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For the following precipitation reaction Complete and balarice the molecular equation (10%) Include labels. and write the net ionic equation. FeCiz(aq) NazS(aq)redo...

Question

For the following precipitation reaction Complete and balarice the molecular equation (10%) Include labels. and write the net ionic equation. FeCiz(aq) NazS(aq)redox reuction. Include labels. net ionic reaclion- Il iS (10%) 10. Balance the following S4Os? (aq) +/ (aq) Ia(s) 52032(aq)

for the following precipitation reaction Complete and balarice the molecular equation (10%) Include labels. and write the net ionic equation. FeCiz(aq) NazS(aq) redox reuction. Include labels. net ionic reaclion- Il iS (10%) 10. Balance the following S4Os? (aq) +/ (aq) Ia(s) 52032(aq)



Answers

Balance the equation for the following precipitation reaction, and then write the net ionic equation. Indicate the state of each species $(s, \ell, \text { aq, or } g$ ). $$\mathrm{CdCl}_{2}+\mathrm{NaOH} \longrightarrow \mathrm{Cd}(\mathrm{OH})_{2}+\mathrm{NaCl}$$

So in this problem, we are looking at a chemical equation where we don't know our products and we would like to first predict our products, then balance our equation and then Writeth e simplified net Ionic equation. So starting off the first want to predict our products, and we can do so by realizing that this reaction is a double displacement reaction. So we're going to switch these two ions right here when predicting our products. So when we swap our ions, we end up with thes two products. And it's important to note here that we need to consider the oxidation numbers on each of these species when we're writing our products and we want to end up with a neutral compound. So we know that nickel has a positive to oxidation number and sulfide has a negative to oxidation number, and ammonia has a positive one oxidation number, and chloride has a negative one. So in both cases, we don't actually have to do any balancing in regards to our oxidation numbers to achieve a neutral compounds. The next step in predicting her products is going to be to determine this state of these products. So we want to look at our scalability table on page 179 to determine this from looking at that table. We see that this compound here is in the Equus fees because it is soluble with water. And this compound here is in these solid V's because it is not soluble with water. In this case, balancing our equation is fairly easy because if we look at the left and right can side the only species that we have an imbalance off our ammonium and chloride. So when we come over here, we see that all we need to do is at a coefficient of to to this compound. And our equation is balanced when writing are not Ionic equation. We want to look at what species in our regular chemically equation are in the acquiesced state and those air going to be the ions that we see showing up in this Net ionic equation. We see here that this nickel sulfide is in the solid state, so that's reflected here in the Net Ionic equation. And when we look at the coefficients that appear in our Net Ionic equation, they can either come from a coefficient that's out front of the entire compound itself. So that's the case here and here. Or it can be reflective of how maney ions are in the original compound, which you can see reflected here with the chloride and also here with the ammonium. So when we want to further simplify our net Ionic equation, we can cancel out ions that show up on both sides of our net ionic equation. So we know that we can cancel thes chloride I arms that appear on both sides. And we can also cancel these ammonium ions that appear on both sides. And after simplifying our ionic equation, we end with this final net Ionic equation here, just like the last chemical equation we looked at. We are also dealing in this case with a double displacement reaction. So as we did last time, we can go ahead and switch these two ions and predict our products by swapping our ions. We end up with this product and this product. And as I mentioned with the last chemical equation that we dealt with, it's important to consider our oxidation numbers and balance our sub scripts here so that the amount of ions that we're dealing with and that total overall oxidation number of the compound itself is equal to zero and then ask me the last time we want to predict the states of each of our compounds that were dealing with. So when we refer to ourselves ability to people, we see that this compound here is not soluble with water, so we can take it to be solid. And this compound here is soluble with water, so we can take it to be in the acquiesced state. So balancing dis equation is a little more complex than balancing the last equation we dealt with. When we look both sides of our equation, we can see that we have an imbalance of our phosphate here so we can go ahead and start are balancing by adding a coefficient of to here. We don't see that the amount up sodium that we have on the left hand side has changed, and we now have six sodium ions on the left hand side and only one sodium ion on the right hand side, so that could be remedied by adding a coefficient of six to that compound over there. Now, the only remaining problem is the imbalance of our nitrate ion on the right hand and left hand side, and we can remedy that by adding a coefficient of three to this compound. And now we see that we have six nitrate on the left hand side and six nitrate on the right hand side. So the Net Ionic equation can be written with thes same method that was used in the last part of this equation. Where were using both are coefficients and the amount of ions in each of our compounds to right the coefficients for these ions that appear in the Net Ionic equation and other Reminder, uh, compounds that are solid liquid or gas don't appear is ions in the Net ionic equation. So when simplifying this equation, we can cancel ions that we see appearing both on the right and left hand side of our equation so we can cancel our nitrate and our sodium from both sides. After making the simplifications, we end up with our final net Ionic equation here

And this question work eminent equation. Grass to balance the equation were also asked to list out the states of matter the components are in, so let's get started. Um, the key to solving this problem is remembering your Salyer building rules and remembering how to balance equations. So first up, we have nickel nitrates. And remember, the nitrates are soluble in water. So right off the bat, we know that this is a twist. Both the nickel and the nitrogen nitrate are going to be iniquitous form. Anytime you see sodium, that's going to be soluble. So that's also Equus. And now we have nickel carbonates. Nickel carbonate is not soluble Sylvia solid. And then we have nickel. I mean, I sodium nitrate, which is also soluble. Okay, so next up is we need to balance the equation. Um, we have to make sure that the number of items on the left hand side of the equation equal the number of items on the right hand side of equation. This is not like algebra in balancing equations. You can add and multiply as you please as long it's you, um, balanced the equation. So songs to stuff on this side. It's equal to that side. You're good to go, so we'll start with looking at nickel. We have one nickel over here. We have one nickel over here. Looks good. We have to. Sodium is over here. Oh, we don't need upon sodium over here. Oh, so let's see what happens if we put a to in front of the sodium on the right. So let's try doing Let's try doing this. So now we have to Sodium to Sodium is good. Um, over here we have two nitrates that this is a nitrate group and we have to of, um and then I just put this to in front of the sodium nitrate on the right hand side. So we have to nitrates on the right hand side to nitrates on the left hand side. So that's good. We have one carbonate on the right hand side. We have one carbon a on the left hand side. So this at this point, everything is balanced. All right, Now we have to look way have to write the next Ionic equation. So next ionic equation. That means we have to identify our spectator ions and get rid of them. All right, So first up, we have nickel and that's going to be an atheist. You don't like that? So we have more room Equus rates than we have, and that's a plus charge. And we have, um and 03 minus and we have two of them. This is also a Quist. Then we have sodium with the plus charge, really two of them. That's also a quis. And then we have a carbon e one minus. Then we have a nickel on the right can side carbonates on the right hand side boots. I'm just kidding because we said that then nickel carbon A is a solid. So we're going to say nickel carbonate is a solid and then we have sodium, which isn't a quickest. And we have to nitrates, which is also in a Prius form. Okay, so this sodium is an A plus form. We have two of them. We started with two sodium cynic This for nothing happened to the sodium so we can drop that from our equation. This nitrate salts on a platform. It started in APIs form. We have two on each side. They're going to cancel out now. We have this, uh, nickel over here and this carbonate over here, both in on the left hand side on a quiz form. But on the right hand side, they form a precipitate. A solid. Okay, so that means that they stay in equation. So are nuts. Ionic equation is nickel in increase form. Plus, you're carbonate ion and I goes form turning into nickel carbonate in salad form. And that's it. This is your necks Ionic equation. Thank you.

Hello, everyone. Thanks for joining. We're going to practice predicting products for a complete for a double replacement precipitation reaction, and we're gonna do the net Ionic equations. I think we'll skip the complete this time. If you need to do that on you when you can try it, Okay? We're going to be given reactant. We're going to predict the products, including states, and then after we have a balanced chemical equation, we're going to write the Net Ionic equation. And we have two problems to Dio to make sure we recording. And we are. Okay, so problem number one Oops, I forgot. Don't have the problem in front of me. There it is. Were given to reactant. And a reactant are for the 1st 1 nickel too. All right. Which is a quiz and ammonium sulphide. Ammonium. So fight. And this has to have to and this is also a quiz. Now we need to predict our products. I like to use the general format, and I'm gonna go and make a little orange up here. Ah x plus b. Why? Where nickel is a chloride is x ammonium is be and sulphide his wife. Our products will be in the format up a why plus B X. Let's go ahead and put her try to predict our products. A is nickel and it is the nickel two. Plus, I am. So I'm gonna put that up there. Why is sulfide and sell fighters a to minus? So that is a good chemical equation, and I'm going to raise my charges. I just need those to write my formula. My good formula should say not equation. Okay, My second product is going to be be which is an age four and X, which is CEO, and this is plus one and minus one. So that is also a good chemical equation. Now I need to assign states, and we know that ammonia and chloride both always contribute to sally ability. So that one's a quiz. And we also know that most soul fights are insoluble, and this is no exception. Now, if you'll recall when we're doing precipitation reactions, we should have three a cues and an ass three accusing an s which we dio now we need to balance this chemical equation, and I'm gonna do this balancing by inspection and I can see that I have to chlorides and to ammonium. So I do too right here. This is a balanced chemical equation. Now we've done the complete Ionic equations in the past, and we've split all of our accuse into its associated ions, crossed off the Spectators and then copied down the equation without the Spectators to get her net Ionic equation. This time we're going to start with putting our precipitate formula right here. And that's this. And then we're going to put the two contributing ions. The cat ion is nickel, which is right there. Got a two plus charge and it's a quiz plus sulfide. It's right here. We have a two minus charge and it's also a quiz. So it's that simple. Okay. For our second problem, I'm going to switch to a different blue. We're given the following chemical equate or brawling reactions A man and all three to Yeah, So we've got manganese to nitrate plus sodium, boss, fake. And these were both a Chris, remember, a x B Why equals equals yields a Why be x So a is manganese. And why is Pio for Magan eases two plus phosphate is three minus. So the formula we're going get will be mn three peel for two and our second product will be in a and in all three, both of which are always soluble. So that's in a queue. Phosphates tend to be insoluble and manganese. Um, to phosphate is no exception. So now again, I have three a cues and in s a a que Let me get here a que a que a que and an s And again we're going to Oh, I better balance this one. This one's gonna be a little tougher to balance that. I'm gonna start by looking at this man Younis three right here. Here, I want to make a three. I'm gonna put a three. So I've got three manganese on each side now, three times to I've got six nitrates, so I'm gonna put a six right here, and I've got three someone to put two right here. My coefficients from left to right are 32 one and six. Let's clean this up a bit, so it's a little bit easier to read even with my very questionable penmanship. And last but not least, we're going to write our Net ionic equation again. We're gonna do this by writing the formula over person to precipitate. And here's my soul, my manganese. And here's my prostate. I have three mn two plus Don't forget the states, plus to your pores three minus. Don't forget the States. So here the Net Ionic equations for each one of our two problems in here are the products and we bounced each equation. Thanks for watching.


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