutions To General Problems 399 Solutions To Strategy Problems 407 Solutions To Cumulative-Skills Problems 410 CHAPTER 12 – SOLUTIONS 413 Solutions To Exercises 413 Answers To Concept Checks 417 Answers To Self-Assessment And Review Questions 418 Answers To Concept Explorations 421 Answers To Conceptual Problems 422 Solutions To Practice Problems 424 Solutions To General Problems 435 Solutions To Strategy Problems 445 Solutions To Cumulative-Skills Problems 448 CHAPTER 13 – RATES OF REACTION 453 Solutions To Exercises 453 Answers To Concept Checks 455 Answers To Self-Assessment And Review Questions 457 Answers To Concept Explorations 460 Answers To Conceptual Problems 461 Solutions To Practice Problems 464 Solutions To General Problems 478 Solutions To Strategy Problems 490 Solutions To Cumulative-Skills Problems 492 CHAPTER 14 – CHEMICAL EQUILIBRIUM 496 Solutions To Exercises 496 Answers To Concept Checks 500 Answers To Self-Assessment And Review Questions 501 Answers To Concept Explorations 504 Copyright © Houghton Mifflin Company All rights reserved Contents vii Answers To Conceptual Problems 505 Solutions To Practice Problems 507 Solutions To General Problems 521 Solutions To Strategy Problems 536 Solutions To Cumulative-Skills Problems 541 CHAPTER 15 – ACIDS AND BASES 543 Solutions To Exercises 543 Answers To Concept Checks 545 Answers To Self-Assessment And Review Questions 545 Answers To Concept Explorations 548 Answers To Conceptual Problems 549 Solutions To Practice Problems 550 Solutions To General Problems 561 Solutions To Strategy Problems 568 Solutions To Cumulative-Skills Problems 571 CHAPTER 16 – ACID-BASE EQUILIBRIA 572 Solutions To Exercises 572 Answers To Concept Checks 582 Answers To Self-Assessment And Review Questions 583 Answers To Concept Explorations 587 Answers To Conceptual Problems 589 Solutions To Practice Problems 591 Solutions To General Problems 624 Solutions To Strategy Problems 648 Solutions To Cumulative-Skills Problems 655 CHAPTER 17 – SOLUBILITY AND COMPLEX-ION EQUILIBRIA 657 Solutions To Exercises 657 Answers To Concept Checks 661 Answers To Self-Assessment And Review Questions 662 Answers To Concept Explorations 664 Answers To Conceptual Problems 666 Solutions To Practice Problems 667 Solutions To General Problems 683 Solutions To Strategy Problems 698 Solutions To Cumulative-Skills Problems 702 Copyright © Houghton Mifflin Company All rights reserved viii Contents CHAPTER 18 – THERMODYNAMICS AND EQUILIBRIUM 706 Solutions To Exercises 706 Answers To Concept Checks 710 Answers To Self-Assessment And Review Questions 710 Answers To Concept Explorations 713 Answers To Conceptual Problems 714 Solutions To Practice Problems 716 Solutions To General Problems 728 Solutions To Strategy Problems 738 Solutions To Cumulative-Skills Problems 743 CHAPTER 19 – ELECTROCHEMISTRY 750 Solutions To Exercises 750 Answers To Concept Checks 756 Answers To Self-Assessment And Review Questions 757 Answers To Concept Explorations 760 Answers To Conceptual Problems 761 Solutions To Practice Problems 764 Solutions To General Problems 795 Solutions To Strategy Problems 808 Solutions To Cumulative-Skills Problems 811 CHAPTER 20 – NUCLEAR CHEMISTRY 816 Solutions To Exercises 816 Answers To Concept Checks 819 Answers To Self-Assessment And Review Questions 820 Answers To Conceptual Problems 822 Solutions To Practice Problems 824 Solutions To General Problems 836 Solutions To Strategy Problems 844 Solutions To Cumulative-Skills Problems 848 CHAPTER 21 – CHEMISTRY OF THE MAIN-GROUP ELEMENTS 852 Answers To Concept Checks 852 Answers To Self-Assessment And Review Questions 852 Answers To Conceptual Problems 860 Solutions To Practice Problems 861 Solutions To General Problems 880 Copyright © Houghton Mifflin Company All rights reserved Contents ix Solutions To Strategy Problems 887 CHAPTER 22 – THE TRANSITION ELEMENTS AND COORDINATION COMPOUNDS 894 Solutions To Exercises 894 Answers To Concept Checks 896 Answers To Self-Assessment And Review Questions 896 Answers To Conceptual Problems 900 Solutions To Practice Problems 901 Solutions To General Problems 912 Solutions To Strategy Problems 915 CHAPTER 23 – ORGANIC CHEMISTRY 918 Solutions To Exercises 918 Answers To Concept Checks 922 Answers To Self-Assessment And Review Questions 922 Answers To Conceptual Problems 925 Solutions To Practice Problems 927 Solutions To General Problems 938 Solutions To Strategy Problems 943 CHAPTER 24 – POLYMER MATERIALS: SYNTHETIC AND BIOLOGICAL 946 Solutions To Exercises 946 Answers To Concept Checks 946 Answers To Self-Assessment And Review Questions 947 Answers To Conceptual Problems 949 Solutions To Practice Problems 951 Solutions To General Problems 956 Solutions To Strategy Problems 959 APPENDIX A – MATHEMATICAL SKILLS 967 Solutions To Exercises 967 Copyright © Houghton Mifflin Company All rights reserved 954 Chapter 24: Polymer Materials: Synthetic and Biological 24.40 Deoxyadenosine consists of deoxyribose and adenine H2N N N O O O P CH2 N N O O H H H OH H H 24.41 Three hydrogen bonds link a guanine-cytosine base pair H N H N N O O H N N Cytosine H N N H Guanine N Because only two hydrogen bonds link an adenine-uracil base pair, the bonding would be expected to be stronger in the guanine-cytosine pair Copyright © Houghton Mifflin Company All rights reserved Chapter 24: Polymer Materials: Synthetic and Biological 955 24.42 Two hydrogen bonds link an adenine-thymine base pair The hydrogen bonding in a uraciladenine base pair is the same as in a thymine-adenine base pair, so the strength of the bonding should be the same CH3 O H N N H O H N N N Thymine N N Adenine There are three hydrogen bonds in a guanine-cytosine base pair, so the bonding should be stronger between guanine and cytosine than between adenine and thymine 24.43 If a codon consisted of two nucleotides, there would be x 4, or 16, possible codons using the four nucleotides Because there are 20 amino acids that must be represented uniquely by a codon for protein synthesis, the codon must be longer than two nucleotides A two-nucleotide codon would not be workable as an amino-acid code 24.44 If the codon were four nucleotides, there would be x x x 4, or 256, possible codons using the four nucleotides This is more than the minimum of 20 codons required for all the amino acids used in protein synthesis A four-nucleotide codon would be workable as an amino-acid code 24.45 When DNA is denatured, the hydrogen bonds between base pairs are broken DNA with a greater percentage composition of guanine and cytosine would be denatured less readily than DNA with a greater percentage composition of adenine and thymine, because there are more hydrogen bonds between the former than between the latter 24.46 RNA of a greater percentage composition of guanine and cytosine would be denatured less readily than RNA with a lower percentage composition because denaturation involves disruption of hydrogen bonds, and the higher the percentage composition of guanine and cytosine, the more hydrogen bonds there are in the RNA 24.47 Mark off the message into triplets beginning at the left GGA│UCC│CGC│UUU│GGG│CUG│AAA│UAG Gly−Ser−Arg−Phe−Gly−Leu−Lys Note that the UAG at the right codes for the end of the sequence Copyright © Houghton Mifflin Company All rights reserved 956 Chapter 24: Polymer Materials: Synthetic and Biological 24.48 Mark the message off into triplets beginning at the left AUU│GGC│GCG│AGA│UCG│AAU│GAG│CCC│AGU Ile−Gly−Ala−Arg−Ser−Asn−Glu−Pro−Ser ■ SOLUTIONS TO GENERAL PROBLEMS 24.49 The codons have bases that are complementary to those in the anticodon Anticodons Codons GAC CUG UGA ACU GGG CCC ACC UGG 24.50 Codons Anticodons UUG AAC CAC GUG ACU UGA GAA CUU 24.51 Consult Table 24.3 to find which nucleotides correspond to the amino acids in the sequence leu−ala−val−glu−asp−cys−met−trp−lys CUU GCU GUU GAA GAU UGU AUG UGG AAA 24.52 tyr−ile−pro−his−leu−his−thr−ser−phe−met UAU AUU CCU CAU CUU CAU ACU UCU UUU AUG 24.53 The addition polymer that forms from cis −1,2-dichloroethene is —CHCH—CHCH—CHCH—CHCH— Cl Cl Cl Cl Cl Cl Cl Cl 24.54 The structural formula of neoprene is Cl Cl CH2C CHCH2CH2C Cl CHCH2CH2C CHCH2 24.55 The monomer units that make up Kevlar are O HOC O COH and H2N NH2 Copyright © Houghton Mifflin Company All rights reserved Chapter 24: Polymer Materials: Synthetic and Biological 24.56 The monomer unit that makes up nylon is O H2N(CH2 )5 COH The reaction is O O NH2–(CH2)5 –COOH + NH2–(CH2)5–COOH → —NH(CH2)5C–NH(CH2)5C— + n H2O 24.57 The amino acid with the nonpolar side chain is NH2 CH3SCH2CH2CHCOOH The other amino acid has a polar SH group in the side chain 24.58 The amino acid with the polar side chain is NH2 H2NCH2CH2CH2CH2CHCOOH 24.59 The zwitterion for serine is NH3+ HOCH2 CHCOO24.60 The zwitterion for leucine is NH3 + CH3 CHCH2 CHCOO– CH3 24.61 The two possibilities are NH2 O NH2 O CH3SCH2 CH2 CH—CNHCHCOOH CH2SH Copyright © Houghton Mifflin Company All rights reserved and HSCH2 CH—CNHCHCOOH CH2CH2 SCH3 957 958 Chapter 24: Polymer Materials: Synthetic and Biological 24.62 The two possibilities are CH3 NH2 O CH3CHCH2CH—CNHCHCOOH CH2CH2 CH2 CH2NH2 and NH2 O H2NCH2 CH2 CH2CH2CH—CNHCHCOOH CH2 CHCH3 CH3 24.63 The number of possible sequences is x x x x x 1, or 720 24.64 arg−pro−glu−gly−asn−gln arg−pro−glu−gly−gln−asn arg−pro−glu−gln−gly−asn NH2 N N O P N N O O CH2 Adenine O O H 24.65 H H OH OH Ribose H Phosphate Copyright © Houghton Mifflin Company All rights reserved Chapter 24: Polymer Materials: Synthetic and Biological 959 O N N O P O N N O H2N CH2 O O H 24.66 H H OH OH H 24.67 Polymers are actually macromolecules consisting of long chains of similar groups of atoms Carothers’s idea is to consider two functional groups that chemically link together This was represented as the linking of a hook and an eye Carothers further suggested finding a molecule having two of the same functional groups at the ends of the molecule; this would be equivalent to a molecule with hooks at both ends Then the reaction of several such molecules would result in a larger molecule With many such molecules, the product would be a long-chain molecule: a macromolecule 24.68 Carothers’s synthesis of nylon, neoprene rubber, and polyesters showed that a synthetic polymer could be made with the properties of fibers, like natural silk, rubber, or wool 24.69 The atomic force microscope is a cousin of a scanning tunneling microscope Both microscopes use a probe to scan a surface; but whereas the scanning tunneling microscope measures an electric current between the probe tip and the sample, the atomic microscope measures the attractive van der Waals force between the probe tip and the sample 24.70 The advantage of the atomic force microscope is that it can be used with almost any surface, whereas a scanning tunneling microscope requires a conductive surface ■ SOLUTIONS TO STRATEGY PROBLEMS 24.71 First, determine the number of moles of NaOH used in the titration (0.2183 M) x (98.6 x 10−3 L) = 0.02152 mol NaOH Since the compound is a dicarboxylic acid, the moles of acid in the titration are 0.02152 mol NaOH x mol acid = 0.01076 mol acid mol NaOH Copyright © Houghton Mifflin Company All rights reserved 960 Chapter 24: Polymer Materials: Synthetic and Biological The molar mass of the compound is 1.273 g = 118.28 g/mol 0.01076 mol acid A dicarboxylic acid has two COOH units, each of which weighs 45 g/mol This accounts for 45 x = 90g/mol so far The other 28 g/mol could be C2H4, giving a dicarboxylic acid with the following possible structural formula O O HOCCH2CH2COH The condensation reaction with ethylene glycol will produce the following polyester O HOCH2CH2OH + O → HOCCH2CH2COH O O O O + nH2O ~OCH2CH2O–CCH2CH2C–OCH2CH2O–CCH2CH2C~ 24.72 The reaction to produce the polyamide is nH2N(CH2)nNH2 → + nHOOC(CH2)8COOH H H H H ~N(CH2)nN–C(CH2)8C–N(CH2)nN–C(CH2)8C~ + nH2O O O O O The general formula of the monomer unit is Cn + 10H2n + 18N2O2 Now use the percentage composition to determine the empirical formula Assume a 100-g sample Then, the moles of each element in the sample are 67.2 g C x mol C = 5.595 mol C 12.01 g 10.5 g H x mol H = 10.41 mol H 1.008 g 10.4 g N x mol N = 0.7423 mol N 14.01 g 11.9 g O x mol O = 0.7437 mol O 16.00 g Copyright © Houghton Mifflin Company All rights reserved Chapter 24: Polymer Materials: Synthetic and Biological 961 Divide each molar quantity by the smallest number of moles, 0.7423 mol N Divide each molar quantity by the smallest number of moles, 0.7423 mol N 5.595 mol C = 7.537 ≅ 7.5 0.7423 mol 10.41 mol H = 14.02 ≅ 14 0.7423 mol 0.7423 mol N = 1.000 ≅ 0.7423 mol 0.7437 mol O = 1.001 ≅ 0.7423 mol Multiplying each mole ratio by gives C15H28N2O2 as the formula for the monomer unit of the polyamide, so n = 24.73 The structure of rubber (poly-cis-isoprene) is CH3 C CH2 H CH3 C C CH2 CH2 H C CH2 The monomer unit of isoprene has the formula C5H8 and contains one double bond that is available to react with Br2 Each C5H8 unit weighs 68.11 g/mol For each mole of monomer units that react, one mole of Br2 also reacts In the process, for one monomer unit, one carbon-carbon double bond is broken, one bromine-bromine bond is broken, one carbon-carbon single bond is formed, and two carbon-bromine bonds are formed The enthalpy change for the reaction can be estimated using the bond energy values from Table 9.5.The monomer unit of isoprene has the formula C5H8 and contains one double bond that is available to react with Br2 Each C5H8 unit weighs 68.11 g/mol For each mole of monomer units that react, one mole of Br2 also reacts In the process, for one monomer unit, one carbon-carbon double bond is broken, one brominebromine bond is broken, one carbon-carbon single bond is formed, and two carbon-bromine bonds are formed The enthalpy change for the reaction can be estimated using the bond energy values from Table 9.5 ΔH°rxn ≅ BE(C=C) + BE(Br−Br) − BE(C−C) − 2BE(C−Br) = (602 kJ) + (190 kJ) − (346 kJ) − 2(285 kJ) = −124 kJ This is the energy released per mole of monomer units Thus, for 10.0 g of rubber, the heat released is 10.0 g x -124 kJ mol monomer units x = −18.20 = −18.2 kJ mol monomer units 68.11 g Copyright © Houghton Mifflin Company All rights reserved 962 Chapter 24: Polymer Materials: Synthetic and Biological 24.74 a When aspartame hydrolyzes, the two amino acids that form are aspartic acid and phenylalanine Their structures are O NH2—CH—C OH CH2 CH2 O and C O NH2 —CH—C—OH b OH Aspartame is the dipeptide asp−phe Another dipeptide that could be prepared from these amino acids is phe−asp It has the following structure CH2 O O NH2—CH—C—NH—CH—C OH CH2 C O c OH There are = possible tripeptides that can be formed using the amino acids valine (val) and alanine (ala) These are val−val−val ala−val−val ala−ala−val val−val−ala val−ala−val vla−ala−ala ala−val−ala ala−ala−ala 24.75 a The number of tripeptides that are possible when four bases are used is x x = 64.The number of tripeptides that are possible when four bases are used is x x = 64 b Yes, this is enough to establish a code for all of the 20 amino acids c The number of possible tetrapeptides when two bases are used is x x x x = 16 d No, this is not enough to establish the code for the 20 amino acids Copyright © Houghton Mifflin Company All rights reserved Chapter 24: Polymer Materials: Synthetic and Biological e 963 If the bases are taken in groups of five, then 32 combinations are possible Therefore, develop a hypothetical code using these combinations For example, using the bases U and T, the 32 possible combinations are UUUUU UTUUU TUUUU TTUUU UUUUT UTUUT TUUUT TTUUT UUUTU UTUTU TUUTU TTUTU UUTUU UTTUU TUTUU TTTUU UUUTT UTUTT TUUTT TTUTT UUTUT UTTUT TUTUT TTTUT UUTTU UTTTU TUTTU TTTTU UUTTT UTTTU TUTTT TTTTT Now, arbitrarily assign amino acids to 20 of these possible code sequences UUUUU (gly) UTUUU (glu) TUUUU (met) TTUUU (leu) UUUUT (ala) UTUUT (his) TUUUT (ser) TTUUT (ile) UUUTU (val) UTUTU (lys) TUUTU (cys) TTUTU (pro) UUTUU (leu) UTTUU (arg) TUTUU (thr) TTTUU (phe) UUUTT (ile) UTUTT (end) TUUTT (asp) TTUTT (trp) UUTUT (pro) UTTUT (gly) TUTUT (glu) TTTUT (met) UUTTU (phe) UTTTU (ala) TUTTU (tyr) TTTTU (ser) UUTTT (trp) UTTTT (val) TUTTT (asn) TTTTT (cys) The base sequence for the tripeptide leu−ala−val would be UUTUU UTTTU UUUTU 24.76 First, use the percentage composition to obtain the molecular formula 66.24 g C x 5.23 g H x 9.06 g Fe x 9.09 g N x 10.38 g O x mol C = 5.5154 mol C 12.01 g mol H = 5.188 mol H 1.008 g mol Fe = 0.1622 mol Fe 55.85 g mol N = 0.6488 mol N 14.01 g mol O = 0.64875 mol O 16.00 g Copyright © Houghton Mifflin Company All rights reserved 964 Chapter 24: Polymer Materials: Synthetic and Biological Divide each molar quantity by the smallest number of moles of Fe, 0.1622 mol 5.5154 mol C = 34.00 ≅ 34 0.1622 mol 5.188 mol H = 31.98 ≅ 32 0.1622 mol 0.1622 mol Fe = 1.000 ≅ 0.1622 mol 0.6488 mol N = 3.997 ≅ 0.1623 mol 0.64875 mol O = 3.997 ≅ 0.1623 mol Therefore, the molecular formula of heme is C34H32FeN4O4, with a molecular mass of 616.49 amu The hemoglobin molecule contains four globin units and four heme units and weighs 64,500 amu Therefore, 64,500 amu = 4(616.49 amu) + 4(weight of globin) Weight of globin = 64,500 amu - 4(616.49 amu) = 15,508.5 = 15,509 amu 24.77 Rearrange the boiling-point-depression formula and solve for the molality of the solution cm = ΔTb Kb = 0.0556°C = 0.1085 m 0.512 °C/m The molar mass of ribose (C5H10O5) is 150.13 g/mol, and for deoxyribose (C5H10O4) the molar mass is 134.13 g/mol Let x be the mass of ribose; then 1.500 g − x is the mass of deoxyribose In 100.0 g of water, the total moles of solute is Moles solute = cm x kg solvent = 0.1085 m x 0.1000 kg = 0.010850 mol Also Moles solute = 1.500 g - x x + = 0.01085 mol 134.13 g/mol 150.13 g/mol Solve in the normal way (134.13 x) + (150.13)(1.500 g − x) = (0.01085 g)(150.13)(134.13) 134.13 x + 225.195 g − 150.13 x = 218.48 g x = 0.419 g The percentage of ribose in the mixture is 0.419 g x 100% = 27.9 = 28% 1.500 g Copyright © Houghton Mifflin Company All rights reserved Chapter 24: Polymer Materials: Synthetic and Biological 965 24.78 The zwitterion form of alanine is H CH3 C COO NH3+ When the zwitterion picks up a proton, the resulting ion has the following structure H CH3 C COOH NH3 + The chemical equation for the reaction is H CH3 C H COO + H3O+ → CH3 NH3+ C COOH + H2O NH3+ When the pH is 4.0, the Henderson-Hasselbalch equation can be used to get the ratio of the substances present pH = pKa = log [base] [acid] Rearranging gives log [base] = pH − pKa = 4.0 − 2.3 = 1.7 [acid] In this equation, the zwitterion is the base and the protonated form is the acid Let HA represent the zwitterion, and let H2A+ represent the protonated form Thus [HA] = 101.7 = 50.1 = x 101 [H A + ] Therefore, the predominant form at this pH is HA, which is the zwitterion form 24.79 From the description, it appears that the monomer units of the material must be a dicarboxylic acid and a dialcohol When the dialcohol is oxidized, it produces another dicarboxylic acid, with a molecular mass of 90.0 amu A carboxylic acid group (COOH) has a molecular mass of 45.02 amu, so the dicarboxylic acid obtained from the alcohol must be HOOCCOOH, which is oxalic acid (H2C2O4) When oxalic acid reacts with NaOH, the following reaction occurs H2C2O4(aq) + 2NaOH(aq) → Na2C2O4(aq) + 2H2O(l) Copyright © Houghton Mifflin Company All rights reserved 966 Chapter 24: Polymer Materials: Synthetic and Biological The mass of oxalic acid that would react with 30.5 mL of 0.1056 M NaOH is (0.1056 M) x (30.5 x 10−3 L) x mol H C O 90.0 g x = 0.1449 = 0.145 g mol NaOH mol H C O This agrees with the data given in the problem When calcium chloride solution is added to the acid solution, a white precipitate forms, which is calcium oxalate, CaC2O4 24.80 a Gutta percha (poly-trans-isoprene) has the following structure CH3 C CH2 C b C H CH2 H CH2 CH3 C CH2 The monomer unit of gutta percha is CH3 C CH2 C CH2 H The formula of this monomer unit is C5H8, which has a molecular mass of 68.11 amu The number of isoprene units in the sample is 250,000 amu = 3670.3 = 3.7 x 103 isoprene units 68.11 amu/unit c Stretching a rubber band evolves heat, so ΔH for the stretching process is negative A stretched rubber band spontaneously contracts, so ΔG is negative for the contraction process For the stretching process, ΔG would be positive (nonspontaneous) Use these two quantities to determine the sign of the entropy change ΔG = ΔH − TΔS Rearrange and solve for the entropy change for the stretching process ΔS = ΔH - ΔG T The denominator (temperature) is always positive Since ΔH is negative and ΔG is positive, the numerator is negative Thus, overall, ΔS is negative, and there is a decrease in entropy when the rubber band is stretched Copyright © Houghton Mifflin Company All rights reserved APPENDIX A Mathematical Skills ■ SOLUTIONS TO EXERCISES a Either leave it as 4.38 or write it as 4.38 x 100 b Shift the decimal point left, and count the number of positions shifted (3) The answer is 4.380 x 103, assuming the terminal zero is significant c Shift the decimal point right, and count the number of positions shifted (4) The answer is 4.83 x 10−4 a Shift the decimal point right three places The answer is 7025 b Shift the decimal point left four places The answer is 0.000897 Express 2.8 x 10−6 as 0.028 x 10−4 Then, the sum can be written (3.142 x 10−4) + (0.028 x 10−4) or (3.142 + 0.028) x 10−4 = 3.170 x 10−4 a (5.4 x 10−7) x (1.8 x 108) = (5.4 x 1.8) x 10−7 x 108 = 9.72 x 101 This rounds to 9.7 x 101 b 5.4 5.4 x 10-7 = x 10−7 x 105 = 0.90 x 10−2 = 9.0 x 10−3 6.0 6.0 x 10-5 a (3.56 x 103)4 = (3.56)4 x (103)4 = 161 x 1012 = 1.61 x 1014 b 4.81 x 102 = 0.481 x 103 = a log 0.00582 = −2.235 b log 689 = 2.838 Copyright © Houghton Mifflin Company All rights reserved 0.481 x 103 = 0.784 x 101 = 7.84 x 100 968 Appendix A: Mathematical Skills a antilog 5.728 = 5.35 x 105 b antilog (−5.728) = 1.87 x 10−6 a ln 9.93 = 2.2955 = 2.296 b e1.10 = 3.004 = 3.0 x= -0.850 ± (0.850) - 4(1.80)(-9.50) -0.850 ± 8.314 = 3.60 2(1.80) The positive root is 7.46 = 2.07 3.60 Copyright © Houghton Mifflin Company All rights reserved ... Complete Solutions Manual provides worked-out answers to all of the problems that appear in General Chemistry, 9th Edition, by Darrell D Ebbing and Steven D Gammon The solutions follow the same order... 716 Solutions To General Problems 728 Solutions To Strategy Problems 738 Solutions To Cumulative-Skills Problems 743 CHAPTER 19 – ELECTROCHEMISTRY ... 764 Solutions To General Problems 795 Solutions To Strategy Problems 808 Solutions To Cumulative-Skills Problems 811 CHAPTER 20 – NUCLEAR CHEMISTRY
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