Bài giải:
- Ta có: số Avogađro(`N`) : `1(mol)≈6.10^{23}` (phân tử)
Bài 4:
`-n_{CO_2}=\frac{9.10^{23}}{N}=\frac{9.10^{23}}{6.10^{23}}=1,5(mol)`
`⇒V_{CO_2}(đktc)=n_{CO_2}.22,4=1,5.22,4=33,6(l)`
`⇒m_{CO_2}=n_{CO_2}.M_{CO_2}=1,5.44=66(g)`
Bài 5:
`-n_{N_2}=\frac{1,5.10^{23}}{N}=\frac{1,5.10^{23}}{6.10^{23}}=0,25(mol)`
`⇒V_{N_2}(đktc)=n_{N_2}.22,4=0,25.22,4=5,6(l)`
`⇒m_{N_2}=n_{N_2}.M_{N_2}=0,25.28=7(g)`
