How long would it take a current of 10A to deposit 6.36g of copper during the electrolysis of copper (ll) tetraoxosulphate VI solution. (Cu = 63.5,1F=96500C) Cu2+ + 2e- Cu

It take 0.54 hours to deposit 6.36g of copper

Faraday's Law I

"The mass of the substance formed at each electrode is proportional to the electric current flowing in the electrolysis

W = e.i.t / 96500

e = equivalent = Ar / valence

i = current, A

t = time, s

W=6.36 g

e = 63.5 : 2 =31.75

i = 10 A

i believe it is 40–40–20%

explanation:

a boiling water reactor does not boil the water. a pressurized water reactor boils the water faster. a pressurized water reactor is heated directly by the core. a pressurized water reactor does not use fission.