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babymother [125]
3 years ago
8

For each of the following substituents, indicate whether it withdraws electrons inductively, donates electrons by hyperconjugati

on, withdraws electrons by resonance, or donates electrons by resonance. (Effects should be compared with that of a hydrogen; remember that many substituents can be characterized in more than one way).
A. Br
1. Withdraws electrons inductively.
2. Donates electrons by hyperconjugation.
3. Withdraws electrons by resonance.
4. Donates electrons by resonance.
B. CH2CH3
1. Withdraws electrons inductively.
2. Donates electrons by hyperconjugation.
3. Withdraws electrons by resonance.
4. Donates electrons by resonance.
C. NHCH31. Withdraws electrons inductively.2. Donates electrons by hyperconjugation.3. Withdraws electrons by resonance.4. Donates electrons by resonance.D. OCH31. Withdraws electrons inductively.2. Donates electrons by hyperconjugation.3. Withdraws electrons by resonance.4. Donates electrons by resonance.E. +N(CH3)31. Withdraws electrons inductively.2. Donates electrons by hyperconjugation.3. Withdraws electrons by resonance.4. Donates electrons by resonance.
Chemistry
1 answer:
aleksandrvk [35]3 years ago
3 0

Answer:

Br- Withdraws electrons inductively

       Donates electrons by resonance

CH2CH3 - Donates electrons by hyperconjugation

NHCH3- Withdraws electrons inductively

              Donates electrons by resonance

OCH3 -  Withdraws electrons inductively

              Donates electrons by resonance

+N(CH3)3 - Withdraws electrons inductively

                   

Explanation:

A chemical moiety may withdraw or donate electrons by resonance or inductive effect.

Halogens are electronegative elements hence they withdraw electrons by inductive effect. However, they also contain lone pairs so the can donate electrons by resonance.

Alkyl groups donate electrons by hyperconjugation involving hydrogen atoms.

-NHCH3  and contain species that have lone pair of electrons which can be donated by resonance. Also, the nitrogen and oxygen atoms are very electron withdrawing making the carbon atom to have a -I inductive effect.

+N(CH3)3 have no lone pair and is strongly electron withdrawing by inductive effects.

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100.00 mL of 0.15 M nitrous acid (HNO2) are titrated with a 0.15 M NaOH solution. (a) Calculate the pH for the initial solution.
wolverine [178]

Answer:

a. pH = 2.04

b. pH = 3.85

c. pH = 8.06

d. pH = 11.56

Explanation:

The nitrous acid is a weak acid (Ka = 5.6x10⁻⁴) that reacts with NaOH as follows:

HNO₂ + NaOH → NaNO₂(aq) + H₂O(l)

a. At the beginning there is just a solution of 0.12M HNO₂. As Ka is:

Ka = [H⁺] [NO₂⁻] / [HNO₂]

Where [H⁺] and [NO₂⁻] ions comes from the same equilibrium ([H⁺] = [NO₂⁻] = X):

5.6x10⁻⁴ = X² / 0.15M

8.4x10⁻⁵ = X²

X = [H⁺] = 9.165x10⁻³M

As pH = -log [H⁺]

<h3>pH = 2.04</h3><h3 />

b. At this point we have HNO₂ and NaNO₂ (The weak acid and the conjugate base), a buffer. The pH of a buffer is obtained using H-H equation:

pH = pKa + log [NaNO₂] / [HNO₂]

<em>Where pH is the pH of the buffer,</em>

<em>pKa is -log Ka = 3.25</em>

<em>And [NaNO₂] [HNO₂] could be taken as the moles of each compound.</em>

<em />

The initial moles of HNO₂ are:

0.100L * (0.15mol / L) = 0.015moles

The moles of base added are:

0.0800L * (0.15mol / L) = 0.012moles

The moles of base added = Moles of NaNO₂ produced = 0.012moles.

And the moles of HNO₂ that remains are:

0.015moles - 0.012moles = 0.003moles

Replacing in H-H equation:

pH = 3.25 + log [0.012moles] / [0.003moles]

<h3>pH = 3.85</h3><h3 />

c. At equivalence point all HNO2 reacts producing NaNO₂. The volume added of NaOH must be 100mL. That means the concentration of the NaNO₂ is:

0.15M / 2 = 0.075M

The NaNO₂ is in equilibrium with water as follows:

NaNO₂(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq) + Na⁺

The equilibrium constant, kb, is:

Kb = Kw/Ka = 1x10⁻¹⁴ / 5.6x10⁻⁴ = 1.79x10⁻¹¹ = [OH⁻] [HNO₂] / [NaNO₂]

<em>Where [OH⁻] = [HNO₂] = x</em>

<em>[NaNO₂] = 0.075M</em>

<em />

1.79x10⁻¹¹ = [X] [X] / [0.075M]

1.34x10⁻¹² = X²

X = 1.16x10⁻⁶M = [OH⁻]

pOH = -log [OH-] = 5.94

pH = 14-pOH

<h3>pH = 8.06</h3><h3 />

d. At this point, 5mL of NaOH are added in excess, the moles are:

5mL = 5x10⁻³L * (0.15mol / L) =7.5x10⁻⁴moles NaOH

In 100mL + 105mL = 205mL = 0.205L. [NaOH] = 7.5x10⁻⁴moles NaOH / 0.205L =

3.66x10⁻³M = [OH⁻]

pOH = 2.44

pH = 14 - pOH

<h3>pH = 11.56</h3>
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Karolina [17]
The answer is false methane is emitted from coal
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noname [10]

Answer:

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7 0
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Evgen [1.6K]

<em>Answer:</em>

4) the one that is reduced, which is the oxidizing agent

<em>Explanation:</em>

<em>An oxidizing agent is one that causes oxidation by gaining electrons from another atom/molecule. </em>

6 0
3 years ago
!HELP ASAP!!!
Snowcat [4.5K]

Answer:

Covalent Bonds

Explanation:

INTERmolecular forces are those that exist between molecules, so you can think of it liek international things taking place between countries. As you are aware, dipoles exist across an entire molecule, so for 2 dipoles to interact, there needs to be 2 molecules. Van der Waals forces also take place between  molecules when there is an uneven distribution of electrons across a molecule, causing a temporary weak dipole. Hydrogen bonding is similar to dipole-dipole forces, but only happen when there is a hydrogen interacting with an atom on another molecule that has a lone pair of electrons.

Covalent bonds, however, are INTRAmolecular, meaning they are present within a molecule. Covalent bonds are the bonds that exist when two atoms, within the same molecule, share electrons so both can have a stable electron configuration.

Hope I helped! xx

6 0
3 years ago
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