Answer:
i is d " The carbons on either side of the double bond are pointed in the
same direction."Explanation:
The heat/enthalpy of vaporization of water represents the energy input required to convert one mole of water into vapor at a constant temperature. Intermolecular forces including hydrogen bondings of significant strength hold water molecules in place under its liquid state. Whereas the molecules experience almost no intermolecular interactions under the gaseous state- consider the way noble gases molecules interact. It is thus necessary to supply sufficient energy to overcome all intermolecular interactions present in the substance under its liquid state to convert the substance into a gas. The heat of vaporization is thus related to the strength of the intermolecular interactions.
Water molecules contain hydrogen atoms bonded directly to oxygen atoms. Oxygen atoms are highly electronegative and take major control of electrons in hydrogen-oxygen bonds. Hydrogen atoms in water molecules thus experience a strong partial-positive charge and would attract lone pairs of electron on neighboring water molecules. "Hydrogen bonds" refer to the attraction between hydrogen atoms bonded to electronegative elements and lone pairs of electrons. The hydrogen-oxygen bonds in water molecules are so polarized that hydrogen bonds in water are stronger than both dipole-dipole interactions and London Dispersion Forces in most other molecules. It thus take high amounts of energy to separate water molecules sufficiently apart such that they no longer experience intermolecular interactions and behave collectively like a gas. As a result, water has one of the highest heat of vaporization among covalent molecules of similar sizes.
Selfmade.ivyy hope this help
Answer:
Yes. The solution would be optically active.
Explanation:
Diastereomer are defined as the image that is non mirror and non -identical. It is made up of two stereoisomers. They are formed when the two stereoisomers or more than two stereoisomers of the compound have the same configuration at the equivalent stereocenters.
In the given context, as the product given is a diastereomeric mixture, the product would have an optical activity in total.
So the answer is Yes.
The trough and the hill part
