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kompoz [17]
3 years ago
5

Which extraction procedure will completely separate an amide from the by-product of the reaction between an amine and excess car

boxylic acid anhydride?
Chemistry
2 answers:
sweet-ann [11.9K]3 years ago
6 0

Answer: A. Add 0.1M NaOH to quench unreacted anhydride. Then add diethyl ether to separate the layers. The amide can be obtained from the ether layer by evaporating the solvent.

The options are

A) Add 0.1 M NaOH to quench unreacted anhydride. Then add diethyl ether and separate the layers. The amid e can be obtained from the ether layer by evaporating the solvent

B) Add 0.1M Hal to quench unreacted anhydride. Then add diethyl ether and separate the layers. The amide can be obtained from the ether layer by evaporating the solvent.

C) Add 0.1M NaOH to quench unreacted anyhyride. Then add diethyl ether and separate the layers. The amide can be obtained from the aqueous layer by neutralizing with HCl.

Explanation:

Extraction with aqueous base will hydrolyze and extract both the acidic carboxylic acid and the excess unreacted starting material to the aqueous layer, leaving the neutral amide in the ether layer.

DedPeter [7]3 years ago
4 0

Answer: You question seems not to be complete.

Complete Question:

Which extraction procedure will completely separate an amide from the by-product of the reaction between an amine and excess carboxylic acid anhydride?

A) Add 0.1 M NaOH to quench unreacted anhydride. Then add diethyl ether and separate the layers. The amid e can be obtained from the ether layer by evaporating the solvent

B) Add 0.1M Hal to quench unreacted anhydride. Then add diethyl ether and separate the layers. The amide can be obtained from the ether layer by evaporating the solvent.

C)Add 0.1M NaOH to quench unreacted anyhyride. Then add diethyl ether and separate the layers. The amide can be obtained from the aqueous layer by neutralizing with HCl.

Explanation:The extraction procedure to use is A.

Add 0.1 M NaOH to quench unreacted anhydride. Then add diethyl ether and separate the layers. The amide can be obtained from the ether layer by evaporating the solvent

This is because the byproduct of the reaction would give an acidic carboxylic acid and unreacted starting material would also be acidic. Therefore, Extraction with aqueous base is necessary which will hydrolyze and extract both of these into the aqueous layer, leaving the neutral amide in the ether layer.

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What is thermal energy?
Finger [1]

Answer:

D-the measure of temperature in a system

4 0
3 years ago
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Classify each element. Note that another term for main group is representative, another term for semimetal is metalloid, and the
NikAS [45]

The question is incomplete, here is the complete question:

Classify each element. Note that another term for main group is representative, another term for semi-metal is metalloid, and the inner transition metals are also called the lanthanide and actinide series.

Hf, Am, In, Ta, As, Se, Rn

<u>Answer:</u>

Hafnium and tantalum are transition elements.

Americium is a inner transition element.

Indium, Selenium and Radon are main group elements.

Arsenic is a metalloid.

<u>Explanation:</u>

Main group elements are the elements which belong to s block and p block. They are also known as representative elements.

S-block elements are defined as the elements whose last electron enters s-sub shell. The general electronic configuration of these elements is ns^{1-2}

P-block elements are defined as the elements whose last electron enters p-sub shell. The general electronic configuration of these elements is np^{1-6}

Metalloids are defined as the elements which show intermediate properties between metals and non-metals. There are 7 metalloids in the periodic table. They are: Boron, Silicon, germanium, Arsenic, Antimony, Tellurium and Polonium.

Transition elements are known as d-block elements. D block elements are defined as the elements whose last electron enters d sub shell. The general electronic configuration of these elements is [(n-1)d^{1-10}ns^{0-2}]

Inner transition elements are known as (f block) elements. (F block) elements are defined as the elements whose last electron enters (f subshell). The general electronic configuration of these elements is [(n-2)f^{1-14}(n-1)d^{0-1}ns^{2}]. They are also known as lanthanide and actinide series.

For the given elements:

  • <u>Option 1:</u> Hf

Hafnium is the 72nd element of the periodic table having electronic configuration of [Xe]4f^{14}5d^26s^2

As, the last electron is entering the d subshell, it is a transition element.

  • <u>Option 2:</u> Am

Americium is the 95th element of the periodic table having electronic configuration of [Rn]5f^{7}6d^07s^2

As, the last electron is entering the (f subshell), it is a inner transition element.

  • <u>Option 3:</u> In

Indium is the 49th element of the periodic table having electronic configuration of [Kr]5s^25p^1

As, the last electron is entering the p subshell, it is a main group element.

  • <u>Option 4:</u> Ta

Tantalum is the 73rd element of the periodic table having electronic configuration of [Xe]4f^{14}5d^56s^2

As, the last electron is entering the d subshell, it is a transition element.

  • <u>Option 5:</u> As

Arsenic is the 33rd element of the periodic table having electronic configuration of [Ar]4s^24p^3

As, the last electron is entering the p subshell, it is a main group element. It shows an intermediate property of metal and non-metal. Thus, it is a metalloid.

  • <u>Option 6:</u> Se

Selenium is the 34th element of the periodic table having electronic configuration of [Ar]4s^24p^4

As, the last electron is entering the p subshell, it is a main group element.

  • <u>Option 7:</u> Rn

Radon is the 86th element of the periodic table having electronic configuration of [Xe]4f^{14}5d^{10}6s^26p^6

As, the last electron is entering the p subshell, it is a main group element.

5 0
3 years ago
At 35 C, a sample of gas has a volume of 256 ml and a pressure of 720.torr. What would the volume
Natalija [7]

Answer: Volume would be 196.15 mL if the temperature were changed to 22^{o}C and the pressure to 1.25 atmospheres.

Explanation:

Given: T_{1} = 35^{o}C = (35 + 273) K = 308 K,     V_{1} = 256 mL,    

P_{1} = 720 torr (1 torr = 0.00131579 atm) = 0.947368 atm

T_{1} = 22^{o}C = (22 + 273) K = 295 K,       P_{2} = 1.25 atm  

Formula used to calculate volume is as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}

Substitute the values into above formula as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{1 atm \times 256 mL}{308 K} = \frac{1.25 atm \times V_{2}}{295 K}\\V_{2} = 196.15 mL

Thus, we can conclude that the volume would be 196.15 mL if the temperature were changed to 22^{o}C and the pressure to 1.25 atmospheres.

4 0
2 years ago
The density of NaCl( s) is 2.165 g cm 3 at 25 C. How will the solubility of NaCl in water be affected by an increase in pressure
sammy [17]

The solubility of NaCl in water will not be affected by an increase in pressure.

We know that the density of NaCl(s) in 2.165 g/cm³ at 25 °C and we want to know how will its solubility in water be affected when the pressure is increased.

<h3>What is solubility?</h3>

Solubility is the maximum mass of a solute that can be dissolved in 100 grams of solvent at a determined temperature.

The solubility of a solid, such as NaCl, in a liquid, is mainly affected by the temperature. However, since solids are not compressible, an increase in pressure will not affect its solubility.

On the other hand, the solubility of gases in water will increase with an increase in pressure, as stated by Henry's law.

The solubility of NaCl in water will not be affected by an increase in pressure.

Learn more about solubility here: brainly.com/question/11963573

7 0
2 years ago
Chapter 2.4.1. The ______ energy lost by the inward movement of nickel-iron within a differentiating molten planetary body would
Gwar [14]

Answer:b. gravitational, kinetic, thermal

Explanation:

The above explains the mechanism of the core forming process on earth/planet.

It is believed that this process might has contributed significantly to a planet's early stages heating. The time when these core-forming event happened is thought to have been mainly completed very early when Earth was young . The type of this event rather than it being seen as a single catastrophic event, it is likely to have been as a result of contractions on the earth severally.

The addition of partially differentiated material from another giant impact the rate of this spasm , and it increases each time the planet's mass is to increased.

This is a little on the history of planetary evolution.

8 0
3 years ago
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