Question

67. If a steady current of 15.0 A is passed through an
aqueous solution of CuSO4,
minutes will
it take to deposit 0.250 mol of Cu at the cathode,
assuming 100% efficiency?
(a) 3.217* 103
(b) 1.613 x 103
(c) 53.62
(d) 0.893​

Answer 1

Answer:

i think that ans is 3

Explanation:

 i explain you-

 n.vf=it/96500

 0.25*2=(15*t)/96500

 0.5*96500=15t

  48250/15=t

 then t=3216sec

and t=3216/60min

t=53.62min

Answer 2

Answer:

Ur answer is here.....!!!!!

Plz Mark as brainliest....!!!!