Suppose that a fully charged lead-acid battery contains 1.50 L of 5.00 M H2SO4. What will be the concentration of H2SO4 in the battery after 2.50 A of current is drawn from the battery for 7.50 hours

Answer : The final concentration of is, 5.47 M

Explanation :

First we have to calculate the charge.

Charge = Current × Time

Charge = 2.50 A × 7.50 hr

Charge = 2.50 A × 7.50 × 3600 s

Charge = 67500 C

Now we have to calculate the moles of electrons.

That means,

Moles of consumed = Moles of electrons = 0.699 mol

Now we have to calculate the initial moles of

Final moles of = 7.5 + 0.699 = 8.199 mol

Now we have to calculate the final concentration of

Therefore, the final concentration of is, 5.47 M

14 days is the half-life of phosphorus-32