Electrolysis

Q = I \times t = 2400\,\text{C}

m = \frac{Q \times M}{n \times F} = 0.79\,\text{g}

Charge passed2400 C

Moles of electrons0.0249 mol

Copper deposited0.790 g

Current2 A

Time20 min

Pass at least 1000 C of charge through the cell (raise the current and the time)
Deposit at least 0.20 g of copper on the cathode
Deposit about 1.0 g of copper
Pass 0.05 mol of electrons through the cell
Now deposit about 2.0 g — double the copper needs double the charge