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MUexam1dvb - Version 001 Make Up Exam 1 VANDEN BOUT(53585...

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Version 001 – Make Up Exam 1 – VANDEN BOUT – (53585) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The phase diagram for a pure substance is given below. Solid Liquid Vapor 50 100 150 200 250 300 100 200 300 400 Temperature, K Pressure, atm What pressure must be applied to liquefy a sample at 425 K? 1. 250 atm 2. 50 atm 3. 350 atm 4. The sample cannot be liquefied at 425 K. correct 5. 150 atm Explanation: 002 10.0 points Everyone should recognize that ? is a strong acid, ? is a weak acid, ? is a strong base, and ? is a salt. 1. HCl; H 2 SO 4 ; NaOH; NaBr 2. HCl; H + ; NaOH; NaCl 3. HNO 3 ; CH 3 COOH; NaOH; MgO 4. HCl; CH 3 COOH; NaOH; NaBr correct 5. HCl; CH 4 ; NaOH; NaCl Explanation: Of the answer choices, hydrochloric acid (HCl), nitric acid (HNO 3 ), and sulfuric acid (H 2 SO 4 ) are strong acids because they dis- sociate completely in water. Acetic acid (CH 3 COOH) is a weak acid due to its par- tial dissociation. Sodium hydroxide (NaOH) and magnesium oxide (MgO) are strong bases, dissociating completely to yield hydroxide (OH ). Sodium bromide (NaBr) and sodium chloride (NaCl) are salts, compounds com- posed of a cation (metal) other than H + and an anion (non-metal) other than OH or O 2 . 003 10.0 points When 20.0 grams of an unknown nonelec- trolyte compound are dissolved in 500 grams of benzene, the freezing point of the result- ing solution is 3.77 C. The freezing point of pure benzene is 5.48 C, and its freezing point depression constant is K f = 5 . 12 C/molality. What is the molecular weight of the unknown compound? 1. 0.334 grams/mole 2. 100 grams/mole 3. 120 grams/mole correct 4. 160 grams/mole 5. 80.0 grams/mole 6. 140 grams/mole 7. 54 grams/mole 8. 240 grams/mole Explanation: m unknown = 20 g T f = 3 . 77 C m benzene = 500 g T 0 f = 5 . 48 C K f = 5.12 C/ m
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Version 001 – Make Up Exam 1 – VANDEN BOUT – (53585) 2 Δ T f = T 0 f - T f = 5 . 48 C - 3 . 77 C = 1 . 71 C Δ T f = K f · m m = Δ T f K f = 1 . 71 C 5 . 12 C / m = 0 . 334 m m = mol compound kg benzene = g cpd / MW cpd kg benzene MW cpd = g cpd kg benzene × m = 20 . 0 g cpd 0 . 500 kg benzene × 1 kg benzene 0 . 334 mol compound = 120 g / mol 004 10.0 points For the reaction POCl 3 (g) POCl(g) + Cl 2 (g) K c = 0 . 30. An initial 0 . 21 moles of POCl 3 are placed in a 2 . 5 L container with initial concentrations of POCl and Cl 2 equal to zero. What is the final concentration of POCl 3 ? ( Note: You must solve a quadratic equation.) 1. final concentration = 0 . 226 M 2. final concentration = 0 . 0311934 M 3. final concentration = 0 . 068 M 4. final concentration = 0 . 142 M 5. final concentration = 0 . 0155967 M cor- rect Explanation: K c = 0.30 V container = 2 . 5 L [POCl 3 ] initial = 0 . 21 mol 2 . 5 L = 0 . 084 M POCl 3 (g) POCl(g) + Cl 2 (g) ini, M 0 . 084 - - Δ, M - x x x eq, M 0 . 084 - x x x K c = [Cl 2 ] [POCl] [POCl 3 ] = x 2 0 . 084 - x = 0 . 3 x 2 = 0 . 0252 - 0 . 3 x x 2 + 0 . 3 x - 0 . 0252 = 0 x = - 0 . 3 ± radicalbig (0 . 3) 2 - 4 (1) ( - 0 . 0252) 2 (1) = 0 . 0684033 or - 0 . 368403 Reject - 0 . 368403 as x because it leads to negative concentrations for POCl and Cl 2 and a concentration larger that the orig- inal concentration for POCl 3 . Therefore x = 0 . 0684033 M and [POCl 3 ] = 0 . 084 M - 0 . 0684033 M = 0 . 0155967 M 005 10.0 points Arrange the bases I) deuterated ammonia (ND 3 ) K b = 1 . 1 × 10 5 ; II) hydroxylamine (NH 2 OH) K b = 1 . 1 × 10
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