Chem302_Laude_Exam1_Spring09

Chem302_Laude_Exam1_Spring09 - Version 295 Exam 1...

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Version 295 – Exam 1 – Laude – (52390) 1 This print-out should have 29 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 6.0 points Write the equilibrium expression for the following reaction: 2 Fe ( s ) + 3 2 O 2 ( g ) ←→ Fe 2 O 3 ( s ) 1. K = 1 P 3 / 2 O 2 correct 2. K = P 2 Fe P O 2 3 / 2 · P Fe 2 O 3 3. K = P Fe P O 2 · P Fe 2 O 3 4. K = 1 P O 2 Explanation: Set up K , products in the numerator, reac- tants in the denominator, all raised to respec- tive stoichiometric coefficients. 002 6.0 points Consider the following statements. Which are true? I) Vapor pressure is an equilibrium phe- nomenon. II) The smaller the IMF, the smaller the vapor pressure. III) The volume of a liquid does not affect the vapor pressure IV) Vapor pressure is temperature depen- dent. 1. I, II, III, and IV 2. III, and IV 3. I, II, and IV 4. I, and III 5. I, II, and III 6. II, III, and IV 7. I, III, and IV correct Explanation: As vapor particles encounter the surface of a solution and condense, their energy released makes another particle move as a vapor. This is an equilibrium process, and it is tempera- ture dependent. As IMF decreases, the va- por pressure increases, since only weak forces will be holding a substance together as a liq- uid. Vapor pressure also only occurs at the surface, so it does not matter how deep the liquid phase is. 003 6.0 points What would be the pH of a solution of hypo- bromous acid (HOBr) prepared by dissolving 9.7 grams of the acid in 20 mL of pure wa- ter (H 2 O)? The Ka of hypobromous acid is 2 × 10 9 1. 6 2. 10 3. 13 4. 1 5. 4 correct Explanation: 9 . 7 g HOBr × 1 mol 97 g = 0 . 1 mol HOBr 0 . 1 mol HOBr 0 . 02 L H 2 O = 5 M HOBr [H + ] = (K a · C a ) 1 / 2 = (2 × 10 9 · 5) 1 / 2 = (10 8 ) 1 / 2 = 10 4 pH = - log[H + ] = - log(10 4 ) = 4 004 6.0 points Assuming all of the following salts dissolve completely in water, which one would be the best to use if you were trying to raise the
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Version 295 – Exam 1 – Laude – (52390) 2 boiling point of the solution? 1. KNO 3 2. (NH 4 ) 2 Cr 2 O 7 3. Al 2 (SO 4 ) 3 correct 4. Na 3 PO 4 5. NaCl Explanation: Al 2 (SO 4 ) 3 is highly soluble and produces 5 ions upon dissolving, has the largest van’t hoff coefficient and is the best choice for at- tempting raise a solution’s boiling point. 005 6.0 points Rank following salts from most to least solu- ble: BiI K sp = 7 . 7 × 10 19 Cd 3 (AsO 4 ) 2 K sp = 2 . 2 × 10 33 AlPO 4 K sp = 9 . 8 × 10 21 CaSO 4 K sp = 4 . 9 × 10 5 1. Cd 3 (AsO 4 ) 2 < CaSO 4 < AlPO 4 < BiI 2. BiI < Cd 3 (AsO 4 ) 2 < CaSO 4 < AlPO 4 3. CaSO 4 < AlPO 4 < BiI < Cd 3 (AsO 4 ) 2 4. AlPO 4 < BiI < Cd 3 (AsO 4 ) 2 < CaSO 4 correct Explanation: Molar solubility can be approximated by taking the n th root of the K sp where n is the number of ions in the salt. Doing so results in approximate molar solubilities of 10 10 , 10 7 , 10 11 and 10 3 for bismuth iodide, cadmium arsenate, aluminum phosphate and calcium sulfate, respectively. Arranging these from least to greatest produces: AlPO 4 < BiI < Cd 3 (AsO 4 ) 2 < CaSO 4 .
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