How much charge flows from a 6.0 v battery when it is connected across a completely discharged 15.7 μf capacitor?

The equation Q=CV (Charge = product of Capacitance and potential difference) tells us that the maximum charge that can be stored on a capacitor is equal to the product of it's capacitance and the potential difference across it. In this case the potential difference across the capacitor will be 6.0V (assuming circuit resistance is negligable) and it has a capacitance of 15.7 μf or 15.7x10^-6f, therefore charge equals (15.7x10^-6) x6=9.42x10^-5C (Coulombs).