I just need to know the formulas in order to solve the following questions because my teacher gave me a review on everything but I don't know the formulas at all. I know how to solve the questions though. I did them before but I just need the formulas in order to solve them... I forgot them.

The questions are:

1) What volume of a 17.4 mol/L acetic acid stock solution is required to make 2.0 L of a 1.5 mol/l acetic acid solution
2) Determine the volume of a balloon at STP assuming that it occupies a volume of 7.75 L at a temperature of 0 degrees Celsius and a pressure of 125 kPa.
3) A pressure reading of 275 kPa is found when the temperature of air in a piston is 35 degrees Celsius and the volume is 2.5 L. What will the temp. reading be, in degrees Celsius, if the volume doubles and the pressure becomes 100 kPa?
4)Black gunpower stored in a barrel has been ignited and 1245g of KNO3 (s) are formed in the exploding mixture. What will the total volume be of all gases produced in this explosive at a temp. of 350 degrees Celsius and a pressure of 2000 kPa?
KNO3 (s) + C(s) + S8 (s) --> N2(g) + CO2(g) + K2S8 (s)

Thanks

#1. You have a solution of concentration 17.4 moles/liter. You want to dilute that to 1.5 moles/liter. You can figure out how many moles will be in the final solution and how much of the original solution to measure out. If you multiply Moles/Liter times Liters, the Liters will cancel out and you'll have moles.

#2. is the ideal gas law. P=pressure, V=volume, n=number of moles, T=absolute temperature. R is the "universal gas constant". You can Google it.

#3. Same equation.

#4. Calculate the number of moles of N2 and CO2. That's what causes the pressure. 1 mole of KNO3 generates 2 moles of gas (from the equation) -- one mole of CO2 and one mole of N2. Calculate the molecular weight of KNO3 (Grams/Mole) and divide the number of grams you have by the molecular weight

You'll find from the ideal gas law (plug in the temperature and pressure and solve for the volume) that at STP, 1 mole occupies 22.4 Liters.