The temperature at which 28g of N, will occupy
a volume of 10.0L at 2.46atm is
(a) 299.6K (b) 0°C (©) 273K (d) 10°C
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Answer:
Option a
Explanation:
Apply ideal gas equation here
PV=nRT(where n is no.of moles)
2.46*10=28/28*0.0821*T
T=2.46*10/1*0.0821
=299.6K
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