Two Faraday of electricityis passed through a solution of CuSO4. the mass of copper deposited at the cathode is (at. mass of cu =63.5 amu)

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Mahima Kanawat · Answer

Dear student No. Of Faraday = no. Of equivalents of Cu deposited 2 = wt of cu deposited /eq wt of cu Wt of cu deposited = 63.5 g WITH REGARDS MAHIMA ASKIITIANS FORUM EXPERT

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Two Faraday of electricityis passed through a solution of CuSO4. the mass of copper deposited at the cathode is (at. mass of cu =63.5 amu)

Two Faraday of electricityis passed through a solution of CuSO4. the mass of copper deposited at the cathode is (at. mass of cu =63.5 amu)

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Two Faraday of electricityis passed through a solution of CuSO4. the mass of copper deposited at the cathode is (at. mass of cu =63.5 amu)

Two Faraday of electricityis passed through a solution of CuSO4. the mass of copper deposited at the cathode is (at. mass of cu =63.5 amu)

1 Answers

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