ceaglex/VisualTTS_Chem · Datasets at Hugging Face

text stringlengths 19 206
f-SJBvBHpuM-014\|So when l is smaller than n, we're in good shape.
f-SJBvBHpuM-015\|So so far, OK.
f-SJBvBHpuM-016\|-1, that's a possible value of m sub l, that's fine.
f-SJBvBHpuM-017\|Because when l is 2, m sub l goes from -2 to 2 in integer steps.
f-SJBvBHpuM-018\|-2, -1, 0, 1, and 2 are all possible values.
f-SJBvBHpuM-019\|And -1 is certainly available.
f-SJBvBHpuM-020\|1/2 is one of the possible values of m sub s, so that's fine.
f-SJBvBHpuM-024\|We have n equal 5, l equal 0.
f-SJBvBHpuM-025\|Those are both fine.
f-SJBvBHpuM-026\|But for m sub l we have -1.
f-SJBvBHpuM-027\|Now when l is 0, all m sub l can be is 0.
f-SJBvBHpuM-028\|So this quantum number here is out of line.
f-SJBvBHpuM-029\|M sub l can only be zero if l is 0.
f-SJBvBHpuM-030\|So that's not an allowed set of quantum numbers.
f-SJBvBHpuM-031\|This final one here, we have n equal 4.
f-SJBvBHpuM-032\|We have l equal 4.
f-SJBvBHpuM-033\|Now the maximum value of l is n -1.
f-SJBvBHpuM-038\|So you have to very carefully apply your rules to see which quantum numbers are allowed.
f-SJBvBHpuM-039\|That determines which orbitals exist around an atom.
R2TrrmdGM2A-000\|Chemical bonds are characterized in a variety of ways.
R2TrrmdGM2A-001\|Ionic bonds, for instance, are where the electron is transferred from one element to the other, and it's a plus-minus Coulombic interaction that holds the bond together.
R2TrrmdGM2A-002\|That occurs when an electronegativity difference is very large between the two elements.
R2TrrmdGM2A-004\|Lithium fluoride is an example of that.
R2TrrmdGM2A-005\|Now you can have bonds where the electron is shared more equally between the two.
R2TrrmdGM2A-006\|You can imagine a perfect sharing.
R2TrrmdGM2A-007\|For instance, covalent bonding is where sharing occurs, and chlorine Cl2 is an example where you have a homonuclear bond.
R2TrrmdGM2A-008\|That is, the electron has a preference for neither chlorine.
R2TrrmdGM2A-009\|chlorines are identical, the electron has equal preference, there's an equal probability of finding the electron on either chlorine atom.
R2TrrmdGM2A-010\|So that bond is perfectly covalent, an equal sharing.
R2TrrmdGM2A-011\|But it's actually more likely that you'll have bonds between different elements.
R2TrrmdGM2A-013\|So those are called polar covalent bonds.
R2TrrmdGM2A-014\|An example, hydrogen chloride.
R2TrrmdGM2A-015\|Here, the chlorine is the more electronegative element.
R2TrrmdGM2A-016\|So it draws the electron toward itself and there's a higher probability of finding the electron around the chlorine.
R2TrrmdGM2A-017\|That gives the chlorine a slight negative charge and the hydrogen a slight positive charge.
R2TrrmdGM2A-018\|So the hydrogen end of the molecule, slightly positive; the chlorine end of the molecule, slightly negative, because there's a higher probability of finding the electron around the chlorine.
R2TrrmdGM2A-020\|A dipole moment is a vector quantity, it has size-- the size of the charge separation-- and the distance between the two.
R2TrrmdGM2A-021\|So in this case, we draw a dipole moment vector from the positive to the negative end of the band.
R2TrrmdGM2A-022\|Now, virtually every bond is going to have a dipole moment because it's very rare you find perfectly equal sharing of electrons.
R2TrrmdGM2A-023\|So if all bonds have dipole moments, it's likely that all molecules will have dipole moments.
R2TrrmdGM2A-024\|But there will be cases where the symmetry of the molecule allows dipole moment vectors to cancel.
R2TrrmdGM2A-026\|There are other symmetry examples where that can happen.
R2TrrmdGM2A-028\|And usually, the molecules will have a permanent dipole moment.
R2TrrmdGM2A-030\|And I can demonstrate the dipole moment in water, because the dipole moment creates a charged end of the molecule that will be attracted to a charged surface.
R2TrrmdGM2A-031\|So let's look at that.
R2TrrmdGM2A-032\|What I have here is a setup for a stream of water, I'm going to bring a stream of water from this burette.
R2TrrmdGM2A-034\|So I'll start a water flow from this burette.
R2TrrmdGM2A-038\|That's a very dramatic demonstration of dipole moment in molecules.
1JrO7QQ1o1M-000\|If you take a particle that behaves like a wave, and you trap it, you put boundaries on it.
1JrO7QQ1o1M-001\|You trap it in a box, it turns out it can only have certain energies.
1JrO7QQ1o1M-002\|It's not like when you take a marble, and you put it in a box, and you shake the marble around.
1JrO7QQ1o1M-003\|It can have any old kinetic energy it wants.
1JrO7QQ1o1M-004\|It can be going fast, it can be going slow.
1JrO7QQ1o1M-005\|It can be going in a continuous amount of velocities in between.
1JrO7QQ1o1M-006\|That's not true for a particle that behaves like a wave.
1JrO7QQ1o1M-007\|Particles that behave like a wave can have only certain energies.
1JrO7QQ1o1M-008\|The energies are quantized.
1JrO7QQ1o1M-009\|Remember wave plus boundaries gives you quantization.
1JrO7QQ1o1M-010\|So I have some quantized levels written here, and quantization is the fundamental property that allows absorption and emission of light.
1JrO7QQ1o1M-011\|Because if you're going to make the system go from one level to the next, you can't absorb just any old wavelength.
1JrO7QQ1o1M-012\|You have to absorb the wavelength that actually fits, that gives you the exact amount of energy to go from low state to high-energy state.
1JrO7QQ1o1M-013\|All other energy levels, and all other energies that the system experiences will be ignored.
1JrO7QQ1o1M-017\|So that is transmitted.
1JrO7QQ1o1M-018\|Here's a wavelength that exactly matches this high-energy transition, that one will be absorbed.
1JrO7QQ1o1M-024\|Some wavelengths are too big, some are too little, some are just right.
1JrO7QQ1o1M-025\|So you have another, just right.
1JrO7QQ1o1M-026\|That will be absorbed and perhaps you'll find a third just right energy level that will be absorbed.
1JrO7QQ1o1M-027\|So you get an absorption spectrum that says, oh, this blue, this yellow, and this red are removed from the continuous light that's hitting this object.
1JrO7QQ1o1M-028\|Or you could have the light emit by the system.
1JrO7QQ1o1M-029\|The system could emit the energy of the highway of length, and then you'd have a single emission.
1JrO7QQ1o1M-030\|It can't emit all wavelengths of light, only certain wavelengths of light are emitted.
OSts9bfX6cA-001\|Do you think that will look like A, B, or C?
OSts9bfX6cA-008\|We're talking about the conductometric titration of barium hydroxide with H2SO4.
OSts9bfX6cA-009\|So we want to know how does the light intensity change with added H2SO4.
OSts9bfX6cA-010\|Will it stay bright?
OSts9bfX6cA-011\|Will it get dim initially and return?
OSts9bfX6cA-012\|Or will it get dim and stay dim?
OSts9bfX6cA-013\|Well, we have this on the bench top.
OSts9bfX6cA-014\|So let's actually do the experiment.
OSts9bfX6cA-015\|Here I have barium hydroxide in this lower flask.
OSts9bfX6cA-016\|And I have H2SO4 in this burette.
OSts9bfX6cA-026\|So the solution has very low ionic strength and very low conductivity.
OSts9bfX6cA-029\|Now, the light returns because H2SO4 forms ions in solution.
OSts9bfX6cA-030\|So initially, I use ions up.
OSts9bfX6cA-031\|And after all the barium is used up, then the extra sulfate that goes into solution stays in the solution and allows the solution to conduct.
OSts9bfX6cA-033\|So H2SO4 lends itself to making ions in solution and gives us a bright light.
OSts9bfX6cA-034\|And after the solid is formed, these ions conduct.
OSts9bfX6cA-035\|The correct answer here is B.
oYVD8MSW8NE-000\|Glucose is oxidized in your body to produce energy.
oYVD8MSW8NE-001\|Now you can oxidize glucose on the benchtop.
oYVD8MSW8NE-002\|You could burn it in oxygen, the combustion of glucose.
oYVD8MSW8NE-003\|You'd burn glucose in oxygen, you'd form carbon dioxide and water, energy would be released.
oYVD8MSW8NE-004\|In your body, the same thing happens.
oYVD8MSW8NE-005\|You oxidize glucose, (INHALING) breathe in oxygen to react with the glucose, (EXHALING) you breathe out carbon dioxide and water.
oYVD8MSW8NE-006\|So the products and reactants are the same.
oYVD8MSW8NE-008\|So thermodynamically, the energetics are the same basically.
oYVD8MSW8NE-009\|Free energy, entropy, enthalpy, they're state functions.
oYVD8MSW8NE-010\|They only depend on the initial and final state.
oYVD8MSW8NE-011\|I have glucose and oxygen, and I form carbon dioxide and water, about the same state, the same energy is released.
oYVD8MSW8NE-013\|In your body, the oxidation occurs more like a battery.

f-SJBvBHpuM-014|So when l is smaller than n, we're in good shape.

f-SJBvBHpuM-015|So so far, OK.

f-SJBvBHpuM-016|-1, that's a possible value of m sub l, that's fine.

f-SJBvBHpuM-017|Because when l is 2, m sub l goes from -2 to 2 in integer steps.

f-SJBvBHpuM-018|-2, -1, 0, 1, and 2 are all possible values.

f-SJBvBHpuM-019|And -1 is certainly available.

f-SJBvBHpuM-020|1/2 is one of the possible values of m sub s, so that's fine.

f-SJBvBHpuM-024|We have n equal 5, l equal 0.

f-SJBvBHpuM-025|Those are both fine.

f-SJBvBHpuM-026|But for m sub l we have -1.

f-SJBvBHpuM-027|Now when l is 0, all m sub l can be is 0.

f-SJBvBHpuM-028|So this quantum number here is out of line.

f-SJBvBHpuM-029|M sub l can only be zero if l is 0.

f-SJBvBHpuM-030|So that's not an allowed set of quantum numbers.

f-SJBvBHpuM-031|This final one here, we have n equal 4.

f-SJBvBHpuM-032|We have l equal 4.

f-SJBvBHpuM-033|Now the maximum value of l is n -1.

f-SJBvBHpuM-038|So you have to very carefully apply your rules to see which quantum numbers are allowed.

f-SJBvBHpuM-039|That determines which orbitals exist around an atom.

R2TrrmdGM2A-000|Chemical bonds are characterized in a variety of ways.

R2TrrmdGM2A-001|Ionic bonds, for instance, are where the electron is transferred from one element to the other, and it's a plus-minus Coulombic interaction that holds the bond together.

R2TrrmdGM2A-002|That occurs when an electronegativity difference is very large between the two elements.

R2TrrmdGM2A-004|Lithium fluoride is an example of that.

R2TrrmdGM2A-005|Now you can have bonds where the electron is shared more equally between the two.

R2TrrmdGM2A-006|You can imagine a perfect sharing.

R2TrrmdGM2A-007|For instance, covalent bonding is where sharing occurs, and chlorine Cl2 is an example where you have a homonuclear bond.

R2TrrmdGM2A-008|That is, the electron has a preference for neither chlorine.

R2TrrmdGM2A-009|chlorines are identical, the electron has equal preference, there's an equal probability of finding the electron on either chlorine atom.

R2TrrmdGM2A-010|So that bond is perfectly covalent, an equal sharing.

R2TrrmdGM2A-011|But it's actually more likely that you'll have bonds between different elements.

R2TrrmdGM2A-013|So those are called polar covalent bonds.

R2TrrmdGM2A-014|An example, hydrogen chloride.

R2TrrmdGM2A-015|Here, the chlorine is the more electronegative element.

R2TrrmdGM2A-016|So it draws the electron toward itself and there's a higher probability of finding the electron around the chlorine.

R2TrrmdGM2A-017|That gives the chlorine a slight negative charge and the hydrogen a slight positive charge.

R2TrrmdGM2A-018|So the hydrogen end of the molecule, slightly positive; the chlorine end of the molecule, slightly negative, because there's a higher probability of finding the electron around the chlorine.

R2TrrmdGM2A-020|A dipole moment is a vector quantity, it has size-- the size of the charge separation-- and the distance between the two.

R2TrrmdGM2A-021|So in this case, we draw a dipole moment vector from the positive to the negative end of the band.

R2TrrmdGM2A-022|Now, virtually every bond is going to have a dipole moment because it's very rare you find perfectly equal sharing of electrons.

R2TrrmdGM2A-023|So if all bonds have dipole moments, it's likely that all molecules will have dipole moments.

R2TrrmdGM2A-024|But there will be cases where the symmetry of the molecule allows dipole moment vectors to cancel.

R2TrrmdGM2A-026|There are other symmetry examples where that can happen.

R2TrrmdGM2A-028|And usually, the molecules will have a permanent dipole moment.

R2TrrmdGM2A-030|And I can demonstrate the dipole moment in water, because the dipole moment creates a charged end of the molecule that will be attracted to a charged surface.

R2TrrmdGM2A-031|So let's look at that.

R2TrrmdGM2A-032|What I have here is a setup for a stream of water, I'm going to bring a stream of water from this burette.

R2TrrmdGM2A-034|So I'll start a water flow from this burette.

R2TrrmdGM2A-038|That's a very dramatic demonstration of dipole moment in molecules.

1JrO7QQ1o1M-000|If you take a particle that behaves like a wave, and you trap it, you put boundaries on it.

1JrO7QQ1o1M-001|You trap it in a box, it turns out it can only have certain energies.

1JrO7QQ1o1M-002|It's not like when you take a marble, and you put it in a box, and you shake the marble around.

1JrO7QQ1o1M-003|It can have any old kinetic energy it wants.

1JrO7QQ1o1M-004|It can be going fast, it can be going slow.

1JrO7QQ1o1M-005|It can be going in a continuous amount of velocities in between.

1JrO7QQ1o1M-006|That's not true for a particle that behaves like a wave.

1JrO7QQ1o1M-007|Particles that behave like a wave can have only certain energies.

1JrO7QQ1o1M-008|The energies are quantized.

1JrO7QQ1o1M-009|Remember wave plus boundaries gives you quantization.

1JrO7QQ1o1M-010|So I have some quantized levels written here, and quantization is the fundamental property that allows absorption and emission of light.

1JrO7QQ1o1M-011|Because if you're going to make the system go from one level to the next, you can't absorb just any old wavelength.

1JrO7QQ1o1M-012|You have to absorb the wavelength that actually fits, that gives you the exact amount of energy to go from low state to high-energy state.

1JrO7QQ1o1M-013|All other energy levels, and all other energies that the system experiences will be ignored.

1JrO7QQ1o1M-017|So that is transmitted.

1JrO7QQ1o1M-018|Here's a wavelength that exactly matches this high-energy transition, that one will be absorbed.

1JrO7QQ1o1M-024|Some wavelengths are too big, some are too little, some are just right.

1JrO7QQ1o1M-025|So you have another, just right.

1JrO7QQ1o1M-026|That will be absorbed and perhaps you'll find a third just right energy level that will be absorbed.

1JrO7QQ1o1M-027|So you get an absorption spectrum that says, oh, this blue, this yellow, and this red are removed from the continuous light that's hitting this object.

1JrO7QQ1o1M-028|Or you could have the light emit by the system.

1JrO7QQ1o1M-029|The system could emit the energy of the highway of length, and then you'd have a single emission.

1JrO7QQ1o1M-030|It can't emit all wavelengths of light, only certain wavelengths of light are emitted.

OSts9bfX6cA-001|Do you think that will look like A, B, or C?

OSts9bfX6cA-008|We're talking about the conductometric titration of barium hydroxide with H2SO4.

OSts9bfX6cA-009|So we want to know how does the light intensity change with added H2SO4.

OSts9bfX6cA-010|Will it stay bright?

OSts9bfX6cA-011|Will it get dim initially and return?

OSts9bfX6cA-012|Or will it get dim and stay dim?

OSts9bfX6cA-013|Well, we have this on the bench top.

OSts9bfX6cA-014|So let's actually do the experiment.

OSts9bfX6cA-015|Here I have barium hydroxide in this lower flask.

OSts9bfX6cA-016|And I have H2SO4 in this burette.

OSts9bfX6cA-026|So the solution has very low ionic strength and very low conductivity.

OSts9bfX6cA-029|Now, the light returns because H2SO4 forms ions in solution.

OSts9bfX6cA-030|So initially, I use ions up.

OSts9bfX6cA-031|And after all the barium is used up, then the extra sulfate that goes into solution stays in the solution and allows the solution to conduct.

OSts9bfX6cA-033|So H2SO4 lends itself to making ions in solution and gives us a bright light.

OSts9bfX6cA-034|And after the solid is formed, these ions conduct.

OSts9bfX6cA-035|The correct answer here is B.

oYVD8MSW8NE-000|Glucose is oxidized in your body to produce energy.

oYVD8MSW8NE-001|Now you can oxidize glucose on the benchtop.

oYVD8MSW8NE-002|You could burn it in oxygen, the combustion of glucose.

oYVD8MSW8NE-003|You'd burn glucose in oxygen, you'd form carbon dioxide and water, energy would be released.

oYVD8MSW8NE-004|In your body, the same thing happens.

oYVD8MSW8NE-005|You oxidize glucose, (INHALING) breathe in oxygen to react with the glucose, (EXHALING) you breathe out carbon dioxide and water.

oYVD8MSW8NE-006|So the products and reactants are the same.

oYVD8MSW8NE-008|So thermodynamically, the energetics are the same basically.

oYVD8MSW8NE-009|Free energy, entropy, enthalpy, they're state functions.

oYVD8MSW8NE-010|They only depend on the initial and final state.

oYVD8MSW8NE-011|I have glucose and oxygen, and I form carbon dioxide and water, about the same state, the same energy is released.

oYVD8MSW8NE-013|In your body, the oxidation occurs more like a battery.