Illustrate an image that showcases the process involved in a chemistry experiment. Picture a laboratory setup with electrochemical cells where a current is passed through a gold salt solution. Do not write any numbers or text. Make sure to showcase the tools typically used in chemistry labs such as an ammeter for current measurement, a beaker filled with a solution of gold salt and electrodes placed into it, gold particles being separated and deposited at the bottom, lab timer and a scale for weighing. The overall ambiance should be scientific and inquisitive.

A current of 45A is passed through a solution of gold salt for 1 hour 45 minutes calculate (a) the mass of gold deposited (b) the numbers of moles of gold deposited (c) if the same current is used, find the time taken for 5.5g of gold to be deposited (Ah=197, 1F=96500C

I assume this is AuCl3 or similar with Au in the +3 state. Also I assume you made a typo with that Ah when you meant to type Au for gold.

1 hr 45 minutes = 60 min + 45 min = 105 min
Coulombs = amperes x seconds = 45 x (105 min x 60 sec/min) = 283,500 C
96,500 C will deposit 197/3 = 65.7 g Au so
65.7 g Au x (283,500/96,500) = ? g Au deposited.
mols Au = grams Au/atomic mass Au = ?
(c) coulombs needed is
96,500 C will deposit 65.7 g Au. To deposit 4.5 g Au will require
96,500 C x 5.5 g/65.7 g = 8,078 C needed, then
C = amperes x seconds.
8078 C = 45 A x seconds. Solve for seconds.
Post your work if you get stuck.

578.8 of Au

0.334G

Not exactly what I got in my solvings

2700sec

127575

To calculate the mass of gold deposited, we can use Faraday's law of electrolysis. According to the law, the mass of a substance deposited during electrolysis is directly proportional to the amount of charge passed through the solution.

(a) The formula to calculate the mass of gold deposited is:

m = (Q * M) / (n * F)

Where:
m = mass of gold deposited
Q = charge passed through the solution (in Coulombs)
M = molar mass of gold (in grams/mol)
n = number of moles of electrons transferred per mole of gold ions (for gold, n = 3)
F = Faraday's constant (96500 C/mol)

Given:
Current (I) = 45A
Time (t) = 1 hour 45 minutes = 1.75 hours
Atomic mass of gold (M) = 197 g/mol
Charge passed (Q) = I * t = 45A * 1.75 hours * 3600s/hour

Let's calculate m:

Q = 45A * 1.75 hours * 3600s/hour = 283,500 C

m = (283,500 C * 197 g/mol) / (3 * 96500 C/mol)

Simplifying the above equation, we get:

m ≈ 0.192 g

Therefore, the mass of gold deposited is approximately 0.192 grams.

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

n = m / M

Where:
n = number of moles of gold deposited
m = mass of gold deposited
M = molar mass of gold

Given:
m = 0.192 g
M = 197 g/mol

Let's calculate n:

n = 0.192 g / 197 g/mol

Simplifying the above equation, we get:

n ≈ 0.000975 moles

Therefore, the number of moles of gold deposited is approximately 0.000975 moles.

(c) Now, we need to find the time taken for 5.5 grams of gold to be deposited using the same current.

We can rearrange the formula used in part (a) to calculate the time:

t = (m * n * F) / (Q * M)

Given:
m = 5.5 g
n = 0.000975 moles
F = 96500 C/mol
M = 197 g/mol

Let's calculate t:

t = (5.5 g * 0.000975 moles * 96500 C/mol) / (45 A * 3600s/hour)

Simplifying the above equation, we get:

t ≈ 0.361 hours or 21.67 minutes

Therefore, if the same current is used, it will take approximately 0.361 hours or 21.67 minutes to deposit 5.5 grams of gold.