Coulombs needed = amperes x seconds
C = Amperes x 1.5 hours x 60 min/hr x 60 sec/min
We need C and we need amperes
We know that 96,485 coulombs will deposit 27/3 = 9 g Al metal. So how many C are necessary to deposit 1.5 g Al. That is
96,485 C x (1.5/9) = about 16,000 C. Then substitute into the above
16,000 = amperes x 1.5 hrs x (60 min/hr ) x (60 sec/min) and solve for amperes.
Post your work if you get stuck.
Calculate the current that must be passed into a solution of aluminum salt for 1 hours 30 minutes, in order to deposit 1.5g of aluminum where aluminum is 27
3 answers
I don't have an answer but I need the answer to the question calculate the current that must be deposited or passed into a solution of aluminium salt for 1hour 30minutes in order to deposit 1.5g of aluminium (Al=27)
To calculate the current required to deposit 1.5g of aluminum, we need to use the Faraday's law of electrolysis, which states that the amount of substance deposited is directly proportional to the charge passed through the solution.
The molar mass of aluminum is 27 g/mol, so 1.5g of aluminum is approximately (1.5/27) = 0.056 mol.
The Faraday's constant is 96,485 C/mol.
Now, we can use the equation:
Coulombs = current (A) x time (s)
We have the time given as 1 hour 30 minutes, which is equivalent to (1 hour + 30 minutes) = (1 x 60 x 60) + (30 x 60) = 5400 s.
So, Coulombs = current (A) x 5400 s.
We know that 1 mole of aluminum requires 96,485 C of charge to deposit. Therefore, 0.056 mol of aluminum will require (0.056 x 96,485) C of charge.
Setting up the equation:
(0.056 x 96,485) = current (A) x 5400
Solving for current:
Current (A) = (0.056 x 96,485) / 5400 ≈ 1 A
Therefore, the current that must be passed into the solution of aluminum salt for 1 hour 30 minutes in order to deposit 1.5g of aluminum is approximately 1 Ampere.
The molar mass of aluminum is 27 g/mol, so 1.5g of aluminum is approximately (1.5/27) = 0.056 mol.
The Faraday's constant is 96,485 C/mol.
Now, we can use the equation:
Coulombs = current (A) x time (s)
We have the time given as 1 hour 30 minutes, which is equivalent to (1 hour + 30 minutes) = (1 x 60 x 60) + (30 x 60) = 5400 s.
So, Coulombs = current (A) x 5400 s.
We know that 1 mole of aluminum requires 96,485 C of charge to deposit. Therefore, 0.056 mol of aluminum will require (0.056 x 96,485) C of charge.
Setting up the equation:
(0.056 x 96,485) = current (A) x 5400
Solving for current:
Current (A) = (0.056 x 96,485) / 5400 ≈ 1 A
Therefore, the current that must be passed into the solution of aluminum salt for 1 hour 30 minutes in order to deposit 1.5g of aluminum is approximately 1 Ampere.