1 12 m 2 3 m 3 6 m 4 18 m 5 0 m correct explanation

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1. 12 M 2. 3 M 3. 6 M 4. 18 M 5. 0 M correct Explanation: 014 (part 1 of 4) 10.0 points Consider the reaction A(g) + 2 B(g) 2 C(g) + D(g) Δ H = 72 kJ/mol at equilibrium. Which way would the equilib- rium shift if a) the volume of the container were de- creased? 1. left 2. right 3. no shift correct Explanation: 015 (part 2 of 4) 10.0 points b) some D were added to the container? 1. right 2. left correct 3. no shift Explanation: 016 (part 3 of 4) 10.0 points c) some A were removed from the container? 1. left correct 2. no shift 3. right Explanation: 017 (part 4 of 4) 10.0 points d) the temperature were increased?
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Version 349 – Exam 1 – Holcombe – (52460) 5 1. no shift 2. right correct 3. left Explanation: 018 10.0 points Catalase (a liver enzyme) dissolves in water. A 11 mL solution containing 0 . 166 g of cata- lase exhibits an osmotic pressure of 1 . 2 Torr at 20 C. What is the molar mass of catalase? 1. 3 . 16 × 10 5 g / mol 2. 1 . 49 × 10 5 g / mol 3. 3 . 61 × 10 5 g / mol 4. 1 . 94 × 10 5 g / mol 5. 2 . 3 × 10 5 g / mol correct 6. 1 . 81 × 10 5 g / mol Explanation: V = 11 mL = 0 . 011 L m = 0 . 166 g T = 20 C + 273 = 293 K Π = 1 . 2 Torr R = 0 . 08206 L · atm / K / mol Π = n RT V = m MW RT V M = mRT V Π MW = (0 . 166 g) (0 . 08206 L · atm / K / mol) (0 . 011 L)(1 . 2 Torr) × (293 K) parenleftbigg 760 Torr 1 atm parenrightbigg = 2 . 29798 × 10 5 g / mol . 019 10.0 points The colligative effects of 1 molal sugar solu- tion would be ? 1 molal sodium chloride solution. 1. greater than 2. less than correct 3. the same as Explanation: Sugar is a non-electrolyte so the concen- tration of particles is the concentration of dissolved sugar. Soduim chloride is an electrolyte which dis- sociates in solution, giving twice as many par- ticles as dissolved: NaCl -→ Na + + Cl Since colligative properties depend on the number of particles in solution, the colligative effect doubles for the NaCl(aq), compared to any identical concentration of sugar. 020 10.0 points Which of the following two solutions will achieve the greatest boiling point increase and what will the new temperature be? Assume that K b = 0 . 52 C / m for water. I) 105 g sucrose (C 6 H 12 O 6 ) dissolved in 0.5 kg water II) 35 g of NaCl dissolved in 0.5 kg of water 1. Solution II with a boiling point of 101 . 25 C correct 2. Solution I with a boiling point of 101 . 92 C 3. Solution II with a boiling point of 100 . 63 C 4. Solution I with a boiling point of 100 . 32 C Explanation: First calculate the molality of each solution, which is mol solute per kg solvent. To do this, you will need to convert from mass to mols, using molar mass. Remember also that i = 2 for NaCl and i = 1 for sucrose. Using Δ T = imK b you will find that Δ T = 0 . 32 C for the sucrose solution and 1.25 for the salt
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Version 349 – Exam 1 – Holcombe – (52460) 6 solution, which means the boiling point of the salt solution will be 101.25 C. 021 10.0 points Given the hypothetical reaction X(g) Y(g) predict what will happen when 1.0 mol Y is placed into an evacuated container.
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