Asked by Rimamtang Josiah
Calcalate the mass of silver deposited when a current of 2.6A is passed through a solution silver salt for 70mins?
Answers
Answered by
DrBob222
coulombs = amperes x seconds
coulombs = 2.6 A x 70 min x (60 sec/min) = approx 11,000 C but you should use a better number than that.
96,485 coulombs will deposit 107.9 grams of Ag. Therefore, 11,000 C will deposit
107.9 g Ag x (11,000/96,485) = ? g Ag deposited.
Remember to start from scratch and recalculate each step above.
coulombs = 2.6 A x 70 min x (60 sec/min) = approx 11,000 C but you should use a better number than that.
96,485 coulombs will deposit 107.9 grams of Ag. Therefore, 11,000 C will deposit
107.9 g Ag x (11,000/96,485) = ? g Ag deposited.
Remember to start from scratch and recalculate each step above.
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