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Answer to Question #26111 in Inorganic Chemistry for salma

Question #26111
A flask contains 2.00 moles of nitrogen and 2.00 moles of helium. How many grams of argon must be pumped into the flask in order to make the partial pressure of argon twice that of helium?
Expert's answer
You can use the following law for solving this question:
PfVf = P1V1 + P2V2 + P3V3 … etc

Pf is final pressure of the mixture
Vf is Final volume of the mixture

P1 = Initial pressure of gas No. 1 before mixing
V1 = Initial volume of gas No. 1 before mixing
P2 = Initial pressure of gas No. 2 before mixing
V2 = Initial volume of gas No. 2 before mixing
The same with P3 and V3

P is proportional to amount
at the beginning P1V1 = P2V2 (2.00 moles of nitrogen and 2.00 moles of helium)

Adding some extra argon will cause decreasing
of nitrogen and helium pressure.

If you have 2.00 moles of nitrogen and 2.00 moles of helium and P is proportional to amount, the amount of argon that must be pumped into the flask in order to make the partial pressure of argon twice that of helium is twice amount of nitrogen. It is 4.00 mol

n = m/Mw;

m = n*Mw = 4.00*40 = 160 g

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Comments

Assignment Expert
02.02.17, 21:53

Dear visitor,
please use panel for submitting new questions

Cecilia
31.01.17, 19:37

Show that κ for the state equation P (Vm-b) = RT are given by
Κ =
1P
(
1
1 + bp / RT).

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