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valina [46]
2 years ago
8

Why do you add hydroxide to your hexanediamine solution? what would occur if you did not add it?.

Chemistry
1 answer:
Leokris [45]2 years ago
8 0
<h3>Why is Hydroxide added ?</h3>

In the polymerization reaction, the lone pair electrons on the NH₂ groups of hexanediamine attack the C=O groups of the dicarboxylic acid in a nucleophilic substitution reaction as shown in the image.

Hydroxide is added to remove any H⁺ ions present and keep the hexanediamine in the deprotonated form, so that the NH₂ lone pair electrons are available for reaction.

<h3 /><h3>What if you don't add it ?</h3>

If hydroxide is not added, the NH₂ groups will get protonated by H⁺  ions present to give NH₃⁺ groups, which cannot react.

To view similar questions based on Hydroxide,refer to:

brainly.com/question/15683618

#SPJ4

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Which 3 of the following is an exception to the law of superposition
Aleksandr [31]

the understanding of dehidatianismistering can be difficult in the modern world created for these specific terms

3 0
3 years ago
In a aqueous solution of 4-chlorobutanoic acid , what is the percentage of -chlorobutanoic acid that is dissociated
lord [1]

Answer:

Explanation:

Let assume that the missing aqueous solution of 4-chlorobutanoic acid = 0.76 M

Then, the dissociation of 4-chlorobutanoic acid can be expressed as:

\mathsf{C_3H_6ClCO_2H }          ⇄      \mathsf{C_3H_6ClCO_2^-}      +      \mathsf{H^+}

The ICE table can be computed as:

                   \mathsf{C_3H_6ClCO_2H }          ⇄      \mathsf{C_3H_6ClCO_2^-}      +      \mathsf{H^+}

Initial              0.76                                 -                           -

Change            -x                                  +x                         +x

Equilibrium   0.76 - x                              x                          x  

K_a = \dfrac{[\mathsf{C_3H_6ClCO_2^-}] [\mathsf{H^+}]}{\mathsf{[C_3H_6ClCO_2H ]}}

K_a = \dfrac{[x] [x]}{ [0.76-x]}

where:

K_a = 3.02*10^{-5}

3.02*10^{-5} = \dfrac{x^2}{ [0.76-x]}

however, the value of x is so negligible:

0.76 -x = 0.76

Then:

3.02*10^{-5}*0.76 = x^2

x=\sqrt{3.02*10^{-5}*0.76 }

x = 0.00479 M

∴

x = \mathsf{[C_3H_6ClCO_2^-] = [H^+]=} 0.00479 M

\mathsf{C_3H_6ClCO_2H }  = (0.76 - 0.00479) M

= 0.75521 M

Finally, the percentage of the acid dissociated is;

= ( 0.00479 / 0.76) × 100

= 0.630 M

7 0
3 years ago
10. a. What is the binding energy released when an alpha particle of mass 6.64 x 10-27 kg escapes from the nucleus of Uranium- 2
shutvik [7]

Answer:

The binding energy released is 1.992 X 10⁻¹⁸ J

Explanation:

Given;

mass of the alpha particle, m = 6.64 x 10⁻²⁷ kg

speed of the alpha particle, c = 3 x 10⁸ m/s

The binding energy released is given by;

E_b = mc^2

where;

m is mass of the particle

c is speed of the particle

E = 6.64 x 10⁻²⁷ (3 x 10⁸)²

E = 1.992 X 10⁻¹⁸ J

Therefore, the binding energy released is 1.992 X 10⁻¹⁸ J

7 0
4 years ago
At the end of the isomerization reaction, what chemical is used to quench the residual bromine?At the end of the isomerization r
valina [46]

Answer: Cyclohexene

Explanation:

Cyclohexane belongs to the Alkenes family. Alkenes react in the cold with pure liquid bromine, or with a solution of bromine in an organic solvent like tetrachloromethane. The double bond breaks, and a bromine atom get attached to each carbon. The bromine loses its original red-brown color to give a colorless liquid. In the case of the reaction with ethene, 1,2-dibromoethane is formed. When bromine is added to cyclohexane in the dark room, there won't be any reaction. If the mixture is exposed to light however, free bromine radicals are generated. In this condition, polybrominated products can be produced as well.

6 0
4 years ago
Hurry up i'll mark you the brainlest
djverab [1.8K]

Answer:

y- axis.

Explanation:

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5 0
3 years ago
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