Final answer:
The current that deposits 0.365 g of silver from a silver nitrate solution in 12960 seconds is calculated using Faraday's laws of electrolysis and is approximately 0.0252 Amperes.
Step-by-step explanation:
To find the electric current that deposits 0.365 g of silver metal from a solution of silver nitrate in 12960 seconds, we can use Faraday's laws of electrolysis. First, we need to calculate the number of moles of silver deposited using its molar mass. Silver has a molar mass of approximately 107.87 g/mol. The number of moles of silver (Ag) can be calculated as follows:
moles of Ag = mass of Ag / molar mass of Ag
= 0.365 g / 107.87 g/mol
= 0.003386 mol
Each mole of Ag requires 1 mole of electrons (because Ag has a +1 charge). To find the number of moles of electrons, we multiply the number of moles of Ag by the charge per mole of electrons (the Faraday constant, approximately 96500 Coulombs/mol):
moles of electrons = moles of Ag
= 0.003386 mol
Now, we need to convert moles of electrons to charge:
charge (Coulombs) = moles of electrons × Faraday constant
= 0.003386 mol × 96500 C/mol
= 326.603 C
Finally, current (I) is the charge flow per unit time (in seconds), which we can calculate using the formula:
I = charge / time
= 326.603 C / 12960 s
= 0.0252 A
Therefore, the current that deposits 0.365 g of silver in 12960 seconds is approximately 0.0252 Amperes (A).