The amount of Faraday required to discharge 4.5 moles of Al\(^{3+}\) is

a

13.5

b

4.5

c

7.5

d

1.5

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Correct Option
a

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Truss
1 year ago

n=q/f
q=nf
n=3 f=4.5
q=3×4.5
q=13.5✓✓

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