Explanation:
The given reaction is as follows.

Initial : 0.160 0.160 0
Change : -x -x 2x
Equilibrium: 0.160 - x 0.160 - x x
It is given that
= [0.160 - x] = 0.036 M
and,
= [0.160 - x] = 0.036 M
so, x = (0.160 - 0.036) M
= 0.124 M
As, [HI] = 2x.
So, [HI] = 
= 0.248 M
As it is known that expression for equilibrium constant is as follows.
![K_{eq} = \frac{[HI]^{2}}{[H_{2}][I_{2}]}](https://tex.z-dn.net/?f=K_%7Beq%7D%20%3D%20%5Cfrac%7B%5BHI%5D%5E%7B2%7D%7D%7B%5BH_%7B2%7D%5D%5BI_%7B2%7D%5D%7D)
= 
= 47.46
Thus, we can conclude that the equilibrium constant, Kc, for the given reaction is 47.46.
Answer:
1st box:4
2nd box:2
3rd box:3
4th box:1
Explanation:
I am not 100% sure on this, but i am pretty confident that this is the answer! BUT If this is somehow wrong, please tell me so i can comment a better answer! :)
Answer:
Oxidation - reduction reaction which is called redox reaction.
Explanation:
The type of reaction that is a part of another type but treated separately is called Oxidation - reduction reaction.
Thus is because they usually involve 2 separate half reactions which are oxidation reactions that involve loss of electrons and then reduction equations which involve gain of electrons.
They are treated separately for example when sodium reacts with chlorine to form sodium chloride.
2Na + Cl2 = 2NaCl
Now, sodium undergoes oxidation by loosing elctrons and it's half reaction is;
2Na → 2Na^(+) + 2e^(-)
Meanwhile chlorine undergoes reduction by gaining electrons and its half reaction is;
Cl2 + 2e^(-) → 2Cl^(-)
Answer:
Hydrogen
Explanation:
A reducing agent is a substance which gives up its electrons to become oxidized. Generally, metals are oxidized (reducing agents) while non-metals are reduced (oxidizing agents).
However, hydrogen which is a non-metal is usually oxidized in the presence of stronger oxidizing non-metals such as fluorine and oxygen.
Hydrogen thus, acts as a reducing agent by giving up its electrons to become oxidized. Even though among all non-metals, Hydrogen has the greatest potential to be oxidized, it is a poor reducing agent compared to reactive metals.