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1. Calculate the volume (in L) of O2 at a pressure of744 mm Hg and 32.0 oC that will be consumed by combusting 10.0 gof C15H21N.2.The CO2 produced from the combust...

Question

1. Calculate the volume (in L) of O2 at a pressure of744 mm Hg and 32.0 oC that will be consumed by combusting 10.0 gof C15H21N.2.The CO2 produced from the combustion of C15H21N iscollected over water at a total pressure of 755 mm Hg, 28.0 oC anda volume of 165 mL. Calculate the partial pressure of CO2 and thegrams of C15H21N consumed

1. Calculate the volume (in L) of O2 at a pressure of 744 mm Hg and 32.0 oC that will be consumed by combusting 10.0 g of C15H21N. 2.The CO2 produced from the combustion of C15H21N is collected over water at a total pressure of 755 mm Hg, 28.0 oC and a volume of 165 mL. Calculate the partial pressure of CO2 and the grams of C15H21N consumed



Answers

Assume that an exhaled breath of air consists of $74.8 \% \mathrm{N}_{2}, 15.3 \% \mathrm{O}_{2}, 3.7 \% \mathrm{CO}_{2},$ and 6.2$\%$ water vapor. (a) If the total pressure of the gases is 0.985 atm, calculate the partial pressure of each component of the mixture. (b) If the volume of the exhaled gas is 455 $\mathrm{mL}$ and its temperature is $37^{\circ} \mathrm{C}$ , calculate the number of moles of $\mathrm{CO}_{2}$ exhaled. (c) How many grams of glucose $\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)$ would need to be metabolized to produce this quantity of $\mathrm{CO}_{2} ?$ (The chemical reaction is the same as that for combustion of $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} .$ See Section 3.2 and Problem 10.57 )

Problem. Party fraser percent. It calls to more percent percent and change the pressure. Do you hide it away mol percent. You can razor present equals two more percent. No change pressure. Do you hide it by more percent? Two mole fraction. Okay, no partial prisoner of each. Gas is more fraction small franks and times total pressure pressure means Pressure of gas x equals two small friction up gas X multiply into total blessing. No place it up nitrogen gas. It calls to more fixing up nitrogen guys that is simply was. We can use pressure present as more more more percent. Now he was 0.78 748 Multiplied by total pressure. That is zero 98 Hi radium. There it is equals two zero point 737 Yeah. Similarly, please. Or to gas equals two more fixing up or two. That is 0.153. Multiplied by total placed so that is zero 985 ATM That is because too 0.1 pipe one ADM. Now, partial pressure drops. You're too again Small fraction of Kyoto. That a zero 037. multiplied by 0.98. Hi bob medium. That equals to 0.0 36 45 ADM. and praise her up at two where? But it calls to 0.062 Mole fraction of H two multiplied by total pressure. That is 0.98. Hi. A team that is close to 0.061 dear. Oh seven 18. This is our final answer for party Now, in part B So in part A Okay. In part B huge. I really guess equation tv calls to an arty therefore pressure is given That is zero 03680 M. We already calculated here. Okay, volume Given questions at a zero poor high five liter Temperature is 37°C,, that is equals two 310 Calvin. Now we can right and he calls to be divided by RT. Now simply put the values Razor is given 0.036 A. T. M. Multiplied by volume, geno point 455 l divided. But value of art that is 0.082 06 Later at um part of mali Calvin multiply by temperature That is 3 10 Calvin. After calculating we get 6.5 in 210 to the power minus port or mall. Sure. This is a number of mold. Uh mold most obscure to Excellent. This is an circle part B. Now in part C chemical reaction glucose react with one more double cross react with six small oxygen gas And it produces six small, obscure to gas Plus six them also as to So okay, in this problem, we need to how in this problem we need to how many grams of glucose would need to materialize to produce this quantity absurd. No, this is a question. So this is a mole substitute produced Excel. So we can compare this equation. Okay means six point hi to zero into 10. To the power minus poor mall. She would do six months 02 produces produces one more one more glucose. No, one more substitute produces how How many moles of glucose. So we need to divide by we can simply use no, six small see or two produces one mole glucose. Now one moves you to produces one mole C6 at 1206. Do behind it by six small C 02 The airport, we already calculated 6.5 into 10 to the power minus poor mole. See you're too well produces one mole C6 at 12 4 6. Do hide it back six small. See You too multiply by 6.5 into 10 to the power minus port mul glucose. Here we You simply You need three method to calculate it. No, this is a mall of blue codes. Mall of glucose. Mm small of glucose metabolize. We need to convert enough. You know, we means graham. So We know that moral equals two. Wait upon molecular weight. Therefore Where it equals two molecular mass. Multiplied by mall. So put the value. Wait up the weight of blue codes. It calls to this small up glucose so 6.5 into 10 to the power minus poor mall six blue coach divided by six mold. See You too and multiply by molecular mass molecular mass of glucose is 1 80 .15 g caramel. After calculating, We get a pro, we get 0.01 nine. Type it Ground that is approximately equals two 0.02 zero g and Blue Coach. This is our final answer for Part C. Sure, so 0.02 zero g and Glucose glucose needs to be met obliged to produce This quantity of CO two.

Pressure after tank is given Bonded and 101 kg. Pastor, 1.1 10 to depart. 5% or one atmosphere. Moeller molds off as to 8.6 point two. That is, for number of words of auto Total number of tools four plus 2 36 PBS call to an arty so volume you can find. And Artie, by pressure, substitute the value number about six. Our is 81314 temperature is given 2 73 plus 1 25 degrees. Searches upon pressure. 1.1 came to the power fight. So it just doing not a minute. Okay? Yeah, yeah. 0.1 night. 657 m, too. Yeah, The park. This is the answer of be parties two h two plus four to each quarter to H two after reduction After reaction, Moles of substance left it left for us to it is zero. And for oxygen, it is zero and edge to will be full. Mm. So PVs go to an arty suppression With me. Substitute is about a number of months. 8.314 Uh huh. So it is 2.673 atmosphere. Mm. He had mhm that son. Thanks for my

The first part of this question talks about the cylinder in an automobile having a particular volume of 524 cm cubed or 524 million liters. If the cylinder then is filled up with Air 74°C and point 98 Atmospheres, you're asked to calculate the moles of oxygen that are present. So we're going to need the ideal gas law and first we'll calculate the moles of total air that are present, rearranging the ideal gas law gives us the moles being equal to pressure times the volume divided by RT. The pressure is given to us at .98 atmospheres. The volume again was 524 million liters, or .5- four L. We divide that by R, which is our constant .08-06 leader atmospheres per kelvin mole Multiplied by the temperature, which will be our 74°C plus 2 73 to convert it into Kelvin. This then gives us moles of .01803. Remember, this is the total moles of air that were introduced in order to figure out the moles that are oxygen. We simply need to multiply by the fraction of air that is oxygen and this is provided at .2095. We then get moles of oxygen being equal to 3.78 times 10 to the -3. Knowing that these are the moles of oxygen that are available for one combustion that occurs inside of the cylinder, you are asked to calculate the grams of octane C eight h 18 that can be combusted from this quantity of oxygen. We need to first then right. The balanced chemical reaction for oxygen with octane C eight H 18 combines with oxygen to make in a combustion reaction carbon dioxide and water. To balance this, we recognize that we have eight carbons here and one here. So we need to put eight in front of carbon dioxide. We then have 18 hydrogen and two hydrogen. So we need to put a nine in front of the water. Now we have eight Times two is 16 plus nine moore's 25 we have 25 oxygen's. So to get 25 oxygen's out of two, we need to multiply two by 25 halves. This, however, is not an appropriately balanced chemical equation, although it is balanced because it has a fraction in it To remove the fraction will multiply everything through by two. Multiply octane by to multiply the 25 halves by two to get 25 Multiply eight x 2 to get 16 and then multiply nine x 2 to get 18. Now it should be balanced and we know the appropriate stock geometry for our calculation of the grams of octane that can be combusted with the moles of oxygen. We just calculated this will then be a conversion calculation. We have the 3.78 times 10 to the -3 moles of oxygen. Knowing the moles of oxygen and the stock geometry of the balanced chemical reaction between oxygen and octane, namely 25 to 2. We can convert the moles of oxygen into the molds of octane that can be combusted, Dividing by the 25. Multiplying by the two. Once we have the moles of octane, we can then use the molar mass of octane to calculate the grams of octane that can be combusted. So you'll need to look up or calculate the molar mass of octane C eight h 18, Which is 1 14.25 or 23 per mole. This then gives us The g of octane that can be combusted from the moles of oxygen that we just calculated in a single combustion in one cylinder of an automobile to be .0345g.

Hi, guys. Let's solve problems by 0.69 if 40 grand a month for York scientists hated in 502 milliliters visit and 20% off mockery. Oxide decomposes, then wanted the pressure off oxygen that warm side 25 sections, first of all, right down the balance chemical equation. And now we say that only to 20% off mercury oxide. Different poses. Therefore, they want the imam. Oh, marketing dockside Monterey Oxide Decomposes is 40. Graham wanted my brain 20% which means eight room No weekend has laid the number of wounds off markedly because I didn't want I'm rough. Morse again commands to rember loneliness. The Mass is 8.0 grim and one of myself. Weaponry oxide ese. 2161 Quiet nine creme for more, which gives the number number of months off. Monthly oxides 0.0 three snakes Line one. Now from the barest equitation chemical equation, we see two more soft monetary oxide decomposed street view on the one off oxygen. Therefore, the number one rule off oxygen produce these how to play by the metal oxide. Let's can relate to that which give some silver win zeal one age for fine. One off oxygen. No, we need to go with the pressure from the ideal gas law. Rina Devi is goingto Artie, or pressure's going to in Artie, divided by volume. It's given that the falling off the container is 502 milliliters and the whole building is being occupied by oxygen. It's converted to the little you need, and the temperature is going to climb. This sheer lives from one to the kill Vinay skin weekend in Central. The information in this situation to get the pressure, which gives us born 899 Jim. Therefore, the pressure off oxygen that homes is their 0.899 18.


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