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AveGali [126]
2 years ago
5

Base your answer on the information below. The hydrocarbon 2-methylpropane reacts with iodine as represented by the balanced equ

ation 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.
Chemistry
1 answer:
Kryger [21]2 years ago
3 0

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.

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3 years ago
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1.645 moles of excess reactant that is of magnesium metal are left over.

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Moles of magnesium metal = 3.29 mol

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According to recation, 2 moles of HCl reacts with 1 mol of magnesium metal, then 3.29 moles of HCl will react with :

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Moles of magnesium metal left = 3.29 mol - 1.645 mol = 1.645 mol

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