One kind of battery used in watches contains mercury (II)oxide. As current is drawn, the mercury oxide is reduced tomercury. HgO(s)+H2O(l)+2e→Hg(l)+2OH−(aq) If a watch draws 1.0 X 10-5 amperes continuouslyfor 365 days, what mass of Hg(l) is formed? _______

Question

Tuthornt · Accepted Answer

The key to this problem is understanding the definitions of theelectrical units: 1 ampere = 1 coulomb of charge per second F = charge on a mol of electrons = 96,500 coulumbs per mol Looking at the balanced equation, we can see that to produce onemol of Hg(l), 2 mols of electrons are transferred. Use factorlabel method (conversion factors) to solve the problem: 1.0×10−5amperes=1.0×10−5coulumbssec 365days×24hrs1day×3600sechr=31536000sec 1.0×10−5coulumbs2sec×31536000sec×1molelectrons96500coulumbs×1molHg2molselectrons×200.59gHgmolHg=0.33gHg(I)

One kind of battery used in watches contains mercury (II)oxide.

Answered question

Answer & Explanation

New Questions in Other