JAMB - Chemistry (2017 - No. 23)
The mass of silver deposited when a current of 10A is passed through a solution of silver salt for 4830s is – (Ag = 108 F = 96500(mol-1)
54.0g
27.0g
13.5g
108.0g
설명
Recall that
mass deposited = \(\frac{MmIt}{96500n}\)
Mm =108, t = 4830s
I = 10A, n = 1
m = 108 × 10 × (\(\frac{4830}{96500}\)) × 1
m = 54.0g
mass deposited = \(\frac{MmIt}{96500n}\)
Mm =108, t = 4830s
I = 10A, n = 1
m = 108 × 10 × (\(\frac{4830}{96500}\)) × 1
m = 54.0g