honestly, who knows. I just want to take a s.h.t.t right now.
(2.03x10^22)/(6.02x10^23) = .033721 mol Li
I hope this helps, if not, i am sorry
Answer:
2.48 mol/L.
Explanation:
- The molarity of the solution can be expressed as <em>the number of moles of solute in 1.0 liter of the solution, </em>(M = n / V).
- It is also can be calculated from the relation:
<em>M = (mass / molar mass) solute x (1000 / V of solution)</em>
The solute is toluene and the solvent is benzene.
mass of toluene (solute) = 57.1 g,
molar mass of toluene (solute) = 92.14 g/mol.
volume of the solution = 250 ml.
∴ M = (mass / molar mass) solute x (1000 / V of solution) = [(57.1 g / 92.14 g/mol) x (1000 / 250 ml)] = 2.48 mol/L.
They differ in their molecular structures and properties.
Answer:
![\frac{[F^{-}]}{[HF]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BF%5E%7B-%7D%5D%7D%7B%5BHF%5D%7D)
is larger
Explanation:
, where 
is the acid dissociation constant.
For a monoprotic acid e.g. HA, ![K_{a}=\frac{[H^{+}][A^{-}]}{[HA]}](https://tex.z-dn.net/?f=K_%7Ba%7D%3D%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
and ![\frac{[A^{-}]}{[HA]}=\frac{K_{a}}{[H^{+}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D%3D%5Cfrac%7BK_%7Ba%7D%7D%7B%5BH%5E%7B%2B%7D%5D%7D)
So, clearly, higher the value , lower will the the 
In this mixture, at equilibrium, ![[H^{+}]](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D)
will be constant.
of HF is grater than of HCN
Hence, ![(\frac{F^{-}}{[HF]}=\frac{K_{a}(HF)}{[H^{+}]})>(\frac{CN^{-}}{[HCN]}=\frac{K_{a}(HCN)}{[H^{+}]})](https://tex.z-dn.net/?f=%28%5Cfrac%7BF%5E%7B-%7D%7D%7B%5BHF%5D%7D%3D%5Cfrac%7BK_%7Ba%7D%28HF%29%7D%7B%5BH%5E%7B%2B%7D%5D%7D%29%3E%28%5Cfrac%7BCN%5E%7B-%7D%7D%7B%5BHCN%5D%7D%3D%5Cfrac%7BK_%7Ba%7D%28HCN%29%7D%7B%5BH%5E%7B%2B%7D%5D%7D%29)
So, is larger
