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# FINAL - Version 086 Final Exam Holcombe(52460 This...

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Version 086 – Final Exam – Holcombe – (52460) 1 This print-out should have 50 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points 0 10 20 30 40 50 60 0 1 2 3 4 5 6 7 8 9 10 11 12 Titration Curve mL of NaOH pH What is the pH at the equivalence point of this titration? 1. 4 . 32 2. 9 . 09 correct 3. 7 . 26 4. 5 . 43 5. 10 . 54 6. 6 . 16 7. 3 . 21 Explanation: The inflection points are shown below. 0 10 20 30 40 50 60 0 1 2 3 4 5 6 7 8 9 10 11 12 Titration Curve mL of NaOH pH (17 . 5, 5 . 43) (35, 9 . 09) 002 (part 2 of 2) 10.0 points What is the p K a of this acid? 1. 5 . 43 correct 2. 4 . 32 3. 6 . 16 4. 9 . 09 5. 7 . 26 6. 3 . 21 7. 10 . 54 Explanation: 003 10.0 points The equilibrium constant (dimensionless) for A B is 2. If the initial concentrations of A and B are each 1.0 M, what are the final concentra- tions of A and B respectively? 1. [A] = 1.00 M; [B] = 1.33 M 2. [A] = 0.67 M; [B] = 0.33 M 3. [A] = 1.00 M; [B] = 2.00 M

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Version 086 – Final Exam – Holcombe – (52460) 2 4. [A] = 0.50 M; [B] = 1.00 M 5. [A] = 0.67 M; [B] = 1.33 M correct Explanation: K = 2 [A] ini = 1 . 0 M [B] ini = 1 . 0 M Q = [B] [A] = 1 . 0 1 . 0 = 1 . 0 < K = 2 Therefore, the equilibrium moves to the right. A B(g) ini, M 1 . 0 1 . 0 Δ, M - x x eq, M 1 . 0 - x 1 . 0 + x K = [B] [A] = 2 1 . 0 + x 1 . 0 - x = 2 1 . 0 + x = 2 . 0 - 2 x x = 0 . 33 [A] = 1 . 0 - x = 0 . 67 M [B] = 1 . 0 + x = 1 . 33 M 004 10.0 points Which of the following gases are more soluble in water when their partial pressure above the solution is increased? HI , SO 3 , NH 3 , O 2 , NO 2 , HF , Cl 2 1. Cl 2 , NO 2 , SO 3 , HI 2. all of them correct 3. HI , NH 3 4. NO 2 , O 2 , HI 5. NO 2 , SO 3 6. SO 3 , NO 2 , HI Explanation: Henry’s Law states that the solubility of gases increase as their partial pressure above the solvent increases. 005 10.0 points For the reaction cyclopropane(g) propene(g) at 500 C, a plot of ln[cyclopropane] vs t gives a straight line with a slope of - 0 . 00067 s 1 . What is the order of this reaction and what is the rate constant? 1. second order; 6 . 7 × 10 4 M 1 · s 1 2. second order; 6 . 7 × 10 2 M 1 · s 1 3. first order; 6 . 7 × 10 2 s 1 4. first order; 6 . 7 s 1 5. None of these 6. second order; 6 . 7 M 1 · s 1 7. first order; 6 . 7 × 10 4 s 1 correct Explanation: 006 10.0 points When the reaction CO 2 (g) + H 2 (g) H 2 O(g) + CO(g) is at equilibrium at 1800 C, the equilib- rium concentrations are found to be [CO 2 ] = 0.24 M, [H 2 ] = 0.24 M, [H 2 O] = 0.48 M, and [CO] = 0.48 M. Then an additional 0 . 45 moles per liter of CO 2 and the same amount of H 2 are added. When the reaction comes to equilibrium again at the same temperature, what will be the molar concentration of CO? 1. 0 . 873 M 2. 0 . 82 M 3. 0 . 86 M 4. 0 . 793 M 5. 0 . 78 M correct
Version 086 – Final Exam – Holcombe – (52460) 3 Explanation: [CO 2 ] = 0.24 M [H 2 ] = 0.24 M [H 2 O] = 0.48 M [CO] = 0.48 M K = [H 2 O] [CO] [CO 2 ] [H 2 ] = (0 . 48 M) (0 . 48 M) (0 . 24 M)(0 . 24 M) = 4 After the addition of the CO 2 and H 2 , [CO 2 ] ini = 0.24 M + 0 . 45 M = 0 . 69 M [H 2 ] ini = 0.24 M + 0 . 45 M = 0 . 69 M CO 2 (g) + H 2 (g) H 2 O(g) + CO(g) 0 . 69 0 . 69 0.48 0.48 - x - x + x + x 0 . 69 - x 0 . 69 - x 0 . 48 + x 0 . 48 + x (0 . 48 + x )(0 . 48 + x ) (0 . 69 - x )(0 . 69 - x ) = 4 0 . 48 + x 0 . 69 - x = 4 = 2 0 . 48 + x = 2(0 . 69 - x ) = 1 . 38 - 2 x 3 x = 0 . 9 x = 0 . 3 [CO] = (0 . 48 + x ) M = 0 . 78 mol / L 007 10.0 points Zinc Copper 1 . 10 V V Voltmeter e e 1 M Zn 2+ (aq) 1 M Cu 2+ (aq) Salt bridge to carry ions In this electrochemical cell, what is the an- ode?

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FINAL - Version 086 Final Exam Holcombe(52460 This...

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