Why is acetone less acidic than cyclohexanol?
if you consider the structure of acetone, it has 2 $\ce{CH3}$ groups and a $\ce{C}$ double bond $\ce{O}$ ($\ce{C=O}$). Therefore, for the $\ce{H}$ to disassociate, it must be removed from the $\ce{CH3}$ group which leaves now a $\ce{C=O}$, $\ce{CH2}$, and a $\ce{CH3}$ group where the negative charge lies on $\ce{C}$ a part of the $\ce{CH2}$ group. Conversely, the cyclohexanol would have the negative charge lying on the $\ce{O}$ because the $\ce{H}$ a part of the $\ce{OH}$ group would be removed.
Just by simple electron negativity rules, $\ce{O}$ is much more favored to possess a (partial) negative charge than $\ce{C}$. Thus, cyclohexanol is a much more stable acid than acetone which is reflected in its $\mathrm{p}K_\mathrm{a}$ value.