please need help with this question, has all tge info mments Question 6 temperat
ID: 519426 • Letter: P
Question
please need help with this question,
has all tge info
Explanation / Answer
air moleucles is
V = 13*12*10 = 1560 ft3 = 44.17428 m3 = 44174.28 L
P = 1 atm, T = 20°C = 293 K
so
P V= nRT
n = PV/(TR)
n = (1)(44174.28)/(293*0.082)
n = 1838.60 moles or air
1 mol = 6.022*10^23 molecules
1838.60 mol --> 1838.60* 6.022*10^23 = 1.10*10^27 molecules of air
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