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M lecule (A) Can IactForm eimnev Preduct (B)Cc) Wimn pot reachensobebis frst Orcler kinetics wt ratc Constants k8 Una Kc bespectively Ke 9'6xlo and kc 1.6 Xlo ...

Question

M lecule (A) Can IactForm eimnev Preduct (B)Cc) Wimn pot reachensobebis frst Orcler kinetics wt ratc Constants k8 Una Kc bespectively Ke 9'6xlo and kc 1.6 Xlo } 5 -| what is mnc Yietd Yar

m lecule (A) Can Iact Form eimnev Preduct (B) Cc) Wimn pot reachens obebis frst Orcler kinetics wt ratc Constants k8 Una Kc bespectively Ke 9'6xlo and kc 1.6 Xlo } 5 -| what is mnc Yietd Yar



Answers

For each of the following, write the equilibrium expression for $K_{\mathrm{sp}}$ :
a) ${Ca}({OH})_{2}$















b) ${Ag}_{3} {PO}_{4}$
c) ${BaCO}_{3}$


















d) ${Ca}_{5}({PO}_{4})_{3} {OH}$

Okay so we have some silver bromide. So let's go ahead and start by writing the equation for that dissolving in water. So silver bromide will be in equilibrium with its ions. Okay so then we'll write the K. S. P. Expression. So that's going to be products. Egg plus times br minus over reacting to what our reaction is a solid. So we're going to ignore that. Okay the second one we have calcium hydroxide. So again we'll go ahead and write that equation for it being an equilibrium with its ions 20. H.- Here. So Kay sp will be the concentration of our calcium ion times the concentration of O. H minus and will be squared. So I have co two as three cobalt three sulfide again. So that will be too Cobalt three plus ions Plus three sulfide ions. And from that equation we can write our K. Sp expression. So that's a concentration of cobalt square it times a concentration of sulfide, cute. And then finally We have f. E three P 042, so iron to phosphate, So we'll have three FE two plus ions and we'll have to phosphate ions. So from there we can go ahead and write the K. S. P. So the concentration of FE two plus, cute times the concentration of phosphate squared.

So this question is asking us to write equilibrium, constant expressions for each of the following hetero genius equilibrium. And to do that, we're going to use the equilibrium. Constant equation K e q equals the concentrations of the products raised to their respective coefficients over the concentrations of the reactant CE raised to their coefficients. Now, one important thing to realize is that these air hetero genius equilibrium, which means different components within the reaction, are within different states of matter. So one important thing that we have to recognize is that when we're writing this cake you equation, we do not include the concentrations of any solids or any liquids because they're concentrations don't change. Over the course of the reaction, they can basically be treated as Constance. So for part A, we have C 10 h eight, which is a solid reacting into see 10 age eight, which is a gas, so we're not going to include the solid here. We're only going to write the concentration of the gaseous product in that equation, so that's going to look like K E. Q equals the concentration of C 10 each. Eight. Is it gas? And it won't be raised to any power because the coefficient out in front of it is a one and anything race to the first power is itself. And so we do the same process for each of the other parts. For part B, you write the cake you expression as the concentration of the products, which in this case is just a TSH to a gas raised to the power the first power over. Well, nothing, because the reactant is a liquid for part C. We only have one gas here. We're not going to take into consideration either of those solids. So we'll write the K E Q as concentration of the product not raised to anything not over anything. For Part D, we have only one solid here. It's a reactant. So we won't include that Que es que is equal to the concentration of C O. Not raised any coefficient multiplied by the concentration of the other product. And again, there's no coefficient there over the concentration of the only gaseous reacted. Which is it, too? Oh, Gus. And there are no coefficients there either. Lastly, we have Part E and R K expression. There is not going to include the solid iron well, the solid iron oxide. It's just going to be written as the concentration of C E o two, divided by the concentration of the reactant See Oh, which is it, CASS? And there you have it. Que es que equations for each of these hetero genius equilibrium on Lee, including the gaseous reactant or products, although it is also worth noting sometimes in equations you'll see things that are the state of matter a que or qu'est ce, and we would also include those. There were none in this problem, but just for future reference, you also include the concentrations of a quickest solutions.

So equilibrium is Thekla dish in of a sister when neither its state of motion nor its internal energy states hands to change within a given time period. So in this podcast, we are writing our equilibrium, constant expressions just as a note here. We need to leave out the solid on liquid states, and we have used the equation that is over here on the right hand side of the screen in order to formulate our expressions where we have our products multiplied by their story key metric coefficients over the starting material multiplied by their strike you metric coefficients. So for our first example, we just have the concentration of seats on each day. Second example, we just have the concentration of water. That's a stream to be in the gas phase. Otherwise it wouldn't be included in this expression. The third example we have cake LeBron equals a concentration of co two second to last year have the concentration of carbon monoxide multiplied by the concentration of diatonic hydrogen over the concentration off Gacy's water. Lastly, we have the concentration of co two over the concentration off C o

I want My name is now I will answer this question by using the appropriate value off KSB. We will use KCB value from table. You have table and table for key air. Uses his value for finds equilibrium for these equations. But right thing is, he's a question here. If we want equilibrium off, eh? We will divide it. The concentration off Bruder see, And six naked before times about selfies Negative too divided by the concentration off season I our c But here p f if he s no booted in the clever so now using the appropriate value for Casey in Kiev to find this. So from Kiev, we have relations. Sees that if he lost too when re act with six and I you see forms of complex if e see and six negative four So every rides this k f it will be the concentration off complex for May, divided by the concentration off iron and plus two times by the concentration off scene I barber six the K f in the table is 1.5 times party in 35. This we can use from Pierre may be related to this case also we can use from Kiesa bees is if we have f p s using Kinkade's be it form F B plus toe last seen night Negative too. And the variety ksb off these value. It will be the concentration off our plus two times of isa concentration off, sir. Fire And the key is be value is 3.7 times fighting negative nine. So how can we lied to relation toe finds? Is he found? If we times is his relation by these relation we get this how a few times his pies is some fine will be a path Foolery is the iron were repeated in bull sign up and down so he can remove two Bizerte Very meaning is a complex and 6 to 9 and some fine So we can see that to get his relation with time Kiev boy K s B. So it will be one boy five times by teen certify times for sleep on seven to times fighting negative 19 and the whole result is fine 0.6 times by then Barber 16. Thank you


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