A 5 ampere current is passed through a solution of zinc sulphate for 40 minutes. The amount of zinc deposited at the cathod is ?
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Answer:
4.065 g
Explanation:
Q= 5×40×60 coulomb
The moles of An liberated by passing 5×40×60 coulomb of electric charge
5×40×60/2×96500 = 0.0622 moles
Weight in grams = 0.0622× 65.4 = 4.065 g
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