a. I would use PV = nRT and solve for P (that gives you initial P). No T is listed but you can use any convenient T. Add mols O2 initially + 0.6 mol O2 and use PV = nRT again to solve for new volume. Use the old P you calculated and don't change T from what you used.
b. Redo PV = nRT with new V = 12.0 L. Solve for n O2 and subtract from total mols O2 to find moles removed.
c. Convert 7.5g He t mols, add to mols O2 available from part b and redo PV = nRT and solve for V.
A sample containing 6.40 g O2 gas has a volume of 15.0 L. Pressure and temperature remain constant.
a. What is the new volume if 0.600 mole O2 gas is added?
b. Oxygen is released until the volume is 12.0L. How many moles of O2 are removed?
c. What is the volume after 7.50 g He is added to the O2 gas already in the container? plss neeed help
3 answers
in a. I got 60 L for volume and .8 moles(.2+.6moles. now in b. i got .16 mole so where do I subtract .16 from should I subtract it from .8 moles or from the initial mole which is .2 mole
From the 0.8; that's where the volume was 60 L.
By the way there is a shorter way of doing this
Notice that if we use PV = nRT and solve for n = (PV/RT). Since P, R, and T are constant, the n = V; therefore,
(n1/V1) = (n2/V2). As an eample, for part a, (0.2/15) = (0.8/V2) and V2 = 60 which can be done in your head.
For part b, (0.8/60) = (n2/12); n2 = 0.16
By the way there is a shorter way of doing this
Notice that if we use PV = nRT and solve for n = (PV/RT). Since P, R, and T are constant, the n = V; therefore,
(n1/V1) = (n2/V2). As an eample, for part a, (0.2/15) = (0.8/V2) and V2 = 60 which can be done in your head.
For part b, (0.8/60) = (n2/12); n2 = 0.16
2 answers