ow many grams of Ag(s) (107.9 g/mol) could be plated out on the cathode of an electrolytic cell by the passage of 5.0 A of current through gNO3 solution for 1.0 hour? Faraday's number = 96485 C/mol e™. elect one: O Mass of Ag(s) = 0.19 g O Mass of Ag(s) = 5.5 x 10³ g O Mass of Ag(s) Mass of Ag(s) = Mass of Ag(s) = 5.2 x 105 g Mass of Ag(s) = 20. g

Chemistry: Principles and Reactions
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Chapter17: Electrochemistry
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Problem 98QAP: Which of the changes below will increase the voltage of the following cell?...
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How many grams of Ag(s) (107.9 g/mol) could be plated out on the cathode of an electrolytic cell by the passage of 5.0 A of current through
AgNO3 solution for 1.0 hour? Faraday's number = 96485 C/mol e.
Select one:
Mass of Ag(s) = 0.19 g
Mass of Ag(s) = 5.5 x 10³ g
Mass of Ag(s) = Mass of Ag(s) = Mass of Ag(s) = 5.2 x 10-5 g
O Mass of Ag(s) = 20. g
Transcribed Image Text:How many grams of Ag(s) (107.9 g/mol) could be plated out on the cathode of an electrolytic cell by the passage of 5.0 A of current through AgNO3 solution for 1.0 hour? Faraday's number = 96485 C/mol e. Select one: Mass of Ag(s) = 0.19 g Mass of Ag(s) = 5.5 x 10³ g Mass of Ag(s) = Mass of Ag(s) = Mass of Ag(s) = 5.2 x 10-5 g O Mass of Ag(s) = 20. g
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