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3 years ago

What is the temperature (°C) of 1.75 g of O2 gas occupying 4.60 L at 1.00 atm?

Chemistry

1 answer:

julsineya [31]3 years ago

4 0

Given- m= 1.75 g

Now moles of O2 gas= Mass/ Molar mass

n = 1.75/32

V= 2L

P= 1atm

R=0.0821.....Constant

As PV=nRT

=> T=PV/nR

= 1* 2/ (1.75/32)*0.0821 K

= 2*32/1.75* 0.0821 K

= 445.45 K

= (445.45- 273.15) degrees celsius

= 172.30 degrees celsius..... ANSWER

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This is a incomplete question. The complete question is:

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