What is the volume occupied by 12.3 g or argon gas at a pressure of 1.17 atm and a temperature of 455K?
So...Dr. Bob...
PV=nRT
1.17V=(12.3)(.08206)(455K)
V=459.25/1.17
V=392.5L ???
No. If you go back and look at my post I said n = number of mols and mols = grams/molar mass
So n = 12.3g/39.948 = ?
The other numbers look ok.