Answer:
92.6%
Explanation:
Hello there!
In this case, according to the first-order kinetics, we can write:
[tex]\frac{A}{Ao}=exp(-kt)[/tex]
Whereas we need to know the rate constant in order to compute the percentages, given the half-life:
[tex]k=\frac{ln(2)}{9hrs}=0.077hrs^{-1}[/tex]
In such a way, we can now compute the fraction in which the reaction has been completed after 3 hrs:
[tex]\frac{A}{Ao}=exp(-0.077hrs^{-1}*3hrs)\\\\\frac{A}{Ao}=0.926[/tex]
And the percent would be 92.6% after multiplying the fraction by 100 %.
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