Answer:
Here you can use the Clausis Clayperon equation: ln P1/P2=-Ea/R-(1/T1 - 1/T2)
where P1 is the pressure at standard condition: 760 mm Hg
P2 is the variable we need to solve
Ea is the activation energy, which in this case is delta H vaporisation: 56.9 kJ/mol
R is the gas constant 8.314 J/mol or 8.314 J/mol /1000 to convert to kJ
T1 is the normal boiling point 356.7 C, but converted to Kelvin: 629.85K
T2 is room temperature 25 C, but converted to Kelvin: 298.15 K
Once you plug everything in, you should get 4.29*10^-3 mmHg
Explanation:
Answer:
SO42 - Trigonal planar ion
Explanation:
The SO4^2- ion is tetrahedral and not trigonal planar because the sulphur atom has four regions of electron density which includes the lone pair of electrons on sulphur atom.
This accounts for the observed tetrahedral arrangement of electron pairs around the central sulphur atom in SO4^2- ion, hence the answer.
Answer:
As the Mentos candy sinks in the bottle, the candy causes the production of more and more carbon dioxide bubbles, and the rising bubbles react with carbon dioxide that is still dissolved in the soda to cause more carbon dioxide to be freed and create even more bubbles, resulting in the eruption
Answer:
Determine the pH of the solution half-way to the end-point on the pH titration curve for acetic acid.
Explanation:
The equation for the ionization of acetic acid is
HA + H₂O ⇌ H₃O⁺ + A⁻
For points between the starting and equivalence points, the pH is given by the Henderson-Hasselbalch equation:
![\text{pH} = \text{pK}_{\text{a}} + \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}](https://tex.z-dn.net/?f=%5Ctext%7BpH%7D%20%3D%20%5Ctext%7BpK%7D_%7B%5Ctext%7Ba%7D%7D%20%2B%20%5Clog%5Cdfrac%7B%5B%5Ctext%7BA%7D%5E%7B-%7D%5D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D)
At the half-way point, half of the HA has been converted to A⁻, so [HA] = [A⁻]. Then,
The pKₐ is the pH at the half-way point in the titration.
