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uysha [10]
3 years ago
8

How many moles of oxygen gas are there in 395 L of oxygen at STP?

Chemistry
1 answer:
kozerog [31]3 years ago
7 0

Answer:

17.6 moles of oxygen gas.

Explanation:

STP Variables:

P=1 atm

R=0.082

T=273 K

Use the PV=nRT, then plug in.

n=PV/RT

n= (1 atm)(395 L)/(0.082) (273 K)

Simplify.

n=17.6 moles of O2

Hoped this helped.

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Answer:

P_1=206.7lbmol/h\\\\P_2=413.3lbmol/h

Explanation:

Hello!

In this case, for the separation of the benzene-toluene mixture, we can use the following mole balances including the given mole fraction at each stream per species:

B:0.36*620=0.98P_1+0.05P_2\\\\T:0.64*620=0.02P_1+0.95P_2

That can be solved by using a solver for P1 (benzene-rich flow) and P2 (toluene-rich flow):

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Explanation:

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