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
Your independent variable would be salt because it doesn't change the only thing that changes is the water.
Answer:
There are three possible chemical equations for the combustion of sulfur:
- 2S (s) + O₂ (g) → 2SO (g)
- 2S (s) + 3O₂ (g) → 2SO₃ (g)
Explanation:
<em>Combustion</em> is a reaction with oxygen. The products of the reaction are oxides, and energy is released in the form of heat and light.
<em>Sulfur</em> iis a nonmetal, so the oxide formed is a nonmetal oxide.
The most common oxidation numbers of sulfur are -2, + 2, + 4, and + 6.
The combination of sulfur with oxygen may be only with the positive oxidation numbers (+2, + 4, and +6).
Then you have three different equations for sulfur combustion:
<u>1) Oxidation number +2:</u>
Which when balanced is: 2S(g) + O₂(g) → 2SO(g)
<u>2) Oxitation number +4:</u>
That equation is already balanced.
<u>3) Oxidation number +6:</u>
Which when balanced is: 2S(s) + 3O₂(g) → 2SO₃(g)
Answer:
SnO2 + 2H2 + energy ➡️ Sn + 2H2O
Explanation
