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Moving to another question wil save this response:In 1ntify the major ionic species present in an aqueous solution of NazCO3 ) A Nat, CO3) B Nat, ct,02Oc Na+ , c4+,...

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Moving to another question wil save this response:In 1ntify the major ionic species present in an aqueous solution of NazCO3 ) A Nat, CO3) B Nat, ct,02Oc Na+ , c4+,0320 D Na2t, CO32DU o another ques save this response: question will Moving to _

Moving to another question wil save this response: In 1 ntify the major ionic species present in an aqueous solution of NazCO3 ) A Nat, CO3 ) B Nat, ct,02 Oc Na+ , c4+,032 0 D Na2t, CO32 DU o another ques save this response: question will Moving to _



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A solution of a mcral ion when trcarctl with KI gives a retl precipitatc which tlissolves in cxccss $\mathrm{KI}$ ro givc a colourless solution. Morcovcr, the solurion of metal ion on trearment wirh a solurion of cobalr(II) thiocyanate gives risc ro a tlecp bluc crystalline precipitarc. The meral ion is (a) $\mathrm{Pb}^{2}$ (b) $11 \mathrm{~g}^{2}$ (c) $\mathrm{Cu}^{2-}$ (d) $\mathrm{Co}^{2+}$

Wow, Let's devise a way to separate the irons by precipitating one of them as an insoluble salt and leaving the other in solution for a We have very, um, two plus and sodium plus. In order to separate thes, we could add uhh some potassium sulfate to this, which would precipitate the berry. Um, and it would leave the, um, sodium iron soluble and H two s 04 is soluble on. So in the solution, we would have an A two s 04 a quick so that would adding sulfate ions, potassium sulfate or anything sulfate ah soluble sulfate would precipitate. This would have to be soluble here for B. We have nickel two plus and lead two plus. And we could precipitate separate these two by adding, uh, sodium chloride because we would form a precipitate of lead chloride. And in solution, the nickel chloride would be soluble. Oh,

Here we have see a three p of four to precipitate on. We're looking for the percentage CEO by Mass, where we have three times 40 point nor 7 8/3 10.18 multiplied by 100 together into percentage while we generate 38.763%. Our equation here is three C A two plus our two h p 04 to minus generate C a P o full add to H. So we do have CEO to bubbles present. We have C 02 at H 12 generates H two c 03 Well, we have see a two plus at H two co three, where we have see a CO three at H two where we have a precipitate formed on. Then we have C A C 03 at h two c 03 generates see a two plus on to H 03 minus, which in fact we dissolves

A similar problem. 89. We simply need to identify what the precipitate is. That precipitate, if there is one, will then allow us to identify what ions are required to form to form that precipitate it. And then we can write the Napoleonic equation. Barium carbonate forms a solid. Therefore it is the berry and two plus coming from the barium chloride in the carbonate coming from the sodium carbonate that produces the barium carbonate. This being the Net Ionic equation for the next one, all reactant and products are soluble. They're all spectator ions. No reaction occurs for the last one. Calcium phosphate formed as a precipitate. Therefore, the Net Ionic equation is three. Calcium two plus is reacting with two phosphates form. Are calcium phosphate solid? Because precipitates forum in the first and the third reaction, these examples of I in exchange

The compound is a substance that contains atoms of two or more different elements and these atoms are chemically joined together. So here we're looking at a series of electrolytes and we're looking at the irons that they create. So K. O. H breaks down into Cape Lots, ohh Minor Science. K two, S 04 breaks down into K plus and S. 04 to minus irons. Next we have n a n 03, which breaks down into any plus an n +03 minus irons. And finally we've got NH four cl which breaks down into NH four plus and cl minus irons.


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