We spent some time talking about significant figures. We've talked about sigfigs and calculations. We've even talked about dimensional analysis at this point. So, let's go ahead and put all of this to practice and see if we can do some more complex examples. Some simple too complex examples. First up, just some more straightforward examples. The first one, convert 0.00275 deciliters to microL. Now some students will be completely comfortable with converting street from desi to micro and if you can do that and you can do it correctly great I don't mind that's not the problem but most students need to convert through the base unit meaning in this case through liters. So I recommend we convert from deciliters to liters first and then liters to microl because most students have those root metric prefix conversions memorized and they can't convert between not through the base. So first thing I want to do dimensional analysis write down what you want to know. I want to know the number of liters. I'm told I have 0275 deciliters. Second thing you do is write down what you want to know. 0275 deciliters. Now you're going to apply conversions to get to your goal unit or your target unit. Again, I am going to convert through liters because that is the conversion. I know I know my all of my conversions compared to my base unit, which in this case is a liter. Your base units are liters, meters, um, my brain is tired, grams, sorry, I was like, what is the third one? Those are your base units. Okay, so that's what I'm going to be converting through. So, I'm going to convert to liters and cancel out my deciliters. Again, you can write this here if you want. There is that is in the numerator, so I can easily put divided by one if I need to see it in the fraction form. I don't mind either way. Another thing students often do is they do what we call railroad tracks is they write their conversions like this. So they would write 0.00275 deciliters here and then the one and the liter and the deciliter here. My multiply sign here is the same thing as this bar right here. Okay? I have no preference on how you guys do it. I do though need to know my conversion between liter and deciliter. I can remember however whatever way makes sense to your brain the most. But for me, I remember it best by remembering that one liter is 10 deciliters. At this point, my deciliters have canceled out. If I would stop the calculation here, I would have liters. But I don't want liters. I want microL. So now I need to cancel out my liters. I know to put in the denominator because it's in the numerator here. The only way to get to cancel is to put in the denominator over here. And I'm trying to get to microL. I know it takes 10 6 microL or a million microL to equal one liter because I have my metric conversions memorized. That cancels out my liters. I've got my micro, which is what I want. That's the goal of the question. So now I just need to plug it in my calculator and I need to plug it in correctly. So in my calculator I'm going to type in 0275. I'm going to hit the divide button. Enter 10. And I'm going to hit the multiply button and I'm going to put 10. And I'm going to look for my little carrot symbol. Most calculators have this. Some will be a little bit different. Some might be x raised to the y. You need to figure out what your calculator does. My calculator does a little carrot and I hit six. And that means 10 6. You should get a numerical answer of 275 microL. So make sure that you are plugging your calculator correctly so that you're getting the right answer here. The next question, the density of liquid mercury is 13.6 g per mill. Express this in units of feet of pounds per feet cubed. So um uh one sorry one thing to back to the last question. The last thing I want analysis is you want to ask yourself does my answer make sense? I have a hard time sometimes with conversions with these like visually seeing if the answer makes sense. But I do know that a microL is smaller than a deciliter. Therefore, if I'm converting to microL, it the number should look bigger. So, because it takes less, it takes a lot of microL to equal to one deciliter. I expect it to be a bigger number in when it is in microl form. So, in that sense, yes, the number does make sense. After that, I have to trust my calculating skills. Okay. So the next question, the density of liquid mercury is 13.6 g per mill. Express this in units of pounds per feet cubed. So the first thing I want to do is write down what I'm trying to find. I'm trying to find the number of pounds per feet cubed. Second thing I do is I write down what I'm given. I'm told I have 13.6 g per mill. What that means is 13.6 g in every 1 mil of sample. So, if I had exactly 1 mil of sample, it would weigh 13.6 g. Or if I go out to a balance and I weigh out 13.6 g of liquid mercury, the volume of that would be exactly 1 milll. But I want that in units of pounds per feet cubed. So, I'm going to be converting the numerator and the denominator. I don't care which one you start with. It makes no difference to me. I'm going to start with the grams to pounds because it's an easy easier conversion for me. It's one one direct conversion. I don't have to do multiple conversions for that one. So, I'm going to do it to get out of the way. I know that one pound is 454 g. It's one of the conversions I have memorized. I also know that this is not exact. This is three sigfigs. This is an exact conversion, so it didn't limit sigfigs. This is exact, so it didn't limit sigfigs. This had one, two, three sigfigs in its um number. So, I kept the question with three sigfigs here. Down here, the 13.6 has three sigfigs. The one is exact. That's just per 1 mil. 1.00 repeating however many um zeros you want on that milliliters. But the pound to um gram conversion is not exact. It is 454 grams. But that is only three sigfigs there. That cancels out my grams. And now I've got my pounds, which is what I was trying to find. So my numerator is done. Now I need to work on that denominator. I need to get the denominator to the right spot. So I'm trying to get to out of mills into feet. I have to think about the conversions I have here. Okay, what conversions do I have that can get me to feet? Well, I do know that 1 milll is equal to 1 cm cubed. That is a conversion I have memorized and that is an exact conversion. We call it 1 cc in the medical field. 1 cm cubed. So 1 milll is exactly 1 centime cubed. Now notice mathematically if I plug in the calculator it's not doing anything to my number but I have to show it because I have to show how I've converted it. This cancels out this conversion here. My milliliters here cancel with my mills here. Notice the mills in the denominator here. So I need it in the numerator over here to cancel it out. But now I've got cime. So again, if I stop the calculation here at this point, I've cancelled out my mills, but I've got centimeters cubed. I don't want centimeters cubed. What I want is feet cubed. So I clearly am not done with my conversions. I also know I also have memorized that 1 in is exactly 2.54 cm. So, I need my centimeters in the numerator, 2.54 cm to 1 in. I need my centimeters in the numerator because I need to cancel it out of my denominator. But be careful there. Notice there's 3 cm here and there's only one here. This 1 cm would cancel this out, but give me centimeters squared down here. That's not what I need. I need to cancel out all three of those. So to do that, what I'm going to do is I'm going to cube this entire quantity. By cubing that entire quantity, that's actually saying 2.54 the value cubed* cime cubed over the value 1 cubed over inches cubed. That looks weird. So what that this is what that actually means down here. This is what it actually means when I write that whole thing in a cube above. So now that I've cubed the whole thing, notice my centimeters cubed will cancel out. So I'm just going to draw it through the centimeters there. I don't want to cross out the cubes. So I remember that I have to cube the number itself. But it does mean I'm getting rid of that cubed centimeter up top. But now I'm in inches cubed. Okay, I don't want inches cubed. I want feet cubed. I know that there is 12 in in every one foot. But again, I have inches cubed on here. I need inches cubed to cancel and I need feet cubed in my final answer. So I am going to cube this whole thing. Again, by doing it like that, what that actually means is the numerical value 12 cubed* in cubed 1 cubed cubed. So, this cancels out my inches and gives me feet cubed. I just move this over so I can fit it. When I do the math here, when I calculate this in the calculator, um, let's think about sigfigs really quick. Okay, I know that I have three sigfigs at 13.6 g for the density. I have three sigfigs in the conversion of pounds to g. I know this is an exact conversion. I know this is an exact conversion because I have it memorized. I It's one of the few numbers like that that's an exact conversion. This is an exact conversion. If you're converting within the same system, like so within the metric system, it's exact or within the US system, it's exact. So overall, I need three sigfigs in my final answer. I'm going to round this to 8.49 pounds per feet cubed. Most of the time I will write that as pounds per feet cubed. It means the same thing. Okay, but let's talk about how I put this in my calculator to get the right answer. First, let's talk about how I got my units to cancel. I saw my grams canceled here. Okay, I got my pounds, which was my ending goal. I saw my mills cancel here. Cime cubed canceled with centimeters cubed. inches cubed cancelled with inches cubed giving me feet cubed. But again, you need to be able to plug this in your calculator. So I highly recommend you practice this to make sure you're plugging in correctly. Orders of operation does not matter if you do all addition or I mean all multiplication first or all division first or mix them up. They are equivalent in the calculator. So what I mean by that is when I plugged this in my calculator, I wrote 13.6 6 / 454 times 2.54 raised to the 3R. You may have a cubed button on your calculator. Some people do * 12 raised to the 3r. You need to make sure that you're cubing those numbers. If you need to do it in parts, that's fine, but you need to make sure that you're cubing those cubing this number only and cubing this number only. And by entering like this in my calculator, I get the right answer. I could have also done all the multiplication and done this at the end. It makes no difference on the actual calculator. So again, I recommend practicing and playing around with it a little bit, making sure you're comfortable using your calculator, but you should get an answer here of 849 pounds per feet cubed. Let's work our next example. If a person weighs 153 lbs on Earth, what is their mass on Jupiter where the force of gravity is 236% that of Earth? So here we're trying to use percent as a conversion factor. What I want to find is the number of pounds. I'm told the person weighs 153 lbs. And there is three sigfigs in this um measurement. Again, you can write divided by one here. That makes no difference to me if you do that or not, but I know some students like because they feel more comfortable with it. percent it's 236%. So what that means is it is 236 out of 100. If it was 78% it would be 78 out of 100. It if it was 159% it would be 159 out of 100. If it was 0.5% it would be 0.5 out of 100. Okay. So, it's that percent over 100. 236 over 100. The 236 has three sigfigs in it. So, for ease, I'm just going to get the 100 to match it. The number underneath. So, like here, if 78 over 100, I would probably say my sigfigs are right here. The 78 is going to limit it. The 100 isn't. The 100 is just the percentage it's out of. Use your new um your percentage to determine your sigfigs of your number here. Overall, 153 has three sigfigs and 236 has three sigfigs. So, my answer will have three sigfigs. So, I will round this to 361 pounds. Okay, what about this next one? An irregularly shaped piece of magnesium with a mass of 11.81 81 g was dropped into a graduated cylinder containing 24.32 ml of water. What is the total volume of the cylinder after the addition of magnesium? Let's give ourselves another piece of paper here so we can actually draw it out because this question is best done when you can see what's happening. So I've got an irregularly piece shape of magnesium. Okay, that's fine. So I've got some magnesium Its mass is 11.81 g. It was dropped into a graduated cylinder that already contained some water. So, let's draw our cylinder. We've got our cylinder. Let's pretend that's straight. Okay, we've got our grad cylinder. And our grad cylinder contains 24.32 mls of water. I add the magnesium to that cylinder. When I add the magnesium to that cylinder, the water is going to increase its volume. So if I add the water in the magnesium into there, let me just get rid of the where I wrote that so we can see it a little bit different. So now that I've added the magnesium into there, I see that the volume of the system is going to increase because it's going to have to account for the, you know, volume of magnesium as well. I want to know the total volume in the cylinder after the addition of magnesium. Well, how am I going to figure that out? Well, this total volume here is due to the volume of water, which I know plus the volume of magnesium, which I have enough information to figure out. I know that my mass of magnesium is 11.81 g. I can use density which is given right here as a conversion factor. Students often want to use their density triangle. I don't recommend it. Okay? Not saying you can't. I'm not saying you won't get the right answer. I'm saying that using density as a conversion factor is much more beneficial to your long-term success because you can keep it in one fell sweep equation. It just makes it easier. And just trust me when I tell you my students perform better if they use density as conversion factor, not the density triangle. So I want the volume of magnesium. I know I have 11.81 g of magnesium. My density is 1 cm cubed is equal to 174 g. So students see this and they really really want to write 1.74 g over 1 cm cubed. That's not incorrect. It is correct. as a conversion factor. But that won't help me here because if I put that here, 1.74 g over 1 cm cubed, my units are going to give me g^2 over cime cubed. And that doesn't help me. That's not what I need. So I need to recognize that it is an equality. Conversion factors are an equality, which means they're equal to one, which is why I can multiply them by things. But that means I can also just flip it upside down and write 1 cm cubed to 17 1.74 g because if I have one if I have 1 cm cubed I have exactly 1.74 g. Now my grams of magnesium have canceled out. Again this is magnesium here just to label what we've got and I've got cime cubed. Now my water is in milliliters and so I need to convert this to ms but I have another conversion I can use. I know that 1 milll is equal to 1 cm cubed. When I do that calculation, so this cancels out my centime cubed. Okay, I get 6.79 ms of magnesium. So I know that 11.81 g magnesium would have a volume of 6.79 mls. Now I can go ahead and add that to the water. So, I've got 24.32 mls of water plus 6.79 mls of magnesium. And in total, that is a capital M. Just trying to really emphasize it's a capital M. I get 31.11 milliliters in the cylinder. So, my total volume is 31.11 ms. Let's get into some examples that are a little bit more difficult because they involve just truthfully some more algebra to do. So, this one is definitely a challenge problem out of the textbook. Okay. Gold is alloyed with other metals to increase its hardness in making jewelry. An alloy is a combination of different metals. So like stainless steel is an alloy of multiple metals. Often times carbon is put into it etc. For the material you um the first sentence though truthfully is useless information. You have a piece of jewelry containing only gold and silver which have pure densities of 19.3 g per cime cubed and 10.5 g per cime cubed respectfully respectively. You measure the piece of jewelry's volume to be 1.25 25 cm cubed and its mass to be 20.5 g. Assuming the total volume of the jewelry is the sum of the volumes of the gold and silver it contains, what is the mass percent of gold in the piece? This question is definitely challenging. But first, let's just talk through a few things. First, the first sentence, completely useless information. I don't really need to know what gold is alloyed with to make it harder. It doesn't matter to the question itself. You have a piece of jewelry. Okay, that fine. It's containing only gold and silver. That part is important. So, let's go ahead and just highlight that contains gold and silver. It tells me that I have pure densities of 19.3 g per centime cubed and 10.5 g per centime cubed respectively. What that means is that this is the density of gold and this is the density of silver. You measure the piece of jewelry's volume to be 1.25 cm cubed. So the jewelry itself, the gold and silver combination is 1.25 cm cubed in volume with a mass of 20.5 g. We assume the total volume of the jewelry is a sum of the volumes of the gold and silver it contains. And I want to know the mass percent of gold in the piece. So what I really need to figure out is I know the total mass of the sample. I need to figure out what mass is just due to the gold. So let's think about what I know here. First, I am going to let X equal the mass of the gold, which again, gold has an atomic symbol of AU. So I'm going to be using AU throughout the work. And I'm going to let y equal the mass of the silver, which is ag. Okay. What I really want to find is the mass of gold divided by the total mass of the sample times 100%. So I can find the mass of the mass percent of gold in the piece. I know that X + Y has to equal 20.5 g because I know that the mass of gold plus the mass of silver has to equal the total piece of the sample, which is 20.5 g. But this also means that I know y, which is the mass of the silver, is equal to 20.5 gus x. Why do I care? because I'm going to have to do some substitution here to do some algebraic substitution. So I know that the mass of my silver has to be equal to the total mass of the system minus the mass of gold. Or vice versa, I could do it the other way as well. I could say that the mass of gold is equal to the total mass of the sample minus the mass of silver. I also know my volumes. Now my volume of gold would be equal to my mass of gold divided by 19.3 g per cime cubed. Remember what density is? Density is g over cime cubed. So what I did here was basically the density triangle, but you can do it in other ways as well. I can just say, you know, if I want just the centimeters cubed, which is the volume of my gold. If I know that the mass of my gold is this in g, then if I use my density as a conversion factor, 1 cm cubed over 19.3 gives me those centime cubed. It's canceling out my grams here. I'm just, you know, again, this is writing in dimensional analysis form. This is writing in density um triangle form. I don't really care which way you pick. I can do the same thing for silver. Same idea. I know that my centimeters cubed of my silver is equal to y which is my gram of my silver times my density of my silver 1 cm cubed over what is my density of silver 10.5 g. Okay, I'm just writing it like the density triangle way because it's going to be easier for my substitution in just a minute. It's the same as saying divided by 10.5 g per cime cubed. Again, this is saying the same thing. This is saying the same thing. Okay. Going to erase these just because I'm going to need this space back. I know that my total volume is the two pieces added together. So my volume total would be equal to my mass of my gold divided by 19.3 g per cime cubed plus my mass of my silver / 10.5 g per cime cubed. And one thing to recognize here is remember the centimeter cubed is in the denominator of a denominator which really makes it the numerator. So I'm just going to go ahead and move that up. Oops. Apologize. Run. But there you go. That's just algebraic manipulation. But this has to equal my total volume, which is 1.25 cm cubed. And again, that's from the question itself. Okay. So now I know that my volume total is my mass of my gold divided by my density of my gold plus the mass of my um silver divided by the density of my silver. That has to equal 1.25 cm. What I'm going to do now is I'm going to substitute in algebraically. I'm going to substitute this value in to my equation for y. So I can solve for x directly. And again x represents the mass of the silver there. So that gives me 1.25 cm cubed equal to x again which is my mass of my gold / 19.3 g. That will give me my cime cubed of my gold. plus and now I'm substituting this in for my y. 20.5 gus x over 10.5 cm cubed because again that represents the mass of my y. I solve for x here. I find a common denominator and I add my fractions. So find your common denominator between these and easiest thing to do is just you know multiply them together to find the common denominator. But so I'm going to solve for x to do so I'm going to find common denominator and add the fractions. When I do that, I have 1.25 cm cubed. This fraction is going to be multiplied by 10.5. This fraction is going to be multiplied by 19.3 to get that common denominator. Okay, so I end up with 10.5x plus let's just do it in steps. 10.5 x over 202.7 plus have to do some distribution here. 20.5 - x * 19.3 that is 19.3 / 202.7. Definitely going to run out of room here. So, I'm going to have to shrink some of this down. One second. It's not working to my advantage. Okay, I'm just going to have to shrink this whole thing down. It is what it is. So the highlighting no longer matches, but I can't help that because I really need some space. I need to shrink this as well because I really need some space to write. Okay, I apologize. Um, now I've got 1.25 cm cubed 10.5x 202.7. So I need to distribute this 19.5 through. So or 19.3 19.3 times 20.5 gives me 395.7 minus 19.3x and again that is over 202.7. So I can add these two things together, these two fractions together and they're all over the same denominator at this point. Just algebraic manipulation here. I'm going to solve for x here. So 10.5x + 395.7 - 19.3x. This gives me -8.8x plus 395.7 over that 202.7. Again, that still equals that 1.25 here. I'm going to go ahead and solve for x. And when I solve for x here, I get a numerical value of 16. But remember what x represents. So I realize this is a very long algebraic problem. Students really freak out. They're like, "Are you going to do this on a test?" You guys can do this. I promise you can do it. It's hard, but you can do it if you think your way through it. Here with subtraction, I get 16. But I'm letting x equal the mass of gold. 16 g. When I calculate this out, 16 grams of silver or sorry, gold is what I just saw for 16 grams of gold and I had a 20.5 g sample multiplied by 100% gives me 79% gold. Next example here. Again, just to back up on this one a little bit, I do know that there's a lot of algebra going on here. Okay, I know it's a lot. I know it's difficult. I also fully know you can do it. And just to clarify your sigfigs here, um, just so everyone's on the same page, when I took the 10.5 and subtracted the 19.3, that's how I got the 8.8. And so because I'm following addition and subtraction rules there, it goes down to two sigfigs there. And that's why I have two sigfigs in my final answer here. I'm not saying it's an easy problem, but I'm saying that you guys do have the algebra skills to be able to do it. Okay? It's just thinking a little bit about it and thinking what represents what. This problem should be a little more straightforward. This one does kind of throw students off because it looks like a lot of words. So let's dissect the wording here. The volume of water in a large scale reservoir is often measured in units of an acre foot, which is the amount of water covering 1 acre of surface to a depth of 1 foot. Okay, we learn something new every day. An acre foot, it covers 1 acre to the depth of 1 ft. Okay, an acre foot is equivalent to 43,560 ft cubed. That's a conversion factor that is telling me 1 acre foot equals 43,560 ft cubed. A particular water tower in town was constructed to hold 2.95 acre feet of water. That's a measurement. What is the volume of water in lers? So that's what I'm trying to find. volume of water in liters that the water tower will hold. And then it gives me a couple more conversions here. It tells me that one US gallon is equal to 231 in cubed and it tells me 3.785 L equals 1 gallon. So what I want to do first is write down what I'm trying to find. What I'm trying to find is number of liters. The sec second thing I'm going to write down is what measurement I'm given. The one measurement I'm given is 2.95 acre feet. I need to get from acet to liters. I don't have a direct conversion here, but I know I need to cancel out my acre feet. So, I'm going to use the one the only conversion I have that deals with acre feet. At least I can get rid of that. 43,560t cubed is one acre foot. So that cancels out my acre feet or acre foot, whatever you want to call it. And now I'm in feet. I don't have any conversions here that have feet, but I do have a conversion here that says inches. And I do know that there is 12 inches in every one foot. But again, like the example we did previously, this shows feet cubed. This is just feet. So I need to cube this whole thing to get the units to cancel properly. Remember cubing the whole thing is the same as saying 12 cubed in cubed over 1 cubed cubed. So that's how it cancels out my feet cubed here. And now what I really have up here in the numerator is inch cubed which is good though because I have a conversion that can deal with that. My next conversion, one gallon is 2 or 231 in cubed. So, one US gallon 231 in cubed. Again, I'm trying to cancel those inches cubed. I just did. I'm not going to cross out the three here because I don't want to forget to calculate that when I put it through my calculator. I don't want to miss the three by crossing out the unit entirely. So, I'm going to leave the three not crossed out. I need to remember to calculate that. But I did cancel the unit and now I'm in gallons. I have one more conversion shown here and it's showing me gallons to liters. And that's perfect cuz that's what I need. I need to cancel out my gallons. They're in the numerator so they cancel if I put in the denominator. And one gallon is 3.785 L. Notice I've canceled all my units and I've gotten to the unit I want, which is liter. I go ahead and calculate this out and I get 3.64 * 10 6 L. So let's spend a second to talk about how to plug in the calculator and then we'll talk about sigfigs. I'm going to plug this in the calculator. I'm going to type in 2.95 and type in the multiplication symbol. I'm going to say 43 560. I'm going to type that multiplication symbol again. I'm going to put 12 carrot 3 again. Some of you guys may have a cubed button or some of you may have an xy button. It depends on your calculator. You have to practice these so you know you're plugging them in correctly. For division, it doesn't matter the order you do it in. Some people like doing all the numerator then the denominator. That doesn't matter. So you could always type in times 3.785. But do not forget to do that divide by 231. That will get you your correct answer. Now, let's talk through your sigfigs. 2.95 is a measurement. It has three sigfigs. 1 acre foot the feet cubed. I doubt that is an exact number, especially cuz it's ending in a zero without a decimal place. There is four sigfigs here. 12 in to 1 ft is exact. It does not limit our sfix. 1 gallon to 231 in is not exact. It is a measurement. It has three sigfigs. 3.785 liters to one gallon is not exact. It is a measurement and it has four sigfigs. So my least number of sigfigs is three. So my final answer has three sigfigs. Hence me rounding to 3.64 * 10 6 lers. Okay. And the galina problem. The galina problem is known among the students. It's probably their least favorite of their um questions, but I promise you, you guys can do this question. You just have to think about what you're doing and calculate it accordingly. So, in this question, we are dealing with another alloy and we're going to be dealing with percentages. And that's what I have shown over here. Sterling silver is a silver alloy containing 92.5% silver by mass. conversion factors that could be used for that are 92.5 grams of silver to 100 grams of alloy or 100 grams of alloy to 92.5 grams of silver. Now, the example we're going to use does not have silver in it, but it does have percentages like that that will be used as a conversion factor. So, I want you to see how that correlates. So, let's read the question. Lead metal can be extracted from a mineral called galina, which contains 86.6% lead by mass. There's my first conversion. A particular ore contains 68.5% galina by mass. If the lead can be extracted with 92.5% efficiency, what mass of ore is required to make a lead sphere with a 5 cm radius? If I asked you to do something with volume with a sphere on an exam or quiz, any sort of assessment, I would give you the formula for a sphere. You do not have to remember the geometric equations from whenever you guys learn those, but you should be able to do everything else. Okay, so let's walk through the question. I need the massive or required. I know that. So what I've got going on here, I've got some sample Okay, I've got this sample of material and it's an ore in which 68% of it or 68.5% is galina which is a mineral out of that galina 86.6% 6% of it is lead and lead is the elemental symbol PB out of that I can extract this with 92.5% efficiency which means that this is the amount of usable lead. So I have what I want is I need enough lead to make a 5 cm radius sphere. So I want a sphere with a 5 cm radius that's made of lead. Okay. I need enough lead to make that. So what I need to figure out is first how much lead is that? Then once I figure that out, I need to say, okay, well, if I need to make that sphere, that's how much usable lead I need to make that sphere, right? But then how much lead do I have to extract knowing I'm going to lose 7.5% of it just to the extraction process itself? But knowing that what I extracted from is galina which is only 68 um 5% of the ore itself which is only um sorry lead metal can be extracted for 86%. We'll write these down a second. So the lead the usable lead compared to the 86.6% by mass of lead in the galina versus how much I'm actually pulling of the ore itself. So, I'm kind of back calculating here to figure out what I actually need. And I'm going to have to overcompensate for how much I need. Sorry, I forgot to lock my iPad in place. I'm going to overcompensate for how much I need to account for the difference so I can get the full sphere out of this. So again, percent by mass is grams of material. per 100 gram of the mixture that you're looking at, whatever you're looking at. Okay, the percent by mass, how much of the material I'm looking at that I need versus how much I'm pulling it out of. It's telling me I have 86.6% lead by mass. What that means is I have 86.6 6 g of lead per 100 g of galina. But that's just lead. That's not extracted lead or usable lead. It's telling me I have 68.5% galina by mass, which means 68.5 g of galina per 100 g of the ore itself. It's also telling me that I have 92.5% efficiency for extraction. I apologize. I cannot write that low on my iPad. What that means is that I have 92.5 g of what I'm going to call usable lead for every 100 g of extracted lead. I know I'm going to make a sphere. And so my sphere I need a radius of 5 cm. My sphere has a volume of 4/3 pi r cubed which is my radius. Okay. The other piece of information that is not stated in here that I need but I will absolutely need is the density of lead. Now on an assessment I would give you the density. Here I need to recognize I'm missing the density. And so my density of lead is 11.29 g per 1 cm cubed. So I got a lot of conversions here. Let's try to shrink them. See if it works better than last time. I'm going to shove these over here right now. Hopefully they'll stay out of my way. I don't remember what else is on this slide or I don't remember what else I have to write on this slide. So, they may or may not be in my way in a minute. We're going to see. Okay. So, first let's figure out how many grams of lead I actually need to make this 5cm sphere. If I figure out the grams of lead needed, I can then convert backwards to figure out the grams of ore that correlates to that. So, first I'm going to figure out the volume of lead needed. I have 4/3 by R cubed. 5.0 00 cm cubed gives me 523.599 cm cubed. My sigfigs are right here, but I don't want to create roundoff area yet. So, I'm just going to carry that number through right now. That is the volume of lead needed. And again, that is usable lead needed. Like I need that to actually make the sphere. So for the grams of lead needed, I've got 523.599 cm cubed of lead needed. I'm going to use that conversion of density to convert to g. I know there is 11.29 g of lead in every 1 cm cubed of lead. So if I have exactly 1 cm cubed, I have it would weigh 11.29 g. This gives me 5911 g of lead. My um I do have three sigfigs in the question. So I could round this to 5910. I am careful about roundoff air, but when it's a zero or a one that I'm dropping, it's not going to create roundoff air here. That's how much lead I need. But I have not yet figured out the grams of the actual ore I need. And that's where the rest of the conversions are going to come into play. And this is where students struggle probably the most with this question is conceptually what's happening. What I need you to remember is that you need to buy more than you actually need. So we're going to be accounting for what we actually need and purchase or you using more. I say buy but like using more or to be able because we're going to be losing some in every part of the process. So, what I'm doing now is working backwards to compensate for that difference. If you use your conversion factors though, you should be able to get there correctly. So, how many grams of ore? I know I need 5910 g of lead, but this is usable lead. Okay, I need to make sure that I can actually use the lead I'm buying. Um, I know that for every 100 g of lead extracted 92.5 g of lead is actually usable. So here I am using this conversion here for every 100 gram of lead extracted I can actually use 92.5 g but notice remember I don't usually write it the divided by one here but this is technically over one so my grams of lead usable is in the numerator so I need to get my grams of lead usable in the denominator so by doing that the conversion version is putting the 100 g of lead extracted in the numerator. That is what's accounting for the fact that I need more lead like I need to pull more to account for the waste that's going to happen in the question. But that gets me the lead extracted. Now I know that for every let's see for every um 86.6 6 g of lead extracted. That takes 100 gram of galina. That is my conversion right here. This lead here is extracted. The amount pulled out of the galina. So for every 100 g of galina, I'm able to pull out 86.6 g of lead. I need to shrink this because I cannot fit this canceled out the grams of lead extracted. Now I've got galina and I've got one conversion left. I know that for every 68.5 gram of galina that takes 100 gram of ore. So that cancels out my grams of galina. I've got my grams of ore, which is the entire goal of the question. Truthfully, I got there. I calculate this through. So 5,910 * 100 / 92.5 * 100 / 86.6 * 100 divided by 68.5. I'll write all that out in a second. This gives me 10773 g of ore, but I need three sigfigs, which I'll discuss in just a sec. So 1080 g of ore, or you could write it as 1.08 08 * 10 4th. I have a zero preference on which one you do there. Okay. So, let's plug this in the calculator. Plug it in the calculator. I'm going to write 5 910 * 100 / 92.5. Notice I'm not using parentheses when I put this in my calculator. Time 100.6* 6 * 100 / 68.5. And then I'm just going to hit enter and it should give you the right answer. Let's talk about your sigfigs really quick. Three sigfigs. And again, that came from above. Three sigfigs here. This had four sigfigs. I had three sigfigs here because of the 5.0 here. So, three sigfigs. The 100 is not limiting sigfigs. That you can match to your um you could write 100.0 here for all I care. Match it to your percentage down here. Your percentage has three sigfigs. Again, the 100 here, not limiting sigfigs, limit it by your percentage. This three sigfigs. three sigfigs and three sigfigs in my final answer. So again, if you ask upper classmen, they don't like the glean a problem. And I get it. It's not easy. But if you they struggle because they don't like that 100 over the percentage. Students like putting the percentage in the numerator and the 100 in the bottom. But in this case, I'm having to account for the fact that I need to use up more of that or because I'm going to lose some every step of the process. And that's what the math is accounting for. So label your things carefully. If it's in the numerator in one spot and the denominator in the other, they cancel out. I absolutely know you guys can do this.