Question

the same current pass through solution of CuSo4 and AgCn. how much silver will be obtained in the same time in which o.35g of Cu was obtained.

Answer 1

Answer:

Explanation:W1/E1=W2/E2

W1=0.35

E1=63.5/2=31.25 {E=molar mass/n factor}

W2=?

E2=107.88

0.35/31.25=W2/107.88

W2=1.188g