What amount of current, in amperes, would be required to deposit 19.7 g of Au metal in 1.00 hour from a solution of Au(NO3)3?

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asked by zak
  1. You can deposit 196.97/3 = approx 66 g Au with 96,485 coulombs of electricity. So how much will be required to deposit 19.7g
    96,485 x (19.7/66) = approx 29,000 coulombs.
    coulombs = amperes x seconds.
    Substitute coulombs and seconds and solve for amperes.
    Note those numbers I used are approximate. You need to redo all of them.

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    posted by DrBob222

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