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Concentraled solulion of & MHLSO hxs moles of HSO dissolved of colutiond 524. The unit of Jg'C is the unit of: reSSutE Ora funshance Specilic heat capaciy ...

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Concentraled solulion of & MHLSO hxs moles of HSO dissolved of colutiond 524. The unit of Jg'C is the unit of: reSSutE Ora funshance Specilic heat capaciy ol a substance Hext enetgy of a substance Guas cunstanl of 4 substance_25. Titration mcthod for tha detertninalion of the unknoun ofa solulion Qualitative mcthod_ volume. Quantitative method Folum QMAnMAMTC conceninnon Qualitalive. conccntralionvolatile liquid liquid that Easily changes solid MalcaOOM Icnpcnlufc Needs be healed -tron

concentraled solulion of & MHLSO hxs moles of HSO dissolved of colution d 5 24. The unit of Jg'C is the unit of: reSSutE Ora funshance Specilic heat capaciy ol a substance Hext enetgy of a substance Guas cunstanl of 4 substance_ 25. Titration mcthod for tha detertninalion of the unknoun ofa solulion Qualitative mcthod_ volume. Quantitative method Folum QMAnMAMTC conceninnon Qualitalive. conccntralion volatile liquid liquid that Easily changes solid MalcaOOM Icnpcnlufc Needs be healed -trongly and long MIe chanee Hardly vaporizes Toomicuncamne Easily vaporizes nahnnt IcmprrJlure_ 27. If pressure applicd - ccnain quantity of gas at 2 constant (emnperature . ils !olume: Incre sc: Decteaies Stays the samc None of the above ~uhshnc= ccd badded to the UItrand the tlask t0 Indicate the endpoint ofan AcId- Mlon Hlrlor delemninalor nendicalor None of thc above 29. 0,5 mole of a g15 hcated 120 "€at ? pressure Of | MI Wni volume will this g4s occups ? (R 0 0821 Lalm mol.K) 16.13 ml: 16.13 4.926 ml: None of the above



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Consider two solutions, the first being 50.0 $\mathrm{mL}$ of 1.00 $\mathrm{MCuSO}_{4}$ and the second 50.0 $\mathrm{mL}$ of 2.00 $\mathrm{M} \mathrm{KOH}$ . When the two solutions are mixed in a constant-pressure calorimeter, a precipitate forms and the temperature of the mixture rises from 21.5 to $27.7^{\circ} \mathrm{C}$ (a) Before mixing, how many grams of Cu are present in the solution of $\mathrm{CuSO}_{4}$ ? (b) Predict the identity of the precipitate in the reaction. (c) Write complete and net ionic equations for the reaction that occurs when the two solutions are mixed. (d) From the calorimetric data, calculate $\Delta H$ for the reaction that occurs on mixing. Assume that the calorimeter absorbs only a negligible quantity of heat, that the total volume of the solution is 100.0 $\mathrm{mL}$ , and that the specific heat and density of the solution after mixing are the same as those of pure water.

Here is an example of a carnot heat engine cycle using an ideal gas. So a reminder that a carnot engine has to ice A thermal processes. One an expansion want a contraction during the expansion at high temperature heat is absorbed during the contraction heat is expelled to the cold reservoir at low temperature. Then there are two idiomatic processes where heat does not have a chance to exchange with the environment. One is again a contraction one. Again, it's an expansion. Usually when analyzing these cycles, a couple of tools are helpful. One is the ideal gas law provided an ideal gas is used. So I usually like to write that down and then find the pressure and volume and temperature At each of the four vortices shown in the cycle. Mm. Uh The other tool that often is useful is the first law of thermodynamics dealt to you is equal to Q. Heat minus work done by the engine. That could be helpful for processes or cycles that involve a idiomatic processes. It is also good to realize that pressure times volume raised to the idiomatic exponent is equal to a constant. It reminder that the automatic exponent is a ratio of cp two C. B. And depends on the nature of the gas. Here we are, given that this constant is 1.3. So this is usually how I like to start. I usually like to start with a table and fill out the three state variables pressure volume and temperature for each of the vertex is in the cycle Here. There are four that are labeled in the diagram for us. Now this problem is challenging because they don't give us outright numbers but they provide clues as to what's going on with this particular sample to put it mildly. Um And so the first thing they tell us is that there's the ambient Environment that the particle is interacting with at the .4. And they give us the ambient temperature as 123 Calvin. Uh now that's good because we now know the temperature at .1 because of the ice a thermal process. They also tell us that we are at one atmosphere at that point but they don't tell us what one atmosphere is other than yeah, it's ambient there. Mhm. Some other clues at .1 we are told there is something called the trigger volume and that's going to become important Because at .2 we get squeezed to half the trigger volume with the radius given As .08cm cubes. That's kind of no not cubed. Sorry. Yeah, that's kind of a weird way to give a clue. But at .2 they also give us the pressure. Yeah, As 20.3 kilo pascal's. So that's an outright calibrated measurement kilo pascal's at 0.3. They give us the clue that No, it's actually a .4. They give us the clue that the final radius expands from the trigger radius by three And since volume is R. Cubed, that tells us that we have 27 times the trigger volume. So we have some clues and I think these are the the most obvious clues. The rest. We're going to have to use some steps with the gas law and the idiomatic exponent to figure out. So looking at the table, the most information appears to be given. Well we'll start with two because there's a nice calibrated point right there. And so at point to what we know is the volume Is 4/3 pi r cubed. And that's going to unravel a lot. So we have 2.14 Times 10 to the -3 m3 figuring that out. I won't show all the math there and then what happens is we can now find the trigger volume By just multiplying it by two and that will give us the Entry for the volume at .1 and it will also give us then The volume at .4 Which is 27 times that. Yeah, 0.116 m3. Yeah. Yes. And we can fill in that entry into the table. Okay, so now we've got to do a little bit more work. Um we see that there's quite a bit of information given in steps one and 2 and they are connected through an idiomatic process. So what we can do between one and 2 because it is idiomatic, We can set the pressure one Times The Volume one raised to 1.3 is equal to the pressure 0.2 times the volume, it too Raised to the 1.3. And let's see that is going to give us of the unknowns. It will give us the pressure at 0.1. So we can solve for that. And we get uh huh 8.24 Killer Pascal's and bingo. Now we've got everything inside appoint one. Uh, we've got, yeah, definitely a bingo type situation. Everything is filled in so we can now use one that everything at that point to solve for the number of moles are, is just the gas constant 8.31 jules per bowl kelvin. And we could solve for that high temperature, which is a very important quantity. Um, once we find the number of moles. Okay, but actually we're using that point to saw for the number of moles. So let me be clear about that. Oh, okay. So we can solve for the number of moles And that's definitely important quantity 3.45 Times 10 to the -2 moles, which I'll just write off to the side because that is an important clue that we can then use in 0.2 to find the high temperature. It's like a mystery that unravels and doing that again, I won't show all the plugging and chugging, But we get a temperature of 151 0.5 kelvin. And that's nice because we now know both the low temperature and the high temperature and that is definitely important things for the operation of a carnot cycle. But while we're at it we have a few more things to fill out so we might as well go ahead and fill those out. There's a lot of information missing at .3. Um and as well as point for But what we see is .3 is connected 2.4 through a uh a dramatic so let's stop with one more step. What about .3? It's connected to four through idiomatic and it's connected to through ice a thermal. So what we can say is the following two relationships the unknown pressure at three Times the unknown volume at three raised to 1.3. That idiomatic exponent is equal to pressure it for volume it for Raised to the 1.3. Okay. Um well we don't know pressure it for, do we? But we can use the fact that four is connected to one through an ice a therm. And so we can say that pressure for volume for is equal to pressure one volume one. And we definitely know both of those at a .1 and we also know the volume at point for so we can solve for pressure for and again, I'll spare the details but we basically get mhm 301 pascal's Okay. Now what do we know? We don't know either pressure three or volume three. So we have to connect it to point to through the ISIS are okay. So we know both Pressure volume at .4. Same with two. We have two equations into unknowns and we can definitely then solve For both pressure three and volume three. And what are the steps to that? Well, I would probably take the bottom relationship And solve it for pressure three and put that back into the top equation. Yeah, 1.3 is the exponent And solving that we get volume three is equal to Okay. Uh huh. Uh huh. 5.84 Times 10 to the -2 cubic meters. And then we can solve for pressure three. Okay. So yeah, we've got the table full at this point. Yeah. So I'll put a little faced by that to show that that one is the most challenging and it's probably the least interesting too on top of it. So at this point, we have all the information about this cycle um we can do some other things with the cycle, basically knowing what the carnot efficiency is. I will use the expression without the percentage because certainly getting the fraction is good enough. But that efficiency is basically the difference between the high and the low temperature divided by the high temperature. So we know both of these temperatures and we can figure out that efficiency, It's about 18.8%. But .188 in terms of a fraction and a reminder that any engine has another definition of efficiency is the work that you get out divided by the heat absorbed at high temperature during the high temperature is a thermal expansion. Yeah. So yes, while the gas is hitting up, it is expanding and pushing against something um whether it's the wheels, the shaft of your car with a piston that's pushing or whether it's this little particle um doing something expanding. Anyway, what we're told is that the work out is 60 killer jewels per hour and that's really a rate at which work is being done, 60 killer jewels per hour Is the same as one killer jewel per minute. Just thinking about How hours translate two minutes. Um and then we can figure out the heat that is absorbed at least the rate. Just turning those into rates does not change the fraction there. And so qh dot the rate at which he dis absorbed is simply uh yeah 5.32 killer jewels per minute. And then we want to use the first law to figure out the difference between um the heat absorbed minus the heat expelled exhaust heat is equal to the output work. And really what you're doing with that is using the first law With Delta U. equals to zero. And solving for qc dot it's the heat absorbed minus the output work, energy must be conserved. And so this is four 32, kill the jewels permit it

So in the first part of this podcast, Wigan, using the T X Y diagrams, which is temperature of us, is more fraction in liquid and vapor pressure phase for a component that is often an easier way to calculate the bullet points the two points in the relative compositions. So we need to look at the T X Y diagram for benzene in the first part here. So we find that the first drop of vapor condenses at 104 degree C and then we move horizontally. From this point to the liquid carbon, determine the mole. Fraction of benzene in the liquid is not 0.0.1. For on that, the more fraction of tolerant and liquid is no point 86 So in the second part here we have the mole fraction of benzene in the liquid and vapor 100 degrees, so we can use the basis of 1.0 mile of original paper, and then we can write equations for the mall balance of fanzine. So what we have is why be not multiplied by NT? People too wide be multiplied by and V x B was by by an Al. So we have not 0.30 multiplied by 1.0 miles, the equal to not 0.46 and be at no point to for an elf. So I will jump onto a fresh page here. So in Vienna on LR numbers of moles of vapor liquid and why the note is a more fraction of benzene in the original vapor. So the total more violence we have NT is equal to M V at N. L. So NT is essentially called one more yourself envy on an L. So we know that an l table 2.73 more and B is not point to seven more. No 0.27 divided by 9.73 is it's not 0.37 when we divide them by one another. So for this last part here, we need to take a look at the T X Y diagram again for benzene. We moved down vertically from the full vapor, with a 9.3 more fraction of benzene at 115 degrees Celsius. That's our temperature now on the line, a intersects with the liquid curve at the temperature of 98 degrees Celsius temperature, so that is the last drop of vapor that condenses at 98 degrees Celsius. So then we move horizontally from that point to the vapor curve, and we can see that the mole fraction of fanzine in the paper is not 0.53 My

Hi guys live. So problem. I find 134. We need to find some other mess off the guests. It's given that mess off. Empty baskets 65.347 gram and Menzel, The flask filled with water is three 27.4 gram. So then Ms O wanted these 3 27.4 minus 65.347 grand, which is due 62.1 grams. Therefore, the rolling off the plastic doing mass off water divided right, the density off watching, which is to 62.1 grams, divided by 0.997 Grandpa Milliliters in front of all the amount of class case to 62.9 millimeter Mum, the mess of the class with the unknown liquid ese 65.7 39 grand. Therefore, myself, the unknown liquid. He's 65 points 739 minus 65.347 grams, which gives us A 0.39 to Graham. The temperature of the system. Given these 99.8 dissensions only cancan wanted to give him a skin, which gives us 3 72.8 the fishing given ese 101 too. In the basket lives Dunwoody do most very Christian by dividing it with 101.3 to 5, he keeps a 0.999 Hey, Jim, no! In this condition, the unknown guest occupies the whole container. Therefore, the world of the bears should be 2 62.9 I really need you all to 62.9 in the 1st 3 And the mess off the guests. We can't believe it. He's zero point 3 92 cream and we know when a mess A grant to mass divided by a number of months. Oh, number of Morse. Is it gonna do, Nancy? Very, very Momus. No. We can use the ideal gas. A convention. It is goingto in D. Oh, baby is going to and using the secretion we company to that one of us I need to see inside. It's inside the values that we know Massive warned. That's constant use Airplanes. Hail storm. No, your son Give me your son In village, it is 72 point do added by the pressure, which is 0.999 It's a dream And the volume please. 2 62.9 to 10 in your 1st 3 leader. Let's check on the You know this, Lieutenant. Everyone will be gone. It was very pressure will be gone. So we elected Graham and Union Verse one. These gives us 45.7 Grandpa which is someone or myself down. And then? Yes.

Okay. Now with the problem. One 1 13. 1st of all, we had to calculate the volume first. So we continue toe, do the problem, Okay? Because we need welcome. So we need the mask off the solution. So 113 the Misha high volume of major equal volume up solution off sodium hydroxide. And I so far. So we have this about this 100 meal, right? Plus five, five, 50 meals. That's equal 1 50 meals on this ego boy. Oh, we don't need We don't need this one. So we stopped right here. Okay. And next we have to calculate them. The more off the sodium hydroxide and sulphuric acid because we have to answer for question. Be so we have n so sodium hydroxide ego. 100 mil high. Okay. All right. That small hitori. Yeah, Yes, that's more on, Then, for suffering as it with all the same, you go one 50 meals we die for, boy. Oh, more here. And this 1000 milk is a good boy off. Five more and nicely do the question a right equation, right? A ballon equation. Okay, so a hey, we arrived by then. Situation, please Yes. Yeah, yeah, yeah. So get it to let me Jane, Let me bride this fall. So easy to see for you Hi. Yeah, we can write software. That's okay. I want to do this way. So easy to see. Okay. And now radio more radio on the equation, More radio. So if we're to more here one more here if you get 10.1, the people or fire, Right? Because the radio the 21 right, the radar to one here. So that's why we answer questions me be no. Okay. And see, because we get by one sodium hydroxide Where you off? And but 05 more up as it's very good you up to, right? No, no, no. As it left. So we answer No, no radio in the question. You have to go away, Cocula. Calculate more radio. Right? So two more So one more on did to Montu. And now, for once or the boy one I and this 12 1 it matched with a number here. So see? And now we don't see that if I kill solution, he usually Lucien okay. Your reaction okay is equal. Let me go live for us. Mhm Yeah, yeah. Mhm. Okay, let me write this Kill reaction equal minor scale solution. Right? And this ego 1 50 meal because the density of solution equal 1 g. So next we have to fight to calculate for hit specific It'd off the water Unless we have to fight to calculate by the type to multiply by the delta t I 31 boy for my minor 22.3 high. So on this new calculated, you will have this about 57 9 and fifties on this echo boy 71 kilo job. Because I got three sit fit. Right. 4711 killer Joe. Okay, Uh, 71 Because I got trace if it right here. Okay. Yeah, on this small piece equal Negative. Negative, E gti. Okay. And now form equation. Yeah, the equation. Off reaction. You see this on the boy? One more off the water. Redo. Right. So we have the hip. What did this equal? They got the Come on, let me. Therefore where is he? Get this. Might be evil. One more water. Oh, and this gave us, uh, You see *** t 31. Okay, so the answers are here, Rock. Yeah,


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