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!
The answer for this would be 69.6
Answer:
Al(NO3)3(s)--------> Al^3+(aq) + 3NO3^-(aq)
Explanation:
The equation shown above describes the dissolution of Al(NO3)3 in water using the lowest coefficients.
This occurs when solid Al(NO3)3 is added to water. It dissolves to give rise to ions as shown. This is a property of all ionic substances.
Explanation:
rbrhrhyhggggsdffffffffffv
The atomic mass of K is 39
from Avogadro's law
39g of K contains 6.02x10^23 atoms
therefore if
39=6.02x19^23
X=5.11×10^22
making X the subject of the formula
X= (5.11×10^22×39)÷6.02×10^23
X= 33g
