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 a is the correct one
Answer:
1. filtration and evaporation
2. i) water is added to the sand and salt mixture
ii) then the mixture is filtrated and so the sand and the salt water was seperated
iii) the salt water is heated with the help of burner until the water gets evaporated
iv) after the water gets evaporated, the salt is remained in the container
3. observation:
- on adding water to the mixture, the salt is completely dissolved in the water
- when filtrated the sand is seperated from the salt water
- now the salt water when heated with the burner until the evaporation, the water is evaporated
- the salt is precipitated and remained in the container
4. cautions:
- while using the burner, we should be cautious with fire
- the container that is heated should be holded with the help of a cloth to avoid heat
Answer:
3–methyl–2–butanol
Explanation:
To name the compound, we must:
1. Identify the functional group.
2. Give the functional group of the compound the lowest possible count.
3. Locate the longest continuous carbon chain. This gives the parent name of the compound.
4. Identify the substituent group attached.
5. Give the substituent group the lowest possible count.
6. Combine the above to get the name of the compound.
Now, let us obtain the name of the compound.
1. The functional group of the compound is Alcohol i.e —OH.
2. The functional group is located at carbon 2.
3. The longest continuous carbon chain is carbon 4 i.e butane. But the presence of the functional group i.e OH will replace the –e in butane with –ol. Therefore, the compound is butanol.
4. The substituent group attached is methyl i.e CH3.
5. The substituent group is located at carbon 3.
6. Therefore, the name of the compound is:
3–methyl–2–butanol.
2 ICl + H2 ----> I2 + 2 HCl
as given that rate is first order with respect to ICl and second order with respect to H2
The rate law will be
Rate = K [ICl] [ H2]^2
b) Given that K = 2.01 M^-2 s^-1
Concentrations are
[ICl] = 0.273 m and [H2] = 0.217 m
Therefore rate = 2.01 X (0.273)(0.217)^2 = 0.0258 M / s
