ExamPlay Light Logo
Registrazione

JAMB - Chemistry (1990 - No. 42)

0.1 faraday of electricity deposited 2.95 g of nickel during electrolysis of an aqueous solution.
Calculate the number of moles of nickel that will be deposited by 0.4 faraday?

(Ni = 58.7)
0.20
0.30
0.04
5.87

Spiegazione

0.1F ↔ 2.95 g Ni = 0.4F ↔ 11.8 g Ni

No of moles = (11.8)/(58.2) = 0.20 moles

Commenti (0)

Accedi per commentare
Annuncio
BrainBehindX Inc Logo
©2026; Offerto da BrainBehindX Inc