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6ive Tan (Xs Balace Relox Rxn uNei 9 cijic ConiiolNaoh +02Nadt + #x 0 Nabh NAD + #'4o 01 4 2#4+ +1 Hz 0...

Question

6ive Tan (Xs Balace Relox Rxn uNei 9 cijic ConiiolNaoh +02Nadt + #x 0 Nabh NAD + #'4o 01 4 2#4+ +1 Hz 0

6ive Tan (Xs Balace Relox Rxn uNei 9 cijic Coniiol Naoh +02 Nadt + #x 0 Nabh NAD + #'4o 01 4 2#4+ +1 Hz 0



Answers

A deliqucscent whitc crysrallinc hydroxide $\mathrm{X}$ rcacts with a nirrarc $\mathrm{Y}$ to form another hydroxide which dccomposcs to give a insoluble brown laycr of irs oxidc. $\mathrm{X}$ is a powerful cautery and brcaks down the protcins of skin flesh ro a pasry mass. $\mathrm{X}$ and $\mathrm{Y}$ arc (a) $\mathrm{NaOll}, 7 \mathrm{n}\left(\mathrm{NO}_{3}\right)_{2}$ (b) $\mathrm{NaOH}, \mathrm{AgNO}_{3}$ (c) $\mathrm{Ca}(\mathrm{OH})_{-} \cdot \mathrm{HgNO}_{3}$ (d) $\mathrm{NaOl} 1, \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}$

In this problem, the reaction, the reaction which aid that the action rich ed given in the problem. The reaction which is given in the problem, is an example of E D. N. Example of, I don't know, is an example of I'll dole condensation under Abdul London Station. So according to the option of some age, correct. So according to the option option age, correct answer for this problem, I hope you understand the solution.

Fruits of the following. Let's complete and balance each equation. If no reaction occurs, will right. No reaction for a There is a reaction taking place as we form F e oh h three precipitate and N a P r. Balancing this equation will need a three in front of the any Ohh and three in front of the n a b r for be reaction occurs product of a GCL precipitate and very, um, nitrates. To balance this equation we needed to in front of the A, G and 03 and a two in front of the GCL. Okay, for C, we do have a reaction forming a precipitate of cobalt carbonates and sodium chloride. To balance this equation, we'll need a two in front of the sodium chloride and for D. Both products are soluble, therefore, there is no reaction.

It's just a really interesting question that asked about how the equilibrium of NH three and water with O. H. Minus and NH four plus so basically an aged reacting as a base, how it's affected by the addition of 0.1 Moles of O. H minus. I think in the form of anyway. So it's that's basically just saying, Oh H. Myers, because it associates 100% in a one leader solution. And so right off the bat, I'm just going to say, because it's in a one leader solution and this is added in solid form, it's not gonna change the volume. So, um, we don't need to worry. Think about it as bowls or something. We can just think of it as 0.1 Moeller, and that's going to be easier in ice tables in mass accents, pressures were always working concentrations. So what we're really saying is for the ZIKA leave room equilibrium, which we're gonna do a nice table for ignore water because it's not part of the mass action expression. The initial concentration off O H minus is going to be 00.1 Moeller, um, for NH four plus is gonna be zero. And for NH three, it's going to be 15 Mueller because that's what we're given Aziz. The initial demonstration of an I street. And so it's just like a normal one, except instead of having this, uh, zero over here 0.1 So now, because we couldn't possibly go to the left because there's no NH four. Plus, we know the reaction's gonna go to the right. So minus X over here is the change plus X plus ax in them 15 minus x 0.1 plus X an ax. And so if we plug us into a mass action expression for so okay is equal to on the top left 0.1 plus X times acts so and on the bottom will have 15 minus X. Now, commonly, we will ignore this value of X from the region of some like this to make the problem simple. However, um, in this case, you actually turn out that, uh, that you can't do that. And the way you can recognize this just like visually, is that we have points here. One is it that big? And since the concentration of any street here is pretty large 15 molars a lot. We might expect that the amount of waste is gonna be added from, like, the amount of 15 over here might be a significant portion of 0.1 on. In fact, find out it's actually bigger than 0.1 So we can't ignore this. And so So we're gonna have a quadratic equation anyways. Might as well not ignore this. We're gonna have to use some calculator, use the quadratic equation. And so if you do, plug in the K for industry, which is 1.8 times since Saturday. Fifth spoke up in a table and plug it into a calculator. You'll get that X is equal 2.1 to Mueller, and so it's also go to negative 0.0 to 2. But there's actually a good point to remember. I bring this up because if you go back to the ice table that has no physical meeting, you can have a negative concentration. But if we're surprised to find the Constitution, which minus, we plug in that 0.12 that should you add those two numbers together, our final concentration is going to be not a negative 20.2 to stay positive, 0.8 point 0 to 2 more so

Okay, So for this question, we want to see if we can balance a reaction. And then, um, figure out which is the limiting reactant. And by how much? Um, this limiting we would need to add to the limiting reactant to get it to be a fully 1 to 1 equivalent. Uh, reaction. Okay, so we'll start off. We have the reaction. H three p 04 That's phosphoric acid. Plus. Hey, sodium hydroxide. Okay, I'll just write this big and take up as little space as possible. And that goes to N a three p 04 sodium phosphate and h 20 water. Right. So they're all liquid or acquiesce. So we're just gonna leave the phase, separate the phase descriptions out of this That's not going to come into effect. Um, but we know that this isn't balanced, right? We can look at that and see that right here We have 100 sodium, and here we have three. Right? So obviously this is not balanced, and it would actually be pretty tricky to count these oxygen's. So what we're gonna do is count these as phosphate groups. So to balance this, we have hydrogen phosphate if hydrogen We have phosphate, we have sodium, and we have oxygen's outside of the phosphate. Right. So first, we have one on the products and three on their reactivates. Sorry. Reactant products. Um, we have one on this side and one on this side for hydrogen. We have three. I have four. Okay, so for here and on the side, we only have two, right? And for oxygen, we have one and one. Okay, So what we're gonna do is start with the sodium, and to get to here, we're gonna multiply the reactions by three for the sodium. So that's going to go right here. So now we have three sodium, three oxygen and six hydrogen. Yeah, sizes in the reactant. So yeah, three sodium, three oxygen, three hydrogen, six hydrogen. Sorry, because we're just adding to six three, and this is three, and we're going to have to mhm, um, balance this oxygen now and the hydrogen, and we could do that. We see if this is a 2 to 1 ratio. So we're just going to be adding water. So we go ahead and make this six, and we make this three, and we get that by putting this here, right? That was kind of a round about way of doing this. But basically, all we're doing here is switching the hydrogen zones and the sodium on the phosphate group and hydroxide group so we can see that it's just gonna need to change the amount of either this or this, but, well, all I need to change that. So, yeah, we have our final reaction. Okay, this up a little bit. Okay. So we can see that this is a one 23 ratio of phosphoric acid, too, uh, sodium hydroxide. So to find the limiting reactant, we should take what we have as our starting amounts and see how many moles of each we have. So for phosphoric acid, we're adding 15 mL because milliliters are the same as cubic centimeters. Yeah, we have. It is a 0.2 molar concentration. Mm. So let me clear this up a little bit. Okay? I'll move it down here. So we have h three p 04 15 mL and volume and polarity. And we have 0.2 moller, right? And for sodium hydroxide, we have 50 mL, and that is equal. And we have that at two molar, right? So to smaller. So 15 mL is equal to 0.15 leaders and 50 is 0.5 Ah, sorry. Zeros, 0.15 liters and 0.5 liters. So polarity are to find the number of moles. We just want to multiply the more clarity by the volume. So that's going to be point 015 times 0.2. That will be 0.3 Add up here we have 0.5 times two and there will be zero point one mall. Mhm. Yeah, So you can see that this is going to be the limiting reactant. But for completion, because we need three times as much hydroxide to find out whether these are equal, we'll have to divide this one by three. So this one divided by three, is going to be zero point as 0.0 three repeating so 0.333 however, many significant figures you would just insert there is the number of threes, and you can see that this is 10 times as much as this. So we have This is our limiting reactive, right? Mhm. Yeah. If we wanted to get them to be equal so that we had the same, um, relative amounts of hydrogen, phosphoric acid and sodium hydroxide we would need to make this would need to be 0.0 one. Mhm. So we would need to add Yeah. So this is an excess by a considerable amount. Um, to this, we would need to add zero point 0 to 7. Mole. Yeah, right. Or if we want to divide that by three to find the excess of the other one. We would say that this is an excess 0.0 09 Small. Okay. Okay. So I hope this makes sense. Um, that's really all there is to solving this problem. You just need to keep in mind pretty constantly what you have and what you have, how much of each thing you have. And, um, the process is going on so that you can figure out the relative proportions of each. Okay, that's it.


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