Answer:
Explanation:
To calculate pH you need to use Henderson-Hasselbalch formula:
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
Where HA is the acid concentration and A⁻ is the conjugate base concentration.
The equilibrium of acetic acid is:
CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka: 4,75
Where <em>CH₃COOH </em>is the acid and <em>CH₃COO⁻ </em>is the conjugate base.
Thus, Henderson-Hasselbalch formula for acetic acid equilibrium is:
pH = 4,75 + log₁₀ ![\frac{[CH_{3}COO^-]}{[CH_{3}COOH]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_%7B3%7DCOO%5E-%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D)
a) The pH is:
pH = 4,75 + log₁₀ ![\frac{[2 mol]}{[2 mol]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B2%20mol%5D%7D%7B%5B2%20mol%5D%7D)
<em>pH = 4,75</em>
<em></em>
b) The pH is:
pH = 4,75 + log₁₀ ![\frac{[2 mol]}{[1mol]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B2%20mol%5D%7D%7B%5B1mol%5D%7D)
<em>pH = 5,05</em>
<em></em>
I hope it helps!
Answer:
80cm3 of water, and 60cm3 carbon IV oxide is formed while 20cm3 of oxygen is left unreacted.
Explanation:
From Gay-Lussac's law, there are five volumes of oxygen, 1 volume if propane, 4 volumes of water and three volumes of CO2. Applying this shows the reacting volumes as we have in the image attached, hence the volumes left after reaction.
D. Making the reactant particles larger
2KMnO4+3Na2SO3+H2O→2MnO2+3Na2SO4+2KOH
In a reaction, the reducing agent is the element or compound that donates electron or the one tht loses electrons. The oxidized species. The opposite is called the oxidizing agent. It is the one who accepts the electrons lost. For this reaction KMnO4 is reduced into MnO2.
The elements in each group have the same number of electrons in the outer orbital. Or also called valence electrons. Khan academy has a great video online explaining why this happens. (It only happens for main group elements). Here is a link (sorry you can’t click it in Brainly) https://www.khanacademy.org/science/chemistry/periodic-table/copy-of-periodic-table-of-elements/v/periodic-table-valence-electrons. Feel free to message me for a better explanation, I would explain now but I’m not sure how much you know about this. If you know how to write an electron configuration you can see how all the electron configurations for the same group (not the transitional metals only the main groups) have the same number of valence electrons. I hope that helped, sorry I was vague about the explanation :)