| Chap | Page | Book | Name | Timestamp | Reference | Description |
| 2 | 14 | PowerPoint Slides (Accessed through MC Study Area) | Lund | 1/19/2012 11:01:12 AM | Slide 14 | The number 2.97x10^5 is incorrectly written as 2.9x710^5 . |
| 2 | 20 [24] | Text | Lund | 1/23/2012 7:51:33 PM | At the top | Answer reads "0.003" should read "0.03" |
| 2 | 37 [41] | Text | Lund | 1/30/2012 1:04:25 PM | 2.15 [15] | The correct conversion should read: 60.0 kg × (1000 g/1 kg) × (1 cm^3/0.752 g) = 7.98×10^4 cm^3, NOT 60.0 kg × (1000 kg/1 kg) × (1 cm^3/0.752 g) = 7.98×10^4 cm^3 |
| 2 | 48 [52] | Text | Lund | 8/29/2011 1:17:31 PM | 49 | The answer in the back of the book, p. A-2 [p. 58] is wrong. The correct answer is 2.231x10^6 not 2.231x10^-6. |
| 2 | 49 [58] | Text | Befus | 12/20/2011 1:49:43 PM | 37a | The answer in the back of the book, p. A-2 [p. 58] is wrong. The correct answer is 32,200,000, not 32,000,000. |
| 4 | 103 [115] | Text | Lund | 2/1/2012 10:40:59 AM | 4.12 [12] | The Periodic Table on this page does not agree with the one on the front cover [p. 661] with regard to polonium. The inside cover [p. 661] shows Po as a metalloid, and on p. 103 as a metal. Checking out a few resources revealed that chemists do not agreee on the matter. |
| 4 | 113 [125] | Text | Lund | 2/1/2012 10:23:59 AM | Calculation Near Top of Page | The second percentage should be 24.23%, NOT 24.33% |
| 4 | 121 [133] | Text | Lutz | 9/13/2011 10:42:54 PM | 75 (c) | They list oxygen with a 2+ charge rather than the 2- charge they intended. As such the answer key gives 8 p 10 e- rather than 8 p 6e- you would have for oxygen 2+. |
| 4 | 122 [134] | Text | Lutz | 9/13/2011 10:40:06 PM | 91 (b) | Radium is atomic number 88 not 28 as shown in the problem |
| 5 | 145 [161] | Text | Befus | 3/2/2012 4:33:44 PM | Table 5.7 | Formula for acetic acid is listed incorrectly as H2C2H3O2. Should be HC2H3O2. |
| 5 | 156 [172] | Text | Lutz | 9/26/2011 8:09:55 PM | 51 (d) | tin(IV) sulfate is reported to be a type I cation with only one known charge, but tin can form tin(II) and tin(IV) compounds |
| 5 | 163 [179] | Text | Lund | 2/13/2013 7:43:24 AM | Skillbuilder 5.1 | The answer is wrong, since the results are NOT consistent with the Law of Constant Composition. The ratio of mass of O to mass of C is 1.3. The ratio for the second sample is 1.6. |
| 5 | 31 | Solutions Manual | Befus | 12/20/2011 9:48:24 PM | 63 | name of compounds not given (correct in appendix) |
| 6 | 174 [194] | Text | Lutz | 9/28/2011 8:25:55 AM | Example 6.5 [5] | in the solution, the book shows 46.1 g/mol in the calculation when they intended 46.01 g/mol. The answer given, however, is correct. |
| 6 | 192 [212] | Text | Befus | 12/20/2011 1:46:02 PM | Example 6.18 [18] | minor error: 59.9 g TiO2, in part of the worked out answer, the .9 g was dropped |
| 7 | A-10 [656] | Text | Lund | 10/25/2011 10:19:44 AM | 65b | Not really an error, since the answer is technically correct. However, the formula for lead(II) acetate is written as Pb(CH3CO2)2, while the students expect Pb(C2H3O2)2. The text does not show the alternative way for writing the acetate ion prior to this point in the text. |
| 9 | A-13 [357] | Text | Befus | 12/20/2011 1:55:12 PM | 51d | Orbital diagram shows 8 electrons, N only has 7 electrons. |
| 9 | A-13 [357] | Text | Lutz | 10/27/2011 3:02:09 PM | 51 (d) | The answer key shows the electron configuration for oxygen rather than nitrogen. The correct answer should have one electron in each 2P orbital for a total of 3 electrons in the 2P orbitals. |
| 10 | A-15 [395] | Text | Lutz | 11/8/2011 9:59:46 PM | 63 | Answer key is mixed up. the answers should be a) 2 bonding groups 2 lone pairs b) 3 bonding groups 1 lone pair c) 2 bonding groups d) 4 bonding groups |
| 11 | 364 | Text | Lund | 3/17/2014 11:49:25 AM | Table 11.1 | The last table entry should read 29.92 in. Hg, NOT 29.92 atm. |
| 11 | 401 [441] | Text | Lutz | 10/25/2011 9:13:40 AM | 37 | The units for the third entry below V2 should be in mL rather than mmHg. |
| 12 | 442 [486] | Text | Lund | 11/26/2012 4:00:01 PM | 61a | Based on Tro's description of polar molecules on p. 344 [p. 380] (1st bullet point), if none of the bonds in the molecule are polar, then the molecule is nonpolar. Therefore, NCl3 is nonpolar and the ONLY intermolecular forces that act between molecules in dispersion forces. |
| 13 | 454 [502] | Text | Lund | 5/10/2012 4:33:53 PM | Conceptual Checkpoint 13.2 [2] | Choice (c) was probably meant to read "Both (a) and (b)", |
| 13 | 466 [514] | Text | Lund | 12/12/2011 10:02:06 AM | Formula for Molality | The word "solvent" is misspelled as "slovent" in the formula. |
| 13 | 469 [517] | Text | Lutz | 12/19/2012 1:51:39 PM | Figure 13.9 [9] | If you drank saltwater is would make the solution inside your intestines more concentrated, not less concentrated. |
| 13 | 472 [520] | Text | Lund | 11/21/2011 7:46:42 AM | Solution Concentration Summary | The last equation in this section is the calculation for molality (m), not molarity (M). In other words, molality (m) equals moles of solute over kg of solvent. |
| 14 | 524 [576] | Text | Lutz | 12/19/2012 2:00:59 PM | problems 63 and 64 | "If the acid is weak..." should read "if the base is weak..." |
| 14 | A-21 [582] | Text | Befus | 12/20/2011 1:18:30 PM | 49b | SrSO4 is a solid based on the solubility chart in the text on p. 217 [p. 241]. The back of the book has it as aqueous. |
| 14 | A-22 [583] | Text | Lund | 12/12/2011 9:21:37 AM | 109 | Answers for pOH are missing. From top-to-bottom, the pOH values are: 10.00, 11.74, 5.50, 5.68, 6.45. |
| 18 | A-28 [638] | Text | Lund | 11/14/2011 8:42:01 PM | 37d | The condensed structural formula is missing. The answer should include CH3CH3 (numbers should be subscripts). |
| 18 | A-29 [640] | Text | Lund | 11/30/2012 2:53:53 PM | 49 - 4th structure | The name for the fourth problem of #49 should be written as 4,4-diethyl-2,3-dimethylhexane (with a hyphen between diethyl and the number 2). |
| 18 | A-29 [640] | Text | Lund | 11/30/2012 2:59:55 PM | 49 - 4th structure | The structure for the fourth problem of #49 should be written as CH3CH(CH3)CH(CH3)C(CH2CH3)2CH2CH3 (there were too many H's in the original structure). |
| 18 | A-29 [640] | Text | Lutz | 11/23/2011 8:25:48 AM | 59 (d) | Question asks for a structure for 4,4-dimethyl-2-hexane, but the solution shows 4,4-dimethyl-2-hexene. Either the 2 should be dropped and the answer changed or the ending should be changed to ene. |
| 18 | A-29 [640] | Text | Lund | 11/30/2012 2:44:10 PM | 49 - 2nd structure | The correct name is 4-isopropylheptane, since the parent chain has 7 carbon atoms. |