CH302 Final - Kanungo Anil Final 1 Due 6:00 pm Inst Donna C...

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Kanungo, Anil – Final 1 – Due: May 12 2007, 6:00 pm – Inst: Donna C Lyon 1 This print-out should have 45 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points If you go into a deep coal mine, say 5000 ft below sea level, the amount of carbon dioxide that will dissolve in water should increase. 1. True correct 2. False Explanation: The carbon dioxide pressure is very great at this depth. 002 (part 1 of 1) 10 points Using the smallest possible integer coefficients to balance the redox equation MnO - 4 + C 2 O - 2 4 Mn +2 + CO 2 (acidic solution), the coefficient for C 2 O 2 - 4 is 1. 2. 2. 4. 3. The correct coefficient is not given. 4. 7. 5. 5. correct Explanation: The oxidation number of C changes from +3 to +4, so C is oxidized. The oxidation num- ber of Mn changes from +7 to +2, so Mn is reduced. We set up oxidation and reduction half-reactions: Red: MnO - 4 Mn 2+ Oxid: C 2 O 2 - 4 CO 2 Mn atoms are balanced. We need 2 CO 2 molecules to balance C: Oxid: C 2 O 2 - 4 2 CO 2 Since this is an acidic solution, we use H 2 O and H + to balance O and H atoms, adding the H 2 O to the side needing oxygen: Red: 8 H + + MnO - 4 Mn 2+ + 4 H 2 O Oxid: C 2 O 2 - 4 2 CO 2 We balance the total charge in each half- reaction by adding electrons. In the preceding reduction reaction there is a total charge of +7 on the left and +2 on the right. Five electrons are added to the left: Red: 5 e - + 8 H + + MnO - 4 Mn 2+ + 4 H 2 O Oxid: C 2 O 2 - 4 2 CO 2 + 2 e - The number of electrons gained by Mn must equal the number of electrons lost by C. We multiply the reduction reaction by 2 and the oxidation reaction by 5 to balance the elec- trons: Red: 10 e - + 16 H + + 2 MnO - 4 2 Mn 2+ + 8 H 2 O Oxid: 5 C 2 O 2 - 4 10 CO 2 + 10 e - Adding the half-reactions gives the overall balanced equation: 5 C 2 O 2 - 4 + 16 H + + 2 MnO - 4 10 CO 2 + 2 Mn 2+ + 8 H 2 O 003 (part 1 of 1) 10 points The following solutions are mixed in equal volumes. I) 0.1 M HCl and 0.1 M NaCl II) 0.1 M HOCl and 0.1 M NaCl III) 0.1 M HOCl and 0.1 M NaOCl IV) 0.1 M HCl and 0.1 M NaOCl Which will give (a) buffer solution(s)? 1. I and II only 2. II and III only 3. III only correct 4. II and IV only 5. IV only Explanation: A buffer must contain a weak acid/base conjugate pair. OCl - is the conjugate base of the weak acid HOCl. HCl/Cl - is a stong acid conjugate pair. 004 (part 1 of 1) 10 points
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Kanungo, Anil – Final 1 – Due: May 12 2007, 6:00 pm – Inst: Donna C Lyon 2 A solution is prepared by mixing equal vol- umes of 0.40 M HF(aq) with 0.40 M KOH(aq). This solution is a buffer. 1. False correct 2. True Explanation: 005 (part 1 of 1) 10 points Ferrous hydroxide is a slightly soluble base. In which of the following would Fe(OH) 2 be most soluble? It is not necessary to know K sp for Fe(OH) 2 .
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