Question
What mass of AL is deposited electronically in 30 minutes by current of 740A?
Answers
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To solve this problem, we need to use the formula for electroplating:
Mass = (Current x Time x Atomic Mass) / (Charge x Faraday's Constant)
First, we need to find the charge using the formula:
Charge = Current x Time
Charge = 740 A x 30 min x 60 s/min = 1,332,000 C
Next, we need to find the atomic mass of aluminum, which is 26.9815 g/mol.
Faraday's constant is 96,485 C/mol.
Now we can plug in the values into our original formula:
Mass = (Current x Time x Atomic Mass) / (Charge x Faraday's Constant)
Mass = (740 A x 30 min x 60 s/min x 26.9815 g/mol) / (1,332,000 C x 96,485 C/mol)
Mass = 1,270.87 g
Therefore, 1,270.87 grams of aluminum is deposited electronically in 30 minutes by a current of 740A.
Mass = (Current x Time x Atomic Mass) / (Charge x Faraday's Constant)
First, we need to find the charge using the formula:
Charge = Current x Time
Charge = 740 A x 30 min x 60 s/min = 1,332,000 C
Next, we need to find the atomic mass of aluminum, which is 26.9815 g/mol.
Faraday's constant is 96,485 C/mol.
Now we can plug in the values into our original formula:
Mass = (Current x Time x Atomic Mass) / (Charge x Faraday's Constant)
Mass = (740 A x 30 min x 60 s/min x 26.9815 g/mol) / (1,332,000 C x 96,485 C/mol)
Mass = 1,270.87 g
Therefore, 1,270.87 grams of aluminum is deposited electronically in 30 minutes by a current of 740A.