Answer :
8.6 mole sample of NO₂ in a 5. 1 L container has 4.6 atmospheric pressure at 33 K.
Given, number of moles of No₂(n) = 8.6 mole
Volume (V) = 5.1 liters
Temperature (T) = 33K
R= 0. 08206 L•atm / (mol•K)
Using ideal gas equation ,
PV=nRT
P (5.1L) = (8.6mol) (0.08206L•atm/(mol•K)) (33K)
P (5.1) = 8.6 × 0.08206 × 33
P (5.1) = 23.3
P = 23.3 ÷ 5.1
P = 4.6 atm
Hence, the pressure of the given gas is 4.6 atmosphere (atm).
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