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0 0 dissolving Which 2NaOH(s) Na (SOalls) MeCl(s) Wtorosm 2 the an (Ilolh snoanbe following 2Na 2 Mg?"(aq) '(aq) (aq) 1 2 RbNO-(s) (aq) sholutfone 2 2 (a...

Question

0 0 dissolving Which 2NaOH(s) Na (SOalls) MeCl(s) Wtorosm 2 the an (Ilolh snoanbe following 2Na 2 Mg?"(aq) '(aq) (aq) 1 2 RbNO-(s) (aq) sholutfone 2 2 (aq) 2 E (lo'h (aq) equations properly describes the listed solid

0 0 dissolving Which 2NaOH(s) Na (SOalls) MeCl(s) Wtorosm 2 the an (Ilolh snoanbe following 2Na 2 Mg?"(aq) '(aq) (aq) 1 2 RbNO-(s) (aq) sholutfone 2 2 (aq) 2 E (lo'h (aq) equations properly describes the listed solid



Answers

All but two of the following solutions yield a precipitate when the solution is also made $2.00 \mathrm{M}$ in $\mathrm{NH}_{3}$ Those two are (a) $\mathrm{MgCl}_{2}(\mathrm{aq}) ;$ (b) $\mathrm{FeCl}_{3}(\mathrm{aq})$ (c) $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(\mathrm{aq}) ;(\mathrm{d}) \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})$ (e) $\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(\mathrm{aq})$

In this video, we're going to practice setting out soluble ity products. Uh, okay, So for this 1st 1 for the balanced equation, we've got P b. The are too solid, put in liquid water and we get dissociation. So we've got he be and that's a two plus obvious. And then we've got to B r. That's a minus obvious so that our Celje bility expression is the concentration of P B. No exponents, because there's no coefficient on that one. You should probably keep the two plus in there. Yeah, let's try that again. Maybe give myself some space two plus and then b R minus, and that one's going to have an exponents on it because off the co vision, the PB B r to the solid does not figure into the soluble ity product. You can put an S P on this, indicate that it's a soluble ity product because it's a solid and pure substances don't figure in to equal every, uh, Constance, do you? The 2nd 1 Let's take a look at this one. We've got a G two s solid dissolving it in water and we get right to to a G with a plus on that obvious plus so far it's changed the color on that sulfur like it's where I said, Change color. There we go. So far Tu minus again. Aw quis and are soluble illit e product here going to be the concentration of a G waas raised to the second power because of the coefficient. And then, yes, tu minus and we're going to go on that one. Let's take a look at the 3rd 1 This one's a little more complicated, so we know that we've got PD e oh, free Saul in dissolved in liquid water and we're gonna get he be two plus and parentheses around this one. See, oh, three tu minus. So that are soluble ity Product is the concentration of mm, the two plus and see 03 Tu minus. No cough. No exponents on either one of those because there's no coefficients. And finally, this last one's fairly complicated. Let's take a look he's got Yes, our should be strong tea on unless I miss my guess. Three. He Oh, for And there's two of those that's a solid in liquid water. And that will give us three strong Thiam Adams and they have a to that two plus charge. Get My awkwardness is in there and p o for those two of those. Did you meet 04 and that I Ahm has a three minus and is again aqua yous. So that gonna move this down here? Hey, sp, for this one is the concentration of S r two plus raised to the third power because of the coefficient times the concentration of the p 04 phosphate I on and that's going to be raised to the second power. And let's make sure we got the three minus hiding in there somewhere. So that's how you set up the balance equations for and calculate the soluble ity products for dissolving ionic solids, these ionic solids in water.

For the first reaction. The potential products are sodium nitrate, which is soluble, and magnesium sulfate, which is soluble unless we get to real high concentrations of magnesium sulfate. And then it is it isn't super soluble. It's kind of moderately soluble, so we'll assume we're at low concentrations. So no reaction occurs and no precipitate forms for the next one with potential products are potassium chloride and iron three phosphate and iron three. Phosphate is insoluble, so this would be the balanced chemical reaction where we produce solid iron three phosphate.

Mhm. So this question asked about expression of the uh queen constant for dissolving process. And in order to answer that, that first review, right. What how to gather? Queen consent from the reaction. So suppose we have a reaction in this form. We have em or a plus and more B. And it produced & two more. See and kim or D. And the queen content of this reaction actually equals two. The concentration of C. Power is a coefficient Q. Right? And the concentration of D. Power, its coefficient P over the same. It's a reactant or the concentration of a power to the coefficient. And in the concentration that be how is the coefficient? And and now we can try to answer the question. So the first question is the dissolving of the love grow might Sony first we need to write down the camp quick question for this dissolving process which is first we have larry permite a solid right? And it dissolves in water to form lad Ireland the increase solution in the interest form and then it's pra might. Uh It has to Right. Um We can just follow the process we mentioned before together K. S. P. Which here is K. S. P. Is the concentration of that at times the concentration that brought life. All right. Power to right? Because the co vision is true over the concentration of a lab bro. Might. But but we know right here in this reaction, the labrum it is a solid. Any case a solid write the constitution. We always regard a solid concentration as warm. So this is the expression of the quick constant for dissolving the bra might. And the second problem is dissolving super. Some fight right? And the same person house. We need to first write down right the chemical equation which here is we first have a super soft fight in a solid form and dissolving the water to form super eilish. Um Remember here we have to write silver in one compliment. So we need to add coefficient as to and a request form and plus sulfide iron you know quick solution And the queen constant equals two right concentration of the sewer iron. Howard too because the coefficient times the sulphide concentration and over same here over warm because the silver sulfide is solid. I always regard the concentration of solid as one. And now let's go to the first question. I think I have to wipe down some part of that hope you don't mind. Yeah. So the third question asked about lad carbon. It's so now we can write down the covenants. So dissolving of lower cabinets. It's a solid form and forms same lad iron. The request form plus the carbon. It's I am you know crystal form. And the queen constant of this reaction is very simple Because there's no coefficient. The coefficient is always one. So it's just the concentration of the light Ireland times the concentration of carbon. Excitement over why and the last one is strong and fast for it. A little bit complicated way saying so strong action phosphates the solid form, we dissolve in the water to form sweet, strong in irons and to phosphate aliens. Mhm. And the queen constant equals to the concentration of the strong some iron power soil. Because the coefficient is three times the concentration of the phosphate iron power two Over saying because inside one and that's all the answer for this question.

Here we are continuing on with our work relating to solvent chemistry and solutions chemistry. And firstly, what we need to do is determine the value of the van half factor. So what we have is delta T freezing point is equal to the freezing point of the solvent seven degrees C. And then what we do is subtract the freezing point of the solution -9.33° C. To get your .33°C. And so with the morality we have, morality is equal to the number of moles of solute, which is the mass divided by the molar mass. 941 g, divided by one of 4.55 g per mole, divide that by the massive solvent. In kilograms. The morality is not .9904 moles per kilogram. And so then we can go back to our equation, that is delta T. F. We just calculated up here which is equal to KF. The constant multiplied by M. The morality calculated here multiplied by I. The band half factor rearranged for the band half factor felt, delta T divided by K F. Multiplied by em. What we get is a value of approximately two. And so because I is equal to two, N a h S 03 gives two irons. So we have an A. Plus at h S +03 miners. That is in equilibrium with an AHS 03. Yeah. And so the correct choice here is a.


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