ceaglex/VisualTTS_Chem · Datasets at Hugging Face

text stringlengths 19 206
W_ngfrFO7f0-016\|So they add to zero.
W_ngfrFO7f0-017\|So this reaction has an enthalpy zero.
RCWYbfXHxXw-000\|Let's look at a sample of sulfur atoms in some form.
RCWYbfXHxXw-001\|And we're going to trap them in 22 liters, about a quarter atmosphere of pressure, and about 546 kelvin.
RCWYbfXHxXw-002\|The question I have for you is, what's the form of sulfur?
RCWYbfXHxXw-010\|We're talking about a mole of sulfur particles.
RCWYbfXHxXw-013\|But I didn't have 1 atmosphere of pressure in my sample.
RCWYbfXHxXw-014\|I only had a quarter atmosphere of pressure.
RCWYbfXHxXw-015\|Well, how can you have that mole of particles give you a quarter atmosphere of pressure?
RCWYbfXHxXw-017\|You have to have fewer particles.
RCWYbfXHxXw-019\|That would give me 1/4 the particles and 1/4 the pressure.
RCWYbfXHxXw-020\|Now, in this sample, the temperature was also twice the standard temperature.
RCWYbfXHxXw-021\|How do I get a factor of two in temperature and stay at a quarter of the pressure in 22 liters?
RCWYbfXHxXw-025\|So those particles must be in groups of eight.
RCWYbfXHxXw-026\|The correct answer here, S8.
4kxMdkzCJGA-001\|Those energy levels will be one to a wave function-- energy level 1, wave function number 1, energy level 2, wave function number 2.
4kxMdkzCJGA-002\|They might be the same energy, that is, wave function 3 may have the same energy as wave function number 2.
4kxMdkzCJGA-003\|But they'll all have different energies associated with different sets of quantum numbers.
4kxMdkzCJGA-004\|So let's look at those quantum numbers and the energy levels for an atom.
4kxMdkzCJGA-005\|The principle quantum number n again runs just as the integers-- 1, 2, 3.
4kxMdkzCJGA-006\|And it gives you the overall energy, as we said, just like in a particle, a box, of the hydrogen wave functions.
4kxMdkzCJGA-007\|So the wave functions energies that actually look quite simple-- Z squared-- that's the charge on the nucleus.
4kxMdkzCJGA-008\|So for a hydrogen atom, that's one proton, so that charge is 1; n, the quantum number; and a constant.
4kxMdkzCJGA-009\|This is Rydberg's constant.
4kxMdkzCJGA-010\|It has a value that we can calculate and a value that we tabulate.
4kxMdkzCJGA-011\|That value is 2.18 times 10 to the minus 18th joules.
4kxMdkzCJGA-013\|So how does this look?
4kxMdkzCJGA-021\|Why do we do this negative energy level?
4kxMdkzCJGA-022\|Well, whenever you have energy, you have to give your energy relative to something.
4kxMdkzCJGA-023\|And the relative to something, we say, for atoms is the electron and the nucleus infinitely far apart and at rest.
4kxMdkzCJGA-024\|So that's the 0 in energy.
4kxMdkzCJGA-025\|When you bring the nucleus and the electron closer together, they become more stable.
4kxMdkzCJGA-026\|They go down in energy.
4kxMdkzCJGA-027\|So when you're at 0, the only way to go down in energy is to go to negative energy values.
4kxMdkzCJGA-028\|So we're going to use a negative system to describe this.
4kxMdkzCJGA-029\|If we bring in n equal 2, we still have 1, the charge of the nucleus.
4kxMdkzCJGA-030\|We're still talking about a hydrogen atom.
4kxMdkzCJGA-031\|The charge of the nucleus is still 1.
4kxMdkzCJGA-032\|n is now 2.
4kxMdkzCJGA-038\|And of course, you don't even have to go to infinite to get ionized.
4kxMdkzCJGA-039\|But the high levels of n are systems where the energy is about 0.
4kxMdkzCJGA-040\|And since it's 1 over n squared, you don't have to go to a very big n for this energy to get very, very small and close to 0 very rapidly.
4kxMdkzCJGA-041\|The ionized state is the 0 energy state for the hydrogen atom.
Li26UtxovcM-000\|Let's look at what happens when you burn a hydrocarbon in oxygen.
Li26UtxovcM-001\|A hydrocarbon is a compound composed of hydrogen and carbon only.
Li26UtxovcM-002\|Now I've written it here generically as HC reacting with O2, oxygen molecules.
Li26UtxovcM-003\|When that occurs, you can take the mass spectrum of the product.
Li26UtxovcM-005\|And they'll have distinct mass spectrum signatures.
Li26UtxovcM-006\|The water, mass 18, and the carbon dioxide at mass 44.
Li26UtxovcM-007\|So the combustion of hydrocarbons gives a very distinctive mass spectrum of the products.
UdINHPWJRYc-000\|Let's apply the first law of thermodynamics to an actual process.
UdINHPWJRYc-001\|Here's an ideal gas expanding isothermally-- that is no temperature change-- against a vacuum.
UdINHPWJRYc-002\|What's the heat flow?
UdINHPWJRYc-003\|Is it less than 0, equal to 0, or greater than 0?
UdINHPWJRYc-013\|We're talking about an ideal gas expanding against a vacuum isothermally.
UdINHPWJRYc-014\|And whenever you hear isothermal for an ideal gas, you have to be excited, because if it's isothermal, the temperature didn't change.
UdINHPWJRYc-015\|Isothermal, delta T is 0.
UdINHPWJRYc-016\|If the temperature doesn't change, the energy of the gas could not have changed.
UdINHPWJRYc-021\|One will be a negative sign.
UdINHPWJRYc-025\|So no heat flows in this process and no work is done.
UdINHPWJRYc-026\|The correct answer here, B, the heat flow is 0.
Mbq2QNVY10k-000\|For higher steric numbers, we hybridize with some of the d orbitals.
Mbq2QNVY10k-001\|So which energy diagram for phosphorous atomic orbitals in PCl5 is true after hybridization?
Mbq2QNVY10k-002\|Here are the standard atomic orbitals for phosphorus.
Mbq2QNVY10k-011\|We're talking about the hybridization in PCl5.
Mbq2QNVY10k-012\|In PCl5 the phosphorus has Steric Number 5 as to accommodate five chlorines.
Mbq2QNVY10k-013\|So we need five equivalent atomic orbitals.
Mbq2QNVY10k-014\|The five equivalent atomic orbitals are going to come from hybridizing the standard atomic orbitals.
Mbq2QNVY10k-018\|So the best answer here is A-- four unchanged d orbitals and five equivalent dsp3's.
8_ZRFjqe6F4-000\|When you talk about ideal gases, we can take a volume and a temperature and a number of moles and calculate a pressure.
8_ZRFjqe6F4-002\|Now, it's interesting to note that the ideal gas law doesn't say, well, what kind of particle is it?
8_ZRFjqe6F4-003\|If it's a hydrogen particle H2, it will exert the same pressure as a bromine particles, so BR2.
8_ZRFjqe6F4-005\|And that's true of gases that behave ideally.
8_ZRFjqe6F4-006\|The pressure is independent of the nature of the particle.
8_ZRFjqe6F4-007\|So why is that?
8_ZRFjqe6F4-008\|Well, we can talk about how pressure originates from first principles.
8_ZRFjqe6F4-010\|A transfer of momentum is a force applied to the walls of the container, and force per unit area is pressure.
8_ZRFjqe6F4-012\|The bromine does it by coming in with a heavy mass and striking the particle, striking the walls of the container.
8_ZRFjqe6F4-013\|The hydrogen particle, though much lighter, just moves very much faster.
8_ZRFjqe6F4-014\|So even though it has a lower mass, it has a higher velocity.
8_ZRFjqe6F4-015\|And it can transfer that same momentum in tiny little hits.
8_ZRFjqe6F4-016\|So two different particles can exert the same pressure on the wall by momentum transfer.
8_ZRFjqe6F4-018\|Either way, I can exert the same pressure on the walls of the container.
Hy9dsigzTiE-000\|Let's look at two gases dissolved in a liquid, helium and nitrogen, aqueous solution for both of them.
Hy9dsigzTiE-001\|Let's say they're both at equal concentrations in the solution.
Hy9dsigzTiE-002\|What does that mean about the partial pressure above the solution when that is at equilibrium?
Hy9dsigzTiE-010\|We're talking about helium and nitrogen dissolved in solution.
Hy9dsigzTiE-011\|Their aqueous concentrations are the same.
Hy9dsigzTiE-012\|So what does that mean about their partial pressures above the solution?
Hy9dsigzTiE-013\|Well, to solve this, we need to look at the equilibrium constants, the Henry's law constants for these gases.
Hy9dsigzTiE-015\|Here's helium, 0.4, and nitrogen 0.7.
Hy9dsigzTiE-020\|The correct answer here is helium partial pressure is greater than nitrogen.
Bl-zUmeYj74-000\|Let's talk more about the properties of electrons.
Bl-zUmeYj74-001\|So far we've talked about one electron systems, a hydrogen atom with only one electron.
Bl-zUmeYj74-002\|We need to add more electrons to get to the rest of the elements on the periodic table.
Bl-zUmeYj74-003\|So we need to talk about groups of electrons.
Bl-zUmeYj74-004\|And electrons have special properties.
Bl-zUmeYj74-005\|So let's look at some of them.
Bl-zUmeYj74-006\|Electrons have a property called spin.
Bl-zUmeYj74-007\|Now, it's called spin for historical reasons because it behaves like a spinning charge.

W_ngfrFO7f0-016|So they add to zero.

W_ngfrFO7f0-017|So this reaction has an enthalpy zero.

RCWYbfXHxXw-000|Let's look at a sample of sulfur atoms in some form.

RCWYbfXHxXw-001|And we're going to trap them in 22 liters, about a quarter atmosphere of pressure, and about 546 kelvin.

RCWYbfXHxXw-002|The question I have for you is, what's the form of sulfur?

RCWYbfXHxXw-010|We're talking about a mole of sulfur particles.

RCWYbfXHxXw-013|But I didn't have 1 atmosphere of pressure in my sample.

RCWYbfXHxXw-014|I only had a quarter atmosphere of pressure.

RCWYbfXHxXw-015|Well, how can you have that mole of particles give you a quarter atmosphere of pressure?

RCWYbfXHxXw-017|You have to have fewer particles.

RCWYbfXHxXw-019|That would give me 1/4 the particles and 1/4 the pressure.

RCWYbfXHxXw-020|Now, in this sample, the temperature was also twice the standard temperature.

RCWYbfXHxXw-021|How do I get a factor of two in temperature and stay at a quarter of the pressure in 22 liters?

RCWYbfXHxXw-025|So those particles must be in groups of eight.

RCWYbfXHxXw-026|The correct answer here, S8.

4kxMdkzCJGA-001|Those energy levels will be one to a wave function-- energy level 1, wave function number 1, energy level 2, wave function number 2.

4kxMdkzCJGA-002|They might be the same energy, that is, wave function 3 may have the same energy as wave function number 2.

4kxMdkzCJGA-003|But they'll all have different energies associated with different sets of quantum numbers.

4kxMdkzCJGA-004|So let's look at those quantum numbers and the energy levels for an atom.

4kxMdkzCJGA-005|The principle quantum number n again runs just as the integers-- 1, 2, 3.

4kxMdkzCJGA-006|And it gives you the overall energy, as we said, just like in a particle, a box, of the hydrogen wave functions.

4kxMdkzCJGA-007|So the wave functions energies that actually look quite simple-- Z squared-- that's the charge on the nucleus.

4kxMdkzCJGA-008|So for a hydrogen atom, that's one proton, so that charge is 1; n, the quantum number; and a constant.

4kxMdkzCJGA-009|This is Rydberg's constant.

4kxMdkzCJGA-010|It has a value that we can calculate and a value that we tabulate.

4kxMdkzCJGA-011|That value is 2.18 times 10 to the minus 18th joules.

4kxMdkzCJGA-013|So how does this look?

4kxMdkzCJGA-021|Why do we do this negative energy level?

4kxMdkzCJGA-022|Well, whenever you have energy, you have to give your energy relative to something.

4kxMdkzCJGA-023|And the relative to something, we say, for atoms is the electron and the nucleus infinitely far apart and at rest.

4kxMdkzCJGA-024|So that's the 0 in energy.

4kxMdkzCJGA-025|When you bring the nucleus and the electron closer together, they become more stable.

4kxMdkzCJGA-026|They go down in energy.

4kxMdkzCJGA-027|So when you're at 0, the only way to go down in energy is to go to negative energy values.

4kxMdkzCJGA-028|So we're going to use a negative system to describe this.

4kxMdkzCJGA-029|If we bring in n equal 2, we still have 1, the charge of the nucleus.

4kxMdkzCJGA-030|We're still talking about a hydrogen atom.

4kxMdkzCJGA-031|The charge of the nucleus is still 1.

4kxMdkzCJGA-032|n is now 2.

4kxMdkzCJGA-038|And of course, you don't even have to go to infinite to get ionized.

4kxMdkzCJGA-039|But the high levels of n are systems where the energy is about 0.

4kxMdkzCJGA-040|And since it's 1 over n squared, you don't have to go to a very big n for this energy to get very, very small and close to 0 very rapidly.

4kxMdkzCJGA-041|The ionized state is the 0 energy state for the hydrogen atom.

Li26UtxovcM-000|Let's look at what happens when you burn a hydrocarbon in oxygen.

Li26UtxovcM-001|A hydrocarbon is a compound composed of hydrogen and carbon only.

Li26UtxovcM-002|Now I've written it here generically as HC reacting with O2, oxygen molecules.

Li26UtxovcM-003|When that occurs, you can take the mass spectrum of the product.

Li26UtxovcM-005|And they'll have distinct mass spectrum signatures.

Li26UtxovcM-006|The water, mass 18, and the carbon dioxide at mass 44.

Li26UtxovcM-007|So the combustion of hydrocarbons gives a very distinctive mass spectrum of the products.

UdINHPWJRYc-000|Let's apply the first law of thermodynamics to an actual process.

UdINHPWJRYc-001|Here's an ideal gas expanding isothermally-- that is no temperature change-- against a vacuum.

UdINHPWJRYc-002|What's the heat flow?

UdINHPWJRYc-003|Is it less than 0, equal to 0, or greater than 0?

UdINHPWJRYc-013|We're talking about an ideal gas expanding against a vacuum isothermally.

UdINHPWJRYc-014|And whenever you hear isothermal for an ideal gas, you have to be excited, because if it's isothermal, the temperature didn't change.

UdINHPWJRYc-015|Isothermal, delta T is 0.

UdINHPWJRYc-016|If the temperature doesn't change, the energy of the gas could not have changed.

UdINHPWJRYc-021|One will be a negative sign.

UdINHPWJRYc-025|So no heat flows in this process and no work is done.

UdINHPWJRYc-026|The correct answer here, B, the heat flow is 0.

Mbq2QNVY10k-000|For higher steric numbers, we hybridize with some of the d orbitals.

Mbq2QNVY10k-001|So which energy diagram for phosphorous atomic orbitals in PCl5 is true after hybridization?

Mbq2QNVY10k-002|Here are the standard atomic orbitals for phosphorus.

Mbq2QNVY10k-011|We're talking about the hybridization in PCl5.

Mbq2QNVY10k-012|In PCl5 the phosphorus has Steric Number 5 as to accommodate five chlorines.

Mbq2QNVY10k-013|So we need five equivalent atomic orbitals.

Mbq2QNVY10k-014|The five equivalent atomic orbitals are going to come from hybridizing the standard atomic orbitals.

Mbq2QNVY10k-018|So the best answer here is A-- four unchanged d orbitals and five equivalent dsp3's.

8_ZRFjqe6F4-000|When you talk about ideal gases, we can take a volume and a temperature and a number of moles and calculate a pressure.

8_ZRFjqe6F4-002|Now, it's interesting to note that the ideal gas law doesn't say, well, what kind of particle is it?

8_ZRFjqe6F4-003|If it's a hydrogen particle H2, it will exert the same pressure as a bromine particles, so BR2.

8_ZRFjqe6F4-005|And that's true of gases that behave ideally.

8_ZRFjqe6F4-006|The pressure is independent of the nature of the particle.

8_ZRFjqe6F4-007|So why is that?

8_ZRFjqe6F4-008|Well, we can talk about how pressure originates from first principles.

8_ZRFjqe6F4-010|A transfer of momentum is a force applied to the walls of the container, and force per unit area is pressure.

8_ZRFjqe6F4-012|The bromine does it by coming in with a heavy mass and striking the particle, striking the walls of the container.

8_ZRFjqe6F4-013|The hydrogen particle, though much lighter, just moves very much faster.

8_ZRFjqe6F4-014|So even though it has a lower mass, it has a higher velocity.

8_ZRFjqe6F4-015|And it can transfer that same momentum in tiny little hits.

8_ZRFjqe6F4-016|So two different particles can exert the same pressure on the wall by momentum transfer.

8_ZRFjqe6F4-018|Either way, I can exert the same pressure on the walls of the container.

Hy9dsigzTiE-000|Let's look at two gases dissolved in a liquid, helium and nitrogen, aqueous solution for both of them.

Hy9dsigzTiE-001|Let's say they're both at equal concentrations in the solution.

Hy9dsigzTiE-002|What does that mean about the partial pressure above the solution when that is at equilibrium?

Hy9dsigzTiE-010|We're talking about helium and nitrogen dissolved in solution.

Hy9dsigzTiE-011|Their aqueous concentrations are the same.

Hy9dsigzTiE-012|So what does that mean about their partial pressures above the solution?

Hy9dsigzTiE-013|Well, to solve this, we need to look at the equilibrium constants, the Henry's law constants for these gases.

Hy9dsigzTiE-015|Here's helium, 0.4, and nitrogen 0.7.

Hy9dsigzTiE-020|The correct answer here is helium partial pressure is greater than nitrogen.

Bl-zUmeYj74-000|Let's talk more about the properties of electrons.

Bl-zUmeYj74-001|So far we've talked about one electron systems, a hydrogen atom with only one electron.

Bl-zUmeYj74-002|We need to add more electrons to get to the rest of the elements on the periodic table.

Bl-zUmeYj74-003|So we need to talk about groups of electrons.

Bl-zUmeYj74-004|And electrons have special properties.

Bl-zUmeYj74-005|So let's look at some of them.

Bl-zUmeYj74-006|Electrons have a property called spin.

Bl-zUmeYj74-007|Now, it's called spin for historical reasons because it behaves like a spinning charge.