hello everyone we're going to go through the uh study guide for the unit 2 test for honors camera so let's jump right into it number one sketch particle diagrams of Cold versus solid Cold versus hot solids like within the gases uh so do my solids over here liquids and gases so if we're doing Cold versus hot a little cold on the left warm on the right so for solids we want to show the particles somewhat close together uh in a nice orally fashion with one wish line and if we're warming it up we want to sew a little bit of thermal expansion where the particles are just a little bit more spaced apart you know try not to draw the particles themselves actually bigger I kind of did that accidentally um but you know the particles will be the same size but they should be spread out now for which lines we're going to go with one small wush line for the cold and one sort of exaggerated whistling for the hot so that shows that thermal expansion um you know think of like railroad buckling or the gaps and bridges for liquid we're going to show the particles a little more spaced out on the bottom of the container and as I did let's see six there so for warm I'm going to again kind of spread them out and really they have only one way to go which is up with two small whistlines on the cold liquid and two more exaggerated wush lines on the warmer liquid so again we see that thermal expansion for gases we're going to show uh evenly distributed throughout the container for both cold and warm so I did six there I'll do another six here so something like that now really without wush lines or pressure areas these are going to be identical they were both evenly distributed but the difference is the cold ones are going to get four small whoosh lines the warm ones are going to get four exaggerated most lines and we do pressure arrows as well so we're going to do small pressure arrows on the cold and larger pressure arrows on the warmth so that shows that difference now remember when it comes to gases we only see thermal expansion if it's a flexible container like a balloon in this case I'm showing a rigid container because it's the same size so we don't see a volume increase but we instead see a pressure increase question two unit conversions so remember One ATM equals 760 mmhg mmhd which also equals 760 torr and these conversions are true by definition or they're exact so we don't need to worry about sig figs for those uh all right so for the first one we have 1802 mmhg intertor and really I don't even need to write any more because well it's one to one well I mean if I want to do my conversions I'm going to do it this way times 760 torr over 760 mmhg or really I could just say one Tor over one mmhg either way the units cancel these numbers cancel and the answer is obviously just going to be 1802 Tor and and that one you could have just done you know without any math Second One ATM tutor so here I'm going to do .0028 ATM I'm going to multiply this by 760 torr per One ATM 0.0028 times 760. now I'm going to go with two sig figs not because of the 760. that's true by definition but because of the point zero zero two wait two sig figs there so I'm going to get an answer of 2.1 [Music] tour and finally for the last one 91.2 inches mercury into ATM and I put a hint there that you can convert inches mercury into millimeters Mercury the same way you convert inches to millimeters so 91.2 inches of mercury we can go directly into millimeters Mercury by using our conversions that we know like the bridge we used across from inches into centimeters so let's get my inches into centimeters so that's going to be times 2.54 centimeters per inch now technically this would be centimeters Mercury which is not a common unit but theoretically that works per inches Mercury cancel cancel and 10 millimeters Mercury per one centimeter Mercury this is just like going from inches to millimeters and now that we have millimeters Mercury back at a familiar unit we can take this into ATM by saying One ATM over 760 mmhg let's see here mmhd cancels mmhg so 91.2 times 2.54 equals times 10 equals divided by 760 equals uh and now I'm gonna go three sig figs because uh my starting point okay the the ten millimeters per centimeter that's true by definition so is the One ATM over 760. so with three sig figs 3.05 ATM okay number three what is the molar mass of propane gas c3h8 so you're going to use a periodic table for this basically we're going to do three carbons and eight hydrogens so carbon is 12.01 times 3. and hydrogen is 1.01 times 8. and we add those together in fact just to be nice and thorough I'll show the workout here so I could do this in my head that's going to be 36.03 and that's going to be 8.08 and now we add these together and I should be able to do this one here so it's a one and that's a 36 and a 44. it's just I am going to double check on the calculator 12.01 times 3 plus 1.01 times 8 yeah so 44.11 and the unit for this is grams per mole and uh you know another way I like to write this is that one mole of c3h8 equals 44.11 grams and that way you can use that in conversion factors question four how many Cobalt atoms are in a 3.25 kilogram block of pure Cobalt so number of atoms well what do we know about the number of atoms like number of well we know the mole right we know that one mole of anything equals 6.022 times 10 to the 23rd of that thing so in this case a mole of cobalt is going to be that many Cobalt atoms what else do we know about Cobalt well from the periodic table which I hopefully have a copy of yes I do from the periodic table we know the molar mass of cobalt is 58.93 grams so 58.93 grams equals one mole for Cobalt so this is really just a conversion factor problem okay so we start with 3.25 kilograms of cobalt now notice you are going to have to turn that into grams because molar mass the number on the periodic table is based in grams not kilograms so times a thousand grams per kilogram cancel cancel and that'll give us grams of cobalt I can go grams into moles so I'm going to use this Factor up here I'm going to say one mole per 58.93 grams and if I solved it out this would give me the number of moles but I'm not going to solve it out I'm going to jump right into number of atoms 6.022 times 10 to the 23rd oops 23rd atoms per mole you can abbreviate Mo m-o-l I just I'm not even sure why I wrote it out I normally don't all right these all cancel we should be able to get our answer 3.25 times a thousand equals divided by 58.93 equals times 6.022 times 10 to the 23rd and checking our sig figs looks like I'm going to go with three sig figs because of this uh 3.32 times 10 to the 25th atoms so that's how many Cobalt atoms are in 3.25 kilograms of cobalt number five what's the volume the steel tank of gas that contains 975 grams of carbon monoxide at a temperature of 19 Celsius and a pressure of 250 Point PSI so this sounds a lot like a gas problem right like a gas law problem so the question is which gas law combined or ideal well remember the combined gas law is used when you have a change you know a starting temperature final temperature or a starting volume final volume ideal gas law is used when you don't have a change and in this problem nothing's changing so I'm going to use the ideal gas law pivnert PV equals nrt oops PV equals nrt all right now what are we solving for uh well we're solving for volume so let's get that right out of the way so V is going to equal nrt over p so let's see what we have uh we need n which is number of moles does it give us moles no but what does it give us that we can turn into moles grams it gives us the grams of carbon monoxide so we're going to turn that I'll just say turn into moles and that's going to go in for n R does it give us r no but we always know R we don't need to be given r r is 0.0821 liters atmospheres over moles Kelvin temperature does it give us that yes it does but is it ready to go no because when you use the ideal gas law temperature must be in Kelvin and finally pressure it gives us pressure 250 PSI is that ready to go no because you have to use ATM in the ideal gas law that's assuming you're using our ideal gas constant of 0.0821 so we're going to do some conversions then we're going to plug in and solve for our answer all right so first of all we're going to take 975 grams of carbon grams of carbon monoxide and turn this into moles so we're going to need to use the molar mass of this so carbon is 12.01 plus oxygen which is 16.00 and the number I get is 28.01 again let's put that in context what is that is that the grams that equals one mole or is that the moles that equals one gram well it's the grams that equals one mole so I'm going to multiply this by one mole over 28.01 grams some people just memorize that if you want to go from grams to moles you divide by molar mass if you want to go from moles to grams you multiply it I like to think of it as a conversion factor so there's no memorizing anything it's the number of grams that equals one mole anyway 975 over this gives me 34.8 moles of Co and that is n we know r T is going to be 19 degrees Celsius we're going to add 273 now I don't even need to worry about the 0.15 because that would be rounded off anyway so if I get a whole number of Celsius I'm just going to get a use a whole number 273 for that 19 plus 273 gives me 292 Kelvin and for p 250 Point PSI that has to be turned into ATM so it gives us a conversion right there times one ATM per 14.7 PSI 250 over 14.7 17.006 so with two three sig figs there I'm going to go with 17.0 ATM and now we have everything we need so volume is going to equal n which is 34.8 moles it's going to be a big tank R which is 0.0821 liters times atmospheres over moles times Kelvin t which is 292 Kelvin over P which is 17.0 ATM now just to make sure I got this all right let's check our units mole cancels mole Kelvin cancels Kelvin ATM cancels ATM I'm left with liters does that make sense if we're solving for volume yeah 34.8 times 0.0821 equals times 292 equals divided by 17.0 equals checking sig figs three sig figs is what I'm going to go with because look at the numbers I wrote In Black all of those have three sig figs now you might say hold on a sec 19 degrees Celsius isn't that two well we didn't use 19 did we we use the Kelvin version of that which is 292. so with three sig figs my answer is 49.1 liters number six all right long question a 39.2 milliliter tank pretty small tank here it's filled with a gas at 35 degrees Celsius the tank is warmed up to 70 degrees celsius if the starting pressure was 1490 torr what's the new pressure again sounds like a gas law problem so which gas law are we using well I noticed that we are warming up a tank and it gives us a starting temperature and a final temperature so I know that this is going to be a combined gas law problem so P1 V1 T2 N2 equals P2 V2 t1n1 now if you're looking at that saying what that's not the combined gas law well that's a way of writing it you you may be used to seeing it this way P1 V1 over t1n1 equals P2 V2 over T2 N2 these are mathematically the same I just like it without the denominators makes it a little easier in my opinion all right now with the combined gas law we can get rid of anything that's held constant so temperature is changing we're solving for the pressure what does it tell you about volume it gives you the starting volume but notice it doesn't say a final volume and in fact I say it's a tank which is kind of a hint that it's not going to change so you can actually get rid of both volumes because it's not changing I know it gives you starting volume but it doesn't give you final so it's not changing now if you want you can plug in 39.2 milliliters as V1 and also plug the same thing in as V2 and you'd end up canceling them out anyway so I'm just going to get rid of them now volume's not changing take V out of there what else doesn't get mentioned number of moles it doesn't say anything about a leak or anything like that so I'm going to get rid of N1 and N2 so this ends up giving us just pressure and temperature changing here uh let's see we're solving for the new pressure so the new pressure is going to be P2 and uh we need to get T1 out of here so it's going to end up being P1 T2 over T1 that's my formula all right now there's only one rule when it comes to this gas law and that's the temperature must be in kelvin so we're going to turn 35 into 308 that's by adding 273 and I'm going to turn 70 oh I realize there's a typo here I'm probably going to change on the sheet that should be 70 point I want that to be to the ones place not the tens place 70 plus 273 is going to give me 343. now technically if I if I just kept it as 70 since that's rounded to the tens place your Kelvin temperature would go down as 340 to the tens place so I'm going to keep that like that uh all right pressure do we have to use ATM for this no you can use Tor and since the question doesn't ask for the new president be reported in any special unit tour is fine so P2 is going to equal 1490 tor times T2 which is 343 Kelvin divided by T1 which is 308 Kelvin Kelvin cancels Kelvin I'm left with Tor which makes sense because that's a pressure unit 1490 times 343 equals divided by 308 equals 1659.3 let's check our sig figs looks like three sig figs in each of those temperatures because remember we don't go by the Celsius we go by the Kelvin because that's what we actually use and pressure S3 so I'm going to go with 1660. tor that's part a Part B explain why on the particle level pressure increases so notice the pressure went from 1490 up to 1660. Why does that make sense well you should know that pressure is the force of the collisions over the area of the container essentially it's how hard and how often the particles are striking the walls of the container so what happens when you warm a gas up well it speeds the particles up which means you're going to have more collisions and stronger collisions with the walls so it makes sense that the Press are going to increase because you're speeding up the particles part C if the temperature doubled from 35 to 70 degrees why didn't the pressure double well the answer to that is zempster really didn't double Celsius is a man-made temperature scale it doesn't really have any fundamental grounding in in speed of particles or anything uh only in the Kelvin scale is a doubling of temperature a true doubling of the speed so in fact you can see that we didn't double the temperature it started at 308 Kelvin and it warmed up to 343 Kelvin so it only increased in temperature by about 10 percent you know like running those numbers in my head and in fact that's the same ratio that it went up by because really if you think about it with this formula it's like the t2 to T1 ratio times the starting pressure so if it went up by 10 percent press is going to go up by 10 so again the short answer is because it really didn't double because in the Kelvin it only went up by a small amount and finally Part D sketch before and after particle diagrams all right so we want to account for all four factors here so I'm going to start with my boxes um for what the temp the the volume did well notice in this problem the volume didn't change so I want to draw two similar size boxes so I get those as similar as I can yeah it's good enough uh now we're going to draw our particles now did the number of particles change no they didn't so we're going to draw the same number of particles that's 5 there was five there uh next we're going to give whoosh lines now wush lines indicate the temperature so did the temperature change yes we went from uh 35 to 70 so we warmed up so I'm going to do four smaller Bush lines on the left and for larger Bush lines on the right and finally pressure pressure arrows uh did pressure increase yes it did so I'm going to do smaller pressure arrows on the left and larger pressure arrows on the right okay so that sketch acts accurately depicts all four factors number seven 40.0 liter weather balloon is released from Earth's surface where the conditions happen to be at STP you should know what STP is STP is zero degrees Celsius and One ATM and that that is true by definition by the way that's not one Sig fig it rises up to a height at which the temperature is negative 114 Celsius and the pressure is 0.17 ATM what's the new volume so hopefully by now you realize this is going to be a combined gas law problem because we are changing things and in this case we are changing two things we are changing the temperature and we're changing the pressure we're solving for the new volume so combined gas law P1 V1 T2 N2 equals P2 V2 T1 and 1. all right is anything held constant in this problem yes number of particles it doesn't tell you whether there's any kind of leak or anything okay now notice in the last two problems n was constant it's not always going to be constant though the problem might tell you that certain number of moles are added or subtracted so don't you know be ready for that uh what else is constant nothing because we're solving for the new volume and there's a pressure change and a temperature change so in this case we're going to plug in five numbers and solve for the sixth one all right we're looking for the new volume so I'm going to rearrange for v2 so getting V2 by itself I need to get P2 and T1 out of there so I'm going to divide those so P1 V1 T2 over P2 T1 uh okay and now we're gonna start plugging in values uh so the starting pressure well we're starting at STP so it's going to be one ATM but remember this is like a perfect infinite Sig fig One ATM so I'm just going to say 1.00 okay now listen you can also report that as 760 mmhg 760 Tor 14.7 PSI 101 300 pascals you can use any presser unit you want why did I choose ATM because I noticed that P2 is an ATM and as long if you're plugging in P1 and P2 you need to be in the same unit uh okay V1 is my starting volume that's given as 40.0 liters uh T2 is the new temperature which is negative 114 but you can't plug that in we have to add 273 to that negative 114 plus 273 159 Kelvin we divide that by P2 which is the new pressure 0.17 ATM and next to that goes T1 which is the starting temperature which is standard temperature but you can't plug in zero because it's not really zero and also you can't divide by zero uh we have to turn that to Kelvin so 273 and really I'll say 0.15 because I know that's true by definition all right let's check our units uh ATM cancels ATM Kelvin cancels Kelvin I'm left with liters which makes sense for volume all right 1.00 times 40.0 equals times 159 equals divided by 0.17 equals divided by 273.15 equals checking sig figs I'm going to go with 2 because of that pressure final answer of 140 liters 140. by the way if you got the same setup but got a drastically different answer in fact I can tell you what that answer would be if you ended up getting something like 10 million liters that's going to be way uh larger than you need uh sorry smoke alarm went off but it sounds like a false alarm we good that sounds like fun um I could tell what you did there that means you divided by 0.17 times 273.15 all at once okay but by doing that you're really telling your calculator to divide by 273.15 sorry divided by 0.17 but then multiplied by 273.15 instead you want to do it one step at a time or put that in print anyway so that's part A Part B if it got colder why did the balloon expand the question right because it it cooled down and when the temperature drops in a flexible container it should shrink so why did the balloon expand well it's because the temperature change wasn't the only Factor here the pressure also changed and the pressure dropped the pressure outside the balloon dropped which allows the balloon to grow in size I mean if you think about it inside a balloon there's like a battle happening there's like a battle it's the air inside is pushing out and the air outside is pushing in now when you have a regular balloon just sitting at somewhere in the ground level these arrows are pushing with the same strength the same pressure and therefore the balloon stays the same size well if you bring this up into the atmosphere and also the atmosphere is pushing with a lot less Force so this Arrow gets much smaller and this Arrow gets much smaller and this Arrow gets much smaller and the error inside is pushing with the same Force what's going to happen to your balloon well it's going to expand and it's going to expand to the point and as the balloon expands the pressure inside is going to drop to equalize with the pressure outside so your balloon is going to end up expanding to offset that pressure so what's going on in this question is yes temperature is decreasing which should shrink the balloon but also pressure is increasing which should grow the balloon now since the balloon ends up growing in size it tells you that the pressure loss is significantly greater than the temperature gain but in percentage Point terms and what I mean by that is let's look how much did our presser drop it dropped from One ATM down to 0.17 that's more than five times loss what about our temperature our temperature went from it's not useful looking at Celsius our temperature went from 273 down to 114. nope sorry that's wrong down to 159. which is like a two times drop it dropped by like half so the pressure went down five times or so the temperature went up by two times notice that the pressure loss offsets the temperature loss and really it makes sense that the balloon grew by like 2.5 times because well actually uh a little more than three my numbers are a little off there so the balloon grew by about three times because of that uh all right part C before and after particle diagrams so uh let's start with our volume so our balloon grew in size so on the left I'm going to do a smaller container on the right I'm going to do a larger container that's my new volume it's larger uh now I'm going to draw the particles now our number of particles didn't change so however many draw on the left let's draw the same on the right so we'll do let me just do one more we'll do five and five like that uh temperature it got colder so I I want larger wuss lines on the left and smaller wash lines on the right and our pressure making sure I'm doing this right our pressure decreased yeah temperature dropped and pressure drop so we should have more pressure on the left so I want bigger presser arrows here and smaller pressure errors there okay so this one's a little trickier you got to think about each one of those all right question eight how many helium atoms are in a 10.0 liter party balloon whose temperature is 21 celsius and pressure is 0.987 ATM okay uh well looks like a gas law problem is anything changing no so this is going to be a PIV nurse problem PV equals nrt all right we're solving for number of helium atoms so that's going to be N and is number of particles however the n in this formula is not going to give you atoms it's going to give you moles so we're going to solve for n which is going to be moles and then we're going to convert moles into atoms using Avogadro's number all right so let's start working through it let's rearrange for n so we need to divide RT so n is going to equal PV over RT uh for pressure we have 0.978 ATM and that is ready to go plug that right into the pivner because it's an ATM volume 10.0 liters again you can plug it right in we divide that by R which is 0.0821 liters times atmospheres over moles times Kelvin and temperature is 21 celsius we have to turn that into Kelvin so that's going to be 294. Kelvin okay ATM cancels a ATM leader cancels leader Kelvin cancels Kelvin and I accidentally crossed out the one thing I don't want to which is moles so moles does not get crossed out it's the denominator within the denominator which means it ends up coming back up to the top as our final unit 0.978 times 10 equals divided by 0.0821 equals divided by 294 equals 0.405 moles and three sig figs looks good with that all right but the problem doesn't ask for moles and asks for number of atoms well that's easy since a mole is the same number of everything it's going to be 6.022 times 10 to the 23rd atoms per mole times 6.022 e23 sig figs Final Answer 2.44 times 10 to the 23rd atoms all right number nine conceptual question imagine a container of gas you want to reduce the pressure inside of it what three things could you do it and explain in the particle level why each of these have reduced the pressure so uh I'd say the mo the thing that comes to my mind quickly uh fastest is temperature change what could you do to reduce the pressure you can cool it down so one thing would be if you drop the temperature the pressure will decrease okay so T down P down uh now on the particle level why is this well remember pressure is all about the force of the collisions on the area of the container or in other words it's how hard and how often the particles are striking the inside walls of the container so when you cool when you cool them down they're going to move slower which means they're going to hit the walls less often and less hard so that's one explanation what else could you do well you could change the number of particles you can drop n which should also drop p this is common sense if you remove some of the air from inside of it uh there's gonna be fewer particles hitting the walls so you can have fewer collisions and less pressure and the last thing you can do is influence volume now what could you do to the volume to reduce the pressure well in this case you want the volume to increase if V goes up p is going to go down now why is that well pressure is all about the force of the collisions exerted over the area of the container so if you stretch out the area of the container you're going to have fewer for collisions per unit area so for that reason the pressure is going to drop so we see two direct proportions temperature pressure is direct where they move in the same direction number of particles and pressure is direct and we see one inverse proportion where one one goes up the other goes down question 10. uh some air pressure demos so we're explaining the world that the atmospheric pressure plays in the phone so first of all how a straw works uh so I'll do a little sketch here so here's a cup here's some liquid here's a straw and when the cups are sitting in the straw there's basically anything of the atmosphere is touching it's exerting atmospheric pressure on now significantly that means here and here but also inside the straw and of course everywhere else atmosphere is touching the walls it's testing you it's touching everything and it's pushing with a force of 14.7 pounds per square inch on everything that it touches so what happens when you put your mouth on a straw so here's here's now you putting your mouth on the straw and you're starting to use the straw now you might think of it as just you're sucking up the liquid but really what's happening is you're utilizing air pressure here so what you do is you're reducing the number of particles inside the straw so if originally this was our setup notice I'm trying to evenly distribute the air if you have this even number of particles before and you use the straw to remove air so let's take away some of these particles inside you're basically causing the number of particles to drop now how you do that is kind of another step to this and it relates to breathing when you breathe you expand your lungs which increases their volume it drops their pressure and as a result the air from outside is forced into your lungs that's really what's happening here you're essentially breathing the air inside the straw into your body so you're creating conditions for the air inside the straw to be forced up into your mouth down your trachea into your lungs but anyway that results in a loss of number of particles which therefore results in a loss of pressure inside the straw now the effect here is that the pressure inside is less so I'm going to drop that pressure error down quite a little bit and as a result the atmosphere is now pushing down on the water outside the straw harder than the air in the straw is pushing down on it so it effectively pushes down and up the straw so you're utilizing the fact that you're reducing the pressure inside by removing particles and this nicely demonstrates the NP relationship where the number of particles is reduced and the pressure is reduced let's talk about the can Crush so in this case you took a can that was filled with very hot gas you flipped it upside down into cold ice water so I'm going to show this as kind of a one step thing here you may have seen a monolith like before versus after but if I just want to show the ice bath here here's our ice bath you know you have your ice cubes and whatnot very cold water and you flip the can where it's closed on the top now and the opening is down here okay so that's like our can well inside the can we had gas gas particles and these gas particles were moving very fast you could even show whoosh lines on them if you want but before they were in equilibrium with the air so if I want to show what it was like before it's that we had pressure pushing out from inside the can and pressure pushing in from outside the can now you might be thinking hold on a sec if the gas was hot inside the can wouldn't it be pushing harder and expand the can well no because the top was open and exposed to the atmosphere so actually you had air particles leaving to offset the pressure increase inside inside so the pressure ended up staying equal before but now you flip it and you effectively close off the opening this is now a closed container because the gas can't really make its way up into the can anymore so what changes well you dunked it in ice water so as a result the gas inside is going to cool down and it's going to cool down fast so you're slowing the gas inside the temperature is dropping inside the can what does that do to the particles well it slows them down it slows them down slows them down so now they're not able to hit the walls nearly as often or as hard so the pressure inside is going to drop as temperature drops pressure drops and notice we're no longer at equilibrium the outside air is pushing with more force over area with more pressure than inside so as a result the can crushes in now it happens to be that the sides are the weak part so it caves in the size it doesn't really cave in down just because that's where the stronger metal is so this nicely shows the p a TP relationship when you drop the temperature inside the pressure inside drops and the can is crushed and the last one is breathing breathing so with this we want to show kind of the human lungs here I'll do my best to to show a human I always botched this image here okay so there's a nose in this person's head the neck okay is the okay that's not too bad all right so here's the person's looks all right good enough open the lungs up so you can see the pathway okay so what happens when you breathe well you know you might think well you just suck in air and then you blow out air well not really what really happens is your diaphragm which is from what I understand down here it pulls your lungs down that's what happens when you breathe in the diaphragm pulls the lungs kind of down and that stretches out your lungs so when you breathe in you're really increasing the volume inside your lungs via your diaphragm so what does that do well with more space the particles now hit the walls less often per area right they're moving at the same speed and the same number of them but since there's now more surface area the pressure inside is dropping so as a result again the pressure inside your lungs drops and we end up with uh with the atmosphere effectively correcting this by forcing air into your lungs to offset it so that's how you breathe in how do you breathe out the exact opposite happens your diaphragm moves back shrinking your lungs down so the volume of your lungs drops all of a sudden you have the same number of particles moving at the same speed but the walls are closer which means you're gonna have more collisions so the pressure increases and since you now have higher pressure in your lungs that higher pressure air is going to be forced out of your lungs into the atmosphere that's what we call exhaling so that shows how a volume change will affect a pressure change and that does it for those 10 let's do number 11. which I even told you a hint at the end this is a dimensional analysis problem AKA conversion factors this is not a gas law problem remember dimensional analysis or conversion factors is a skill that we're going to be using all year long so it's important that you know this methane CH4 is a natural gas that's utilized as an oil alternative when it undergoes combustion it reacts as follows so you can see the equation there don't be thrown off by the fact that we haven't done chemical equations like this you can still figure this one out a student decides to combust 32.0 grams of methane in the lab and once used as knowledge of dimensional analysis connected the volume of CO2 that can be produced he knows that 6.022 times 10 to the 23rd molecules has a mass of 16.0 grams in fact you could have worked that one out because we know that that's one mole and if you look on the periodic table CH4 essentially 12 plus 1 1 1 that's 16. and one molecule CH4 produces one molecule of CO2 you can see that here where one of these produces one of these he also knows that three molecules of CO2 have a mass of 2.19 times 10 to the negative 22nd grams and that the density of CO2 at this temperature is 1.13 grams per liter what volume in liters of CO2 can be produced from 32.0 grams of CH4 so this is really a conversion factor problem our starting point is 32.0 grams of CH4 and we're looking to turn that into liters so what can we do with our grams first well it tells you that 16 grams uh here we go this many molecules has a mass of 16 grams so we can turn grams into molecules so 6.022 times 10 to the 23rd molecules per 16.0 grams grams cancels grams and now we have molecules well what can we do with molecules says here he knows that three molecules of CO2 let me change the color there three molecules of CO2 has a mass of this so I'm going to multiply this by 3 uh not three molecules whoops times 2.19 times 10 to the negative 22nd grams per three molecules and that's going to cancel molecules molecules what else do we know well we know the density 1.13 grams per liter now I would strongly encourage you to be able to use density as a conversion factor and if I see 1.13 grams per liter I can rate that as 1.13 grams equals one liter and I can now use this conversion factor as one liter per 1.13 grams and notice I can now solve this just like a conversion factor 32.0 times 6.022 e23 equals divided by 16. don't forget that equals times 2.19 times 10 to the negative 22nd equals divided by 3 equals divided by 1.13 I get a final answer checking all my sig figs looks like three my answer is 77.8 liters okay that's as we do it