The Boiling Point of 2-methylpropane is approximately -11.7 °C, while, Boiling Point of <span>2-iodo-2-methylpropane is approximately 100 </span>°C.
As both compounds are Non-polar in nature, So there will be no dipole-dipole interactions between the molecules of said compounds.
The Interactions found in these compounds are London Dispersion Forces.
And among several factors at which London Dispersion Forces depends, one is the size of molecule.
Size of Molecule:
There is direct relation between size of molecule and London Dispersion forces. So, 2-iodo-2-methylpropane containing large atom (i.e. Iodine) experience greater interactions. So, due to greater interactions 2-iodo-2-methylpropane need more energy to separate from its partner molecules, Hence, high temperature is required to boil them.
First, we will get the average pH of the two given values:
average pH = (6.4+8) / (2) = 7.2
At this average pH, the concentration of the acid from the phenol red is equal to the concentration of the base.
pH = 7.2
[H+] = 10^(-7.2) = 6.3 * 10^-8
Phenol red has the general formula HA, this gives us:
HA <.......> H+ + A-
At pH = 7.2, [H+] = [A-]
<span>Ka = [H+][A-]/ [HA]
</span>Ka = [H+] = <span>6.3 x 10^-8</span>
Answer:
D. Molecules are able to move as they heat up to melt
Explanation:
PH=-log[H⁺]
pH=-log(1.87×10⁻¹³)
pH=12.72
I hope this helps. Let me know if anything is unclear.
