Stability of α''-Fe16N2 in hydrogenous atmospheres

Protons and neutrons, collectively called nucleons, are packed together tightly in a nucleus. With a radius of about 10−15 meters, a nucleus is quite small compared to the radius of the entire atom, which is about 10−10 meters. Nuclei are extremely dense compared to bulk matter, averaging 1.8 × 1014 grams per cubic centimeter. If the earth’s density were equal to the average nuclear density, the earth’s radius would be only about 200 meters.
To hold positively...

The free energy change for a process taking place with reactants and products present under nonstandard conditions (pressures other than 1 bar; concentrations other than 1 M) is related to the standard free energy change according to this equation:

The relative stability of alkenes can be determined by comparing their heats of hydrogenation. The lower heat of hydrogenation indicates the more stable alkene.  The three main factors determining the relative stability of alkenes are i) the number of substituents attached to the double-bond carbon atoms, ii) hyperconjugation, and iii) the stereochemistry of the double bond.

Near absolute zero temperatures, in the presence of a magnetic field, the majority of nuclei prefer the lower energy spin-up state to the higher energy spin-down state. As temperatures increase, the energy from thermal collisions distributes the spins more equally between the two states. The Boltzmann distribution equation gives the ratio of the number of spins predicted in the spin −½ (N−) and spin +½ (N+) states.

This lesson illustrates the role of deuterium substitution in simplifying the NMR spectrum of compounds comprising labile protons. One method employed is the use of deuterium. Amongst the three isotopes of hydrogen, deuterium (2H) has a nucleus composed of one proton and one neutron. When the D2O solvent is added to a pure dry ethanol solution, its labile proton is substituted with deuterium.

The difference between the calculated and experimentally measured masses is known as the mass defect of the atom. In the case of helium-4, the mass defect indicates a “loss” in mass of 4.0331 amu – 4.0026 amu = 0.0305 amu. The loss in mass accompanying the formation of an atom from protons, neutrons, and electrons is due to the conversion of that mass into energy that is evolved as the atom forms. The nuclear binding energy is the energy produced when the atoms’ nucleons...