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Runiccamplesgas described hc tabla bclowordorncreisino Jutmqespeld ol the atom smoll:u LSThat " select nert sampl chlch tne atons molecules haycthenext slowcst...

Question

Runiccamplesgas described hc tabla bclowordorncreisino Jutmqespeld ol the atom smoll:u LSThat " select nert sampl chlch tne atons molecules haycthenext slowcst avcrade *peed, #noTnane culesVowcst uverae specd Uanlelctnertto tho cmoleInich thetatomgcueraontoted alonle molecut02t Lmple15 molaroon qj Ulam JnoCharttonal24 nolargon gas &f /.2 nim andcroose DalMintolneonat Zlaun ano72 'ICtEim and 71 'C(Chooso 2741.7 molhellum 905

Runic camples gas described hc tabla bclow ordor ncreisino Jutmqespeld ol the atom s moll:u LS That " select nert sampl chlch tne atons molecules haycthenext slowcst avcrade *peed, #no Tnane cules Vowcst uverae specd Uanlelct nertto tho cmole Inich thetatomg cueraontoted alonle molecut 02t Lmple 15 mol aroon qj Ulam Jno Charttonal 24 nol argon gas &f /.2 nim and croose Dal Mintol neon at Zlaun ano 72 ' ICt Eim and 71 'C (Chooso 274 1.7 mol hellum 905



Answers

$\Lambda$ closed container of volume $0.02 \mathrm{~m}^{2}$ contains a mixture of neon and argon gases at a temperature of $27^{\circ} \mathrm{C}$ and pressure of $1 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}$. The lotal mass of the mixture is $28 \mathrm{~g}$. If molar masses of neon and argon are 00 and $40 \mathrm{~g} \mathrm{~mol}^{-1}$ respectively, find the masses of individual gases in \mathrm{\{} ~ t h e ~ c o n t a i n e r , ~ a s s u m i n g ~ t h a n ~ t o ~ b e ~ i d e a l . ~ $(R=8.314 \mathrm{~J} / \mathrm{mol}-\mathrm{K})$

As we know that blue color of copper sulfate blue, good lad of corporal sulfate solution of copper self aid solution does not Faith does not states in access of ammonia in access of ammonia. It will remain blue in color. So here all the other statement are correct, but his statement B is incorrect, therefore, according to the option. Therefore, for the option A. See the are correct.

Let's ah, calculate the fraction of space occupied by the Neon Adams. They were told that the atomic radius of neon is 69 PICO meters. Let's change this in two centimeters. So we know that, um, tenderly 12th PICO meters in one meter and in one meter we have 100 centimeters. This would be equivalent to 6.9 times 10 to the minus nine centimeters. Now it's all for the volume occupied by a single I me on Adam 4/3 pi r cubed So 4/3 pi radius is 6.9 times 10 to the negative nine centimeters. That would be cube sore volume would work out to be, uh, to seek figs 1.4 times 10 to the minus 24 centimeters cubed. Um, let's turn this into leaders. So 1.4 times 10 to the minus 24 centimeters cube and there is 1000 centimeters cubed in one leader. This would work out to 1.4 times 10 to the minus 27 leaders. Therefore, if we have 1.4 times 10 to the negative 27 leaders per Adam present and we're told that we have in this question, Um, to 0.69 times 10 to the 22 Adams. This would work out to 3.7 comes tend the negative five leaders and therefore are percent by volume of neon is equal to the volume of neon off the volume Of the total comes 100% the blame of the neon 3.7 times 10 to the negative five leaders over the total wine, which is one leader 100%. This would work up to 3.7 times 10 to the negative three percent. That would be the volume the present by volume of the neon in the um in this space. This reveals that the separation between atoms in the gas phase is very, very large.

So the average amount of kinetic energy in a gas molecule would be equal to three halves, multiplied by cable, spins constant, multiplied by by T the temperature. My apologies for part A. We for an ideal gas, the average kinetic energy per molecule rests upon only the temperature because Holtzmann constant is of course simply a constant. So for part A we can say that the kinetic energy for neon kinetic energy for krypton and the kinetic energy for Rayden would be equal to simply 12121 again, if all of them are at the same temperature. Now, for part B, we want to find the root, mean squared root, mean squared velocity, or speed of a molecule. This would be the square root, or rather Equalling square root of three times the ideal gas constant times the temperature divided by the molar mass of the gas. And so for we can calculate this for neon root mean square speed for neon Would be equal to the square root of three times 8.3145. This would be jewels per mole Calvin multiplied by the temperature here. And we could just simply keep this as T divided by the Mueller mass At 20.18 we can say kilograms per mole or grams per mole. It doesn't really matter because at the end it's all going to be um it's all going to be divided by one another. So actually given that it's all going to be divided, we really don't even need to substitute and for our so we can just keep this as our and this would be giving us then point 22 26 Radical three R. T. We really just want this coefficient here and then for root mean squared of neon, I'm sorry, I mean squared of krypton rather this would be equal to then again square root of three R. T. But this would be divided by here 83 .8 g per mole. So this is giving us then point 1092 radical three Rt. And then for right on we have Radical three RT. And this will be divided by here 222 g per mole. Giving us then .067 one radical three R. T. And at this point we can simply divide so root mean squared for neon root mean squared for krypton root mean square speed for raid on this would be equal to then .2226 .1092 .0671. And in order to get the whole number, we can simply divide all of these by .061. So divided by the smallest, essentially the smallest coefficient. And we find that this is going to be equal to then 3.32 To 1.63 21 So we can see that the rate root mean square root mean square speed of raden is the least And then root mean square of Krypton will be 1.63 times that of raden root. Mean squared of Neon will be 3.32 times that of raden. That is the end of the solution. Thank you for watching.

Between two. The evidence. Six. Wish it's more so. A simple question. Um, and you're supposed in turning the factual speech that then you want Adams occupy so long your musical to four birds Time que named. So before turns pry 69 terms 10 to the minus Tune Cube, The sequel to 1 23 7 trying to turn to a minus 27 Dustin rivers, too. You mean you shall have your 1.37 terms to into the minus 27 Dustin eaters, too, times your two point 69 times 10 to the 22nd, which pierce you 3.6 money terms 10 to the minus this immediate cube. And then you get 316 lying times two the minus five dosimeters to over one distant, meager que, which will give you three points. Recline strings tend to the English five or one older, 27 zero. Touring weren't 13 I'm when


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