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5. [10 pts] Is it this reaction between aqueous AgNOs and aqueous LizSO4 & precipitation reaction? If yes, determine 4 Net ionic equation and Spectator_ions?...

Question

5. [10 pts] Is it this reaction between aqueous AgNOs and aqueous LizSO4 & precipitation reaction? If yes, determine 4 Net ionic equation and Spectator_ions?

5. [10 pts] Is it this reaction between aqueous AgNOs and aqueous LizSO4 & precipitation reaction? If yes, determine 4 Net ionic equation and Spectator_ions?



Answers

Write the net ionic equation for a chemical reaction that
occurs in an aqueous solution and produces water.

Here we are, considering the Net Ionic equation. So what I mean by the Net Ionic equation is that we have taken out our spectator ions that do not partake in the reactivity. So, for example, we have to f e three plus the pancreas phase at three C 03 to minus in the acquis phase gives us f e two c o three three that is a solid. In the next example, we have another net Ionic equations. We have HD 22 plus the acquis phase two CR minus gives us hg two cl two and the solid states we have former precipitate next example of our Net Ionic equation where we've removed a spectator and to see you two plus yeah, s two minus gives a C. U S in solid state. Next example, PB two plus the two cl minus gives us PBC l two and the solid states were gamma forming. Precipitate that. Next we have see a two plus the acres phase at C 03 to minus and the increased phase goes with C o. C. 03 Any solid state from the Ionic equation. Next we have G. So we have a U three plus at 30 H minus gives us a you oh H three in the solid states were forming precipitator.

So this question wanted us to write the equation for the precipitation of cadmium phosphate and also to write the net on equation for this reaction and identify the spectator ions. Um, And before we start writing the, um, before you start writing, I mean it on equation. Um, it is useful to first balance this regular equation as it is, and we can start by balancing the number of cadmium. Um, Adams. So there are three on the right hand side, but only one on the left hand side. Um, have over with this, there are now an uneven number of nitrate ions. There are six on the left hand side, so you must make it such that there six on the right hand side. Um, and now the number of phosphate ions, as well as the number of sodium ions are uneven. Um, but we can fix this by of doubling the amount of sodium phosphate we have on the reactant side. And with this, we have a balanced equation, and our Net Ionic equation, or at least on Ionic equation, can be written as follows, and we will be multiplying the number of, um, each eye on by the coefficient proceeding. It, um, just for ease of writing down the ions. So the ionic equation will be six sodium atoms and to phosphate Adams and three cadmium, um, ions and six nitrate ions. Um, And on the product side, it will be six sodium ions, six nitrate ions, and our precipitate the cadmium prostate. And with this ionic equation, we can start determine what are the spectator ions. Which ones are unchanged. And those are thesis odium ions as well as our nitrate ions. And so are net. Ionic equations is as follows. It is just the reaction of to, um, ions phosphate with three I owns of cadmium. They're reacting to form our cadmium phosphate, Um, and with our coefficients in front of our ions, um, we have a balanced net ionic equation.

So here we have a scenario where each of the following pairs of Aquarius solutions are mixed on. We're determining whether precipitation reactions occur. So if a precipitation reaction does in fact occur, we will write out the balance. Molecular total Ionic on net Ionic equation. So in our first example, we have sodium nitrate and copper to sulfate. So no, no precipitation occurs for this reaction. However, next we have ammonium bromide and silver nitrate. So we do have precipitation occur in this reaction. And so I molecular equation is on h four b r a. G and 03 is an equilibrium with NH four No. Three at a G b r. We then look at our catatonic on an ionic counterpart of our ionic species in the molecular equation, and we display them with their formal charges. And then we take a look at our Net ionic equation where what we have done is condensed on our total ionic equation and removed are spectator irons

So when the following powers of chemicals are mixed, were just determining if a precipitation will occur on, we'll just write out a few different equations. If they do so at first relapse. Sodium sulfide are nickel to sulfate. So, yes, we do have a precipitation. So therefore, I've written our our molecular total ionic on net Ionic. So my molecular equation, we have our neutral ionic compounds. And then this is broken down in our total ionic equation, where we have our positively charged Catalans on Alan's. And then we condense down this total ionic equation into a net ionic equation where we remove our spectator on ions. I'm cat ions in the second equation we're looking at, we have a reaction of lead two nitrate and potassium bromide. So, again, yes, we do have another precipitation reaction occurring here. So just like the first example, we've written our our molecular total ionic our Net ionic well again, we have all mutual ionic compounds which is broken down into a catatonic, and an ionic counterparts with former positive and former negative charges. And then we condense this down to a net ionic equation where we remove our spectator islands


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