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Archy [21]
4 years ago
13

Once the ionic solid has dissolved, the anion that is formed is able to react as a base, with water as the acid. Write the net a

cid-base reaction that occurs when dissolved NaClO reacts with water. (Use the lowest possible coefficients. Omit states-of-matter in your answer.)
Chemistry
1 answer:
Kay [80]4 years ago
8 0

<u>Answer:</u> The net acid-base reaction of the anion formed and water is written below.

<u>Explanation:</u>

Ionization reaction is defined as the reaction in which an ionic compound dissociates into its ions when dissolved in aqueous solution.

The chemical equation for the ionization of NaClO follows:

NaClO(s)\rightarrow Na^+(aq.)+ClO^-(aq.)

Now, the anion formed which is ClO^- reacts with water to form conjugate acid.

The chemical equation for the reaction of anion with water follows:

ClO^-(aq.)+H_2O(l)\rightarrow OH^-(aq.)+HClO(aq.)

Hence, the net acid-base reaction of the anion formed and water is written above.

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Answer:

See explanation below

Explanation:

First, you are not providing any data to solve this, so I'm gonna use some that I used a few days ago in the same question. Then, you can go and replace the data you have with the procedure here

The concentration of liquid sodium will be 8.5 MJ of energy, and I will assume that the temperature will not be increased more than 15 °C.

The expression to calculate the amount of energy is:

Q = m * cp * dT

Where: m: moles needed

cp: specific heat of the substance. The cp of liquid sodium reported is 30.8 J/ K mole

Replacing all the data in the above formula, and solving for m we have:

m = Q / cp * dT

dT is the increase of temperature. so 15 ° C is the same change for 15 K.

We also need to know that 1 MJ is 1x10^6 J,

so replacing all data:

m = 8.5 * 1x10^6 J / 30.8 J/K mole * 15 m = 18,398.27 moles

The molar mass of sodium is 22.95 g/mol so the mass is:

mass = 18,398.27 * 22.95 = 422,240.26 g or simply 422 kg rounded.

5 0
4 years ago
What is the answer to-6 + 1 = -5/
Jlenok [28]
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A children's liquid medicine contains 100mg of active ingredient in 5 mL. If a child should receive 200 mg of the active ingredi
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0.4g/ml

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