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Dahasolnce [82]
3 years ago
6

What is the chemical formula and net ionic equations for all three solutions.

Chemistry
1 answer:
timofeeve [1]3 years ago
4 0

Answer:

See answer below

Explanation:

As you are asking for chemical formula and ionic equation, then, I will assume that after the station #4 below, are the solutions you are requiring.

You are also not specifing if you want for example, result of solution 1 + solution 3. If you need that, please post that on another question.

Now for the chemical formula, you need to identify the elements in all 3 solutions, and also the type of compound.

<u>1. Solution 2 Potassium Chloride: </u>

In this case we have Potassium on one side, and Chlorine on the other side, the symbol for those are K and Cl. As Potassium have the +1 oxidation state, cause is the only one that it can have, when it's next to an halide like chlorine or bromine, it will form a binary salt. The halides, usually work with the lowest oxydation state. In the case of Chlorine it will be -1, so, the formula will be:

KCl

And the net ionic equation will be the chemical equation that shows how the charges and atoms are balanced. In this case it would be:

K⁺ + Cl⁻ ------> KCl

<u>2. Solution 1 Copper(II) sulfate: </u>

In this case we have a tertiary salt, The copper's symbol is Cu, and is working with it oxydation state +2. Sulfate is an anion and it's formula is SO₄ and works with oxydation state -2 instead.

The chemical formula and ionic equation will be:  

Copper(II) sulfate: CuSO₄

And the net equation:

Copper sulfate: Cu²⁺ + SO₄²⁻ -------> CuSO₄

<u>3. Solution 3 Sodium hydroxide:</u>

In this case, we have a compound that it's usually used in acid base reactions. This is a strong base or hydroxide, and we have the element of Sodium (Na) with the oxydation state +1, is the only one it can have, and for the other side we have the oxydrile anion OH, and together is working with the oxydation state -1. So the chemical formula will be:

NaOH

And the net ionic equation:

Na⁺ + OH⁻ -------> NaOH

Hope this helps

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