Answer: After 4710 seconds, 1/8 of the compound will be left
Explanation:
Using the formulae
Nt/No = (1/2)^t/t1/2
Where
N= amount of the compound present at time t
No= amount of compound present at time t=0
t= time taken for N molecules of the compound to remain = 4710 seconds
t1/2 = half-life of compound = 1570 seconds
Plugging in the values, we have
Nt/No = (1/2)^(4710s/1570s)
Nt/No = (1/2)^3
Nt/No= 1/8
Therefore after 4710 seconds, 1/8 molecules of the compound will be left
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.
Answer: d
Explanation: I got it right
