Answer:
The functional group
Explanation:
A functional group defines a hydrocarbon. an ester the functional group is an O- alkyl group while that one of an alcohol is -OH group. An aldehyde is a functional group on its own.
I think it could be C maybeee though
Addition of water to an alkyne gives a keto‑enol tautomer product and that is the product changed into 2-pentanone, then the alkyne need to had been 1-pentyne. 2-pentyne might have given a combination of 2- and 3-pentanone.
<h3>
What is the keto-enol means in tautomer?</h3>
They carries a carbonyl bond even as enol implies the presence of a double bond and a hydroxyl group. The keto-enol tautomerization equilibrium is depending on stabilization elements of each the keto tautomer and the enol tautomer.
- The enol that could provide 2-pentanone might had been pent-1- en - 2 -ol. Because an equilibrium favors the ketone so greatly, equilibrium isn't an excellent description.
- If the ketone have been handled with bromine, little response might be visible because the enol content material might be too low.
- If a catalyst have been delivered, NaOH for example, then formation of the enolate of pent-1-en - 2 - ol might shape and react with bromine.
- This might finally provide a bromoform product. Under acidic conditions, the enol might desire formation of the greater substituted enol constant with alkene stability.
Answer: The value of 
for the half-cell reaction is 0.222 V.
Explanation:
Equation for solubility equilibrium is as follows.
Its solubility product will be as follows.
![K_{sp} = [Ag^{+}][Cl^{-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BAg%5E%7B%2B%7D%5D%5BCl%5E%7B-%7D%5D)
Cell reaction for this equation is as follows.
Reduction half-reaction: 
, 
Oxidation half-reaction: 
, 
= ?
Cell reaction: 
So, for this cell reaction the number of moles of electrons transferred are n = 1.
Solubility product,
= 
Therefore, according to the Nernst equation
![E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%20%3D%20E%5E%7Bo%7D_%7Bcell%7D%20-%20%5Cfrac%7B0.0592%20V%7D%7Bn%7D%20log%20%5Cfrac%7B%5BAgCl%5D%7D%7B%5BAg%5E%7B%2B%7D%5D%5BCl%5E%7B-%7D%5D%7D)
At equilibrium, 
= 0.00 V
Putting the given values into the above formula as follows.
![0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}](https://tex.z-dn.net/?f=0.00%20%3D%20E%5E%7Bo%7D_%7Bcell%7D%20-%20%5Cfrac%7B0.0592%20V%7D%7B1%7D%20log%20%5Cfrac%7B1%7D%7B%5BAg%5E%7B%2B%7D%5D%5BCl%5E%7B-%7D%5D%7D)
= 
= 0.577 V
Hence, we will calculate the standard cell potential as follows.
= 0.222 V
Thus, we can conclude that value of for the half-cell reaction is 0.222 V.
