In the problem, we are tasked to solved for the amount of carbon (C) in the acetone having a molecular formula of C 3 H 6 O. We need to find first the molecular weight if Carbon (C), Hydrogen (H), Oxygen (O).
Molecular Weight:
C=12 g/mol
H=1 g/mol
O=16 g/mol
To calculate for the percent by mass of acetone, we assume 1 mol of acetone.
%C=
%C=62.07%
Therefore, the percent by mass of carbon in acetone is 62.07%
Sodium. 11
Carbon. 12
Hydrogen 1
Oxygen 2
Fluuorine. 14
Boron. 5
Lithium. 6
Helium 3
Phosphorus 15
Sulfur 6
Answer: 10L
Explanation:
Given that:
Initial pressure P1 = 1 atm
New pressure P2 = 3 atm
Initial volume V2 = 30 L
New volume V2 = ?
Since pressure and volume are involved, apply the formula for Boyle's law
P1V1 = P2V2
1 atm x 30L = 3 atm x V2
30 atm L = 3 atm x V2
V2 = (30 atm L / 3 atm)
V2 = 10L
Thus, volume changed to 10 liters
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.
