Known Typos and Errors in Chem 105/106 Text

Spring and Fall 2004, first printing of Chemistry: the science in context
Compiled with the help of Spring 2004 Chemistry 105 students
and Fall 2004 Chemistry 106 students.
Chapter 1 | Chapter 2 | Chapter 3 | Chapter 4 | Chapter 5 | Chapter 6 | Chapter 7 | Chapter 8 | Chapter 9 | Chapter 10 | Chapter 11 | Chapter 12 | Chapter 13 | Chapter 14 | Chapter 15 | Chapter 16 | Chapter 17 | Chapter 18

Chapter 1:
  1. Speed of light pages 24 and 25 should be 2.99792458 x 108 m/s.
Chapter 2:
  1. Page 60 where the math problem is worked out, the mass of the proton is wrong it should be
    1.67263x10-24g but the book has 1.67267x10-24g.
  2. Page 73 the practice exercise 2.4 either antimony is wrong or the answer in the back of rubidium is wrong as the product of the decay.  The possible overall fission, assuming Sb is correct is: 23592U + 10n ---> ???51Sb + 410n + ???44Ru.  They do not provide enough information to find the isotope masses.
  3. Page A-46 "Practice Tests" should read "Practice Exercises".
  4. Page A-60 43d should read 0+1e since we have lost one + charge.  The other possibility is that this is a case of electron capture.
  5. Page 92 equation for t in left margin is missing a negative sign or has the fraction within the ln inverted.  Should read either: t=  {-t1/2/ln(2)}ln{At/Ao} or {t1/2/ln(2)}ln{Ao/At}.
Chapter 4:
  1. Page 177, last equation on the page should read [(1 mol Na x 22.99 g/mol) + (1 mol Cl x 33.45 g/mol)] /mol NaCl=, not [(1 mol Na x 22.99 g/mol) X(1 mol Cl x 33.45 g/mol)] /mol NaCl=
  2. Practice excercise 4.2 answers PCl5 name should be phosphorus pentachloride and As2O3 should be diarsenic trioxide, based on the rules we know.  Phosphorus and arsenic can take on multiple charges so sometimes people do name them using roman numerals to specify the charge on the positive ion.  As long as the charge on the negative ion is known either way clearly identifies the compound.
Chapter 5:
  1. Equation 5.3, p. 220, which shows the relation between the concentration of two dilutions has a typo.  Since the total moles of solute is constant n= M*V in both cases.  So  the relation should be MiVi=MfVf , not MiVi=VfVf.
  2. Answer to end of chapter problem 73 is wrong.  It should be c since im is biggest for this compound.
Chapter 6:
  1. Page 314, example 6.9, the right hand figure needs a double bond to the oxygen with a formal charge of zero.
  2. Page 314, Answer to Practice exercise, the first resonance structure for NO2+ should have a triple bond on the right hand side.
Chapter 7:
  1. Figure 7-27 has too many dashed lines.  The erroneous and corrected figures are shown below:
Wrong
 Correct
Erroneous Figure
 Corrected Figure

Chapter 8:
  1. Practice 8.11, P. 408 answer in back is wrong.  Should be 0.0808 mole fraction O2 and 0.9920 mole fraction He.
  2. Practice 8.14 rounding error in answer it should be 9.5 x 10-5 mol/L.
  3. End of Chapter Problem #27.  The answer is backwards.  If you read carefully you will note that the questions asks which line is "not consistent"  thus the answer is curve #2 not #1 as stated in the back.
Chapter 11:
  1. Page 529. Incorrect: (1419 mol)(75.3J/mol C)(-25 ˚C) = -2.67 x 107 J.  Correct: (1419 mol)(75.3 J/mol C)(-25 ˚C) = -2.67 x 106 J.
  2. Page 529, bottom.  Incorrect: 2.69 x 104 kJ = (x) (6.31 kJ/mol). Correct: 2.69 x 103 kJ = (x)(6.31 kJ/mol).
  3. Page 530, top. Incorrect:4.26 x 103.  Correct: 4.26 x 102.  Incorrect: 75 pounds -> Correct: 16.9 pounds.
  4. Homework problem 11-63 a. Incorrect: 2 N2  ->  Correct: N2.
Chapter 12:
  1. Page 601.  Section 12.9 is part of section 12.6.  Section 12.10 is actually 12.9.  Section 12.11 is actually 12.10.

Chapter 13:
  1. Page 613.  In sample exercise 13.1 q = -573 J  not -264 J.  This results in ∆Hsoln= 57.3 kJ/mol.
  2. Answer to Practice Problem 13.1 (page A-52).  The correct answer is 20.66 degrees Celcius not 71 degrees.
Chapter 14:
1. Question 67: NOCl2 should be 2 NOCl.

Chapter 15:
1. Pages 745-746 the equilibrium constants have units which they should not.
2. Answer to problem 15 in the appendix is for some other problem.  The answer should be: KP = 7.22 x 10-19.
3. Answer to practice problem 15.7 on page 750 should be K = 6.08 x 105.
4. Answer to practice problem 15.10 on page 762 should be [N2O4] = 0.00158 M (0.002 w/SF) and [NO2] = 0.08648 M (0.086 w/SF).
5. Page 774 problem 32 the answer is Kp is greater than Kc.
6. Page 775 problem 49a the answer is 5.08 kJ/mol.
7. Page 775 problem 51 N2(g)O2(g) <––>2NO(g) should read N2(g) + O2(g) <––>2NO(g).
8. Page 776 problem 56.  The answer is "yes, stays the same" not "no".
9. Page 776 problem 61 answers are: a) both [O3] and [O2] will increase, reaction shifts towards products to relieve stress.  b) both [O3] and [O2] will increase, reaction shifts towards reactants to relieve stress.  c) Since volume decreases all concentrations will increase, but the concentration of O3 will increase more to relieve stress.
10. Page 776 problem 63 the answer should be the equilibrium will shift towards reactants.

Chapter 17:
1. Page 849 the calculation of ∆G has the wrong result. It should be -212 kJ.
2. Page A-28 (table of standard reduction potentials)  11th item from top  (Cu+ + 2e- ––> Cu) does not belong or is a typo for some other reaction.
3. Page A-28 (table of standard reduction potentials)  13th item from the bottom should be PbSO4 + 2e- ––> Pb + SO42-.
4. Page A-27 (table of standard reduction potentials) 12th item from the top.  PbO2 reduction has a reduction potential which is slightly inconsistent with the rest of the text.  The value should probably be: 1.685 V rather than 1.6913 V.
5. Page 868, figure 17.10.  The calculation of the cell potential ignores the fact that H2SO4 is such a strong acid that it is nearly 100% dissociated in solution at those concentrations.  This means that the H+ and  SO42- must be included separately in the reaction quotient.  Thus the calculation using the Nernst equation should look like what we did in class on 11/30/2004: Ecell = 2.041 -(RT/{nF} )ln(1/{[SO42-]2[H+]4}).  Because the ion concentration is so high the assumption that we can use concentrations actually breaks down, but we will ignore that in this class.  For 4.8 M sulfuric acid in the electrolyte the cell voltage should be 2.20 V per cell.  This corresponds well to a typical car battery which produces a little more than 13 V when fully charged (6 x 2.20 V = 13.2 V).

Chapter 18:
  1. Typo in answer to problem 71.  The structure for polybutadiene is shown.  The structure for the addition polymer formed from vinyl acetate is:  polyvinylacetate structure

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Last updated: December 12, 2004
Expires: December 17, 2004