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Flauer [41]
3 years ago
9

How many moles of oxygen must be placed in a 3.00 L container to exert a pressure of 2.00 atm at 25.0°C? Which variables are giv

en?
Chemistry
2 answers:
elixir [45]3 years ago
7 0
PV = nRT
n = PV/RT
n = (2)(3)/(.0821)(273*25)
n = 6/24.466 = 0.245
liubo4ka [24]3 years ago
5 0

Answer: 0.24 moles

Explanation: Using IDEAL GAS LAW

PV=nRT

where,

P = pressure of the gas= 2.0 atm

V = volume of the gas= 3.00L

T =Temperature of gas=25^0C=(25+273)K=298K

n = number of moles of gas=?

R = Gas constant = 0.0821 Latm/moleK

n=\frac {PV}{RT}=\frac {2.0\times 3.0}{0.0821\times 298}=0.24moles


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