Answer:
4) Titration
Explanation:
Titration is a common process used to determine the concentration of acids. It does this by adding a solution of base with a known concentration to the acid until it reaches neutralization.
The graduated cylinder is used to measure the volume of KOH and H2SO4 when accurate volume measurement is not required.
In the laboratory certain graduated apparatus are used to measure liquids. These graduated apparatus used to measure liquids include;
- burette
- pipette
- measuring cylinder
- Erlenmeyer flask
Sometimes, we are not really looking for a strictly accurate volume of liquid and we can use a graduated cylinder to measure the volume of liquid in such cases.
However, when we need to have strictly accurate volume measurement, we need a pipet or a buret.
Learn more: brainly.com/question/15670537
Answer:
![[SO_2Cl_2] = 0.09983 M](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D%20%3D%200.09983%20M)
Explanation:
Write the balance chemical equation ,

initial concenration of 
lets assume that degree of dissociation=
concenration of each component at equilibrium:
![[SO_2Cl_2] = 0.1-0.1\alpha](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D%20%3D%200.1-0.1%5Calpha)
![[SO_2] = 0.1\alpha](https://tex.z-dn.net/?f=%5BSO_2%5D%20%3D%200.1%5Calpha)
![[Cl_2] = 0.1\alpha](https://tex.z-dn.net/?f=%5BCl_2%5D%20%3D%200.1%5Calpha)


as
is very small then we can neglect 
therefore ,



Eqilibrium concenration of ![[SO_2Cl_2] = 0.1-0.1\alpha = 0.1-0.1\times 0.00173](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D%20%3D%200.1-0.1%5Calpha%20%3D%200.1-0.1%5Ctimes%200.00173)
![[SO_2Cl_2] = 0.09983 M](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D%20%3D%200.09983%20M)
Answer:
0.1
Explanation:
We must first put down the equation of the reaction in order to guide our solution of the question.
2HNO3(aq) + Sr(OH)2(aq) -------> Sr(NO3)2(aq) + 2H2O(l)
Now from the question, the following were given;
Concentration of acid CA= ??????
Concentration of base CB= 0.299M
Volume of acid VA= 17.8ml
Volume of base VB= 24.7ml
Number of moles of acid NA= 2
Number of moles of base NB= 1
From;
CAVA/CBVB= NA/NB
CAVANB= CBVBNA
CA= CBVBNA/VANB
SUBSTITUTING VALUES;
CA= 0.299 × 24.7 ×2 / 17.8×1
CA= 0.8298 M
But;
pH= -log[H^+]
[H^+] = 0.8298 M
pH= -log[0.8298 M]
pH= 0.1