The question is incomplete, here is the complete question:
At elevated temperature, nitrogen dioxide decomposes to nitrogen oxide and oxygen gas

The reaction is second order for
with a rate constant of
at 300°C. If the initial [NO₂] is 0.260 M, it will take ________ s for the concentration to drop to 0.150 M
a) 1.01 b) 5.19 c) 0.299 d) 0.0880 e) 3.34
<u>Answer:</u> The time taken is 5.19 seconds
<u>Explanation:</u>
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 = ?
[A] = concentration of substance after time 't' = 0.150 M
= Initial concentration = 0.260 M
Putting values in above equation, we get:

Hence, the time taken is 5.19 seconds
Neon I think. Go to the periodic table and see which one is the 11th
MgCl2(s) + H2O(l) → MgO(s) + 2 HCl(g)
Using the standard enthalpies of formation given in the source below:
(−601.24 kJ) + (2 x −92.30 kJ) − (−641.8 kJ) − (−285.8 kJ) = +141.76 kJ
So:
MgCl2(s) + H2O(l) → MgO(s) + 2 HCl(g), ΔH = +141.76 kJ