[H+] = ?
pH = -log[H+]
[H+] = 10^-pH
[H+] = 10^-11,3
[H+] = 5,01 × 10^-12 mol/l
Answer: H+ concentration in this solution equal 5,01 × 10^-12 mol/l.
:-) ;-)
Answer:
<em>The pH of the solution is 7.8</em>
Explanation:
The concentration of the solution is 0.001M and the dye could be in its protonated and deprotonated forms. If the concentration of the protonated form [HA] is 0.0002 M the concentration of the deprotonated form will be the subtraction between the concentration of the bye and the concentration of the protonated form:
[A-] = 0.001M - 0.0002M = 0.0008M
Also, the Henderson-Hasselbalch equation is
![pH = pKa + Log \frac{[A^{-}]}{[HA]}](https://tex.z-dn.net/?f=pH%20%3D%20pKa%20%2B%20Log%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
this equation shows the dependency between the pH of the solution, the pKa and the concentration of the protonated and deprotonated forms. Thus, replacing in the equation
Explanation:
A compound that contains more stronger bonds will need more amount of heat in order to break the bonds so that it changes into vapor state.
In 1-propanol, there is hydrogen bonding and it is stronger in nature. As a result, more amount of heat is required to break the bonds between molecules of 1-propanol.
Whereas in propanone, there will be dipole-dipole interactions which are less stronger than hydrogen bonding. Hence, propanone molecule will need less amount of energy than 1-propanol.
On the other hand, pentane will need more amount of heat as it has longer chain of carbon atoms as compared to methane.
Thus, we can conclude that given compounds are arranged in order of increasing vapor pressure as follows.
methane < pentane < propanone < 1-propanol
I think it's either B or D
Explanation:
it really depends on how differently shaped the beakers are sorry if this doesn't help very much
