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How many hours are required for the deposition of 15.0 grams of Al from molten AlCl3 by a current of 15.0 amps?

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It will require 96,485 coulombs to deposit 27/3 = 9 g Al. Therefore, it will require 96,485 x (15/9) coulombs to deposit 15 g.
1 C = Amp x sec. You know C and amps solve for seconds.

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