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Explanation:
Two events involving electrons are gain and loss of electrons.
When there is gain or loss of electrons between two atoms then it results in the formation of ionic bond.
Whereas when there is sharing of electrons between two atoms then it results in the formation of covalent bond.
Therefore, the chemical bonds formed can be ionic or covalent bonds.
Answer:
The number of moles of benzaldehyde = 0.0253 moles
Explanation:
The molecular formula of benzaldehyde is C₇H₆O
Its molecular mass is calculated from the atomic masses of the constituent atoms.
C = 12.0 g: H = 1.0 g; O = 16.0 g
Molecular mass = ( 12 * 7) + (1 * 6) + (16 * 1) = 106.0 g/mol
Number of moles of substance = mass of substance/ molar mass of the substance
mass of benzaldehyde = 2.68; molar mass = 106.0 g/mol
Number of moles of benzaldehyde = 2.68 g/ 106 g/mol = 0.0253 moles
Therefore, the number of moles of benzaldehyde = 0.0253 moles
Vol.250 before its to much pressure
Answer:
Product: ethyl L-valinate
Explanation:
If we want to understand what it is the molecule produced we have to an<u>alyze the reagents</u>. We have valine an <u>amino acid</u>, in this kind of compounds we have an <em>amine group</em> (
) and a <em>carboxylic acid</em> group (
). Additionally, we have an <u>alcohol </u>(
) in the presence of HCl (a <u>strong acid</u>) in the first step, and a base (
).
When we have an acid and an alcohol in a vessel we will have an <u>esterification reaction</u>. In other words, an ester is produced. As the <em>first step,</em> the oxygen in the C=O (in the carboxylic acid group) would be protonated. In the <em>second step</em>, the ethanol attacks the carbon in the C=O of the carboxylic acid group producing a new bond between the oxygen in the ethanol and the carbon in the carboxylic acid. In <em>step 3</em>, a proton is transferred to produce a better leaving group (
). In <em>step 4</em>, a water molecule leaves the main structure to produce again the double bond C=O. <em>Finally</em>, a base () removes the hydrogen from the C=O bond to produce ethyl L-valinate
See figure 1
I hope it helps!
