Can you calculate how many moles of the gas was finally present in the inflated bag?
ok there were 1.21moles of gas in the fully inflated bag, and 3.94x10^-1moles of nitrogen initially, so would that mean that if I substract the 1.21moles of gas minus the 3.94x10^-1 moles of nitrogen then I get the moles of oxygen... I hope I'm right, therefore, there were 0.816 moles of oxygen in the fully inflated bag. Please tell me if I'm doing the correct thing here... if I use the moles of oxygen to find P, will it be the oxygen partial pressure? like this:
P=nrt/v which'd be P=(0.816mol O)(0.0821 atm L/ mol K) (298K)/ 30L
P=0.665 atm final partial pressure of O??? makes sense since there is 1/3 of nitrogen in the bag, 1/3 of .990atm, and there are 2/3 of oxygen in the bag, 2/3 of .990 which makes sense for the number to be 0.665...
am I right?