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Kris Lord S.
Chemist currently taking up a master's degree in biochemistry
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Organic Chemistry
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Question:

Why do the rates of bimolecular nucleophilic substitution (SN2) reactions depend on the concentration of both the nucleophile and the starting material.

Kris Lord S.
Answer:

This is because the rate-limiting step of SN2 reactions involves the simultaneous departure of the leaving group (bond breaking) and attack of the nucleophile (bond formation). Since the bond-breaking and bond-forming steps are concerted, both the concentration of the starting material (in which the bond is broken) and the nucleophile (that which forms the bond) affect the rate of reaction.

Inorganic Chemistry
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Question:

Explain the "anomalous" trend for the first ionization energies of nitrogen and oxygen.

Kris Lord S.
Answer:

The general trend for the first ionization energy is that it increases from left to right of the periodic table however, this is not observed for nitrogen (Group 15) and oxygen (Group 16). This is due to the stability of half-filled p orbitals. When oxygen loses one electron, a half-filled p orbital where 3 electrons occupy separate sub-shells resulting in minimal repulsion. On the other hand, this particularly stable half-filled shell is "destroyed" when you remove an electron from Nitrogen thus requiring a higher amount of energy.

Biochemistry
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Question:

Why does the melting point of a triglyceride sample decrease when the percentage of unsaturated fatty acid increases?

Kris Lord S.
Answer:

The melting point of a triglyceride sample, or any substance, is highly dependent on the intermolecular forces of attraction present. For triglycerides, it is due mainly to the van der Waals forces which is dependent on the surface area of molecules available for interaction. An unsaturated fatty acid is characterized by having a "bent" structure as a consequence of the carbon-carbon double bond as compared to a saturated fatty acid which is "linear". This bent structure makes the molecules more unlikely to align or pack which effectively decreases the total surface area of contact. Therefore, van der Waals forces are weakened which results to a lower melting point.

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