ceaglex/VisualTTS_Chem · Datasets at Hugging Face

text stringlengths 19 206
qpmTRXIJhFo-031\|Now, two p electrons are very well shielded by s electrons, because the electrons have access to the nucleus.
qpmTRXIJhFo-032\|There's no nodes in an s orbital at the nucleus.
qpmTRXIJhFo-033\|But the p electrons do have a node at the nucleus, an angular node.
qpmTRXIJhFo-036\|So shielding is very effective of low l-- l equals 0-- s electrons on p and higher electrons.
pJ9_mvvB98Y-000\|I'd like to show you two simple chemical reactions, the dissolving of salts in water.
pJ9_mvvB98Y-014\|But the energy was not a good predictor.
pJ9_mvvB98Y-015\|One absorbed energy as it proceeded forward, the other released energy as it proceeded forward.
pJ9_mvvB98Y-016\|So there has to be another thermodynamic parameter that will predict what's the forward direction for chemical reactions?
pJ9_mvvB98Y-017\|What's the preferred way that chemical reactions will go?
pJ9_mvvB98Y-018\|We need to find a thermodynamic parameter, and that's the subject of this lesson.
Mh8MzONj6sM-000\|Let's look at the reaction between two weak acids, HA1 and HA2.
Mh8MzONj6sM-001\|And p Ka1 is smaller than p Ka2.
Mh8MzONj6sM-003\|What is the expression for the equilibrium constant?
Mh8MzONj6sM-004\|Is it KA1 plus KA2, KA1 times KA2, or KA1/KA2?
Mh8MzONj6sM-011\|We're talking about the reaction of HA1, a weak acid, and A 2-, the conjugate base of another weak acid.
Mh8MzONj6sM-012\|So how do those two react?
Mh8MzONj6sM-013\|Well, here's HA1 reacting with water.
Mh8MzONj6sM-017\|But what we want is HA1 reacting with A 2-.
Mh8MzONj6sM-018\|So let's reverse this lower reaction.
Mh8MzONj6sM-023\|Now, you can tell which is the stronger acid to which side this equilibrium will lie.
Mh8MzONj6sM-024\|And you'd probably say that intuitively.
Mh8MzONj6sM-025\|You'd say if HA1 is the stronger acid, then we'll favor the products here.
Mh8MzONj6sM-026\|If HA2 is the stronger acid, the reaction would be pushed back this way, and we'd favor the reactants.
Mh8MzONj6sM-027\|And this expression tells you the same thing, because if HA1 is the stronger acid, then KA1 is larger than KA2.
Mh8MzONj6sM-028\|And that gives you a K bigger than 1, or a reaction that favors the products.
Mh8MzONj6sM-029\|If HA2 is the stronger acid, then KA2 is larger.
Mh8MzONj6sM-030\|And this has a K less than 1, or one that favors the reactants.
FCnePMnxfF0-000\|Paramagnetic species are species that have at least one unpaired electron.
FCnePMnxfF0-001\|For which species does the paramagnetism increase upon ionization?
FCnePMnxfF0-002\|Carbon, nitrogen, or neon?
FCnePMnxfF0-009\|We're looking for a species that increases its paramagnetism upon ionization.
FCnePMnxfF0-011\|Neon, all the electrons are paired.
FCnePMnxfF0-012\|Nitrogen, three unpaired electrons.
FCnePMnxfF0-013\|Carbon, two unpaired electrons.
EgZZXsWKM7o-000\|Molecular geometry and individual bond dipole moments often lead to molecules having dipole moments.
EgZZXsWKM7o-008\|We're determining dipole moments for molecules.
EgZZXsWKM7o-010\|So I've done those steps for these molecules.
EgZZXsWKM7o-011\|And I've presented just the final molecular geometry.
EgZZXsWKM7o-012\|So for CHCl2, a tetrahedral confirmation, and this tetrahedral confirmation is going to lead to a dipole moment.
EgZZXsWKM7o-013\|The Cl atoms are more electronegative than the carbon.
EgZZXsWKM7o-014\|There'll be a dipole moment pointing away from the Cl atoms.
EgZZXsWKM7o-015\|And the H atoms, there will also be a small dipole moment.
EgZZXsWKM7o-016\|Those won't cancel out.
EgZZXsWKM7o-017\|The whole molecule will have a dipole moment.
EgZZXsWKM7o-021\|And for BrF3, now this molecule is not trigonal.
EgZZXsWKM7o-022\|If it were perfectly trigonal and symmetric, the dipoles might cancel out.
EgZZXsWKM7o-026\|And indeed, it's common-- more common for molecules to have dipole moments than to not have them, because remember, every individual bond has a dipole moment.
EgZZXsWKM7o-027\|And for them to cancel out, you need a very exact symmetry in the molecule.
EgZZXsWKM7o-028\|These molecules, not symmetric, so they have dipole moments.
Q9oGYNqs5TU-000\|Let's look at some bond angles in two different molecules.
Q9oGYNqs5TU-002\|Well, looking at it here, it looks like those angles are both 90 degrees.
Q9oGYNqs5TU-003\|And of course, that's just because we've drawn them straight out on paper.
Q9oGYNqs5TU-004\|In order to determine the actual bond angle, you have to go to a good Lewis dot structure, get steric numbers, and go to the molecular geometry.
Q9oGYNqs5TU-005\|We can actually do that.
Q9oGYNqs5TU-006\|So about the carbon in question, I've drawn the appropriate Lewis dot structure.
Q9oGYNqs5TU-007\|I haven't drawn the full Lewis dot structure, but I've drawn the full one, and then taken just the important part.
Q9oGYNqs5TU-008\|That allows me to determine the steric number around this carbon and this carbon.
Q9oGYNqs5TU-009\|So the steric number here around this carbon, 1, 2, 3, 4.
Q9oGYNqs5TU-010\|That carbon has to accommodate four things.
Q9oGYNqs5TU-011\|This carbon has to accommodate 1, 2, 3 things.
Q9oGYNqs5TU-012\|Now, again, it's not the number of bonds.
Q9oGYNqs5TU-013\|It's the number of things bonded to that carbon, an oxygen, a hydrogen, and a carbon.
Q9oGYNqs5TU-014\|This has two hydrogens, an oxygen, and a carbon.
Q9oGYNqs5TU-015\|Steric number four here, so the bond angle is going to be tetrahedral.
Q9oGYNqs5TU-017\|So the bond angle increases as I do that oxidation.
6V6JgmfTpyY-000\|Let's think about hydrocarbon combustion in terms of a ChemQuiz.
6V6JgmfTpyY-001\|I'm going to take a hydrocarbon, burn it in oxygen, and take the product, inject it into my mass spectrometer, and generate a mass spectrum.
6V6JgmfTpyY-002\|The question I have is, which of these hydrocarbons would give this mass spectrum after reaction with oxygen?
6V6JgmfTpyY-009\|Let's look at each of those hydrocarbons, their reaction with oxygen, and the mass spectrum of the products.
6V6JgmfTpyY-014\|So the mass spectrum of the products would show a one to one peak height at 18 and 44 from the carbon dioxide in the water.
SswF_wnX1Pg-000\|In electrochemistry, we tabulate the various half-cell reactions relative to the standard hydrogen electrode.
SswF_wnX1Pg-001\|And here I've done that.
SswF_wnX1Pg-003\|I've arranged them from high potential to low potential, from large positive potential to large negative potential.
SswF_wnX1Pg-004\|In doing that, that tells me a few things.
SswF_wnX1Pg-006\|That means copper ions are relatively strong oxidizing agents.
SswF_wnX1Pg-007\|They're willing to accept electrons.
SswF_wnX1Pg-009\|When they withdraw electrons, they're oxidizing that metal.
SswF_wnX1Pg-010\|The copper ions themselves are being reduced.
SswF_wnX1Pg-011\|Conversely, this zinc system would work the opposite way.
SswF_wnX1Pg-012\|When zinc metal is in contact with ions, the tendency is for zinc metal to give up electrons to those ions.
SswF_wnX1Pg-013\|So the zinc metal is a reducing agent.
SswF_wnX1Pg-014\|It tends to donate electrons, to give up its electrons, and thereby reduce something while the zinc metal itself is oxidized.
SswF_wnX1Pg-016\|So I can write down now the reaction.
SswF_wnX1Pg-018\|I could look at it the other way and say the zinc metal is going to donate electrons to that copper ion.
SswF_wnX1Pg-024\|So the potential for this reaction to go should remind you of another thermodynamic parameter-- the free energy.
SswF_wnX1Pg-025\|The potential for this reaction to go, the fact that it favors the products here, sounds like the free energy.
SswF_wnX1Pg-026\|And indeed, the potential is related to the standard state free energy difference.
SswF_wnX1Pg-027\|The standard state free energy difference is given by minus nF delta E, the potential difference.
SswF_wnX1Pg-028\|In this case, n is the number of electrons transferred in the process, in that case, two electrons.
SswF_wnX1Pg-029\|F is Faraday's constant, that's the charge on a mole of electrons.
SswF_wnX1Pg-030\|So what we see is the free energy difference related to the voltage in a way we understand.
SswF_wnX1Pg-031\|A positive voltage, the potential for this reaction to go as positive, relates to a negative free energy difference.
fryL1jN2OHY-000\|Certain bacteria can live around oceanic thermal vents.
fryL1jN2OHY-002\|High pressure means water has a higher boiling point, and you can have liquid water above 100 degrees C.
fryL1jN2OHY-003\|Now, temperatures above 100 degrees C normally kill bacteria.
fryL1jN2OHY-004\|In fact, you can use hot water and steam to disinfect surfaces, to remove bacteria.
fryL1jN2OHY-005\|These bacteria, however, thrive there.
fryL1jN2OHY-021\|Well, what about at higher temperatures?
3tl5gAwHj6w-000\|Molecular shape and geometry is motivated by trying to keep everything apart in space.
3tl5gAwHj6w-001\|So if you have lone pairs or if you have a bonded atom, you'd like to arrange them as far apart as possible.

qpmTRXIJhFo-031|Now, two p electrons are very well shielded by s electrons, because the electrons have access to the nucleus.

qpmTRXIJhFo-032|There's no nodes in an s orbital at the nucleus.

qpmTRXIJhFo-033|But the p electrons do have a node at the nucleus, an angular node.

qpmTRXIJhFo-036|So shielding is very effective of low l-- l equals 0-- s electrons on p and higher electrons.

pJ9_mvvB98Y-000|I'd like to show you two simple chemical reactions, the dissolving of salts in water.

pJ9_mvvB98Y-014|But the energy was not a good predictor.

pJ9_mvvB98Y-015|One absorbed energy as it proceeded forward, the other released energy as it proceeded forward.

pJ9_mvvB98Y-016|So there has to be another thermodynamic parameter that will predict what's the forward direction for chemical reactions?

pJ9_mvvB98Y-017|What's the preferred way that chemical reactions will go?

pJ9_mvvB98Y-018|We need to find a thermodynamic parameter, and that's the subject of this lesson.

Mh8MzONj6sM-000|Let's look at the reaction between two weak acids, HA1 and HA2.

Mh8MzONj6sM-001|And p Ka1 is smaller than p Ka2.

Mh8MzONj6sM-003|What is the expression for the equilibrium constant?

Mh8MzONj6sM-004|Is it KA1 plus KA2, KA1 times KA2, or KA1/KA2?

Mh8MzONj6sM-011|We're talking about the reaction of HA1, a weak acid, and A 2-, the conjugate base of another weak acid.

Mh8MzONj6sM-012|So how do those two react?

Mh8MzONj6sM-013|Well, here's HA1 reacting with water.

Mh8MzONj6sM-017|But what we want is HA1 reacting with A 2-.

Mh8MzONj6sM-018|So let's reverse this lower reaction.

Mh8MzONj6sM-023|Now, you can tell which is the stronger acid to which side this equilibrium will lie.

Mh8MzONj6sM-024|And you'd probably say that intuitively.

Mh8MzONj6sM-025|You'd say if HA1 is the stronger acid, then we'll favor the products here.

Mh8MzONj6sM-026|If HA2 is the stronger acid, the reaction would be pushed back this way, and we'd favor the reactants.

Mh8MzONj6sM-027|And this expression tells you the same thing, because if HA1 is the stronger acid, then KA1 is larger than KA2.

Mh8MzONj6sM-028|And that gives you a K bigger than 1, or a reaction that favors the products.

Mh8MzONj6sM-029|If HA2 is the stronger acid, then KA2 is larger.

Mh8MzONj6sM-030|And this has a K less than 1, or one that favors the reactants.

FCnePMnxfF0-000|Paramagnetic species are species that have at least one unpaired electron.

FCnePMnxfF0-001|For which species does the paramagnetism increase upon ionization?

FCnePMnxfF0-002|Carbon, nitrogen, or neon?

FCnePMnxfF0-009|We're looking for a species that increases its paramagnetism upon ionization.

FCnePMnxfF0-011|Neon, all the electrons are paired.

FCnePMnxfF0-012|Nitrogen, three unpaired electrons.

FCnePMnxfF0-013|Carbon, two unpaired electrons.

EgZZXsWKM7o-000|Molecular geometry and individual bond dipole moments often lead to molecules having dipole moments.

EgZZXsWKM7o-008|We're determining dipole moments for molecules.

EgZZXsWKM7o-010|So I've done those steps for these molecules.

EgZZXsWKM7o-011|And I've presented just the final molecular geometry.

EgZZXsWKM7o-012|So for CHCl2, a tetrahedral confirmation, and this tetrahedral confirmation is going to lead to a dipole moment.

EgZZXsWKM7o-013|The Cl atoms are more electronegative than the carbon.

EgZZXsWKM7o-014|There'll be a dipole moment pointing away from the Cl atoms.

EgZZXsWKM7o-015|And the H atoms, there will also be a small dipole moment.

EgZZXsWKM7o-016|Those won't cancel out.

EgZZXsWKM7o-017|The whole molecule will have a dipole moment.

EgZZXsWKM7o-021|And for BrF3, now this molecule is not trigonal.

EgZZXsWKM7o-022|If it were perfectly trigonal and symmetric, the dipoles might cancel out.

EgZZXsWKM7o-026|And indeed, it's common-- more common for molecules to have dipole moments than to not have them, because remember, every individual bond has a dipole moment.

EgZZXsWKM7o-027|And for them to cancel out, you need a very exact symmetry in the molecule.

EgZZXsWKM7o-028|These molecules, not symmetric, so they have dipole moments.

Q9oGYNqs5TU-000|Let's look at some bond angles in two different molecules.

Q9oGYNqs5TU-002|Well, looking at it here, it looks like those angles are both 90 degrees.

Q9oGYNqs5TU-003|And of course, that's just because we've drawn them straight out on paper.

Q9oGYNqs5TU-004|In order to determine the actual bond angle, you have to go to a good Lewis dot structure, get steric numbers, and go to the molecular geometry.

Q9oGYNqs5TU-005|We can actually do that.

Q9oGYNqs5TU-006|So about the carbon in question, I've drawn the appropriate Lewis dot structure.

Q9oGYNqs5TU-007|I haven't drawn the full Lewis dot structure, but I've drawn the full one, and then taken just the important part.

Q9oGYNqs5TU-008|That allows me to determine the steric number around this carbon and this carbon.

Q9oGYNqs5TU-009|So the steric number here around this carbon, 1, 2, 3, 4.

Q9oGYNqs5TU-010|That carbon has to accommodate four things.

Q9oGYNqs5TU-011|This carbon has to accommodate 1, 2, 3 things.

Q9oGYNqs5TU-012|Now, again, it's not the number of bonds.

Q9oGYNqs5TU-013|It's the number of things bonded to that carbon, an oxygen, a hydrogen, and a carbon.

Q9oGYNqs5TU-014|This has two hydrogens, an oxygen, and a carbon.

Q9oGYNqs5TU-015|Steric number four here, so the bond angle is going to be tetrahedral.

Q9oGYNqs5TU-017|So the bond angle increases as I do that oxidation.

6V6JgmfTpyY-000|Let's think about hydrocarbon combustion in terms of a ChemQuiz.

6V6JgmfTpyY-001|I'm going to take a hydrocarbon, burn it in oxygen, and take the product, inject it into my mass spectrometer, and generate a mass spectrum.

6V6JgmfTpyY-002|The question I have is, which of these hydrocarbons would give this mass spectrum after reaction with oxygen?

6V6JgmfTpyY-009|Let's look at each of those hydrocarbons, their reaction with oxygen, and the mass spectrum of the products.

6V6JgmfTpyY-014|So the mass spectrum of the products would show a one to one peak height at 18 and 44 from the carbon dioxide in the water.

SswF_wnX1Pg-000|In electrochemistry, we tabulate the various half-cell reactions relative to the standard hydrogen electrode.

SswF_wnX1Pg-001|And here I've done that.

SswF_wnX1Pg-003|I've arranged them from high potential to low potential, from large positive potential to large negative potential.

SswF_wnX1Pg-004|In doing that, that tells me a few things.

SswF_wnX1Pg-006|That means copper ions are relatively strong oxidizing agents.

SswF_wnX1Pg-007|They're willing to accept electrons.

SswF_wnX1Pg-009|When they withdraw electrons, they're oxidizing that metal.

SswF_wnX1Pg-010|The copper ions themselves are being reduced.

SswF_wnX1Pg-011|Conversely, this zinc system would work the opposite way.

SswF_wnX1Pg-012|When zinc metal is in contact with ions, the tendency is for zinc metal to give up electrons to those ions.

SswF_wnX1Pg-013|So the zinc metal is a reducing agent.

SswF_wnX1Pg-014|It tends to donate electrons, to give up its electrons, and thereby reduce something while the zinc metal itself is oxidized.

SswF_wnX1Pg-016|So I can write down now the reaction.

SswF_wnX1Pg-018|I could look at it the other way and say the zinc metal is going to donate electrons to that copper ion.

SswF_wnX1Pg-024|So the potential for this reaction to go should remind you of another thermodynamic parameter-- the free energy.

SswF_wnX1Pg-025|The potential for this reaction to go, the fact that it favors the products here, sounds like the free energy.

SswF_wnX1Pg-026|And indeed, the potential is related to the standard state free energy difference.

SswF_wnX1Pg-027|The standard state free energy difference is given by minus nF delta E, the potential difference.

SswF_wnX1Pg-028|In this case, n is the number of electrons transferred in the process, in that case, two electrons.

SswF_wnX1Pg-029|F is Faraday's constant, that's the charge on a mole of electrons.

SswF_wnX1Pg-030|So what we see is the free energy difference related to the voltage in a way we understand.

SswF_wnX1Pg-031|A positive voltage, the potential for this reaction to go as positive, relates to a negative free energy difference.

fryL1jN2OHY-000|Certain bacteria can live around oceanic thermal vents.

fryL1jN2OHY-002|High pressure means water has a higher boiling point, and you can have liquid water above 100 degrees C.

fryL1jN2OHY-003|Now, temperatures above 100 degrees C normally kill bacteria.

fryL1jN2OHY-004|In fact, you can use hot water and steam to disinfect surfaces, to remove bacteria.

fryL1jN2OHY-005|These bacteria, however, thrive there.

fryL1jN2OHY-021|Well, what about at higher temperatures?

3tl5gAwHj6w-000|Molecular shape and geometry is motivated by trying to keep everything apart in space.

3tl5gAwHj6w-001|So if you have lone pairs or if you have a bonded atom, you'd like to arrange them as far apart as possible.