This page has all of the required homework for the material covered in the second exam of the first semester of General Chemistry. The textbook associated with this homework is CHEMISTRY The Central Science by Brown, LeMay, et.al. The last edition I required students to buy was the 12th edition (CHEMISTRY The Central Science, 12th ed. by Brown, LeMay, Bursten, Murphy and Woodward), but any edition of this text will do for this course.
Note: You are expected to go to the end of chapter problems in your textbook, find similar questions, and work out those problems as well. This is just the required list of problems for quiz purposes. You should also study the Exercises within the chapters. The exercises are worked out examples of the questions at the back of the chapter. The study guide also has worked out examples.
These are bare-bones questions. The textbook questions will have additional information that may be useful and that connects the problems to real life applications, many of them in biology.
When HCl is added to the solution a precipitate is formed. The precipitate is filtered from the solution. When K2SO4 is added to the filtered solution no precipitate is formed. When NaOH is then added to the filtered solution a precipitate is formed.
Ba2+ | Ni2+ | Pb2+ | |
HCl | BaCl2(aq) | NiCl2(aq) | PbCl2(s) |
K2SO4 | BaSO4(s) | NiSO4(aq) | -- |
NaOH | -- | Ni(OH)2(s) | -- |
A precipitate with HCl means that Pb2+ must be present. No precipitate with K2SO4 means that Ba2+ can not be present. A precipitate with NaOH means that Ni2+ must be present.
Answer: Pb2+ and Ni2+ are in the solution.
When KOH is added to the solution a precipitate is formed. The precipitate is filtered from the solution. When Na2SO4 is added to the filtered solution a precipitate is again formed. This new precipitate is then filtered from the solution. When (NH4)2CO3 is then added to the filtered solution a precipitate is not formed.
Sr2+ | Fe2+ | Ca2+ | |
KOH | Sr(OH)2(aq) | Fe(OH)2(s) | Ca(OH)2(aq) |
Na2SO4 | SrSO4(s) | -- | CaSO4(aq) |
(NH4)2CO3 | -- | -- | CaCO3(s) |
A precipitate with KOH means that Fe2+ must be present. A precipitate with Na2SO4 means that Sr2+ is present. No precipitate with NaOH means that Ca2+ is not present.
Answer: Sr2+ and Fe2+ are in the solution.
molarity = |
| = |
| = 0.5 M |
( | 5 g KMnO4 | ) | ( |
| ) | = 0.0316 mole KMnO4 |
molarity = |
| = |
| = 0.0633 M KMnO4 |
( | 0.25 L soln | ) | ( |
| ) | = 0.3 moles KCl |
( | 1.1 mole LiBr | ) | ( |
| ) | = 0.5 L soln |
( | 0.5 L soln | ) | ( |
| ) | ( |
| ) | = 29.99 g NaOH |
( | 0.250 L soln | ) | ( |
| ) | = 0.05 mole solute |
| = 5 M |
( | 0.1 L soln | ) | ( |
| ) | ( |
| ) | = 0.033 L stock soln |
( | 4.7 L blood | ) | ( |
| ) | ( |
| ) | = 14.05 g Na (a) |
( | 4.7 L blood | ) | ( |
| ) | ( |
| ) | = 12.97 g Na |
( | 4.7 L blood | ) | ( |
| ) | ( |
| ) | = 3.79 g C2H5OH (a) |
( | 3.79 g C2H5OH | ) | ( |
| ) | = 0.19 can Only one fifth of a can! (b) |
( | 0.015 L NiCl2 solution | ) | ( |
| ) | = 0.012 mole NiCl2 |
( | 0.012 mole NiCl2 | ) | ( |
| ) | ( |
| ) | = 1.3 g Na3PO4 |
( | 0.1 L HCl soln | ) | ( |
| ) | = 0.02 mole HCl |
( | 0.02 mole HCl | ) | ( |
| ) | ( |
| ) | = 0.04 L or 40 mL NaOH soln |
( | 0.015 L NaOH soln | ) | ( |
| ) | = 0.0015 mole NaOH |
( | 0.0015 mole NaOH | ) | ( |
| ) | = 0.00075 mole H2SO4 |
[H2SO4] = |
| = | = 0.0375 M H2SO4 |
( | 0.2 L AgC2H3O2 solution | ) | ( |
| ) | = 0.28 mole AgC2H3O2 |
( | 0.28 mole AgC2H3O2 | ) | ( |
| ) | ( |
| ) | = 16.4 g NaCl |
( | 0.1 L AgC2H3O2 soln | ) | ( |
| ) | = 0.14 mole AgC2H3O2 |
( | 0.2 L NaCl soln | ) | ( |
| ) | = 0.12 mole NaCl |
( | 0.12 mole NaCl | ) | ( |
| ) | = 0.12 mole AgC2H3O2 needed |
We need 0.12 mole AgC2H3O2, but have 0.14 mole. We have extra AgC2H3O2 and so NaCl is the limiting reactant. Use the amount of limiting reactant that we have to determine the amount of product produced.
( | 0.12 mole NaCl | ) | ( |
| ) | ( |
| ) | = 17.2 g AgCl |
[Ag+] = |
| = |
| = 0.067 M Ag+ |
[Na+] = |
| = |
| = 0.4 M Na+ |
[C2H3O2-] = |
| = |
| = 0.467 M C2H3O2- |
( | 0.1 L KOH soln | ) | ( |
| ) | = 0.12 mole KOH |
( | 0.15 L ZnCl2 soln | ) | ( |
| ) | = 0.15 mole ZnCl2 |
( | 0.12 mole KOH | ) | ( |
| ) | = 0.06 mole ZnCl2 needed |
We need 0.06 mole ZnCl2, but have 0.15 mole. We have extra ZnCl2 and so KOH is the limiting reactant. Use the amount of limiting reactant that we have to determine the amount of product produced.
( | 0.12 mole KOH | ) | ( |
| ) | ( |
| ) | = 5.96 g Zn(OH)2 |
[Zn2+] = |
| = |
| = 0.36 M Zn2+ |
[K+] = |
| = |
| = 0.48 M K+ |
[Cl-] = |
| = |
| = 1.2 M Cl- |