What is the volume occupied by 2.00 moles of Ar(g) at STP? Another one I am stuck on...please help!

Our only unknown is the volume of H2(g) . Our known variables are P,n,R, and T. At STP, the temperature is 273K and the pressure is 1 atm. Rearrange the equation to solve for V: n×R×TP. V=2.20mol ×0.0821L×atm mol×K ×273K 1atm. V=49.3L

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Tringa0 Tringa0 · Answer 2 · 2021-09-23T14:08:31+02:00

The volume occupied by 2.00 moles of argon gas at STP is 44.8 Liters.

Given:

The 2.00 moles of argon gas at standard temperature and pressure:

To find:

The volume occupied by 2.00 moles of argon gas.

Solution:

At STP. the value of the temperature =T = 273.15 K

At STP. the value of the pressure=P =1 atm

The volume of 2.00 moles of argon gas at STP = V

The moles of argon gas = n = 2.00 mol

Using an ideal gas equation:

[tex]PV=nRT\\1atm\times V=2.00mol\times 0.0821 atm L/mol K\times 273.15 K\\V=\frac{2.00mol\times 0.0821 atm L/mol K\times 273.15 K}{1 atm}\\V=44.8 L[/tex]

The volume occupied by 2.00 moles of argon gas at STP is 44.8 Liters.

Learn more about the ideal gas here:

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