Exam 1 Solutions

# Exam 1 Solutions - Version 406 – Exam 1 – flowers...

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Unformatted text preview: Version 406 – Exam 1 – flowers – (53885) 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Use 2 pencils. No programmable calcula- tors. No books. No notes 001 10.0 points What is the temperature change when 22 . 7 g of MgBr 2 (with- 185 . 6 kJ / mol en- thalpy of solution) is dissolved in 75 g of wa- ter? Assume that the specific heat capacity of the solution is 4 . 18 J K · g . 1. 21.2227 2. 18.811 3. 78.138 4. 54.986 5. 79.1027 6. 52.092 7. 41.3199 8. 65.5973 9. 17.8463 10. 72.9931 Correct answer: 72 . 9931 ◦ C. Explanation: m MgBr 2 = 22 . 7 g m H 2 O = 75 g C = 4 . 18 J K · g MW MgBr 2 = 184 . 113 g / mol Δ H =- 185 . 6 kJ / mol =- 1 . 856 × 10 5 J / mol The number of moles is n = 22 . 7 g 184 . 113 g / mol = 0 . 123294 mol , so the heat released is obtained from n Δ H = mC Δ T Δ T = n Δ H mC =- (0 . 123294 mol) (- 1 . 856 × 10 5 J / mol) (4 . 18 J K · g ) (75 g) = 72 . 9931 K = 72 . 9931 ◦ C . 002 10.0 points Consider the reaction NH 4 (NH 2 CO 2 )(s) → 2 NH 3 (g) + CO 2 (g) . What is the relationship between K and K c ? 1. K c = ( RT ) 3 K 2. K = ( RT ) 2 K c 3. K = RT K c 4. K c = ( RT ) 2 K 5. K = ( RT ) 3 K c correct Explanation: 003 10.0 points A 1 . 47 g sample of a protein of molar mass 1 × 10 5 g / mol is dissolved in 1 kg of water. What is the expected freezing point of the solution? k f = 1 . 86 K · kg / mol. 1. -3.9246e-08 2. -3.3108e-08 3. -2.4738e-08 4. -2.7342e-08 5. -5.0778e-08 6. -2.5296e-08 7. -2.6412e-08 8. -4.1478e-08 9. -4.8546e-08 10. -3.5526e-08 Correct answer:- 2 . 7342 × 10 − 8 ◦ C. Explanation: k f = 1 . 86 K · kg / mol MM = 1 × 10 5 g / mol m = 1 kg m solvent = 1 . 47 g = 0 . 00147 kg n protein = . 00147 kg 1 × 10 5 g / mol = 1 . 47 × 10 − 8 mol . Δ T f = k f m = k f n CO(NH 2 ) 2 m = (1 . 86 K · kg / mol) (1 . 47 × 10 − 8 mol) 1 kg = 2 . 7342 × 10 − 8 K = 2 . 7342 × 10 − 8 ◦ C , Version 406 – Exam 1 – flowers – (53885) 2 and the freezing point will be FP = 0 . ◦ C- 2 . 7342 × 10 − 8 ◦ C =- 2 . 7342 × 10 − 8 ◦ C 004 10.0 points Solutions in which intermolecular forces are stronger in the solution than in the pure com- ponents have negative enthalpies of mixing. 1. True correct 2. False Explanation: 005 10.0 points The reaction N 2 (g) + 3 H 2 (g) ⇀ ↽ 2 NH 3 has an equilibrium constant 0 . 50 at 467 ◦ C. What will eventually happen if 4.50 moles of N 2 , 8.00 moles of H 2 , and 4.00 moles of NH 3 are put in a 10 L vessel and held at 467 ◦ C? 1. Nothing, the system is at equilibrium. 2. More NH 3 will be formed. 3. More N 2 and H 2 will be formed. correct Explanation: [N 2 ] = 4 . 50 mol 10 L [H 2 ] = 8 . 00 mol 10 L [NH 3 ] = 4 . 00 mol 10 L K c = 0 . 50 Concentrations are 0.450 M N 2 , 0.800 M H 2 , and 0.400 M NH 3 . This problem requires that the reaction quotient Q be calculated and compared to the equlibrium constant K c ....
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Exam 1 Solutions - Version 406 – Exam 1 – flowers...

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