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1 year ago

The electrophilic bromination or chlorination of benzene requires ______ along with the halogen.

Chemistry

1 answer:

Oksanka [162]1 year ago

8 0

The electrophilic bromination or chlorination of benzene requires Lewis acid along with the halogen.

<h3>What is bromination of benzene?</h3>

The bromination or chlorination of benzene is an example of an electrophilic aromatic substitution reaction.

During the reaction, the bromine forms a sigma bond to the benzene ring, yielding an intermediate. Subsequently a a proton is removed from the intermediate to form a substituted benzene ring.

This reaction is achieved with the help of Lewis acid as catalysts.

Thus, the electrophilic bromination or chlorination of benzene requires Lewis acid along with the halogen.

Learn more about bromination of benzene here: brainly.com/question/26428023

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Hey please answer this thanks.

Aleks04 [339]

Explanation:

Percentage composition of oxygen = (80/134) * 100% = 59.7%.

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2 years ago

When 250 ml of water is added to 35 ml of 0.2 M HCl

Paladinen [302]

Assuming that you’re looking for the concentration of water in the solution, then it would be 0.028 M.

You would have to use the formula:
c1v1 = c2v2, where c =concentration and
v = volume

C1 = ?
V1 = 250 mL
C2 = 0.2 M
V2 = 35 mL

C1 x 250 mL = 0.2 M x 35 mL

C1 = (0.2 M x 35 mL) / 250 mL

C1 = 0.028 M of water added to 35mL of 0.2M HCl

Therefore, there is 0.028 M of water added to 35mL of 0.2M HCl

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2 years ago

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Laney left a nail in water for two weeks, and the nail became rusted from sitting in the water.

Simora [160]

B) A chemical change because the nail reacts with water/oxygen to create rust (a type of oxide)

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2 years ago

If 21.00 mL of a 0.68 M solution of C6H5NH2 required 6.60 mL of the strong acid to completely neutralize the solution, what was

Andru [333]

Answer:

pH = 2.46

Explanation:

Hello there!

In this case, since this neutralization reaction may be assumed to occur in a 1:1 mole ratio between the base and the strong acid, it is possible to write the following moles and volume-concentrations relationship for the equivalence point:

$n_{acid}=n_{base}=n_{salt}$

Whereas the moles of the salt are computed as shown below:

$n_{salt}=0.021L*0.68mol/L=0.01428mol$

So we can divide those moles by the total volume (0.021L+0.0066L=0.0276L) to obtain the concentration of the final salt:

$[salt]=0.01428mol/0.0276L=0.517M$

Now, we need to keep in mind that this is an acidic salt since the base is weak and the acid strong, so the determinant ionization is:

$C_6H_5NH_3^++H_2O\rightleftharpoons C_6H_5NH_2+H_3O^+$

Whose equilibrium expression is:

$Ka=\frac{[C_6H_5NH_2][H_3O^+]}{C_6H_5NH_3^+}$

Now, since the Kb of C6H5NH2 is 4.3 x 10^-10, its Ka is 2.326x10^-5 (Kw/Kb), we can also write:

$2.326x10^{-5}=\frac{x^2}{0.517M}$

Whereas x is:

$x=\sqrt{0.517*2.326x10^{-5}}\\\\x=3.47x10^-3$

Which also equals the concentration of hydrogen ions; therefore, the pH at the equivalence point is:

$pH=-log(3.47x10^{-3})\\\\pH=2.46$

Regards!

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2 years ago

student plans to isolate chlorophyll, mix it in a solution of carbon dioxide and water, and then shine light on the mixture. Do

Inga [223]

Answer:

Explanation:

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<em>Once the chlorophyll is isolated, it becomes separated from the enzymes necessary for the completion of photosynthesis, and the process is truncated. </em>When light is shined on the mixture, the majority would instead be lost as heat while some cause the chlorophyll molecules to glow.

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2 years ago