WEBVTT
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so from a chemical equation, if you're given a
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mass or a model of one molecule, you confined
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theme Mueller Mass. The moles, the number of
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particles or the number of atoms than any other molecules
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in the equation. And whenever you're working with these
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problems, you want to make sure that your stoke
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geometry is correct. So S O. E com
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tree is very important because the number of rules that
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you're working with will change the values drastically on DSO
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. For this video will be answering a couple of
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questions that are based off of the combustion of sugar
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based on the fact that you are given a known
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quantity of sugar. And so for these problems,
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we'll be working with the fact that you are given
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to 0.77 g of shirt. And so, for
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the first problem, we want to figure out the
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number of molecules of co two that you make in
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your product set. So to do this, we
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need to figure out the number of moles, love
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, sugar, the number of moles of carbon dioxide
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and from their use. Avocado is never to calculate
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the number of molecules. So first, let's start
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with figuring out the number of bulls. Fortunate.
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So from our previous problems, we know that sugar
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has a molar mass of 342.2 g per mole.
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So let's use this to solve this problem. So
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again, whenever you're using dimensional analysis, you want
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to start with what you're given and then manipulates this
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so that you get the moles that you want.
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So here we want to multiply this. I think
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so. So the grams are on the bottom and
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the bulls are on the top. And when we
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do this, grams with cancel, so then you're
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left with moles. And when you do at the
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math, you should get zero point 008 09 walls
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of shirt. So now we figure out the bowls
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of sugar, but we need the moles of carbon
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dioxide. So then you need to multiply this by
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12 to get the number of moles for carbon dioxide
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which ends up being 0.0 971 moles. And now
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you want Thio Europe, the number of well,
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girls LCU to. So then you can use this
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by multiplying this by ABBA God Rose number which is
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6.0 to to times 10 to the 23rd particles.
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And in this case, we're working with molecules of
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CO two, so we can just write molecules to
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be more specific over one more. So then,
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if you do that out, you should get 5.8
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five times on to the 22nd molecules. Terrible molecules
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we'll see. And this should be your final answer
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to this problem. So here we show that if
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you're given the mass for one of the molecules in
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the equation, you can use it to find the
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number of molecules for another molecule in the same equation
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. Now, for the second part, we want
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to figure out the total number of hydrogen atoms on
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the product side. So we'll be doing something very
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similar to the top. Except this time we have
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an additional step where you want to figure out the
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number of hydrogen atoms, um, that you made
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on the product side. So here we're going to
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start this very similarly eso again. You have to
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0.77 g. Sure, and you want to apply
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this to multiply this with the more mass in this
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way. So This is technically dividing the grams by
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the molar mass to obtain your moles. And in
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this case, we have 11 for the U coefficient
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instead of 12. So you'll be multiplying this by
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11, and we should end up with is zero
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point 08 he nine. Sorry, that's not right
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. Um, Shana, with 0.890 malls of each
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too well and then from there, you want to
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multiply this, my Alma God Rose number to make
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sure that the units cancel out and again instead of
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just writing particles, we can write molecules, be
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more specific. And when you do this, the
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moles should cancel out. And what you're left with
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is I'm going to write the answer here 5.36 times
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10 to the 22nd molecules of water. And it's
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important to know that that's not what you're looking for
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. We have one more step, and so we
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want to figure out the number the total number of
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hydrogen atoms on the product side. Then we need
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to multiply that number by two because for every one
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molecule of water, there are two atoms of hydrogen
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, and so your final answer should be 1.7 times
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turn to the 23rd bottoms of hydrogen. And so
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, for the last problem, we can figure out
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the number of grams of a molecule and the chemical
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reaction given the number of grams for a different molecule
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and the chemical reaction, Um and so here will
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be starting out the problem the same way. So
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you want to always figure out the number of moles
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before you calculate different quantities. And so again,
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we start off with dividing the grams by the molar
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mass off what you were given to get your moles
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. And here, like the problem, we live
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. We're working with water. So then you won't
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apply this value by 11 to get 0.890 And then
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from there right now are units are in moles of
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water and you actually want to get grams, then
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two that will use dimensional analysis. And so when
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you do this by multiplying this by theme Mueller mass
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of water So the molar mass of water eyes about
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18 grams Permal. So then when we do out
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the calculation, we should get one point 60 g
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off water. And so from this video, we've
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learned that by starting out with the mass, uh
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, one component off one molecule, we can figure
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out the number of molecules or any other molecule that
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is in the same equation. In addition, we
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can also figure out the number of atoms and the
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equation based on this information. And in addition,
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we can also figure out the number of grams for
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any of the molecules in the chemical equation just from
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the mass of one molecule. And so when you
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do these problems, it's really important to keep your
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units very clear because you always want to cancel out
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units to obtain the ones that you want by using
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specific ratios. And you always want to keep in
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mind that whenever you want thio, get number of
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particles, or if your answer involves getting that you
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always want to use avocados number. And whenever you're
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looking for Grams, you want to make sure that
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you include some kind of Moeller mass in your final
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equation. And whenever you're working with atoms, you
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want to make sure that you are indeed working with
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atoms and not just molecules. Eso again, you
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want to make sure that your units are clear and
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your ratios are clear to make sure you don't make
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any silly mistakes.