Explanation:
The given data is as follows.
= 0.042 M,
for 
According to the given situation
acts as a base.The reaction equation will be as follows.

Relation between
and
are as follows.


= 
= 
Also,
Let us take
= x
So, ![K_{b} = \frac{[HP^{-}][OH^{-}]}{P^{2-}}](https://tex.z-dn.net/?f=K_%7Bb%7D%20%3D%20%5Cfrac%7B%5BHP%5E%7B-%7D%5D%5BOH%5E%7B-%7D%5D%7D%7BP%5E%7B2-%7D%7D)
x = 
= 
pOH = - log![[OH^{-}]](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%5D)
= - log (
)
= 4.99
As it is known that pH + pOH = 14
so, pH + 4.99 = 14
pH = 9.01
Thus, we can conclude that pH of the solution is 9.01.
The answer is flunking a class. Flunking a class during
college years is the most stressful item in the College Undergraduate Stress
Scale (CUSS). Flunking a class means failing it, which results to more stress,
it will make you take it again, but it depends on that subject’s availability.
You’re chances of graduating early may be hindered because of a flunked class. There is nothing more stressful than flunking a class.
Answer:
This question is incomplete
Explanation:
This question is incomplete as the volume of the base that was used during the titration was not provided. However, the completed question is in the attachment below.
The formula to be used here is CₐVₐ/CbVb = nₐ/nb
where Cₐ is the concentration of the acid = unknown
Vₐ is the volume of the acid used = 25 cm³ (as seen in the question)
Cb is the concentration of the base = 0.105 mol/dm³ (as seen in the question)
Vb is the volume of the base = 22.13 cm³ (22.1 + 22.15 + 22.15/3)
nₐ is the number of moles of acid = 1 (from the chemical equation)
nb is the number of moles of base = 2 (from the chemical equation)
Note that the Vb was based on the concordant results (values within the range of 0.1 cm³ of each other on the table) of the student
Cₐ x 25/0.105 x 22.13 = 1/2
Cₐ x 25 x 2 = 0.105 x 22.13 x 1
Cₐ x 50 = 0.105 x 22.13
Cₐ = 0.105 x 22.13/50
Cₐ = 0.047 mol/dm³
The concentration of the sulfuric acid is 0.047 mol/dm³
Based on Pauling Scale, electro negativity of Cl = 3.2, Na = 0.9 and H = 2.1
Thus, Electronegativity difference in

= 3.2 -3.2 = 0
Electronegativity difference in NaCl = 3.2-0.9 = 2.3
Similarly, Electronegativity difference in HCl = 3.2 - 2.1 = 1.1
Thus, among the listed molecules following is the decreasing order of electronegativity difference: NaCl> HCl >