Answer:
i am pretty sure its A
Explanation:
they are etheir together like black and white or some sort of pattern if notice.
If they are like B,C,D like that they are mixtures which are NOT PURE SUBSTANCE'S.
Question:
<em>For an exothermic reaction at equilibrium, how will increasing the temperature affect Keq?</em>
Answer:
<em>The reaction will proceed towards the liquid phase. Heat is on the reactant side of the equation. Lowering temperature will shift equilibrium left, creating more liquid water. A reaction that is exothermic releases heat, while an endothermic reaction absorbs heat.</em>
<em>If you increase the temperature, the position of equilibrium will move in such a way as to reduce the temperature again. It will do that by favouring the reaction which absorbs heat. In the equilibrium, that will be the back reaction because the forward reaction is exothermic.</em>
Hope this helps, have a good day. c;
25. C
26. B
27. A
28. D
29. D
30. A
31. B
32. D
33. B
34. A
Answer: Equilibrium concentration of ![[Cl^-]](https://tex.z-dn.net/?f=%5BCl%5E-%5D)
at 
is 4.538 M
Explanation:
Initial concentration of 
= 0.056 M
Initial concentration of 
= 4.60 M
The given balanced equilibrium reaction is,
![COCl_2+2Cl^-\rightleftharpoons [CoCl_4]^{2-}+6H_2O](https://tex.z-dn.net/?f=COCl_2%2B2Cl%5E-%5Crightleftharpoons%20%5BCoCl_4%5D%5E%7B2-%7D%2B6H_2O)
Initial conc. 0.056 M 4.60 M 0 M 0 M
At eqm. conc. (0.056-x) M (4.60-2x) M (x) M (6x) M
The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[CoCl_4]^{2-}\times [H_2O]^6}{[CoCl_2]^2\times [Cl^-]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCoCl_4%5D%5E%7B2-%7D%5Ctimes%20%5BH_2O%5D%5E6%7D%7B%5BCoCl_2%5D%5E2%5Ctimes%20%5BCl%5E-%5D%5E2%7D)
Given : equilibrium concentration of ![[CoCl_4]^{2-}](https://tex.z-dn.net/?f=%5BCoCl_4%5D%5E%7B2-%7D)
=x = 0.031 M
Concentration of = (4.60-2x) M = 
=4.538 M
Thus equilibrium concentration of at is 4.538 M
