Final exam 3-21-07

Final exam 3-21-07 - Chemistry 1B Final Examination VERSION...

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Chemistry 1B, Final Examination VERSION A March 23, 2007 Professor Watts Instructions: Your exam booklet should contain 50 multiple choice questions on 12 numbered pages. Check your booklet to be sure that it includes all 50 questions and 12 numbered pages. If not, return it to a proctor for a replacement. The 50 multiple choice questions are each worth 4 points credit for a correct answer. Questions must be answered on the scan tron exam sheet that is provided with the exam. Fill out the bubbles for the 50 questions on the answer sheet. Be sure to mark your perm number in the bubbles on the sheet and to print your name and perm number on the scan tron . The version of your exam must be marked on the scan tron answer sheet to assure that it is graded correctly. The exam version of this booklet is VERSION A. When you have completed your exam, turn in only the scan tron answer sheet. Keep the exam booklet with questions for your own records. Information sheets with formulas, values of constants, standard half cell potentials and a periodic table are included in your test booklet YOUR EXAM IS VERSION A Mark “A” on the scan tron in the section called “test form”
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Formulas and Constants, Final Exam, 2007 electron mass, m e = 9.10939 × 10 -31 kg proton mass, m p = 1.67262 × 10 -27 kg neutron mass, m n = 1.67493 × 10 -27 kg E R (Rydberg constant) = -2.178 ×10 -18 J h = 6.626 x 10 -34 J·s h/2 п = 1.055 x 10 -34 J·s c = 2.997 x 10 8 m/s ( ) S Δ T H Δ G Δ ln k S Δ V V ln nR S Δ T q S Δ T Δ nC H Δ T Δ nC E Δ V Δ P w T Δ nC q T Δ nC q T Δ nC q V Δ P E Δ H Δ PV Δ E Δ H Δ w q E Δ B i f rev P V ext P P V V = = = = = = = = = = + = + = + = Coulombs 500 , 96 F K / J 10 x 38 . 1 k J 184 . 4 cal 1 J 101 atm L 1 mol K atm L 0821 . 0 R mol K J 31 . 8 R 23 B = = = = = = G = G o + RT lnQ G o = -RT lnKeq E = E o –(RT/nF) lnQ E = E o – (0.059/n) logQ E o =( 0.059/n) log Keq [ ] [ ] inside outside mem x x log mV Z 62 ) x ( E × = E 0 cell = E 0 1/2 (cathode) – E 0 1/2 (anode) [A] = [A] o e – k t ln [A] = ln [A] o – k t kt A 1 A 1 0 = k t 1/2 = ln 2 = 0.693 2 4 h p x h = π Δ Δ m 2 v x h Δ Δ × = 2 18 1 10 178 . 2 n J E n λν = c E h ν = E binding + E kinetic E = h ν 2 2 / 1 mv E kinetic = mv h = λ binding kinetic E h E = ν E n = E R (Z 2 /n 2 ) T K nm 10 x 898 . 2 6 max = λ 2 2 4 2 2 n e Z m E e n h =
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Final Exam - Chem 1B - Version A 3-23-07 1. 1.00 L of Helium gas is placed in a closed container that is insulated to prevent heat exchange. The container is fitted with a piston to allow the pressure to be changed. The initial pressure of the gas is 2.00 atm and it is initially at a temperature of 298 K. The pressure of the gas is then reduced in 100 small steps until a final pressure of 1.00 atm is reached. The internal energy ( E), work (w) and heat (q) changes for this process are a) E negative, w negative, and q positive b) E no change, w negative, and q positive c) E positive, w negative and q no change d) E no change, w positive, and q negative e) None of the above
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