Answer:
6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.
Explanation:
![Rate = k[AB]^2](https://tex.z-dn.net/?f=Rate%20%3D%20k%5BAB%5D%5E2)
The order of the reaction is 2.
Integrated rate law for second order kinetic is:
![\frac{1}{[A_t]} = \frac{1}{[A]_0}+kt](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA_t%5D%7D%20%3D%20%5Cfrac%7B1%7D%7B%5BA%5D_0%7D%2Bkt)
Where, ![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration = 1.50 mol/L
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the final concentration = 1/3 of initial concentration = 
= 0.5 mol/L
Rate constant, k = 0.2 L/mol*s
Applying in the above equation as:-
<u>6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.</u>
Answer:
i think snowball, it sounds weird but its true (i think im sorry if its wrong)
Explanation:
Answer: the answer is X.
Explanation:
Think of it as a graph with potential and kinetic energy
To solve this we use the
equation,
M1V1 = M2V2
where M1 is the concentration of the stock solution, V1 is the
volume of the stock solution, M2 is the concentration of the new solution and
V2 is its volume.
2 M x V1 = 1.50 M x 100 mL
<span>V1 = 75 mL of the 2 M KNO3 solution</span>
