Worked example: Relating reaction stoichiometry and the ideal gas law

WEBVTT
Kind: captions
Language: en

00:00:00.000 --> 00:00:01.460
- [Instructor] So we're
told that silver oxide

00:00:01.460 --> 00:00:04.260
decomposes according to
the following equation.

00:00:04.260 --> 00:00:06.070
So for every two moles of silver oxide

00:00:06.070 --> 00:00:08.940
it decomposes into four moles of silver

00:00:08.940 --> 00:00:11.764
and one mole of molecular oxygen.

00:00:11.764 --> 00:00:15.670
How many grams of silver
oxide are required to produce

00:00:15.670 --> 00:00:19.640
1.50 liters of oxygen gas

00:00:19.640 --> 00:00:23.690
at 1.22 atmospheres
and 30 degrees Celsius?

00:00:23.690 --> 00:00:25.440
So I want you to pause this video

00:00:25.440 --> 00:00:26.740
and see if you can figure this out.

00:00:26.740 --> 00:00:28.530
And I'll give you a little bit of a hint.

00:00:28.530 --> 00:00:31.310
So you're used to saying,
well, if I know how many moles

00:00:31.310 --> 00:00:33.710
of oxygen I need to produce,

00:00:33.710 --> 00:00:36.910
then I need twice as many
moles of silver oxide

00:00:36.910 --> 00:00:38.870
because the ratio of moles of silver oxide

00:00:38.870 --> 00:00:41.440
to molecular oxygen is two to one.

00:00:41.440 --> 00:00:43.670
But they don't tell us how many moles

00:00:43.670 --> 00:00:45.620
of molecular oxygen we're producing.

00:00:45.620 --> 00:00:48.450
They give us volume and
pressure and temperature.

00:00:48.450 --> 00:00:51.270
But a little bit of a
hint is the ideal gas law.

00:00:51.270 --> 00:00:56.060
It tells us that PV is equal to NRT.

00:00:56.060 --> 00:00:57.330
And if we solve for N,

00:00:57.330 --> 00:00:59.060
remember N is just the number of moles.

00:00:59.060 --> 00:01:01.360
So we divide both sides by RT.

00:01:01.360 --> 00:01:06.290
We get PV over RT is equal to N.

00:01:06.290 --> 00:01:08.900
And it looks like they
gave us all of this stuff.

00:01:08.900 --> 00:01:11.130
The pressure is right over here.

00:01:11.130 --> 00:01:13.820
The volume is right over here.

00:01:13.820 --> 00:01:16.380
The ideal gas constant,
we can look that up,

00:01:16.380 --> 00:01:18.120
and then temperature,
we just have to convert

00:01:18.120 --> 00:01:21.110
this right over here into Kelvin.

00:01:21.110 --> 00:01:22.500
And to help you there,

00:01:22.500 --> 00:01:26.320
I will give you some of the
constants and the conversions.

00:01:26.320 --> 00:01:28.863
And so see if you can
have a go at this now.

00:01:29.740 --> 00:01:32.520
All right, now let's work
through this together.

00:01:32.520 --> 00:01:35.960
So the number of moles of oxygen

00:01:35.960 --> 00:01:39.630
is going to be equal to
the pressure of our oxygen.

00:01:39.630 --> 00:01:44.320
So 1.22 atmospheres

00:01:44.320 --> 00:01:46.160
times the volume of oxygen,

00:01:46.160 --> 00:01:49.970
times 1.50 liters

00:01:49.970 --> 00:01:52.520
divided by the ideal gas constant.

00:01:52.520 --> 00:01:54.100
And we have to use the right one

00:01:54.100 --> 00:01:57.260
that is going to deal with
atmospheres, liters and Kelvin.

00:01:57.260 --> 00:02:01.290
So if we're dealing with
atmospheres, liters and Kelvin,

00:02:01.290 --> 00:02:04.960
then we'll use this ideal
gas constant right over here.

00:02:04.960 --> 00:02:06.953
So divided by 0.08206.

00:02:10.180 --> 00:02:11.130
I'll write the units here.

00:02:11.130 --> 00:02:13.763
Liters, atmospheres,

00:02:15.370 --> 00:02:18.340
divided by moles,

00:02:18.340 --> 00:02:21.330
and also divided by Kelvin.

00:02:21.330 --> 00:02:23.070
So this is our ideal gas constant,

00:02:23.070 --> 00:02:24.850
and then we're going to multiply that

00:02:24.850 --> 00:02:27.960
times the temperature in Kelvin.

00:02:27.960 --> 00:02:30.130
Now they only gave us two
significant figures here.

00:02:30.130 --> 00:02:31.700
They're only going to the ones place.

00:02:31.700 --> 00:02:33.530
So let's only go only go to the ones place

00:02:33.530 --> 00:02:34.950
when we convert to Kelvin.

00:02:34.950 --> 00:02:37.700
So let's just add 273 to this.

00:02:37.700 --> 00:02:40.290
So this is going to be times 303 Kelvin.

00:02:43.420 --> 00:02:45.010
And let's make sure the units work out.

00:02:45.010 --> 00:02:47.880
That cancels with that,
that cancels with that,

00:02:47.880 --> 00:02:49.130
that cancels with that.

00:02:49.130 --> 00:02:52.310
And if we have a dividing
by moles in the denominator,

00:02:52.310 --> 00:02:53.960
then that's just going to become a moles

00:02:53.960 --> 00:02:55.360
eventually in the numerator.

00:02:56.231 --> 00:02:58.000
So this is going to be
approximately equal to

00:02:58.000 --> 00:03:03.000
1.22 times 1.5 divided by 0.08206.

00:03:07.880 --> 00:03:10.770
And then we're going to
also divide that whole thing

00:03:10.770 --> 00:03:13.270
by 303.

00:03:13.270 --> 00:03:15.040
Is equal to this.

00:03:15.040 --> 00:03:16.510
And let's see how many
significant figures.

00:03:16.510 --> 00:03:18.960
We have three, we have three,

00:03:18.960 --> 00:03:22.250
we have a lot more than
three right over here.

00:03:22.250 --> 00:03:24.730
And then we have three right over here.

00:03:24.730 --> 00:03:27.160
So I'm going to round to
three significant figures.

00:03:27.160 --> 00:03:29.523
So 0.0736.

00:03:33.200 --> 00:03:38.200
So 0.0736 moles

00:03:39.530 --> 00:03:43.780
of molecular oxygen is how many we need,

00:03:43.780 --> 00:03:46.060
how much we need to
produce to get this volume

00:03:46.060 --> 00:03:47.990
at this pressure at this temperature.

00:03:47.990 --> 00:03:50.780
And so we're going to need two times this

00:03:50.780 --> 00:03:53.390
number of moles of silver oxide,

00:03:53.390 --> 00:03:55.700
because notice for every one mole

00:03:55.700 --> 00:03:57.010
of molecular oxygen we produce

00:03:57.010 --> 00:03:59.890
we need twice as many
moles of silver oxide.

00:03:59.890 --> 00:04:02.190
So let's multiply this times two.

00:04:02.190 --> 00:04:04.920
So, times two gets us,

00:04:04.920 --> 00:04:08.477
and once again, three
significant figures, 0.147.

00:04:10.420 --> 00:04:15.250
So this is approximately 0.147 moles

00:04:16.320 --> 00:04:21.320
of silver oxide that we need to produce.

00:04:21.650 --> 00:04:23.330
But they're not asking us how many moles

00:04:23.330 --> 00:04:24.630
of silver oxide are needed.

00:04:24.630 --> 00:04:26.440
They're asking us how many grams.

00:04:26.440 --> 00:04:30.400
So we just have to multiply
this times the molar mass.

00:04:30.400 --> 00:04:33.480
So let's first figure out the
molar mass of silver oxide.

00:04:33.480 --> 00:04:34.380
I'll write it right over here.

00:04:34.380 --> 00:04:39.180
So silver oxide's molar
mass is going to be

00:04:39.180 --> 00:04:42.540
whatever the molar mass
of silver is times two,

00:04:42.540 --> 00:04:45.290
plus the molar mass of oxygen.

00:04:45.290 --> 00:04:48.553
And so let me get the periodic
table of elements out.

00:04:49.920 --> 00:04:54.503
Molar mass of silver is
107.87, oxygen, 16.00.

00:04:55.500 --> 00:05:00.500
So that gives us 107.87
for each of the silvers.

00:05:01.570 --> 00:05:06.150
And then plus 16.00 for
the oxygen, which gives us

00:05:08.409 --> 00:05:13.409
107.87 times two is equal to that.

00:05:14.930 --> 00:05:16.935
Let's see it goes to the hundredth place.

00:05:16.935 --> 00:05:20.440
And then plus 16.00 also
goes to the hundreds place.

00:05:20.440 --> 00:05:22.960
So we'll go to 231.74.

00:05:22.960 --> 00:05:24.030
Also going to the hundreds place.

00:05:24.030 --> 00:05:25.680
I'm just keeping track of
the significant figures

00:05:25.680 --> 00:05:26.880
when we're adding.

00:05:26.880 --> 00:05:31.880
So this is 231.74 grams
per mole of silver oxide.

00:05:35.110 --> 00:05:37.380
And so if we take the
moles of silver oxide

00:05:37.380 --> 00:05:42.380
and we multiply that times
231.74 grams per mole,

00:05:43.860 --> 00:05:45.480
and notice the moles cancel out,

00:05:45.480 --> 00:05:47.840
and we're just left with the
grams, which is what we want.

00:05:47.840 --> 00:05:49.810
This is going to be
approximately equal to,

00:05:49.810 --> 00:05:51.580
and we're going to end up
with three significant figures

00:05:51.580 --> 00:05:54.070
'cause we have three here and five here.

00:05:54.070 --> 00:05:57.863
So we're going to take that
and multiply it by 0.147.

00:06:00.890 --> 00:06:01.940
Gives us this.

00:06:01.940 --> 00:06:06.460
And three significant
figures would be 34.1.

00:06:06.460 --> 00:06:10.170
Approximately 34.1 grams

00:06:10.170 --> 00:06:15.170
of silver oxide is required
to produce this much oxygen.


Worked example: Calculating the amount of product formed from a limiting reactant

WEBVTT
Kind: captions
Language: en

00:00:00.950 --> 00:00:02.540
- [Instructor] So right
here, we have a reaction

00:00:02.540 --> 00:00:05.265
where you can take some
carbon monoxide gas

00:00:05.265 --> 00:00:09.070
and some hydrogen gas,
and when they react,

00:00:09.070 --> 00:00:10.660
you're gonna produce methanol,

00:00:10.660 --> 00:00:11.910
and this is actually pretty interesting.

00:00:11.910 --> 00:00:13.410
Methanol has many applications.

00:00:13.410 --> 00:00:16.818
One of them it's actually race car fuel.

00:00:16.818 --> 00:00:20.210
But we're gonna do is
study how much methanol

00:00:20.210 --> 00:00:22.170
we can produce if we have a certain amount

00:00:22.170 --> 00:00:25.430
of carbon monoxide and molecular hydrogen.

00:00:25.430 --> 00:00:27.950
So let's say, we have 356
grams of carbon monoxide

00:00:27.950 --> 00:00:31.310
and 65.0 grams of molecular hydrogen.

00:00:31.310 --> 00:00:32.900
Pause this video, and based on this,

00:00:32.900 --> 00:00:37.023
figure out how many grams
of methanol will we produce?

00:00:38.130 --> 00:00:42.030
Well, a good place to start
is by converting these numbers

00:00:42.030 --> 00:00:44.277
of carbon monoxide, this
amount of carbon monoxide,

00:00:44.277 --> 00:00:48.040
and molecular hydrogen into moles.

00:00:48.040 --> 00:00:49.810
And so to do that, we can take out

00:00:49.810 --> 00:00:52.280
a periodic table of elements

00:00:52.280 --> 00:00:54.766
and the molar mass of carbon monoxide,

00:00:54.766 --> 00:00:58.490
you can look at the molar
masses of carbon and oxygen

00:00:58.490 --> 00:00:59.780
and add them together.

00:00:59.780 --> 00:01:04.780
So 12.01 plus 16, that is going
to be 28.01 grams per mole.

00:01:09.790 --> 00:01:12.860
And if we want to convert to moles,

00:01:12.860 --> 00:01:17.550
we're gonna have to multiply
this times moles per gram.

00:01:17.550 --> 00:01:19.690
And so for every one mole, we have,

00:01:19.690 --> 00:01:24.690
we just figured it out 28.01 grams.

00:01:24.790 --> 00:01:27.920
And this is going to
be approximately equal

00:01:27.920 --> 00:01:32.920
to 356 divided by 28.01
is equal to, let's see,

00:01:34.477 --> 00:01:37.520
and we have three significant
figures here and four here.

00:01:37.520 --> 00:01:42.120
So I'll round to 12.7 moles,
approximately 12.7 moles.

00:01:44.280 --> 00:01:45.578
And then we could do the same thing

00:01:45.578 --> 00:01:48.390
for the molecular hydrogen.

00:01:48.390 --> 00:01:53.050
And here we're going, for every one mole,

00:01:53.050 --> 00:01:56.288
how many grams or what's our molar mass

00:01:56.288 --> 00:01:58.360
of our molecular hydrogen?

00:01:58.360 --> 00:02:03.360
Well, each hydrogen atom
is 1.008 grams per mole,

00:02:04.150 --> 00:02:06.950
but each molecule of hydrogen
has two hydrogens in it,

00:02:06.950 --> 00:02:11.348
so it's gonna be two times six, so 2.016.

00:02:11.348 --> 00:02:16.348
2.016 grams per mole or one
mole for every 2.016 grams.

00:02:17.140 --> 00:02:19.930
And so this is going to
be approximately equal to,

00:02:19.930 --> 00:02:21.303
get the calculator out again,

00:02:21.303 --> 00:02:26.303
65 divided by 2.016 is equal is to that.

00:02:28.020 --> 00:02:29.460
And we have three significant figures,

00:02:29.460 --> 00:02:31.890
four significant figures,
so if I round to three,

00:02:31.890 --> 00:02:36.890
it's approximately 32.2
moles, so 32.2 moles.

00:02:40.180 --> 00:02:42.385
And so the first thing to think about is,

00:02:42.385 --> 00:02:46.056
in our reaction for every
one mole of carbon monoxide,

00:02:46.056 --> 00:02:49.420
we use two moles of molecular hydrogen,

00:02:49.420 --> 00:02:53.630
and then that produces one mole
of methanol right over here.

00:02:53.630 --> 00:02:56.053
And so however much
carbon monoxide we have

00:02:56.053 --> 00:02:58.970
in terms of moles, we need
twice as much hydrogen.

00:02:58.970 --> 00:03:01.350
And so we see here a molecular hydrogen.

00:03:01.350 --> 00:03:05.710
And so two times 12.7 is going to be 25.4.

00:03:05.710 --> 00:03:08.837
So we actually have more than
enough molecular hydrogen.

00:03:08.837 --> 00:03:13.833
And so we are going to use 25.4
moles of molecular hydrogen.

00:03:18.400 --> 00:03:19.420
How did I do that?

00:03:19.420 --> 00:03:20.550
Well, it's gonna be twice the number

00:03:20.550 --> 00:03:21.780
of moles of carbon monoxide,

00:03:21.780 --> 00:03:25.930
twice this number right over
here is this right over here.

00:03:25.930 --> 00:03:27.010
And we can immediately see

00:03:27.010 --> 00:03:28.760
how much we're gonna have leftover.

00:03:28.760 --> 00:03:30.483
We're going to have leftover,

00:03:31.420 --> 00:03:36.420
leftover 32.2 minus 25.4 is 6.8,

00:03:36.513 --> 00:03:40.650
6.8 moles of molecular hydrogen.

00:03:40.650 --> 00:03:43.840
And how many moles of
methanol we're gonna produce?

00:03:43.840 --> 00:03:45.220
Well, the same number of moles

00:03:45.220 --> 00:03:46.920
of carbon monoxide that we're using up.

00:03:46.920 --> 00:03:48.336
It's a one-to-one ratio.

00:03:48.336 --> 00:03:53.080
So we're going to produce
12.7 moles of methanol.

00:03:53.080 --> 00:03:54.310
And so let me write that a year.

00:03:54.310 --> 00:03:59.310
So if I have 12.7 mole of methanol, CH3OH,

00:04:02.221 --> 00:04:04.310
how do I convert this to grams?

00:04:04.310 --> 00:04:06.554
Well, I have to multiply
this times a certain number

00:04:06.554 --> 00:04:11.554
of grams per mole so that
we can cancel out the moles

00:04:12.910 --> 00:04:15.320
or essentially the molar mass of methanol,

00:04:15.320 --> 00:04:18.030
and to figure out the
molar mass of methanol,

00:04:18.030 --> 00:04:19.606
we'll get our calculator out again.

00:04:19.606 --> 00:04:22.870
So we have four hydrogens here.

00:04:22.870 --> 00:04:27.870
So four times 1.008 is going to be that.

00:04:28.920 --> 00:04:30.230
And then to that,

00:04:30.230 --> 00:04:32.970
we're going to add the
molar mass of carbon

00:04:32.970 --> 00:04:36.370
'cause we have one carbon plus 12.01,

00:04:36.370 --> 00:04:39.512
and then plus one oxygen
in that methanol molecule

00:04:39.512 --> 00:04:40.830
is equal to that.

00:04:40.830 --> 00:04:44.240
And let's see, we will
round to the hundreds place

00:04:44.240 --> 00:04:46.810
because our oxygen and carbon molar mass

00:04:46.810 --> 00:04:49.310
is only went to the hundreds
place here, so 32.04,

00:04:52.557 --> 00:04:54.307
32.04 grams per mole.

00:04:57.000 --> 00:05:02.000
So we have 12.7 moles
times 32.04 grams per mole

00:05:04.812 --> 00:05:07.810
will tell us that we are going to produce

00:05:07.810 --> 00:05:09.250
that much methanol.

00:05:09.250 --> 00:05:11.300
And let's see, we have three
significant figures, four,

00:05:11.300 --> 00:05:12.690
so I'll round to three.

00:05:12.690 --> 00:05:15.740
So approximately 407 grams of methanol,

00:05:15.740 --> 00:05:20.643
407 grams of CH3OH.

00:05:22.840 --> 00:05:24.600
Now the next question is,

00:05:24.600 --> 00:05:28.870
what's the mass of hydrogen
that we have leftover?

00:05:28.870 --> 00:05:31.620
Well, we just have to
convert our moles of hydrogen

00:05:31.620 --> 00:05:34.100
that we have leftover to grams,

00:05:34.100 --> 00:05:39.100
6.8 moles of molecular hydrogen times,

00:05:39.210 --> 00:05:41.090
the molar mass here in grams per mole

00:05:41.090 --> 00:05:43.430
is just gonna be the reciprocal
of this right over here,

00:05:43.430 --> 00:05:45.293
so times 2.016,

00:05:49.152 --> 00:05:53.150
2.016 is going to give
us this right over here.

00:05:53.150 --> 00:05:57.240
And if we were rounding to
two significant figures,

00:05:57.240 --> 00:05:59.220
which I have right over here,

00:05:59.220 --> 00:06:03.770
that is going to give us
approximately 14 grams,

00:06:03.770 --> 00:06:07.810
approximately 14 grams
of molecular hydrogen

00:06:07.810 --> 00:06:09.833
is leftover, leftover.

00:06:11.250 --> 00:06:12.440
So we used a good bit of it.

00:06:12.440 --> 00:06:16.980
We used about 51 grams and
we have 14 grams leftover

00:06:16.980 --> 00:06:18.580
and it was carbon monoxide

00:06:18.580 --> 00:06:20.830
that was actually the
limiting reactant here.


Worked example: Calculating amounts of reactants and products

WEBVTT
Kind: captions
Language: en

00:00:01.270 --> 00:00:02.980
- [Instructor] We're told that glucose,

00:00:02.980 --> 00:00:06.570
C6H12O6, reacts with oxygen

00:00:06.570 --> 00:00:09.960
to give carbon dioxide and water.

00:00:09.960 --> 00:00:13.100
What mass of oxygen, in grams,

00:00:13.100 --> 00:00:15.950
is required for complete reaction

00:00:15.950 --> 00:00:20.380
of 25.0 grams of glucose?

00:00:20.380 --> 00:00:22.830
What masses of carbon dioxide and water,

00:00:22.830 --> 00:00:25.190
in grams, are formed?

00:00:25.190 --> 00:00:27.340
So, pause this video and see
if you can have a go at this

00:00:27.340 --> 00:00:29.660
and then we'll work through this together.

00:00:29.660 --> 00:00:33.090
All right, now first let's
just set up the reaction.

00:00:33.090 --> 00:00:34.500
So, this is going to be,

00:00:34.500 --> 00:00:37.010
we have glucose, so that is C6H12O6,

00:00:40.370 --> 00:00:42.200
is going to react with oxygen.

00:00:42.200 --> 00:00:45.633
Now, oxygen in its molecular
form is going to be O2.

00:00:46.500 --> 00:00:51.500
And what it gives is
carbon dioxide and water.

00:00:52.640 --> 00:00:54.970
Plus water.

00:00:54.970 --> 00:00:56.890
Now the next question is are we balanced.

00:00:56.890 --> 00:00:58.710
Do we have a conservation of mass here?

00:00:58.710 --> 00:01:01.370
And let's just go element by element.

00:01:01.370 --> 00:01:03.750
So first let's focus on the carbons.

00:01:03.750 --> 00:01:05.250
So, we have six carbons

00:01:05.250 --> 00:01:07.300
on the left-hand side of this reaction.

00:01:07.300 --> 00:01:10.330
How many carbons do we have
on the right-hand side?

00:01:10.330 --> 00:01:12.800
Well, right now we only have one carbon

00:01:12.800 --> 00:01:15.130
in this carbon dioxide molecule.

00:01:15.130 --> 00:01:17.240
So if we want the carbons to be conserved

00:01:17.240 --> 00:01:20.090
we need to have six carbons on
the right-hand side as well.

00:01:20.090 --> 00:01:23.530
So let me see what'll happen
when I throw a six there.

00:01:23.530 --> 00:01:25.070
So now my carbons are balanced:

00:01:25.070 --> 00:01:27.170
six on the left, six on the right.

00:01:27.170 --> 00:01:29.690
Now let's look at the other elements.

00:01:29.690 --> 00:01:34.130
So on the left-hand side I
have 12 hydrogens in total.

00:01:34.130 --> 00:01:37.650
On the right-hand side I
only have two hydrogens.

00:01:37.650 --> 00:01:38.840
So if I want to balance that

00:01:38.840 --> 00:01:40.800
I could multiply the water molecule.

00:01:40.800 --> 00:01:44.530
Instead of just one
here, I could have six,

00:01:44.530 --> 00:01:47.800
and now this would be 12 hydrogen atoms

00:01:47.800 --> 00:01:50.560
so both the carbons and the
hydrogens are now balanced.

00:01:50.560 --> 00:01:53.110
By putting that six there I
haven't messed with the carbons.

00:01:53.110 --> 00:01:54.890
And now last, but not least,

00:01:54.890 --> 00:01:57.850
let's think about the oxygens here.

00:01:57.850 --> 00:02:01.500
So on the left, we have six oxygens there

00:02:01.500 --> 00:02:03.500
and then another two oxygens there.

00:02:03.500 --> 00:02:05.532
So that's a total of eight oxygens.

00:02:05.532 --> 00:02:10.532
And on the right, I have
six times two, I have 12

00:02:10.780 --> 00:02:13.580
plus another six oxygens.

00:02:13.580 --> 00:02:16.120
So I have 18 oxygens
on the right-hand side,

00:02:16.120 --> 00:02:19.230
and I only have eight oxygens
on the left-hand side.

00:02:19.230 --> 00:02:20.840
So I have to increase
the number of oxygens

00:02:20.840 --> 00:02:22.450
on the left-hand side.

00:02:22.450 --> 00:02:23.976
So let's see.

00:02:23.976 --> 00:02:27.060
This six you can imagine
matches up with this six.

00:02:27.060 --> 00:02:29.740
So if I could somehow
make this into 12 oxygens,

00:02:29.740 --> 00:02:31.240
I'll be good.

00:02:31.240 --> 00:02:33.260
So the best way to make
this into 12 oxygens

00:02:33.260 --> 00:02:37.610
is to multiply this by
six, so let me do that.

00:02:37.610 --> 00:02:41.920
So if I put a six right over
here, I think I'm all balanced.

00:02:41.920 --> 00:02:43.740
I have six carbons on both sides,

00:02:43.740 --> 00:02:45.750
I have 12 hydrogens on both sides,

00:02:45.750 --> 00:02:50.070
and it looks like I have
18 oxygens on both sides.

00:02:50.070 --> 00:02:52.610
So I'm a fully balanced equation here.

00:02:52.610 --> 00:02:55.360
So let me get a periodic table of elements

00:02:55.360 --> 00:02:58.060
and I only need to think about
carbon, hydrogen, and oxygen.

00:02:58.060 --> 00:03:00.960
So for the sake of space, let
me scroll down a little bit.

00:03:01.990 --> 00:03:05.690
I just need to look at
this stuff over here.

00:03:05.690 --> 00:03:07.470
And so, let's first,

00:03:07.470 --> 00:03:09.280
let's see hydrogen's right up here.

00:03:09.280 --> 00:03:12.590
And we can see in this
periodic table of elements,

00:03:12.590 --> 00:03:14.970
it gives the average atomic mass,

00:03:14.970 --> 00:03:16.800
but you can also view that number

00:03:16.800 --> 00:03:21.300
1.008, as its molar mass.

00:03:21.300 --> 00:03:26.300
So it's 1.008 grams per mole of hydrogen.

00:03:27.770 --> 00:03:30.160
Now we can move on to carbon.

00:03:30.160 --> 00:03:35.160
Carbon is 12.01, 12.01
grams per mole of carbon.

00:03:39.920 --> 00:03:43.340
And then last but not
least, oxygen over here.

00:03:43.340 --> 00:03:48.340
That is 16.00 grams per mole of oxygen.

00:03:51.600 --> 00:03:53.660
And now we can use that information.

00:03:53.660 --> 00:03:56.600
I can now remove our
periodic table of elements.

00:03:56.600 --> 00:03:58.410
We can use this information to figure out

00:03:58.410 --> 00:04:01.760
the molar masses of
each of these molecules.

00:04:01.760 --> 00:04:05.270
So for example, glucose right over here,

00:04:05.270 --> 00:04:07.903
if we're talking about C6H12O6,

00:04:11.610 --> 00:04:15.540
how many grams per mole
is that going to be?

00:04:15.540 --> 00:04:19.570
How many grams is a mole
of glucose going to be?

00:04:19.570 --> 00:04:23.320
Well, it's six carbons, 12
hydrogens, and six oxygens.

00:04:23.320 --> 00:04:24.880
So one way to think about is going to be

00:04:24.880 --> 00:04:27.398
six times this, so it's going to be

00:04:27.398 --> 00:04:29.720
six times 12.01

00:04:29.720 --> 00:04:34.640
plus we have 12 times 1.008

00:04:36.640 --> 00:04:40.473
plus six times 16.00,

00:04:42.080 --> 00:04:45.720
and then this is going
to be grams per mole.

00:04:45.720 --> 00:04:47.440
And this is going to
be equal to, let's see,

00:04:47.440 --> 00:04:52.393
72.06 plus 12.096

00:04:55.820 --> 00:04:59.903
plus six times 16 is 96.00.

00:05:02.140 --> 00:05:06.310
Of course all of this
is in grams per mole.

00:05:06.310 --> 00:05:08.720
And so this is going to
be equal to, let's see,

00:05:08.720 --> 00:05:12.180
72 plus 12 is 84

00:05:12.180 --> 00:05:15.950
plus 96 is 180,

00:05:15.950 --> 00:05:18.510
180 point, and let's see,

00:05:18.510 --> 00:05:23.140
we have 60 thousandths
plus 96 thousandths,

00:05:23.140 --> 00:05:25.760
which would be 156 thousandths,

00:05:25.760 --> 00:05:30.050
so 156 thousandths grams per mole.

00:05:30.050 --> 00:05:32.100
Let me make sure I got the
significant figures right.

00:05:32.100 --> 00:05:34.070
The six, the 12 and six are pure numbers,

00:05:34.070 --> 00:05:35.870
so I'm still good with this.

00:05:35.870 --> 00:05:40.080
And then when I add
these numbers together,

00:05:40.080 --> 00:05:44.120
I need to round to as
much precision as I have

00:05:44.120 --> 00:05:45.210
in the least one.

00:05:45.210 --> 00:05:47.920
So here I go up to the hundredths,

00:05:47.920 --> 00:05:50.060
up to the hundredths,
up to the thousandths.

00:05:50.060 --> 00:05:52.740
So I need to round this,
actually, to the hundredths place.

00:05:52.740 --> 00:05:57.740
So this is going to be 180.16
grams per mole of glucose.

00:06:02.690 --> 00:06:04.780
And now let's think about
the other molecules.

00:06:04.780 --> 00:06:06.230
If we think about oxygen,

00:06:06.230 --> 00:06:08.380
I'll do that over here to save space,

00:06:08.380 --> 00:06:10.690
oxygen, that's pretty straightforward.

00:06:10.690 --> 00:06:14.170
A molecular oxygen molecule
just has two oxygen atoms,

00:06:14.170 --> 00:06:15.410
so it's going to be two times this,

00:06:15.410 --> 00:06:20.410
so it's going to be 32.00 grams per mole.

00:06:20.440 --> 00:06:23.220
And then carbon dioxide.

00:06:23.220 --> 00:06:26.280
Carbon dioxide is going to be,

00:06:26.280 --> 00:06:28.370
it's two oxygens plus one carbon.

00:06:28.370 --> 00:06:30.620
So it's going to be this plus one carbon,

00:06:30.620 --> 00:06:32.900
so that's this plus 12.01,

00:06:32.900 --> 00:06:37.750
so that's 44.01 grams per mole.

00:06:37.750 --> 00:06:41.570
And then last but not
least we have the water.

00:06:41.570 --> 00:06:43.723
And so, H2O.

00:06:45.130 --> 00:06:47.910
This is going to be two hydrogens,

00:06:47.910 --> 00:06:49.377
so that's two times 1.008,

00:06:50.320 --> 00:06:54.730
so that's 2.016 plus an oxygen.

00:06:54.730 --> 00:06:59.600
So that's going to be 2.016 plus 16

00:06:59.600 --> 00:07:01.357
is going to be 18.016.

00:07:05.290 --> 00:07:07.770
Once again, we only go to
the hundredths place here,

00:07:07.770 --> 00:07:10.170
so I'm going to round to
the hundredths place here.

00:07:10.170 --> 00:07:15.170
So 18.02 grams per mole.

00:07:15.940 --> 00:07:18.070
Now the next step is to think about,

00:07:18.070 --> 00:07:22.600
all right we're reacting
with 25 grams of glucose.

00:07:22.600 --> 00:07:25.780
How many moles of glucose is that?

00:07:25.780 --> 00:07:28.770
So we have 25.0,

00:07:28.770 --> 00:07:33.770
25.0 grams of glucose.

00:07:34.347 --> 00:07:37.030
And we want to turn this
into moles of glucose.

00:07:37.030 --> 00:07:41.060
And we know the molar mass
is 180.16 grams per mole,

00:07:41.060 --> 00:07:43.900
or we could multiply and say that this is

00:07:43.900 --> 00:07:48.900
for every one mole, per
mole, it is 180.16 grams.

00:07:50.940 --> 00:07:53.290
All I did is take the
reciprocal of this over here.

00:07:53.290 --> 00:07:55.320
And notice, the grams
will cancel with the grams

00:07:55.320 --> 00:07:57.963
and I'll be left with moles of glucose.

00:07:58.850 --> 00:08:02.490
That's going to be equal to 25.0

00:08:02.490 --> 00:08:07.490
divided by 180.16 moles of glucose.

00:08:08.660 --> 00:08:13.660
25 divided by 180.16 is equal to this.

00:08:15.800 --> 00:08:17.820
And let's see, I have
three significant figures

00:08:17.820 --> 00:08:19.590
divided by five significant figures.

00:08:19.590 --> 00:08:21.850
So I'm going to round to
three significant figures.

00:08:21.850 --> 00:08:23.593
So 0.139.

00:08:25.740 --> 00:08:30.740
So this is approximately equal
to 0.139 mole of glucose.

00:08:34.990 --> 00:08:36.840
Now, they say the first question is

00:08:36.840 --> 00:08:39.100
what mass of oxygen is required

00:08:39.100 --> 00:08:41.360
for a complete reaction of this.

00:08:41.360 --> 00:08:43.430
Well, for every mole of glucose we need

00:08:43.430 --> 00:08:46.720
six moles of molecular oxygen.

00:08:46.720 --> 00:08:48.210
And so, we're going to need,

00:08:48.210 --> 00:08:50.410
let me just multiply that times six.

00:08:50.410 --> 00:08:53.950
So times six is equal to that.

00:08:53.950 --> 00:08:56.230
And remember, I had three
significant figures.

00:08:56.230 --> 00:08:57.903
So, 0.833.

00:09:00.660 --> 00:09:01.760
So we're going to need

00:09:04.785 --> 00:09:05.863
0.833 moles of molecular oxygen.

00:09:12.490 --> 00:09:14.070
And then I just multiply that

00:09:14.070 --> 00:09:17.220
times the molar mass of molecular oxygen.

00:09:17.220 --> 00:09:22.220
So, times 32.00 grams per
mole of molecular oxygen.

00:09:26.180 --> 00:09:28.847
0.833 times 32 is equal to that.

00:09:33.020 --> 00:09:35.208
If you go three significant figures,

00:09:35.208 --> 00:09:36.041
it's 26.7.

00:09:38.988 --> 00:09:42.488
26.7 grams of oxygen, of molecular oxygen.

00:09:47.260 --> 00:09:50.100
And so we've answered the
first part of the question.

00:09:50.100 --> 00:09:53.240
What mass of oxygen is
required for complete reaction?

00:09:53.240 --> 00:09:56.130
So, that's this right over here.

00:09:56.130 --> 00:09:58.130
And then the next question is what masses

00:09:58.130 --> 00:10:01.570
of carbon dioxide and
water in grams are formed.

00:10:01.570 --> 00:10:03.610
Remember, for every one mole of glucose,

00:10:03.610 --> 00:10:05.960
we needed six moles of molecular oxygen

00:10:05.960 --> 00:10:08.070
and we produce six moles of carbon dioxide

00:10:08.070 --> 00:10:10.940
and we produce six moles of water.

00:10:10.940 --> 00:10:13.100
So this is the number of moles of glucose

00:10:13.100 --> 00:10:15.020
we input into the reaction.

00:10:15.020 --> 00:10:16.710
Six times that was this.

00:10:16.710 --> 00:10:19.020
So this is actually
also the number of moles

00:10:19.020 --> 00:10:22.860
of carbon dioxide or water
that we're going to produce.

00:10:22.860 --> 00:10:27.007
So, if we take our original 0.833 mole

00:10:29.750 --> 00:10:32.100
of carbon dioxide and I multiply that

00:10:32.100 --> 00:10:34.060
times carbon dioxide's molar mass,

00:10:34.060 --> 00:10:38.350
44.01 grams per mole,

00:10:38.350 --> 00:10:41.173
this is going to be
approximately equal to,

00:10:41.173 --> 00:10:45.870
.833 times 44.01

00:10:45.870 --> 00:10:48.340
is equal to three significant figures,

00:10:48.340 --> 00:10:51.430
36.7 grams.

00:10:51.430 --> 00:10:55.440
36.7 grams of carbon dioxide.

00:10:55.440 --> 00:10:56.600
And once again, you can see

00:10:56.600 --> 00:10:58.796
that the moles cancel with the moles,

00:10:58.796 --> 00:11:02.110
just as they did before, to
give us grams of carbon dioxide.

00:11:02.110 --> 00:11:04.520
And then last but not least,
we can do that with the water.

00:11:04.520 --> 00:11:09.520
0.833 mole of H2O times its molar mass,

00:11:12.120 --> 00:11:17.120
times 18.02 grams per mole of water

00:11:19.370 --> 00:11:21.577
is going to give us approximately

00:11:21.577 --> 00:11:22.967
.833 times 18.02, gives us,

00:11:28.800 --> 00:11:33.800
three significant figures
is going to be 15.0.

00:11:34.060 --> 00:11:39.060
15.0 grams of water.

00:11:39.470 --> 00:11:41.040
And once again, those moles canceled out

00:11:41.040 --> 00:11:42.350
to give us the right units.

00:11:42.350 --> 00:11:44.510
And we're done, which is pretty cool.

00:11:44.510 --> 00:11:47.840
This is really useful in chemistry
to be able to understand,

00:11:47.840 --> 00:11:50.410
based on a balanced chemical equation,

00:11:50.410 --> 00:11:51.680
to be able to understand,

00:11:51.680 --> 00:11:54.750
hey, if I have a certain
mass of one of the inputs,

00:11:54.750 --> 00:11:57.000
one of the things that
are one of the reactants,

00:11:57.000 --> 00:11:58.270
how much do I need of the other?

00:11:58.270 --> 00:12:00.920
And then how much mass of the products

00:12:00.920 --> 00:12:02.610
am I actually going to produce?

00:12:02.610 --> 00:12:04.663
All of which we just figured out.


Physical and chemical changes

WEBVTT
Kind: captions
Language: en

00:00:00.200 --> 00:00:01.950
- [Instructor] So what we have
are three different pictures

00:00:01.950 --> 00:00:05.880
of substances undergoing
some type of change.

00:00:05.880 --> 00:00:08.230
And we're going to focus on in this video

00:00:08.230 --> 00:00:11.950
is classifying things as
either being physical changes

00:00:11.950 --> 00:00:14.050
or chemical changes.

00:00:14.050 --> 00:00:15.540
And you might already have already thought

00:00:15.540 --> 00:00:18.790
about this or seen this in
a previous science class.

00:00:18.790 --> 00:00:21.120
But when we talk about a physical change,

00:00:21.120 --> 00:00:23.500
we're talking about where
there could be a change

00:00:23.500 --> 00:00:25.900
in properties but we're
not having a change

00:00:25.900 --> 00:00:29.140
in the actual composition
of what we're talking about,

00:00:29.140 --> 00:00:32.820
while in a chemical change
you actually do have a change

00:00:32.820 --> 00:00:36.210
in composition, how the
different constituent atoms

00:00:36.210 --> 00:00:39.420
and elements match up or
connect or bond to each other

00:00:39.420 --> 00:00:41.680
might be different.

00:00:41.680 --> 00:00:43.970
So my first question to
you is pause this video

00:00:43.970 --> 00:00:45.670
and we have some ice melting here.

00:00:45.670 --> 00:00:49.060
We have some propane
combusting or burning here,

00:00:49.060 --> 00:00:52.400
and we have some iron rusting here.

00:00:52.400 --> 00:00:53.233
And I want you to think

00:00:53.233 --> 00:00:54.970
about which of these are physical changes,

00:00:54.970 --> 00:00:57.370
and which of these are
chemical changes and why.

00:00:58.700 --> 00:01:01.550
All right, now let's first
think about this water,

00:01:01.550 --> 00:01:02.940
this ice melting.

00:01:02.940 --> 00:01:06.285
And if we wanted to write it
in fancy chemical language,

00:01:06.285 --> 00:01:09.920
or chemistry language, we
could write this as H2O,

00:01:09.920 --> 00:01:14.920
going from its solid form to
H2O going into its liquid form.

00:01:16.890 --> 00:01:19.500
Now, we don't have a
change in composition,

00:01:19.500 --> 00:01:21.010
in either state whether you're looking

00:01:21.010 --> 00:01:22.430
at this liquid water here,

00:01:22.430 --> 00:01:25.000
or whether you're looking
at the solid water there,

00:01:25.000 --> 00:01:29.240
you'll see a bunch of water molecules.

00:01:29.240 --> 00:01:33.490
Each oxygen is still
bonded to two hydrogens.

00:01:33.490 --> 00:01:36.450
And so you're not having
a change in composition.

00:01:36.450 --> 00:01:41.410
And so this over here is a
physical, physical change.

00:01:41.410 --> 00:01:42.890
And if we kept heating that water up

00:01:42.890 --> 00:01:44.330
and it started to vaporize

00:01:44.330 --> 00:01:46.710
that would also be a physical change.

00:01:46.710 --> 00:01:49.010
Whereas it turns into water vapor.

00:01:49.010 --> 00:01:52.150
You have your intermolecular
forces being overcome

00:01:52.150 --> 00:01:53.660
but the covalent bonds

00:01:53.660 --> 00:01:55.630
between the oxygens and the hydrogens,

00:01:55.630 --> 00:01:58.720
those aren't breaking
or forming in some way.

00:01:58.720 --> 00:02:01.010
So once again, when you
go from ice to water,

00:02:01.010 --> 00:02:02.900
physical change, from water to vapor

00:02:02.900 --> 00:02:04.840
or you could say from liquid to gas

00:02:04.840 --> 00:02:08.350
that is also going to
be a physical change.

00:02:08.350 --> 00:02:10.872
One general rule of thumb
when you think about

00:02:10.872 --> 00:02:13.000
what's going on on a microscopic level,

00:02:13.000 --> 00:02:14.100
and this is a general rule of thumb.

00:02:14.100 --> 00:02:15.120
It doesn't always apply.

00:02:15.120 --> 00:02:17.230
And we'll think about an
edge case in a little bit

00:02:17.230 --> 00:02:21.340
is when you're overcoming
intermolecular forces

00:02:21.340 --> 00:02:23.930
that tends to be a physical change.

00:02:23.930 --> 00:02:27.351
But if you have chemical
bonds forming or breaking

00:02:27.351 --> 00:02:30.564
that would be a chemical change.

00:02:30.564 --> 00:02:33.840
Now let's think about what's
going on here with the propane.

00:02:33.840 --> 00:02:35.980
If you were to write the
chemical reaction here

00:02:35.980 --> 00:02:40.700
it would be propane C3H8 in gas form.

00:02:40.700 --> 00:02:43.020
It needs oxygen to combust.

00:02:43.020 --> 00:02:44.350
So for every mole of propane

00:02:44.350 --> 00:02:48.880
we have five moles of
molecular oxygen in gas form.

00:02:48.880 --> 00:02:51.760
And then when it combusts

00:02:51.760 --> 00:02:54.770
you're going to produce
three for every one mole

00:02:54.770 --> 00:02:57.320
of propane and five moles
of molecular oxygen,

00:02:57.320 --> 00:03:02.140
you're gonna produce three
moles of carbon dioxide gas

00:03:02.140 --> 00:03:07.140
and four moles of water
in vapor form as well.

00:03:07.790 --> 00:03:10.170
And so what you actually have is the bonds

00:03:10.170 --> 00:03:11.900
in those molecules are actually breaking

00:03:11.900 --> 00:03:13.900
and then re forming.

00:03:13.900 --> 00:03:16.960
So you don't just have
physical change going on here.

00:03:16.960 --> 00:03:21.670
You have chemical change, chemical change.

00:03:21.670 --> 00:03:22.520
One way to think about it.

00:03:22.520 --> 00:03:25.860
You had propane here before, C3 H8,

00:03:25.860 --> 00:03:29.223
after the reaction you no
longer have the propane here.

00:03:29.223 --> 00:03:32.300
When you actually see as
fire, which is fascinating.

00:03:32.300 --> 00:03:34.487
This is just very hot gas,

00:03:34.487 --> 00:03:36.540
and that very hot air that you're seeing,

00:03:36.540 --> 00:03:38.600
and there's gonna be some
carbon dioxide in there

00:03:38.600 --> 00:03:41.190
and there's gonna be some
water vapor in there.

00:03:41.190 --> 00:03:42.900
The reason why it's getting so hot

00:03:42.900 --> 00:03:46.480
is because this releases a lot of energy.

00:03:46.480 --> 00:03:50.370
Now let's think about what's
going on here with this iron.

00:03:50.370 --> 00:03:53.640
If I were to write this
as a chemical reaction,

00:03:53.640 --> 00:03:57.100
for every four moles of iron in solid form

00:03:57.100 --> 00:04:01.160
plus three moles of
molecular oxygen in gas form,

00:04:01.160 --> 00:04:04.160
and that would just be the
ambient oxygen around this iron.

00:04:04.160 --> 00:04:09.160
It is going to produce two
moles of iron oxide as a solid

00:04:13.740 --> 00:04:15.750
and that's what you see
there in the orange.

00:04:15.750 --> 00:04:17.190
That is the iron oxide.

00:04:17.190 --> 00:04:21.030
So notice this reaction
is forming new ionic bonds

00:04:21.030 --> 00:04:22.920
in that ferrous oxide.

00:04:22.920 --> 00:04:24.810
And to undergo the reaction we had to

00:04:24.810 --> 00:04:28.280
break the metallic bonds of the solid iron

00:04:28.280 --> 00:04:31.710
and the covalent bonds
in the molecular oxygen.

00:04:31.710 --> 00:04:36.130
So anytime we are breaking and
making these chemical bonds

00:04:36.130 --> 00:04:38.160
we have a chemical change.

00:04:38.160 --> 00:04:40.150
Now, let me give you an
interesting question.

00:04:40.150 --> 00:04:44.260
What about the dissolving of
sodium chloride or table salt?

00:04:44.260 --> 00:04:48.010
So you have NaCl in its solid form,

00:04:48.010 --> 00:04:50.100
when you place it into water,

00:04:50.100 --> 00:04:54.330
you get sodium chloride in aqueous form.

00:04:54.330 --> 00:04:57.160
And how does that look when this happens?

00:04:57.160 --> 00:04:59.430
Well, sodium chloride in solid form,

00:04:59.430 --> 00:05:02.580
you know it forms this lattice

00:05:02.580 --> 00:05:04.810
because they have these ionic bonds.

00:05:04.810 --> 00:05:06.660
I'll try to draw a few of them.

00:05:06.660 --> 00:05:08.910
So it looks something like this,

00:05:08.910 --> 00:05:10.640
where we've seen this multiple times,

00:05:10.640 --> 00:05:13.940
chlorine nabs electrons becomes
chloride, becomes negative.

00:05:13.940 --> 00:05:18.170
Sodium loses electron
becomes a positive ion.

00:05:18.170 --> 00:05:19.790
And so they get attracted to each other

00:05:19.790 --> 00:05:21.600
and then they can form this

00:05:21.600 --> 00:05:23.760
three-dimensional lattice structure,

00:05:23.760 --> 00:05:25.860
when it is in solid form.

00:05:25.860 --> 00:05:27.410
And I'm trying my best to draw it

00:05:27.410 --> 00:05:30.300
in three dimensions over here.

00:05:30.300 --> 00:05:33.720
And actually I could put one
right over here like this,

00:05:33.720 --> 00:05:35.900
but when you dissolve it into water

00:05:35.900 --> 00:05:38.400
the ions get split apart.

00:05:38.400 --> 00:05:40.780
You get the ions that all get dissolved,

00:05:40.780 --> 00:05:42.810
they get split apart in the water.

00:05:42.810 --> 00:05:47.810
And so the ionic bonds
themselves are being broken.

00:05:48.210 --> 00:05:49.810
Now you could argue that, hey, you know

00:05:49.810 --> 00:05:52.280
the constituent ions are
just being separated.

00:05:52.280 --> 00:05:54.720
So maybe this is a physical change

00:05:54.720 --> 00:05:56.360
or you could say, no, ionic bonds.

00:05:56.360 --> 00:05:57.810
Those are legit bonds.

00:05:57.810 --> 00:06:01.550
Those are strong bonds, and
those are getting broken.

00:06:01.550 --> 00:06:04.020
And so you're also
getting chemical change.

00:06:04.020 --> 00:06:07.670
And it actually turns out
that this is a gray area

00:06:07.670 --> 00:06:10.770
when you're talking about
dissolving a salt like this

00:06:10.770 --> 00:06:13.511
into water, whether you're
talking about a physical change

00:06:13.511 --> 00:06:16.180
or you're talking about a chemical change.

00:06:16.180 --> 00:06:18.620
And actually just this past
weekend, I was at the beach

00:06:18.620 --> 00:06:21.780
and we ran out of salt
and I did this in reverse.

00:06:21.780 --> 00:06:25.140
I went to the ocean and I
got a pot full of salt water.

00:06:25.140 --> 00:06:27.460
And it actually took me
about 30 minutes to boil it.

00:06:27.460 --> 00:06:30.630
And I was able to evaporate
all of the water out of it.

00:06:30.630 --> 00:06:33.200
And essentially the salt
in that water was able

00:06:33.200 --> 00:06:35.550
to reform these ionic bonds.

00:06:35.550 --> 00:06:39.240
And so one could debate whether
I was creating a physical

00:06:39.240 --> 00:06:40.880
or a chemical change.

00:06:40.880 --> 00:06:44.360
I was definitely doing a
physical change with the water.

00:06:44.360 --> 00:06:48.670
I was evaporating it, but I was
actually forming ionic bonds

00:06:48.670 --> 00:06:49.840
between the sodium and chloride.

00:06:49.840 --> 00:06:51.690
So one could argue that
that was maybe a physical

00:06:51.690 --> 00:06:53.283
and a chemical change.


Drawing particulate models of reaction mixtures

WEBVTT
Kind: captions
Language: en

00:00:00.260 --> 00:00:01.400
- [Instructor] In a previous video,

00:00:01.400 --> 00:00:04.000
we used a particulate model like this

00:00:04.000 --> 00:00:05.310
to understand a reaction,

00:00:05.310 --> 00:00:07.477
not just to understand the reaction,

00:00:08.490 --> 00:00:11.120
but to balance the
chemical reaction as well.

00:00:11.120 --> 00:00:14.390
And when I hand drew these particles,

00:00:14.390 --> 00:00:17.300
the atoms in this particulate model here,

00:00:17.300 --> 00:00:19.190
I tried to draw it pretty close

00:00:19.190 --> 00:00:21.620
to their actual relative sizes.

00:00:21.620 --> 00:00:24.140
Carbon atoms are a little
bit bigger than oxygen atoms,

00:00:24.140 --> 00:00:27.220
and they're both a lot
bigger than hydrogen atoms.

00:00:27.220 --> 00:00:28.360
What we're gonna do in this video

00:00:28.360 --> 00:00:32.190
is extend our understanding
using a particulate model

00:00:32.190 --> 00:00:35.000
to start to visualize
what actually might go on

00:00:35.000 --> 00:00:36.220
in a mixture

00:00:36.220 --> 00:00:40.860
of some of these reactant molecules.

00:00:40.860 --> 00:00:43.320
So what I have here on the left-hand side

00:00:43.320 --> 00:00:45.120
are the various molecules.

00:00:45.120 --> 00:00:47.530
I have two methane molecules here.

00:00:47.530 --> 00:00:49.500
I have three water molecules.

00:00:49.500 --> 00:00:51.670
And what I wanna do with you

00:00:51.670 --> 00:00:56.360
is draw what we would expect
to see after the reaction.

00:00:56.360 --> 00:00:58.790
And I encourage you, like
always, pause this video

00:00:58.790 --> 00:01:00.190
and see if you can have a go at that,

00:01:00.190 --> 00:01:01.200
maybe with a pencil and paper,

00:01:01.200 --> 00:01:03.210
at least just try to
imagine it in your head,

00:01:03.210 --> 00:01:04.693
before I do this with you.

00:01:05.550 --> 00:01:07.700
All right, now, let's do this together.

00:01:07.700 --> 00:01:11.410
Now, we know that for every
methane and every water,

00:01:11.410 --> 00:01:15.060
we're going to produce one carbon monoxide

00:01:15.060 --> 00:01:17.610
and three molecular hydrogens.

00:01:17.610 --> 00:01:19.470
And each of those molecules of hydrogen

00:01:19.470 --> 00:01:21.280
have two hydrogens in them.

00:01:21.280 --> 00:01:26.280
So let's just say that this
one and this one react.

00:01:26.320 --> 00:01:28.980
They're going to produce
one carbon monoxide.

00:01:28.980 --> 00:01:31.660
I'm gonna try to draw the
relative sizes roughly, right?

00:01:31.660 --> 00:01:35.150
So one carbon monoxide.

00:01:35.150 --> 00:01:38.770
And then they're gonna
produce six hydrogen atoms

00:01:38.770 --> 00:01:41.530
that are going to be in
three hydrogen molecules.

00:01:41.530 --> 00:01:43.200
So let's do,

00:01:43.200 --> 00:01:47.700
that's two and four,

00:01:47.700 --> 00:01:50.020
and then I'll just do one here

00:01:50.020 --> 00:01:53.270
and then six.

00:01:53.270 --> 00:01:56.370
All right, so I took care
of this one and this one.

00:01:56.370 --> 00:02:00.700
And now we can imagine that
maybe this water molecule

00:02:00.700 --> 00:02:03.400
reacts with this methane molecule,

00:02:03.400 --> 00:02:07.080
and so that would produce
another carbon monoxide.

00:02:07.080 --> 00:02:08.280
Let me draw that

00:02:08.280 --> 00:02:09.800
roughly at the right size.

00:02:09.800 --> 00:02:14.030
Another carbon monoxide molecule

00:02:15.660 --> 00:02:19.340
and three more hydrogen molecules

00:02:19.340 --> 00:02:21.130
for a total of six more hydrogens.

00:02:21.130 --> 00:02:25.740
So that's one and two

00:02:26.770 --> 00:02:28.830
and three.

00:02:28.830 --> 00:02:31.810
And now we have this water right over here

00:02:31.810 --> 00:02:34.560
that had no one to react
with in this situation,

00:02:34.560 --> 00:02:35.850
had no partner.

00:02:35.850 --> 00:02:40.100
And so that's just going to be
a leftover reactant molecule.

00:02:40.100 --> 00:02:42.220
So let me just draw it right over here.

00:02:42.220 --> 00:02:45.885
So that water could be

00:02:45.885 --> 00:02:49.830
right over right over here.

00:02:49.830 --> 00:02:51.660
And so this was a useful way

00:02:51.660 --> 00:02:53.880
of starting to visualize
what might be going on.

00:02:53.880 --> 00:02:55.610
Remember, this is happening
at a very high temperature.

00:02:55.610 --> 00:02:58.060
They're all bouncing around, et cetera.

00:02:58.060 --> 00:03:00.220
And then when they react,
you might get this,

00:03:00.220 --> 00:03:02.640
but then this water molecule
has no one to react to,

00:03:02.640 --> 00:03:05.840
so it is, you could view it as a leftover

00:03:05.840 --> 00:03:07.453
after the reaction.


Visualizing chemical equations using particulate models

WEBVTT
Kind: captions
Language: en

00:00:00.260 --> 00:00:02.180
- [Instructor] A question that
some of you might have asked,

00:00:02.180 --> 00:00:05.560
or maybe haven't asked is where
do we get our hydrogen from?

00:00:05.560 --> 00:00:09.030
Because molecular hydrogen,
if it was just in the air,

00:00:09.030 --> 00:00:12.050
it is lighter than the other
things that make up the air,

00:00:12.050 --> 00:00:14.540
so it would just float to
the top of the atmosphere.

00:00:14.540 --> 00:00:16.120
So how would we get it?

00:00:16.120 --> 00:00:18.660
Well, this reaction right over here

00:00:18.660 --> 00:00:21.860
is actually one of the
most cost-effective ways

00:00:21.860 --> 00:00:24.660
of getting molecular hydrogen,

00:00:24.660 --> 00:00:27.600
which you can see right
over here on the right.

00:00:27.600 --> 00:00:29.740
What do you do at a very high temperature?

00:00:29.740 --> 00:00:31.120
What I would consider a high temperature,

00:00:31.120 --> 00:00:34.880
roughly between 700 and
1,000 degrees Celsius,

00:00:34.880 --> 00:00:38.620
you get some methane gas
in the presence of water.

00:00:38.620 --> 00:00:39.453
And of course,

00:00:39.453 --> 00:00:41.820
water at that temperature
is going to be a gas,

00:00:41.820 --> 00:00:43.370
we're talking about steam.

00:00:43.370 --> 00:00:47.210
And then they will react
to produce carbon monoxide

00:00:47.210 --> 00:00:50.410
and molecular hydrogen.

00:00:50.410 --> 00:00:53.180
Now something might be
feeling a little off

00:00:53.180 --> 00:00:55.770
when I wrote this reaction like this.

00:00:55.770 --> 00:00:58.900
So pause this video and
think about what is off here.

00:00:58.900 --> 00:01:00.710
And I'll give you a little bit of a hint.

00:01:00.710 --> 00:01:02.850
Think about what are we inputting?

00:01:02.850 --> 00:01:04.890
What are the atoms and the number of atoms

00:01:04.890 --> 00:01:06.650
that we're inputting into the reaction?

00:01:06.650 --> 00:01:09.290
And then what are the number
and the types of atoms

00:01:09.290 --> 00:01:10.470
that we are outputting?

00:01:10.470 --> 00:01:13.110
For example, we have one
carbon that we are inputting

00:01:13.110 --> 00:01:14.670
between the methane and the water.

00:01:14.670 --> 00:01:17.210
And we have one carbon
that we are getting out

00:01:17.210 --> 00:01:18.300
on the other side.

00:01:18.300 --> 00:01:20.490
Think about that for the
oxygen and the hydrogens,

00:01:20.490 --> 00:01:23.223
and see whether it all makes sense.

00:01:24.210 --> 00:01:26.980
All right, now let's work
through this together.

00:01:26.980 --> 00:01:29.210
And actually to help us visualize,

00:01:29.210 --> 00:01:31.350
instead of just writing it in this form,

00:01:31.350 --> 00:01:34.650
I'm also gonna try to visualize
the various molecules.

00:01:34.650 --> 00:01:36.860
So this right over here
is a methane molecule.

00:01:36.860 --> 00:01:40.190
You have one carbon that is
bonded to four hydrogens.

00:01:40.190 --> 00:01:42.993
You can see that up there, CH4, CH4.

00:01:43.990 --> 00:01:45.620
Here we have a water molecule.

00:01:45.620 --> 00:01:49.800
You have an oxygen that is
bonded to two hydrogens.

00:01:49.800 --> 00:01:51.520
And then they react.

00:01:51.520 --> 00:01:53.400
You get a carbon monoxide molecule,

00:01:53.400 --> 00:01:55.300
or this is how I've visualized it.

00:01:55.300 --> 00:01:57.690
So you have a carbon and an oxygen.

00:01:57.690 --> 00:01:59.760
And then I draw the molecular hydrogen.

00:01:59.760 --> 00:02:03.080
Molecular hydrogen has two
hydrogens bonded to each other.

00:02:03.080 --> 00:02:05.790
And that is what I have depicted here.

00:02:05.790 --> 00:02:07.940
Now based on the hint I gave you before

00:02:07.940 --> 00:02:10.090
I asked you to pause the video,

00:02:10.090 --> 00:02:14.260
you will notice that we have
a carbon on the input side,

00:02:14.260 --> 00:02:15.340
you have it right there.

00:02:15.340 --> 00:02:17.510
And we have one carbon on the output side.

00:02:17.510 --> 00:02:21.450
So that seems to obey
conservation of mass.

00:02:21.450 --> 00:02:23.830
Now what about for the oxygens?

00:02:23.830 --> 00:02:28.410
Well, we have one oxygen between
the methane and the water

00:02:28.410 --> 00:02:30.170
that we're inputting into the reaction

00:02:30.170 --> 00:02:31.780
and we have it drawn right over here.

00:02:31.780 --> 00:02:35.094
And then we have one oxygen
that we are outputting

00:02:35.094 --> 00:02:39.410
on the output side of our
reaction right over here.

00:02:39.410 --> 00:02:40.880
Now what about the hydrogens?

00:02:40.880 --> 00:02:43.330
Well, on the left side of our
reaction, right over here,

00:02:43.330 --> 00:02:47.210
we have four hydrogens plus
another two or six hydrogens.

00:02:47.210 --> 00:02:48.580
You can also count them here:

00:02:48.580 --> 00:02:52.430
one, two, three, four,
five, six hydrogens.

00:02:52.430 --> 00:02:53.640
While on the right hand side,

00:02:53.640 --> 00:02:55.630
we only have two hydrogens,

00:02:55.630 --> 00:02:57.850
and they're in one hydrogen molecules.

00:02:57.850 --> 00:03:00.010
So what happen to the
other four hydrogens?

00:03:00.010 --> 00:03:01.790
They can't just disappear.

00:03:01.790 --> 00:03:04.520
We have to have conservation of mass.

00:03:04.520 --> 00:03:07.280
So we need to have another four hydrogens

00:03:07.280 --> 00:03:09.890
on the right-hand side of this equation.

00:03:09.890 --> 00:03:12.490
Well, how can we have
another four hydrogens?

00:03:12.490 --> 00:03:16.340
Well, that's if we have two
more molecules of hydrogen.

00:03:16.340 --> 00:03:19.960
So that's one, and then that is two.

00:03:19.960 --> 00:03:24.290
So instead of just having one
molecule of molecular hydrogen

00:03:24.290 --> 00:03:27.550
that has two hydrogen atoms
in it, we now have three.

00:03:27.550 --> 00:03:30.930
So to balance this chemical equation,

00:03:30.930 --> 00:03:32.640
all we have to do is say, okay,

00:03:32.640 --> 00:03:36.020
we don't just have one, one
molecule of hydrogen here,

00:03:36.020 --> 00:03:39.660
we have three molecules of hydrogen.

00:03:39.660 --> 00:03:42.870
And what I have just done is
balance the chemical equation.

00:03:42.870 --> 00:03:45.930
It's just making sure that we
have a conservation of mass,

00:03:45.930 --> 00:03:49.160
that we don't have constituent
atoms on the left-hand side

00:03:49.160 --> 00:03:52.420
that somehow disappear
on the right hand side,

00:03:52.420 --> 00:03:55.080
or we don't have constituent atoms

00:03:55.080 --> 00:03:57.290
that somehow appear on the right hand side

00:03:57.290 --> 00:04:00.663
without ever being
input into the reaction.


Khan Academy Needs Your Help To Keep Going

WEBVTT
Kind: captions
Language: en

00:00:00.250 --> 00:00:02.780
- Hi, Sal Khan here from Khan Academy.

00:00:02.780 --> 00:00:04.550
And I'm just here to remind everyone

00:00:04.550 --> 00:00:06.830
that Khan Academy is a
not-for-profit organization

00:00:06.830 --> 00:00:09.230
with a mission of providing
a free world-class education

00:00:09.230 --> 00:00:10.600
for anyone anywhere.

00:00:10.600 --> 00:00:13.740
And we can only do that work
through philanthropic donations

00:00:13.740 --> 00:00:16.120
from folks like yourself.

00:00:16.120 --> 00:00:19.470
And if you need any extra
motivation, I will remind you

00:00:19.470 --> 00:00:22.550
that there's actually a good
body of scientific research

00:00:22.550 --> 00:00:25.550
that one of the top ways
to become a happier person,

00:00:25.550 --> 00:00:28.600
which we all want to do,
is to give to others.

00:00:28.600 --> 00:00:30.180
And when you give to Khan Academy,

00:00:30.180 --> 00:00:31.960
the social return on investment

00:00:31.960 --> 00:00:36.080
is really one of the highest
impact philanthropy things

00:00:36.080 --> 00:00:37.430
that you can do.

00:00:37.430 --> 00:00:39.240
So when you give those dollars,

00:00:39.240 --> 00:00:41.239
not only will it help
tens of millions of folks

00:00:41.239 --> 00:00:45.690
around the world, but it will
also make you a happier person

00:00:45.690 --> 00:00:49.263
as we go through the back
half to 2020 and into 2021.

00:00:50.280 --> 00:00:51.630
To make it even better,

00:00:51.630 --> 00:00:55.050
every dollar you give will
be matched dollar for dollar

00:00:55.050 --> 00:00:56.830
by a matching gift

00:00:56.830 --> 00:01:00.670
that we have received from a
very generous philanthropist.

00:01:00.670 --> 00:01:03.260
So, as we make our way to the end of 2020,

00:01:03.260 --> 00:01:04.560
which has been a tough year,

00:01:04.560 --> 00:01:07.440
and more folks are
dependent on Khan Academy

00:01:07.440 --> 00:01:10.230
and being able to keep
learning than ever before,

00:01:10.230 --> 00:01:12.210
I hope you seriously consider

00:01:12.210 --> 00:01:15.650
thinking about how powerful
a donation to Khan Academy

00:01:15.650 --> 00:01:18.763
can be, and how happy
that's going to make you.


Limits of composite functions: internal limit doesn't exist

WEBVTT
Kind: captions
Language: en

00:00:00.566 --> 00:00:01.500
- [Instructor] All right, let's get

00:00:01.500 --> 00:00:03.140
a little bit more practice taking limits

00:00:03.140 --> 00:00:04.890
of composite functions.

00:00:04.890 --> 00:00:06.980
Here, we want to figure
out what is the limit

00:00:06.980 --> 00:00:11.740
as x approaches negative
one of g of h of x?

00:00:11.740 --> 00:00:14.300
The function g, we see
it defined graphically

00:00:14.300 --> 00:00:15.800
here on the left, and the function h,

00:00:15.800 --> 00:00:18.140
we see it defined graphically
here on the right.

00:00:18.140 --> 00:00:20.090
Pause this video and have a go at this.

00:00:21.060 --> 00:00:23.790
All right, now your first
temptation might be to say,

00:00:23.790 --> 00:00:26.780
all right, what is the limit
as x approaches negative one

00:00:26.780 --> 00:00:31.780
of h of x, and if that limit
exists, then input that into g.

00:00:32.250 --> 00:00:34.710
If you take the limit as
x approaches negative one

00:00:34.710 --> 00:00:37.430
of h of x, you see that
you have a different limit

00:00:37.430 --> 00:00:39.350
as you approach from the right

00:00:39.350 --> 00:00:41.830
than when you approach from the left.

00:00:41.830 --> 00:00:45.560
So your temptation might be
to give up at this point,

00:00:45.560 --> 00:00:47.870
but what we'll do in
this video is to realize

00:00:47.870 --> 00:00:51.030
that this composite limit actually exists

00:00:51.030 --> 00:00:53.660
even though the limit as
x approaches negative one

00:00:53.660 --> 00:00:56.920
of h of x does not exist.

00:00:56.920 --> 00:00:58.900
How do we figure this out?

00:00:58.900 --> 00:01:00.990
Well, what we could do
is take right-handed

00:01:00.990 --> 00:01:02.920
and left-handed limits.

00:01:02.920 --> 00:01:06.180
Let's first figure out what is the limit

00:01:07.110 --> 00:01:11.940
as x approaches negative
one from the right hand side

00:01:11.940 --> 00:01:14.937
of g of h of x?

00:01:19.600 --> 00:01:22.380
Well, to think about that,
what is the limit of h

00:01:22.380 --> 00:01:25.280
as x approaches negative one
from the right hand side?

00:01:25.280 --> 00:01:28.640
As we approach negative one
from the right hand side,

00:01:28.640 --> 00:01:32.140
it looks like h is
approaching negative two.

00:01:32.140 --> 00:01:35.359
Another way to think about
it is this is going to be

00:01:35.359 --> 00:01:40.359
equal to the limit as h of
x approaches negative two,

00:01:42.580 --> 00:01:45.630
and what direction is it
approaching negative two from?

00:01:45.630 --> 00:01:47.820
Well, it's approaching
negative two from values

00:01:47.820 --> 00:01:49.470
larger than negative two.

00:01:49.470 --> 00:01:52.820
H of x is decreasing down to negative two

00:01:52.820 --> 00:01:55.730
as x approaches negative
one from the right.

00:01:55.730 --> 00:01:59.910
So it's approaching from
values larger than negative two

00:01:59.910 --> 00:02:02.380
of g of h of x.

00:02:02.380 --> 00:02:05.960
G of h of x.

00:02:05.960 --> 00:02:09.100
I'm color coding it to be
able to keep track of things.

00:02:09.100 --> 00:02:12.050
This is analogous to
saying what is the limit,

00:02:12.050 --> 00:02:16.010
if you think about it as
x approaches negative two

00:02:16.010 --> 00:02:18.630
from the positive direction of g?

00:02:18.630 --> 00:02:21.810
Here, h is just the input into g.

00:02:21.810 --> 00:02:25.430
So the input into g is
approaching negative two

00:02:25.430 --> 00:02:28.610
from above, from the right I should say,

00:02:28.610 --> 00:02:30.010
from values larger than negative two,

00:02:30.010 --> 00:02:32.890
and we can see that g
is approaching three.

00:02:32.890 --> 00:02:36.930
So this right over here is
going to be equal to three.

00:02:36.930 --> 00:02:41.443
Now, let's take the limit
as x approaches negative one

00:02:44.580 --> 00:02:49.580
from the left of g of h of x.

00:02:51.720 --> 00:02:54.490
What we could do is first think
about what is h approaching

00:02:54.490 --> 00:02:57.830
as x approaches negative
one from the left?

00:02:57.830 --> 00:03:00.800
As x approaches negative
one from the left,

00:03:00.800 --> 00:03:03.860
it looks like h is
approaching negative three.

00:03:03.860 --> 00:03:06.000
We could say this is the limit

00:03:07.060 --> 00:03:11.660
as h of x is approaching negative three,

00:03:11.660 --> 00:03:13.620
and it is approaching negative three

00:03:13.620 --> 00:03:15.860
from values greater than negative three.

00:03:15.860 --> 00:03:20.200
H of x is approaching
negative three from above,

00:03:20.200 --> 00:03:23.850
or we could say from values
greater than negative three,

00:03:23.850 --> 00:03:28.850
and then of g of h of x.

00:03:31.340 --> 00:03:32.173
Another way to think about it,

00:03:32.173 --> 00:03:34.790
what is the limit as the input to g

00:03:34.790 --> 00:03:38.550
approaches negative three from the right?

00:03:38.550 --> 00:03:42.600
As we approach negative
three from the right,

00:03:42.600 --> 00:03:46.300
g is right here at three,

00:03:46.300 --> 00:03:51.230
so this is going to be
equal to three again.

00:03:51.230 --> 00:03:54.890
So notice the right hand
limit and the left hand limit

00:03:54.890 --> 00:03:58.100
in this case are both equal to three.

00:03:58.100 --> 00:03:59.850
So when the right hand
and the left hand limit

00:03:59.850 --> 00:04:02.450
is equal to the same thing,
we know that the limit

00:04:02.450 --> 00:04:04.540
is equal to that thing.

00:04:04.540 --> 00:04:06.480
This is a pretty cool example,

00:04:06.480 --> 00:04:09.830
because the limit of, you
could say the internal function

00:04:09.830 --> 00:04:12.520
right over here of h of x, did not exist,

00:04:12.520 --> 00:04:16.293
but the limit of the composite
function still exists.


Limits of composite functions: external limit doesn't exist

WEBVTT
Kind: captions
Language: en

00:00:00.000 --> 00:00:01.580
- [Instructor] So over
here I have two functions

00:00:01.580 --> 00:00:03.950
that have been visually
or graphically defined.

00:00:03.950 --> 00:00:06.950
On the left here I have
the graph of g of x,

00:00:06.950 --> 00:00:09.920
and on the right here I
have the graph of h of x.

00:00:09.920 --> 00:00:11.760
And what I want to do is figure out

00:00:11.760 --> 00:00:16.760
what is the limit of g of
h of x as x approaches one.

00:00:17.690 --> 00:00:20.783
Pause this video and see
if you can figure that out.

00:00:21.690 --> 00:00:23.030
All right, now let's do this together.

00:00:23.030 --> 00:00:24.950
Now the first thing that
you might try to say is,

00:00:24.950 --> 00:00:27.530
all right, let's just figure out first,

00:00:27.530 --> 00:00:32.530
the limit as x approaches one of h of x.

00:00:33.180 --> 00:00:35.310
And when you look at that,
what is that going to be?

00:00:35.310 --> 00:00:38.210
Well, as we approach one from the left,

00:00:38.210 --> 00:00:40.650
it looks like h of x is approaching two.

00:00:40.650 --> 00:00:42.100
And as we approach from the right,

00:00:42.100 --> 00:00:44.260
it looks like h of x is approaching two.

00:00:44.260 --> 00:00:46.680
So it looks like this
is just going to be two.

00:00:46.680 --> 00:00:47.513
And let me see, okay,

00:00:47.513 --> 00:00:50.510
well maybe we can then
just input that into g.

00:00:50.510 --> 00:00:52.120
So what is g of two?

00:00:52.120 --> 00:00:54.830
Well, g of two is zero,

00:00:54.830 --> 00:00:56.870
but the limit doesn't seem defined.

00:00:56.870 --> 00:00:58.870
It looks like when we
approach two from the right,

00:00:58.870 --> 00:01:00.280
we're approaching zero.

00:01:00.280 --> 00:01:01.640
And when we approach two from the left,

00:01:01.640 --> 00:01:03.440
we're approaching negative two.

00:01:03.440 --> 00:01:06.400
So maybe this limit doesn't exist.

00:01:06.400 --> 00:01:07.800
But if you're thinking that,

00:01:07.800 --> 00:01:10.120
we haven't fully thought through it,

00:01:10.120 --> 00:01:12.550
because what we could do
is think about this limit

00:01:12.550 --> 00:01:16.190
in terms of both the left-handed
and right-handed limits.

00:01:16.190 --> 00:01:17.940
So let's think of it this way.

00:01:17.940 --> 00:01:21.540
First, let's think about what is the limit

00:01:22.420 --> 00:01:27.420
as x approaches one from the
left-hand side of g of h of x.

00:01:31.180 --> 00:01:32.980
All right, when you
think about it this way,

00:01:32.980 --> 00:01:37.520
if we're approaching one from
the left right over here,

00:01:37.520 --> 00:01:41.380
we see that we are
approaching two from the left,

00:01:41.380 --> 00:01:44.620
I guess you could say, we're
approaching two from below.

00:01:44.620 --> 00:01:48.360
And so the thing that we
are inputting into g of x

00:01:48.360 --> 00:01:51.180
is approaching two from below.

00:01:51.180 --> 00:01:54.270
So the thing that we are inputting into g

00:01:54.270 --> 00:01:56.730
is approaching two from below.

00:01:56.730 --> 00:01:59.220
So if you approach two from below,

00:01:59.220 --> 00:02:01.000
right over here, what is g approaching?

00:02:01.000 --> 00:02:04.290
It looks like g is
approaching negative two.

00:02:04.290 --> 00:02:07.450
So this looks like it is going
to be equal to negative two,

00:02:07.450 --> 00:02:09.620
at least this left-handed limit.

00:02:09.620 --> 00:02:11.460
Now let's do a right-handed limit.

00:02:11.460 --> 00:02:14.960
What is the limit as x approaches one

00:02:14.960 --> 00:02:19.670
from the right hand of g of h of x?

00:02:19.670 --> 00:02:21.670
Well, we can do the same exercise.

00:02:21.670 --> 00:02:24.480
As we approach one from the right,

00:02:24.480 --> 00:02:28.050
it looks like h is
approaching two from below,

00:02:28.050 --> 00:02:30.660
from values less than two.

00:02:30.660 --> 00:02:33.820
And so if we are
approaching two from below,

00:02:33.820 --> 00:02:34.730
because remember,

00:02:34.730 --> 00:02:38.110
whatever h is outputting
is the input into g.

00:02:38.110 --> 00:02:40.900
So if the thing that
we're inputting g into g

00:02:40.900 --> 00:02:43.190
is approaching two from below,

00:02:43.190 --> 00:02:45.640
that means that g, once again,

00:02:45.640 --> 00:02:49.130
is going to be approaching negative two.

00:02:49.130 --> 00:02:52.370
So this is a really, really,
really interesting case,

00:02:52.370 --> 00:02:57.370
where the limit of g of
x as x approaches two

00:02:57.470 --> 00:02:59.010
does not exist.

00:02:59.010 --> 00:03:00.440
But because on h of x,

00:03:00.440 --> 00:03:03.220
when we approach from both the
left and the right hand side,

00:03:03.220 --> 00:03:05.490
h is approaching two from below.

00:03:05.490 --> 00:03:07.770
We just have to think
about the left-handed limit

00:03:07.770 --> 00:03:11.900
as we approach two from
below or from the left on g,

00:03:11.900 --> 00:03:15.210
because in both situations, we
are approaching negative two.

00:03:15.210 --> 00:03:17.930
And so that is going to be our limit.

00:03:17.930 --> 00:03:19.860
When the left-handed and
the right-handed limit

00:03:19.860 --> 00:03:22.150
are the same, that is
going to be your limit.

00:03:22.150 --> 00:03:24.163
It is equal to negative two.


Finding decreasing interval given the function

WEBVTT
Kind: captions
Language: en

00:00:00.510 --> 00:00:02.330
- [Instructor] Let's
say we have the function

00:00:02.330 --> 00:00:06.080
F of X is equal to X to the sixth,

00:00:06.080 --> 00:00:08.540
minus three X to the fifth.

00:00:08.540 --> 00:00:10.630
And my question to you

00:00:10.630 --> 00:00:13.740
is using only what we
know about derivatives.

00:00:13.740 --> 00:00:16.670
Try to figure out over
what interval or intervals

00:00:16.670 --> 00:00:18.930
is this function decreasing?

00:00:18.930 --> 00:00:21.480
Pause the video and
try to figure that out.

00:00:21.480 --> 00:00:24.130
All right, now let's do this together.

00:00:24.130 --> 00:00:26.650
So we know that a function is decreasing

00:00:26.650 --> 00:00:28.530
when its derivative is negative.

00:00:28.530 --> 00:00:29.780
Or another way to say it,

00:00:29.780 --> 00:00:30.990
It's going to be decreasing

00:00:30.990 --> 00:00:33.560
when F prime of X

00:00:33.560 --> 00:00:36.800
is less than zero.

00:00:36.800 --> 00:00:39.010
So what is F prime of X?

00:00:39.010 --> 00:00:40.770
Well, we could use the derivative rules

00:00:40.770 --> 00:00:42.420
and drew the properties we know.

00:00:42.420 --> 00:00:45.000
We apply the power rule
to X, to the sixth,

00:00:45.000 --> 00:00:46.300
We bring the six out front

00:00:46.300 --> 00:00:48.430
or multiply the one coefficient here,

00:00:48.430 --> 00:00:51.510
times six, to get six,

00:00:51.510 --> 00:00:54.000
X to the fifth

00:00:54.000 --> 00:00:55.570
detriment that exponent,

00:00:55.570 --> 00:00:58.550
minus bring the five times the three

00:00:58.550 --> 00:01:01.040
minus 15, X to the,

00:01:01.040 --> 00:01:02.330
we detriment the five,

00:01:02.330 --> 00:01:04.410
so X to the fourth.

00:01:04.410 --> 00:01:05.940
And we need to figure out

00:01:05.940 --> 00:01:10.310
over what intervals is this
going to be less than zero.

00:01:10.310 --> 00:01:13.160
And now let's see how we can
simplify this a little bit.

00:01:13.160 --> 00:01:16.730
Both of these terms are
divisible by X to the fourth,

00:01:16.730 --> 00:01:19.160
and they're both divisible by three.

00:01:19.160 --> 00:01:23.480
So let's factor out a
three X to the fourth

00:01:23.480 --> 00:01:25.830
times, you factor out of
three X to the fourth year,

00:01:25.830 --> 00:01:29.090
you're left with a two X,

00:01:29.090 --> 00:01:32.060
and then over here you have minus five

00:01:32.060 --> 00:01:34.650
has to be less than zero.

00:01:34.650 --> 00:01:36.240
Any interval where this is true,

00:01:36.240 --> 00:01:38.570
we are going to be decreasing.

00:01:38.570 --> 00:01:40.710
Now, how do we get this
to be less than zero?

00:01:40.710 --> 00:01:42.580
Well, if I take the product of two things

00:01:42.580 --> 00:01:43.970
and it's less than zero,

00:01:43.970 --> 00:01:46.530
that means that they have
to have different signs,

00:01:46.530 --> 00:01:48.250
either one's positive
and the other's negative

00:01:48.250 --> 00:01:50.390
or one's negative and
the other's positive.

00:01:50.390 --> 00:01:52.030
So we have two situations.

00:01:52.030 --> 00:01:54.200
So we could say either,

00:01:54.200 --> 00:01:57.810
either, three X to the fourth

00:01:57.810 --> 00:01:59.600
is greater than zero

00:01:59.600 --> 00:02:04.260
and, and two X minus five

00:02:04.260 --> 00:02:06.200
is less than zero.

00:02:06.200 --> 00:02:07.870
So that's one situation.

00:02:07.870 --> 00:02:09.520
I'll do some in a different color,

00:02:09.520 --> 00:02:12.700
or I'll do this one in a different color,

00:02:12.700 --> 00:02:16.920
three X to the fourth is less than zero

00:02:16.920 --> 00:02:20.620
and two X minus five

00:02:20.620 --> 00:02:22.810
is greater than zero.

00:02:22.810 --> 00:02:25.170
Actually let me stay on
the second case first.

00:02:25.170 --> 00:02:26.420
Are there any situations

00:02:26.420 --> 00:02:29.270
where three X to the fourth
can be less than zero?

00:02:29.270 --> 00:02:30.320
You take any number,

00:02:30.320 --> 00:02:31.480
you take it to the fourth power

00:02:31.480 --> 00:02:32.480
even if it's a negative,

00:02:32.480 --> 00:02:33.940
it's going to become a positive.

00:02:33.940 --> 00:02:37.310
So you can't get a negative
expression right over here.

00:02:37.310 --> 00:02:40.790
So actually the second condition
is impossible to obtain.

00:02:40.790 --> 00:02:43.190
You can't get any situation for any X

00:02:43.190 --> 00:02:45.410
where three X to the
fourth is less than zero.

00:02:45.410 --> 00:02:47.330
So we can rule this one out.

00:02:47.330 --> 00:02:49.310
And so this is our best hope.

00:02:49.310 --> 00:02:52.170
So under what conditions
is three X to the fourth,

00:02:52.170 --> 00:02:54.630
greater than zero.

00:02:54.630 --> 00:02:56.570
Well, if you divide both sides by three,

00:02:56.570 --> 00:03:01.530
you get X to the fourth
is greater than zero.

00:03:01.530 --> 00:03:02.950
And if you think about it,

00:03:02.950 --> 00:03:05.930
this is gonna be true for any X value

00:03:05.930 --> 00:03:08.110
that is not equal to zero.

00:03:08.110 --> 00:03:09.410
Even if you have a negative value there,

00:03:09.410 --> 00:03:10.243
if you have a negative one,

00:03:10.243 --> 00:03:11.980
you take it to the fourth
power becomes a positive one.

00:03:11.980 --> 00:03:14.100
Only zero will be equal to zero

00:03:14.100 --> 00:03:15.570
when you take it to the fourth power.

00:03:15.570 --> 00:03:16.510
So one way we could say this

00:03:16.510 --> 00:03:19.090
is going to be true for any non zero X,

00:03:19.090 --> 00:03:22.130
or we could just say,
X does not equal zero,

00:03:22.130 --> 00:03:25.140
and, this is a little
bit more straightforward,

00:03:25.140 --> 00:03:26.750
we add five to both sides

00:03:26.750 --> 00:03:29.240
we get two X is less than five,

00:03:29.240 --> 00:03:31.380
divide both sides by two,

00:03:31.380 --> 00:03:35.250
you get X is less than five halves.

00:03:35.250 --> 00:03:37.180
So it might be tempting to say, all right,

00:03:37.180 --> 00:03:38.860
the intervals that matter are

00:03:38.860 --> 00:03:41.930
all the exes less than five halves,

00:03:41.930 --> 00:03:45.710
but X cannot be equal to zero.

00:03:45.710 --> 00:03:48.210
Now, is that the entire interval

00:03:48.210 --> 00:03:50.540
where our function is decreasing?

00:03:50.540 --> 00:03:53.370
Well, let's think about
what happens at zero itself.

00:03:53.370 --> 00:03:57.990
We're decreasing over the
interval from negative infinity

00:03:57.990 --> 00:03:59.710
all the way up to zero.

00:03:59.710 --> 00:04:03.920
And we're also decreasing
from zero to five halves.

00:04:03.920 --> 00:04:06.220
And so for decreasing
right to the left of zero,

00:04:06.220 --> 00:04:07.940
we're decreasing right
to the right of zero,

00:04:07.940 --> 00:04:10.020
we're actually going to
be decreasing at zero

00:04:10.020 --> 00:04:14.010
at we're also gonna be
decreasing at zero as well.

00:04:14.010 --> 00:04:15.640
So there's something interesting here,

00:04:15.640 --> 00:04:18.800
even though the derivative
at X equals zero,

00:04:18.800 --> 00:04:20.950
is going to be equal to zero,

00:04:20.950 --> 00:04:23.510
we are still decreasing.

00:04:23.510 --> 00:04:25.810
And so the interval that we care about

00:04:25.810 --> 00:04:27.550
the interval over which we're decreasing

00:04:27.550 --> 00:04:31.090
is just X is less than five halves.

00:04:31.090 --> 00:04:33.340
And we can see that by
graphing the function,

00:04:33.340 --> 00:04:35.790
I graphed it on Desmos,

00:04:35.790 --> 00:04:37.620
and you can see here

00:04:37.620 --> 00:04:40.580
that the function is decreasing
from negative infinity,

00:04:40.580 --> 00:04:42.710
it's decreasing at a
slower and slower rate

00:04:42.710 --> 00:04:45.790
we get to zero still
decreasing to the left of zero,

00:04:45.790 --> 00:04:48.410
and then it continues to
decrease to the right of zero.

00:04:48.410 --> 00:04:51.570
So any, any value, any X
value to the left of zero

00:04:51.570 --> 00:04:52.940
the value of the function is going to be

00:04:52.940 --> 00:04:54.980
larger than F of zero,

00:04:54.980 --> 00:04:56.610
and X to the right of zero.

00:04:56.610 --> 00:04:57.770
The value of the function is going to be

00:04:57.770 --> 00:04:59.760
less than the function at zero.

00:04:59.760 --> 00:05:02.180
So it's actually decreasing through zero,

00:05:02.180 --> 00:05:06.110
even though the slope of the
tangent line at zero is zero,

00:05:06.110 --> 00:05:07.800
even though it's non negative

00:05:07.800 --> 00:05:09.530
and then we keep decreasing

00:05:09.530 --> 00:05:11.970
so we're decreasing for all values of X,

00:05:11.970 --> 00:05:14.890
less than five halves,

00:05:14.890 --> 00:05:16.693
which you can see visually here.


Example: Graphing y=-cos(π⋅x)+1.5

WEBVTT
Kind: captions
Language: en

00:00:00.740 --> 00:00:01.710
- [Instructor] We're told to graph

00:00:01.710 --> 00:00:05.850
y is equal to negative
cosine of pi times x plus 1.5

00:00:06.998 --> 00:00:09.190
in the interactive widget.

00:00:09.190 --> 00:00:11.810
So, pause this video and think
about how you would do that.

00:00:11.810 --> 00:00:13.980
And just to explain how this widget works

00:00:13.980 --> 00:00:15.780
if you're trying to do it on Khan Academy,

00:00:15.780 --> 00:00:19.610
this dot right over here
helps define the midline.

00:00:19.610 --> 00:00:21.180
You can move that up and down.

00:00:21.180 --> 00:00:23.090
And then this one right over here

00:00:23.090 --> 00:00:25.070
is a neighboring extreme point.

00:00:25.070 --> 00:00:28.850
So either a minimum or a maximum point.

00:00:28.850 --> 00:00:32.240
So, there's a couple of ways
that we could approach this.

00:00:32.240 --> 00:00:33.073
First of all,

00:00:33.073 --> 00:00:37.420
let's just think about what
would cosine of pi x look like,

00:00:37.420 --> 00:00:38.881
and then we'll think about
what the negative does

00:00:38.881 --> 00:00:40.740
and the plus 1.5.

00:00:40.740 --> 00:00:42.900
So, cosine of pi x.

00:00:42.900 --> 00:00:45.390
When x is equal zero,

00:00:45.390 --> 00:00:48.500
pi times zero, is just going to be zero,

00:00:48.500 --> 00:00:51.750
cosine of zero is equal to one.

00:00:51.750 --> 00:00:54.010
And if we're just talking
about cosine of pi x,

00:00:54.010 --> 00:00:56.570
that's going to be a maximum
point when you hit one.

00:00:56.570 --> 00:00:58.320
Just cosine of pi x would oscillate

00:00:58.320 --> 00:01:00.550
between one and negative one.

00:01:00.550 --> 00:01:02.130
And then what would its period be

00:01:02.130 --> 00:01:05.190
if we're talking about cosine of pi x?

00:01:05.190 --> 00:01:06.300
Well, you might remember,

00:01:06.300 --> 00:01:09.530
one way to think about the
period is to take two pi

00:01:09.530 --> 00:01:12.730
and divide it by whatever
the coefficient is

00:01:12.730 --> 00:01:14.530
on the x right over here.

00:01:14.530 --> 00:01:17.850
So two pi divided by pi would tell us

00:01:17.850 --> 00:01:20.940
that we have a period of two.

00:01:20.940 --> 00:01:23.770
And so how do we construct
a period of two here?

00:01:23.770 --> 00:01:27.420
Well, that means that as we
start here at x equals zero,

00:01:27.420 --> 00:01:30.500
we're at one, we want to get
back to that maximum point

00:01:30.500 --> 00:01:33.830
by the time x is equal to two.

00:01:33.830 --> 00:01:36.180
So let me see how I can do that.

00:01:36.180 --> 00:01:38.660
If I were to squeeze it a little bit,

00:01:38.660 --> 00:01:40.120
that looks pretty good.

00:01:40.120 --> 00:01:42.100
And the reason why I worked
on this midline point

00:01:42.100 --> 00:01:45.020
is I liked having this
maximum point at one

00:01:45.020 --> 00:01:47.440
when x is equal to zero,

00:01:47.440 --> 00:01:49.550
because we said cosine of pi times zero

00:01:49.550 --> 00:01:50.670
should be equal to one.

00:01:50.670 --> 00:01:53.480
So that's why I'm just
manipulating this other point

00:01:53.480 --> 00:01:55.620
in order to set the period right.

00:01:55.620 --> 00:01:56.510
But this looks right.

00:01:56.510 --> 00:01:58.100
We're going from this maximum point

00:01:58.100 --> 00:01:59.420
and we're going all the way down

00:01:59.420 --> 00:02:01.120
and then back to that maximum point,

00:02:01.120 --> 00:02:03.800
and it looks like our
period is indeed two.

00:02:03.800 --> 00:02:07.600
So this is what the graph of
cosine of pi x would look like.

00:02:07.600 --> 00:02:09.920
Now, what about this negative sign?

00:02:09.920 --> 00:02:12.650
Well, the negative would
essentially flip it around.

00:02:12.650 --> 00:02:14.820
So, instead of whenever
we're equaling one,

00:02:14.820 --> 00:02:15.940
we should be equal to negative one.

00:02:15.940 --> 00:02:17.510
And every time we're
equal to negative one,

00:02:17.510 --> 00:02:18.670
we should be equal to one.

00:02:18.670 --> 00:02:20.100
So what I could is I could just take that

00:02:20.100 --> 00:02:21.520
and then bring it down here,

00:02:21.520 --> 00:02:23.300
and there you have it,
I flipped it around.

00:02:23.300 --> 00:02:27.090
So this is the graph of y
equals negative cosine of pi x.

00:02:27.090 --> 00:02:29.773
And then last but not least,
we have this plus 1.5.

00:02:30.669 --> 00:02:33.840
So that's just going to
shift everything up by 1.5.

00:02:33.840 --> 00:02:36.670
So I'm just going to
shift everything up by,

00:02:36.670 --> 00:02:41.660
shift it up by 1.5 and shift it up by 1.5.

00:02:43.530 --> 00:02:44.700
And there you have it.

00:02:44.700 --> 00:02:49.700
That is the graph of negative
cosine of pi x plus 1.5.

00:02:49.710 --> 00:02:53.170
And you can validate
that that's our midline.

00:02:53.170 --> 00:02:56.560
We're still oscillating
one above and one below.

00:02:56.560 --> 00:02:59.940
The negative sign, when
cosine of pi time zero,

00:02:59.940 --> 00:03:01.800
that should be one, but then
you take the negative that,

00:03:01.800 --> 00:03:03.340
we get to negative one.

00:03:03.340 --> 00:03:07.050
You add 1.5 to that,
you get to positive .5.

00:03:07.050 --> 00:03:09.593
And so this is all looking quite good.


Example: Graphing y=3⋅sin(½⋅x)-2

WEBVTT
Kind: captions
Language: en

00:00:00.220 --> 00:00:01.450
- [Instructor] So we're asked to graph

00:00:01.450 --> 00:00:04.440
y is equal to three times sine of 1/2x

00:00:04.440 --> 00:00:06.900
minus 2 in the interactive widget.

00:00:06.900 --> 00:00:08.250
And this is the interactive widget

00:00:08.250 --> 00:00:10.460
that you would find on Khan Academy.

00:00:10.460 --> 00:00:14.410
And it first bears mentioning
how this widget works.

00:00:14.410 --> 00:00:16.330
So this point right over here,

00:00:16.330 --> 00:00:19.670
it helps you define the midline,

00:00:19.670 --> 00:00:20.800
the thing that you could imagine

00:00:20.800 --> 00:00:23.540
your sine or cosine
function oscillates around,

00:00:23.540 --> 00:00:26.400
and then you also define a
neighboring extreme point,

00:00:26.400 --> 00:00:29.240
either a maximum or a minimum point

00:00:29.240 --> 00:00:32.060
to graph your function.

00:00:32.060 --> 00:00:33.700
So let's think about how we would do this,

00:00:33.700 --> 00:00:35.470
and like always, I encourage
you to pause this video

00:00:35.470 --> 00:00:37.810
and think about how you
would do it yourself.

00:00:37.810 --> 00:00:39.510
But the first way I like to think about it

00:00:39.510 --> 00:00:42.010
is what would a regular,
just, if this just said

00:00:42.010 --> 00:00:45.370
y is equal to sine of x,
how would I graph that?

00:00:45.370 --> 00:00:47.560
Well, sine of 0 is 0.

00:00:47.560 --> 00:00:52.560
Sine of pi over 2 is 1.

00:00:52.770 --> 00:00:55.530
And then sine of pi is 0, again.

00:00:55.530 --> 00:00:59.550
And so this is what just regular
sine of x would look like.

00:00:59.550 --> 00:01:01.770
But let's think about
how this is different.

00:01:01.770 --> 00:01:04.300
Well, first of all,
it's not just sine of x,

00:01:04.300 --> 00:01:06.820
it's sine of 1/2x.

00:01:06.820 --> 00:01:10.730
So what would be the graph
of just sine of 1/2x?

00:01:10.730 --> 00:01:11.940
Well, one way to think about it,

00:01:11.940 --> 00:01:13.660
there's actually two
ways to think about it,

00:01:13.660 --> 00:01:16.720
is a coefficient right
over here on your x term

00:01:16.720 --> 00:01:19.160
that tells you how fast

00:01:19.160 --> 00:01:22.070
the thing that's being
inputted into sine is growing.

00:01:22.070 --> 00:01:24.720
And now it's going to grow half as fast.

00:01:24.720 --> 00:01:26.680
And so one way to think about it is

00:01:26.680 --> 00:01:29.920
your period is now going
to be twice as long.

00:01:29.920 --> 00:01:31.210
So one way to think about it is

00:01:31.210 --> 00:01:33.940
instead of getting to this next
maximum point at pi over 2,

00:01:33.940 --> 00:01:35.620
you're going to get there at pi.

00:01:35.620 --> 00:01:36.770
And you could test that.

00:01:36.770 --> 00:01:39.210
If you at, when x is equal to pi,

00:01:39.210 --> 00:01:42.120
this will be 1/2 pi, sine of 1/2 pi,

00:01:42.120 --> 00:01:44.930
is indeed equal to 1.

00:01:44.930 --> 00:01:47.100
Another way to think about it is

00:01:47.100 --> 00:01:48.640
you might be familiar with the formula,

00:01:48.640 --> 00:01:49.920
although I always like you to think

00:01:49.920 --> 00:01:51.790
about where these formulas come from,

00:01:51.790 --> 00:01:55.490
that to figure out the period
of a sine or cosine function,

00:01:55.490 --> 00:01:57.080
you take 2 pi and you divide it

00:01:57.080 --> 00:01:59.370
by whatever this coefficient is.

00:01:59.370 --> 00:02:02.040
So 2 pi divided by 1/2
is going to be 4 pi.

00:02:02.040 --> 00:02:03.160
And you could see the period here,

00:02:03.160 --> 00:02:07.440
we go up, down, and back
to where we were over 4 pi.

00:02:07.440 --> 00:02:08.400
And that makes sense,

00:02:08.400 --> 00:02:10.360
because if you just had
a 1 coefficient here,

00:02:10.360 --> 00:02:12.950
your period would be 2 pi, 2 pi radians.

00:02:12.950 --> 00:02:15.820
You make one circle around the unit circle

00:02:15.820 --> 00:02:17.430
is one way to think about it.

00:02:17.430 --> 00:02:21.670
So right here we have the
graph of sine of 1/2x.

00:02:21.670 --> 00:02:23.370
Now what if we wanted to, instead,

00:02:23.370 --> 00:02:27.180
think about 3 times the
graph of sine of 1/2x,

00:02:27.180 --> 00:02:29.560
or 3 sine 1/2x?

00:02:29.560 --> 00:02:31.010
Well then our amplitude's just going to be

00:02:31.010 --> 00:02:32.450
three times as much.

00:02:32.450 --> 00:02:35.450
And so instead of our
maximum point going from,

00:02:35.450 --> 00:02:37.030
instead of our maximum point being at 1,

00:02:37.030 --> 00:02:38.910
it will now be at 3.

00:02:38.910 --> 00:02:40.500
Or another way to think about it is

00:02:40.500 --> 00:02:44.500
we're going 3 above the midline
and 3 below the midline.

00:02:44.500 --> 00:02:48.560
So this right over here is
the graph of 3 sine of 1/2x.

00:02:48.560 --> 00:02:50.620
Now we have one thing left to do,

00:02:50.620 --> 00:02:52.330
and this is this minus 2.

00:02:52.330 --> 00:02:55.960
So this minus 2 is just going
to shift everything down by 2.

00:02:55.960 --> 00:02:57.810
So we just have to shift everything down.

00:02:57.810 --> 00:02:59.780
So let me shift this one down by 2

00:02:59.780 --> 00:03:01.760
and let me shift it this one down by 2.

00:03:01.760 --> 00:03:03.460
And so there you have it.

00:03:03.460 --> 00:03:07.250
Notice our period is still 4 pi.

00:03:07.250 --> 00:03:09.770
Our amplitude, how much we oscillate

00:03:09.770 --> 00:03:12.670
above or below the midline, is still 3.

00:03:12.670 --> 00:03:14.600
And now we have this minus 2.

00:03:14.600 --> 00:03:17.180
Another way to think about
it, when x is equal to 0,

00:03:17.180 --> 00:03:19.140
this whole first term is going to be 0,

00:03:19.140 --> 00:03:21.530
and y should be equal to negative 2?

00:03:21.530 --> 00:03:22.563
And we're done.