Дано:
$CH_{4}$
$NH_{3}$
$SO_{2}$
$O_{3}$
Найти:
$D_{H_{2}} (CH_{4}$) − ?
$D_{H_{2}} (NH_{3}$) − ?
$D_{H_{2}} (SO_{2}$) − ?
$D_{H_{2}} (O_{3}$) − ?
Решение:
$D_{H_{2}}$ = $\frac{M_{r}}{M_{r} (H_{2})}$
$M_{r}$ ($CH_{4}$) = 12 + 4 * 1 = 16
$M_{r}$ ($NH_{3}$) = 14 + 3 * 1 = 17
$M_{r}$ ($SO_{2}$) = 32 + 2 * 16 = 64
$M_{r}$ ($O_{3}$) = 3 * 16 = 48
$M_{r}$ ($H_{2}$) = 2 * 1 = 2
$D_{H_{2}} (CH_{4}$) = $\frac{16}{2}$ = 8
$D_{H_{2}} (NH_{3}$) = $\frac{17}{2}$ = 8,5
$D_{H_{2}} (SO_{2}$) = $\frac{64}{2}$ = 32
$D_{H_{2}} (O_{3}$) = $\frac{48}{2}$ = 24
Ответ: $D_{H_{2}} (CH_{4}$) = 8; $D_{H_{2}} (NH_{3}$) = 8,5; $D_{H_{2}} (SO_{2}$) = 32; $D_{H_{2}} (O_{3}$) = 24.