Calculate the amount of silver deposited at cathod when 5 ampere of current is passed for 50 minutes through a solution of AgNO3
7 answers
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.
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.
We know 96,485 coulombs will deposit 107.9 g Ag so
107.9 x (150,000/96,500) = ?g g deposited.
1.554
No answer is here whyy
My answer is 35.5 is this question correct please response.๐ค๐คจ
My answer is 35.5 is this question correct ๐ฏ
These answers are not correct because they use the wrong formula.