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exam 4 - Version 326 – Exam 4 – Holcombe –(52460 This...

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Version 326 – Exam 4 – Holcombe – (52460) 1 This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points What is the average current generated in the Cu(s) | Cu 2+ (aq) || Fe 3+ (aq) | Fe(s) electrochemical cell if 50 g of Cu(s) are used up in a 24 hour period? Cu 2+ + 2 e Cu E red = +0 . 22 V Fe 3+ + 3 e Fe E red = - 0 . 04 V 1. 1.76 amp correct 2. 13.00 amp 3. 2.64 amp 4. 111.85 amp 5. 42.17 amp Explanation: m Cu = 50 g t = 24 h The half equation of interest is Cu(s) Cu 2+ (aq) + 2 e Find the moles of Cu dissolved: (50 g Cu) 1 mol Cu 63 . 546 g Cu = 0 . 786832 mol Cu Find the amount of charge carried by the electrons: (0 . 786832 mol Cu) × 2 mol e 1 mol Cu × 96485 C 1 mol e = 1 . 51835 × 10 5 C Find the average current in 24 hours: Colombs = amp × second amp = Coulomb second = 1 . 51835 × 10 5 C 24 h × 1 h 60 min × 1 min 60 s = 1 . 75735 C / s = 1 . 75735 A 002 10.0 points Magnesium metal is placed in contact with an underground steel storage tank to inhibit the corrosion of the tank because 1. magnesium is cheaper than gold. 2. magnesium metal is a better reducing agent than iron. correct 3. magnesium functions as the cathode in the electrochemical cell instead of the iron. 4. magnesium has a higher reduction poten- tial than iron. 5. magnesium has a lower oxidation poten- tial than iron. Explanation: 003 10.0 points Biochemical reactions are most often cat- alyzed by 1. monomers 2. prostaglandins 3. isomers 4. NSAIDs 5. heat energy 6. enzymes correct Explanation: Enzymes are biological catalysts. 004 10.0 points For the multiple step reaction mechanism slow NO + Br 2 NOBr 2 fast NOBr 2 + NO 2 NOBr net 2 NO + Br 2 2 NOBr which would be the correct rate law? The constant k may represent a combination of elementary reaction rate constants.
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Version 326 – Exam 4 – Holcombe – (52460) 2 1. Rate = k [NO] 2 [Br 2 ] 2. Rate = k [NOBr 2 ] [NO] 3. Rate = k [NO] [Br 2 ] correct 4. Rate = k [NOBr 2 ] 2 [NO] 2 [Br 2 ] 5. Rate = k [NOBr] 2 [NOBr 2 ] [NO] Explanation: The correct rate law must reflect the rate law for the slowest step which is the rate determining step: Rate = k [NO] [Br 2 ] 005 10.0 points What would be the E cell of an electrolytic cell made from the following two half reac- tions? Half reaction E AgCl( s ) + e -→ Ag( s ) + Cl ( aq ) +0 . 22 Al 3+ ( aq ) + 3 e -→ Al( s ) - 1 . 66 1. - 1 . 44 2. 1 . 88 3. - 1 . 88 correct 4. 1 . 44 Explanation: E = E cathode - E anode = - 1 . 66 - 0 . 22 = - 1 . 88 006 10.0 points Consider the data below: [CH 4 ] [O 2 ] initial rate M M M · s 1 Exp 1 0 . 5 1 . 3 2 . 5 × 10 3 Exp 2 0 . 5 2 . 6 1 . 0 × 10 2 Exp 3 1 . 0 1 . 3 5 . 0 × 10 3 Which of the following is a correct rate law for the reaction?
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