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what will be the volume of o 2 at ntp liberated by 5A current flowing for 193 sec through acidulated water.

what will be the volume of oat ntp liberated by 5A current flowing for 193 sec through acidulated water.

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1 Answers

Pankaj
askIITians Faculty 131 Points
8 years ago
Mass of Oxygen liberated(w) = EQ/(96500) = 8x965/965000
32 g of O2 = 22400mL
8x965/965000 g of O2 = [22400/32 ][ 8x965/965000] = 56mL

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