Question

Ce numar de particule contin:
a) 0,01 Kmoli CO
b) 0,224 dm la a 3 O2
c) 2 kg C2H2 ( ACETILENA)

Answer 1

a) 0.01Kmoli=10 moli CO

1mol CO................6.022*10^23 particule/mol
10 mol..................x=6.022*10^24=6022*10^27 particule/mol

b) n=V/Vmolar=0.224/22.4=0.01 moli O2

1mol....................6.022*10^23 particule/mol
0.01mol...............x=6.022*10*21 particule/mol


c)  MC2H2=26g/mol=26kg/kmol

n=m/M=2/26=0.076923 kmoli=76.923 moli C2H2

1mol....................6.022*10^23 particule/mol
76.923 mol.........x=463.23*10^23=46323*10^21 particule/mol