The question is incomplete, the complete question is;
When a lead acid car battery is recharged by the alternator, it acts essentially as an electrolytic cell in which solid lead(II) sulfate PbSO₄ is reduced to lead at the cathode and oxidized to solid lead(II) oxide PbO at the anode.
Suppose a current of 96.0 A is fed into a car battery for 37.0 seconds. Calculate the mass of lead deposited on the cathode of the battery. Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol.
Answer:
3.81 g of lead
Explanation:
The equation of the reaction is;
Pb^2+(aq) + 2e ---->Pb(s)
Quantity of charge = 96.0 A * 37.0 seconds = 3552 C
Now we have that 1F = 96500 C so;
207 g of lead is deposited by 2 * 96500 C
x g of lead is deposited by 3552 C
x = 207 * 3552/2 * 96500
x = 735264/193000
x = 3.81 g of lead
Hi how are you Henson’s dndndndndndndn
Answer is: pOH = 3,29.
Kb(NH₃) = 1,5·10⁻⁶.
c(NH₃) = 0,175M.
pOH = ?
Chemical reaction: NH₃ + H₂O ⇄ NH₄⁺ + OH⁻.
Kb = c(NH₄⁺) · c(OH⁻) ÷ c(NH₃).
c(NH₄⁺) = c(OH⁻) = x.
x² = Kb · c(NH₃)
x² = 1,5·10⁻⁶ · 0,175 = 2,625 ·10⁻⁷.
x = c(OH⁻) = √2,625 ·10⁻⁷ = 5,12 · 10⁻⁴.
pOH = -log(c(OH⁻)) = 3,29.