How many milli moles are in 45.0 mL of 0.295 M zinc nitrate( Zn(NO3)2)?

The molarity is simply the number of moles per litre. So that's (45.0 x 10^-3 mL) x (0.295 mol / L) x (1000 mmol / 1 mol) = 13.275 mmol