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Unformatted text preview: McCord 11AM ONLY ! ! ! ! Final exam solutions will be available at 6 PM on the HWS. Also refer to our web page for any last minute notices about the exam. c = 3 . 00 10 8 m/s h = 6 . 626 10 34 J s m e = 9 . 11 10 31 kg R = 3 . 29 10 15 s 1 N A = 6 . 022 10 23 mol 1 k = 1 . 381 10 23 J/K g = 9 . 81 m/s 2 d water = 1 . 00 g/mL 1 atm = 1 . 01325 10 5 Pa 1 lbs = 453.6 g 1 in = 2.54 cm E = h c = = h p = h mv 1 2 mv 2 = h = R 1 n 2 1 1 n 2 2 n ( x ) = 2 L 1 2 sin nx L E n = n 2 h 2 8 mL 2 n = 1 , 2 , 3 , P = dhg PV = nRT P total = P A + P B + P C + x A = P A /P total P + a n 2 V 2 ( V nb ) = nRT v rms = 3 RT M 1 / 2 rate of effusion r T M E = q + w H = E + PV E = H P V G = H T S E = H nRT E = q v = nC v T H = q p = nC p T w = P V w = nRT w = nRT ln V 2 V 1 q = + nRT ln V 2 V 1 S = nR ln V 2 V 1 S = q rev /T S = nC v ln T 2 T 1 S = nC p ln T 2 T 1 H rxn = X n H f (prod) X n H f (react) G rxn = X n G f (prod) X n G f (react) S rxn = X nS (prod) X nS (react) S = k ln W S trans = H trans /T trans Monteiro, Dalton Final 1 Due: Dec 14 2007, 6:00 pm Inst: McCord 2 This printout should have 56 questions. Multiplechoice 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 How many electrons can possess this set of quantum numbers: principal quantum number n = 4, magnetic quantum number m = 1? 1. 14 2. 8 3. 4 4. 5. 10 6. 16 7. 18 8. 12 9. 6 correct 10. 2 Explanation: Use the rules for the quantum numbers: If n = 4 then = 0 , 1 , 2 , 3; however, for m = 1, = 1 , 2 , 3. Each of these permit ted sets of values of n , and m specifies ONE orbital: n = 4, = 1, m = 1: 4 p n = 4, = 2, m = 1: 4 d n = 4, = 3, m = 1: 4 f and each orbital can have m s = 1 2 ; i.e. , can hold two electrons. 002 (part 1 of 1) 12 points A mixture of oxygen and helium is 92.3% by mass oxygen. It is collected at atmospheric pressure (745 torr). What is the partial pressure of oxygen in this mixture? 1. 688 Torr 2. 299 Torr 3. 412 Torr 4. 446 Torr correct 5. 333 Torr Explanation: 003 (part 1 of 1) 11 points 500 g of water at 80 C is mixed with 200 g of water at 25 C in an insulated container. Find the final temperature once the mixture has come to equilibrium. 1. 67.3 C 2. 50.5 C 3. 55.5 C 4. 48.2 C 5. 64.3 C correct Explanation: For 500 g, T initial = 80 C For 200 g, T initial = 25 C H = m SH ( T final T initial ) H lost + H gained = 0 0 = (500 g) 4 . 184 J g C ( T f 80) C +(200 g) 4 . 184 J g C ( T f 25) C 0 = 2092 J C T f 167360 J + 836 . 8 J C T f 20920 J 0 = T f (2092 + 836 . 8)J C 188280 J Monteiro, Dalton Final 1 Due: Dec 14 2007, 6:00 pm Inst: McCord...
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This note was uploaded on 03/28/2008 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas at Austin.
 Fall '07
 Fakhreddine/Lyon
 Chemistry

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