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the balanced chemical equation representing the reaction between represent sodium metal (Na) and chlorine gas (cl2) is as below
2 Na + Cl2→ 2 NaCl
2 moles of sodium metal ( Na) reacted with 1 mole of gas chlorine gas (cl2) and 2 moles of sodium chloride (NaCl)
Answer : The concentration of HI (g) at equilibrium is, 0.643 M
Explanation :
The given chemical reaction is:

Initial conc. 0.10 0.10 0.50
At eqm. (0.10-x) (0.10-x) (0.50+2x)
As we are given:

The expression for equilibrium constant is:
![K_c=\frac{[HI]^2}{[H_2][I_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BHI%5D%5E2%7D%7B%5BH_2%5D%5BI_2%5D%7D)
Now put all the given values in this expression, we get:

x = 0.0713 and x = 0.134
We are neglecting value of x = 0.134 because the equilibrium concentration can not be more than initial concentration.
Thus, we are taking value of x = 0.0713
The concentration of HI (g) at equilibrium = (0.50+2x) = [0.50+2(0.0713)] = 0.643 M
Thus, the concentration of HI (g) at equilibrium is, 0.643 M
Explanation:
The ideal gas law or equation can also be reduced to the general gas law or the combined gas law.
Let us assume that n = 1;
From PV = nRT ideal gas law
= R (constant) if n = 1
Therefore, the combined gas law is;

The expression is a combination of Boyle's Law and Charles's law.
learn more:
Boyle's law brainly.com/question/8928288
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