Final - Version 070 Final sutcliffe (51060) 1 This...

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Unformatted text preview: Version 070 Final sutcliffe (51060) 1 This print-out should have 41 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Calculate the [OH ] concentration in a 0.5 M solution of potassium fluoride (KF). K a for HF is 7 . 2 10 4 . 1. 0.5 M 2. 6 . 10 6 M 3. 3 . 6 10 11 M 4. 2 . 64 10 6 M correct 5. 6 . 94 10 12 M Explanation: 002 10.0 points NOTE: The mercury ion in this question is Hg 2+ 2 . Treat this unusual ion as a polyatomic ion with a +2 charge. The K sp for mercury(I) iodide is 1.2 10 28 . What is the solubility of mercury(I) iodide? 1. 3 . 1 10 10 correct 2. 1 . 1 10 14 3. 3 . 9 10 10 4. 5 . 2 10 8 Explanation: 003 10.0 points A B Reaction progress Energy(kJ) 250 300 350 Graph I A B Reaction progress Energy(kJ) 250 300 350 Graph II Which graph corresponds to an exothermic process? 1. Both Graph I and Graph II 2. Cannot be determined with the informa- tion given 3. Neither Graph I nor Graph II 4. Graph I 5. Graph II correct Explanation: Exothermic processes end with less energy than they began with; the energy was trans- ferred to the surroundings. 004 10.0 points Yes, its (Ru)thenium. When a ruthenium chloride solution was electrolyzed for 500 s with a 120 mA current, 31 mg of ruthenium was deposited. What is the oxidation number of ruthenium in ruthe- nium chloride? 1. 2 correct 2. 3 3. 5 4. 6 5. 1 6. 4 Explanation: Version 070 Final sutcliffe (51060) 2 t = 500 s m = 31 mg = 0 . 031 g I = 120 mA = 0 . 12 A = 0 . 12 C / s F = 96485 C Ru n + (aq) + ne Ru(s) The amount of Ru is n Ru = (0 . 031 g Ru) parenleftbigg 1 mol Ru 101 . 07 g Ru parenrightbigg = 0 . 000306718 mol Ru . The total charge is q = I t = (0 . 12 C / s) (500 s) = 60 C . n e = (60 C) parenleftbigg 1 mol e 96485 C parenrightbigg = 0 . 000621858 mol e . n = . 000621858 mol e . 000306718 mol = 2 . 02746 mol charge 1 mol e . Therefore, the oxidation number is +2; i.e. , Ru 2+ (aq) + 2 e Ru(s). 005 10.0 points When the rate of the reaction 2 NO(g) + O 2 (g) 2 NO 2 (g) was studied, the rate was found to double when the O 2 concentration alone was doubled but to quadruple when the NO concentration alone was doubled. What is the rate expres- sion for this reaction? 1. rate = k [NO 2 ] [NO] 2. rate = k [NO 2 ] [O 2 ] 2 3. rate = k [NO 2 ] [O 2 ] 4. rate = k [NO] [O 2 ] 2 5. rate = k [NO 2 ] 2 [NO] 6. rate = k [NO 2 ] 2 [O 2 ] 7. rate = k [NO 2 ] [NO] 2 8. rate = k [NO] 2 [O 2 ] correct 9. rate = k [NO] [O 2 ] Explanation: The rate increases in proportion to [O 2 ] 1 and to [NO] 2 , so the rate expression should be rate = k [NO] 2 [O 2 ] ....
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This note was uploaded on 07/25/2011 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas at Austin.

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Final - Version 070 Final sutcliffe (51060) 1 This...

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