Дано:
$Cl_{2}$
$HCl$
Найти:
$D_{H_{2}}$ ($Cl_{2}$) − ?
$D_{H_{2}}$ (HCl) − ?
$D_{возд}$ ($Cl_{2}$) − ?
$D_{возд}$ (HCl) − ?
Решение:
$M_{r}$ ($Cl_{2}$) = 2 * 35,5 = 71
$M_{r}$ ($HCl$) = 1 + 35,5 = 36,5
$M_{r}$ ($H_{2}$) = 2 * 1 = 2
$D_{H_{2}}$ ($Cl_{2}$) = $\frac{M_{r} (Cl_{2})}{M_{r} (H_{2})}$ = $\frac{71}{2}$ = 35,5
$D_{H_{2}}$ (HCl) = $\frac{M_{r} (HCl)}{M_{r} (H_{2})}$ = $\frac{36,5}{2}$ = 18,25
$D_{возд}$ ($Cl_{2}$) = $\frac{M_{r} (Cl_{2})}{M_{r} (возд)}$ = $\frac{71}{29}$ = 2,5
$D_{возд}$ (HCl) = $\frac{M_{r} (HCl)}{M_{r} (возд)}$ = $\frac{36,5}{29}$ = 1,26
Ответ:$D_{H_{2}}$ ($Cl_{2}$) = 35,5; $D_{H_{2}}$ (HCl) = 18,25; $D_{возд}$ ($Cl_{2}$) = 2,5; $D_{возд}$ (HCl) = 1,26.