Answer:
The answer is 5.7 minutes
Explanation:
A first-order reaction follow the law of ![Ln [A] = -k.t + Ln [A]_{0}](https://tex.z-dn.net/?f=Ln%20%5BA%5D%20%3D%20-k.t%20%2B%20Ln%20%5BA%5D_%7B0%7D)
. Where <em>[A]</em> is the concentration of the reactant at any <em>t</em> time of the reaction, ![[A]_{0}](https://tex.z-dn.net/?f=%5BA%5D_%7B0%7D)
is the concentration of the reactant at the beginning of the reaction and <em>k</em> is the rate constant.
Dropping the concentration of the reactant to 6.25% means the concentration of A at the end of the reaction has to be ![[A]=\frac{6.25}{100}.[A]_{0}](https://tex.z-dn.net/?f=%5BA%5D%3D%5Cfrac%7B6.25%7D%7B100%7D.%5BA%5D_%7B0%7D)
. And the rate constant (<em>k</em>) is 8.10×10−3 s−1
Replacing the equation of the law:
![Ln \frac{6.25}{100}.[A]_{0} = -8.10x10^{-3}s^{-1}.t + Ln[A]_{0}](https://tex.z-dn.net/?f=Ln%20%5Cfrac%7B6.25%7D%7B100%7D.%5BA%5D_%7B0%7D%20%3D%20-8.10x10%5E%7B-3%7Ds%5E%7B-1%7D.t%20%2B%20Ln%5BA%5D_%7B0%7D)
Clearing the equation:
![Ln [A]_{0}.\frac{6.25}{100} - Ln [A]_{0} = -8.10x10^{-3}s^{-1}.t](https://tex.z-dn.net/?f=Ln%20%5BA%5D_%7B0%7D.%5Cfrac%7B6.25%7D%7B100%7D%20-%20Ln%20%5BA%5D_%7B0%7D%20%3D%20-8.10x10%5E%7B-3%7Ds%5E%7B-1%7D.t)
<em>Considering the property of logarithms: </em>
Using the property:
![Ln \frac{[A]_{0}}{[A]_{0}}.\frac{6.25}{100} = -8.10x10^{-3}s^{-1}.t](https://tex.z-dn.net/?f=Ln%20%5Cfrac%7B%5BA%5D_%7B0%7D%7D%7B%5BA%5D_%7B0%7D%7D.%5Cfrac%7B6.25%7D%7B100%7D%20%3D%20-8.10x10%5E%7B-3%7Ds%5E%7B-1%7D.t)
Clearing <em>t </em>and solving:
The answer is in the unit of seconds, but every minute contains 60 seconds, converting the units:
to see if its copatible with our environment or to see if there is any unkown element, otherwise i dont know
if wrong very sorry
1 mole of an alkane gives 2 moles is correct answer
Oxidation of D -Ribose in presence of hypobromous acid gives D-Ribonic acid
1.00
Explanation:
the density of water is always 1
