Cứu tui vs ạ xia xìa gấc nhiều

$\rm n_{Cl_2} = \dfrac{9,033\times10^{21}}{6,022\times 10^{23}} = 0,015$ mol

$\rm n_{O_2} = \dfrac{12,044\times 10^{21}}{6,022 \times 10^{23}} = 0,02$ mol

a)$\rm m_X = m_{Cl_2} + m_{O_2}$

$\rm \to m_X = 0,015 \times 71+0,02 \times 32 = 1,705g$

$\rm V_X = (0,015 + 0,02) \times 24,79 = 0,86765L$

b)$\rm M_X = \dfrac{1,705}{0,015 + 0,02} = \dfrac{341}{7}g/mol$

$\rm d_{X/O_3} = \dfrac{ \dfrac{341}{7}}{48} \approx 1,015$

Vậy khí $\rm X$ năng gấp $\rm 1,015$ lần khí Ozone

$\rm n_{Cl_2}=\dfrac{9,033.10^{21}}{6.10^{23}}=0,015(mol)$ 

$\rm n_{O_2}=\dfrac{12,044.10^{21}}{6.10^{23}}=0,02(mol)$ 

`a)` 

$\rm m_{Cl_2}=0,015.71=1,065(g)$ 

$\rm m_{O_2}=0,02.32=0,64(g)$ 

$\rm V_{X}=(0,015+0,02).24,79=0,86765(l)$ 

`b)` 

$\rm M_{O_3}=16.3=48(g/mol)$ 

$\rm M_X=\dfrac{0,015.71+0,02.32}{0,035}=48,71(g/mol)$ 

Tỉ lệ $\rm \dfrac{M_X }{M_{O_3}}=\dfrac{48,71}{48}=1,015(lần)$