The amount of Faraday required to discharge 4.5 moles of Al\(^{3+}\) is
13.5
4.5
7.5
1.5
n=q/f q=nf n=3 f=4.5 q=3×4.5 q=13.5✓✓
ceoofwahala
20th June, 2026
ASSAAS
infinitehoaxx
21st May, 2026