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Unformatted text preview: McCord 10am Please read the following!! If all goes according to plan this evening, your scores for the final should be available by midnight tonight on Quest. Solu tions will not be available until 9AM on Sunday morning. Double check your ver sion number and make sure it (and your UTEID) are bubbled correctly. R = 8 . 314 J/mol K R = 62 . 36 L torr/mol K 1 L atm = 101.325 J 1 F = 96485 C G = H T S T f = k f m T b = k b m P A = x A P A , pure = cRT ln parenleftbigg P 2 P 1 parenrightbigg = H vap R parenleftbigg 1 T 1 1 T 2 parenrightbigg ln parenleftbigg K 2 K 1 parenrightbigg = H rxn R parenleftbigg 1 T 1 1 T 2 parenrightbigg ln parenleftbigg k 2 k 1 parenrightbigg = E a R parenleftbigg 1 T 1 1 T 2 parenrightbigg G = nFE G = RT ln K anode  anode solution  cathode solution  cathode E = E  RT nF ln Q E = E  . 05916 n log Q I t n F = moles ln parenleftbigg [A] [A] parenrightbigg = kt 1 [A] 1 [A] = kt [A] [A] = kt Version 208 Final Exam McCord (53575) 2 This printout should have 52 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points The first order rate constant for A products is 7 . 2 s 1 . If [A] = 2 . 9 M at t = 0, what is the value of [A] at t = 0 . 5 s? 1. 1 . 58478 M 2. . 0158478 M 3. 3 . 96194 10 5 M 4. 1 . 98097 10 5 M 5. . 0792388 M correct Explanation: [A] = 2 . 9 M t = 0 . 5 s k = 7 . 2 s 1 If the reaction is first order, the integrated rate equation is ln parenleftbigg [A] [A] parenrightbigg = ak t ln[ A ] ln[ A ] = ak t ln[ A ] = ln[ A ] ak t = ln(2 . 9 M) (1) ( 7 . 2 s 1 ) (0 . 5 s) = 2 . 53529 M [ A ] = 0 . 0792388 M 002 10.0 points If you go to the top of a mountain, the amount of carbon dioxide that will dissolve in water should increase compared to sea level. 1. False correct 2. True Explanation: Gas pressures are low at high altitudes. 003 10.0 points What is K sp for Hg 2 Cl 2 , if its molar solu bility is 5 . 2 10 7 mol / L? 1. 6 . 9 10 9 2. 5 . 6 10 19 correct 3. 5 . 2 10 19 4. 1 . 4 10 21 5. 8 . 3 10 17 6. 0.0053 7. 1 . 7 10 14 Explanation: S = 5 . 2 10 7 mol / L The solubility equilibrium is Hg 2 Cl 2 (s) Hg 2+ 2 (aq) + 2 Cl (aq) [Hg 2+ 2 ] = S = 5 . 2 10 7 mol / L [Cl ] = 2 S = 1 . 04 10 6 mol / L K sp = [Hg 2+ 2 ][Cl ] 2 = (5 . 2 10 7 ) ( 1 . 04 10 6 ) 2 = 5 . 62432 10 19 004 10.0 points Which of the following indicators would be most suitable for the titration of 0.10 M (CH 3 ) 3 N(aq) with 0.10 M HClO 4 (aq)? For trimethyamine, p K b = 4.19....
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This note was uploaded on 02/21/2010 for the course CH 5 taught by Professor Staff during the Spring '08 term at University of Texas.
 Spring '08
 Staff

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