Posted by Joe on Monday, December 3, 2012 at 12:21am.
It will require 96,485 coulombs to deposit 27/3 = 9 g Al. Therefore, it will require 96,485 x (15/9) coulombs to deposit 15 g.
1 C = Amp x sec. You know C and amps solve for seconds.
Related Questions
chemistry - consider the electrolysis of molten AlCl3 what must the current be ...
Chemistry - I did a lab where I put Copper (II) Chloride in water with Alumnim. ...
College Chemistry - How many grams of Al would be deposited from an aqueous ...
Chemistry - A current of 0.80 A was applied to an electrolytic cell containing ...
Chemistry - The antacid Amphogel contains aluminum hydroxide Al(OH}3 How many ...
Chemistry - I don't know how to set up this problem...can you help? How many...
Chemistry - 1) Determine the equilibrium constant, Keq, at 25°C for the ...
chemistry - how many milliliters of 6.00 M HCl are required to react with 60.0 ...
Chemistry - What mass of AlCl3 can be formed from 5.000 mol NiCl2 and 2.000 mol ...
Chemistry - Determine how many grams of active antacid ingredient AL(OH)3 would ...
For Further Reading