calculate the mass of silver deposited when a current of 2.6a is passed through a solution of a silver salt for 70mins ag=108,1 faraday 96500

Answer 1

Convert 70 min to seconds.
coulombs = C = Amps x seconds. C = approx 1E4 C but you need to do it more accurately than that.
Then 96,500 C will deposit 108.1 g (are you sure about the atomic mass Ag???. More like 107.9 I thought). So
108.1 x (1E4/96,500) = ? g Ag deposited.

Answer 2

12.22

Answer 3

calculate the Mass of silver deposited when a current of 2.6A is passed through a solution of a silver salt for 70 minutes (Ag =108 1 Faraday=96500)

Answer 4

calculate the mass of silver deposited when a current of 2.6A is passed through solution of a silver salt for 70 minutes(Ag=108,IF=96500 c)

Answer 5

in this question Q=it. i=2.6A t=70min. Q=2.6*70*60=10920C
108g yield 96500F
Xg will yield 10920C
cross multiply you will have
10920*108/96500=12.22g

Answer 6

Calculated the mass of silver deposited when a current of 2.6A is passed through a solution of a silver for 70minutes (Ag=108;1 Faraday=96500c)

Answer 7

How will you calculate it after changing the minutes to second and finding your Coulombs which is current × time

Answer 8

You will cross multiply

Answer 9

M=Q * m.m / f * c
M=10920 * 108 / 96500 * 1
M= 12.22g