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Chemistry 301 Final - McCord 11AM ONLY Final exam solutions...

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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 = c = λν λ = h p = h mv 1 2 mv 2 = - Ψ ν = R 1 n 2 1 - 1 n 2 2 ψ n ( x ) = 2 L 1 2 sin nπx 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 Δ E = q + w H = E + PV Δ E = Δ H - P Δ V Δ G = Δ H - T Δ S Δ E = Δ H - Δ nRT Δ E = q v = n C v Δ T Δ H = q p = n C p Δ T w = - P Δ V w = - Δ nRT w = - nRT ln V 2 V 1 q = + nRT ln V 2 V 1 Δ S = nR ln V 2 V 1 Δ S = q rev /T Δ S = n C v ln T 2 T 1 Δ S = n C p ln T 2 T 1 Δ H rxn = X n Δ H f (prod) - X n Δ H f (react) Δ G rxn = X n Δ G f (prod) - X n Δ G f (react) Δ S rxn = X nS (prod) - X nS (react) S = k ln W Δ S trans = Δ H trans /T trans
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Barkley, Thane – Final 1 – Due: Dec 14 2007, 6:00 pm – Inst: McCord 2 This print-out should have 56 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 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. 0 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 Δ H = m · SH · ( T final - T initial ) Δ H lost + Δ 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
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Barkley, Thane – Final 1 – Due: Dec 14 2007, 6:00 pm – Inst: McCord 3 T f = 188280 J (2092 + 836 . 8) J / C = 64 . 3 C 004 (part 1 of 1) 10 points An electron in a 3 d orbital could have which of the following quantum numbers? 1. n = 3; = 1; m = - 1 2. n = 3; = 3; m = 1 3. n = 3; = 0; m = 0 4. n = 3; = 2; m = 0 correct 5. n = 2; = 3; m = 0 6. n = 3; = 2; m = - 3 7. n = 2; = 2; m = 2 Explanation: 3 refers to the principal quantum number n . d corresponds to the subsidiary quantum number = 2. Since = 2, m could be - 2 , - 1 , 0 , 1 or 2.
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