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Cansiceelanced chemlcul equabion 2H" (wq) 1, (mn)- 2H,O() Hos(wn) 3 (m) concanlrailon & [ drops trom 1.350 M lo 0.844 M In Ueu Iet Goconda 0l Ina qnctlon I...

Question

Cansiceelanced chemlcul equabion 2H" (wq) 1, (mn)- 2H,O() Hos(wn) 3 (m) concanlrailon & [ drops trom 1.350 M lo 0.844 M In Ueu Iet Goconda 0l Ina qnctlon Inh tcucilon thia Ilme Interval Calculbla Lte avutogu Express your answer #rith approprlate unite.

Cansicee lanced chemlcul equabion 2H" (wq) 1, (mn)- 2H,O() Hos(wn) 3 (m) concanlrailon & [ drops trom 1.350 M lo 0.844 M In Ueu Iet Goconda 0l Ina qnctlon Inh tcucilon thia Ilme Interval Calculbla Lte avutogu Express your answer #rith approprlate unite.



Answers

$$ \begin{aligned} &\eta=\frac{\text { concetration of } H_{2}}{\text { concentration of } N_{2}}=\eta_{0} \frac{e^{-M_{H_{2}} g h / R T}}{e^{-M_{N_{2}}-g h / R T}}=\eta_{0} e^{\left(M_{N_{2}}-M_{H_{2}}\right)} g h / R T\\ &\text { So more } N_{2} \text { at the bottom, }\left(\frac{\eta}{\eta_{0}}=1.39 \text { here }\right) \end{aligned} $$

In this question we've been given a reaction in which two malls off and or two are creating with um all of F. Two to produce two malls off and all to act. So this is the reaction that we're looking at and the rate law for this reaction is going to be in the form of great is the quantum Okay The concentration of one or 2 raised to the power of em multiplied by the concentration of F. Troop raised to the power of end. So the main goal here is to determine the order and the right expression. But since we want to determine our um our end and finally our K. For us to have a full rate expression of this reaction. So for us to determine the quarter of reaction with respect to an autumn, We can substitute the two initial concentrations of n. 0. 2 and the corresponding right that is let's say we have afraid being a courtroom. Okay prime multiplied by the concentration of an auto is to the pope. And so here our K. Prime is representing K multiplied by the concentration of F. Troop. So we're going to leave that for a moment. So we want to determine the rate the order of the reaction with respect to an auto. In other words, we want to determine our camp. So let's take for example a certain condition where we have our right one Corresponding to a concentration number one and we have our late true corresponding to the concentration of an autumn. It points to raised to the power of em. We have been given all this information. For example When the rate is equal 0.0-6, remember our key. Okay prime this is the same. So when the rate is equal 0.0-6 The concentration is going to be equal to a 7.1. And we raise this to the power of um And again when the rate is 0.051 the concentration is going to be equal to self control again or raise this to the power of him. So what we here at the end of the day is an expression of 7.5 B equal to 7.5. Place to the power of em. And if we introduce logarithms we're going to have our law 0.5 Being equal to M Milk. So we can conclude that our aim is equal to one achieves this reaction is first order first order with respect to with respect to and or two then walking on we now need to determine the reaction order with respect to F two. And for us to do that we're going to apply the similar approach to say our right Let's look at .2 being equal to K. Prime. The concentration of F two raised to the power And at point #2 and our rate say it point blank it's going to be equal to K. Multiplied by the concentration of afternoon. 8.3 Ways to the power of end Here our K. Is going to be equal to care the concentration of an auto raised to the power of em. Now that we all have this information, remember this is the same reaction. So these kinds of So we know that when the rate is 0.051 concentration is 0.1 and we raised this to the power of and And when the rate is 0.103 the concentration of F two is equal to 0.2. And we also raise this to the power of mm. What we now have is we also have a situation where we have 7.5 billion equal to 7.5 raised to the power And And again if we introduce logarithms, the log of 7.5 is equal to end log officer 15. Again our L is equal to one. That is it is first order with respect to F two. So now that we have this information we can then look at our overall expression and if we look at the overall right expression we have the rate being equal to this is the overwrite expression. Now that we have our and our end this rate is going to be returned in the form of great is equal to k concentration of an autumn, Multiplied by concentration of F two. And the overall right over order of this reaction is M plus M which is one plus one and this is true. So the overall order of this reaction is while trump. Now we have to move on to determine our cake K. Part of this expression. So to determine the value of K. Let's look for example, we know that our K. Is going to be equal to right Divided by the concentration of N. or two, multiplied by the concentration of F Troop. So at any point let's just take any point The same .1 When the rate is zero, our concentration of one or two is going to be served with one And that of F two is also going to the airport one. So our K. here is there is 2.6 and let's take another point. The rate is 0.051. The concentrations are going to be equal to 0.2 by 7.1 and we have K two of 2.5. And let's take another point. If our rate is 0.103 the concentrations are going to be 0.2 by 0.2 and this is Equal Truth 2.5. Okay. And another point they say careful, The rate is equal to 7.411 Concentrations are going to be a 0.4 x 0.4 and this is equal to two seven and if we look at these these units are pay more could be lead to pick Tessa meter or later the second Or 2.7. I am yes, they can. Now, if we look at this they are relatively the same. So what we can do is to just take the average of these K values to get an average off K being equal to if we make the average here, They say 2.6 plus 2.55 Plus right up to 2.7 and we divide this by four. Our key Is going to be 2.61 am this again. So this is our value. Okay, now that we have this, we can then write the full rate expression to say the rate of this reaction. The rate is equal to 2.7, multiplied by the concentration of N or two. And the concentration of actually, so this is the final solution And over order of this reaction is 2nd order. That is the order is true.

So we have a, uh, sulfurous acid solution at 0.1 Moeller and we are given the two k is for the two protons associations. And we've got five choices for what would be true here. Now, uh, that first Proton is not gonna be strong acid. So it is not going to disassociate 2.1. So we're not gonna wind up with, um, Mawr than 0.1. Can't store geometrically wind up with more than 0.1 of the hydrogen sulfide ions. So that one is out. Um, this one, huh? Wonder about that is the hydro Nia, my on 0.1. Moeller. Well, I don't think so, but let's calculated and see if we wrote out the standard equilibrium and made the assumption that X would be small. Then we would wind up with X, which is the Hydro Nehemiah concentration equal to the square root of Casey and the hydro name ion. Concentration would be 0.36 If you did the quadratic formula and actually worked out, you'll find that it's actually 0.30 So yeah, we're not going to get point when Moeller uh hi. Drawing on my own. Ah, and In fact, we're above already 0.13 with the first proton. Looks like about 20 room three now. So we're left with these two choices. Are we going to get another, uh, 0.6 This again? That's 36% association. We calculated that assuming we dropped the minus X on the 360.10 Ah, but we can only do that if we have less than 5%. Association is 36%. So that is not the answer. But let's see why this is theano, sir. Now, when you take account of the first association, it produces again by the quadratic 0.30 of the conjugal base and 0.30 of the hydro Nia, my on. So when we start looking here at the K two equilibrium, those air our initial concentrations and then take the ice table from there and as we have seen before, then once we plug in, uh, we read the K expression from the reaction we plug in the expressions from the Ellen into the K expression well, where X is added or subtracted. It's probably small. We always check that we found here that it wasn't small, but here X is 6.3 times 10 to the minus eight Moeller, and that is the sole fight concentration. And that is the answer to our question. Both the additional age plus as well as the contra get base. Ah, in the second, equilibrium tend to be equal to or do you come out to be equal to the value of the equilibrium constant?

Just expressed indicate, given that are so basically in the first part even to tm and visual can that to introgen to devour off to millimeter and which we conducted is a 200. Really, meat that And in the next part it is Yeah, it is two hours. It was 21 20 minutes so obviously to our hours equals to 1 20 minutes. So there are 1 30 minutes total. So 1 30 minutes equals two 1 30 minutes. Educ wants to just 1 30 time self. No. 60 really wants to. Seven night, Gino. Zero seconds Now in the next spot off this question, we have given the difference in the 16 Graham musical students toe 16 into 10 to the power off six microgram my program or we can eat right a 1.6 injured, 10 to the power of seven in the more scientific way. So and now, moving back to the next part off this question. So, in the spot where we have given the 0.75 kilometer, so which we can i d take the 0.75 into tender, devout off TV engine is out off two centimeter which began, darted a 0.75 into genders level five, Sending me toe and in the next part of this question, we haven't given judo point 675 milligram, which we can guard a 0.675 into 10 to the power off Mina's todo good. Um, and in the more appropriate rated, this camp region is a 6.75 into tender, devout off minus of food. Good, I'm and then I'm going back to the next part which we know that one micro meter, which is equal student of the power off and it's four centimeter and this is equivalent to renting out off my 6 m. So yeah, 462 micro meter, which will be nothing but 46 to indigenous out of miners. Poor centimeter. And we can also 34.62 into 10 to the power off, minus two. Sending me that and now moving back to the next part off this question which we have given that we know we have to convert the pillow Mitchell proper 2 m per second. So we know that kilometer, but our which we can donated a kilometer upon our into thousands off meter upon upon one kilometer because I really wanted him to 1002 m in tow. One hour upon 36 off second. So our castle gm gm, get canceled. Those you know, the ones you know, Good cancel and the kilometer by our let me tell upon our which basically become one upon 3.6 meter per second. So we can say that No, you can say that to convert the kilometers per second and multiply so to combat killam capacity and we multiply one point to the sixth through the And if I had my which so we cannot just 35 do you know, meter but our musical So 30 fire upon 30.6 meter parts again. No, we can Badgett todo 30 ft now begin died It is there 35 35 35 kilometer But Howard, she calls to 9.7 m per second. So this will be nine point seven

Good day. In this question we will be solving the instantaneous speed out there and let sir. So to solve this, we have the instantaneous speed is equal to X divided by time. Where our I am given is then Milliseconds. So time is equal to 10 million seconds. And according to our graph, The graph or figure to 25. At this time the eggs is approximately 4.3 mm. So we are also given the generation factories of one M. S. Is for the Y 001 2nd. And one is why 001 m. So solving where the instant thing is being, we have eggs over tea. She is equal to 4.3 m. m. Converting two m. We have one. It's 1001 m divided by then M. S. Times one M. S. is for in Syria 01 seconds. Therefore The instantaneous speed is equal to .43 meters per second. Thank you.


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