winter final exam 2005

winter final exam 2005 - CHEM 14B-1 Instructor Dr Laurence...

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CHEM 14B-1 YOUR NAME ................................ Instructor: Dr. Laurence Lavelle ID# ............................... WINTER 2005 FINAL EXAM (Total number of pages = 13) (Total points = 100) (Total time = 3 hours) You may Carefully remove pages 2 and 3 (Constants & Formulas, Periodic Table). Check that you have 13 pages. If you want your final exam mailed to you, then clearly print your name and address and affix three 37 cent stamps on the back of the last page. Or Final exams will be available next quarter from 3056 Young Hall. YOUR DISCUSSION SECTION. ..................................... YOUR TA IS: John-Carl Olsen Sarah Angelos Cathy Skibo Sanaz Kabehie Multiple choice questions may have one or more answers and points will be deducted for incorrect answers. Write in pen. Show all your work. Check your units and significant figures. Check that all reactions are balanced. Think clearly. Good Luck
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Constants and Formulas Planck's constant, h = 6.63 x 10 –34 J · s Boltzmann's constant, k = 1.38 x 10 –23 J.K -1 Avogadro's constant, N A = 6.02 x 10 23 mol -1 Faraday's constant, F = 96,485 C.mol -1 Gas constant, R = 8.314 J.K -1 .mol -1 = 8.206 x 10 -2 L.atm.K -1 .mol -1 = 62.364 L.Torr.K -1 .mol -1 Speed of light, c = 3.0 x 10 8 m.s 1 Specific heat capacity of water = 4.18 J. ° C -1 .g -1 log 10 (X) = 2.303 ln(X) 1 kcal = 4.18 kJ 1 A = 1 C.s -1 For water: H fus = 40.7 kJ.mol -1 H vap = 6.02 kJ.mol -1 Cp(vap) = 36.4 J.K -1 .mol -1 Cp(liq) = 75.3 J. K -1 .mol -1 Cp(solid) = 36.0 J. K -1 .mol -1 0 ° C = 273.15 K 1L = 1 dm 3 1 atm = 101.325 kPa π = 3.14 E = h ν c = λ v E = q + w q = n C T W = - P x V S = k B ln W P n e -E n k B T PV = nRT w = - V1 V2 PdV = -nRT ln V 2 V 1 S = q REV T S T1 ---> T2 = T1 T2 dq REV T = n C ln T 2 T 1 G ° = H ° - T S ° G ° = - RT ln K G = G ° + RT ln Q G ° = - n F E ° E CELL = E ° - 0.0592 n LOG K d[A] [A] = -k dt ln [A] = -k t + ln [A] o t 1/2 = 0.693 k - d[A] [A] 2 = k dt 1 [A] = k t + 1 [A] o t 1/2 = 1 k [A] o d[A] = - k dt [A] = - k t + [A] 0 t 1/2 = [A] o 2 k k = A exp( -E A RT ) ln k = - E A RT + ln A Fe 3+ (aq) + e - Æ Fe 2+ (aq) Eº = 0.77 V I 2 (s) + 2 e - Æ 2 I - (aq) Eº = 0.54 V V 2+ (aq) + 2e - Æ V(s) Eº = -1.18 V Sn 2+ (aq) + 2e Æ Sn(s) Eº = -0.14 V 2 H + (aq) + 2 e Æ H 2(g) Eº = 0.0 V
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Periodic Table
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Do not write on this page. QUESTION SCORE 1 2 3 4 5 6 7 8 TOTAL
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Q1A. When 1.00 L of 1.00 M Ba(NO 3 ) 2 solution at 25.0 o C is mixed with 1.00 L of 1.00 M Na 2 SO 4 solution at 25.0 o C in a calorimeter, the white solid BaSO 4 forms (at 100% yield) and the temperature of the mixture increases to 29.2 o C. Assuming that the calorimeter absorbs only a negligible quantity of heat, that the specific heat capacity of the solution is 4.18 J/ o
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winter final exam 2005 - CHEM 14B-1 Instructor Dr Laurence...

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