Sucrose decompose in acid solution into glucose and fructose according to the first order rate law, with `t_(1/2)=3.00` hours. What fraction of sample of sucrose remains after 8 hours ?

Sucrose decompose in acid solution into glucose and fructose according

1.	Sucrose decompose in acid solution into glucose and fructose according to the first order rate law, with `t_(1/2)=3.00` hours. What fraction of sample of sucrose remains after 8 hours ?
Answer» For `1^(st)` order reactions `k=(0.693)/(t_(1//2))=(0.693)/((3.h))` ` t=(2.303)/(k)log""([A]_(0))/([A])` `log ""([A]_(0))/([A])=(kxxt)/(2.303)` `log ""([A]_(0))/([A])=(0.693)/(3) xx((8h))/(2.303)=0.8024` `([A]_(0))/([A])="Antilog"0.8024=6.345` `[A]_(0)=1M,` `[A]=([A]_(0))/(6.345)=(1M)/(6.345)=0.1576M` After 8 hours sucrose left `=0.1576M`

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Sucrose decompose in acid solution into glucose and fructose according to the first order rate law, with `t_(1/2)=3.00` hours. What fraction of sample of sucrose remains after 8 hours ?

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