1.

The number of electrons required to reduced4.5 xx 10^(-5)g of Al is

Answer»

` 1.03xx10^(18)`
`3.01 XX 10^(18)`
`4.95 xx 10^(26)`
`7.31 xx 10^(20)`

Solution :`underset(27)(AL^(3+)) +3e^(-) to underset(27g)(Al)`
27 g of Al is reduced by `=3 xx 6.023xx10^(23)E^(-)s`
` 4.5xx10^(-5)` g of Al will be reduced by
`=(3xx6.023xx10^(23) xx 4.5xx10^(-5))/(27)`
`=3.01 xx 10^(18)` electrons


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