<span>a. x and y are atoms of the same element.
If both atoms contain the same amount of protons, they are always the same element.
</span>
Answer:
By absorbing energy electron is jump into higher energy level. This is called excitation.
Explanation:
The electron is jumped into higher level and back into lower level by absorbing and releasing the energy.
The process is called excitation and de-excitation.
Excitation:
When the energy is provided to the atom the electrons by absorbing the energy jump to the higher energy levels. This process is called excitation. The amount of energy absorbed by the electron is exactly equal to the energy difference of orbits. For example if electron jumped from K to L it must absorbed the energy which is equal the energy difference of these two level. The excited electron thus move back to lower energy level which is K by releasing the energy.
De-excitation:
When the excited electron fall back to the lower energy levels the energy is released in the form of radiations. this energy is exactly equal to the energy difference between the orbits. The characteristics bright colors are due to the these emitted radiations. These emitted radiations can be seen if they are fall in the visible region of spectrum.
Answer:
The value of the equilibrium constant for reaction asked is 
.
Explanation:
![K_{goal}=\frac{[C][O_2]}{[CO_2]}](https://tex.z-dn.net/?f=K_%7Bgoal%7D%3D%5Cfrac%7B%5BC%5D%5BO_2%5D%7D%7B%5BCO_2%5D%7D)
..[1]
![K_1=\frac{[CH_3COOH][O_2]^2}{[CO_2]^2[H_2O]^2}](https://tex.z-dn.net/?f=K_1%3D%5Cfrac%7B%5BCH_3COOH%5D%5BO_2%5D%5E2%7D%7B%5BCO_2%5D%5E2%5BH_2O%5D%5E2%7D)
..[2]
![K_2=\frac{[H_2O]^2}{[H_2]^2[O_2]}](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B%5BH_2O%5D%5E2%7D%7B%5BH_2%5D%5E2%5BO_2%5D%7D)
..[3]
![K_3=\frac{[C]^2[H_2]^2[O_2]}{[CH_3COOH]}](https://tex.z-dn.net/?f=K_3%3D%5Cfrac%7B%5BC%5D%5E2%5BH_2%5D%5E2%5BO_2%5D%7D%7B%5BCH_3COOH%5D%7D)
[1] + [2] + [3]
( on adding the equilibrium constant will get multiplied with each other)
![K=\frac{[C]^2[O_2]^2}{[CO_2]^2}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BC%5D%5E2%5BO_2%5D%5E2%7D%7B%5BCO_2%5D%5E2%7D)
On comparing the K and 
:
The value of the equilibrium constant for reaction asked is .
