Răspuns:
MHNO3 =1+14+(3×16)=15+48=63g/mol
MNO=14+16=30g/mol
63gHNO3......,....14gN
6,3gHNO3............X
X=(6.3×14)÷63=1,4gN
numărul de moli n=m÷M
n=6,3÷63=0,1 moli acid azotic HNO3
numărul de atomi
1mol HNO3......6,022×10 ^23(10la puterea 23)
0.1moli................Y
Y=0,1×6,022×10^23
M NO =14+16=30g/mol
30g NO...............14gN
60g NO................X
X=(60×14)÷30=28gN azot
numărul de moli
n=m÷M
n=60÷30=2 moli NO
numărul de atomi
1mol.....6,022×10^23 atomi
2 moli...........X
X=(2×6,022×10^23)÷1
X=12,044×10^23 atomi azot N