Base your answer on the information below. The hydrocarbon 2-methylpropane reacts with iodine as represented by the balanced equation below. At standard pressure, the boiling point of 2-methylpropane is lower than the boiling point of 2-iodo-2-methylpropane. Explain the difference in the boiling points of 2-methylpropane and 2-iodo-2-methylpropane in terms of both molecular polarity and intermolecular forces.

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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.

m=9

step-by-step explanation:

the best answer is d because the strong cohesion is caused by the ability of water molecules to undergo hydrogen bonding.

a is incorrect because when water solidifies, it becomes   ice, which expands and occupies a bigger volume for the same mass of water in its liquid state.

b is incorrect because water has high surface tension. besides bonding to itself it hydrogen bonds to other molecules at the top decreasing possibility for breaking of attractive forces.

c is incorrect because the water molecule has oxygen which is highly electronegative and so pulls the electron pair towards itself creating a dipole moment, making the molecule polar.

e is incorrect because water has strong adhesion, ability to stick to other molecules such as glass because of this molecules having stronger hydrogen bonding than water.