Explain how the structure of ammonium lauryl sulfate, as described in parts E and F, produces the properties identified in part C. 

Part E Description: Here are lauryl sulfate and the ammonium ions, the lauryl sulfate contains lauric acid which is represented by the black and grey, in the red is the lauric acid that is formed by the covalent bonds between c-c attaches to the hydrogens, and the sulfate ions attached to the lauric acid with c-s colavent bond. Sulfur attaches to oxygen by colavent bonds. Ammounium ions are surrounded by 4 hydrogen atoms as well.

Part F description: The fatty acid present is nonpolar, meaning it will not disolve easily in water. What will dissolve in water is the sulfate ions and ammonium ions, so we then know that the fatty acid will bind to nonpolar substances in dirt and then the ammonium/sulfate ions will bind to polar water, they together will work and bond to help wash away the dirt produced. So nonpolar molecules are present and are attracting to one another.

Part C properties: The molecule consists of a long non-polar hydrocarbon chain and a polar sulphate end group, this combination makes the material of a surfactant. These two components allow the compound to dissolve between the two polar and non-polar molecules. ALS is an alkyl sulfate and is an anionic surfactant found primarily in shampoos and body washes as a foaming agent. Lauryl sulfates are very high foaming surfactants that disrupt the surface tension of water by forming micelles around the highly polar water molecules at the surface air interface. ALS has a slightly lower pH than 7 which is a slightly acidic. The active components of ALS and SLS are chemically identical and must act identically as well.