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Answer:
Chloroform is expected to boil at 333 K (60 
).
Explanation:
For liquid-vapor equilibrium at 1 atm, 
= 0.
We know, 
, where T is temperature in kelvin scale.
Here both 
and 
are corresponding to vaporization process therefore T represents boiling point of chloroform.
So, ![0=(31.4\times 10^{3}\frac{J}{mol})-[T\times (94.2\frac{J}{mol.K})]](https://tex.z-dn.net/?f=0%3D%2831.4%5Ctimes%2010%5E%7B3%7D%5Cfrac%7BJ%7D%7Bmol%7D%29-%5BT%5Ctimes%20%2894.2%5Cfrac%7BJ%7D%7Bmol.K%7D%29%5D)
or, T = 333 K
So, at 333 K (60 ) , chloroform is expected to boil.
Bonding MO's have lower energy than antibonding MO's. The bonding MO's lower energy, even lower than its constituent atomic orbitals, accounts for the stability of a molecule in relation to its individual atoms. However, the sum of energy of the MO's must equal the sum of energy of the AO's.
<h3>What is atomic orbital?</h3>
An atomic orbital is a function in atomic theory and quantum mechanics that describes the location and wave-like behavior of an electron in an atom. This formula can be used to calculate the likelihood of locating any atom's electron in any given location surrounding the nucleus. The phrase atomic orbital can also refer to the actual region or place where the electron is projected to be present given the orbital's mathematical form.
Each orbital in an atom is defined by a set of values of the three quantum numbers n, l, and ml, which correspond to the energy, angular momentum, and an angular momentum vector component of the electron, respectively (magnetic quantum number).
To learn more about atomic orbital visit:
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