Uranium-236 is ____.
2 answers:
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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
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This is an incomplete question, here is a complete question.
The conversion of cyclopropane to propene occurs with a first-order rate constant of 2.42 × 10⁻² hr⁻¹. How long will it take for the concentration of cyclopropane to decrease from an initial concentration 0.080 mol/L to 0.053 mol/L?
Answer : The time taken will be, 17.0 hr
Explanation :
Expression for rate law for first order kinetics is given by:
where,
k = rate constant =
t = time passed by the sample = ?
a = initial concentration of the reactant = 0.080 M
a - x = concentration left = 0.053 M
Now put all the given values in above equation, we get
Therefore, the time taken will be, 17.0 hr