I need help with with number 1515. How many moles of NO2 are in a 37.058 L container at 101.28 kPa and -139.86
1 answer:
Explanation:
To solve this question, we will use the Clayperon Equation:
P.V = n.R.T
where:
P = 101.28 kPa
1 atm = 101,325 Pa
x atm = 101,280 Pa
x = 1 atm
V = 37.058 L
n = we don't know
R = 0.082 atm.L/K.mol
T = -139.88 ºC = -139.88+273.15 = 133.27 K
1*37.058 = n*0.082*133.27
n = 0.29 moles
Answer: 0.29 moles
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