What volume is occupied by 0.25 moles of nitrogen gas at 273 K and 1.50 atm.

asked Dec 25, 2021 22.2k views

3 votes

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1 vote

Answer:

3.74L

Step-by-step explanation:

PV=nRT

V=nRT/P

= 0.25mol*0.0821LatmK^-1*273k/1.50atm

=3.74L

Master Chief answered Dec 31, 2021

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