Answer:
sorry i'm not a scientist but i think it's C
Explanation:
sorry again if i get it wrong for u :(
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.
Bacteria can’t live without humans
Here’s what I found:
It takes very little energy to remove that outermost electron from an alkali metal. Thus, alkali metals easily lose their outermost electron to become a +1 ion. ... In fact, as you go down the 1A column, the first ionization energies get lower and lower, making cesium the most easily ionized element on the periodic table.
So basically it’s because part of what makes alkali metals so reactive is that they have one electron in their outermost electron layer.
