a current of 0.75 A is passed through acidic solution of CuSO4 for 10 minutes. The volume of oxygen at anode (at STP) will be

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vanshith · Answer

2cu+2e- ---->Cu 4OH - +4e - ----->2H 2 O+O 2 See Q=0.75×10×60=450C Now 4×96500c deposite 32g of Oxygen 450c will deposite=32×450/4×96500=0.0373g O 2 at STP 32g of oxygen has volume 22.4L thus 0.373g will have 22.4×0.373/32=0.261L=26.1ml

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a current of 0.75 A is passed through acidic solution of CuSO4 for 10 minutes. The volume of oxygen at anode (at STP) will be

a current of 0.75 A is passed through acidic solution of CuSO4 for 10 minutes. The volume of oxygen at anode (at STP) will be

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a current of 0.75 A is passed through acidic solution of CuSO4 for 10 minutes. The volume of oxygen at anode (at STP) will be

a current of 0.75 A is passed through acidic solution of CuSO4 for 10 minutes. The volume of oxygen at anode (at STP) will be

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