Answer:
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Answer:
Explanation:
![\rm MX(s) $\, \rightleftharpoons \,$ M$^{+}$(aq) + $^{-}$(aq); $K_{\text{sp}}$ = [M$^{+}$][X$^{-}$]\\\\\text{$K_{\text{sp}}$ gives us information on}\\\\\boxed{\textbf{ the equilibrium between the solid and its ions in solution}}](https://tex.z-dn.net/?f=%5Crm%20MX%28s%29%20%24%5C%2C%20%5Crightleftharpoons%20%5C%2C%24%20M%24%5E%7B%2B%7D%24%28aq%29%20%2B%20%24%5E%7B-%7D%24%28aq%29%3B%20%24K_%7B%5Ctext%7Bsp%7D%7D%24%20%3D%20%5BM%24%5E%7B%2B%7D%24%5D%5BX%24%5E%7B-%7D%24%5D%5C%5C%5C%5C%5Ctext%7B%24K_%7B%5Ctext%7Bsp%7D%7D%24%20gives%20us%20information%20on%7D%5C%5C%5C%5C%5Cboxed%7B%5Ctextbf%7B%20the%20equilibrium%20between%20the%20solid%20and%20its%20ions%20in%20solution%7D%7D)
It tells us nothing about the amount of precipitate that will form or the temperature at which the equilibrium occurs.
Answer:
At equilibrium, the concentration of the reactants will be greater than the concentration of the products. This does not depend on the initial concentrations of the reactants and products.
Explanation:
The value of Kc gives us an idea of the extent of the reaction. A big Kc (Kc > 1) means that in the equilibrium there are more products than reactants, and the opposite happens for a small Kc (Kc < 1). The equilibrium is reached no matter what the initial concentrations are.
The value of the equilibrium constant is relatively SMALL; therefore, the concentration of reactants will be GREATER THAN the concentration of products. This result is INDEPENDENT OF the initial concentration of the reactants and products.
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>
