current and resistance problem...?

Question

3.00*10^-3 kg of gold is deposited on the negative electrode of an electrolytic cell in a period of 2.65 h. Calculate the current in the cell during that period

i think this helps but what goes where?
I=changeQ/change t = ne / change t

Answer 1

number of atoms A= km N
m - mass of gold (Au)
N - Avogadro's number of atoms per mole
k -proportionality constant to convert mole/kg for Au

n - number of electrons per atom
q- 1.602 E-19 C (charge per electron)
Q- total charge

So as you have said I=change Q/change t

Q= A n q
I=k m N n q/(change t)
Just put the numbers in and remember that 1 h = 3600 sec.