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!
Quick answer: Exothermic reactions mean that heat has escaped the system. The answer is A. Heat is released.
First find the oxidation states of the various atoms:
<span>in Cr2O2 2- Cr @ +1; In NH3 N @ +3; in CrO3 Cr @ +3, N2 N @ 0 </span>
<span>Note that N gained electrons, ie, was reduced; Cr was oxidized </span>
<span>Now there is a problem, because B has NH4+ which the problem did not, and is not balanced, showing e- in/out </span>
<span>B.NH4+ → N2 </span>
<span>Which of the following is an oxidation half-reaction? </span>
<span>A.Sn 2+ →Sn 4+ + 2e- </span>
<span>Sn lost electrons so it got oxidized</span>
Explanation:
(a) As the given chemical reaction equation is as follows.
So, when we double the amount of hypochlorite or iodine then the rate of the reaction will also get double. And, this reaction is "first order" with respect to hypochlorite and iodine.
Hence, equation for rate law of reaction will be as follows.
Rate = ![K \times [OCl^{-}] \times [l^{-}]](https://tex.z-dn.net/?f=K%20%5Ctimes%20%5BOCl%5E%7B-%7D%5D%20%5Ctimes%20%5Bl%5E%7B-%7D%5D)
(b) Since, the rate equation is as follows.
Rate = ![K [OCl^{-}][l^{-}]](https://tex.z-dn.net/?f=K%20%5BOCl%5E%7B-%7D%5D%5Bl%5E%7B-%7D%5D)
Let us assume that (![[OCl^{-}] = [l^{-}]](https://tex.z-dn.net/?f=%5BOCl%5E%7B-%7D%5D%20%3D%20%5Bl%5E%7B-%7D%5D)
)
Putting the given values into the above equation as follows.
K = 
= 
Hence, the value of rate constant for the given reaction is
.
(c) Now, we will calculate the rate as follows.
Rate =
= 
= 
Therefore, rate when ![[OCl^{-}] = 1.8 \times 10^{3}](https://tex.z-dn.net/?f=%5BOCl%5E%7B-%7D%5D%20%3D%201.8%20%5Ctimes%2010%5E%7B3%7D)
M and ![[I^{-}]= 6.0 \times 10^{4}](https://tex.z-dn.net/?f=%5BI%5E%7B-%7D%5D%3D%206.0%20%5Ctimes%2010%5E%7B4%7D)
M is .
