Use a sheet of paper to answer the following question. Take a picture of your answers and attach to this assignment. Draw the st
ructure of 2,2,4,4-tetramethyl-3-pentanone and explain why it does not form an enol tautomer Explain why cyclohexa-2,4-dien-1-one (also called 2,4-cyclohexadienone) exists exclusively in the enol form
1 answer:
Answer:
Explanation:
Structure of the 2,2,4,4-tetramethyl-3-pentanone is give in the attachment
In 2,2,4,4-tetramethyl-3-pentanone, no alpha hydrogen is present, therefore, enol form is not possible and hence, exist only in keto form.
Explanation for existence of cyclohexa-2,4-diene-1-one only in enol form:
keto form of cyclohexa-2,4-diene-1-one not aromatic and hence less stable.
Whereas enol form it is aromatic which makes it highly stable. that's why cyclohexa-2,4-diene-1-one exists only in enol form.
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Answer:
1. 0.02 M
2. 0.01 M
3. 4×10⁻⁶
Explanation:
We know that V₁S₁ = V₂S₂
1.
Concentration of HCl = 0.05 M
end point comes at = 10 ml
So, concentration of OH⁻(aq) = [OH⁻(aq)] ⇒ (0.05 × 10) ÷ 25 ⇒ 0.02 M
2.
2mol of OH⁻(aq) ≡ 1 mole of Ca²⁺(aq)
[Ca²⁺] = 0.02 ÷ 2 = 0.01 M
3.
= [Ca²⁺(aq)] [OH⁻(aq)]²
Ca(OH)₂ (aq) ⇄ Ca²⁺ (aq) + 2OH⁻ (aq)
= [0.01 × (0.02)²] = 4×10⁻⁶
4.
If reaction is exothermic which means heat energy will get evolved as a result temperature of the reaction media will get increased during the course of the reaction. If temperature is externally increased, the reaction will go backward to accumulate extra heat energy.
5.
value describes the solubility of a particular ionic compound. The higher the
value, the higher the Solubility will be.
6.
This may be due to uncommon ion effect. The process of other ions (K⁺ or Na⁺) may increase the solubility
<u>Answer:</u> The amount of water required to prepare given amount of salt is 398.4 mL
<u>Explanation:</u>
To calculate the volume of solution, we use the equation used to calculate the molarity of solution:
We are given:
Molarity of solution = 0.16 M
Given mass of manganese (II) nitrate tetrahydrate = 16 g
Molar mass of manganese (II) nitrate tetrahydrate = 251 g/mol
Putting values in above equation, we get:
Volume of water = Volume of solution = 398.4 mL
Hence, the amount of water required to prepare given amount of salt is 398.4 mL