One kind of battery used in watches contains mercury(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 ameres flows continuously 1095 days, what mass of Hg(l) is produced?

(2.5x10^-5)(1095 days)(24hours/day)(3600seconds/hour) = (xg Hg)(1 mol Hg/200.59g Hg)(2mol e^-/1 mol Hg)(96,000)

2365.2 = (x)(957.18) = .4046g Hg

I would have used 96,485 but I think your math is still off. I'm getting something like 2.5 g.

It appears that you are dividing the wrong way.
2365.2 = x(957.18)
Divide both sides by 957.18.
(2365.2/957.18)=x(957.18/957.18)
2.47 = x (You divided 957.18/2365.2) = 0.4046)
Again, if you intend to use digits like 2365.2 and 957.18, I would also use 96,485 instead of rounding that to 96,000.

To solve this problem, let's break it down step by step:

1. Start by calculating the total number of seconds in 1095 days:
1095 days * 24 hours/day * 3600 seconds/hour = 95,040,000 seconds

2. Next, calculate the total charge (in coulombs) that flows:
2.5x10^-5 A * 95,040,000 seconds = 2,376 coulombs

3. Since we know the balanced equation for the reduction of mercury(II) oxide, we can relate the number of electrons (e^-) to the amount of mercury produced.
From the equation, we see that 2 moles of electrons are required to produce 1 mole of mercury (Hg(l)).

4. Convert the charge in coulombs to moles of electrons:
2,376 C * 1 mol e^- / 96,485 C = 0.0246 mol e^-

5. Now, use the stoichiometry of the balanced equation to determine the moles of mercury produced:
0.0246 mol e^- * 1 mol Hg / 2 mol e^- = 0.0123 mol Hg

6. Finally, calculate the mass of Hg produced using the molar mass of mercury:
0.0123 mol Hg * 200.59 g/mol = 2.47 g Hg

So, the mass of mercury produced (Hg(l)) is approximately 2.47 grams.