Answer:
![T_b=239.7K=-33.49\°C](https://tex.z-dn.net/?f=T_b%3D239.7K%3D-33.49%5C%C2%B0C)
Explanation:
Hello!
In this case, since the relationship between entropy and enthalpy for any process is defined below:
![S=\frac{H}{T}](https://tex.z-dn.net/?f=S%3D%5Cfrac%7BH%7D%7BT%7D)
For the vaporization of ammonia or any liquid, we can write:
![\Delta S_{vap}=\frac{\Delta H_{vap}}{T_{vap}}](https://tex.z-dn.net/?f=%5CDelta%20S_%7Bvap%7D%3D%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7BT_%7Bvap%7D%7D)
In such a way, solving the temperature of vaporization, or boiling point, we have:
![T_{vap}=\frac{\Delta H_{vap}}{\Delta S_{vap}}](https://tex.z-dn.net/?f=T_%7Bvap%7D%3D%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7B%5CDelta%20S_%7Bvap%7D%7D)
Plugging in the given enthalpy and entropy of vaporization we obtain:
![T_{vap}=T_b=\frac{23350\frac{J}{mol} }{97.43\frac{J}{mol*K}} \\\\T_b=239.7K=-33.49\°C](https://tex.z-dn.net/?f=T_%7Bvap%7D%3DT_b%3D%5Cfrac%7B23350%5Cfrac%7BJ%7D%7Bmol%7D%20%7D%7B97.43%5Cfrac%7BJ%7D%7Bmol%2AK%7D%7D%20%5C%5C%5C%5CT_b%3D239.7K%3D-33.49%5C%C2%B0C)
Best regards!
A molecule that can h-bond will not always necessarily and does not have guarantee to have a higher boiling point than one than cannot have h-bond.
we can take an example of Pentan-2-one that cannot h-bond but instead of this it has a high boiling point that is 102.3 °C, while propan-1-ol can h-bond but it has a boiling point of 97.2°C, that is lower than the boiling point of Pentan-2-one.
Answer:
![[C_6H_{12}O_6]](https://tex.z-dn.net/?f=%5BC_6H_%7B12%7DO_6%5D)
Explanation:
Hello there!
In this case, since the concept of solution concentration is majorly used in terms of moles per liters of solution or molarity (M), it is also possible to represent this chemical unit by using squared brackets, [ ].
In such a way, when focused on the concentration of glucose, C6H12O6, we can use:
![[C_6H_{12}O_6]](https://tex.z-dn.net/?f=%5BC_6H_%7B12%7DO_6%5D)
Best regards!
The false statement from the above is that: Temporary charge imbalances in the molecules lead to London dispersion forces.
<h3>What are the factors that affect London dispersion forces?</h3>
Generally, the factors which affects the London dispersion forces a dispersion force are as follows:
- Shape of the molecules
- Distance between molecules
- Polarizability of the molecules
However, London dispersion forces simply refers to a sort of temporary attractive force formed when electrons in two adjacent atoms occupy positions that make the atoms form dipoles.
So therefore, temporary charge imbalances in the molecules lead to London dispersion forces is a false statement
Learn more about London dispersion forces:
brainly.com/question/1454795