Answer:
The electronengativity values of given elements is as follows.
Fluorine - 4
Chlorine -3
Bromine - 2.9
Iodine- 2.5
Explanation:
Electronegativity =consant (I.E-E.A)
The electron affinity and ionization energy values of the given elements is as follows.
(In attachment)
First we have to find the value of constant by using the fluorine atom to whom the electronengativity taken as "4".
<u>Fluorine:</u>
![4=constant[1678-(-327.8)]](https://tex.z-dn.net/?f=4%3Dconstant%5B1678-%28-327.8%29%5D)

By using this constant values we can find electronegatvity values of remaining elements.
<u>Chlorine:</u>
![Electronegativity=0.0019942168[1255+348.7]=3.1980\sim 3](https://tex.z-dn.net/?f=Electronegativity%3D0.0019942168%5B1255%2B348.7%5D%3D3.1980%5Csim%203)
Therefore, electronegativity of chlorine is 3.
<u>Bromine:</u>
![Electronegativity=0.0019942168[1138+324.5]=2.91\sim 2.9](https://tex.z-dn.net/?f=Electronegativity%3D0.0019942168%5B1138%2B324.5%5D%3D2.91%5Csim%202.9)
Therefore, electronegativity of bromine is 2.9.
<u>Iodine:</u>
![Electronegativity=0.0019942168[1007+295.7]=2.59\sim 2.5](https://tex.z-dn.net/?f=Electronegativity%3D0.0019942168%5B1007%2B295.7%5D%3D2.59%5Csim%202.5)
Therefore, electronegativity of iodine is 2.5.
Answer:
Mg + O2 ---> MgO
Explanation: V2O5 + CaS → CaO + V2S5. 5V2O5 + 5CaS → 10CaO + 5V2S5. 3V2O5 + 3CaS → 3CaO + 3V2S5.
Answer: 
Explanation: The given chemical reaction is:

Equilibrium constant (Kc) in general is written as:
![K_c=\frac{[products]}{[reactants]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5Bproducts%5D%7D%7B%5Breactants%5D%7D)
Note:- Coefficients are written as their powers
So, the Kc expression for the above reaction will be:
![K_c=\frac{[NO]^2[Br_2]}{[NOBr]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO%5D%5E2%5BBr_2%5D%7D%7B%5BNOBr%5D%5E2%7D)
Equilibrium moles are given for all of them. Let's divide the moles by given liters to get the concentrations.
= 0.040 M
= 0.0658 M
= 0.0217 M
Plug in the values in the equilibrium expression to calculate Kc.


Answer:
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