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Use the following information to identify element $mathrm{A}$ and compound $mathrm{B}$, then answer questions a and $mathrm{b}$.An empty glass container has a mass ...

Question

Use the following information to identify element $mathrm{A}$ and compound $mathrm{B}$, then answer questions a and $mathrm{b}$.An empty glass container has a mass of $658.572 mathrm{~g} .$ It has a mass of $659.452 mathrm{~g}$ after it has been filled with nitrogen gas at a pressure of 790 . torr and a temperature of $15^{circ} mathrm{C}$. When the container is evacuated and refilled with a certain element (A) at a pressure of 745 torr and a temperature of $26^{circ} mathrm{C}$, it has a mass o

Use the following information to identify element $mathrm{A}$ and compound $mathrm{B}$, then answer questions a and $mathrm{b}$. An empty glass container has a mass of $658.572 mathrm{~g} .$ It has a mass of $659.452 mathrm{~g}$ after it has been filled with nitrogen gas at a pressure of 790 . torr and a temperature of $15^{circ} mathrm{C}$. When the container is evacuated and refilled with a certain element (A) at a pressure of 745 torr and a temperature of $26^{circ} mathrm{C}$, it has a mass of $660.59 mathrm{~g}$ Compound $mathrm{B}$, a gaseous organic compound that consists of $85.6 %$ carbon and $14.4 %$ hydrogen by mass, is placed in a stainless steel vessel $(10.68 mathrm{~L})$ with excess oxygen gas. The vessel is placed in a constant-temperature bath at $22^{circ} mathrm{C}$. The pressure in the vessel is $11.98 mathrm{~atm}$. In the bottom of the vessel is a container that is packed with Ascarite and a desiccant. Ascarite is asbestos impregnated with sodium hydroxide; it quantitatively absorbs carbon dioxide: $$ 2 mathrm{NaOH}(s)+mathrm{CO}_{2}(g) longrightarrow mathrm{Na}_{2} mathrm{CO}_{3}(s)+mathrm{H}_{2} mathrm{O}(l) $$ The desiccant is anhydrous magnesium perchlorate, which quantitatively absorbs the water produced by the combustion reaction as well as the water produced by the above reaction. Neither the Ascarite nor the desiccant reacts with compound $mathrm{B}$ or oxygen. The total mass of the container with the Ascarite and desiccant is $765.3 mathrm{~g}$ The combustion reaction of compound $mathrm{B}$ is initiated by a spark. The pressure immediately rises, then begins to decrease, and finally reaches a steady value of $6.02 mathrm{~atm} .$ The stainless steel vessel is carefully opened, and the mass of the container inside the vessel is found to be $846.7 mathrm{~g}$. $mathrm{A}$ and $mathrm{B}$ react quantitatively in a $1: 1$ mole ratio to form one mole of the single product, gas $mathrm{C}$. a. How many grams of $mathrm{C}$ will be produced if $10.0 mathrm{~L} mathrm{~A}$ and $8.60 mathrm{~L}$ $mathrm{B}$ (each at STP) are reacted by opening a stopcock connecting the two samples? b. What will be the total pressure in the system?



Answers

The heating value of combustible fuels is evaluated based on the quantities known as the higher heating value (HHV) and the lower heating value (LHV). The HHV has a higher absolute value and assumes that the water formed in the combustion reaction is formed in the liquid state. The LHV has a lower absolute value and assumes that the water formed in the combustion reaction is formed in the gaseous state. The LHV is therefore the sum of the HHV (which is negative) and the heat of vaporization of water for the number of moles of water formed in the reaction (which is positive). The table on the right lists the enthalpy of combustion-which is equivalent to the HHV-for several closely related hydrocarbons.
Use the information in the table at right to answer the following questions:
a. Write two balanced equations for the combustion of C3H8;= one equation assuming the formation of liquid water and the other equation assuming the formation of gaseous water.
b. Given that the heat of vaporization of water is 44.0 kJ/mol, what is Hrxn for each of the reactions in part a? Which quantity is the HHV? The LLV?
c. When propane is used to cook in an outdoor grill, is the amount of heat released the HHV or the LLV? What amount of heat is released upon combustion of 1.00 kg of propane in an outdoor grill?
d. For each CH2 unit added to a linear alkane, what is the average increase in the absolute value of Hcomb?

But formula of mag nation nitrate iss This balanced reaction Yes, and the three and two plus three vegetable because I wanted to NGO and understood driving force for this reaction is production off among young guests. So you're on Calvert. In the initial massive funded, we just point, don't in 3 g. So let's see the myself and you convert a 20 year old on Why did the muscles and she converted 23 203 and two does extend regulated as point point 293 g minus off. Why so now to go through the massive friendly in and victory and to muscles, NGO is eggs eso 4th 3.3. So Fangio by 24.3 from so far energy on, um, also 103 and two, which is supposed to whine, uh, white off 100.9 g of empty three to divide advice 73 72.9 70. But as we know myself and your plus months Gangotri into is going to a +747 video grams. So, uh, my second shoe is this on myself in G three center is this which is the first of going seven video going for 70 g. So after calculating after her calculating, I have got Hawaii, is this point the 579 so massive Andrian and veteran toe this point verify pictograms. Uh, now because play the most percentage job, I'm J three and two most of family and recreation. 30.58 g Marshall 102 years Things Certify waiting 200.9 g of 103 and two divided US geo 0.9 g of country, which is find there is 0 to 7, only to find out more percentage off the train, too, because I maintain 0.8%. The teapot balanced reaction is three m Jesus, the industry which funds in between two on and now that's a sex literature. When the three over here to determine the limiting reaction will amass, a furnished is 17.3 g. Remember, he is since indifferent to this point program for most well trained took 24 points of your family into the orcs with 27 tingling military and Vanish Street. The 670 industry has to live 2.4 g of vanish since we're providing provided to 2.57 g of finished three Does industry is I am inconvenienced int calculate the mass of fish toe found, uh, there's, uh Those 2.57 g of ministry produces this much from massive victories 0.45 g of h two. Now to calculate the standard and still be changed for the reaction. So which the formula is a trash, not reaction is supposed to. Delta is not off products minus our national powerful. The reactions after putting all the values we got, a Delta ash not off The action is minus three honest 38.7 village.

Podcast. We're taking a look at density. So against is often denoted. Is this strange little P that is the mass of the substance w divided by the volume feet. It can also be expressed as P is equal to p M. Divided by R T. You can rearrange for em. That is density R T overpay. LaPierre's pressure are is 8.31 40 is temperature. And so in the fast part. Well, starting page the pressure of the gas in the hot air balloon show at the opening off. The entire chapter is equal to the pressure of the atmosphere outside of the balloon because the balloon is free to expand until the pressures are equalized. So moving on to the next partner, the density of the gas in the hot air balloon is less than the density of the atmosphere. So due to the last density of gas present inside the balloon than the surrounding air, that the hot air balloon will rise as a result of this. So in part seal, start fresh page because you've got a small calculation. So but the molecular mass of the dryer was calculated, so is equal to 1.2256 grams per liter, minus one multiplied by 9.98 to 1 atmosphere per liter per MOL minus one Calvin's the minus warm that is multiplied by 293 Calvin. And then that is all divided by one atmosphere, so M equal to 29.48 grams. Put mall to the minus one in the next part. The density of the gas inside the hot air balloon is calculated as follows we have. The density is equal to 29.48 g per mole, minus one multiplied by one atmosphere. Let me divide that by 9.98 to 1 atmosphere. Elita multi minus one. Calvin's minus one multiplied by 1.30 times 10 square part two degrees Fahrenheit. So what we need to be doing is changing up some units is so we need to come vote this Calvin. So we have 3 to 7.59 Calvin, So density is even 1.96 g per liter. So what we've got in the next part because the lifting capacity of the higher balloon is equal to the difference in the mass of the cool air displaced by the balloon and the mass of the gas in the balloon. So are lifting capacity equal to 1.2256 g. Take away 1.9 666 g. That's equal to not 0.12894 g, so the lifting capacity is equal to not put 1 to 894 g. Next ramped apart F. So we're calculating the volume in liters and the multiplying the volume by the density difference to find the lifting capacity of the balloon. Then we subtract the way of the balloon after converting £2 so volume in liters is equal to 3.11 times 10 to the six liters. Then the mass of the balloon is calculated. That is 4.1 times 10 to the 5 g, so that stands to multiplied by volume than in pounds. That is 884 pounds lifting capacity people to 884 pounds. Subtract £500. Lifting capacity is £384 Female parts to go in this podcast. So G, we're calculating the volume of gas produced from the combustion off propane. So fast we calculate the mass of propane. So that is 75 point 78 g. The moles of propane people to 1.72 miles. We take a look at our reaction. We look at the total volume standard temperature pressure, that is 269 0.6 liters. One last part h. So we have our reaction. Delta reach not a combustion is equal to negative 2043.96 kg joules per mole. And so that we multiply this value Our 2043.96 by 1.72 models to get 3.52 times 10 to the 3 kg jewels. So the heat released, divided by 90 minutes. So the heat lost from the hot air bag people to study 9.1 killer jewels per minute.

So for this problem, it starts off with the reaction with an unknown gas and under most gusty and says that the, um total volume of the product if gases is 500 milliliters, total temperature, total pressure and also has to find the total moles of the guests. So we need to use PV equals NRT ideo Gasol on the R value has units of leaders times HTM over moves Tom's Kelvin. So we need to get our, um, are knowns into the right units so we'll divide by 1000 to get out of milliliters into leaders. So we get 0.5. Leaders will add 273 to get 293 Coben. I will divide by 760 to get a T M and then solving for end. We get 0.704 Walls, then the next problem. So this is our answer. Nice problem. He gives us in another reaction. It says specifically for gas be the volume, the temperature of the pressure. So again we'll do the same thing. So we need to divide by 1000 we get 0.25 Leaders meet 273. So we had 293. Calvin, you should abide by 760. So we get 0.452 HTM. And then if we solve for N, we get 0.0 47 most. So then this problem or this part tells us the mass of gas. Be with 0.2 anagrams and ask justifying the density of be in grams per leader. So let's just follow what the unit say. So it's 20.218 grams and about it by the leaders. So let's use the volume of gas. Be so 0.25 leaders from right here and then we get, um, zero 0.872 grams per liter. So then, for the next port assets to find me molar mass of gas be. And what is the formula? So for this molar masses, grams per mole. So let's do the same strategy we did to find the density. So we took Just follow the units. So zero point you wanna grams divided by V Moles of gas be which were to sell for 0.47 moves and that is 46 grams per mole. This is sort of tricky, but, um, but you just have to know dot that forms and no to if you add that together, um, and then just balance the equation. So Seo three, whenever you see, see how it three just look out for that. It's a common pattern that it will form carbon dioxide as a guess, as it does in this situation. And then two further bounce the equation. Like the questions asking. So first, let's bounce H. G. So we need out of two. And then for the next part, you just sort of have to fiddle with it. So we know that we need balance the ends. So do L. It's out to here a six here and then we need to balance the oxygen's and hydrogen. And so we need you at three. There

Okay. So to start with two given information here is that when one mole of nitro Glazer and decomposes, we end up with a collection of gases that at one atmosphere and 1950 degrees Celsius take up 1323 leaders. And in the first part of the question is, if we have 0.4 moles of nitro glycerin, how many moles of gas are we going to create eso to start off? Let's figure out how many moles of gas we got from one mole of nitro glycerin. So one atmosphere 13 23 leaders 1950 Celsius. Let's go ahead and use PV equals NRT are ideal gas law we're solving for end So P V over r t equals end. And so what that tells us is our pressure is one atmosphere are volume is 1323 leaders are our value is 13230.8 to 1. And of course, as always, we need to convert into Kelvin S O. That would be 22 23. Uh, kill them. So let's go ahead and do that math out. So 13. 23 divided by 230.8 to 1 divide by to 2 to 3. That gives a 7.25 moles of gas created for every one mole of nitro glycerine. So if we take our 7.25 and multiply it by 0.4 we get 0.29 moles of hot gas being created. So that is part a of the question. Uh, part B is telling us that we're going to Ah, I would do this in a 500 milliliters flask on. Cool it to negative 10 degrees Celsius. Ah, and then the pressure inside the flask is 623 after product, A solidifies. So, uh, what we're gonna do here is we're going to look at product a solidifying some. First of all, it is how many old moles of a were present. And what is its likely identity? Well, we know that whenever we have hydrocarbons, combusting or exploding, usually we get co two and H 20 as products. And so h 20 is a very likely product that would solidify by the time we got to negative 10 degrees Celsius. But what we do need to figure out then is how many moles of gas we have left. We're going to assume that the ice, if it is ice, has no volume. It's pretty common to assume solids have no volume on and we're going to at a new page. We're going to use P V over r T equals and to figure out how many moles of gas we have left from our original 0.2 nights. So the pressure now is 623 millimeters of mercury. The volume is 0.5 leaders. The our value since now we're in millimeters of mercury is going to be 62.4 and the temperature is negative. 10 Celsius, which is 263 killed in so are in value now is zero point zero one nine. So originally it was 0.29 Now it's 0.19 which tells us that we lost zero point 01 moles of H 20 Uh uh make sure that's what the question is. Actually asked how many moles of a were present? Ah yes. So we have 0.1 moles of a were present. Uh, part C is asking us eso when when these air passed through, when what's left is passed through a tube of powdered ally to oh, gas be reacts to form Ally to CEO. Three. The remaining gas to see India collected another 500 milliliters flask, a found of a pressure of 260 millimeters of mercury at 25 degrees Celsius. How many moles of B were present and what is its likely identity? OK, well, uh, let's start with the identity cause that's pretty easy If we have alli to oh, we have to add something to it to get Ally to CEO three. Then clearly what we are adding to it is CEO, too. So for adding CO two, that's that's the identity of be there. And the question is, ah, when. So the question is how many moles of of that co two did we have. So we now have a 500 milliliters flask, 260 mil millimeters of mercury and 25 degrees Celsius. We can use the same equation using our new pressure of 260 millimeters of mercury. Another 500 millimeter flask are. Our value is still is still 62.4 and our temperature is now 25 degrees Celsius, which is 298 Kelvin. And so what we have left then to 60 times 600.5. Uh, divide 62.4. Divide to 98 gives us 0.7 is what's left. Which means that what we lost is 0.3 Moles of co two. Okay, so that's part C. Finally, we have part D on and then I guess we have party afterward. But let's move on to part D, Sephardi says, When gases cnd were passed through a hot tube of powdered copper gas, see reacted to form copper two oxide, the remaining gas D was collected in 1/3 500 militar flask found to have a mass of 5000.168 grams and a pressure of 223 millimeters of mercury, a 25 degrees Celsius. How many moles of CND were present and what are their likely identities? Okay, so moving on to a new page s So we know that when we added copper to something, we ended up with copper two oxide. So obviously this gas had to be 02 That's what gas see is and what we are going to remember here is that the remaining gas. So the nitrogen gas that's left in a 500 military flask Ah, a lot of pressure of 223 millimeters of mercury, it 25 Celsius. So going back and using our same equation. 223 millimeters millimeters of mercury millimeters of mercury s times are 0.5 leaders divide by our 62.4 and by our 298 gives us to 23 times 0.5 divide 62.4 divide to 98 point 006 moles of whatever D is, which means that since we had 0.7 before, we have point 00 one moles of oh too. And 0.6 moles of D that 0.6 moles of d we also know takes up a mass of 0.168 grams 0.168 grams. Divide by 0.6 Moles gives us a Moeller mass of 28 grams per mole and based on the ah, compounds that we have, the elements that we have in our original that tells us that are 28 grams from all is probably in two. So then the last thing we have to do is we have to add it all together. We have to get our, uh, our full balanced equation here. So we know that we end up with 0.6 moles. Pretty much everything as we've been doing this has has come out very close to the to the thousands, starting with the 0.4 moles of nitro glycerin. So if we just multiply everything through by 1000 we should be able to get our equation. So we have our four moles of the Nitro glycerin, which is C three h five in 309 And what do we get out of it? We get ah, first the water vapor and how much water vapour did we get? We got 0.10 moles, so multiply that by 1000 and we have 10 h 20 and then be waas points Here is there are three moles of CO two So that's three co two, see, was our one. Oh, too. I'm not gonna write the one. I'm just gonna right 02 And then finally six into and nice and convenient that we finished right with that end to right there. And now, let's double check. Always good to double check. Make sure that we have proper balancing here. So our carbon is not balanced properly. Why not? Did we end up with the wrong amount of carbon there? Yes, I see what I did. Okay, I subtracted this 0.0 Zahra seven from the amount of water instead of from the amount of Ah, the amount that was left the 0.19 So I should have got 0.0. Let me go ahead and fix that right here on this part. Here. It's not 0.3 It is actually 0.12 moles of carbon dioxide. And so what that's going to do here is it's gonna do same thing instead of three carbon dioxides. Of course, it is going to be 12 carbon dioxides eso Now we can see that we've got 12 carbons and we've got 12 carbons right here. We've got 20 hydrogen ins, and here they all are 20. Hydrogen is right there. We've got 12 nitrogen ins and here's 12. Nitrogen is right here and then we've got 36 oxygen's and out of those 36 Oxygen's there's 10 of them here, plus another 24 that's 34 plus another two here. That's 36. So this balances out. That is the correct equation for the explosion of nitro glycerine.


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