No units are provided in the data table, though one would usually assume that 0.244 is molarity of NaOH, and buret data of 14.7 is probably the final volume of NaOH in mL. No identification is given for the 38, but maybe this is the initial buret volume in mL.
If the buret was rinsed with water, it should have been left to dry for some time, or else the added water may slightly dilute the NaOH solution placed in it.
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
MAke two piles and spread them apart
Answer:
What forms of energy does a burning candle give out?
- The burning candle is converting chemical energy—stored in the hydrocarbon bonds in the wax—into thermal (heat) and radiant (light) energy.
What questions do you have about energy?
- Can momentum be hidden to human eyes like how kinetic energy can be hidden as heat?
- Can you go fast enough to get enough mass to become a black hole?
- How can a material at a certain temperature have all of its molecules at the same energy?
- How can we travel to the past?
Nacl = (. 23+35. 5)
= 58.5g
.
1 mol of Nacl = 58.5g
X mol Of Nacl = 25g
X Mol of Na Cl =25 ÷ 58.5
X mol of Nacl = 0.4 mol
