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What volume of O_(2)at NTP liberated by 5 A current flowing for 193 and through acidulated water ? |
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Answer» Solution :`W = ( I xx t xx E)/(F) = (5 xx 193 xx 8)/(96500)= 0.08 G ` At NTP , volume of 32 g of `O_(2)` = 22400 mL `therefore 0.08` g = x ` x = (22400 xx 0.08)/(32) = 56 ` mL |
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