The reaction mechanism for an alpha,beta-unsaturated ketone to react with basic peroxide to form an epoxide is shown below with a general ketone. The basic hydroxide is used to deprotonate the peroxide molecule to create a strong HOO- nucleophile. The peroxide then attacks the beta-carbon of the alkene and this pushes the electrons up to the oxygen of the carbonyl. This is the first intermediate that is formed during this reaction.
After the intermediate is formed, the lone pair from the oxygen pushes back down to form the carbonyl once more and this breaks a carbon-carbon bond which attacks the oxygen of the peroxy group, ultimately substituting an -OH group and forming the final epoxide ketone product.
The answer would have to be letter b. Energy
Answer:
1.20 V
Explanation:

Here Pb undergoes oxidation by loss of electrons, thus act as anode. Bromine undergoes reduction by gain of electrons and thus act as cathode.

Where both
are standard reduction potentials.
Given,

![E^0_{[Pb^{2+}/Pb]}= -0.13\ V](https://tex.z-dn.net/?f=E%5E0_%7B%5BPb%5E%7B2%2B%7D%2FPb%5D%7D%3D%20-0.13%5C%20V)

![E^0_{[Br_2/Br^{-}]}=+1.07\ V](https://tex.z-dn.net/?f=E%5E0_%7B%5BBr_2%2FBr%5E%7B-%7D%5D%7D%3D%2B1.07%5C%20V)
![E^0=E^0_{[Br_2/Br^{-}]}- E^0_{[Pb^{2+}/Pb]}](https://tex.z-dn.net/?f=E%5E0%3DE%5E0_%7B%5BBr_2%2FBr%5E%7B-%7D%5D%7D-%20E%5E0_%7B%5BPb%5E%7B2%2B%7D%2FPb%5D%7D)

Answer:
again same questions
i think its from 1st chapter
The envelope of gases surrounding the earth or another planet.