a)1 atm = 1,013 * 10^5 Pa
pV=nRT=m/μ*RT=> V = 168*0.082*273/1*28 = 134,316 l
b)28x+44y=516
x/y=3/2 => x=3y/2
28*3y/2+44y=516 => y = 6 moli
x = 3*6/2 = 9 moli
nr total moli = 6+9=15 moli
V c.n = 22,4 l
V=n*22,4 = 15*22.4=336 l
c)1 atm = 1,013 * 10^5 N/m2
pV=nRT=> n = pV/RT = 224*1/273*0.082= 10 moli amestec
echimolecular = raport 1:1 => 5 moli O2 si 5 moli N2
masa amestec =5*32+5*28 = 300 g
d) V=n*22,4=>n=112/22,4=5 moli
n=m/μ=>μ=355/5=71 g/mol
e) 760 mmHg = 1 atm
1 atm si 273K = conditii normale
22,4 l............2*6.022*10^23 atomi
0,0336 l ......................x
x = 18 * 10 ^20 atomi