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Question 23 (1 point) Suppose balloon is filled s0 that its volume is when the pressure 750 torr and the

Question:

Question 23 (1 point) Suppose balloon is filled s0 that its volume is when the pressure 750 torr and the tempcrature is 249C. What volume will occupy if it rises to an elevation Ihere the pressure 375 tOrr and the temperature is 12*C? Enter only number (unit is L) Question 24 (1 point) What [s the mass of BaClz in a 0.250 L bottle of 0.200 V BaCl2? In gram; enter only number


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The vapor pressure of $\mathrm{I}_{2}(s)$ at $30^{\circ} \mathrm{C}$ is $0.466 \mathrm{~mm} \mathrm{Hg}$. (a) How many milligrams of iodine will sublime into an evacuated 750.0 -mL flask? (b) If $3.00 \mathrm{mg}$ of $\mathrm{I}_{2}$ are used, what will the final pressure in the flask be? (c) If $7.85 \mathrm{mg}$ of $\mathrm{I}_{2}$ are used, what will the final pressure in the flask be?

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For this problem, we are gonna be using this balanced chemical equation. And for this question, they are wanting us to find the final pressure in tour as well as the partial pressure of I two in Tor. For us to solve this problem, they tell us that the volume is equal to 2.50 leaders of the container. The temperature is equal to 250 Calvin, and they tell us that the partial pressure of F two is equal to 350 tour, and this is equal to 0.461 atmospheres. They also tell us that we have a mass of I two equal to 2.50 g.

All right, so to start out this problem, we need to figure out what are limiting. Reactant is gonna are limiting region is gonna be whether it be that I two r f two. So let's calculate the molds of F two. So n is gonna equal p v over r t So n is gonna equal are partial pressure and atmosphere. So 0.161 atmosphere were in times that by the 2.50 leaders we're gonna be using our is equal to 0.8 to 1 leaders atmosphere per mole, Calvin and were at times is by 250 Calvin.

So our moles of F two is going to be 0.562 moles of F two. And so now that we have the most F two, let's figure out the moles of I to All right, so we know that the mass that we have by two is equal to 2.50 g, and we also know that the molar mass is equal to 253 point 81 g per mole. So to get the number off moles, we're just gonna take our mass that we have 200 or 2.5 g and we're divide that by its Moeller mass 253.8 1 g per more. So our moles of I two is going to be 0.0 old 985 moles of I to are right. So let's figure out the limiting region.

So if I to was the limiting re agent, I'm just gonna abbreviate that l r moles of product would equal 0.0 nine a five moles I to all right. And if we go back up here to our balanced chemical equation, we see that we have one mole of I two for every two moles of product that we have. So we would take this and we would divide that by two moles of our product and divide that by one mole of I to all right, So our moles of product would equal 0.19 seven moles of I F seven. Now let's see what it would be if I have to was the limiting region. So this would equal moles Product would equal 0.562 moles of F two and for every two moles of product two moles of our product we know from our bounce chemical equation We have seven moles of F two so our moles of our product would equal 0.161 Moles are I f seven product.

So if we look at the most of product, we get from I to and the moles of product that we get from F two F two, we see that we have less product than we would have. It was I too. So f two is the limiting region. So let's write that out. F two is limiting Region are right.

So now that we found our limiting re agent, we need to figure out how many moles of I two are actually reacting. So most I to that are reacting. And this is gonna equal the moles of F two 0.562 moles of F two. And we know that for one mole I two, we have seven moles of F two. And this is just to figure out how many moles of I two are reacting with F two.

So when we calculate this out, we have 0.803 moles of I two reacting with F two for this problem. All right, I'm gonna write out are balanced chemical equation again. All right, so I'm gonna set up a nice table so ice I and I stands for initial C stands for change, and I see usually stands for equilibrium. But I'm gonna put F because it would go so far to the right that it wouldn't go back into equilibrium. So we know that initially we have 0.56 two moles of F two in the reaction.

We also know that we have 0.98 five moles of I two. We calculated those in the beginning and we know that in the beginning, like initially, we only we don't have any product. So that zero for the change Well, we know that after his eliminate reagent, so all of that is gonna be used up, so that would be our change. And we figured out how many moles would be reacting with off of I two with f two. So 0.0 a 03 moles would be reacting with f two, so that would be the change.

And we're gonna do for every two moles of our product, we have one mole of our I too. So we're gonna take two times that part of our I to so that would be the change for our product. So, for our final R F two is gonna be zero for I two, we're gonna take the difference from this first initial with the change. So we do that it's going to be 0.182 and then we're just going to multiply this out for our product and that will be the final. So we calculate that out it 0.1606 So that is how we set up our ice table.

So now that we know the changes, we need to find out how many moles total we have at the end. So moles total and this is going to equal the moles of I to that. We have so 0.0 182 moles of I to that we have plus are 0.1606 moles of our product are wrecked. So are total number of moles is gonna equal zero point 0.1788 moles. All right, so now we want to find our final pressure.

So we're just gonna use P is equal to an R T over our volume. So P is gonna equal that number of moles that we just calculated 0.1788 moles. We're in a times that by our value so 0.8 more to one leaders atmosphere per more Calvin, or in times that by our temperature. And they tell us that this product or this reaction is heated up at 550 degree Calvin. So where do you use that temperature? And we're going to divide this entire thing by the volume, which is 2.50 leaders.

All right, so our pressure final is gonna equals 0.323 atmospheres. We're just gonna times this by 760 tour per atmosphere. So our pressure is going to be 245 point 48 tour, and this is our final pressure for the reaction, and they're also wanting us to find the partial pressure of I to So the partial pressure of I two is gonna equal the total pressure times the mole fraction of I to all right. So the mole fraction of I two is just the moles of AII to that we have at the end of the reaction divided by the total number of moves, so are partial pressure of I two is going to equal the total pressure, so I'm just gonna keep it in atmospheres. 0.3 to 3 atmospheres.

I went times that by 0.182 moles of I two and divide that by the total number of moles, which is your 20.17 88 moles total that we have at the end of our ice table. So we calculate this out, are partial pressure of I two is gonna be 0.3 to 9 atmospheres again. Just to get this into tour, we're just gonna times it by 760 tour per atmosphere. So we have 25.0 tour as our partial pressure for I to all right, and that's how we solve this problem..

For this question. Well, first ignore the presence of nitrogen and just calculate the pressure associated with the decomposition of the silver oxide. When one mole of silver oxide decomposes, we get to, I went to molds of silver oxide decomposes, We get one mole of oxygen. So let's take the gram silver oxide converted to moles. Then use the stock geometry to go from old silver oxide, two moles oxygen.

This will be the moles of oxygen introduced into the 75 millim container. Knowing the moles of oxygen and the ideal gas law with the temperature and volume, Add 2-73 to the 320 to get it in, Kelvin convert the 75 ml into leaders And we get 7-7 atmospheres. Now let's not forget. We started with nitrogen that won't affect the partial pressure of the oxygen just introduced. But when we heat it up The silver oxide to produce the oxygen from 32°C to 320°C..

We also changed the pressure associated with nitrogen. It was originally 7 60 tour, or one atmosphere, multiplying that by the new temperature divided by the old temperature Gives us the new pressure 1.94 atmospheres. According to dalton's law, partial pressures will just some these two together to get the total pressure, Which will be 9.15 atmospheres. Or, if you like, divided by multiply it by 760 to get 6.96 times 10 to the three tour..

Solving party of this problem. So Boyle's law from Boyle's law I can write even be, one is equal to B two, B two. So here we have to calculate the value of P. Two which is equal to even be even by B two, which is equal to 29.4 tall multiplication 35. Later.

Bye 1 50 ml multiplication, one letter by 1000 ml. I just changed the unit which is equal to 13860 So changing the unit, I just multiplied by 1 80 M. By 7 16. God we can't simplification. Finally I get the value age 18.28 P.

M. Now going forward and solving part B. So here I can like B two is equal to B. One B. One by P two which is equal to 59.4.

So multiplication 18 PM by 7 60 tall, multiplication, 25 liters by 10 80 M. On solving it, I get the answer at 0.27 gold leader..

When were given pressure and volume of a gas, and we changed one of the variables. We can see how that affects the other variable. This is called Boyle's Law. In the equation that we use is P. One V one equals P two V two, where P one is the initial pressure.

The one is the initial volume. P two is the new or final pressure, and V two is the new or final volume. So if we're given an initial pressure and volume and in this case a new pressure, we can find the new volume. So if we know that my initial volume is 50 leaders and my initial pressure this 760 millimeters of mercury, if I change the pressure to 725 millimeters of mercury, I can calculate the new volume by multiplying my initial pressure times my initial volume and this will equal my new pressure times. My new volume dividing both sides by 725.

I find that my new volume is 52 point for leaders if my pressure is given in different units. So if, for example, the pressure changes to two atmospheres I have to first change that two millimeters of mercury, since that's what my initial pressure was in. And I know that every atmosphere is equal to 760 millimeters of mercury. So I just multiply two by 760 to give me pressure of 1520 millimeters of mercury. And then I can solve it again.

Using P one V one equals P two v two. My initial pressure times. My initial volume is equal to my new pressure times, my new volume surviving both sides. By 15 20 I knew volume Miss 25 leaders. Here's another example of a pressure and atmospheres.

We reduced the pressure 2.500 atmospheres changing at 1st 2 millimeters of mercury by multiplying by 760 or 380 millimeters of mercury. Then I could solve the problem using my initial pressure times. My initial volume equals my new pressure times. My new volume dividing both sides by 380 gives me that my new volume is 100 leaders. And finally, 1/3 unit of pressure is the tour.

So I've changed the pressure 850 tour. So I have to change that two millimeters of mercury before I do my calculation and one millimeter of mercury is equal to one tour, so it's simply the same number. So my final pressure is 850 millimeters of mercury. So my calculation is 760 times 50 equals my new pressure. Sometimes my new volume divide both sides by 850 gives me a new volume of 45 leaders..

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