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1 year ago

Identify the Lewis acid and Lewis base in each reaction:Na⁺ + 6H₂O ⇆ Na (H₂O)⁺₆

Chemistry

1 answer:

klemol [59]1 year ago

4 0

In the reaction, Na⁺ + 6H₂O ⇆ Na (H₂O)⁺₆, Na⁺ is a Lewis acid and H₂O is a Lewis base.

<h3>What is a Lewis acid-base reaction?</h3>

According to the Lewis theory of acid-base reactions, acids accept pairs of electrons and bases donate pairs of electrons.
Any substance like H+ ion, which is capable of accepting a pair of nonbonding electrons or an electron-pair acceptor is known as a Lewis acid.
Any substance, like the OH- ion, that is capable of donating a pair of nonbonding electrons or an electron-pair donor is a Lewis base.
Here Na⁺ that is electron deficient accepts electrons from the electron donor, H₂O
From the Lewis theory, with no change in the oxidation numbers of any atoms, acids react with bases to share a pair of electrons.

To learn more about Lewis acid-base reactions: brainly.com/question/14861040

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vladimir2022 [97]

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3 years ago

Use dimensional analysis to convert 14.5mi/hr to km/s

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Answer:

0.006 48 km/s

Explanation:

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14.5 mi × (1.609 km/1 mi) = 23.33 km

2. Convert hours to seconds

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3 years ago

Fill in the left side of this equilibrium constant equation for the reaction of benzoic acid (HC_6H_5CO_2) with water.

il63 [147K]

Answer:

[C₆H₅COO⁻][H₃O⁺]/[C₆H₅COOH] = Ka

Explanation:

The reaction of dissociation of the benzoic acid in water is given by the following equation:

C₆H₅-COOH + H₂O ⇄ C₆H₅-COO⁻ + H₃O⁺ (1)

The dissociation constant of an acid is the measure of the strength of an acid:

HA ⇄ A⁻ + H⁺ (2)

$K_{a} = \frac{[A^{-}][H^{+}]}{[HA]}$ (3)

<em>Where the dissociation constant of the acid (Ka) is equal to the ratio of the concentration of the dissociated forms of the acid, [A⁻][H⁺], and the concentration of the acid, [HA]. </em>

So, starting from the equations (2) and (3), the constant equation for the dissociation reaction of benzoic acid in water, of the equation (1), is:

$K_{a} = \frac{[C_{6}H_{5}COO^{-}][H_{3}O^{+}]}{[C_{6}H_{5}COOH]}$

I hope it helps you!

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