Answer:
Phenols do not exhibit the same pka values as other alcohols;
They are generally more acidic.
Using the knowledge that hydrogen acidity is directly related to the stability of the anion formed, explain why phenol is more acidic than cyclohexane.
Explanation:
According to Bromsted=Lowry acid-base theory,
an acid is a substance that can release
ions when dissolved in water.
So, acid is a proton donor.
If the conjugate base of an acid is more stable then, that acid is a strong acid.
In the case of phenol,
the phenoxide ion formed is stabilized by resonance.

The resonance in phenoxide ion is shown below:
Whereas in the case of cyclohexanol resonance is not possible.
So, cyclohexanol is a weak acid compared to phenol.
Answer : The reaction rate at
are 
Solution : Given,
Mass of tablet = 1000 mg
Volume of water = 0.200 L
The given formula will be,

Now we have to calculate the reaction rate at different temperatures and reaction time.




Therefore, the reaction rate at
are 
Answer:
high electronegativity difference between oxygen and hydrogen
Explanation:
The electronegativity value of O and H are 3.5 and 2.1 respectively. The electronegativity difference of the O-H bond in H₂O is 1.4.
- When the elctronegativity difference between two atoms in bond is below 0.5 the bond is nonpolar covalent.
- When the elctronegativity difference between two atoms in bond is between 0.5 and 2.0 the bond is polar covalent.
- When the elctronegativity difference between two atoms in bond is above 0.5 the bond is ionic.
Thus the O-H is a polar covalent bond. This explains the polarity in water molecules.
Answer : The value and unit of the rate constant for this reaction is,
and
respectively.
Explanation :
As we know that half-life of zero-order and second-order reaction depend on the concentration of reactant while the half-life of first-order reaction is independent of the concentration of the reactant.
The expression of half-life for zero order reaction:
![t_{1/2}=\frac{[A_o]}{2k}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B%5BA_o%5D%7D%7B2k%7D)
The expression of half-life for first order reaction:
The expression of half-life for second order reaction:
![t_{1/2}=\frac{1}{[A_o]k}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B1%7D%7B%5BA_o%5Dk%7D)
The given reaction can not be first order reaction because it is independent of the concentration of the reactant.
So, the given reaction can be zero order or second order.
First we have to calculate the value of rate constants for zero order.
![t_{1/2}=\frac{[A_o]}{2k}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B%5BA_o%5D%7D%7B2k%7D)


and,
![t_{1/2}=\frac{[A_o]}{2k}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B%5BA_o%5D%7D%7B2k%7D)


From this we conclude that the reaction can not be zero order.
Now we have to calculate the value of rate constants for second order.
![t_{1/2}=\frac{1}{[A_o]k}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B1%7D%7B%5BA_o%5Dk%7D)


and,
![t_{1/2}=\frac{1}{[A_o]k}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B1%7D%7B%5BA_o%5Dk%7D)


From this we conclude that the reaction is a second order.
Therefore, the value and unit of the rate constant for this reaction is,
and
respectively.