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Nadia1999
@Nadia1999
March 2022
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задание во вложении 3 и 4
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mrvladimir2
Verified answer
C4H10 => C2H6 + C2H4
C2H4 + Br2 => C2H4Br2
C2H4Br2 => C2H2 + 2HBr (р-я идет в две стадии)
4. М(УВ)=29 г/моль*3,93=114 г/моль
m(C) = 114*0.8421=96 г
n(C) = 96/12=8
m(H) = 114-96=18 г
n(H) = 18/1=18
Следовательно, ф-ла С8Н18
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Answers & Comments
Verified answer
C4H10 => C2H6 + C2H4C2H4 + Br2 => C2H4Br2
C2H4Br2 => C2H2 + 2HBr (р-я идет в две стадии)
4. М(УВ)=29 г/моль*3,93=114 г/моль
m(C) = 114*0.8421=96 г
n(C) = 96/12=8
m(H) = 114-96=18 г
n(H) = 18/1=18
Следовательно, ф-ла С8Н18