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Required is Hg mm 8 and 3 at oxygen of STP)? volume (at What monoxide...

Question

Required is Hg mm 8 and 3 at oxygen of STP)? volume (at What monoxide

required is Hg mm 8 and 3 at oxygen of STP)? volume (at What monoxide



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A 100.0 -mL sample of air is collected at $21{ }^{\circ} \mathrm{C}$ and $772 \mathrm{~mm} \mathrm{Hg} .$ What is the volume of air at STP?

This question here tells us that we have, um, helium gas were to retreat as ideal at a volume of 25.2 milliliters and at 95 degrees Celsius. And where has to figure and also 892 millimeters of mercury, which is a unit of pressure and were asked to figure out what volume this would be at STP. In other words, what volume would it be at 2 73 Calvin and one atmosphere. And so the first thing we need to do is we need to convert all these units is three units that would typically used to work with for gases. And so this 95 degrees Celsius, we're gonna add 273 to that to turn into Kelvin. And so we have 368 Calvin as our temperature and for millimeters of mercury. If we divide this by 760 millimeters of mercury, which is equal to one atmosphere, the millimeters America units will cancel out there and get the number of atmospheres that we have. And that is equal to 1.173 atmospheres. And so now we have the volume, the temperature and the pressure. And so we can use the ideal gas law to figure out the amount of moles of helium gas that we have. So I'm gonna stroll down. I'm going to rewrite the ideal gas law, but solving for n the number of most. So that's PV over our team. The pressure we have right here, 1.173 atmospheres. The volume we haven't in mill leaders 25.2. But we're not divide by 1000 to get into leaders. So 10000.252 leaders we want in leaders because the units of our that we're gonna use our 0.8 to 1 leaders atmospheres per mole, Calvin and then our temperature. We have a right. Here's 3 68 Calvin and so and so Using that information, we can get the number of moles of helium gas that we have, which I plugged into a calculator and got 9.78 times 10 to the negative Fourth mulls. Great. Now we're asked to figure out the volume if we know that the temperature and pressure at STP meaning temperatures to 73 Calvin and the pressure is one atmosphere And so now we again rewrite the ideal gas law this time solving from volume. So volume equals and Artie overpay volume is gonna equal this number right here. 9.78 time sense negative fourth times 0.8 to 1 times to 73 all over one. That's just plug in all the information we have above. And that will equal a volume of 0.219 leaders, which is equal to 21.9 No.

Problem. Number 49 on Experiment is designed to build 44.5 little oxygen. Make sure that SDP So what is made by Estee Pee? That it's slathered complexion impression. The standard temperature is zero different substances, or it equal to 273. Kelly and started Pressure is 18 PM Only one atmospheric pressure pressure on the volume is given its 44.5. Since this is like the first stage I'm going like that does he won t one p one. Everyone on their goingto get it until it's 28. Yes, because on now, if you think give you substances so you're going to write 273 to get killed in Valley, which is 32. And Peter is also given the 0.8948 p. M. But we don't know we to. We have to figure it out on things. You have temperature, pressure and volume. All three parameters. We're going to a combined gas law. Andi, when you get the question for me to get the one we wanted to divided by t one and we two on, then you're going to subject or your values here on cancels out your value units here on. Ultimately, you get the under in leader, which is the 54.9 leader.

So we're given an equation here to mercury to oxide yields to mercury plus oxygen gas. And we're told that we have 10.57 grams of mercury and the question is, what is the volume of the oxygen gas that is released it STP. So we need to do a little bit of strike Yama tree followed by converting this into leaders. So first thing I would do is I would start out with our 10.57 grams of mercury oxide, and I know that I'm going from grams to moles of mercury oxide and then two moles of 02 And I don't have to go to grams of 02 because we're trying to get to leaders instead. So I'm gonna go from grams two moles. I'm using the ladder method here. So this ladder just it's really just multiplication. But it's it's written in a different way, and then two moles of 02 So ah, in this first set of and in this this middle column here, we need to find the molar mass of mercury oxide. So the molar mass of mercury is 200.6 and the molar mass for oxygen is 16. So we add those together we get 216.6 grams of mercury oxide per one mole and then for the moles of 02 vs Moles of H. Geo. We just have to look at these coefficients. So we've got two moles of mercury oxide. Every two moles of mercury oxide gives us one mole of 02 So now if we just multiply that out So we do 10.57 times one divide by 2 16.6 and then we do time is 1/2, and that gives us zero point 244 moles of 02 And from here, all we have to do is we know that the molar volume for an ideal gas is 22.414 leaders Permal. So we multiply that by 22.414 and we get 0.5 for six nine leaders or 500 46.9 milliliters of oxygen

Question 1 13 is a slight geometry question with a reaction, a combustion reaction of octane with oxygen to answer to determine the volume of oxygen and STP required to react with 18.9 kg of octane. The first thing that we need to do is obtain a balanced chemical reaction to get the balance. Chemical reaction will draw the chemical or will write the chemical reaction first. After writing the chemical reaction, we will then place coefficients of as appropriate. We balance this already in question 1 12. So now we see that if we have 18.9 kg of octane, we could go from kilograms 2 g, which is 1000 to 1, and then from grams to moles there's 114.2 g of octane per mole of octane. And then we'll go from moles of octane. Two moles of oxygen. Recognizing the relationship is 2 to 25. Then when we have moles of oxygen at S T P. One mole of oxygen is 22.4 leaders. So we get a total of 4.63 times 10 to the four leaders


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