2 lose two electrons and are oxidized 3 gain two

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2. lose two electrons and are oxidized. 3. gain two electrons and are reduced. cor- rect 4. gain two electrons and are oxidized. Explanation: For this question Cu 2+ (aq) Cu(s) For this to happen, the charge must be bal- anced by the Cu 2+ reacting with 2 electrons: Cu 2+ + 2 e Cu(s) This is a reduction reaction (represented by the gain of electrons). 017 10.0 points NOTE: Use a=1 for this question. For a first-order reaction, after 230 s, 33% of the reactants remain. Calculate the rate constant for the reaction. 1. 207 s 1 2. 0.000756 s 1 3. 0.00482 s 1 correct 4. 0.00209 s 1 5. 0.00174 s 1 Explanation: 018 10.0 points A catalyst 1. is used up in a chemical reaction. 2. is always a solid. 3. lowers the activation energy of the reac- tion. correct 4. does not take part in the reaction in any way. 5. changes the value of Δ G of the reaction. Explanation: A catalyst speeds up a chemical reaction by providing an alternate mechanism which re- quires a lower energy of activation. Although the catalyst takes part in the reaction it is not used up. Catalysts may be in solid, liquid, gaseous or aqueous phase and only a small amount is used. 019 10.0 points Consider the voltaic cell In | In 3+ (0 . 010 M) || Ru 3+ (1 . 0 M) , Ru 2+ (0 . 010 M) | C In 3+ + 3 e In E 0 = 0 . 34 V Ru 3+ + 1 e Ru 2+ E 0 = 0 . 08 V The electron flow in the external circuit is from 1. In to Ru 3+ . 2. Ru 3+ to In. 3. In to C. correct 4. In 3+ to C. 5. In 3+ to Ru 2+ .
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Version 024 – Exam 4 – Sutcliffe – (52410) 7 Explanation: 020 10.0 points The solubility of PbF 2 in water is 0 . 0021 mol / L. K sp for PbF 2 is closest to which of the following values? 1. K sp = 9 . 2 × 10 9 2. K sp = 3 . 7 × 10 8 correct 3. K sp = 2 . 1 × 10 3 4. K sp = 4 . 4 × 10 6 5. K sp = 2 . 0 × 10 4 Explanation: solubility of PbF 2 = 0 . 0021 mol / L . PbF 2 Pb 2+ + 2 F 2 [Pb 2+ ] = [F ] K sp = [Pb 2+ ] [F ] 2 = [Pb 2+ ] ( 2 [Pb 2+ ] ) 2 = 4 x 3 = 4 (0 . 0021) 3 = 3 . 7 × 10 8 . 021 10.0 points The un-ionized form of an acid indicator is yel- low and its anion is blue. The K a of this indi- cator is 10 6 . What will be the approximate pH range over which this indicator changes color? 1. 4 6 2. 9 11 3. 5 7 correct 4. 8 10 5. 3 5 Explanation: K a = 10 6 The p K a of this indicator is 6, so the indica- tor will change color around pH 6. Thus you would expect a color change between a pH of 5 and 7. 022 10.0 points NOTE: A strong complex is one in which ev- erything is very tightly bound together. An- other way to put this would be to say that K d for FeEDTA is VERY tiny. Fe(OH) 3 (s) is very insoluble in water ( K sp = 1 . 6 × 10 39 ; however, Fe 3+ forms a strong complex with EDTA (FeEDTA , K f = 1 . 3 × 10 25 ). For a solution which is at equilibrium and contains Fe(OH) 3 (s) precipi- tate, which of the following occurs if EDTA is added to the solution? 1. Nothing happens because K sp is much smaller than K f . 2. More Fe(OH) 3 (s) precipitates because [Fe 3+ ] decreases. 3. More Fe(OH) 3 (s) dissolves. correct 4. No more Fe(OH) 3 (s) dissolves, but Fe 3+ complexes with EDTA. Explanation: 023 10.0 points When we titrate a weak base with a strong acid, the pH at the equivalence point will be 1. pH < 7. correct 2. pH = 7.
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