Question 8 of 25 < } AD View Policies Current Attempt in Progress Calculate E for the reaction: 31₂+5 Cr₂0, +34H₂0-610+1

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Question 8 of 25 < } AD View Policies Current Attempt in Progress Calculate E for the reaction: 31₂+5 Cr₂0, +34H₂0-610+1

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Question 8 Of 25 Ad View Policies Current Attempt In Progress Calculate E For The Reaction 31 5 Cr 0 34h 0 610 1 1
Question 8 Of 25 Ad View Policies Current Attempt In Progress Calculate E For The Reaction 31 5 Cr 0 34h 0 610 1 1 (15.16 KiB) Viewed 42 times
Question 8 Of 25 Ad View Policies Current Attempt In Progress Calculate E For The Reaction 31 5 Cr 0 34h 0 610 1 2
Question 8 Of 25 Ad View Policies Current Attempt In Progress Calculate E For The Reaction 31 5 Cr 0 34h 0 610 1 2 (22.54 KiB) Viewed 42 times
Question 8 Of 25 Ad View Policies Current Attempt In Progress Calculate E For The Reaction 31 5 Cr 0 34h 0 610 1 3
Question 8 Of 25 Ad View Policies Current Attempt In Progress Calculate E For The Reaction 31 5 Cr 0 34h 0 610 1 3 (19.57 KiB) Viewed 42 times
Question 8 Of 25 Ad View Policies Current Attempt In Progress Calculate E For The Reaction 31 5 Cr 0 34h 0 610 1 4
Question 8 Of 25 Ad View Policies Current Attempt In Progress Calculate E For The Reaction 31 5 Cr 0 34h 0 610 1 4 (20.27 KiB) Viewed 42 times
Question 8 Of 25 Ad View Policies Current Attempt In Progress Calculate E For The Reaction 31 5 Cr 0 34h 0 610 1 5
Question 8 Of 25 Ad View Policies Current Attempt In Progress Calculate E For The Reaction 31 5 Cr 0 34h 0 610 1 5 (17.79 KiB) Viewed 42 times
Question 8 Of 25 Ad View Policies Current Attempt In Progress Calculate E For The Reaction 31 5 Cr 0 34h 0 610 1 6
Question 8 Of 25 Ad View Policies Current Attempt In Progress Calculate E For The Reaction 31 5 Cr 0 34h 0 610 1 6 (22.08 KiB) Viewed 42 times
Question 8 Of 25 Ad View Policies Current Attempt In Progress Calculate E For The Reaction 31 5 Cr 0 34h 0 610 1 7
Question 8 Of 25 Ad View Policies Current Attempt In Progress Calculate E For The Reaction 31 5 Cr 0 34h 0 610 1 7 (15.55 KiB) Viewed 42 times
Question 8 Of 25 Ad View Policies Current Attempt In Progress Calculate E For The Reaction 31 5 Cr 0 34h 0 610 1 8
Question 8 Of 25 Ad View Policies Current Attempt In Progress Calculate E For The Reaction 31 5 Cr 0 34h 0 610 1 8 (24.82 KiB) Viewed 42 times
Question 8 Of 25 Ad View Policies Current Attempt In Progress Calculate E For The Reaction 31 5 Cr 0 34h 0 610 1 9
Question 8 Of 25 Ad View Policies Current Attempt In Progress Calculate E For The Reaction 31 5 Cr 0 34h 0 610 1 9 (19.09 KiB) Viewed 42 times
Question 8 Of 25 Ad View Policies Current Attempt In Progress Calculate E For The Reaction 31 5 Cr 0 34h 0 610 1 10
Question 8 Of 25 Ad View Policies Current Attempt In Progress Calculate E For The Reaction 31 5 Cr 0 34h 0 610 1 10 (17.73 KiB) Viewed 42 times
Question 8 of 25 < } AD View Policies Current Attempt in Progress Calculate E for the reaction: 31₂+5 Cr₂0, +34H₂0-610+10 Cr³+51H₂O E100+1.195, EcoH+1232 Round your answer to 3 significant figures. 0.03 V eTextbook and Media Save for Later Last saved 45 minutes ago. Saved work will be auto-submitted on the due date. Auto submission can take up to 10 minutes. F 11 Attempts: 0 of 3 used Submit Answer
Question 9 of 25 < > Use standard reduction potentials to calculate Eº for the following reactions: Reaction 1: Hs10, +Cr 10+ Cr³+ (acidic) (unbalanced) ED(V) 10+4H₂0 +1.601 -0.744 HslO,+ H₂O+20 Cr³+3e Cr Reaction 2: Se + Cr(OH)3-Cr+SeO₂2 (basic) (unbalanced) E (V) SeO² + 3H₂O+40 Se+60H -0.366 Cr(OH), 3e Cr+30H -1.48 Reaction 3: NO₂+H₂O₂-NO+O₂ (basic) (unbalanced) ED(V) NO+2H₂O 3e NO+40H +0.109) O₂+2H₂O+2e H₂O₂+20H -0.146 -/4 = 1
Question 9 of 25 Part 1 What is Eº for (the balanced) Reaction 1? eTextbook and Media Hint Save for Later Part 2 What is E for (the balanced) Reaction 2? Save for Later -/4 E 1 Attempts: 0 of 3 used Submit Answer Attempts: 0 of 3 used Submit Answer
Question 9 of 25 Part 2 What is E for (the balanced) Reaction 2? Save for Later Part 3 < > Save for Later V What is E for (the balanced) Reaction 3? Attempts: 0 of 3 used Submit Answer Attempts: 0 of 3 used Submit Answer
Question 11 of 25 < View Policies Current Attempt in Progress The following half-reactions occur during use of the rechargeable nickel-cadmium battery: E- > eTextbook and Media This battery has a potential of 1.35 V under standard conditions, with nickel as the cathode. What potential does this battery produce if its hydroxide ion concentration is 1.89 x 102 M? Save for Later Cd(OH)₂() +2e-Cd(s) + 2OH(aq) NIO(OH)(s) + H₂O(+eNi(OH)₂(s) + OH(aq) -/4 = 1 Attempts: 0 of 3 used Submit Answer
Question 12 of 25 View Policies Show Attempt History Current Attempt in Progress Your answer is partially correct. A galvanic cell consists of a Pt electrode immersed in a solution containing Fe2 at 1.00 M and Feat unknown concentration, as well as a Cu electrode immersed in a 1.00 M solution of Cu². The cell voltage is 0.00 V. What is the concentration of Fe³+ ions? 5.4 eTextbook and Media Save for Later x 108 2/4 M Attempts: 1 of 3 used Submit Answer
Question 14 of 25 View Policies Current Attempt in Progress Use standard reduction potentials to calculate AG" for the reaction: UMO, MO, OH- +0.595, UCNOCN. OH-0.970 eTextbook and Media Hint Save for Later 3 CN+2 MnO₂ +H₂0-3 CNO+2 MnO₂ + 2OH kJ -/4 E ! Attempts: 0 of 3 used Submit Answer
Question 15 of 25 Current Attempt in Progress Suppose that automobiles were equipped with "thermite" batteries that operate according to the following half-reactions: 2AI +60H-Al₂O₂ + 3H₂O+6e Fe₂O3 + 3H₂O+6e-2Fe +60H What masses of Al and Fe₂O3 would be consumed if it takes 15 sec to start your car engine and the battery provides 5.3 A of current to the starter motor. Faraday's constant is 96485 C/mol. Part 1 Mass of Al consumed? eTextbook and Media Hint Save for Later -/4 E Part 2 Attempts: 0 of 3 used Submit Answer
Question 15 of 25 Part 1 Mass of Al consumed? eTextbook and Media Hint Save for Later Part 2 < Mass of Fe₂O₂ consumed? Save for Later 8 -/4 E ! Attempts: 0 of 3 used Submit Answer Attempts: 0 of 3 used Submit Answer
Question 16 of 25 < > View Policies Current Attempt in Progress A digital watch draws 0.20 mA of current provided by a mercury battery, whose net reaction is: O₂(g) +2 Zn(s)-2 ZnO(s) If a partially used battery contains 1.36 g of each of zinc, for how many more hours will the watch run? The Faraday constant is 96485 C/mol. eTextbook and Media Hint -/4 = 1 Save for Later: Attempts: 0 of 3 used Submit Answer
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