A 3.00 L flask contains 2.33 g of argon gas at 312 mm Hg What is the temperature of the gas

T= 257.36 k

Explanation:

using the ideal gas law

pv=nRt

first, convert pressure from mmhg to kpa

312 x (101.3\760)= 41.58 kpa

R is constant= 8.31

to get n(number of moles)

n=m\M          (m is mass, M is molar mass)

molar mass of argon is 39.948

n= 2.33\39.948

n=0.0583

substitute;

41.58 x 3 = 0.0583 x 8.31 x T

T= (41.58 x 3)\ (0.0583 x 8.31)

T= 257.36 k

the correct answer to this question is 2

the correct answer would be b. amelia, because the particals of gas move really fast (charged particals) and gas also has a lot of room (particals move independently) so the only answer that has both of those is amelia's.