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Question

Tnnt astonautt_ Dronelied bxkrbacbnack PlLtard Gulor 1 [5koetenod tcntn Eohandod 60*- Whut b Uhe (9} mjor TUce ard (B) an3 = nrt) Ea fatet tnoon I tha {316, nltn 5 , 1-Jin,71= SsN,F = Jon0, = J0" (Mejturcd nlatiyc to tne potinve J retton 0{ the * &I5s m 04 [no-0 (-1Fo" Iej )) cithe #rtenced $ Jceeler ion;

Tnnt astonautt_ Dronelied bxkrbacbnack PlLtard Gulor 1 [5koetenod tcntn Eohandod 60*- Whut b Uhe (9} mjor TUce ard (B) an3 = nrt) Ea fatet tnoon I tha {316, nltn 5 , 1-Jin,71= SsN,F = Jon0, = J0" (Mejturcd nlatiyc to tne potinve J retton 0{ the * &I5s m 04 [no-0 (-1Fo" Iej )) cithe #rtenced $ Jceeler ion;



Answers

Complete this table: $$\begin{array}{|lcllll|}\hline V_{1} & T_{1} & P_{1} & V_{2} & T_{2} & P_{2} \\ \hline 546 \mathrm{L} & 43^{\circ} \mathrm{C} & 6.5 \mathrm{atm} & & 65^{\circ} \mathrm{C} & 1.9 \mathrm{atm} \\ \hline 43 \mathrm{mL} & -56^{\circ} \mathrm{C} & 865 \mathrm{torr} & & 43^{\circ} \mathrm{C} & 1.5 \mathrm{atm} \\ \hline 4.2 \mathrm{L} & 234 \mathrm{K} & 0.87 \mathrm{atm} & 3.2 \mathrm{L} & 29^{\circ} \mathrm{C} & \\ \hline 1.3 \mathrm{L} & 25^{\circ} \mathrm{C} & 740 \mathrm{mm} \mathrm{Hg} & 0^{\circ} \mathrm{C} & 1.0 \mathrm{atm} \\ \hline\end{array}$$

This question asks us to convert the values of casing the values of K p or the values of K P to the values of Casey. And so therefore, parts of broken them up in the individual slides. Speaking with part, eh? Here we have sealed to and b r two going to be our C. L tells us that the equilibrium concept perspective concentration is 4.7 times 10 to the negative second at 25 degrees Celsius. So then we're gonna be converting this to KP or the equilibrium constant with respect to pressure. So that equation looks like this k p equals K. C. Times are the ideal gas constant times t the temperature raised to delta end, which is the change in the number of moles in the reaction as it's written. And so this is how you find that Delta and equals the number of moles of gas and product minus the number of moles of gas and reacted. And so if there are two moles of gas in the products and there are two moles of gas in the reactant ce one here and one here, So to total, that means the Delta and is zero. And so now we can plug in each of the pieces Casey was given to us in the question. The ideal gas constant. It's 0.8 to 06 The temperature in Kelvin is 298 and Delta is zero. And, you know, when you raise something to the zero power, this whole thing becomes one. So now KP actually just equals Casey or 4.7 times 10 to the negative. Second you go. And for B, we have s 02 and no to going to s 03 This time it gives us KP the equilibrium constant with respect to pressure at 500 degrees Celsius. And so it's similar setup. But the equation looks a little bit different to find Casey, since we're given KP, that equals KP, divided by Artie Delta and so delta and again, number of bowls a product to most of gas and products minus the number of bowls of gasoline. Actives There three here. So two months three is negative one you can plug in all of our parts. K c equals K P was given to us 48.2 divided by the ideal gas. Constant times. The temperature, which says 500 degree Celsius converted to Calvin, is 773 raised to the negative one, and this is 3.6 at times 10 to the third. That's Casey. Here we have see a seal to with six waters around it, going to see a seal to, and Water tells us KP is five points here nine times 10 to the negative, 44th and Delta end. Number of bulls in of gas and products. They're six, and there are no Mel's of gas and reactant. So Delta is going to equal six. Now we're gonna do again. Casey to KP. I'm sorry. Calculating Casey from K P. That's KP over. Artie dealt, and we can plug in all of the pieces now 5.9 times 10 and they get a 44th. Was KP given to us by the by 0.206 times the temperature it was 25 converted to Calvin to 98 raised a delta and which we just have a six and this is too 0.3 eight times 10 the negative. 52nd. That is Casey for this reaction. And finally, we have each to a liquid. Going to each to a gas tells us KP is 0.0 point 196 at 60 degrees Celsius Delta end. There's one mole of gas in the products and none in the reactors and said one minus. Here was one you. Same question we did for Parts B and C So K C equals 0.196 divided by 0.206 times the temperature now in Kelvin, 333 raised to the delta and which is one. And this is 7.17 times 10 to the negative. Third, that is Casey for this reaction.

This problem were asked to complete this table shown up top. But before we can do that, we need to make sure everything is in the correct units. So I drum table again and copy down everything that's already in the right units, and we're gonna convert everything else. So for volume, we want our units to be leaders for temperature. We want Calvin's. And for pressure, we want atmospheres. So the View one column is already in leaders. So that's great. T one, uh, is all in Celsius. We need to be Calvin. So to convert from Celsius to Calvin, we're gonna add 273. So for 10 degrees Celsius, that gives us 2 83 Calvin zero gives those to 73 Calvin and 75 degrees Celsius. Gives us 348. Calvin R P. One column is all in atmosphere is already so. That's great. Um, t two, we have to convert from Celsius to Calvin again. So we'll have to 73 k 308 k and to 73. Okay. And then, uh, P two. We have one pressure in tour 1 80 Em is equal to 760 tour. So we will take that 7 35 and divide it by 7 16 And that will give us zero point 97 a t. M. Now to fill in the missing boxes in the table. We're going to use the combined gas law. So this tells us that p one times V one over t one equals p two times the two over t two. So for the first column here, r p one is 0.75 a t m b one is 6.35 leaders over T one of 2 83 Calvin, that is equal to 1.0 a t m for Pete to be, too, is what we're looking for in our teach you is to 73. Calvin. So to do this math first, we're gonna start with the left side. We're gonna most 5.75 by six point 35 and then divide it by 2 83 and we'll get 0.17 leaders a t m. Over Calvin equal to um one divided by 2 73 is Syria 730.37 a. T m. Over Calvin times be, too. Now we can divide that point. 017 by the 0.0, 37 And that will give us a B two of 4.6 leaders. That's what fills in that box for the second rover and do the exact same thing. So we'll set it up the same way. Just make some room. So P one is 1.8 e. M. Times V one is 75.6 leaders over to 73 K that is equal Teoh Syria 0.97 80 m times be to over the Rio eight k And if we solve that you doing the math the exact same way we'll get V two equal to 87.9 leaders and or last column we're gonna do the exact same thing. But this time will be solving for P two instead of be too. So, uh, on the left side we have 0.55 80 m for P one. B one is 1.6 leaders over 3 48 k for t one. We're looking for P two times 3.2 Leaders be to over to 73 k t two. So again we'll do the math the same away and we will get 0.14 80 m four p two.

Hi there in problem No. We have some pressure conversions to calculate. So each of these will just be a one step dimensional analysis once we have the correct conversion. So we are starting with point in letter a. With .19912 atmospheres. And we want to express that in terms of tor so in one atmosphere There are 760 tour. So these two values are equal so I can use that as a conversion factor, Calculate my answer rounded to three significant figures And I get 693 tour. That's the answer for party. Okay. In part B We're starting out with bar we have .685. Yeah. Uh huh. So .685. Bar oh. And the conversion or the equality there Is that one bar Is equal to 100 kilo pascal's. Mhm. So we get 68.5 kilo pascal's or K. P. A. That's our answer. Mhm. Okay. The letter C. We have 655 millimeters of mercury and we want to convert that to atmospheres. So the conversion here the equality 760 mm of Mercury are equal to one atmosphere. So completing the division here and rounding to three significant figures Get .862 atmospheres. Yeah. Yeah. Mhm. Mhm. For a letter D. We have 1.3-3 Times 10 to the 5th Pascal's. Yeah we want to cover that to atmospheres. So the conversion one atmosphere Is equal to 101,325 past scouts. Okay. And this one can be in four significant digits since our original measurement was in four. So we get 1.306 atmospheres. Yeah. Okay. And finally let her eat. Okay For a letter E We are given atmospheres 2.50 atmospheres. And asked to convert £2 per square inch or see. So one atmosphere Is equal to 1470 P. S. I. that means 2.50 atmospheres is the same As 36.8 pounds per square inch for C. Yeah, that is the final answer of this series of conversions. Thank you so much for watching. Yeah.

Okay, so this is another confirmed problem basically it has our knowledge on how to convert between the concentration equipment constant towards the pressure equipment constant and Vice Visa. So in order to solve this kind of question, we need to first know their mathematical relations. It's actually equals two K. C equals two. So K. P over the R. T. Power delta. And and here or is just gassing consent. And delta end is just the difference between the coefficient coefficient difference between the reactant the product and reactant. So once you know that the problem will be very easy to solve. So you can first look at the first reaction which is chlorides react with Pro might and you will form bombing. Cool, right. The question gives us the concentration equipment constant which is 4.7 times 10 to minus two. So now I want to do is to calculate the pressure couldn't constant which is equal to K. C. Times Artie Power Delta. And also it says it's 25 degrees C. So here the data and as I said is the difference of the coefficient between the products and reactant. So here the coefficient of the product is too and reactant is one plus one. So here the delta end is equal to zero which means here if delta and is equal to zero, this whole part would be just one which means the pressure constant which is equal to the concentration constant which is the same value 4.7 times 10 to minus two. And now we can look at the second chemical equation which is the to S. 03 gas face reform or decompose. Oh sorry it's reversed. It's actually two S. 02 gas phase react with oxygen and they were formed so sweet as gas face here. It gives us the pressure equitable and constant which is 40 eight coin to the temperature is 500 degrees C. So obviously we need to calculate the concentration constant which were equal to K. P. Over artie power delta and and here the delta and just equal to two minus one minus two, which is just minus one. So then it just becomes K. P. Times R. T. And now we can just plug in the number KTS KPs 48.2. And our is point oh 6 to 1. And the temperature is since this is Celsius, we need to convert into kelvin which is seven 173.15 Now we can use our calculator to calculate the values and the final value which is equal to 3.6 times, tend to sweet. And this is the answer. And now we can look at the third equipment, which is calcium chlorides. But with water actually decompose to form pure calcium chlorides, clause six. Water which is in the gas face. Sure. And it gives us the pressure equivalent constant which is pretty small. Five point all nine times 10 to minus 44. It's a 25 degrees C. So here we will calculate the concentration queen constant which equals two K. P. Over R. T. Delta. And here is the delta and equals to six plus zero minus zero here because there are solid. So the coefficient can be just regarded as zero. So here is a six. Now we can just plug in the number which equals to five point oh nine times 10 to minus 44 over point oh 8 to 1 times the temperature which is 25 degrees C. In Calvin. It's 298.15 Power six. Now we can try to do the calculation and the answer would just be equal to around 237 times 10 to minus 52 which is really, really small number, which means distraction proceeds very tiny in the right direction. And now we can go to the last equation which is again water liquid face gas. If I into water vapor and gives us the pressure constant which is point 196 at 60 degrees C. Absolutely. We need to calculate the concentration constant which were equal to the K. P. Over R. T. Delta. And that end here is just equal to warm minus zero because in the reacting part, the water is liquid phase. So then just becomes K. P over R. T. So the KPs 0.196 R. S point oh 8 to 1 times the temperature is 60 degree C. So it's around 333.15 kelvin, and now we can calculate the answer, which is equal to around 7.1 seven times, tend to minor sweet, and this is the answer.


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