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What is the boiling point of a substance with a molecular formula of C6H14 if it experiences dipole-dipole interactions and London dispersion forces? Show your calculation and explain your reasoning.

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C6H14, also known as hexane, is a nonpolar molecule. Nonpolar molecules do not have significant dipole-dipole interactions because they do not have permanent dipoles. Instead, they experience London dispersion forces, which are temporary dipoles caused by the random movement of electrons.To estimate the boiling point of hexane, we can use the molecular weight and the strength of the London dispersion forces. The molecular weight of hexane can be calculated as follows:Molecular weight of C6H14 =  6 * 12.01  +  14 * 1.01  = 72.06 + 14.14 = 86.20 g/molThe strength of London dispersion forces increases with the molecular weight and surface area of the molecule. Since hexane is a relatively large molecule with a molecular weight of 86.20 g/mol, it will have stronger London dispersion forces compared to smaller molecules.To estimate the boiling point, we can compare hexane to other similar nonpolar molecules and their boiling points. For example, pentane  C5H12  has a boiling point of 36.1C, and heptane  C7H16  has a boiling point of 98.4C. Hexane, being in between pentane and heptane, will have a boiling point between these two values.In fact, the boiling point of hexane  C6H14  is approximately 69C. This value is consistent with the trend of increasing boiling points with increasing molecular weight and stronger London dispersion forces.In conclusion, the boiling point of hexane  C6H14  is approximately 69C, which can be estimated based on its molecular weight and the strength of London dispersion forces. Dipole-dipole interactions are not significant in this case, as hexane is a nonpolar molecule.
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