Answer:
0.049 g of Ag
Step-by-step explanation:
From the question;
Ag^+ (aq) + e ----->Ag(s)
1 e = 96500 C
Note that Q= It
Where;
I = current and t= time taken
108 g of silver is deposited by 96500 C of electricity
x g is deposited by (0.49 * 90) C
x = 108 * (0.49 * 90) /96500
x= 0.049 g of Ag