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Вычислите плотность: а) по кислороду; б) по азоту, в) по воздуху следующих газов:
1) аммиака NH3; сернистого газа SO2;
2) иодоводорода HI; оксида азота (I);
3) оксида азота (IV), фтороводорода HF.

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Ответ

Ответ:
1) `D_(O_2)(NH_3) = 0.531`, `D_(N_2)(NH_3) = 0.607`, `D_("возд.")(NH_3) = 0.586`, `D_(O_2)(SO_2) = 2`, `D_(N_2)(SO_2) = 2.285`, `D_("возд.")(SO_2) = 2.207`
2) `D_(O_2)(HI) = 4`, `D_(N_2)(HI) = 4.571`, `D_("возд.")(HI) = 4.414`, `D_(O_2)(N_2O) = 1.375`, `D_(N_2)(N_2O) = 1.571`, `D_("возд.")(N_2O) = 1.517`
3) `D_(O_2)(NO_2) = 1.438`, `D_(N_2)(NO_2) = 1.643`, `D_("возд.")(NO_2) = 1.586`, `D_(O_2)(HF) = 1.375`, `D_(N_2)(HF) = 1.571`, `D_("возд.")(HF) = 1.517`.

Дано: Решение

`"1) "NH_3, SO_2;`

`"2) "HI, N_2O;`

`"3) "NO_2, HF.`

1)

`D_(O_2)(NH_3) = 17/32 = 0.531`

`D_(N_2)(NH_3) = 17/28 = 0.607`

`D_("возд.")(NH_3) = 17/29 = 0.586`

`D_(O_2)(SO_2) = 64/32 = 2`

`D_(N_2)(SO_2) = 64/28 = 2.285`

`D_("возд.")(SO_2) = 64/29 = 2.207`

2)

`D_(O_2)(HI) = 128/32 = 4`

`D_(N_2)(HI) = 128/28 = 4.571`

`D_("возд.")(HI) = 128/29 = 4.414`

`D_(O_2)(N_2O) = 44/32 = 1.375`

`D_(N_2)(N_2O) = 44/28 = 1.571`

`D_("возд.")(N_2O) = 44/29 = 1.517`

3)

`D_(O_2)(NO_2) = 46/32 = 1.438`

`D_(N_2)(NO_2) = 46/28 = 1.643`

`D_("возд.")(NO_2) = 46/29 = 1.586`

`D_(O_2)(HF) = 44/32 = 1.375`

`D_(N_2)(HF) = 44/28 = 1.571`

`D_("возд.")(HF) = 44/29 = 1.517`

`D_(O_2)("газа") = ?`

`D_(N_2)("газа") = ?`

`D_("возд.")("газа") = ?`

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