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Ljesceonea Diloton"4 charnlst muet duute 6417 RL Mauoous (0np0t (II) Quonde (CuF,) solution unlii tho concrntauon (allt 427, MM . #c"W thle by Lolnq disti...

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Ljesceonea Diloton"4 charnlst muet duute 6417 RL Mauoous (0np0t (II) Quonde (CuF,) solution unlii tho concrntauon (allt 427, MM . #c"W thle by Lolnq distilled Watcr to Itt solltion unti (rol7es 0 cartaln nal voluing:Caic Iete 'nls fnal volume. Muiecir round Yolr answcr I0 venficant digtmMml:D?

Ljesceonea Diloton" 4 charnlst muet duute 6417 RL Mauoous (0np0t (II) Quonde (CuF,) solution unlii tho concrntauon (allt 427, MM . #c"W thle by Lolnq distilled Watcr to Itt solltion unti (rol7es 0 cartaln nal voluing: Caic Iete 'nls fnal volume. Muiecir round Yolr answcr I0 venficant digtm Mml: D?



Answers

A solurion, when dilurod with $\mathrm{H}_{2} \mathrm{O}$ and boilct, givcs a whire prccipirate. On addition of cxccss N11 Cl/NIIOll, the volume of prccipirate chcercascs lcaying bchind a white gclatinous precipirare. Idcntify the prccipitare which dissolvcs in $\mathrm{N} 1 \mathrm{l}_{1} \mathrm{OH} I / \mathrm{N} \mathrm{I} \mathrm{I}_{4} \mathrm{Cl}$ (a) $7 \mathrm{n}\left[\mathrm{O} 1 \mathrm{l}_{3}\right.$ (b) Al(OII) (c) $\mathrm{M}_{\mathrm{B}}(\mathrm{OH})_{2}$ (d) $\left(\mathrm{ca}(\mathrm{OH})_{2}\right.$

Today we'll start with the 3.8 milliliters of nitric acid unconverted two leaders. Then we will use the polarity of the nitric acid 1.25 In order to convert the leaders into malls, we recognize that reaction is one mole of nitric acid toe, one mole sodium hydroxide. Then we'll go from bull sodium hydroxide to leader sodium hydroxide using the polarity 0.215 man that will convert the leaders into middle leaders. 22.1. Male leaders. The next one will start with the 8.5 milliliters converted to leaders of phosphoric acid, then leaders phosphoric acid to mold phosphoric acid using polarity of phosphoric acid. We then see the relationship is one mole phosphoric acid to three moles. Sodium hydroxide. Because of the three hydrogen bombs in the acid phosphoric acid, they won't go to moles. Sodium hydroxide using the polarity. I am a little leaders, sodium hydroxide from old sodium hydroxide using polarity and then leaders to middle leaders. 97.8 million leaders of sodium hydroxide

So the question is a solution when diluted with at two and boy will give a white precipitate now on addition of access and then for seal or and it's for sale or and it's four words the volume or participate increases. Living behind a white gelatinous precipitate. Now we need to identify the precipitate it dissolve and it's four words or NH four cl. So we have a L. O. H three. You have a long ways three. This is a gelatinous. This is a gelatinous white precipitate quite crisp. It owned by, owned by the reaction. Now, why did he gets enough L three plus and and it's forwards in the presence of NH four cm. To the correct answer is obscene. E.

Each of these is a relatively simple conversion. We'll start with the mill leaders of phosphoric acid. First they covered the middle leaders. Two leaders then use the polarity of phosphoric acid to get most phosphoric acid. Recognize that one mole phosphoric acid reacts with three moles of looking hydroxide uncovered the molds looking my drug side to leaders. Lithium hydroxide using polarity, the leaders to mill leaders, 69.3 mil leaders looking hydroxide. The next one will start with 15 milliliters of sulfuric acid converted to leaders, and you similarity sulfuric acid to get mold sulfuric acid. Recognize that one. Most of uric acid, reacts with two moles within my drug side and then use the polarity of lithium hydroxide to go from moles within my drop side to leaders and then leaders to mill leaders. 50.4 milliliters. Lithium hydroxide

In this problem, we're combining 200 milliliters of 0.24 Mueller potassium sulphate with an unknown volume of 0.16 Mohler potassium nitrate. The resulting solution has a molar concentration of potassium ions equal two point for Mueller. We were asked to find the unknown volume to determine the concentration of our potassium ions that are enter original to solutions that are being mixed. We can look at the amount of potassium that's present in the original compounds. So we look at this first solution here and we see the potassium is Onley present once in this compound. So as a result, we can say that the polarity of the potassium ions will be equal to that of the original compound. Now, when we look at our potassium sulfate here we have two potassium involved in our molecule. So it means that we're going to take the small arat e and multiply it by two because we have two moles of potassium for every one mole of our entire compound. So the polarity of our potassium ions in this solution will be equal 2.48 Mueller. Now that we've determined our respective polarities of our potassium ions. We can go ahead and plug into our dilution equation shown here. So we know in our potassium nitrate solution, the polarity of our potassium ions is equal to that of the regular solution. And we haven't unknown volume that we're solving for. And that's being mixed with our other solution, which we know has a polarity equal 2.48 and a volume of 200 milliliters. And we're gonna set that equal to our resulting final solution. So that has similarity equal 2.4 Moeller. And the volume will be equal to whatever unknown volume is plus the 200 that came from our solution of our potassium sulfate. When we solve for X, we find that are unknown. Volume is equal to 66.7 milliliters.


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