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
The element used in advertising signs is neon.
Answer:
-162,5 kJ/mol
Explanation:
Cl(g) + 2O2(g) --> ClO(g) + O3(g) ΔH = 122.8 kJ/mol (as we used the reaction in the opposite direction, it will turn the enthalpy from exothermic to endothermic)
2O3(g) --> 3O2(g) ΔH = -285.3 kJ/mol
Cl(g) + O2(g) --> ClO(g) + O3(g) ΔH = 122.8 kJ
+ 2O3 (g) --> 3O2(g) ΔH = - 285.3 kJ
O3(g) + Cl(g) --> ClO(g) + 2O2(g) ΔH = 122.8 + (-285.3) = -162,5 kJ
