#Chem 161-2005 Exam II + answers

#Chem 161-2005 Exam II + answers - Chemistry 161 Exam II...

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Chemistry 161 Exam II October 30, 2005 Student Name (Print): ___TAVSS_________________ Recitation Section Number: t Recitation Instructor: _____TAVSS_____________ The exam booklet has 25 questions for credit and one additional question to check the color of your exam booklet. Please answer all 26 questions on the OpScan sheet. There is no penalty for guessing. Answer each question with the best choice from those provided. At the end of the 80-minute exam period, please hand in only this top sheet and your OpScan form. If you finish early, please do not disturb your fellow students. A proctor will check your picture ID, OpScan form, signature and calculator during the exam. The use of calculators with permanent memories (graphing calculators), cell phones, pagers, PDAs or other electronic devices other than a basic scientific calculator is expressly forbidden. The last page of the booklet contains a periodic table along with other useful data. The use of any other notes or information on this test will be considered a violation of the Academic Honesty provisions of the student code. Exam scores will be posted as soon as possible. ON THE OpScan FORM (Use a #2 pencil or darker) Sec Per. Instr Sec Per. Instr 1. SIGN your name across the top of the form. 01 M3 LH 22 W1 ET 2. Code the following information ( blacken circles) 02 M5 LH 23 W1 ME Your Name (LAST NAME FIRST) 03 M5 RA 24 W3 NM Your SOCIAL SECURITY NUMBER 04 W3 RA 25 Th1 ET [Start under Box A and continue to Box I ] 05 W5 RA 26 Th1 ME Your RECITATION SECTION NUMBER in K & L 06 W5 LH 27 Th3 ME [Sections 01-08, code a 0 under box K ] 07 F3 HS 28 Th3 ET Your EXAM FORM NUMBER under box P 08 F5 HS 29 Th3 NM 09 F5 TH 30 M6 RA LH = L. Huebner, RP = R. Porcja, RA = R. Agarwal 18 M1 NM 31 M6 TH HS = H. Sangari, TH = T. Hyde, NM = N. Marky 19 M1 HS 32 M6 RP ME = M. Esguerra, ET = E. Tavss 20 M3 NM 33 W6 RP Periods: 1 = 9:30-10:25, 3 = 12:15-1:10 or 12:50-1:45 21 M3 HS 35 W8 RP 5 = 3:35-4:30, 6 = 5:15-6:10, 8 = 8:25p-9:20p 36 W8 TH Your EXAM FORM is C161f05e2v1 1
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1 Chem 161-2005 Hourly Exam II Chapter 6 – Thermodynamics Enthalpy H 2 (g) + Br 2 (g) 2HBr(g) H = -104 kJ The information given in the equation above means that: A. 104 kJ is released for each mole of HBr formed B. 104 kJ is absorbed for each mole of HBr formed C . 52 kJ is released for each mole of HBr formed D. 52 kJ is absorbed for each mole of HBr formed Ε. ∆ H f for HBr(g) = -104 kJ A negative H means that the reaction is exothermic, i.e., heat is released. H of -104 kJ means that 104 kJ of heat is released when 1 mol of H 2 reacts with 1 mol of Br 2 to form 2 moles of HBr. Therefore, 52 kJ of heat is released when 1 mol of HBr is formed. 2 Chem 161-2005 Hourly Exam II Chapter 5- Gases Kinetic molecular theory/Effusion and Diffusion/Real gases Under identical conditions, gas X escapes through a small hole 1.48 times faster than CO 2 ? Which one of the following is gas X?
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