Answer:
true
because with the both states we increase the surface of reaction
Answer: (i) F = 2
(ii) F = 3
(iii) F = 2
Explanation:
We would be applying the famous Gibbs Phase Rule to explaining this problem;
By applying the formula;
F+P = C +2
Where P = this represent the phase
F = this is called the degree of freedom
C = this represent the component in the system
Ok let us begin;
(i). from this we can see that there are 2 components i.e. (water + ethanol) and the phase in question is a vapor phase + liquid phase.
So from the formula;
F = C-P+2
F = 2 – 2 + 2 = 2
Therefore, F = 2.
(ii). Also, from the statement, we can figure there are 3 components, while the phases are two like the previous one above, i.e. liquid + vapor
F = 3 – 2 + 2 = 5 – 2 = 3
F = 3
(iii). From this statement, we can figure there are 3 components, and the phases are 3 i.e. (2 liquid phases + 1 vapor phase)
From the formula;
F = 3 – 3 + 2 = 0 + 2
F = 2
Answer:
See explanation
Explanation:
The boiling point of a substance is affected by the nature of bonding in the molecule as well as the nature of intermolecular forces between molecules of the substance.
2-methylpropane has only pure covalent and nonpolar C-C and C-H bonds. As a result of this, the molecule is nonpolar and the only intermolecular forces present are weak dispersion forces. Therefore, 2-methylpropane has a very low boiling point.
As for 2-iodo-2-methylpropane, there is a polar C-I bond. This now implies that the intermolecular forces present are both dispersion forces and dipole interaction. As a result of the presence of stronger dipole interaction between 2-iodo-2-methylpropane molecules, the compound has a higher boiling point than 2-methylpropane.
Most elements on group 18 are the Noble Gases. They already have a complete last level with 8 electrones. Actually they can form compounds but only on the lab and they will not even last half a second.
