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

HgO(s) + H2O( ) + 2 e–  Hg( ) + 2 OH–(aq)
If 2.3  10–5 amperes flows continuously for 1200 days, what mass of Hg( ) is produced?
a. 2.5 g
b. 5.0 g
c. 9.9 g
d. 13 g
e. 15 g

coulombs = amperes x seconds.

C = 2.3E-5 x 1200 d x (24 hr/d) x (60 m/hr) x(60 s/min) = ?
96,485 coulombs will produce (200.6/2)g Hg; therefore,
g Hg = (200.6/2) x (?coulombs/96,485)

a. 2.5 G

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a

Well, well, well, it seems like we have a battery situation on our hands. Or should I say, on our wrists? Watch batteries can be quite shocking sometimes, don't you think? Anyway, let's calculate the mass of Hg produced using the given information.

First, we need to calculate the total charge (Q) that flows over 1200 days. We can use the formula I = Q/t, where I is the current and t is the time. Plugging in the values, we get:

2.3 × 10-5 A = Q / (1200 days × 24 hours/day × 60 minutes/hour × 60 seconds/minute)

Now, let's solve for Q. So, Q = (2.3 × 10-5 A) × (1200 × 24 × 60 × 60 seconds)

Calculating this gives us Q ≈ 1,872 C.

Now, we know that for every 2 electrons flowing, one Hg atom is produced. And in our chemical equation, we see that it takes 2 electrons to reduce one HgO molecule to Hg. So, to find the number of Hg atoms produced, we divide Q by the number of electrons per mole (Avogadro's number × 2).

N = (Q / (2 × 6.02 × 10^23 electrons/mol))

Calculating this gives us N ≈ 1.56 × 10^-4 mol of Hg.

Lastly, we can calculate the mass of Hg produced using the molar mass of Hg, which is approximately 200.59 g/mol.

Mass = N × molar mass = (1.56 × 10^-4 mol) × (200.59 g/mol)

Calculating this gives us a mass ≈ 3.12 × 10^-2 g, which is not an option!

So, it seems like my calculations have let me down this time. I apologize for the confusion. Maybe the battery in this watch is drained, just like my sense of humor when faced with chemistry problems.

To find the mass of Hg produced, we need to use Faraday's law of electrolysis, which relates the amount of substance produced during an electrolysis process to the electric current passing through the electrolyte and the time.

The equation given shows that 2 electrons are required to reduce 1 molecule of HgO to Hg. Therefore, the number of moles of Hg produced will be directly proportional to the number of electrons passed.

First, calculate the number of moles of electrons passed:
(2.3 × 10^-5 A) × (1 C/1 A) × (1 mol e^-/96485 C) × (24 h/1 d) × (1 day/3600 s) × (3600 × 24 × 1200 s) = X moles

The formula breaks down as follows:
- Convert Amperes (A) to Coulombs (C) using the definition of Ampere: 1 A = 1 C/s
- Convert Coulombs (C) to moles of electrons using Faraday's constant: 1 F = 96485 C/mol e^-
- Convert days (d) to seconds (s): 1 d = 24 h × 3600 s
- Multiply by the total number of seconds in 1200 days.

Next, calculate the molar mass of Hg:
Hg = 200.59 g/mol

Finally, calculate the mass of Hg produced:
mass of Hg = X moles × molar mass of Hg

Plug in the calculated value for X into the equation above and calculate the mass. The correct answer choice will correspond to the calculated mass.

Let's calculate it step by step:

Number of moles of electrons passed:
X = (2.3 × 10^-5) × (1) × (1/96485) × (24/1) × (1/3600) × (3600 × 24 × 1200)

Solving these values, we get X ≈ 0.0214 moles

Mass of Hg produced:
mass of Hg = 0.0214 moles × 200.59 g/mol

Solving this, we get the mass of Hg produced ≈ 4.286 g

The correct answer choice closest to 4.286 g is option b) 5.0 g.