Answer:
900 J/mol
Explanation:
Data provided:
Enthalpy of the pure liquid at 75° C = 100 J/mol
Enthalpy of the pure vapor at 75° C = 1000 J/mol
Now,
the heat of vaporization is the the change in enthalpy from the liquid state to the vapor stage.
Thus, mathematically,
The heat of vaporization at 75° C
= Enthalpy of the pure vapor at 75° C - Enthalpy of the pure liquid at 75° C
on substituting the values, we get
The heat of vaporization at 75° C = 1000 J/mol - 100 J/mol
or
The heat of vaporization at 75° C = 900 J/mol
A molecular orbital that decreases the electron density between two nuclei is said to be <u>antibonding.</u>
The bonding orbital, which would be more stable and encourages the bonding of the two H atoms into
, is the orbital that is located in a less energetic state than just the electron shells of the separate atoms. The antibonding orbital, which has higher energy but is less stable, resists bonding when it is occupied.
An asterisk (sigma*) is placed next to the corresponding kind of molecular orbital to indicate an antibonding orbital. The antibonding orbital known as * would be connected to sigma orbitals, as well as antibonding pi orbitals are known as
* orbitals.
Therefore, molecular orbital that decreases the electron density between two nuclei is said to be <u>antibonding.</u>
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Hence, the correct answer will be option (b)
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Answer:
it is a replacement reaction
Answer:
A condensate is a liquid formed by condensation.
Explanation:
I believe a photon is the answer you are looking for.