Answer: The pH at the equivalence point for the titration will be 0.65.
Solution:
Let the concentration of ![[OH^-]](https://tex.z-dn.net/?f=%5BOH%5E-%5D)
be x
Initial concentration of ![[CH3NH_2]](https://tex.z-dn.net/?f=%5BCH3NH_2%5D)
, c = 0.230 M
at eq'm c-x x x
Expression of 
:
![K_b=\frac{[CH_3NH_3^+][+OH^-]}{[CH_3NH_2]}=\frac{x\times x}{c-x}=\frac{x^2}{c-x}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%7B%5BCH_3NH_3%5E%2B%5D%5B%2BOH%5E-%5D%7D%7B%5BCH_3NH_2%5D%7D%3D%5Cfrac%7Bx%5Ctimes%20x%7D%7Bc-x%7D%3D%5Cfrac%7Bx%5E2%7D%7Bc-x%7D)
Since ,methyl-amine is a weak base,c>>x so 
.
Solving for x, we get:
Given, HCl with 0.230 M , it dissociates fully in water which means ![[H^+]](https://tex.z-dn.net/?f=%5BH%5E%2B%5D)
= 0.230 M
![[OH^-]=[H^+]](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5BH%5E%2B%5D)
will result in neutral solution, since ![[OH^-]](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3C%5BH%5E%2B%5D)
Remaining after neutralizing ions
![[H^+]_{\text{left in solution}}=[H^+]-[OH^-]=0.230-1.07\times 10^{-2}=0.2193 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D_%7B%5Ctext%7Bleft%20in%20solution%7D%7D%3D%5BH%5E%2B%5D-%5BOH%5E-%5D%3D0.230-1.07%5Ctimes%2010%5E%7B-2%7D%3D0.2193%20M)
![pH=-log{[H^+]_{\text{left in solution}}=-log(0.2193)=0.65](https://tex.z-dn.net/?f=pH%3D-log%7B%5BH%5E%2B%5D_%7B%5Ctext%7Bleft%20in%20solution%7D%7D%3D-log%280.2193%29%3D0.65)
The pH at the equivalence point for the titration will be 0.65.
1.) MgO + Fe --> FeO + Mg
2.) H + I --> HI
3.) Na + I --> NaI
4.) NaO + H2O --> NaOH + H
The rate of effusion of ammonia (NH₃) in the same apparatus is 63.3 cm/min
<h3>Graham's law of diffusion </h3>
This states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass i.e
R ∝ 1/ √M
R₁/R₂ = √(M₂/M₁)
<h3>How to determine the rate of ammonia (NH₃) </h3>
- Rate of HCl (R₁) = 43.2 cm/min
- Molar mass of HCl (M₁) = 1 + 35.5 = 36.5 g/mol
- Molar mass of NH₃ (M₂) = 14 + (3×1) = 17 g/mol
R₁/R₂ = √(M₂/M₁)
43.2 / R₂ = √(17 / 36.5)
Cross multiply
43.2 = R₂ × √(17 / 36.5)
Divide both side by √(17 / 36.5)
R₂ = 43.2 / √(17 / 36.5)
R₂ = 63.3 cm/min
Thus, the rate of effusion of ammonia is 63.3 cm/min
Learn more about Graham's law of diffusion:
brainly.com/question/14004529
Your instructor is probably expecting the answer B) III only.
III) The evolution or <em>absorption of heat</em> is evidence of a <em>chemical change</em>.
I) Boiling water and IV) mixing paint are <em>physical changes</em>.
II) A change in colour on mixing is usually an indication of a reaction (but not when mixing paint). However, in this case, the lighter blue colour might have happened simply because you are diluting the solution. You would have to do an experiment which is correct.
Answer:
0.52 L.
Explanation:
Let P be the initial pressure.
From the question given above, the following data were obtained:
Initial pressure (P1) = P
Initial volume (V1) = 1.04 L
Final pressure (P2) = double the initial pressure = 2P
Final volume (V2) =?
The new volume (V2) of the gas can be obtained by using the the Boyle's law equation as shown below:
P1V1 = P2V2
P × 1.04 = 2P × V2
1.04P = 2P × V2
Divide both side by 2P
V2 = 1.04P /2P
V2 = 0.52 L
Thus, the new volume of the gas is 0.52 L.
