Chemistry, 31.01.2021 14:00 lexxa58

What is the Chemical Formula and Net Ionic Equations for all three solutions.

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Answer from: EwTrTw4584

See answer below

Explanation:

As you are asking for chemical formula and ionic equation, then, I will assume that after the station #3 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.

1. Solution 2 Potassium Iodide:

In this case we have Potassium on one side, and Iodine on the other side, the symbol for those are K and l. 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 Iodide it will be -1, so, the formula will be:

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⁺ + I⁻ > KI

2. Solution 1 and 3, Lead (II) nitrate and Sodium carbonate:

In this case I will work with both, because both of the solution are tertiary compounds. In this cases, we have two tertiary salt, The Sodium symbol is Na, and is working with it oxydation state +1. Carbonate is an anion and have the formula CO₃ working with the oxydation state -2. Lead can work with oxidation state +2 and +4. It's symbol is Pb. Nitrate is NO₃ and works with oxydation state -1 instead.

The chemical formula and ionic equation for both will be:

Lead(II) nitrate: Pb(NO₃)₂

Sodium Carbonate: Na₂CO₃

And the net equations:

Lead nitrate: Pb²⁺ + 2NO₃⁻ > Pb(NO₃)₂

Copper sulfate: 2Na⁺ + CO₃²⁻ > Na₂CO₃

Hope this helps

Answer from: Quest

The answer is c) pyrite

Answer from: Quest

Answerthe half-reaction having the greatest tendency to occur at the anode: i₂(aq) + 2e⁻ ⇄ 2i⁻ ; e = +0.53 v====== $\begin{aligned} \text{i}_2\text{\footnotesize (aq)}+2\,\text{e}^{-} & \rightleftharpoons 2\,\text{i}^{-}\text{\footnotesize (aq)}& \text{e} & = +0.53\text{ v} \\ \text{co}^{3+}\text{\footnotesize (aq)} + \text{e}^{-} & \rightleftharpoons \text{co}^{2+}\text{\footnotesize (aq)} & \text{e} & = +1.80\text{ v} \\ \text{o}_2\text{\footnotesize (g)}+4\,\text{h}^{+}\text{\footnotesize (aq)} + 4\,\text{e}^{-} & \rightleftharpoons 2\,\text{h}_2\text{o}\text{\footnotesize(l)}& \text{e} & = +1.23\text{ v} \\ \end{aligned}$ oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be i₂(aq) + 2e⁻ ⇄ 2i⁻ e = +0.53 vbecause its reduction potential is the lowest out of the three.