Here is the answer for your question which the mole of chlorine is 1.34
Empirical Formula = C20 H60 HG1 S1 O4 (numbers should be in subscript)
Percent Composition =
MG = 43%
S = 56.8%
If you would draw the Lewis structures of these atoms, you would see that A has 2 electron pairs and 2 lone electrons (that can bond). For B you’d see that you only have 1 electron that can form a bond. This means that 1 atom of A (2 lone electrons) can bond with 2 atoms of B. To know the kind of bond you have to know wether or not there will be a ‘donation’ of an electron from one atom to another. This happens when the number of electrons on one atoms is equal to the number of electrons another atom needs to reach the noble gas structure. As you can see, this is not the case here. This means that you get an AB2 structure with covalent character.
Answer : The concentration of NO is, 
Solution : Given,
Concentration of 
and 
= 152 M
Equilibrium constant, 
= 
The given equilibrium reaction is,
The expression of will be,
![K_c=\frac{[NO]^2}{[N_2][O_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO%5D%5E2%7D%7B%5BN_2%5D%5BO_2%5D%7D)
Now put all the given values in this expression, we get:
![2.0\times 10^{-9}=\frac{[NO]^2}{152\times 152}](https://tex.z-dn.net/?f=2.0%5Ctimes%2010%5E%7B-9%7D%3D%5Cfrac%7B%5BNO%5D%5E2%7D%7B152%5Ctimes%20152%7D)
![[NO]=6.8\times 10^{-3}M](https://tex.z-dn.net/?f=%5BNO%5D%3D6.8%5Ctimes%2010%5E%7B-3%7DM)
Therefore, the concentration of NO is,
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