Answer:
pH at equivalence point is 8.52
Explanation:
1 mol of HCOOH reacts with 1 mol of NaOH to produce 1 mol of 
So, moles of NaOH used to reach equivalence point equal to number of moles produced at equivalence point.
As density of water is 1g/mL, therefore molarity is equal to molality of an aqueous solution.
So, moles of produced = 
Total volume of solution at equivalence point = (25+29.80) mL = 54.80 mL
So, at equivalence point concentration of = 
At equivalence point, pH depends upon hydrolysis of . So, we have to construct an ICE table.
I: 0.1940 0 0
C: -x +x +x
E: 0.1940-x x x
So, ![\frac{[HCOOH][OH^{-}]}{[HCOO^{-}]}=K_{b}(HCOO^{-})=\frac{10^{-14}}{Ka(HCOOH)}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BHCOOH%5D%5BOH%5E%7B-%7D%5D%7D%7B%5BHCOO%5E%7B-%7D%5D%7D%3DK_%7Bb%7D%28HCOO%5E%7B-%7D%29%3D%5Cfrac%7B10%5E%7B-14%7D%7D%7BKa%28HCOOH%29%7D)
species inside third bracket represent equilibrium concentrations
So, 
or,
So, 
So, 
So, ![pH=14-pOH=14+log[OH^{-}]=14+logx=14+log(3.285\times 10^{-6})=8.52](https://tex.z-dn.net/?f=pH%3D14-pOH%3D14%2Blog%5BOH%5E%7B-%7D%5D%3D14%2Blogx%3D14%2Blog%283.285%5Ctimes%2010%5E%7B-6%7D%29%3D8.52)
It is very cold there 90 degrees celsius so dress very warm!
The atomic number is the number of protons in the nucleus
We need to know the value of van't hoff factor.
The van't hoff factor is: 2.66 or 2.7 (approximately)
(NH₄)₂SO₄ is an ionic compound, so it dissociates in solution and produces 3 ionic species. Therefore van't hoff factor is more than one.
From the equation: Δ
=i 
.m, where Δ= elevation of boiling point=102.5 - 100=2.5°C.
m=molality of solute=1.83 m (Given)
= Ebullioscopic constant or Boiling point elevation constant= 0.512°C/m (Given)
i= Van't Hoff factor
So, 2.5= i X 0.512 X 1.83
i=
i=2.66= 2.7 (approx.)
False, because physical change does not involve the formation of substances. An example of physical change is change of state e.g water freezing .It is still water if you were to melt it.
