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!
Answer: The rate constant is 
Explanation ;
Expression for rate law for first order kinetics is given by:
where,
k = rate constant = ?
t = age of sample = 4.26 min
a = initial amount of the reactant = 2.56 mg
a - x = amount left after decay process = 2.50 mg
Now put all the given values in above equation to calculate the rate constant ,we get
Thus rate constant is [tex]0.334s^{-1}
Answer: the ability to be dissolved, especially in water.
Explanation: I think the answer you've picked is right
Hope this helps
