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QUESTION 10Calculate the partial pressure of nitrogen when the atmospheric pressure is _ 3o0mmkg 6300 mmhg B, 1429 mmHg C.63 MmHB D. 234 mmHg 384 mmHgQUESTION 11Whi...

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QUESTION 10Calculate the partial pressure of nitrogen when the atmospheric pressure is _ 3o0mmkg 6300 mmhg B, 1429 mmHg C.63 MmHB D. 234 mmHg 384 mmHgQUESTION 11Which ofthe following accurately describes Henry's Law? the pressure volume of a Ras are directly proportional (for example; Ivolume Fisos prcssure rises) the amount of gas dissolved solution solution directly proportional to the partial pressure of that gas In thc air above thatthe tolal pressure Iturn ofgas Lala Partial pressures

QUESTION 10 Calculate the partial pressure of nitrogen when the atmospheric pressure is _ 3o0mmkg 6300 mmhg B, 1429 mmHg C.63 MmHB D. 234 mmHg 384 mmHg QUESTION 11 Which ofthe following accurately describes Henry's Law? the pressure volume of a Ras are directly proportional (for example; Ivolume Fisos prcssure rises) the amount of gas dissolved solution solution directly proportional to the partial pressure of that gas In thc air above that the tolal pressure Iturn ofgas Lala Partial pressures cocn Bas (amneture the amount of gas dissolved in solulon Inversely proportiona partial Dnessune Yoution of that Gas in the Jr Aoove tnal the pressurg and volurne Gas are inversely onortiona



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52. Nitrogen gas can be obtained by decomposing ammonium nitrate at high temperatures. The nitrogen gas is collected over water in a $500-\mathrm{mL}$ (three significant figures) flask at $19^{\circ} \mathrm{C}$. The ambient pressure is $745 \mathrm{~mm}$ Hg. (Vapor pressure of water at $19^{\circ} \mathrm{C}$ is $\left.16.48 \mathrm{~mm} \mathrm{Hg} .\right)$ (a) What is the partial pressure of nitrogen? (b) How many moles of water are there in the wet gas? (c) How many moles of dry gas are collected? (d) If $0.128 \mathrm{~g}$ of $\mathrm{Ne}$ are added to the flask at the same temperature, what is the partial pressure of neon in the flask? (e) What is the total pressure after Ne is added?

Let us discuss the person this composition of trade at sea level. In comparison with the given sample this table gives the information of person days competition of trade at sea level gas and the person there's the volume of each gas is given To calculate the partial pressure of Co two. Let us write the given data. Total pressure is 7:40 a.m. A veggie. Yeah, Partial pressure of Bento is 5 75 mmorpg. Yeah, partial pleasure of organ is 6.9 mmorpg. Yeah. Mhm Yeah, Partial pressure of the water is 0.2 mm. RPG. Yeah. Yeah, Partial pressure of vegetable use four mmorpg. Yes. No partial pressure of water can be calculated by using the formula total pressure is equal to some of the partial projects of the given gases. Total pressure is equal to partial pressure of enter place partial pressure of seawater. Place partial pressure of argon. Place partial pressure of vegetable place partial pressure of photo. Yeah. Yeah. Yeah. Yeah. Yeah. Stop shooting all the values total treasure is 740 is equal to partial pressure of employees. A 5 75 plus Partial pressure of 02 is 0.2 plus partial pressure of argon is a 6.9 plus partial pressure of water is a four mmorpg plus partial pleasurable for you too. From this Partial Pressure of Co two is equal to 7 40 minus 5 75 plus 0.2 Plus 6.9 Plus four. Yeah. The final value equals to 1:50 for Mmorpg. Okay, let us discuss the second part of the problem which is more fraction of each gas, mole fraction of each gas can be calculated by using the formula partial pressure is equal to total pressure into mole fraction. Yeah From this mole fraction of N two is equal to partial pressure of and to buy total pressure some shooting the values partial pressure of N two is 5 75 A veggie. Yeah. Do I buy Total pressure is 7:40mmorpg. Yeah The value equals to 0. 7. Similarly we can calculate the mole fraction of for other gases, mole fraction of organ is equal to partial pressure of argonne divided by total pressure. Partial pressure of organ is a 6.9 um Ahmad Fiji divide by total pressure is 7:40 a.m. A. Fiji. The value equals to zero point double 0932 Mole fraction of 02 is equal to Yeah partial pressure. Roxy go to buy total pressure. Uh huh. Partial pressure of the water is 0.2 mm. A Fiji divide by Total pressure is 7:40mmorpg. Mhm. The value equals to 0.00 three mole fraction of vegetable is equal to Yeah Partial pressure off its two or by total fresh air. Yeah. Yeah. Yeah partial pressure on vegetable is 1 54 mmorpg. Mhm. Uh huh divide by 745 G. Which is a total pressure. The value is zero point do not eat Similarly. We can calculate the mole fraction of four. To by using the same formula. Mhm. More fractional photo is equal to partial pressure of CO two by total pressure. Yeah partial pressure of photo is a 1 50. For mmorpG divide by Total pressure is 7:40mmorpg. The value equals to 0.208. Yeah. Let us discuss the 3rd Point of the Problem. Mhm. Person days off and two is equal to mole fraction of entering. 200%. Yeah. Yeah mole fraction of N 20. Is a zero point triple seven in 200. The value equals to 77.7%. Person days of organ is equal to Mole fraction of are gone in 200 Mole fraction of organ is a 0.093 in 200. Yeah The value equals to 0.93 percenters of seawater is equal to More fractional for moto in 200. Yeah mole fraction of c. Photo is zero point triple 03 in 200 The value equals to 0.03 percentage of photo is equal to Mole fraction of water in 200%. Yeah more fractional photo is zero point do not. Eight in 200. The value equals to 20.8%. Finally percenters of water is equal to more flags. Knowledge to walk into hundreds. Yeah More fractional vegetable is 0.05.4 in 200. The value equals to 0.54%. so that we can compare the percentage composition of the dry air at sea level from the table with a given sample, the sample given his vet, and the table given is dry A's.

So this question. I hate to say it is actually a little bit tricking. For the first part, it asked us to calculate the partial pressure that is nitrogen. So Dalton's law tells us the total pressure is equal to some of all the partial pressures. We know the total pressure want all the partial pressure of helium so we can calculate the partial pressure of nitrogen to be 7 13 millimeters of mercury ing. However, the volume is going to be defined by the total pressure and the volume of nitrogen and the volume of helium will be exactly the same. They will be the volume of the mixture. Both gases will occupy the total volume, so we need to use the total pressure we use P one B one over T one equals P. T. V two over t two and recognize that the final volume is the same for the mixture nitrogen and helium. So the initial pressure of four of 7 45 is what we need to use the initial pressure of everything because we need to calculate the volume of everything and then the initial volume of 250 milliliters multiplied by the final temperature to 73 at STP, divided by the initial temperature to 73 plus of 30 degrees Celsius, and then get to the final pressure will be the pressure of STP One atmosphere or 7 60 tour gives us 220 mil milliliters of the mixture, which is also the volume of nitrogen and also the volume of helium.

The first thing we can start with here is stone contextual information before we jump into the content. So just to define partial pressure for us all this is the pressure that would be exerted by one of the gas is within a mixture if it occupies the same volume volume on its own. So the equation we're looking at here is the ideal gas equation. So we've got P equals NRT over V. So I've just rearranged that slightly moved volume onto the right hand side. So I've plugged in my numbers. You can see we've got one multiplied by one more event. 20 5/1 08 g multiplied by not point nor 8 to 06 multiplied by 273 at 65. So this is just to get our temperature in Calvin, which is S I units, This is just our volume. Then we're multiplying the outcome of the equation above by 7 60 mmh g over one atmosphere. So then we get 13 mmh g so moving on the half life for the first order, decomposition off end to five. Gas is 2.38 minutes. Therefore the initial pressure will decrease to half of the calculated value. So what we take is the 13 that we originally calculated on simply divided by two. This gives us a value off 6.5. So moving on to the last portion here from the balanced chemical equation, the reaction off two miles off and 205 generates four miles of an 02 on one mole of 02 So, by the consumption of two, moles of reacting gas were able to generate five miles of product gas. So if we consider the same temperature and volume in this case, the pressure will behave as amounts. So the reaction off to mmh G off reactant generates five mmh g of product. So we have 13 take away 6.5, add 6.5 times five over to So then I've simplified that down to 13. Take away 6.5, take away 16.25 And that gives us a final figure off 22.75 H g

So just to start off with some contextual information fast, so we'll just run over the definition of partial pressure. So this is the pressure that would be exerted by one off. The gas is within a mixture if it occupies the same volume on its own. So the equation that we're using it is the ideal gas equation. So it's PV equals NRT. However, I've rearranged this and put volume on the right hand side. I've then plugged in my numbers and then multiplied the outcome by the above equation Uh huh. By 7. 60 mmh g over one atmosphere, we generate figure of 13 mm H g and then moving on. If the half life of the first order decomposition of N 205 gas is 2.38 minutes, the initial pressure will decrease the half of it the calculated value. Therefore, we simply take the figure that we determined in the previous part of this podcast and divided by two. So we get figure of 6.5 now. So just moving on to the last part from the balanced chemical equation, the reactant off two moles of n 205 generates four Moles of CO two and one mole of 02 So the consumption of two moles of reactant gas generates five moles of product gas in this reaction. So if we consider the same temperature and volume in this case, the pressure will behave as amounts. So the reactant off to mmh G off reactant will generate five mmh g off product. So we have 13 take away 6.5. I had 6.5 multiplied by five over to I've simplified that down to 13. Take away 6.5 take away 16.25 That generates a figure off 22.75 where just one correction needs to be made in there to accurately portray the final figure off 22.75


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