Question

A 9.5 v battery supplies a 3.5 ma current to a circuit for 6.0 h . part a how much charge has been transferred from the negative to the positive terminal?

Answer 1

The total time is (1h=3600 s): 
$\Delta t=6.0 h=21600 s$
The current intensity I is the amount of charge Q that passes through a certain point in a time $\Delta t$:
$I= \frac{Q}{\Delta t}$
Since we know the current, $I=3.5 mA=0.0035 A$, we can find how much charge has been trasferred from one terminal to the other in 6 hours:
$Q= I \Delta t=(0.0035 A)(21600 s)=75.6 C$