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sergey [27]
2 years ago
11

Based on the products obtained, rank the functional groups (acetamido, amino, and methoxy) in order of increasing ability to act

ivate the benzene ring (make the aromatic ring susceptible to bromination).
Chemistry
1 answer:
I am Lyosha [343]2 years ago
7 0

Answer:

Amino >Methoxy > Acetamido

Explanation:

Bromination is of aromatic ring is an electrophilic substitution reaction. The attached functional group to the benzene ring activates or deactivate the aromatic ring towards electrophilic substitution reaction.

The functional group which donates electron to the benzene ring through inductive effect or resonance effect activates the ring towards electrophilic substitution reaction.

The functional group which withdraws electron to the benzene ring through inductive effect or resonance effect deactivates the ring towards electrophilic substitution reaction.

Among given, methoxy and amino are electron donating group. Amino group are stronger electron donating group than methoxy group. Acetamido group because of presence of carbonyl group becomes electron withdrawing group.

Therefore, decreasing order will be as follows:

Amino >Methoxy > Acetamido

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Which of these statements describes a limitation of the Treaty on Plant Genetic Resources?
saw5 [17]

Answer:

C. It does not address threats to genetic diversity.

Another APEX Answer:

<em>It ignores threats to genetic diversity.</em>

6 0
2 years ago
Read 2 more answers
You are performing a simple distillation of roughly 50:50 liquid solution containing two components, hexane and nonane You place
kifflom [539]

Answer:

Explanation:

Refractive index is a undimensional measurement that describes how fast light travels through the material. It is used to see the liquid purity of compounds.

In this problem you have in the beginning a 50:50 solution oh hexane:nonane and realize distillation.

In distillation the collect is enriched with the most volatille substance (hexane). So you will have in the final sample more hexane than nonane.

Assuming the refractive index varies linearly with mole fraction, you will have:

1,375 → 100% hexane

1,407 → 100% nonane

1,385 → ???

molar fraction of hexane:

100 - (1,385 - 1,375) / (1,407 - 1,375) × 100 = 69 %

molar fraction of nonane:

100 - (1,407 - 1,385) / (1,407 - 1,375) × 100 = 31 %

The relation of the distillating fraction is (69:31) of (hexane:nonane)

You can see that distillation is a technique that allows the separation of a liquid mixture, for example.

I hope it helps!

6 0
2 years ago
A student prepared an unknown sample by making a dilute solution of the unknown sample. The dilute sample was prepared by adding
telo118 [61]

Answer:

0.27 mM

Explanation:

According to the law of Lambert Beer:

A = C × E × L

Where:

A = Absorbance

C = Concentration

E = molar absorptivity

L = Step Length

If me make C the subject of the formula, we have = A / E × L

We know Absorbance and we can assume L = 1 cm, but being an unknown substance, we must know the molar Absorptivity of it, therefore, only having that value we could calculate the concentration of the mixture using the previous equation.

Then we could use the dilution ratio:

Cc * Vc = Cd * Vd

from the above formula, to fin the  concentrated Concentration, we have:

Cc = Cd * Vd / Vc

we then replace the known values.

Vd = Diluted volume 25 mL

Vc = Volume concentrated 5 mL

and we would not know the diluted concentration (Cd), which is what we had to calculate in the first section. Substituting it in the previous equation, we can obtain the initial concentration.

Assuming a molar absorptivity of 5000 M-1 * cm -1 we would have:

Cd = 0.270 / 5000 * 1

= 5.4 x 10 ^ -5 M * (1000 mM / 1 M)

= 0.054 mM

and initial concentration:

Cc = 0.054 mM * (25 mL / 5 mL)

= 0.27 mM.

4 0
2 years ago
What volume of 0.500 M HNO₃(aq) must completely react to neutralize 100.0 milliliters of 0.100 M KOH(aq)?
USPshnik [31]

Answer:

v = 500 milliliters

Explanation:

HNO_{3}  ⇒  H^{+}  + NO^{3-}

KOH  ⇒  K^{+}  + OH^{-}

1 H^{+} to 1 OH^{-}

\frac{0,5}{V}  = \frac{0,1}{100} \\\\v * 0,1 = 50\\v = 500 milliliters

4 0
2 years ago
Consider 2.4 moles of a gas contained in a 4.0 L bulb at a constant temperature of 32°C. This bulb is connected to an evacuated
Luden [163]

Complete Question

The complete question is shown on the first uploaded image

Answer

a

As the valve is opened , the gas will flow into the empty container until the both containers have the same pressure

b

\Delta H = 0\ , \Delta E = 0 , q= 0 , w= 0

c

The driving force for this process is the increase in entropy this is because the movement of the internal energy of the gas into a larger volume, what this does is that it increases the amount of disorder(entropy).

Explanation

In order to obtain the parameter in the part B of the question we are first obtain the initial pressure, using the ideal gas equation  

                      P = \frac{nRT}{V}

                     P = \frac{(2.4mol)(0.0821\frac{1 atm}{K \cdot \ mol} )}{4.0L}

                         P =15 \ atm

The next thing is to obtain the new pressure of the gas , using boyle's law

              P_1V_1 = P_2V_2

                  P_2 = \frac{P_1 V_1}{V_2}

                  P_2 = \frac{(15 \ atm)(4.0L)}{24.0 L}

                  P_2 = 2.5 \ atm    

Since the this process is isothermal , the change in heat is equal to zero

                      i.e  q = 0 J

  The workdone to move  the gas to the other container is zero because the  the pressure at this second container is zero due to the fact that it is a vacuum

    i.e  w = -P_{external} \Delta V

              =-(0 \ atm) (24.0 - 4.0L)

              = 0L \cdot atm

  Since the change in heat is zero and the workdone is zero then the change in internal energy is equal to 0

     This is because the change in internal energy is equal to a summation of change in heat and the workdone

                i.e \Delta E = q + w

                            = 0J

Generally the change in enthalpy is mathematically represented as

                \Delta H = n C_p \Delta T

Since the temperature is zero this means that the change in temperature is zero , substituting this value for change in temperature into the equation for  change in enthalpy

           \Delta H = n C_p (0)

                   = 0J

 

7 0
2 years ago
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