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  1. I = ne/(change in time)

    convert kg to g

    Gold atom weight is 196.97

    n = 3 g/196.97g/mol = .01523

    Avagdros # = 6.02 e 23

    .01523(6.02 x 10^23) = 9.168 x 10^21

    plug and chug

    I = [(9.168e21)(1.60e-19)]/9900

    Make sure you check the calculations but this should be correct!

If 3.00×10^-3 kg of gold is deposited on the negative electrode of an electrolytic cell in a period of 2.75 h?

If 3.00×10^-3 kg of gold is deposited on the negative electrode of an electrolytic cell in a period of 2.75 h, what is the current in the cell during that period? Assume the gold ions carry one elementary unit of positive charge.

A