Calculate the current that must be passed into a solution of aluminium salt for 30mins in order to deposit 1.5g of aluminium (A1=27).
2.98ampere
Well, first things first, I must say that "A1" sounds more like a college course than a chemical symbol. But don't worry, I'm here to help with some enlightening humor!
To calculate the current needed to deposit 1.5g of aluminum, we need to employ some fancy equations. Let's channel our inner nerds for a moment:
First, we need to find the number of moles of aluminum deposited. Given that the molar mass of aluminum is approximately 27 g/mol, we can divide 1.5g by 27 g/mol to get the number of moles.
1.5g / 27 g/mol = 0.0556 mol
Now, to find the total charge needed, we multiply the number of moles by the Faraday constant, which is 96,485 C/mol (Coulombs per mole). So, let's crunch those numbers:
0.0556 mol * 96,485 C/mol ≈ 5,366 C
Lastly, we divide the total charge by the time to get the current. Since you want to do this in 30 minutes (or 0.5 hours):
5,366 C / 0.5 h ≈ 10,732 C/h
Or if you prefer, approximately 10.7 kiloCoulombs/hour, but who's counting?
So, to deposit 1.5g of aluminum, you'll need to pass a current of about 10,732 C/h into the solution. Now you're ready to light up the lab with your electrifying calculations!
To calculate the current required to deposit a specific amount of aluminum from a solution, we can use Faraday's law of electrolysis. This law states that the amount of substance deposited on an electrode is directly proportional to the quantity of electricity passed through it.
First, we need to determine the amount of charge (Q) required to deposit 1.5g of aluminum. We can use the equation:
Q = m * F / M
Where:
Q is the charge required in coulombs (C)
m is the mass of aluminum to be deposited in grams (1.5g in this case)
F is the Faraday constant, which is approximately 96,485 C/mol
M is the molar mass of aluminum (27 g/mol)
Substituting in the values, we have:
Q = 1.5g * 96,485 C/mol / 27 g/mol
Now, we need to determine the current (I) required to pass this charge in 30 minutes (t = 30 minutes = 1800 seconds). The relationship between charge, current, and time is given by:
Q = I * t
Rearranging the equation:
I = Q / t
Substituting the values obtained earlier:
I = (1.5g * 96,485 C/mol / 27 g/mol) / 1800 s
Simplifying:
I ≈ 2.68 A
Therefore, approximately 2.68 Amperes of current must be passed into the solution of aluminum salt for 30 minutes in order to deposit 1.5g of aluminum.
Yes please
Calculate the current that must be passed into a solution of aluminum salt for,1 hour 30 minutes in order to deposit 1.5g of aluminum A =27
96,485 coulombs of electricity will deposit 27 g Al so you want how many coulombs? That will be 96,485 x (1.5/27) = ?
Then coulombs = amperes x seconds. Plug in coulombs and seconds [30 min x (60 sec/min)] and solve for coulombs. Post your work if you get stuck.