Answer : The molarity after a reaction time of 5.00 days is, 0.109 M
Explanation :
The integrated rate law equation for second order reaction follows:
![k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7Bt%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%5BA%5D_o%7D%5Cright%29)
where,
k = rate constant = 
t = time taken = 5.00 days
[A] = concentration of substance after time 't' = ?
![[A]_o](https://tex.z-dn.net/?f=%5BA%5D_o)
= Initial concentration = 0.110 M
Now put all the given values in above equation, we get:
![9.7\times 10^{-6}=\frac{1}{5.00}\left (\frac{1}{[A]}-\frac{1}{(0.110)}\right)](https://tex.z-dn.net/?f=9.7%5Ctimes%2010%5E%7B-6%7D%3D%5Cfrac%7B1%7D%7B5.00%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%280.110%29%7D%5Cright%29)
![[A]=0.109M](https://tex.z-dn.net/?f=%5BA%5D%3D0.109M)
Hence, the molarity after a reaction time of 5.00 days is, 0.109 M
Answer:
Which line represents travel at the fastest speed? Justify your answer.__
b. Describe the line that should be added for an object that is not moving with respect to the chosen reference point.__
Answer the question using 5 complete sentences and the vocabulary terms; velocity, position, and distance
Which line represents travel at the fastest speed? Justify your answer.__
b. Describe the line that should be added for an object that is not moving with respect to the chosen reference point.__
Answer the question using 5 complete sentences and the vocabulary terms; velocity, position, and distance
Explanation:
Answer:
5 1 2 4and 3 this is correct way
Answer:
<h2>All Group 1 metals form halides that are white solids at room temperature. The melting point is correlated to the strength of intermolecular</h2>
Answer:
Most common insulation materials work by slowing conductive heat flow and--to a lesser extent--convective heat flow. Radiant barriers and reflective insulation systems work by reducing radiant heat gain. To be effective, the reflective surface must face an air space.
Explanation:
To be effective, the reflective surface must face an air space.
