Дано:
w (C) = 14,12%
w (Cl) =
83,53%
D(возд) =
2,93
Найти:
CxHyClz
Решение:
w
(H) = 100% - w (Cl + C) = 100% - (83,53 + 14,12%) = 2,35%
Mr
(в-ва) = 2,93∙ 29 = 85
n (э) = w(э) ∙ Mr(в-ва) \Ar(э) ∙ 100%
n (C) = 14,12%
∙ 85 \12 ∙ 100% = 1
n (Cl) =
83,53% ∙ 85 \35,5 ∙ 100% = 2
n (H) = 2,35%
∙ 85\1 ∙ 100% = 2
n (C) : n
(H) : n (CL) = 1:2:2 = CxHyClz = С1H2Cl2
ответ: С1H2Cl2
– дихлорметан
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Verified answer
Дано:
w (C) = 14,12%
w (Cl) =
83,53%
D(возд) =
2,93
Найти:
CxHyClz
Решение:
w
(H) = 100% - w (Cl + C) = 100% - (83,53 + 14,12%) = 2,35%
Mr
(в-ва) = 2,93∙ 29 = 85
n (э) = w(э) ∙ Mr(в-ва) \Ar(э) ∙ 100%
n (C) = 14,12%
∙ 85 \12 ∙ 100% = 1
n (Cl) =
83,53% ∙ 85 \35,5 ∙ 100% = 2
n (H) = 2,35%
∙ 85\1 ∙ 100% = 2
n (C) : n
(H) : n (CL) = 1:2:2 = CxHyClz = С1H2Cl2
ответ: С1H2Cl2
– дихлорметан