Answer: The molar solubility of barium fluoride is 0.0183 moles/liter.
Explanation:
The equation for the reaction will be as follows:
By Stoichiometry,
1 mole of 
gives 2 moles of 
and 1 mole of 
Thus if solubility of is s moles/liter, solubility of is s moles/liter and solubility of is 2s moles/liter
Therefore,
![K_sp=[Ba^{2+}][F^{-}]^2](https://tex.z-dn.net/?f=K_sp%3D%5BBa%5E%7B2%2B%7D%5D%5BF%5E%7B-%7D%5D%5E2)
![2.45\times 10^{-5}=[s][2s]^2](https://tex.z-dn.net/?f=2.45%5Ctimes%2010%5E%7B-5%7D%3D%5Bs%5D%5B2s%5D%5E2)
Thus the molar solubility of barium fluoride is 0.0183 moles/liter.
Checking if it is balanced:
C
3 3
H
8 4*2 = 8
O
8 3*2 + 4 = 6 + 4 = 10
8 10
Only O is not balanced.
So O is violating the law. Option C.
<span>C. O</span>
Hydrazine has a higher boiling point than water because hydrazine exhibits a stronger intermolecular forces than water. Although both compounds exhibit hydrogen bonding which is a strong intermolecular force resulting to a high boiling point, hydrogen bonding is stronger in hydrazine. This is because in a hydrazine molecule, there are two atoms of nitrogen available for hydrogen bonding as compared to water which has only 1 oxygen atom for hydrogen bonding.
