Answer:
115g/mol
Explanation:
To get the molar mass, we know that the it is equal to the mass divided by the number of moles. We have the mass but we do not have the number of moles.
We get this by working through the solution information. Firstly, we need to know the number of moles in 750ml for a molarity of 0.29m
Now, since 0.29 moles is present in 1000ml, x moles will be present in 750ml
The value of x is obtained as follows:
x = (750 * 0.29)/1000 = 0.2175 moles
Now since we have the number of moles, we can then obtain the molar mass.
Molar mass = mass/number of moles = 25.0g/0.2175 = 114.94 approximately 105g/mol
the answer is 0.000097 KM
Because the ring is hollow
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!
