a 2.48 liter sample of a He at 739 torr is mixed with a 6.08 L of Ar at 325 torr

Question

a 2.48 liter sample of a He at 739 torr is mixed with a 6.08 L of Ar at 325 torr. Both at 25 degrees Celsius is placed in a 4.80 liter container at 25 degrees Celsius. What is the partial pressure, ATM, of He (helium)? I need step by step clearly legible answer please thank you

4.803 24S L sample of He at 73 to k mised with 6 0S L of A at 325 tor a 4. SOL container at 25CThis es. A 2 4S L sample of He at 739 torr ki smixed with 60 L of t 325to the partial pressure (atm) ef d) 0.50 He a) 0.4S b) 0 54 c) 042 Ethylene, C H reacts with O according to the following equation. at STP is needed to react with 1.50 moles of C.H?

Explanation / Answer

a 2.48 liter sample of a He at 739 torr is mixed with a

6.08 L of Ar at 325 torr.

Both at 25 degrees Celsius is placed in a 4.80 liter container at 25 degrees Celsius.

What is the partial pressure, ATM, of He (helium)? I need step by step clearly legible answer please thank you

First calculate the moles of both gases:

He

PV =n RT

n = PV/RT

P = 739 torr = 0.972 atm

n = 0.972 * 2.48/0.08206*298

= 0.0986 mol He

Ar

PV =n RT

n = PV/RT

P = 325 torr = 0.428 atm

n = 0.428 * 6.08 /0.08206*298

= 0.106 mol Ar

Total number of moles = 0.106+0.0986

= 0.2046 Moles

Now calculate the pressure of mixture:

P = nRT/ v

= 0.206*0.08206*298/4.80

= 1.05 atm

Mole fraction of He = 0.0986/0.2046

= 0.48

partial pressure of He = Mole fraction x total pressure

= 0.48* 1.05 atm

= 0.504 atm

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