Question 5 1.00 g of He, 14.0g F2, and 19.0 g Ar are placed in a 13.0-L container at 20.0 °C.

The total pressure (in atm) in the container is atm.

1 point

2.03 atm

Explanation:

Number of moles of He = 1g/4g/mol = 0.25 moles

Number of moles of F2 = 14.0g/38 g/mol = 0.37 moles

Number of moles of Ar=19.0g/40g/mol = 0.48 moles

Total number of moles = 0.25 + 0.37 + 0.48 = 1.1 moles

From;

PV=nRT

P= pressure of the gas mixture

V= volume of the gas mixture

n= total number of moles of the gas mixture

R= gas constant

T= temperature of the gas mixture

P= nRT/V

P= 1.1 × 0.082 × 293/13

P= 2.03 atm

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