The initial number of moles of the unknown acid (HA) is equal to the final amount of the number of moles of the known base (NaOH) because they fully consumed each other.
<h3>What is neutralization reaction?</h3>
Neutralization reactions are those reactions in which acids and bases reacts with each other for the formation of water and salt.
Acids reacts with base to form salt means all gets consumed by each other so the initial moles of an unknown acid is equal to the final moles of the known base.
And reaction between them is shown as:
HA + NaOH → NaA + H₂O
Hence acids and bases fully consumed each other that's why we can calculate the moles of unknown substance.
To know more about neutralization reaction, visit the below link:
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Answer:
Colloidal can not be separated through filtration.
Suspension can be separated through filtration.
Explanation:
Colloidal:
Colloidal consist of the particles having size between 1 - 1000 nm i.e, 0.001- 1μm. While the pore size of filter paper is 2μm. That's why we can not separate the colloidal through the filtration. However it can be separated through the ultra filtration. In ultra filtration the pore size is reduced by soaking the filter paper in gelatin and then in formaldehyde. This is only in case of when solid colloidal is present, if colloid is liquid , there is no solid particles present and ultra filtration can not be used in this case.
Suspension:
The particle size in suspension is greater than 1000 nm. The particles in suspension can be separated through the filtration. These particles are large enough and can be seen through naked eye.
<span>The answer is the first option. Enzyme and substrate bind using a lock-and-key mechanism. Enzymes act on a specific substrate and a substrate needs a specific enzyme, this is what is called a lock-and-key mechanism. Enzymes and substtates are like a key and a lock, one is for each other.</span>
Explanation:
It is known that 
value of acetic acid is 4.74. And, relation between pH and is as follows.
pH = pK_{a} + log ![\frac{[CH_{3}COOH]}{[CH_{3}COONa]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_%7B3%7DCOOH%5D%7D%7B%5BCH_%7B3%7DCOONa%5D%7D)
= 4.74 + log 
So, number of moles of NaOH = Volume × Molarity
= 71.0 ml × 0.760 M
= 0.05396 mol
Also, moles of 
= moles of 
= Molarity × Volume
= 1.00 M × 1.00 L
= 1.00 mol
Hence, addition of sodium acetate in NaOH will lead to the formation of acetic acid as follows.
Initial : 1.00 mol 1.00 mol
NaoH addition: 0.05396 mol
Equilibrium : (1 - 0.05396 mol) 0 (1.00 + 0.05396 mol)
= 0.94604 mol = 1.05396 mol
As, pH = pK_{a} + log ![\frac{[CH_{3}COONa]}{[CH_{3}COOH]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_%7B3%7DCOONa%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D)
= 4.74 + log 
= 4.69
Therefore, change in pH will be calculated as follows.
pH = 4.74 - 4.69
= 0.05
Thus, we can conclude that change in pH of the given solution is 0.05.
Answer:
The value of dissociation constant of the monoprotic acid is 
.
Explanation:
The pH of the solution = 2.46
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![2.46=-\log[H^+]](https://tex.z-dn.net/?f=2.46%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=0.003467 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.003467%20M)
Initially
0.0144 0 0
At equilibrium
(0.0144-x) x x
The expression if an dissociation constant is given by :
![K_a=\frac{[A^-][H^+]}{[HA]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BA%5E-%5D%5BH%5E%2B%5D%7D%7B%5BHA%5D%7D)
![x=[H^+]=0.003467 M](https://tex.z-dn.net/?f=x%3D%5BH%5E%2B%5D%3D0.003467%20M)
The value of dissociation constant of the monoprotic acid is .
