This problem is incomplete. Luckily, I found a similar problem from another website shown in the attached picture. The data given can be made to use through the Clausius-Clapeyron equation:
ln(P₂/P₁) = (-ΔHvap/R)(1/T₂ - 1/T₁)
where
P₁ = 14 Torr * 101325 Pa/760 torr = 1866.51 Pa
T₁ = 345 K
P₂ = 567 Torr * 101325 Pa/760 torr = 75593.78 Pa
T₂ = 441 K
ln(75593.78 Pa/1866.51 Pa) = (-ΔHvap/8.314 J/mol·K)(1/441 K - 1/345 K)
Solving for ΔHvap,
<em>ΔHvap = 48769.82 Pa/mol or 48.77 kPa/mol</em>

Let's solve ~
Initial concentration of weak acid HA = 0.200 M
and dissociation constant (
) is :


Now, at initial stage :
At equilibrium :
Now, we know :
![\qquad \sf \dashrightarrow \: { K_a = \dfrac{[H+] [A-]}{[HA]}}](https://tex.z-dn.net/?f=%5Cqquad%20%5Csf%20%20%5Cdashrightarrow%20%5C%3A%20%7B%20K_a%20%3D%20%5Cdfrac%7B%5BH%2B%5D%20%5BA-%5D%7D%7B%5BHA%5D%7D%7D%20)
( big brackets represents concentration )




Answer:
The symbol for an atom can be written to show its mass number at the top, and its atomic number at the bottom. To calculate the numbers of subatomic particles in an atom, use its atomic number and mass number: number of protons = atomic number. number of electrons = atomic number.
I answered that in your last question refer to your answer in section 2
The conjugate base of ch3ch2co2h or propanoic acid would be CH3CH2COO- or propanoate. The conjugate base is what is left after an acid has donated a proton in a reaction. Hope this answers the question. Have a nice day.