ceaglex/VisualTTS_Chem · Datasets at Hugging Face

text stringlengths 19 206
sihnvASQFSw-022\|So how long is a half-life?
WitrD0bFL24-000\|We know electrons can absorb energy and change energy levels, go from one orbital to another, in an atom.
WitrD0bFL24-009\|We're talking about electronic transitions, going from n equal 1 to n equal 2.
WitrD0bFL24-010\|Which of these possible species can do that?
WitrD0bFL24-011\|Let's look at their electronic configurations.
WitrD0bFL24-012\|Here's fluorine, 1s2 2s2 2p5.
WitrD0bFL24-013\|Fluorine minus has an extra electron, one more electron than fluorine, so it has electronic configuration 1s2 2s2 2p6.
WitrD0bFL24-016\|So 1s2 2s2 2p6.
WitrD0bFL24-017\|Notice fluorine minus and neon have the same electronic configuration, plus it has an electron in the three s orbital.
Zfx54dYEDBs-007\|We're looking at the oxidation of NO2 minus to NO3 minus and looking at how the formal charge on nitrogen changes in that process.
Zfx54dYEDBs-012\|And it'll share one of these and two of those.
Zfx54dYEDBs-013\|So one, two, three, four, five electrons in a sharing mode for nitrogen in that molecule.
Zfx54dYEDBs-014\|This molecule, nitrogen shares all the electrons around it.
Zfx54dYEDBs-015\|It shares one from this bond, one from this bond, and two from this bond.
Zfx54dYEDBs-016\|One, two, three, four.
Zfx54dYEDBs-017\|Four electrons.
Zfx54dYEDBs-018\|Nitrogen as an atom has five electrons.
4blDQFbly-Y-000\|Let's look at the Gibbs free energy for a system.
4blDQFbly-Y-001\|The Gibbs function is defined as the enthalpy minus the temperature times the entropy.
4blDQFbly-Y-004\|You can talk about that for any system in general or you can talk about specifically the standard states of the system.
4blDQFbly-Y-014\|But the free energy is always a predictor of the direction of the chemical reaction.
4blDQFbly-Y-019\|So I have predictive power now, whether a reaction is likely to go based on the free energy of the system.
TPQjDIPwJ-A-000\|Let's do a calculation involving the photoelectric effect, ejected photo electrons and photons of a certain energy hitting a metal.
TPQjDIPwJ-A-002\|So this work function for chromium metal, 4.37 electron volts, is how strongly chromium holds on to its electrons.
TPQjDIPwJ-A-003\|So we have to say, well, what is the photoelectric effect?
TPQjDIPwJ-A-004\|The photoelectric effect, we have to balance the energies.
TPQjDIPwJ-A-006\|So the metal holds on to the electron, but if we bring in a high enough energy photon we can eject an electron with a certain kinetic energy.
TPQjDIPwJ-A-007\|So that's what we have to find.
TPQjDIPwJ-A-010\|Now, we do that to keep all the things we multiply together consistent unit wise.
TPQjDIPwJ-A-011\|We're dimensionally consistent.
TPQjDIPwJ-A-012\|We always use kilograms for mass, so don't just put your masses into your equations with grams or pounds or some random mass unit.
TPQjDIPwJ-A-013\|If you come across a mass, convert it to the kilograms.
TPQjDIPwJ-A-014\|If you come across a distance, a length, convert that to meters.
TPQjDIPwJ-A-015\|A time, seconds.
TPQjDIPwJ-A-016\|An energy, joules.
TPQjDIPwJ-A-022\|That's going to be an energy, a kinetic energy, so this will be joules.
TPQjDIPwJ-A-023\|In fact, that's how I remember the SI units of joules.
TPQjDIPwJ-A-030\|Tiny, tiny, tiny number of joules.
TPQjDIPwJ-A-031\|Of course, it's an electron.
TPQjDIPwJ-A-035\|Now electron volts in a unit of energy, it's the kinetic energy that an electron gains as you accelerate it across the potential of 1 volt.
TPQjDIPwJ-A-037\|So we can do that product, so the work function in terms of joules, 7.04 times 10 to the minus 19th joules.
TPQjDIPwJ-A-046\|And that's what we want.
TPQjDIPwJ-A-047\|It's always good to check, do the units that we have left makes sense for the quantity that we're solving for?
TPQjDIPwJ-A-048\|We're solving for wave length, a length.
TPQjDIPwJ-A-049\|Do I have meters?
TPQjDIPwJ-A-050\|In this case, I do.
TPQjDIPwJ-A-053\|109 nanometers is, as we recall, in the UV, the ultraviolet region.
TPQjDIPwJ-A-054\|We know visible went from 700 down to 400 nanometers.
wA_IvsdRjpM-000\|Let's look at the root mean squared velocity for bromine, at 300 Kelvin in a sample of gas, in our Nuts and Boltz section.
wA_IvsdRjpM-001\|And in honor of Maxwell Boltzmann distribution, we'll spell bolts with a z this time.
wA_IvsdRjpM-006\|And I can solve for that.
wA_IvsdRjpM-007\|I get 216 meters per second.
wA_IvsdRjpM-008\|Now, 216 meters per second, that's about 800 feet per second.
wA_IvsdRjpM-011\|That's amazing speeds for gas particles at modest temperatures.
ktySzUONBUU-000\|So how do we determine the exact mass of an atom?
ktySzUONBUU-001\|Well, one way is to use a mass spectrometer.
ktySzUONBUU-003\|A mass spectrometer, you take a sample and you ionize it.
ktySzUONBUU-004\|And by ionization, we mean stripping off electrons.
ktySzUONBUU-005\|In this case, we strip off all of the electrons.
ktySzUONBUU-006\|The sample is then accelerated towards an electric field and directed through a magnetic field.
ktySzUONBUU-007\|Now when charged particles hit a magnetic field, they are deflected by the magnetic field.
ktySzUONBUU-008\|And they're deflected according to their mass.
ktySzUONBUU-009\|Heavier particles are not as deflected as lighter particles.
ktySzUONBUU-010\|So you have this stream of particles and it hits that magnetic field, and it's fanned out based on their mass.
ktySzUONBUU-015\|Let's look at a larger mass coming through the same system.
ktySzUONBUU-016\|We're going to ionize it, it's not deflected as much.
ktySzUONBUU-017\|So you have a distribution of masses that you can resolve using a mass spectrometer.
ktySzUONBUU-019\|And the vertical axis will tell us about how many particles there were, and that's very useful.
ktySzUONBUU-020\|Remember when we had our carbon, we said that one in hundred of these particles was a carbon-13.
ktySzUONBUU-021\|The other 99 were carbon-12.
ktySzUONBUU-022\|That's a difference of one mass unit.
ktySzUONBUU-024\|So, let's look at a few masses.
ktySzUONBUU-025\|We have, for instance, what would show up at mass 1?
ktySzUONBUU-029\|So you can tell by the mass spectrometer you can get the mass.
ktySzUONBUU-030\|But remember, mass does not determine the identity of the element-- certainly a clue, but it does not determine the identity.
ktySzUONBUU-031\|We need to know the number of protons in the nucleus to get the identity of an element.
ktySzUONBUU-032\|So mass three, that could be tritium or it could be a molecule of deuterium and hydrogen hooked together to form a hydrogen molecule.
ktySzUONBUU-036\|And you can go on down the line.
ktySzUONBUU-037\|Here is oxygen 16.
ktySzUONBUU-038\|At mass 18, you could have a water molecule with oxygen 16 and two protons, two hydrogen atoms.
ktySzUONBUU-044\|So a mass spectrometer, exquisitely sensitive makes.
ktySzUONBUU-046\|The e equals mc squared mass loss can be determined by modern mass spectrometer.
ktySzUONBUU-047\|That's how accurate they are.
ktySzUONBUU-048\|Incredibly tiny mass losses can be determined by this method.
ktySzUONBUU-049\|It's a beautiful instrument and it's very, very, very powerful for measuring atomic masses.
ksZoOde-7zY-000\|We noticed that the trend in ionization energies is to kind of increase across a row in the periodic table.
ksZoOde-7zY-001\|Let's look at that in more detail.
ksZoOde-7zY-003\|That's a 1s electron.
ksZoOde-7zY-004\|For helium, about 2,400 kilojoules per mole for that ionization.
ksZoOde-7zY-006\|So increasing that positive charge on the nucleus was much more dramatic than adding another electron.
ksZoOde-7zY-007\|It's much harder to pull off an electron from helium than hydrogen.
ksZoOde-7zY-011\|And lithium is remarkably easy to ionize.
ksZoOde-7zY-018\|It's a general trend.
ksZoOde-7zY-019\|And we can kind of rationalize that.
ksZoOde-7zY-020\|Because what we're doing is we're adding more nuclear charge, but we're staying in the same principle quantum level.
ksZoOde-7zY-021\|So electrons in about the same region of space but more positive charge holding them.
ksZoOde-7zY-022\|Of course, those will be slightly more difficult to ionize.
ksZoOde-7zY-023\|As you continue across, there's a slight dip here at oxygen.
ksZoOde-7zY-024\|And I think we can understand that too in terms of the fact that nitrogen now is that half-filled state.
ksZoOde-7zY-025\|Oxygen has that paired electron.

sihnvASQFSw-022|So how long is a half-life?

WitrD0bFL24-000|We know electrons can absorb energy and change energy levels, go from one orbital to another, in an atom.

WitrD0bFL24-009|We're talking about electronic transitions, going from n equal 1 to n equal 2.

WitrD0bFL24-010|Which of these possible species can do that?

WitrD0bFL24-011|Let's look at their electronic configurations.

WitrD0bFL24-012|Here's fluorine, 1s2 2s2 2p5.

WitrD0bFL24-013|Fluorine minus has an extra electron, one more electron than fluorine, so it has electronic configuration 1s2 2s2 2p6.

WitrD0bFL24-016|So 1s2 2s2 2p6.

WitrD0bFL24-017|Notice fluorine minus and neon have the same electronic configuration, plus it has an electron in the three s orbital.

Zfx54dYEDBs-007|We're looking at the oxidation of NO2 minus to NO3 minus and looking at how the formal charge on nitrogen changes in that process.

Zfx54dYEDBs-012|And it'll share one of these and two of those.

Zfx54dYEDBs-013|So one, two, three, four, five electrons in a sharing mode for nitrogen in that molecule.

Zfx54dYEDBs-014|This molecule, nitrogen shares all the electrons around it.

Zfx54dYEDBs-015|It shares one from this bond, one from this bond, and two from this bond.

Zfx54dYEDBs-016|One, two, three, four.

Zfx54dYEDBs-017|Four electrons.

Zfx54dYEDBs-018|Nitrogen as an atom has five electrons.

4blDQFbly-Y-000|Let's look at the Gibbs free energy for a system.

4blDQFbly-Y-001|The Gibbs function is defined as the enthalpy minus the temperature times the entropy.

4blDQFbly-Y-004|You can talk about that for any system in general or you can talk about specifically the standard states of the system.

4blDQFbly-Y-014|But the free energy is always a predictor of the direction of the chemical reaction.

4blDQFbly-Y-019|So I have predictive power now, whether a reaction is likely to go based on the free energy of the system.

TPQjDIPwJ-A-000|Let's do a calculation involving the photoelectric effect, ejected photo electrons and photons of a certain energy hitting a metal.

TPQjDIPwJ-A-002|So this work function for chromium metal, 4.37 electron volts, is how strongly chromium holds on to its electrons.

TPQjDIPwJ-A-003|So we have to say, well, what is the photoelectric effect?

TPQjDIPwJ-A-004|The photoelectric effect, we have to balance the energies.

TPQjDIPwJ-A-006|So the metal holds on to the electron, but if we bring in a high enough energy photon we can eject an electron with a certain kinetic energy.

TPQjDIPwJ-A-007|So that's what we have to find.

TPQjDIPwJ-A-010|Now, we do that to keep all the things we multiply together consistent unit wise.

TPQjDIPwJ-A-011|We're dimensionally consistent.

TPQjDIPwJ-A-012|We always use kilograms for mass, so don't just put your masses into your equations with grams or pounds or some random mass unit.

TPQjDIPwJ-A-013|If you come across a mass, convert it to the kilograms.

TPQjDIPwJ-A-014|If you come across a distance, a length, convert that to meters.

TPQjDIPwJ-A-015|A time, seconds.

TPQjDIPwJ-A-016|An energy, joules.

TPQjDIPwJ-A-022|That's going to be an energy, a kinetic energy, so this will be joules.

TPQjDIPwJ-A-023|In fact, that's how I remember the SI units of joules.

TPQjDIPwJ-A-030|Tiny, tiny, tiny number of joules.

TPQjDIPwJ-A-031|Of course, it's an electron.

TPQjDIPwJ-A-035|Now electron volts in a unit of energy, it's the kinetic energy that an electron gains as you accelerate it across the potential of 1 volt.

TPQjDIPwJ-A-037|So we can do that product, so the work function in terms of joules, 7.04 times 10 to the minus 19th joules.

TPQjDIPwJ-A-046|And that's what we want.

TPQjDIPwJ-A-047|It's always good to check, do the units that we have left makes sense for the quantity that we're solving for?

TPQjDIPwJ-A-048|We're solving for wave length, a length.

TPQjDIPwJ-A-049|Do I have meters?

TPQjDIPwJ-A-050|In this case, I do.

TPQjDIPwJ-A-053|109 nanometers is, as we recall, in the UV, the ultraviolet region.

TPQjDIPwJ-A-054|We know visible went from 700 down to 400 nanometers.

wA_IvsdRjpM-000|Let's look at the root mean squared velocity for bromine, at 300 Kelvin in a sample of gas, in our Nuts and Boltz section.

wA_IvsdRjpM-001|And in honor of Maxwell Boltzmann distribution, we'll spell bolts with a z this time.

wA_IvsdRjpM-006|And I can solve for that.

wA_IvsdRjpM-007|I get 216 meters per second.

wA_IvsdRjpM-008|Now, 216 meters per second, that's about 800 feet per second.

wA_IvsdRjpM-011|That's amazing speeds for gas particles at modest temperatures.

ktySzUONBUU-000|So how do we determine the exact mass of an atom?

ktySzUONBUU-001|Well, one way is to use a mass spectrometer.

ktySzUONBUU-003|A mass spectrometer, you take a sample and you ionize it.

ktySzUONBUU-004|And by ionization, we mean stripping off electrons.

ktySzUONBUU-005|In this case, we strip off all of the electrons.

ktySzUONBUU-006|The sample is then accelerated towards an electric field and directed through a magnetic field.

ktySzUONBUU-007|Now when charged particles hit a magnetic field, they are deflected by the magnetic field.

ktySzUONBUU-008|And they're deflected according to their mass.

ktySzUONBUU-009|Heavier particles are not as deflected as lighter particles.

ktySzUONBUU-010|So you have this stream of particles and it hits that magnetic field, and it's fanned out based on their mass.

ktySzUONBUU-015|Let's look at a larger mass coming through the same system.

ktySzUONBUU-016|We're going to ionize it, it's not deflected as much.

ktySzUONBUU-017|So you have a distribution of masses that you can resolve using a mass spectrometer.

ktySzUONBUU-019|And the vertical axis will tell us about how many particles there were, and that's very useful.

ktySzUONBUU-020|Remember when we had our carbon, we said that one in hundred of these particles was a carbon-13.

ktySzUONBUU-021|The other 99 were carbon-12.

ktySzUONBUU-022|That's a difference of one mass unit.

ktySzUONBUU-024|So, let's look at a few masses.

ktySzUONBUU-025|We have, for instance, what would show up at mass 1?

ktySzUONBUU-029|So you can tell by the mass spectrometer you can get the mass.

ktySzUONBUU-030|But remember, mass does not determine the identity of the element-- certainly a clue, but it does not determine the identity.

ktySzUONBUU-031|We need to know the number of protons in the nucleus to get the identity of an element.

ktySzUONBUU-032|So mass three, that could be tritium or it could be a molecule of deuterium and hydrogen hooked together to form a hydrogen molecule.

ktySzUONBUU-036|And you can go on down the line.

ktySzUONBUU-037|Here is oxygen 16.

ktySzUONBUU-038|At mass 18, you could have a water molecule with oxygen 16 and two protons, two hydrogen atoms.

ktySzUONBUU-044|So a mass spectrometer, exquisitely sensitive makes.

ktySzUONBUU-046|The e equals mc squared mass loss can be determined by modern mass spectrometer.

ktySzUONBUU-047|That's how accurate they are.

ktySzUONBUU-048|Incredibly tiny mass losses can be determined by this method.

ktySzUONBUU-049|It's a beautiful instrument and it's very, very, very powerful for measuring atomic masses.

ksZoOde-7zY-000|We noticed that the trend in ionization energies is to kind of increase across a row in the periodic table.

ksZoOde-7zY-001|Let's look at that in more detail.

ksZoOde-7zY-003|That's a 1s electron.

ksZoOde-7zY-004|For helium, about 2,400 kilojoules per mole for that ionization.

ksZoOde-7zY-006|So increasing that positive charge on the nucleus was much more dramatic than adding another electron.

ksZoOde-7zY-007|It's much harder to pull off an electron from helium than hydrogen.

ksZoOde-7zY-011|And lithium is remarkably easy to ionize.

ksZoOde-7zY-018|It's a general trend.

ksZoOde-7zY-019|And we can kind of rationalize that.

ksZoOde-7zY-020|Because what we're doing is we're adding more nuclear charge, but we're staying in the same principle quantum level.

ksZoOde-7zY-021|So electrons in about the same region of space but more positive charge holding them.

ksZoOde-7zY-022|Of course, those will be slightly more difficult to ionize.

ksZoOde-7zY-023|As you continue across, there's a slight dip here at oxygen.

ksZoOde-7zY-024|And I think we can understand that too in terms of the fact that nitrogen now is that half-filled state.

ksZoOde-7zY-025|Oxygen has that paired electron.