Calculate the amount of silver deposited at cathod when 5 ampere of current is passed for 50 minutes through a solution of AgNO3

Answer 1

Faraday's Law... 1 mole electrons = 96,500 amp-sec. => ?gAg = 5amps(1mol e/96,500amp-sec)(1 mole Ag/1 mole e)(107g Ag/1 mole Ag)(60sec/min)(50min) = You punch it into your calculator.

Answer 2

coulombs = amperes x seconds = approx 50 x 50min x (60 sec/min) = about 150,000 but check that.

We know 96,485 coulombs will deposit 107.9 g Ag so
107.9 x (150,000/96,500) = ?g g deposited.

Answer 3

1.554

Answer 4

No answer is here whyy

Answer 5

My answer is 35.5 is this question correct please response.🤔🤨

Answer 6

My answer is 35.5 is this question correct 💯

Answer 7

These answers are not correct because they use the wrong formula.