ceaglex/VisualTTS_Chem · Datasets at Hugging Face

text stringlengths 19 206
FeJ1LQe6Zh4-000\|Let's look at a cell potential of 0.5 volts for the oxidation of bromide by permanganate.
FeJ1LQe6Zh4-001\|The question I have is what pH would cause this voltage?
FeJ1LQe6Zh4-011\|We're looking at the oxidation of bromide by permanganate.
FeJ1LQe6Zh4-012\|And we find that cell is operating at 0.5 volts.
FeJ1LQe6Zh4-013\|So how can that happen?
FeJ1LQe6Zh4-014\|Well, let's look back at the standard reaction.
FeJ1LQe6Zh4-017\|Well, you could raise the reactant concentration.
FeJ1LQe6Zh4-018\|Raising the reactant concentrations would increase the potential for the reaction to go.
FeJ1LQe6Zh4-019\|So let's look at that.
OkoFU7hocpc-000\|If I take hydrogen molecules and oxygen molecules and put them in a balloon, I can pretty much wait forever and they don't form water molecules.
OkoFU7hocpc-001\|That reaction is favorable, but there is a barrier in between that prevents the rapid room temperature conversion.
OkoFU7hocpc-002\|It turns out though that hydrogen and oxygen can be made to form water, release the same amount of energy at room temperature.
OkoFU7hocpc-003\|You can do with a process called catalysis.
OkoFU7hocpc-004\|Catalysis lowers this energy barrier.
OkoFU7hocpc-005\|When you lower that energy barrier, that allows particles-- maybe even at room temperature kinetic energies-- to get over the barrier and form products.
OkoFU7hocpc-008\|So catalysis lowers the energy barrier between products and reactants.
OkoFU7hocpc-009\|It allows a low road path between the products and reactants.
OkoFU7hocpc-010\|And allows reactions to go at much lower temperatures and much more quickly.
wcMCDbnuNms-000\|When an electron is bound around an atom, an electron has wave-like properties.
wcMCDbnuNms-001\|It's bounded, and when you bounce something with a wave-like property, you naturally get quantization, different energy levels.
wcMCDbnuNms-002\|Now, our energy levels are described by wave functions with three quantum numbers-- n, l, and m sub l.
wcMCDbnuNms-003\|And there's a myriad of possibilities.
wcMCDbnuNms-004\|So what we want to do is catalog them and see which energy levels an electron will exist in about an atom.
wcMCDbnuNms-005\|So let's start with the n equal 1 state.
wcMCDbnuNms-006\|n gives the total energy of the system.
wcMCDbnuNms-007\|Now, you can also have m equal 2 and keep on going, up to the ionized state, n equal infinity.
wcMCDbnuNms-011\|The lowest energy orbital about an atom is n equal 1, and the only possible value of l is s.
wcMCDbnuNms-013\|And m sub l is also 0.
wcMCDbnuNms-014\|And the only possibility is 0, so we just leave that off.
wcMCDbnuNms-015\|If you go to n equal 2, you can also have an s.
wcMCDbnuNms-016\|So n equals 1 and n equals 2 both give rise to a spherically symmetric s orbital.
wcMCDbnuNms-017\|But when n equals 2, you can also have l equal 1.
wcMCDbnuNms-018\|l equal 1 has three values of m sub l-- m sub l minus 1, 0, and 1.
wcMCDbnuNms-019\|These are the three equivalent p orbitals.
wcMCDbnuNms-023\|Now, we can continue and go to principal quantum level 3.
wcMCDbnuNms-024\|And a 3 s orbital is also possible.
wcMCDbnuNms-025\|It'll also be spherically symmetric.
wcMCDbnuNms-027\|So the number of nodes increase, but the overall spherical symmetry remains the same.
wcMCDbnuNms-028\|And of course, when n equals 2, you can still have l equal 1.
wcMCDbnuNms-029\|So you'll have a set of 3s orbitals-- 3px, 3py, and 3pz.
wcMCDbnuNms-030\|But when n equals 3, you can also have l equal 2.
wcMCDbnuNms-032\|That gives rise to a set of five different shape and orientation, but equal in energy orbitals.
wcMCDbnuNms-033\|And I have some models of them right here.
wcMCDbnuNms-034\|So one, two, three, four, five d orbitals.
wcMCDbnuNms-036\|And again, when you look in your textbook, these will be given geometric notations-- z squared, x squared minus y squared, et cetera, rather than the m sub l integer values.
wcMCDbnuNms-037\|We do that for purely geometric reasons.
wcMCDbnuNms-038\|The important thing is there's five equivalent orbitals in a set of d orbitals.
wcMCDbnuNms-039\|So you can see an electron can exist in a myriad of different states around the atom.
wcMCDbnuNms-040\|And when you have more than one electron they need to share these states.
wcMCDbnuNms-041\|So that's what we'll look at next.
ER7lUmsy9m4-000\|So let's do that phosphorus oxide reaction in a vessel where we have a fixed amount of oxygen and we add differing initial amounts of phosphorus.
ER7lUmsy9m4-010\|You add a little bit more phosphorus, you'll produce a little bit more phosphorus oxide.
ER7lUmsy9m4-011\|You've used a little bit more of your oxygen.
ER7lUmsy9m4-013\|You cannot produce any more phosphorus oxide.
ER7lUmsy9m4-014\|Oxygen limits the reaction.
ER7lUmsy9m4-015\|It's called the limiting reagent.
ER7lUmsy9m4-016\|So, product will build up until all the oxygen is consumed.
lItaSuF6Uqg-000\|Let's do a calculation where we calculate ionization energy.
lItaSuF6Uqg-001\|We're going to look at a system where we bathe rubidium atoms in ultraviolet light.
lItaSuF6Uqg-002\|That will eject electrons and those electrons will travel off with a certain kinetic energy, and we'll measure that.
lItaSuF6Uqg-003\|So knowing the kinetic energy of the electrons and the amount of energy from the photons of light, we can calculate the kinetic energy.
lItaSuF6Uqg-005\|Let's do a calculation where we calculate ionization energy.
lItaSuF6Uqg-006\|What we're going to do is we're going to take rubidium atoms and bathe them in ultraviolet light.
lItaSuF6Uqg-007\|That ultraviolet light will ionize the rubidium atoms.
lItaSuF6Uqg-008\|Electrons will leave with a certain kinetic energy.
lItaSuF6Uqg-009\|We can measure that kinetic energy.
lItaSuF6Uqg-010\|Now the ultraviolet radiation that comes in has to do two things.
lItaSuF6Uqg-011\|Some of the energy goes into ionizing the rubidium atoms.
lItaSuF6Uqg-012\|The rest of the energy goes into the kinetic energy.
lItaSuF6Uqg-013\|So I know the sum of the kinetic energy and the ionization energy is the photon energy.
lItaSuF6Uqg-017\|So the photon is split into two things, ionizing plus kinetic energy.
lItaSuF6Uqg-020\|Now this is called a photo electron spectroscopy experiment.
lItaSuF6Uqg-021\|Ultraviolet radiation measure the kinetic energy of electrons emitted.
lItaSuF6Uqg-022\|So I know that kinetic energy and the ionization energy add to give the photon energy.
lItaSuF6Uqg-023\|So I can calculate the ionization energy from the photon energy and the kinetic energy.
lItaSuF6Uqg-024\|Kinetic energy of electron at 2,450 kilometers per second, we've done these kind of things before.
lItaSuF6Uqg-025\|1/2 half times the mass of the electron times the velocity squared.
lItaSuF6Uqg-026\|And I write the velocity in meters per second to always operate in kilometers, excuse me, meters, kilograms, and seconds.
lItaSuF6Uqg-029\|So from that and the photon energy, I can calculate the ionization energy.
lItaSuF6Uqg-030\|So the photon energy, I'll cast in terms of the wavelength because that's the parameter I have.
lItaSuF6Uqg-032\|And now I can calculate.
P_e4n9XSjUw-000\|Particles, electrons, small atoms have wave-like properties.
P_e4n9XSjUw-001\|Let's look at that in terms of a chem quiz.
P_e4n9XSjUw-003\|The sodium atom, impacted by the wavelength of light, behaving like a particle, behaving like a photon.
P_e4n9XSjUw-004\|About how many will stop that?
P_e4n9XSjUw-005\|About 1, about 100, about 10,000?
P_e4n9XSjUw-014\|Matter has both wave and particle characteristics.
P_e4n9XSjUw-015\|We've seen the momentum and the wavelength are related by the de Broglie relationship, and we calculated it for several different objects.
P_e4n9XSjUw-016\|Now we're saying, well, a sodium atom is going to be traveling at about 300 meters per second.
P_e4n9XSjUw-017\|It's going to encounter photons of 600 nanometer wavelength.
P_e4n9XSjUw-018\|About how many photons have to strike those sodium atoms to get them to slow down, and about stop?
P_e4n9XSjUw-021\|Those photons of yellow light coming in at 600 nanometers.
P_e4n9XSjUw-022\|We can say, well, I want the momenta of these two systems to be equal.
P_e4n9XSjUw-023\|I want to transfer enough momenta from these waves to stop the particle.
P_e4n9XSjUw-024\|So the wave has a particle nature, it has a tiny little momentum.
P_e4n9XSjUw-025\|I need to transfer it to this larger momentum.
P_e4n9XSjUw-026\|And here comes the sodium atom, I want to bap, bap, bap, bap, bap, bap, bap, bap, bap.
P_e4n9XSjUw-027\|Keep hitting it with photons until-- bap, bap, bap, bap, bap, bap-- that sodium about stops.
P_e4n9XSjUw-028\|And I can do that by just calculating the momenta of each.
P_e4n9XSjUw-029\|And as I do so, I see that the momenta are related by their wavelength by the de Broglie relationship.

FeJ1LQe6Zh4-000|Let's look at a cell potential of 0.5 volts for the oxidation of bromide by permanganate.

FeJ1LQe6Zh4-001|The question I have is what pH would cause this voltage?

FeJ1LQe6Zh4-011|We're looking at the oxidation of bromide by permanganate.

FeJ1LQe6Zh4-012|And we find that cell is operating at 0.5 volts.

FeJ1LQe6Zh4-013|So how can that happen?

FeJ1LQe6Zh4-014|Well, let's look back at the standard reaction.

FeJ1LQe6Zh4-017|Well, you could raise the reactant concentration.

FeJ1LQe6Zh4-018|Raising the reactant concentrations would increase the potential for the reaction to go.

FeJ1LQe6Zh4-019|So let's look at that.

OkoFU7hocpc-000|If I take hydrogen molecules and oxygen molecules and put them in a balloon, I can pretty much wait forever and they don't form water molecules.

OkoFU7hocpc-001|That reaction is favorable, but there is a barrier in between that prevents the rapid room temperature conversion.

OkoFU7hocpc-002|It turns out though that hydrogen and oxygen can be made to form water, release the same amount of energy at room temperature.

OkoFU7hocpc-003|You can do with a process called catalysis.

OkoFU7hocpc-004|Catalysis lowers this energy barrier.

OkoFU7hocpc-005|When you lower that energy barrier, that allows particles-- maybe even at room temperature kinetic energies-- to get over the barrier and form products.

OkoFU7hocpc-008|So catalysis lowers the energy barrier between products and reactants.

OkoFU7hocpc-009|It allows a low road path between the products and reactants.

OkoFU7hocpc-010|And allows reactions to go at much lower temperatures and much more quickly.

wcMCDbnuNms-000|When an electron is bound around an atom, an electron has wave-like properties.

wcMCDbnuNms-001|It's bounded, and when you bounce something with a wave-like property, you naturally get quantization, different energy levels.

wcMCDbnuNms-002|Now, our energy levels are described by wave functions with three quantum numbers-- n, l, and m sub l.

wcMCDbnuNms-003|And there's a myriad of possibilities.

wcMCDbnuNms-004|So what we want to do is catalog them and see which energy levels an electron will exist in about an atom.

wcMCDbnuNms-005|So let's start with the n equal 1 state.

wcMCDbnuNms-006|n gives the total energy of the system.

wcMCDbnuNms-007|Now, you can also have m equal 2 and keep on going, up to the ionized state, n equal infinity.

wcMCDbnuNms-011|The lowest energy orbital about an atom is n equal 1, and the only possible value of l is s.

wcMCDbnuNms-013|And m sub l is also 0.

wcMCDbnuNms-014|And the only possibility is 0, so we just leave that off.

wcMCDbnuNms-015|If you go to n equal 2, you can also have an s.

wcMCDbnuNms-016|So n equals 1 and n equals 2 both give rise to a spherically symmetric s orbital.

wcMCDbnuNms-017|But when n equals 2, you can also have l equal 1.

wcMCDbnuNms-018|l equal 1 has three values of m sub l-- m sub l minus 1, 0, and 1.

wcMCDbnuNms-019|These are the three equivalent p orbitals.

wcMCDbnuNms-023|Now, we can continue and go to principal quantum level 3.

wcMCDbnuNms-024|And a 3 s orbital is also possible.

wcMCDbnuNms-025|It'll also be spherically symmetric.

wcMCDbnuNms-027|So the number of nodes increase, but the overall spherical symmetry remains the same.

wcMCDbnuNms-028|And of course, when n equals 2, you can still have l equal 1.

wcMCDbnuNms-029|So you'll have a set of 3s orbitals-- 3px, 3py, and 3pz.

wcMCDbnuNms-030|But when n equals 3, you can also have l equal 2.

wcMCDbnuNms-032|That gives rise to a set of five different shape and orientation, but equal in energy orbitals.

wcMCDbnuNms-033|And I have some models of them right here.

wcMCDbnuNms-034|So one, two, three, four, five d orbitals.

wcMCDbnuNms-036|And again, when you look in your textbook, these will be given geometric notations-- z squared, x squared minus y squared, et cetera, rather than the m sub l integer values.

wcMCDbnuNms-037|We do that for purely geometric reasons.

wcMCDbnuNms-038|The important thing is there's five equivalent orbitals in a set of d orbitals.

wcMCDbnuNms-039|So you can see an electron can exist in a myriad of different states around the atom.

wcMCDbnuNms-040|And when you have more than one electron they need to share these states.

wcMCDbnuNms-041|So that's what we'll look at next.

ER7lUmsy9m4-000|So let's do that phosphorus oxide reaction in a vessel where we have a fixed amount of oxygen and we add differing initial amounts of phosphorus.

ER7lUmsy9m4-010|You add a little bit more phosphorus, you'll produce a little bit more phosphorus oxide.

ER7lUmsy9m4-011|You've used a little bit more of your oxygen.

ER7lUmsy9m4-013|You cannot produce any more phosphorus oxide.

ER7lUmsy9m4-014|Oxygen limits the reaction.

ER7lUmsy9m4-015|It's called the limiting reagent.

ER7lUmsy9m4-016|So, product will build up until all the oxygen is consumed.

lItaSuF6Uqg-000|Let's do a calculation where we calculate ionization energy.

lItaSuF6Uqg-001|We're going to look at a system where we bathe rubidium atoms in ultraviolet light.

lItaSuF6Uqg-002|That will eject electrons and those electrons will travel off with a certain kinetic energy, and we'll measure that.

lItaSuF6Uqg-003|So knowing the kinetic energy of the electrons and the amount of energy from the photons of light, we can calculate the kinetic energy.

lItaSuF6Uqg-005|Let's do a calculation where we calculate ionization energy.

lItaSuF6Uqg-006|What we're going to do is we're going to take rubidium atoms and bathe them in ultraviolet light.

lItaSuF6Uqg-007|That ultraviolet light will ionize the rubidium atoms.

lItaSuF6Uqg-008|Electrons will leave with a certain kinetic energy.

lItaSuF6Uqg-009|We can measure that kinetic energy.

lItaSuF6Uqg-010|Now the ultraviolet radiation that comes in has to do two things.

lItaSuF6Uqg-011|Some of the energy goes into ionizing the rubidium atoms.

lItaSuF6Uqg-012|The rest of the energy goes into the kinetic energy.

lItaSuF6Uqg-013|So I know the sum of the kinetic energy and the ionization energy is the photon energy.

lItaSuF6Uqg-017|So the photon is split into two things, ionizing plus kinetic energy.

lItaSuF6Uqg-020|Now this is called a photo electron spectroscopy experiment.

lItaSuF6Uqg-021|Ultraviolet radiation measure the kinetic energy of electrons emitted.

lItaSuF6Uqg-022|So I know that kinetic energy and the ionization energy add to give the photon energy.

lItaSuF6Uqg-023|So I can calculate the ionization energy from the photon energy and the kinetic energy.

lItaSuF6Uqg-024|Kinetic energy of electron at 2,450 kilometers per second, we've done these kind of things before.

lItaSuF6Uqg-025|1/2 half times the mass of the electron times the velocity squared.

lItaSuF6Uqg-026|And I write the velocity in meters per second to always operate in kilometers, excuse me, meters, kilograms, and seconds.

lItaSuF6Uqg-029|So from that and the photon energy, I can calculate the ionization energy.

lItaSuF6Uqg-030|So the photon energy, I'll cast in terms of the wavelength because that's the parameter I have.

lItaSuF6Uqg-032|And now I can calculate.

P_e4n9XSjUw-000|Particles, electrons, small atoms have wave-like properties.

P_e4n9XSjUw-001|Let's look at that in terms of a chem quiz.

P_e4n9XSjUw-003|The sodium atom, impacted by the wavelength of light, behaving like a particle, behaving like a photon.

P_e4n9XSjUw-004|About how many will stop that?

P_e4n9XSjUw-005|About 1, about 100, about 10,000?

P_e4n9XSjUw-014|Matter has both wave and particle characteristics.

P_e4n9XSjUw-015|We've seen the momentum and the wavelength are related by the de Broglie relationship, and we calculated it for several different objects.

P_e4n9XSjUw-016|Now we're saying, well, a sodium atom is going to be traveling at about 300 meters per second.

P_e4n9XSjUw-017|It's going to encounter photons of 600 nanometer wavelength.

P_e4n9XSjUw-018|About how many photons have to strike those sodium atoms to get them to slow down, and about stop?

P_e4n9XSjUw-021|Those photons of yellow light coming in at 600 nanometers.

P_e4n9XSjUw-022|We can say, well, I want the momenta of these two systems to be equal.

P_e4n9XSjUw-023|I want to transfer enough momenta from these waves to stop the particle.

P_e4n9XSjUw-024|So the wave has a particle nature, it has a tiny little momentum.

P_e4n9XSjUw-025|I need to transfer it to this larger momentum.

P_e4n9XSjUw-026|And here comes the sodium atom, I want to bap, bap, bap, bap, bap, bap, bap, bap, bap.

P_e4n9XSjUw-027|Keep hitting it with photons until-- bap, bap, bap, bap, bap, bap-- that sodium about stops.

P_e4n9XSjUw-028|And I can do that by just calculating the momenta of each.

P_e4n9XSjUw-029|And as I do so, I see that the momenta are related by their wavelength by the de Broglie relationship.