Rank the following compounds in order of increasing heat of combustion. Be sure to answer all parts. Drag each item to the correct box. A) pentane B) propane C) butane

The compounds are arranged from the least heat of combustion to the highest (in order of increasing heat of combustion).

Propane--> butane --> pentane.

Explanation:

Hydrocarbons are compounds which contains carbon and hydrogen only. Alkane is called a saturated hydrocarbon because all four bonds of carbon are single bonds. Examples of alkanes include: pentane, propane and butane. One of the chemical properties of alkanes includes combustion reaction.

In COMBUSTION REACTION, alkanes burns in air (oxygen) to produce carbon dioxide and water.The heat released on complete combustion of one mole of a substance is called HEAT OF COMBUSTION.

The heat of combustion of saturated hydrocarbon ( alkanes) increases with increase in number of carbon atoms in a molecule. This is due to more carbon available for burning and more number of bonds undergoing changes. Therefore in the arrangement of the given compounds, PENTANE has the highest heat of combustion while PROPANE is the least.

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.

where are the choice options?