How many hour are required for a current of `3.0` ampere to decompose

1.	How many hour are required for a current of `3.0` ampere to decompose `18 g` water?
Answer» `H_(2)O rarr H_(2)+1//2O_(2) [(2H^(+)+2e rarr H^(2)),(O^(2-) rarr (1)/(2)O_(2)+2e)]` `:.` Eq. of `H_(2)O = (i.t)/(96500)` Equivalent weight of `H_(2)O = 18//2` as two electrons are used for `1` mole `H_(2)O` to decompose in `H_(2)O` and `O_(2)`. `:. (18)/(18//2) = (3 xx t)/(96500)` `t = 64333.3 sec` `= 1072.2` minute `= 17.87` hr

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How many hour are required for a current of `3.0` ampere to decompose `18 g` water?

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