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Question 113 ptsWhat would be the precipitate in the following reaction? Pb(CzH3Ozlz (aq) 2 KI (aq) Pblz (s) 2 KCzH3Oz (aq)Lead (II) acetatePotassium iodidePotassiu...

Question

Question 113 ptsWhat would be the precipitate in the following reaction? Pb(CzH3Ozlz (aq) 2 KI (aq) Pblz (s) 2 KCzH3Oz (aq)Lead (II) acetatePotassium iodidePotassium acetateLead (II) iodide

Question 11 3 pts What would be the precipitate in the following reaction? Pb(CzH3Ozlz (aq) 2 KI (aq) Pblz (s) 2 KCzH3Oz (aq) Lead (II) acetate Potassium iodide Potassium acetate Lead (II) iodide



Answers

For the following chemical reactions, determine the precipitate produced when the two reactants listed below are mixed together. Indicate “none” if no precipitate will form
$\mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{K}_{3} \mathrm{PO}_{4}(a q) \longrightarrow$__________________(s)
$\mathrm{K}_{2} \mathrm{CO}_{3}(a q)+\mathrm{AgNO}_{3}(a q) \longrightarrow$__________________(s)
$\mathrm{NaCl}(a q)+\mathrm{KNO}_{3}(a q) \longrightarrow$__________________(s)
$\mathrm{KCl}(a q)+\mathrm{AgNO}_{3}(a q) \longrightarrow$__________________(s)
$\mathrm{FeCl}_{3}(a q)+\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q) \longrightarrow$__________________(s)

So in this video, we're gonna go over Question 52 from Chapter four, which size when the following solutions are mixed together. What precipitate if any will form. So when a we have ht to ino 32 plus C u S 04 Um, So what products could we possibly form? Well, we could form copper to nitrate or mercury one sulfate those air the two products we could form. And since most nitrate salts are soluble, coppers to nitrate is soluble and most soul fights assaults air soluble. But the exception is mercury one sulfate. So this is insoluble. So we are going to form a precipitate, and it will be the mercury one sulfate and B. We're given nickel to nitrate plus calcium chloride. So, um, what can we what products can we possibly form? Will of our nickel can tie on, combines with their chloride, and I on we get nickel to chloride. And if our calcium cat eye on combines with their nitrate and I on, we get called on calcium nitrate. So since most of our chloride salts are soluble, um, then our nickel to chloride is valuable. And since most of our nitrate salts, air soluble are calcium nitrate is also soluble, So we're not gonna form any precipitate and B and see we're given Que tu sio three plus m g. I too. So potassium carbonate plus magnesium iodide. So what products can reform will live far Potassium cat eye on combines with our iodide and I only get potassium iodide. And since most of our iodide salt are soluble than this is valuable, um And then if we combine our magnesium cat eye on with our carbonate and I on, we get magnesium carbonate and most of our carbonate salts are only slightly soluble, except for those with certain cat ions. But my museum is not one of those cat eye on, so this is only slightly soluble, so we'll form some purchase, um, precipitate and that precipitate will be magnesium carbonate in d were given, um, sodium chrome eight plus aluminum bromide. So what products can we form? Well, if our sodium cat eye on combined with their bromide, and I only get sodium bromine And since most of our bromide salts are soluble, this is soluble, um And then if we combine our aluminum cat eye on with our crow, my and I on. Then we get aluminum crow, mate. Um, and most of our chrome eight salts are only slightly soluble except where those with certain cat I owns. But aluminum is not one of those cat eye on, so this is only slightly soluble, so we'll form some precipitate on that precipitate will be aluminum crow, mate.

Chapter six Problem 48 gives us a variety of solutions mixed together and asks which will form precipitate so to solve these. First, we'll look at the different products that could form and then see if any of them are solid or it's in soluble in water. Those that are insoluble will be the ones that form a precipitate. So let's start with this first problem in order to see what the potential products could be, we will exchange the ions here. Essentially H g will switch with See you. We do that. We find that we have HD s 04 and see you and no three to now we need to determine if either of these is solid or insoluble in water. If we look at this eligibility table, we find that H G s 04 is in fact solid. Therefore, we do have a precipitate forming specifically the precipitate HD as so four solid. Now let's look at the next problem here we have and I know 32 and c a c l two. So again it will switch our ions here. And that will give us and I, c e o or C A and No. Three. If we look at a soluble ity chart, however, we see that neither of the's is a solid precipitate were insoluble in water. Therefore, for this problem, we have no precipitate. Now let's move on to see again we'll switch our ions So we have MGI c 03 plus k I. And if we look at a soluble ity table, we find that MG CEO three doesn't fact form a solid precipitate. So we have a precipitate of m G c 03 Finally, let's look at the last one again. We will switch our ions giving us a l c R O for three with a little too. And again To get those numbers, we will have to be able to balance out all of our charges for each molecule. And that also gives us and a BR once again looking at our soy ability table, we find that ailed to C R O for three does form a solid for

So will now work on Problem one hundred from chapter seventeen. So this question here is asking us two. There is telling us that a solution containing potassium bromide is mix with one containing lead acid tate to form a solution that is a certain concentration of potassium bromide and lead acid ate. Now they're asking us does a precipitate form, and if so, I don't find the precipitous. So when we have problems like this, the first thing that we wanted the first thing that we want to check is the cross products. So the possible cross products for caviar and it has lead acid Tate. So if we mix, the Catalans were going to get potassium acid. Tate the Let me read it all together. Sometimes it's better to read it out sometimes not so see to H three o two. So potassium acid tate and we know from our cell ability rules that this should be side or are highly sought. Now the other one. If we combined the lead with bromide, we give lead. PBB are too, and this has a cast P, which is equal to four point six seven times ten to the minus six. So after we do that, we identify the possible cross product as being led bromide. If a precipitate forms, we need to calculate cute. So cue is equal to the concentration of lead to us. Time's a concentration of bromide minus squared because of the sub skirt. Two. So from the concentrations were given in the problem. We have zero point zero three five Buller lead and Sarah point zero one three of bromide, and we square that. And when we do this calculation, we get five point nine two times ten to the minus seven, so we can see from here that Q is less than K S P. And so what this means is that we have no precipitation. Q. Has to be at least Precipitation will start when Q equals caspi and will continue when Q is greater than caspi.

So for the for the given question the complete ionic equation will be to potash um positive plus C. 03 to negative plus. See you too positive plus two N. +03 negative leads to formation of ceo. See you see oh three less to petition positive plus two N. 03 negative. Whereas the net my neck equation will be See You too positive plus ceo tree to negative leads to formation of see you C. 03 See that the sum of charges on both the sides of net equation is equal. We have Plus two and -2 Which leads to overall charge on the product side as zero. And the name of the precipitate formed will be corporal two Carboni.


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