1.

Calculate the mass of silver deposited from silver nitrate solution by a current of 2 amperes flowing for 30 minutes?

Answer»

SOLUTION :`M= Z l t`
`M=(108 times 2 times 30 times 60)/(96,500)`
`M=4.029g `
OR
`Q = l t`
`Q=2 times 30 times 60`
=3600 C
M=ZQ
`=108/(96,500) times 3600`
=4.029g
Or
`Q= l t`
`Q=2 times 30 times 60=3600C`
FOr 96,500 C mass of silver DEPOSITED is 108g
`THEREFORE` for 3600... it is
`(3600 times 108)/(96,500)=4.029g`


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