Asked by Steve

One kind of battery used in watches contains mercur(II) oxide. As current flows, the mercury oxide is reduced to mercury.

HgO(s) + H2O(l) + 2e^- ==> Hg(l) + 2OH^-(aq)

If 2.5x10^-5 amperes flows continuously 1095 days, what mass of Hg(l) is produced?

(2.5x10^-5)(1095days)(24hours/day)(3600seconds/hour) = (mass of Hg)(1 mol Hg/mw Hg)(2 mol e^-/1 mol Hg)

2365.2 = (x grams Hg)(1 mol Hg/200.59gHg)(2 mol e-/1 mol Hg)

= 23.58 grams

Answers

Answered by DrBob222
Check your work. It looks to me as if you have omitted the 96,485 factor. The 2365.2 looks ok.
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