Explanation:
The given reaction will be as follows.

So, equilibrium constant for this equation will be as follows.
![K_{c} = \frac{[CH_{3}OH]}{[CO][H_{2}]^{2}}](https://tex.z-dn.net/?f=K_%7Bc%7D%20%3D%20%5Cfrac%7B%5BCH_%7B3%7DOH%5D%7D%7B%5BCO%5D%5BH_%7B2%7D%5D%5E%7B2%7D%7D)
As it is given that concentration of all the species is 2.4. Therefore, calculate the value of equilibrium constant as follows.
![K_{c} = \frac{[CH_{3}OH]}{[CO][H_{2}]^{2}}](https://tex.z-dn.net/?f=K_%7Bc%7D%20%3D%20%5Cfrac%7B%5BCH_%7B3%7DOH%5D%7D%7B%5BCO%5D%5BH_%7B2%7D%5D%5E%7B2%7D%7D)
= 
= 0.173
Thus, we can conclude that equilibrium constant for the given reaction is 0.173.
The energy released when electron move from n=4 to n=3 is 0.66 eV
We know that in an atom energy of nth state is
eV
where n is the energy level
Therefore,

Thus,
= -0.85eV
= -1.51eV
Therefore, total mount of energy released in moving electron from n=4 to n=3 is given by -

= -0.85 - ( -1.51)
= 0.66eV
To know more about energy released in electron transition
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By giving the dog 2 or more different dog foods and seeing what one he eats the must of and the variables u control is how much you give the dog and the hypothesis is if I give the dog two or more different kinds of dog food than we will see what one he/she like the most
Negatively charged particles in atom - Electrons