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Stels [109]
3 years ago
10

In most cases of fischer esterification, the carboxylic acid is the more "valuable" piece and the alcohol is used in excess. let

's say you are making the ethyl ester of an acid and use the ethanol in excess. what technique would be used to remove the excess ethanol?
Chemistry
1 answer:
Daniel [21]3 years ago
4 0
There are a lot of separation processes. To name a few, these can be distillation, centrifugation, extraction, membrane or sorption process and many other. To know which is the best technique, you should know the property between two substances that have a stark difference. In this case, it is the polarity. Ethyl alcohol is more polar than ethyl ester and less dense. Thus, these two won't mix. So, take advantage of their density difference by decantation or centrifugation.
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The activation energy for proline isomerization of a peptide depends on the identity of the preceding residue and obeys Arrheniu
MAVERICK [17]

Answer:

Activation energy of phenylalanine-proline peptide is 66 kJ/mol.

Explanation:

According to Arrhenius equation-     k=Ae^{\frac{-E_{a}}{RT}}    , where k is rate constant, A is pre-exponential factor, E_{a} is activation energy, R is gas constant and T is temperature in kelvin scale.

As A is identical for both peptide therefore-

                                   \frac{k_{ala-pro}}{k_{phe-pro}}=e^\frac{[E_{a}^{phe-pro}-E_{a}^{ala-pro}]}{RT}

Here \frac{k_{ala-pro}}{k_{phe-pro}}=\frac{0.05}{0.005} , T = 298 K , R = 8.314 J/(mol.K) and E_{a}^{ala-pro}=60kJ/mol

So, \frac{0.05}{0.005}=e^{\frac{[E_{a}^{phe-pro}-(60000J/mol)]}{8.314J.mol^{-1}.K^{-1}\times 298K}}

   \Rightarrow E_{a}^{phe-pro}=65705J/mol=66kJ/mol (rounded off to two significant digit)

So, activation energy of phenylalanine-proline peptide is 66 kJ/mol

7 0
2 years ago
The shielding of electrons gives rise to an effective nuclear charge, Zeff, which explains why boron is larger than oxygen. Esti
Vesna [10]

Zeff = Z - S

Here, Z is the number of protons in the nucleus, that is, atomic number, and S is the number of nonvalence electrons.

For boron, the electronic configuration is 1s₂ 2s₂ 2p₄

Z = 5, S = 2

Zeff = 5-2 = +3

For O, electronic configuration is 1s₂ 2s₂ 2p₄

Z = 8, S = 2

Zeff = 8-2 = +6

Hence, the correct answer is second option, that is, +3 and +6, the Zeff of boron is smaller in comparison to O, thus, boron exhibits a bigger size than O.

3 0
3 years ago
What the three injuries which can happen in the lab when chemicals are used incorrectly?
stepladder [879]

Answer:

Eye injuries,Burns,Respiratory injuries

7 0
3 years ago
Please help me. I'll give brainliest. I am stuck. True or false Q.
belka [17]
I believe your answer is true! Hope this helps!
5 0
3 years ago
Draw the Lewis dot structure for water.
Gekata [30.6K]
It would be HO2 Around there
3 0
2 years ago
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