Based on the type or types of intermolecular forces, predict the substance in each pair that has the higher boiling point Given the molecules propane (C3H8) and n-butane (C4H10) has the higher boiling point mainly due to Given the molecules diethyl ether (CH3 CH2OCH2 CH3) and 1-butanol (CH3 CH2CH2 CH2OH) higher boiling point mainly due to .

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Explanation:

-) Given the molecules propane (C3H8) and n-butane (C4H10) n-butane has the higher boiling point mainly due to a larger chain of carbons.

In this question, in propane, we have a chain of three carbons. In butane, we have fourth carbons. If we have more carbons we will have more interactions. If we have more interactions we have to give more energy to go from liquid to gas, therefore we will have a higher boiling point.

-) Given the molecules diethyl ether (CH3 CH2OCH2 CH3) and 1-butanol (CH3 CH2CH2 CH2OH) 1-butanol has a higher boiling point mainly due to hydrogen bonding.

In the case of butanol, we have the "OH" group. If we have a hydrogen bond to a heteroatom (O, S, P, or N) we will have the presence of this type of interaction between molecules. If we have more interactions we have to give more energy to go from liquid to gas, therefore we will have a higher boiling point.

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answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.