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weeeeeb [17]
3 years ago
13

Any groups present on a benzene ring can impact the success and regioselectivity of an electrophilic aromatic substitution. Dete

rmine which group from the list best fits each activation and directing description.
Strongly deactivating meta- director __________
Moderately deactivating meta- director ________
Strongly activating ortho-/para- director _________
Weakly deactivating ortho-/para- director _________
Weakly activating ortho-/para- director ________
Chemistry
1 answer:
r-ruslan [8.4K]3 years ago
4 0

Answer:

Explanation:

In electrophilic aromatic substitution, the substituent(s) that favors electrophilic attack is said to be ortho or para to the substituent. Meta directors are deactivators. Altogether, these directors have an integral role to play in the success and regioselectivity of electrophilic aromatic substitution reactions.

From the information given: The;

Strongly deactivating meta- director  ⇒ NO₂ (nitro group)

Moderately deactivating meta- director  ⇒ CN (cyano group)

Strongly activating ortho-/para- director ⇒ OH (hydroxy; group)

Weakly deactivating ortho-/para- director ⇒ -X (halide group)

Weakly activating ortho-/para- director ⇒ - Ph  (phenyl group)

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8 0
3 years ago
How many molecules in 3.4 moles of NH4NO3?
aliya0001 [1]

There are  20.5 x 10^24 molecules are present in 3.4 moles of NH4NO3.

<h3>How many molecules in 3.4 moles of NH4NO3?</h3>

We know that one mole of a substance has  6.022 × 10²³ molecules so in 3.4 moles of NH4NO3, we have 20.5 x 10^24 molecules if we multiply the  6.022 × 10²³ with 3.4.

So we can conclude that there are  20.5 x 10^24 molecules are present in 3.4 moles of NH4NO3.

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5 0
1 year ago
Mg(OH)2 + 2HNO3 → Mg(NO3)2 + 2H20
LenaWriter [7]

Answer:

A. 1350

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3 0
3 years ago
A buffer solution contains 0.496 M hydrocyanic acid and 0.399 M sodium cyanide . If 0.0461 moles of sodium hydroxide are added t
pochemuha

Answer : The pH of the solution is, 9.63

Explanation : Given,

The dissociation constant for HCN = pK_a=9.31

First we have to calculate the moles of HCN and NaCN.

\text{Moles of HCN}=\text{Concentration of HCN}\times \text{Volume of solution}=0.496M\times 0.225L=0.1116mole

and,

\text{Moles of NaCN}=\text{Concentration of NaCN}\times \text{Volume of solution}=0.399M\times 0.225L=0.08978mole

The balanced chemical reaction is:

                          HCN+NaOH\rightarrow NaCN+H_2O

Initial moles     0.1116       0.0461     0.08978

At eqm.       (0.1116-0.0461)    0       (0.08978+0.0461)

                        0.0655                       0.1359

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

Now put all the given values in this expression, we get:

pH=9.31+\log (\frac{0.1359}{0.0655})

pH=9.63

Therefore, the pH of the solution is, 9.63

4 0
3 years ago
H2(g) + F2(g) → 2 HF(g) ΔH=-546 kJ
Arisa [49]

Answer: help please !!

Explanation:

4 0
3 years ago
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