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Picture illustrating how the potassium Kur easlv dssolves In water. Show av Iwronide ions € stv dlssolved Show the relevant attractions.Wlwt do we call these ...

Question

Picture illustrating how the potassium Kur easlv dssolves In water. Show av Iwronide ions € stv dlssolved Show the relevant attractions.Wlwt do we call these attractions? Why are these attraction so strong?at 87 K and HCI boils at 188 K Explain Let" $ look at boiling points. Ar boils terms of attractions.

picture illustrating how the potassium Kur easlv dssolves In water. Show av Iwronide ions € stv dlssolved Show the relevant attractions. Wlwt do we call these attractions? Why are these attraction so strong? at 87 K and HCI boils at 188 K Explain Let" $ look at boiling points. Ar boils terms of attractions.



Answers

The lattice enthalpy of potassium chloride is 717 $\mathrm{kJ} / \mathrm{mol}$; the heat of solution in making up $1 \mathrm{M} \mathrm{KCl}(a q)$ is $+18.0 \mathrm{~kJ} / \mathrm{mol}$. Using the value for the heat of hydration of $\mathrm{Cl}^{-}$ given in Problem $12.133,$ obtain the heat of hydration of $\mathrm{K}^{+}$. Compare this with the value you obtained for $\mathrm{Na}^{+}$ in Problem 12.133. Explain the relative values of $\mathrm{Na}^{+}$ and $\mathrm{K}^{+}$.

So we have strong electrolytes that are going to completely break up into their component. I resolving a wire. So let's start with sodium. Oh my. So we have sodium attached to grow might in the market we put it into solution it's going to basically dissolve and give us sodium grow my eye on. So if we were to draw these ionic solutions we want to show that We have a 1-1 ratio. So each molecule will give us one. So do you have one night? So the next one we're going to do magnesium fluoride. So for magnesium chloride you have magnesium attached to chlorides in this molecule. So as you would expect we have one magnesium and then to chloride ions in solution. So we have to also show that for each molecule we're going to get Wine, Magnesium and one and 2 Chlorides. So then for the next one we are doing aluminum nitrate. So aluminum is attached to three nitrate and i on. So this is a policy atomic ion. So when this goes in solution will have one aluminum and then we'll have three of the nitrate ions. Or we can just draw either showing the three different or throughout showing the three oxygen's on the nitrogen. Or we can draw that as one unit if we want. And when we show that we get one three ofi nitrates. So in reality of course we will get more. So we can draw another one just the same exact thing. And then we would I have to show that for each one we get three Of the nitrate and one. So that's what will happen. So now let's do the next one. So the next one as you can see, we have to pollen talk guys, we have to ammonia Pneumonia means and one sulfate. So let's say this is ammonium. So I'm not going to actually draw out the entire thing. But we'll just say that this one represents pneumonia. And then we have sulfate. So then we're just going to have another ammonia attached in the same molecule when this is in solution, we just have this and that and then we have this solution. So again, if you were to draw more of those molecules we would have more in solution. But we need to make sure that the ratios are right. So if we have to of these on the side where we have molecules need two of them in the solution. And we have one here. We need one here. So one molecule is always going to give us to ammonia and one soft food. Okay, so the next one would just basically be exactly like a except now we have hydroxide. So you can show this is sodium and then for the hydroxide you can draw it out or you can just station of it the label as hydroxide. So this will break up and we're gonna have sodium here and then we will have the hydroxide. So it was 1-1. So one molecule gives you one of each now for the iron sulfite, I'll let you do that on your own because that is similar to C. And H. Is also similar because you can see that we'll get one potassium at one of the poly atomic ions. The permanganate, mn oh four. Umno four is one unit. And then for age we have hydrogen and the other part of it will be the Pourquoi I on the c. l. 04. And then for the last one, let's do that. Together. We have ammonia here and then we have we have acetate. So we're going to label that. So we don't again, we don't really have to show exactly just to show how it's breaking up to age. Well, so then we will get an ammonium. Then we're going to get One of the escape ions. So again, it's just 1-1 for the mole for the vulgar ratio. So we have one Ammonia and one asset and then we get one ammonia. Yes.

In reviewing the description of the chemical reaction, all three states of matter are present. We have solid potassium, we have liquid water and we have gaseous hydrogen that is produced be observed. Change is a chemical change because we are describing a chemical reaction that is occurring, and any chemical reaction induces a chemical change. The reactant of this chemical reaction are potassium and water, and the products of this chemical reaction are potassium hydroxide and hydrogen gas. Qualitative observations that can be made by just looking at the picture would be that smoke and light appeared to be evolving from the chemical reaction, and there appears to also be a purple flame that is produced.

To describe how K. C. L. Dissolves in water and why ion disciple forces helped to bring the potassium chloride solid into solution. We need to understand what is happening between the water and the solid potassium chloride. There's a first a strong attraction between the potassium cat I and and the chloride. An ion. In order for this to dissolve as free potassium and chloride ions. The water molecules need to overcome this strong force of attraction. They can do that by surrounding the chloride with their partial positive ends. The partial positive ends of the water molecules are the hydrogen ins. These then come in and create these ion dipole forces the disciple of the water and the ion in this case chloride. If we get enough water molecules to come in and create the strong the create enough of an ion disciple attraction to overcome the ion ion attraction, then the chloride will come into solution, leaving behind the potassium which will then also be surrounded with water molecules, creating additional ion disciple forces of attraction. Where the negative ends of the water molecules are attracted to the positive potassium ion.

Okay, so we're gonna calculate the Ionic cohesive energy and were given or not an M. So let us recall the the Ionic cohesive energy is given by this equation here. Notice that it will give you units of jewels, and we would like our energies in units of electron volts. So one jewell is approximately 6.24 times 10 to the 18th electron volts. So that makes part a just in algebra problem at this point. So you should get 7.13 electron volts. Since we know the value of Alfa, we know the value of K e squared were given are not and were given em 7.13 electron volts for part A and part B. It gives us the ionization energy of potassium as 4.34 electron volts. And this is the ionization energy and the electron affinity of chlorine as 3.61 electron volts. Okay, so that makes the atomic cohesive. Energy is what we found. 47.13 Electron volts and potassium will gain energy to convert. So you're positive potassium will gain energy to convert into neutral potassium and your negative Klein will lose energy to convert into a neutral chlorine thio 7.13 minus 4.34 electron volts plus 3.61 electron volts. And this gives you an atomic, cohesive energy of 6.58 electron volts. That's the problem.


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