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CHM 116 Practice Exam 3 w KEY

CHM 116 Practice Exam 3 w KEY - CHM 115/116 Exam 3 Practice...

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CHM 115/116 Exam 3 Practice Exam Chapters 20 and 24 1. You are responsible for the information on this page. Please read it carefully. 2. Print and code both your name and 10-digit affiliate ID on the scantron sheet. The affiliate ID is the second sequence of numbers on your University ID card. 3. Use only a #2 pencil. 4. Do all calculations on the exam pages. Do not make any unnecessary marks on the answer sheet. 5. This exam consists of 25 multiple choice questions worth 4 points each and a periodic table. Make sure you have them all. 6. Choose the best answer to each of the questions and answer it on the computer-graded answer sheet. Read all responses before making a selection. 7. When you are finished, turn in only your scantron. You my keep the exam questions. Answer keys will be posted on Blackboard and in the display case this afternoon. Potentially useful information: T = t + 273.15 K c = k f /k r PV = nRT aA + bB cC + dD K c = [C] c [D] d /[A] a [B] b Δ E = q + w, Δ E = Δ H - P Δ V K p = K c (RT) Δ n(gas) w = -P Δ V = - Δ n gas RT R = 0.08206 L atm/mol K = 8.314 J/mol K ln(K 1 /K 2 ) = ( Δ H ° /R)(T 2 -1 - T 1 -1 ) Δ S vap = Δ H vap /T boil , Δ S fus = Δ H fus /T melt ax 2 + bx + c = 0, x = {-b ± (b 2 - 4ac) 1/2 }/2a Δ S ° = Σ nS ° products - Σ nS ° reactants Q = π [products] a / π [reactants] b Δ H ° = Σ n Δ H ° f (products) - Σ n Δ H ° f (reactants) Q = π P products a / π P reactants b Δ G ° = Σ n Δ G ° f (products) - Σ n Δ G ° f (reactants) K w =[H 3 O + ][OH - ] = 1.0x10 -14 at 25 ° C Δ G = Δ H - T Δ S, Δ G ° = Δ H ° - T Δ S ° pH = - log [H 3 O + ]; pX= - log X Δ G = Δ G ° + RT ln Q pH + pOH = pK w = 14 Δ G ° = -RT ln K K w = K a K b Zero order: - Δ [A]/ Δ t = k, [A] = [A] o - kt, pH = pKa + log ([A - ]/[HA]) t ½ = [A] o /2k E ° rxn =E ° cell = E ° ox + E ° red First order: - Δ [A]/ Δ t = k[A], ΔG ° = - n F E ° ; ΔG= - n F E ln[A] = ln[A] o - kt, = 96,500coul/mol e - t ½ = 0.693/k E = E ° - 0.05916/n log Q Second order: - Δ [A]/ Δ t = k[A] 2 , N t =N o e -kt ; ln(N t /N o ) = -kt; t 1/2 = 0.693/k [A] -1 = [A] o -1 + kt, E=mc 2 t ½ = 1/k[A] o c =2.9979x10 8 m/s Δ H ° = E a,f - E a,r 1 amu = 1.66056 x10 -27 kg k = Ae -Ea/RT 1 J= 1 kg m 2 /s 2 ln (k 1 /k 2 ) = (E a /R)(T 2 -1 - T 1 -1 ) 1 g = 6.022x10 23 amu Mass of a proton = 1.00728 amu Mass of a neutron = 1.00866 amu
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