Answer:
![[H_2]_{eq}=0.183M](https://tex.z-dn.net/?f=%5BH_2%5D_%7Beq%7D%3D0.183M)
![[I_2]_{eq}=0.183M](https://tex.z-dn.net/?f=%5BI_2%5D_%7Beq%7D%3D0.183M)
![[HI]_{eq}=0.025M](https://tex.z-dn.net/?f=%5BHI%5D_%7Beq%7D%3D0.025M)
Explanation:
Hello.
In this case, for this equilibrium problem, we first realize that at the beginning there is just HI, it means that the reaction should be rewritten as follows:

Whereas the law of mass action (equilibrium expression) is:
![Kc=\frac{[H_2][I_2]}{[HI]^2}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BH_2%5D%5BI_2%5D%7D%7B%5BHI%5D%5E2%7D)
That in terms of initial concentrations and reaction extent or change
turns out:
![Kc=\frac{x*x}{([HI]_0-2x)^2}\\\\54.3=\frac{x^2}{(0.391M-2x)^2}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7Bx%2Ax%7D%7B%28%5BHI%5D_0-2x%29%5E2%7D%5C%5C%5C%5C54.3%3D%5Cfrac%7Bx%5E2%7D%7B%280.391M-2x%29%5E2%7D)
And the solution via solver or quadratic equation is:

Whereas the correct answer is 0.183 M since the other value yield a negative concentration of HI at equilibrium (0.391-2*0.210=-0.029M).This, the equilibrium concentrations are:
![[H_2]_{eq}=0.183M](https://tex.z-dn.net/?f=%5BH_2%5D_%7Beq%7D%3D0.183M)
![[I_2]_{eq}=0.183M](https://tex.z-dn.net/?f=%5BI_2%5D_%7Beq%7D%3D0.183M)
![[HI]_{eq}=0.391M-2*0.183M=0.025M](https://tex.z-dn.net/?f=%5BHI%5D_%7Beq%7D%3D0.391M-2%2A0.183M%3D0.025M)
Regards.
Answer:The kinetic-molecular theory of gases assumes that ideal gas molecules (1) are constantly moving; (2) have negligible volume; (3) have negligible intermolecular forces; (4) undergo perfectly elastic collisions; and (5) have an average kinetic energy proportional to the ideal gas's absolute temperature.
Explanation:
Answer:
1.750 M
Explanation:
In case of titration , the following formula can be used -
M₁V₁ = M₂V₂
where ,
M₁ = concentration of acid ,
V₁ = volume of acid ,
M₂ = concentration of base,
V₂ = volume of base .
from , the question ,
M₁ = ? M
V₁ = 7.14 mL
M₂ = 0.250 M
V₂ = 50 mL
Using the above formula , the molarity of acid , can be calculated as ,
M₁V₁ = M₂V₂
M₁ * 7.14 mL = 0.250 M * 50mL
M₁ = 1.750 M
Hence, the molarity of the hydrochloric acid = 1.750 M