exam 4 - Version 326 Exam 4 Holcombe (52460) 1 This...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 =- . 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. 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- . 22 =- 1 . 88 006 10.0 points Consider the data below: [CH 4 ] [O 2 ] initial rate M M M s 1 Exp 1 . 5 1 . 3 2 . 5 10 3 Exp 2 . 5 2 . 6 1 . 10 2 Exp 3...
View Full Document

This note was uploaded on 09/14/2010 for the course C E 301 taught by Professor Staff during the Spring '08 term at University of Texas at Austin.

Page1 / 8

exam 4 - Version 326 Exam 4 Holcombe (52460) 1 This...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online