At 1.26 A, how long (in minutes) would it take to deposit 1.00 g of copper?

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Multiple Choice

At 1.26 A, how long (in minutes) would it take to deposit 1.00 g of copper?

Explanation:
During electrolysis, the amount of copper that plates out is determined by Faraday's law, which relates the mass deposited to the total charge passed: m = (M I t)/(n F). Here, copper’s molar mass M is 63.55 g/mol, the deposited copper corresponds to Cu2+ gaining 2 electrons (n = 2), the current is 1.26 A, and we want m = 1.00 g. Solve for time: t = (m n F)/(M I) = (1.00 g × 2 × 96485 C/mol) / (63.55 g/mol × 1.26 A). This yields t ≈ 2413 s ≈ 40.2 minutes.

During electrolysis, the amount of copper that plates out is determined by Faraday's law, which relates the mass deposited to the total charge passed: m = (M I t)/(n F). Here, copper’s molar mass M is 63.55 g/mol, the deposited copper corresponds to Cu2+ gaining 2 electrons (n = 2), the current is 1.26 A, and we want m = 1.00 g. Solve for time: t = (m n F)/(M I) = (1.00 g × 2 × 96485 C/mol) / (63.55 g/mol × 1.26 A). This yields t ≈ 2413 s ≈ 40.2 minutes.

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