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# chexam3 - Version 167 Exam 3 Holcombe(52460 This print-out...

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Version 167 – Exam 3 – Holcombe – (52460) 1 This print-out should have 33 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Consider the cell Ag(s) | Ag + (aq , 0 . 100 M) || Ag + (aq , 0 . 100 M) | Ag(s) . What is the voltage of this cell? 1. 0 correct 2. +0.0296 V 3. +0.80 V 4. +0.0592 V Explanation: 002 10.0 points A battery formed from the two half re- actions below dies (reaches equilibrium). If [Fe 2+ ] was 0 . 24 M in the dead battery, what would [Cd 2+ ] be in the dead battery? Half reaction E Fe 2+ -→ Fe - 0 . 44 Cd 2+ -→ Cd - 0 . 40 1. 0 . 01 M correct 2. 5 . 4 M 3. 120 . 3 M 4. 0 . 0 . 0005 M Explanation: E = +0 . 04 E cell = E cell - 0 . 05916 N e log Q 0 = 0 . 04 - 0 . 05916 2 log 0 . 24 [Cd 2+ ] log 0 . 24 [Cd 2+ ] = 1 . 35 0 . 24 [Cd 2+ ] = 10 1 . 35 [Cd 2+ ] = 0 . 24 10 1 . 35 = 0 . 0107 M 003 10.0 points What is the E cell of Zn(s) | Zn 2+ (aq) || Ce 4+ (aq) | Ce 3+ (aq) Zn 2+ + 2 e Zn E red = - 0 . 76 Ce 4+ + e Ce 3+ E red = +1 . 61 1. - 2 . 37 2. +2 . 37 correct 3. - 0 . 85 4. +0 . 85 5. +1 . 61 Explanation: 004 10.0 points 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 equi- librium and contains Fe(OH) 3 (s) precipitate, which of the following occurs if EDTA is added to the solution? 1. No more Fe(OH) 3 (s) dissolves, but Fe 3+ complexes with EDTA. 2. More Fe(OH) 3 (s) precipitates because [Fe 3+ ] decreases. 3. More Fe(OH) 3 (s) dissolves. correct

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Version 167 – Exam 3 – Holcombe – (52460) 2 4. Nothing happens because K sp is much smaller than K f . Explanation: 005 10.0 points Consider the cell Zn | Zn 2+ (1 M) || Cu 2+ (1 M) | Cu What reaction is occurring at the anode? 1. Cu 2+ + 2 e Cu 2. Zn 2+ + 2 e Zn 3. Zn Zn 2+ + 2 e correct 4. Cu Cu 2+ + 2 e Explanation: 006 10.0 points Consider the addition of 0.130 moles of solid NaOH to 1.0 liter of 0.230 M HCl. What will be the pH of the resulting solution? As- sume no volume change due to the addition of NaOH. 1. pH = 0.50 2. pH = 2.00 3. pH = 0.75 4. pH = 1.00 correct 5. pH = 1.50 Explanation: 007 10.0 points Which of the following acid-base indicators should be used for the titration of CH 3 COOH with KOH? For acetic acid, K a = 1 . 8 × 10 5 . 1. Neutral red, color change red/yellow 6 . 8 < pH < 8 . 0 2. None of these is suitable. 3. Phenolphthalein, color change colorless/red-violet 8 . 0 < pH < 10 . 0 correct 4. Methyl red, color change red/yellow at 4 . 4 < pH < 6 . 2 5. Any of these is suitable. Explanation: The endpoint of the weak acid (CH 3 COOH) - strong base (KOH) titration is above 7. An appropriate indicator has color change range pH = p K a ± 1. 008 10.0 points The following reaction is not balanced: VO 3 + Zn + H + V 2+ +Zn 2+ + H 2 O Balance the reaction. How many moles of H + are needed for every mole of VO 3 that is reduced? 1. 2 2. 6 correct 3. 12 4. 4 5. 1 Explanation: Balanced reaction is: 2VO 3 + 3Zn + 12H + 2V 2+ +3Zn 2+ + 6H 2 O So there are 6 H + for every 1 VO 3 . 009 10.0 points What is the molar solubility of CaF 2 ? ( K sp = 3 . 9 × 10 11 .) 1. 3 . 4 × 10 4 2. 3 . 9 × 10 11 3. 6 . 2 × 10 6
Version 167 – Exam 3 – Holcombe – (52460) 3 4. 4 . 4 × 10 6 5. 2 . 1 × 10 4 correct Explanation: CaF 2 Ca 2+ + 2 F K sp = [Ca 2+ ] [F ] 2 3 . 9 × 10 11 = ( x ) (2 x ) 2 = 4 x 3 x = 2 . 1 × 10 4 010 10.0 points If 100 mL of 0.040 M NaOH solution is added

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chexam3 - Version 167 Exam 3 Holcombe(52460 This print-out...

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