1.

For a chemical reaction aAtoB,log[-(d[A])/(dt)]=log[(d[B])/(dt)]+0.3 then find the approximate ratio and of a and b

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Solution :`aAtobB-(d[A])/(dt)=(_d[B])/(dt)xx(a/b)`
`LOG((dA)/(dt))=log((DB)/(dt))+log(a/b)` So `log (a/b)=0.3=log2,a/b=2`


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