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greatlorde
@greatlorde
August 2022
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w%(Br) в монобромпроизводном алкана 52,98%. Определите м.ф алкана
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alima1958
Verified answer
СnH2n+1 Br 52.98%-----Br 80г
100%------xг
х=100%*80/52,98% = 151г /моль 151-80=71 СnH2n+1=71
12n +2n +1=71
14n=70
n=5 M(C5H11Br)=151г/моль
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Answers & Comments
Verified answer
СnH2n+1 Br 52.98%-----Br 80г100%------xг
х=100%*80/52,98% = 151г /моль 151-80=71 СnH2n+1=71
12n +2n +1=71
14n=70
n=5 M(C5H11Br)=151г/моль