Question

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

Answer 1

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