Answer:
I don't know this language please send me in English please
Answer:
a) Chloride dioxide

b) Di nitrogen tetra oxide

c) Potassium phosphide

d) Silver -Ag
e) Aluminium nitride - AlN
f) Silicon dioxide

I) Sulfide

Answer:
Equilibrium constant of the given reaction is 
Explanation:
....
....
The given reaction can be written as summation of the following reaction-


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
Equilibrium constant of this reaction is given as-
![\frac{[NOBr]^{2}}{[N_{2}][O_{2}][Br_{2}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BNOBr%5D%5E%7B2%7D%7D%7B%5BN_%7B2%7D%5D%5BO_%7B2%7D%5D%5BBr_%7B2%7D%5D%7D)
![=(\frac{[NOBr]}{[NO][Br_{2}]^{\frac{1}{2}}})^{2}(\frac{[NO]^{2}}{[N_{2}][O_{2}]})](https://tex.z-dn.net/?f=%3D%28%5Cfrac%7B%5BNOBr%5D%7D%7B%5BNO%5D%5BBr_%7B2%7D%5D%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D%29%5E%7B2%7D%28%5Cfrac%7B%5BNO%5D%5E%7B2%7D%7D%7B%5BN_%7B2%7D%5D%5BO_%7B2%7D%5D%7D%29)


Ethanol is polar because the oxygen atoms attract electrons because of their higher electronegativity than other atoms in the molecule. Thus the -OH group in ethanol has a slight negative charge. Ammonia (NH3) is polar. Sulfur dioxide (SO2) is polar.
B because it’s got the least connections