A current of 4.5A passed through a gold salt for 1h45min,(a) calculate the mass of gold deposited,(b) calculate the number of mole of gold deposited

Answer 1

coulombs = amperes x seconds = 4.5 x 105 = 472.5
96,485 coulombs will deposit APPROXIMATELY (200/3) g Au so
(200/3) x 472.5/96,485 = g Au metal deposited. Note that the 200/3 is an estimate. It should be atomic mass Au/3 = ?

mols Au = grams/atomic mass = ?
Post your work if you get stuck.

Answer 2

I don't know it please send the ans to me

Answer 3

Please send the answer to me

Answer 4

57.88g and0.294mol

Answer 5

To calculate the mass of gold deposited, we can use Faraday's law which states that the amount of substance produced at an electrode is directly proportional to the amount of electricity passed through the electrode.

The formula for calculating the mass of gold deposited is:

mass = (current × time × atomic mass) / (Faraday's constant × 1000)

where current is in amperes, time is in seconds, atomic mass is in grams per mole, and Faraday's constant is 96,485 coulombs per mole.

(a) Substituting the given values, we get:

mass = (4.5 × 1, 45 × 60 × 197) / (96,485 × 1000)
mass = 57.88 grams

Therefore, the mass of gold deposited is 57.88 grams.

(b) To calculate the number of moles of gold deposited, we can use the formula:

moles = mass / atomic mass

Substituting the values, we get:

moles = 57.88 / 197
moles = 0.294 mol

Therefore, the number of moles of gold deposited is 0.294 mol.