[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.00,0:00:01.22,Default,,0000,0000,0000,,
Dialogue: 0,0:00:01.22,0:00:03.48,Default,,0000,0000,0000,,I think it makes sense to keep\Ndoing a few more problems just
Dialogue: 0,0:00:03.48,0:00:06.62,Default,,0000,0000,0000,,so you really get intimately\Ncomfortable with PV equals
Dialogue: 0,0:00:06.62,0:00:08.95,Default,,0000,0000,0000,,nRT, or the ideal\Ngas equation.
Dialogue: 0,0:00:08.95,0:00:15.24,Default,,0000,0000,0000,,So we have 98 milliliters\Nof an unknown gas.
Dialogue: 0,0:00:15.24,0:00:17.98,Default,,0000,0000,0000,,
Dialogue: 0,0:00:17.98,0:00:20.75,Default,,0000,0000,0000,,They say weighs in the problem,\Nbut I'll use mass.
Dialogue: 0,0:00:20.75,0:00:26.13,Default,,0000,0000,0000,,Its mass is 0.081 grams. You\Nshouldn't say weighs 0.081
Dialogue: 0,0:00:26.13,0:00:27.17,Default,,0000,0000,0000,,grams. This is mass.
Dialogue: 0,0:00:27.17,0:00:29.95,Default,,0000,0000,0000,,Weight would be in newtons if\Nwe're in the metric system at
Dialogue: 0,0:00:29.95,0:00:32.87,Default,,0000,0000,0000,,standard temperature\Nand pressure.
Dialogue: 0,0:00:32.87,0:00:37.74,Default,,0000,0000,0000,,Calculate the molar\Nmass of the gas.
Dialogue: 0,0:00:37.74,0:00:39.67,Default,,0000,0000,0000,,So they want to know\Nthe mass per mole.
Dialogue: 0,0:00:39.67,0:00:49.75,Default,,0000,0000,0000,,So the molar mass, or\Nthe mass per mole.
Dialogue: 0,0:00:49.75,0:00:55.54,Default,,0000,0000,0000,,Now, we could just substitute\Nthis into PV is equal to nRT.
Dialogue: 0,0:00:55.54,0:00:57.76,Default,,0000,0000,0000,,We know what standard\Ntemperature and pressure are.
Dialogue: 0,0:00:57.76,0:01:01.22,Default,,0000,0000,0000,,Standard temperature is\N273 degrees Kelvin.
Dialogue: 0,0:01:01.22,0:01:06.41,Default,,0000,0000,0000,,Standard pressure\Nis 1 atmosphere.
Dialogue: 0,0:01:06.41,0:01:09.94,Default,,0000,0000,0000,,
Dialogue: 0,0:01:09.94,0:01:15.22,Default,,0000,0000,0000,,And then, of course, they're\Ngiving us 98 milliliters.
Dialogue: 0,0:01:15.22,0:01:20.30,Default,,0000,0000,0000,,And then you can just solve for\Nhow many moles we have.
Dialogue: 0,0:01:20.30,0:01:21.73,Default,,0000,0000,0000,,And actually, maybe\NI'll do that.
Dialogue: 0,0:01:21.73,0:01:24.21,Default,,0000,0000,0000,,But the other way to think about\Nit is that at standard
Dialogue: 0,0:01:24.21,0:01:27.01,Default,,0000,0000,0000,,temperature and pressure, an\Nideal gas-- and we did this
Dialogue: 0,0:01:27.01,0:01:27.90,Default,,0000,0000,0000,,two videos ago.
Dialogue: 0,0:01:27.90,0:01:34.60,Default,,0000,0000,0000,,We said 1 mole of an ideal gas\Nat standard temperature and
Dialogue: 0,0:01:34.60,0:01:39.77,Default,,0000,0000,0000,,pressure will fill up a\Nvolume of 22.4 liters.
Dialogue: 0,0:01:39.77,0:01:42.75,Default,,0000,0000,0000,,And I'm normally not a big fan\Nof memorizing things, but this
Dialogue: 0,0:01:42.75,0:01:45.16,Default,,0000,0000,0000,,might be something handy to\Nmemorize if you want to get
Dialogue: 0,0:01:45.16,0:01:47.40,Default,,0000,0000,0000,,through your chemistry test\Nfast. But you can always
Dialogue: 0,0:01:47.40,0:01:49.18,Default,,0000,0000,0000,,derive this if you know what\Nstandard temperature and
Dialogue: 0,0:01:49.18,0:01:51.72,Default,,0000,0000,0000,,pressure is by PV equals nRT.
Dialogue: 0,0:01:51.72,0:01:54.39,Default,,0000,0000,0000,,But if you know that 1 mole is\Ngoing to take up that much
Dialogue: 0,0:01:54.39,0:02:03.65,Default,,0000,0000,0000,,space, so one mole is to 22.4\Nliters, as how many moles,
Dialogue: 0,0:02:03.65,0:02:05.43,Default,,0000,0000,0000,,let's say x moles in\Nthis question.
Dialogue: 0,0:02:05.43,0:02:08.49,Default,,0000,0000,0000,,We want to figure out how many\Nmoles of the gas we have and
Dialogue: 0,0:02:08.49,0:02:14.14,Default,,0000,0000,0000,,we know we are at 98\Nmilliliters, so 0.098, so we
Dialogue: 0,0:02:14.14,0:02:16.27,Default,,0000,0000,0000,,set up this proportional\Nequation and we could figure
Dialogue: 0,0:02:16.27,0:02:18.06,Default,,0000,0000,0000,,out how many moles we're\Ndealing with.
Dialogue: 0,0:02:18.06,0:02:22.24,Default,,0000,0000,0000,,If 1 mole takes up 22.4 liters,\Nthen our number of
Dialogue: 0,0:02:22.24,0:02:25.03,Default,,0000,0000,0000,,moles are going to take up\N0.098, and this is an ideal
Dialogue: 0,0:02:25.03,0:02:26.90,Default,,0000,0000,0000,,gas in both circumstances.
Dialogue: 0,0:02:26.90,0:02:34.25,Default,,0000,0000,0000,,So we could say 22.4x\Nis equal to 0.098.
Dialogue: 0,0:02:34.25,0:02:40.63,Default,,0000,0000,0000,,And then we have x is equal\Nto 0.098 divided by 22.4.
Dialogue: 0,0:02:40.63,0:02:43.31,Default,,0000,0000,0000,,And this, of course,\Nis in moles.
Dialogue: 0,0:02:43.31,0:03:05.48,Default,,0000,0000,0000,,So 0.098 divided by 22.4 is\Nequal to 0.004375 moles.
Dialogue: 0,0:03:05.48,0:03:10.54,Default,,0000,0000,0000,,And they're telling us that\Nthis amount has a mass of
Dialogue: 0,0:03:10.54,0:03:15.52,Default,,0000,0000,0000,,0.081 grams. So let\Nme get the number.
Dialogue: 0,0:03:15.52,0:03:18.03,Default,,0000,0000,0000,,So how many grams are\Nthere per mole?
Dialogue: 0,0:03:18.03,0:03:24.39,Default,,0000,0000,0000,,So we take 0.081 grams and we\Njust did the math to figure
Dialogue: 0,0:03:24.39,0:03:30.95,Default,,0000,0000,0000,,out that we're dealing\Nwith 0.004375 moles.
Dialogue: 0,0:03:30.95,0:03:34.14,Default,,0000,0000,0000,,So how many grams do\Nwe have per mole?
Dialogue: 0,0:03:34.14,0:03:35.61,Default,,0000,0000,0000,,Let's take the calculator out.
Dialogue: 0,0:03:35.61,0:03:50.53,Default,,0000,0000,0000,,So we have 0.081 divided by\N0.004375 is equal to 18.51.
Dialogue: 0,0:03:50.53,0:03:58.84,Default,,0000,0000,0000,,So it's equals 18.5\Ngrams per mole.
Dialogue: 0,0:03:58.84,0:04:01.46,Default,,0000,0000,0000,,
Dialogue: 0,0:04:01.46,0:04:03.33,Default,,0000,0000,0000,,So now this is an interesting\Nquestion.
Dialogue: 0,0:04:03.33,0:04:06.45,Default,,0000,0000,0000,,So we figured out the molar mass\Nof our mystery substance
Dialogue: 0,0:04:06.45,0:04:11.68,Default,,0000,0000,0000,,that took up 98 milliliters and\Nhad a mass of 0.081 grams
Dialogue: 0,0:04:11.68,0:04:14.16,Default,,0000,0000,0000,,at standard temperature and\Npressure, and we figured out
Dialogue: 0,0:04:14.16,0:04:19.17,Default,,0000,0000,0000,,its molar mass, or its mass per\Nmole, is 18.5 grams. So
Dialogue: 0,0:04:19.17,0:04:23.70,Default,,0000,0000,0000,,any guess as to what molecule\Nwe're dealing with?
Dialogue: 0,0:04:23.70,0:04:25.63,Default,,0000,0000,0000,,18.5 grams.
Dialogue: 0,0:04:25.63,0:04:28.99,Default,,0000,0000,0000,,And probably it's not going to\Nbe exact, but at least in my
Dialogue: 0,0:04:28.99,0:04:31.68,Default,,0000,0000,0000,,brain, water seems to\Nbe a good candidate.
Dialogue: 0,0:04:31.68,0:04:33.39,Default,,0000,0000,0000,,Water is H20.
Dialogue: 0,0:04:33.39,0:04:35.37,Default,,0000,0000,0000,,Maybe I should do it in blue\Nbecause it's water.
Dialogue: 0,0:04:35.37,0:04:38.03,Default,,0000,0000,0000,,
Dialogue: 0,0:04:38.03,0:04:40.26,Default,,0000,0000,0000,,The mass of each\Nhydrogen is 1.
Dialogue: 0,0:04:40.26,0:04:43.70,Default,,0000,0000,0000,,Remember, hydrogen, at least\Nin its most standard form,
Dialogue: 0,0:04:43.70,0:04:45.61,Default,,0000,0000,0000,,doesn't have a neutron, so it's\Nreally just a proton and
Dialogue: 0,0:04:45.61,0:04:48.81,Default,,0000,0000,0000,,and an electron, so it has an\Natomic mass of 1 or a molar
Dialogue: 0,0:04:48.81,0:04:50.29,Default,,0000,0000,0000,,mass of 1 gram.
Dialogue: 0,0:04:50.29,0:04:53.70,Default,,0000,0000,0000,,And oxygen has a molar mass of\N16 grams. So you have two
Dialogue: 0,0:04:53.70,0:04:57.72,Default,,0000,0000,0000,,hydrogens, so it's 2 plus\N16 is equal to 18.
Dialogue: 0,0:04:57.72,0:05:01.97,Default,,0000,0000,0000,,So it looks like our mystery\Nsubstance is water.
Dialogue: 0,0:05:01.97,0:05:02.94,Default,,0000,0000,0000,,