A current of 5 amperes is passed for 1930 seconds through an aqueous solution of copper(II) sulphate. The mass, in grams, of copper deposited will be:

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Vikas TU · Answer

From Faraday’s first la, W = Z*Q 5*1930*2 = W W = 19300 gm or 19.3 kgs. Z =2 as the CuSO4 has the valence factor as 2.

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A current of 5 amperes is passed for 1930 seconds through an aqueous solution of copper(II) sulphate. The mass, in grams, of copper deposited will be:

A current of 5 amperes is passed for 1930 seconds through an aqueous solution of copper(II) sulphate. The mass, in grams, of copper deposited will be:

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A current of 5 amperes is passed for 1930 seconds through an aqueous solution of copper(II) sulphate. The mass, in grams, of copper deposited will be:

A current of 5 amperes is passed for 1930 seconds through an aqueous solution of copper(II) sulphate. The mass, in grams, of copper deposited will be:

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