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.
E = mct
Energy = (mass) x (specific heat capacity of water) x (change in temp)
585.24 = 53.2 x 4.2 x (X-24.15)
585.24 divided by 53.2 divided by 4.2 = X - 24.15
2.62 = X - 24.15
X= 26.77degrees C
(Specific heat capacity for water is 4.2 but is different for other liquids)
Answer:
The volume of the balloon will be 5.11L
Explanation:
An excersise to solve with the Ideal Gases Law
First of all, let's convert the pressure in mmHg to atm
1 atm = 760 mmHg
760 mmHg ___ 1 atm
755.4 mmHg ____ (755.4 / 760) = 0.993 atm
922.3 mmHg ____ ( 922.3 / 760) = 1.214 atm
T° in K = 273 + °C
28.5 °C +273 = 301.5K
26.35°C + 273= 299.35K
P . V = n . R .T
First situation: 0.993atm . 6.25L = n . 0.082 . 301.5K
(0.993atm . 6.25L) / 0.082 . 301.5 = n
0.251 moles = n
Second situation:
1.214 atm . V = 0.251 moles . 0.082 . 301.5K
V = (0.251 moles . 0.082 . 301.5K) / 1.214 atm
V = 5.11L
Ookay
1) T
2) T
3) F
Hope i helped :)
Which solution is the least concentrated?
O 2 moles of solute dissolved in 4 liters of solution
