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The molecules CH2F2 and CH2Cl2. Both of these molecules can participate in London dispersion and dipole-dipole intermolecular forces. (2 points) If you were to only consider dipole-dipole intermolecular forces which molecule would have a higher boiling point, CH2F2 or CH2Cl2? (2 points) If you were to only consider London dispersion forces which molecule would have a higher boiling point, CH2F2 or CH2Cl2? (2 points) The actual boiling point CH2F2 is -52 ºC and CH2Cl2 is 39.6 ºC. Which intermolecular force appears to be the most predominate force between these molecules? Please explain. (4 points) Why are London dispersion typically weaker than dipole-dipole interactions? In your explanation, please relate all of this to Coulomb’s law. (6 points) Consider the following balanced chemical equation then determine how much CO2 (in grams) can be produced if 52.10 g of CH3OH is allowed to react with excess O2. Show all of your work here. 2 CH3OH (l) + 3 O2 (g) à 2 CO2 (g) + 4 H2O (l)

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