Answer: as copper has lower electrode potential value than hydrogen, it could be reduced by hydrogen.
Explanation: hydrogen has zero reduction potential while Cu has +0.34V and Al has -1.66 V .
SO in electrochemical series who has most negative or less reduction potential value tends to be a good reducing agent than the other.
Hope it helps...
6.4 * 6.02 * 10^23 = 3.8528*10^24 atoms
Don't let the fact that it's vanadium throw you off, avagadros constant stays the same for all elements
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.
