A volume of 122 mL of argon gas was collected at 25.0 degrees celsius and 756 torr. What does this sample weigh?
My friend and I are working on the same problem and he switched his weight to pounds and I want to know why he did that?
I know you would use the ideal gas law which is PV=nRT
so far I have (.99atm)(.122L) / (298K) [(0.082L)(1atm) / (1K)(1mole)]= .0049 moles.
which then goes into (.0049 moles)(39.95g/mole of Ar) = .20 g
grams is a measurement of weight, but my friend insists that it's supposed to be in pounds... is there a reason why it's supposed to be in pounds??
My friend and I are working on the same problem and he switched his weight to pounds and I want to know why he did that?
I know you would use the ideal gas law which is PV=nRT
so far I have (.99atm)(.122L) / (298K) [(0.082L)(1atm) / (1K)(1mole)]= .0049 moles.
which then goes into (.0049 moles)(39.95g/mole of Ar) = .20 g
grams is a measurement of weight, but my friend insists that it's supposed to be in pounds... is there a reason why it's supposed to be in pounds??