Answer:
See answer below
Explanation:
As you are asking for chemical formula and ionic equation, then, I will assume that after the station #5 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 1 Barium Chloride:</u>
In this case we have Barium on one side, and Chlorine on the other side, the symbol for those are Ba and Cl. As Barium have the +2 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:
BaCl₂
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:
Ba²⁺ + 2Cl⁻ ------> BaCl₂
2. Solution 2 and 3, Copper (II) nitrate and Copper(II) sulfate:
In this case I will work with both, because both of the solution have copper and the only difference is the anion, which is nitrate in one, and sulfate on the other. In this cases, we have a tertiary salt, The copper's symbol is Cu, and is working with it oxydation state +2. Nitrate is an anion and have the formula NO₃ working with the oxydation state -1. Sulfate is SO₄ and works with oxydation state -2 instead.
The chemical formula and ionic equation for both will be:
Copper(II) nitrate: Cu(NO₃)₂
Copper(II) sulfate: CuSO₄
And the net equations:
Copper nitrate: Cu²⁺ + 2NO₃⁻ ------> Cu(NO₃)₂
Copper sulfate: Cu²⁺ + SO₄²⁻ -------> CuSO₄
Hope this helps
