Answer:
activity coefficient 
activity coefficient 
The change in pH in part A = 0.092
The change in pH in part B = 0.102
Explanation:
From the given information:
pH of HCl solution = 1.092
Activity of the pH solution [a] = 
[a] = 0.0809 M
Recall that [a] =
× C
where;
= activity coefficient
C = concentration
Making the activity coefficient the subject of the formula, we have:
![\gamma = \dfrac{[a]}{C}](https://tex.z-dn.net/?f=%5Cgamma%20%3D%20%5Cdfrac%7B%5Ba%5D%7D%7BC%7D)


B.
The pH of a solution of HCl and KCl = 2.102
[a] = 
[a] = 0.00791 M
activity coefficient 

C. The change in pH in part A = 1.091 - 1.0 = 0.092
The change in pH in part B = 2.102 -2.00 = 0.102
Answer:
Follows are the solution to this question:
Explanation:
- Fact: In this, the ice is melted at 0 ° C.
- law: It is used to repeated experiments consistently showed which objects marked to both the contrary attract each law.
- Hypothesis: If carbohydrates and nitrogen are combined at 1500 ° C, they interact with one another.
- Theory: Protons, nuclei, or ions are made of Atoms. As these molecules could not be seen explicitly, such legal structure several experimental observations.
Answer is: 0,133 mol/ l· atm.
T(chlorine) = 10°C = 283K.
p(chlorine) = 1 atm.
V(chlorine) = 3,10 l.
R - gas constant, R = 0.0821 atm·l/mol·K.
Ideal gas law: p·V = n·R·T
n(chlorine) = p·V ÷ R·T.
n(chlorine) = 1atm · 3,10l ÷ 0,0821 atm·l/mol·K · 283K = 0,133mol.
Henry's law: c = p·k.
k - <span>Henry's law constant.
</span>c - solubility of a gas at a fixed temperature in a particular solvent.
c = 0,133 mol/l.
k = 0,133 mol/l ÷ 1 atm = 0,133 mol/ l· atm.
Answer:
0.3023 M
Explanation:
Let Picric acid = 
So,
+
⇄
+ 
The ICE table can be given as:
+
⇄
+ 
Initial: 0.52 0 0
Change: - x + x + x
Equilibrium: 0.52 - x + x + x
Given that;
acid dissociation constant (
) = 0.42
![K_a = \frac{[H_3O^+][Picric^-]}{H_{picric}}](https://tex.z-dn.net/?f=K_a%20%3D%20%5Cfrac%7B%5BH_3O%5E%2B%5D%5BPicric%5E-%5D%7D%7BH_%7Bpicric%7D%7D)
![0.42 = \frac{[x][x]}{0.52-x}}](https://tex.z-dn.net/?f=0.42%20%3D%20%5Cfrac%7B%5Bx%5D%5Bx%5D%7D%7B0.52-x%7D%7D)
![0.42 = \frac{[x]^2}{0.52-x}}](https://tex.z-dn.net/?f=0.42%20%3D%20%5Cfrac%7B%5Bx%5D%5E2%7D%7B0.52-x%7D%7D)
0.42(0.52-x) = x²
0.2184 - 0.42x = x²
x² + 0.42x - 0.2184 = 0 -------------------- (quadratic equation)
Using the quadratic formula;
; ( where +/- represent ± )
= 
= 
=
OR 
=
OR 
=
OR 
= 0.30225 OR - 0.72225
So, we go by the +ve integer that says:
x = 0.30225
x = [
] = [
] = 0.3023 M
∴ the value of [H3O+] for an 0.52 M solution of picric acid = 0.3023 M (to 4 decimal places).