ceaglex/VisualTTS_Chem · Datasets at Hugging Face

text stringlengths 19 206
ESX5q4FtkAo-001\|Now the gases could be very different.
ESX5q4FtkAo-003\|Now they do it differently.
ESX5q4FtkAo-004\|That is bromine comes in with large, massive shots to the walls to exert pressure, where hydrogen comes in with lighter shot, but more rapidly.
ESX5q4FtkAo-005\|The property that differentiates the gases is their velocity.
ESX5q4FtkAo-006\|They have different velocities based on their masses.
ESX5q4FtkAo-007\|So the velocity is the distinguishing characteristic, and it would be nice if we could solve for that velocity.
ESX5q4FtkAo-008\|Our expression for the kinetic energy has the velocity squared, the mean velocity squared, in it.
ESX5q4FtkAo-009\|So we can solve for the mean velocity squared, take the square root, and we'll have a quantity called the root mean squared velocity.
ESX5q4FtkAo-010\|Now it's not exactly the mean velocity, but it's very close, and it depends on fundamental properties like the temperature and the nature of the gas.
ESX5q4FtkAo-011\|So we can easily solve for that quantity.
ESX5q4FtkAo-016\|Here we'll express R in joules, so our velocities come out in meters per second.
ESX5q4FtkAo-018\|So I'll take the square root of these three quantities, and I'll find a very simple expression for an important distinguishing property for the gases-- their actual velocity.
ESX5q4FtkAo-019\|The root mean squared velocity for a mole of gas is 3RT over the molar mass, square root.
VY3yL2nDHuQ-000\|We all know glass is transparent, so let's look at absorption over a broader range of the electromagnetic spectrum.
VY3yL2nDHuQ-001\|Transparent glass should have an absorption spectrum that looks like either 1, 2, or 3.
VY3yL2nDHuQ-008\|The absorption spectrum of transparent glass is what we're talking about.
VY3yL2nDHuQ-009\|And we're talking about a rather broad region of the electromagnetic spectrum from ultraviolet wavelengths down to infrared wavelengths.
VY3yL2nDHuQ-010\|And I've overlaid the common visible colored wavelengths in the center.
VY3yL2nDHuQ-013\|But in order for the glass to be transparent, all visible wavelengths must pass through.
VY3yL2nDHuQ-014\|Transparent glass will past white light directly through it.
VY3yL2nDHuQ-015\|You can see all colors through transparent glass, so all colors must pass through.
VY3yL2nDHuQ-016\|So we have to choose one that has no absorption spectra in the visible range.
VY3yL2nDHuQ-017\|That is option 1.
VY3yL2nDHuQ-018\|So the correct answer here is A.
p26UwTYXXF0-000\|Let's calculate the heat involved for a physical process.
p26UwTYXXF0-001\|How much energy is required to heat 250 grams of water from 50 degrees C to 110 degrees C?
p26UwTYXXF0-002\|Now, the way you do this, is to break it up into distinct energy changes.
p26UwTYXXF0-008\|The sum of those three enthalpies will give me the enthalpy for the total process.
p26UwTYXXF0-009\|So we can calculate each one of them.
YttroH0ftiA-001\|And the energy amount is h times nu, Planck's constant times the frequency of the light.
YttroH0ftiA-002\|Now a particle has momentum.
YttroH0ftiA-003\|And we've seen the momentum of the particle, that photon, can cause an electron to be ejected from a metal.
YttroH0ftiA-004\|We saw in the photoelectric effect a incoming photon ejecting electrons from a metal.
YttroH0ftiA-005\|So how does that particle nature and wave nature reconcile themselves?
YttroH0ftiA-006\|Well, let's talk about that.
YttroH0ftiA-007\|The light wave particle duality.
YttroH0ftiA-009\|Now the wave particle, the light particle, that we call a photon has a momentum.
YttroH0ftiA-010\|We've seen it can transfer momentum from the photon to the electron, but the momentum is we often associate with mass.
YttroH0ftiA-011\|But the photon has no mass.
YttroH0ftiA-012\|The photon is a particle and it massless, moving at the speed of light.
YttroH0ftiA-014\|Now we have two expressions for the energy, the energy of the photon and the relativistic energy, m c squared.
YttroH0ftiA-016\|So the momentum is m times c times another c gives you m c squared.
YttroH0ftiA-020\|Waves and particles acting the way they choose, sometimes light will behave like a wave, sometimes it behaves like a particle.
YttroH0ftiA-021\|There's a duality between them expressed by this beautiful relationship between the wavelength and the momentum.
YttroH0ftiA-022\|Planck's constant, again, an extremely small number, is the proportionality constant between the momentum and the wavelength.
YttroH0ftiA-023\|So waves, particles, light, we have to think of them all at the same time.
JnNVdsrkKV8-000\|Let's do some calculations with weak acids and weak bases.
JnNVdsrkKV8-001\|We're going to talk about the salt of hydrocyanic acid and formic acid.
JnNVdsrkKV8-002\|Now, the salt is the compound that's formed when you react the acid with a strong base.
JnNVdsrkKV8-003\|You form the salt. You take the counter ion.
JnNVdsrkKV8-005\|The conjugate ions from the base and the acid.
JnNVdsrkKV8-006\|Sodium coming from the strong base, and cyanide coming from the acid.
JnNVdsrkKV8-007\|So the salt of acids contains the weak base.
JnNVdsrkKV8-008\|If I did that with formate, I'd take formic acid and react it with sodium hydroxide.
JnNVdsrkKV8-009\|I'd form sodium formate.
JnNVdsrkKV8-010\|And sodium formate contains the formate ion.
JnNVdsrkKV8-011\|That's the conjugate base of formic acid.
JnNVdsrkKV8-013\|So oh actually I've said, which is less basic?
JnNVdsrkKV8-014\|So which has the lower pH?
JnNVdsrkKV8-024\|So let's do that.
JnNVdsrkKV8-028\|So the formate ion is the weaker base.
JnNVdsrkKV8-029\|It will be less basic in solution.
JnNVdsrkKV8-030\|It will have the lower pH.
JnNVdsrkKV8-031\|So let's calculate that.
JnNVdsrkKV8-037\|So formate ion plus water forms OH minus, the base, and of course the conjugate acid of the formate ion.
JnNVdsrkKV8-038\|So by forming this base, that's what raises the pH above 7 when you put sodium formate, a salt, in water.
JnNVdsrkKV8-039\|How much base is formed?
JnNVdsrkKV8-040\|Well, we can do the calculations.
JnNVdsrkKV8-041\|We can say, well, I took 0.1 Molar of this ion from the salt and put it in water.
JnNVdsrkKV8-042\|Initially, there was no OH minus.
JnNVdsrkKV8-060\|And you see it's 2.37 times 10 to the minus 6.
JnNVdsrkKV8-061\|And indeed, that's much smaller than 10 to the minus 1.
JnNVdsrkKV8-062\|Our original concentration, 10 to the minus 1, the amount that associates, about 10 to the minus 6.
JnNVdsrkKV8-063\|So it's indeed true that x is small compared to 0.1.
JnNVdsrkKV8-064\|I have the OH minus concentration.
JnNVdsrkKV8-065\|And in water, H3O plus times OH minus is always 10 to the minus 14.
JnNVdsrkKV8-069\|Now, I solved for H3O plus because that's how we get the pH.
JnNVdsrkKV8-070\|We take minus log of H3O plus.
JnNVdsrkKV8-071\|I could have taken minus log of OH minus.
JnNVdsrkKV8-072\|That would be the pOH, and then subtracted that from 14.
JnNVdsrkKV8-073\|That's the same mathematics basically.
JnNVdsrkKV8-074\|Or say the pH is minus log of H3O plus and 8.37.
JnNVdsrkKV8-075\|So here, we have a salt of a weak acid, formate ion, added to water.
JnNVdsrkKV8-076\|It makes the water slightly basic.
rvH5eufKoe8-009\|We're comparing two acids, HF, a weak acid, and HBr, a strong acid.
rvH5eufKoe8-010\|We know the relative bond enthalpies go like this.
rvH5eufKoe8-014\|Now, that's really the indicator of what we're looking for here because, if K is less than 1, then the standard state free energy difference must be positive somewhere.
rvH5eufKoe8-015\|It must have some positive values.

ESX5q4FtkAo-001|Now the gases could be very different.

ESX5q4FtkAo-003|Now they do it differently.

ESX5q4FtkAo-004|That is bromine comes in with large, massive shots to the walls to exert pressure, where hydrogen comes in with lighter shot, but more rapidly.

ESX5q4FtkAo-005|The property that differentiates the gases is their velocity.

ESX5q4FtkAo-006|They have different velocities based on their masses.

ESX5q4FtkAo-007|So the velocity is the distinguishing characteristic, and it would be nice if we could solve for that velocity.

ESX5q4FtkAo-008|Our expression for the kinetic energy has the velocity squared, the mean velocity squared, in it.

ESX5q4FtkAo-009|So we can solve for the mean velocity squared, take the square root, and we'll have a quantity called the root mean squared velocity.

ESX5q4FtkAo-010|Now it's not exactly the mean velocity, but it's very close, and it depends on fundamental properties like the temperature and the nature of the gas.

ESX5q4FtkAo-011|So we can easily solve for that quantity.

ESX5q4FtkAo-016|Here we'll express R in joules, so our velocities come out in meters per second.

ESX5q4FtkAo-018|So I'll take the square root of these three quantities, and I'll find a very simple expression for an important distinguishing property for the gases-- their actual velocity.

ESX5q4FtkAo-019|The root mean squared velocity for a mole of gas is 3RT over the molar mass, square root.

VY3yL2nDHuQ-000|We all know glass is transparent, so let's look at absorption over a broader range of the electromagnetic spectrum.

VY3yL2nDHuQ-001|Transparent glass should have an absorption spectrum that looks like either 1, 2, or 3.

VY3yL2nDHuQ-008|The absorption spectrum of transparent glass is what we're talking about.

VY3yL2nDHuQ-009|And we're talking about a rather broad region of the electromagnetic spectrum from ultraviolet wavelengths down to infrared wavelengths.

VY3yL2nDHuQ-010|And I've overlaid the common visible colored wavelengths in the center.

VY3yL2nDHuQ-013|But in order for the glass to be transparent, all visible wavelengths must pass through.

VY3yL2nDHuQ-014|Transparent glass will past white light directly through it.

VY3yL2nDHuQ-015|You can see all colors through transparent glass, so all colors must pass through.

VY3yL2nDHuQ-016|So we have to choose one that has no absorption spectra in the visible range.

VY3yL2nDHuQ-017|That is option 1.

VY3yL2nDHuQ-018|So the correct answer here is A.

p26UwTYXXF0-000|Let's calculate the heat involved for a physical process.

p26UwTYXXF0-001|How much energy is required to heat 250 grams of water from 50 degrees C to 110 degrees C?

p26UwTYXXF0-002|Now, the way you do this, is to break it up into distinct energy changes.

p26UwTYXXF0-008|The sum of those three enthalpies will give me the enthalpy for the total process.

p26UwTYXXF0-009|So we can calculate each one of them.

YttroH0ftiA-001|And the energy amount is h times nu, Planck's constant times the frequency of the light.

YttroH0ftiA-002|Now a particle has momentum.

YttroH0ftiA-003|And we've seen the momentum of the particle, that photon, can cause an electron to be ejected from a metal.

YttroH0ftiA-004|We saw in the photoelectric effect a incoming photon ejecting electrons from a metal.

YttroH0ftiA-005|So how does that particle nature and wave nature reconcile themselves?

YttroH0ftiA-006|Well, let's talk about that.

YttroH0ftiA-007|The light wave particle duality.

YttroH0ftiA-009|Now the wave particle, the light particle, that we call a photon has a momentum.

YttroH0ftiA-010|We've seen it can transfer momentum from the photon to the electron, but the momentum is we often associate with mass.

YttroH0ftiA-011|But the photon has no mass.

YttroH0ftiA-012|The photon is a particle and it massless, moving at the speed of light.

YttroH0ftiA-014|Now we have two expressions for the energy, the energy of the photon and the relativistic energy, m c squared.

YttroH0ftiA-016|So the momentum is m times c times another c gives you m c squared.

YttroH0ftiA-020|Waves and particles acting the way they choose, sometimes light will behave like a wave, sometimes it behaves like a particle.

YttroH0ftiA-021|There's a duality between them expressed by this beautiful relationship between the wavelength and the momentum.

YttroH0ftiA-022|Planck's constant, again, an extremely small number, is the proportionality constant between the momentum and the wavelength.

YttroH0ftiA-023|So waves, particles, light, we have to think of them all at the same time.

JnNVdsrkKV8-000|Let's do some calculations with weak acids and weak bases.

JnNVdsrkKV8-001|We're going to talk about the salt of hydrocyanic acid and formic acid.

JnNVdsrkKV8-002|Now, the salt is the compound that's formed when you react the acid with a strong base.

JnNVdsrkKV8-003|You form the salt. You take the counter ion.

JnNVdsrkKV8-005|The conjugate ions from the base and the acid.

JnNVdsrkKV8-006|Sodium coming from the strong base, and cyanide coming from the acid.

JnNVdsrkKV8-007|So the salt of acids contains the weak base.

JnNVdsrkKV8-008|If I did that with formate, I'd take formic acid and react it with sodium hydroxide.

JnNVdsrkKV8-009|I'd form sodium formate.

JnNVdsrkKV8-010|And sodium formate contains the formate ion.

JnNVdsrkKV8-011|That's the conjugate base of formic acid.

JnNVdsrkKV8-013|So oh actually I've said, which is less basic?

JnNVdsrkKV8-014|So which has the lower pH?

JnNVdsrkKV8-024|So let's do that.

JnNVdsrkKV8-028|So the formate ion is the weaker base.

JnNVdsrkKV8-029|It will be less basic in solution.

JnNVdsrkKV8-030|It will have the lower pH.

JnNVdsrkKV8-031|So let's calculate that.

JnNVdsrkKV8-037|So formate ion plus water forms OH minus, the base, and of course the conjugate acid of the formate ion.

JnNVdsrkKV8-038|So by forming this base, that's what raises the pH above 7 when you put sodium formate, a salt, in water.

JnNVdsrkKV8-039|How much base is formed?

JnNVdsrkKV8-040|Well, we can do the calculations.

JnNVdsrkKV8-041|We can say, well, I took 0.1 Molar of this ion from the salt and put it in water.

JnNVdsrkKV8-042|Initially, there was no OH minus.

JnNVdsrkKV8-060|And you see it's 2.37 times 10 to the minus 6.

JnNVdsrkKV8-061|And indeed, that's much smaller than 10 to the minus 1.

JnNVdsrkKV8-062|Our original concentration, 10 to the minus 1, the amount that associates, about 10 to the minus 6.

JnNVdsrkKV8-063|So it's indeed true that x is small compared to 0.1.

JnNVdsrkKV8-064|I have the OH minus concentration.

JnNVdsrkKV8-065|And in water, H3O plus times OH minus is always 10 to the minus 14.

JnNVdsrkKV8-069|Now, I solved for H3O plus because that's how we get the pH.

JnNVdsrkKV8-070|We take minus log of H3O plus.

JnNVdsrkKV8-071|I could have taken minus log of OH minus.

JnNVdsrkKV8-072|That would be the pOH, and then subtracted that from 14.

JnNVdsrkKV8-073|That's the same mathematics basically.

JnNVdsrkKV8-074|Or say the pH is minus log of H3O plus and 8.37.

JnNVdsrkKV8-075|So here, we have a salt of a weak acid, formate ion, added to water.

JnNVdsrkKV8-076|It makes the water slightly basic.

rvH5eufKoe8-009|We're comparing two acids, HF, a weak acid, and HBr, a strong acid.

rvH5eufKoe8-010|We know the relative bond enthalpies go like this.

rvH5eufKoe8-014|Now, that's really the indicator of what we're looking for here because, if K is less than 1, then the standard state free energy difference must be positive somewhere.

rvH5eufKoe8-015|It must have some positive values.