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Aleksandr [31]
2 years ago
4

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

boxylic acid anhydride?
A) Add 0.1M NaOH (aq) 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.
B) Add 0.1 M HCl (aq) to quench unreacted anhydride. Then add diethyl ether and separate the layers. The amide can be obtained from ether layer by evaporating the solvent.
C) Add 0.1M NaOH (aq) to quench unreacted anhydride. Then add diethyl ether and separate the layers. The amide can be obtained from the aqueous layer by neutralizing with HCl (aq).
D) Add 0.1M HCl (aq) to quench the unreacted anhydride. Then add diethyl ether and separate the layers. The amide can be obtained from the aq. layer by neutralizing the NaOH (aq).
Chemistry
1 answer:
lyudmila [28]2 years ago
3 0

Answer:

Option C is the correct answer

Explanation:

Amines generally are bases, as such they react or dissolve in acids. as explained from lewis theory, they also behave as lewis bases because they are electron pair donors. Amines are organic compounds that have nitrogen atom bonded to an organic radical. they have the general formula RCH3NH2.

from the question and options, since they are bases, they reacts and dissolves in strong acids like HCL, as such to completely separate an amide from the by-product of the reaction between an amine and excess carboxylic acid anhydride,  proceed by adding  molar concentration of NaOH (aq) to quench unreacted anhydride. Then add diethyl ether and separate the layers. The amide can be obtained from the aqueous layer by neutralizing with HCl (aq), since they dissolve easily in HCL.

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A block of aluminum weighing 140 g is cooled from 98.4°C to 62.2°C with the release of 1080 joules of heat. From this data, calc
GrogVix [38]

<u>Answer:</u>

Specific heat of a substance is the value that describe how the added heat energy of substance has the impact on its temperature.

Unit is <em>(\frac {J}{Kg.K})</em>

<em>C = Q/m. ∆T</em>

<em>C – Specific heat (\frac {J}{Kg.K})</em>

<em>Q- heat energy (J)</em>

<em>M – Mass (Kg)</em>

<em>∆T- change in temperature (K) </em>

<u>Explanation:</u>

<em>Given data:</em>

<em>M= 140 g = 0.14 Kg</em>

<em>Q – 1080 Joules.</em>

<em>∆T – 98.4 – 62.2 = 36.2</em>

Substituting  the given data in Equation

<em>Specific heat of Aluminium  = \frac {1080}{(0.14 \times 36.2)} = 213.10 (\frac {J}{Kg.K})</em>

3 0
2 years ago
Read 2 more answers
Insoluble sulfide compounds are generally black in color. Which of the following combinations could yield a black precipitate? C
Lubov Fominskaja [6]

Answer:

Li₂S(aq) + Pb(NO₃)₂(aq)

2 K₂S(aq) + Sn(NO₃)₄(aq)

Explanation:

For these reactions, the products are:

1. Na₂S(aq) + 2 KCl(aq) → K₂S + 2 NaCl

2. Li₂S(aq) + Pb(NO₃)₂(aq) →  PbS + 2 Li(NO₃)

3. Pb(ClO₃)₂(aq) + 2 NaNO₃(aq) → Pb(NO₃)₂ + 2 Na(ClO₃)

4. AgNO₃(aq) + KCl(aq) → AgCl + KNO₃

5. 2 K₂S(aq) + Sn(NO₃)₄(aq) → SnS₂ + 4 KNO₃

K₂S is a soluble sulfide that, in solid state, is white. As K has a low electronegativity, the relative polar bond K-S allows its dissolution in water.

PbS is a black and insoluble compound. As Pb electronegativity is higher than K electronegativity, The bond is less-polar and its dissolution will not be allowed.

For the third reaction, all nitrates are soluble and Na(ClO₃) is also a very soluble compound.

AgCl is an insoluble white compound.

As Sn electronegativity is high, the solubility of this compound is very low. Also, this compound is black!

Thus, combinations could yield a black precipitate are:

Li₂S(aq) + Pb(NO₃)₂(aq)

2 K₂S(aq) + Sn(NO₃)₄(aq)

I hope it helps!

6 0
2 years ago
An unknown salt is either NaF, NaCl, or NaOCl. When 0.050 mol of the salt is dissolved in water to form 0.500 L of solution, the
victus00 [196]

<u>Answer:</u> The unknown salt is NaF

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles of salt = 0.050 moles

Volume of solution = 0.500 L

Putting values in above equation, we get:

\text{Molarity of salt}=\frac{0.050mol}{0.500L}\\\\\text{Molarity of salt}=0.1M

  • To calculate the hydroxide ion concentration, we first calculate pOH of the solution, which is:

pH + pOH = 14

We are given:

pH = 8.08

pOH=14-8.08=5.92

  • To calculate pOH of the solution, we use the equation:

pOH=-\log[OH^-]

Putting values in above equation, we get:

5.92=-\log[OH^-]

[OH^-]=10^{-5.92}=1.202\times 10^{-6}M

The unknown salt given are formed by the combination of weak acid and strong acid which is NaOH

The chemical equation for the hydrolysis of X^- ions follows:

                    X^-(aq.)+H_2O(l)\rightleftharpoons HX(aq.)+OH^-(aq.);K_b

<u>Initial:</u>              0.1

<u>At eqllm:</u>        0.1-x                           x              x

Concentration of OH^-=x=1.202\times 10^{-6}M

The expression of K_b for above equation follows:

K_b=\frac{[OH^-][HX]}{[X^-]}

Putting values in above expression, we get:

K_b=\frac{(1.202\times 10^{-6})\times (1.202\times 10^{-6})}{(1-(1.202\times 10^{-6}))}\\\\K_b=1.445\times 10^{-11}M

  • To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:  

K_w=K_b\times K_a

where,

K_w = Ionic product of water = 10^{-14}

K_a = Acid dissociation constant

K_b = Base dissociation constant = 1.445\times 10^{-11}

Putting values in above equation, we get:

10^{-14}=1.445\times 10^{-11}\times K_a\\\\K_a=\frac{10^{-14}}{1.445\times 10^{-11}}=6.92\times 10^{-4}

We know that:

K_a\text{ for HF}=6.8\times 10^{-6}

K_a\text{ for HCl}=1.3\times 10^{6}

K_a\text{ for HClO}=3.0\times 10^{-8}

So, the calculated K_a is approximately equal to the K_a of HF

Hence, the unknown salt is NaF

6 0
2 years ago
Chemists use activity series to determine whether which type of reaction will occur?
bazaltina [42]

Answer:

The correct answer is "single replacement".

Explanation:

The activity series is a representation of a series of metals that are organized according to their order of reactivity. They start from the highest order of reactivity to the lowest order of reactivity.

The purpose of the series is to determine which products are formed in single-displacement reactions. The higher-order metals replace the lower order metals if there are two of them in the same solution. If metal is higher in the series it is more reactive than a metal that is below it in the activity series.

Have a nice day!

5 0
2 years ago
1) Analysis subquestions (7 points): (a) Draw the mechanism of the reaction - remember, there are two main parts to the aldol co
scZoUnD [109]

Find the attachments

5 0
2 years ago
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