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Strontium chloride sodium carbonate arc mixednitric acid calcium carbonate are mixedzinc barium bromide are mixedarmonium chromate harium chloride are maredAnef (II...

Question

Strontium chloride sodium carbonate arc mixednitric acid calcium carbonate are mixedzinc barium bromide are mixedarmonium chromate harium chloride are maredAnef (II) chloride ammonum phosphate arc mixedsodium hydroxide copper(II) nitrate are mixed172 Lab Erercise #9 Chemhicall Reactions and Net Fontc Equations

strontium chloride sodium carbonate arc mixed nitric acid calcium carbonate are mixed zinc barium bromide are mixed armonium chromate harium chloride are mared Anef (II) chloride ammonum phosphate arc mixed sodium hydroxide copper(II) nitrate are mixed 172 Lab Erercise #9 Chemhicall Reactions and Net Fontc Equations



Answers

If aqueous solutions of potassium carbonate and copper(II) nitrate are mixed, a precipitate is formed. Write the complete and net ionic equations for this reaction, and name the precipitate.

So, so when I trade is AGN three and strong shuhm Nike was a strange and Parade is a star steel too. Okay, so if you have the right molecular equation, well, if you're equation would look something like this, you would have silver nitrate. A cool solution reacting with strong chin fluoride. Michael was forming Gcl solid last strong jim nitrate of course. Right? So if you need to balance this balancing will be to wage in all three gives to a tcl. Okay then you have net ionic equation. So not ionic equation will be eggy positive reacting with C and negative forming A G C. L. Solid. Next one is gcl is silver two. Um All right and strong him. S. R. N. 03 World wise is drawn shuhm nitrate. So the equation will be AGC. L. 04 Across, reacting with any cl back was forming gcl solid. Does any cl four, of course.

Here we are taking a look at net ionic equations for precipitation reactions. So firstly we have seen are three plots At 30 H-. That yields cr Oh H three as a solid. And in the second example, what we have is following where we have to a G plus uh co three to minus, And this yields AG to CO three. In the third example, we have to h G At S 04 to minus MS, yields h g two, S 04. And now, in the final reaction, there is no precipitation.

Okay, what we want to do is try to write net ionic equations um for these different reactions I believe they're all going to be double replacement reactions. And so the first thing we need to do is we have chromium three chloride reacting with sodium hydroxide. So the first thing we need to do is write the equation to begin with. Right? So um we need to make sure that we have our periodic table with our oxidation states. Um some of these are oxy and ion. So we also need to have a list of your oxy and ions with your oxidation states as well. So chromium three means that we have an oxidation number of plus three and we know chlorine has a one minus. Right? So that means we need um three chlorine. Right? And these are all going to be awkward to solutions. Um and then sodium of course has one plus or positive one oxidation and hydroxide is an oxy ion. He has a -1. So this is going to be in A. O. H. And remember they're all accurate solutions. And so basically they switch partners. Double replacement switch partners. So typically your chromium, your first element in this one will go to your second element in the second um compound. So this is going to be chromium hydroxide and it is going to also have sodium chloride. Okay, so now we need to make sure it's balanced. Um so we have three chlorine is here but we only have one over here. So I need three here. That means my sodium Is now three. My hydroxide or three. So I should be okay just by putting a three here. Okay, so now um that's the first step. Now, the second step is to go ahead and kind of notice are solid ability rules. So these are probably pretty good to have handy um have a cheat sheet in front of you. Get to know these solid ability rules. Um because uh this double replacement reaction is only gonna occur if we have an insoluble part, which means we will have a precipitate insoluble means that it will not turn to acquiesce it will um come out as a solid. So um we do know that hydroxide or slightly soluble. Um So he's going to be a questionable so salt with hydroxide or slightly soluble but we do know that salts with um chlorine are very soluble. So we definitely know that this will say obvious. Now slightly soluble are that means they can um stay as an adequate solution or they can also precipitate out. Um These are slightly soluble. Um and so we're just going to say this is going to um come out as a solid. Okay, so now for our um before we get into our net ionic equation um we need to kind of look at what we're gonna do is our ionic equation is only going to be um the solid that is the product and whatever is also. First of all, let's do this. First of all, we're gonna change this entire equation into its ionic form ion form. So we have um the chromium three plus and it's obvious Plus um three, we won't have any of the subscript. So you're subscript come up as coefficient coefficient stay as coefficient and we're just gonna break them apart. And the only thing that is not broken apart is our and don't forget to put your states in here are are solid. So anything that is a solid over here that you're going to keep intact and then but we're gonna keep breaking everything apart into its ions. Okay so now that I have that done now we're going to take a look at our solid and we didn't look underline are solid. And then we're going to look on the left hand side and what which ones of our ions made up? The solid to the chromium ion and hydroxide are so those are the only ones that are going to be dropped down. So we have our chromium i on plus our three hydroxide ions to make the chromium hydroxide precipitate. And this last part is our net ionic equation. Okay so we're going to practice again, we're gonna go down and we're going to do another one and this one is silver nitrate With ammonia carbonate. Um so the first thing is silver nitrate has a oxidation state of -1. Silver has a one positive ammonia um has a Positive one and a carbonate has a two minus. Okay, so those are kind of critical things to do. Okay, so then silver nitrate is Going to be silvers eggy and nitrate is n. 0. 3. And remember you're you're oxidation numbers have to add up to zero. Um And of course this is going to be obvious. Plus now Ammonia is NH four and I need two of those to balance out to make him a 22 plus to balance out your carbonate. And of course this is going to be a double replacement. Um So um silver is going to go with carbonate. So we have silver. Um and I need there once again is um oxidation number is a one plus carbonates, the tu minus. So I need two of those. And then we will have um actually ammonia ammonia nitrate. So this is going to be And Ammonia is one plus a nitrate to one minus. So this is in age four, you know three, you know three. Okay, so now we need to balance, so let's go ahead and balance. Um I have to ammonia. So I need to make it to here. And uh to hear, let's see that worked out. So to two, yep, so that should be okay, so now I have to figure out which one if any of my products are going to be the precipitate. So um any um anything typically with a nitrate is going to be soluble. So salts containing nitrate or generally soluble Um This is ammonia nitrate. So that's not really assault. But definitely anything that has a carbon A. Is going to be insoluble. So this will be our precipitate. And this will say obvious. Okay so now we're gonna break our balanced equation into its ionic equation. So we have two silvers and of course always carry your states with you. Um These are oxygen and ion so they stay kind of intact when you write them down. So we have two nitrates plus we have to ammonia ins And Ammonia was a one plus. Okay and of course is Aquarius and then we have a carbonate and that's going to produce and of course your solid stays intact. And I might have to put him down a little bit. Okay so now now what I'm gonna do is plus um two of our ammonia. It's um which are Aquarius. And then plus two of our nitrates which are also Aquarius. Okay so then we look at our solid and which of our I ions made up our solid. So we had our silver ions and we had our carbonate ions. And so that is the only one that I actually write down. So too silver ions plus a carbonate ion made our solid of silver nitrate. And there we have our ionic equation. Okay so let's do another one and once you get the hang of it is pretty easy. It's pretty nicely going I'm gonna move this down, Give me a more room I think is what I need. Okay so we have um cop copper um to sulphate. So copper too has a too positive self is a tu minus. Um Mercury one is a one positive but nitrate is a one negative. Now there is something a little bit tricky about the night. The mercury one nitrate. Um Which we'll find out in a minute. It's kind of a weird. Um Okay so copper copper sulfate. So uh see you S. 04. And remember these are obvious. Plus mercury is aged G. But we actually even though we have mercury since it's the liquid comes in um die atomic always. Um So we have a two here. Okay and remember this is a double replacement. So we're going to get a copper nitrate. Um And so this would be a copper nitrate. And of course I need the two of them Because copper is too positive nitrates to 1-. Um And then are mercury whips him and then our self. Okay so now we gotta figure out okay um which if any of our products are precipitate out um So salts with nitrates are genuinely soluble um And then um salts with sulfate most salts with sulfate um are also soluble except mercury is very insoluble. So even though salts with sulfate are generally soluble. He's combined with mercury and mercury are is insoluble. So this is going to be our precipitate and this would be our acquiesce solution. Okay so now we do our ionic equation. So we have our copper and that's acquis yeah yeah. Plus our sulfate ion which is acquis plus our um to mercury ions Plus our two nitrate ions. And remember anything that is a solid you keep intact and then we have our copper ion Plus our two nitrate ions. And then um remember always even though we tend to a lot of writing we don't necessarily want to do it um it's very good habit to carry all of your states around with you. Okay And then of course keep your solid in tact. Okay so then um circle or underline your solid and then underlying the elements that made that's solid and that comprises our that comprises our now ionic equation lips and then plus our sulfate. Okay and by writing tool decides to kind of freak out on me. Okay so here we go. And there is our net ionic equation. Okay Last one. Okay so we have strontium nitrate and reacting with potassium iodide. So john tm has a and Positive to nitrate has a one potassium of course is a one plus and I died is a one minus. Okay so that's the first thing is kind of get to know your oxidation numbers. So it's drawn Thiam is S R. And nitrate in 03 and we're going to have to have two of them. And remember these are requests Plus potassium K. I. And are you serious? Okay. Um double replacement. Um so strong. Tm. I died yeah. Plus a potassium nitrate. Okay. Okay. So I need to hear and a two here. Okay, so now I need to know my cell viability rules. Okay, so salts with iodine are iodine are genuinely soluble um except if they're kind of with um with silver, lead mercury. Um They tend to be more insoluble. Um So these are generally soluble um and anything salts with nitrate. I lost my things are generally soluble as well. So both of these and Sean team is just a typical um alkaline earth metal. So I would seem this would be typically soluble. The only time they tend to not be soluble would be insoluble is when you have kind of some of your transition metals and so forth. So this would be soluble and nitrate soluble. And of course potassium in that same category as the alkaline alkaline earth metals. So both of these are soluble. So both of these are going to be acquiesce. So no precipitate forms. So we have no net ionic equation. Okay, um so really no reaction occurs, they just say as Aquarius solutions, I hope this helps

Chapter six Problem 54 asks us to write net ionic equations for each of these reactions given. So remember that a Net Ionic equation will only have the reactant is that are actively part of forming solid precipitate spectator ions, or I owns that aren't in the precipitate won't be written in our Net Ionic equation. So let's look at A for A. We started with chromium three chlorate and sodium hydroxide, So let's convert days from words into chemical acquaintance. So chromium three chloride. We know that we have chromium and chloride. The three after chromium tells us that the charge on it is three. And since the charge on C. L is one, if we swap those numbers or cross them, that gives us our sub scripts. So we'll have C r. One c l three and sodium hydroxide is an a o. H. So first, let's see what our hypothetical products would be. Weaken. Do this by swapping the ions so we see that we would have c r O H and thats three to account for the charge, plus an a c O. For these equations were not going to balance them at this step will balance when we hit our net Ionic equation. Okay, so let's consider which of these will form a solid precipitate. Well, most chloride salts are soluble, and an a C. L, in fact, is regular table salt, which we know is soluble. However, the C R O H three here is a solid product. So are net. Ionic equation will only have this as our product. Now let's look at the ions that made up CR. Ohh three. Here we have that CR three plus ion plus the O H minus I on and that gave us CR o h. Three solid. Now let's make sure this is balanced and to balance it out, we'll add a three in front of our O h i o. So this is the net Ionic equation for reaction A next let's look a reaction. Beef. Here we have silver nitrate and ammonium carbonate. Silver is a G and nitrate is No. Three and the charges here are already balanced. Next for ammonium carbonate, ammonium is an H four and carbonate is CEO three, and the balance of the charges will need to ammonium here. So let's look at the potential products that conform by switching the ions so we can have a g C 03 and to balance out that charge with me, too, of our silver. Or we could have any each four and No. Three or ammonium nitrate. Most nitrates are soluble. Therefore, a G C 03 is our solid precipitate. Now let's write our net Ionic equation for the formation of a G N 03 So we have 80 plus acquiesced ion plus C +03 to minus a quickest ion forming a G to C 03 solid. To balance this out, we'll need two of those silver molecules. Okay, Great. Now let's look at sea for see? We start with copper to Seoul. Pete and Mercury, one nitrate. So copper. See you sulfate s 04 plus mercury for HD nitrate and no. Three. And we'll need to balance up the charges here. Now, let's see what the potential products are. Well, we could have see you and no 32 or H g to s 04 Remember, however, that most nitrates are soluble, therefore are solid. Here would be the HD to s 04 molecule. Alice. What? The Net Ionic equation for the production of that molecule. So we have HD plus plus s +04 tu minus yielding H G two s 04 solid. And to balance this out, we will need two of our copper mode of our mercury molecules. Finally, let's look at deep here. We have strong Tiemann nitrate as our and No 32 plus potassium I or K I now notice Here, here we have nitrate and iodide. If you remember your scalability bulls, the vast majority of nitrate and iodide salts will be soluble. Therefore, we don't even really need to look at what the potential products are because whatever they are, they will be soluble because they either have a nitrate or in iodide. So here, because there's no precipitate that forms. We will also have no net ionic equation. Instead, everything in the reaction will stay in its


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