Asked by Mark
A gas mixture in a 1.55L at 298K container contains 10.0g Ne and 10.0g Ar. Calculate the partial pressure " in atm" of Ne and Ar in the container
Answers
Answered by
DrBob222
n Ne = grams/molar mass = ?
n Aar = grams/molar mass = ?
Then PV = mRT and use n Ne to solve for pNe. Use n for Ar to solve for pAr.
n Aar = grams/molar mass = ?
Then PV = mRT and use n Ne to solve for pNe. Use n for Ar to solve for pAr.
Answered by
urWelcome
10.0 g Ne / 20.18 g/mol = 0.496 mol
P*1.55 L = 0.496 mol * 0.08206 L*atm/mol*K * 298 K
P of Ne = 7.82 atm
10.0 g Ar / 39.948 g/mol = 0.25 mol
P*1.55 L = 0.25 mol * 0.08206 L*atm/mol*K * 298 K
P of Ar = 3.94 atm
P*1.55 L = 0.496 mol * 0.08206 L*atm/mol*K * 298 K
P of Ne = 7.82 atm
10.0 g Ar / 39.948 g/mol = 0.25 mol
P*1.55 L = 0.25 mol * 0.08206 L*atm/mol*K * 298 K
P of Ar = 3.94 atm
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.