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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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Read 2 more answers
t 745 K, the reaction below has an equilibrium constant (Kc) of 5.00 × 102. H2 (g) + I2 (g) ⇌ 2 HI (g) If a mixture of 0.10 mol
spayn [35]

Answer : The concentration of HI (g) at equilibrium is, 0.643 M

Explanation :

The given chemical reaction is:

                        H_2(g)+I_2(g)\rightarrow 2HI(g)

Initial conc.    0.10        0.10      0.50

At eqm.        (0.10-x)  (0.10-x)   (0.50+2x)

As we are given:

K_c=5.00\times 10^2

The expression for equilibrium constant is:

K_c=\frac{[HI]^2}{[H_2][I_2]}

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

5.00\times 10^2=\frac{(0.50+2x)^2}{(0.10-x)\times (0.10-x)}

x = 0.0713  and x = 0.134

We are neglecting value of x = 0.134 because the equilibrium concentration can not be more than initial concentration.

Thus, we are taking value of x = 0.0713

The concentration of HI (g) at equilibrium = (0.50+2x) = [0.50+2(0.0713)] = 0.643 M

Thus, the concentration of HI (g) at equilibrium is, 0.643 M

8 0
3 years ago
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