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aikenbeiker
@aikenbeiker
July 2022
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При сжигании углеводорода получено 15,4 г оксида углерода(4) и 7,2 г воды. Определите формулу, если плотность паров по водороду равна 50
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bearcab
Verified answer
N(C)=n(CO2)=15.4/44=0.35моль
2n(H)=n(H2O)=7.2*2/18=0.8моль
C:H=0.35/0.35:0.8/0.35=(1:2.285)7=7:16
Mr(CnH2n+2)=2*50=100
14n+2=100
14n=100-2
n=98/14
n=7
C7H16
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Thanks 2
bearcab
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Answers & Comments
Verified answer
N(C)=n(CO2)=15.4/44=0.35моль2n(H)=n(H2O)=7.2*2/18=0.8моль
C:H=0.35/0.35:0.8/0.35=(1:2.285)7=7:16
Mr(CnH2n+2)=2*50=100
14n+2=100
14n=100-2
n=98/14
n=7
C7H16