ceaglex/VisualTTS_Chem · Datasets at Hugging Face

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Hx9dFbQ2zhk-052\|So the molecular orbital picture more powerful in determining a molecular property than the Lewis electron dot structure.
VabAZE71dH0-000\|Let's look at the chemical reaction hydrogen plus oxygen go into water under three different sets of circumstances.
VabAZE71dH0-001\|One, will ignite the hydrogen and oxygen. Two, will add platinum.
VabAZE71dH0-002\|And three, will just let the reaction go.
VabAZE71dH0-011\|We're looking at the chemical reaction hydrogen and oxygen forming water under three sets of circumstances.
VabAZE71dH0-012\|One, we ignite the reaction.
VabAZE71dH0-013\|Two, we add platinum.
VabAZE71dH0-014\|And three, we allow the reaction to proceed.
VabAZE71dH0-015\|Now, we've seen each of these in the demo lab.
VabAZE71dH0-016\|And when we ignite the chemical reaction it goes very rapidly.
VabAZE71dH0-017\|When you add platinum, the reaction goes at room temperature.
VabAZE71dH0-018\|Platinum is a catalyst for this chemical reaction and it lowers the activation energy.
VabAZE71dH0-021\|Reaction C and reaction A have the same activation energy.
VabAZE71dH0-022\|This one will go faster because we add energy, but their activation energies are the same.
X0nCoOX0LAg-000\|Let's look at a real gas that we're all familiar with, water vapor.
X0nCoOX0LAg-006\|Above that temperature, you'll always have a single phase.
X0nCoOX0LAg-007\|That is, you can compress the gas down.
X0nCoOX0LAg-008\|That phase might get very dense, but I won't see distinct liquid and gas phase.
X0nCoOX0LAg-009\|Below the critical temperature, I'll see distinct liquid and gas phases.
X0nCoOX0LAg-013\|That's 100 degrees C.
X0nCoOX0LAg-014\|Where does this flat line occur at 100 degrees C?
X0nCoOX0LAg-015\|Well, I think you can guess.
X0nCoOX0LAg-016\|It's going to be one atmosphere of pressure.
X0nCoOX0LAg-020\|So it's the partial pressure of the gas phase water above the liquid for that temperature.
X0nCoOX0LAg-021\|The normal boiling point is defined as where the vapor pressure equals atmospheric pressure.
X0nCoOX0LAg-026\|That's near room temperature.
X0nCoOX0LAg-027\|So the vapor pressure of water at room temperature is [? 0.31 ?] of an atmosphere.
X0nCoOX0LAg-028\|So that's the partial pressure of water above liquid water at room temperature.
X0nCoOX0LAg-029\|So the total pressure, if the sample is open to the atmosphere, you'll have a atmosphere of the atmospheric gases, nitrogen and oxygen.
X0nCoOX0LAg-030\|And [? 0.31 ?] of an atmosphere of water vapor pressure.
X0nCoOX0LAg-031\|So the vapor pressure can be defined as the pressure at which the gas spontaneously condenses for a given temperature.
X0nCoOX0LAg-032\|Or it can be defined as the partial pressure of the gas above a liquid sample.
100n-E7o9Ug-000\|When we make molecular orbitals from atomic orbitals, we take the atomic orbital wave functions and use linear combinations.
100n-E7o9Ug-001\|We add and subtract them with constant coefficients to make molecular orbitals.
100n-E7o9Ug-005\|So a helium molecule has four total electrons, and they need to occupy the molecular orbitals.
100n-E7o9Ug-009\|And I find that both the bonding and antibonding orbital for this molecule are completely full.
100n-E7o9Ug-010\|Now the bonding orbital is favorable for bonding.
100n-E7o9Ug-011\|The antibonding orbital is unfavorable for bonding.
100n-E7o9Ug-012\|A lower energy interaction and a higher energy interaction.
100n-E7o9Ug-013\|So those two cancel each other out.
100n-E7o9Ug-017\|Now I'm looking at a 2s electron.
100n-E7o9Ug-018\|A 2s electron I can take the 2s orbitals and use this same procedure.
100n-E7o9Ug-019\|Add and subtract them to take a linear combination, form a sigma 2s and a sigma 2s star antibonding orbital.
100n-E7o9Ug-020\|Those molecular orbitals are filled with the electrons from lithium.
100n-E7o9Ug-021\|Lithium has one electron in its 2s orbital, so two lithiums, two electrons, they fill up the bonding orbital for the lithium molecule.
100n-E7o9Ug-022\|Li2 we would predict to be a stable molecule.
100n-E7o9Ug-023\|It has a bond order of one.
100n-E7o9Ug-024\|Now the bond order will define as the sum of the bonding electrons minus the antibonding electrons.
100n-E7o9Ug-027\|Be2 has four electrons.
100n-E7o9Ug-028\|I'll put two electrons in the bonding of sigma and the two electrons in the antibonding sigma star orbital.
100n-E7o9Ug-029\|I'll calculate the bond order, two electrons here for bonding minus two antibonding electrons.
100n-E7o9Ug-030\|The bond order is zero.
100n-E7o9Ug-031\|So Be2, the beryllium 2 molecule, doesn't form.
100n-E7o9Ug-032\|So my molecular orbital theory has predictive power and it can predict the bond order and whether or not bonds form in molecules.
c-yoRAiBQ1k-001\|So to which energy level scheme-- A, B, or C-- does this emission spectrum correspond?
c-yoRAiBQ1k-007\|Let's look at the relationship between emission spectra and energy-level spacing in the matter.
c-yoRAiBQ1k-008\|So if you have an energy-level spacing that looks like A, what would the emission spectrum look like?
c-yoRAiBQ1k-009\|Well, we have to look at every possible transition in the system.
c-yoRAiBQ1k-010\|And here you can see three high energy level transitions-- those would give you high frequency lines-- and three low.
c-yoRAiBQ1k-011\|You could have this tiny transition, this tiny transition, and this tiny transition.
c-yoRAiBQ1k-012\|These two of equal energy, so they're degenerate.
c-yoRAiBQ1k-013\|They would fall right on top of each other and give you only one line even though there's two transitions.
c-yoRAiBQ1k-014\|But the two transitions have the same energy, so we can't resolve them in terms of energy.
c-yoRAiBQ1k-015\|So you get just two lines, one for this transition and one corresponding to both of these transitions.
c-yoRAiBQ1k-016\|So that doesn't look like the right answer.
c-yoRAiBQ1k-018\|They would fall right on top of each other.
c-yoRAiBQ1k-019\|So you'd have a very high energy level, a medium and a medium at the same energy, and a low in your high band.
c-yoRAiBQ1k-020\|That's a total of three from these four transitions since two are exactly the same.
c-yoRAiBQ1k-021\|And then two tiny low energy transitions, but again, they are of equal energy, so that would give you one line.
c-yoRAiBQ1k-022\|That looks like the spectrum we've seen, and if you look at C, that, of course, isn't anything like what we see.
c-yoRAiBQ1k-024\|So this pair gives you one line, this pair gives you one line, two intermediate lines, and then a small energy transition.
c-yoRAiBQ1k-028\|And remember, you can't look at tiny little matter in your microscope.
c-yoRAiBQ1k-030\|So here the correct answer is B.
Ie1qr8XGi6g-000\|When you form compounds from elements, there'll be an enthalpy associated with that.
Ie1qr8XGi6g-001\|So let's start with the elements in their standard states and form compounds.
Ie1qr8XGi6g-002\|What are the standard states of the elements?
Ie1qr8XGi6g-003\|Well, that's how you would find an element at 1 atmosphere of pressure and 25 degrees C.
Ie1qr8XGi6g-004\|So some elements from the periodic table will be a solid.
Ie1qr8XGi6g-005\|Some will be liquids.
Ie1qr8XGi6g-006\|Some will be gases.
Ie1qr8XGi6g-007\|Some will be diatomic.
Ie1qr8XGi6g-012\|Those are lower on enthalpy hill.
Ie1qr8XGi6g-013\|Overall, those compounds are a little more likely to be found because they're lower in energy than the elements.
Ie1qr8XGi6g-014\|So we have standard enthalpies of formation are the formation of compounds from their elements in their standard states.
9sDdIaBhtgk-000\|Let's look at weak acid, weak base, strong acid, and strong base solutions.
9sDdIaBhtgk-004\|So the pH is about one.
9sDdIaBhtgk-006\|The HAc weak acid dissociates to a lesser extent.
9sDdIaBhtgk-007\|So the pH is higher, indicating a lower concentration of H3O plus.
9sDdIaBhtgk-008\|Water, concentrations of H3O pluses at around 10 to the minus 7.
9sDdIaBhtgk-009\|NH3.
9sDdIaBhtgk-010\|Now we're getting to a base.
9sDdIaBhtgk-011\|So the H3O plus concentration drops even further, so I get a higher pH, 10 to the 10.
9sDdIaBhtgk-013\|The OH minus and the H3O plus concentrations, the product is always 10 to the minus 14 in water.
9sDdIaBhtgk-014\|So I can look at these, and we can put a little indicator in each.
9sDdIaBhtgk-016\|There's my strong acid solution.
9sDdIaBhtgk-017\|Here's my weak acid solution.
9sDdIaBhtgk-018\|Let's add a little more so that we get a little darker color.
9sDdIaBhtgk-019\|Here's my strong acid solution.
9sDdIaBhtgk-020\|Here's my weak acid solution.
9sDdIaBhtgk-021\|I already added a little indicator to the water, so we'll skip over that one.

Hx9dFbQ2zhk-052|So the molecular orbital picture more powerful in determining a molecular property than the Lewis electron dot structure.

VabAZE71dH0-000|Let's look at the chemical reaction hydrogen plus oxygen go into water under three different sets of circumstances.

VabAZE71dH0-001|One, will ignite the hydrogen and oxygen. Two, will add platinum.

VabAZE71dH0-002|And three, will just let the reaction go.

VabAZE71dH0-011|We're looking at the chemical reaction hydrogen and oxygen forming water under three sets of circumstances.

VabAZE71dH0-012|One, we ignite the reaction.

VabAZE71dH0-013|Two, we add platinum.

VabAZE71dH0-014|And three, we allow the reaction to proceed.

VabAZE71dH0-015|Now, we've seen each of these in the demo lab.

VabAZE71dH0-016|And when we ignite the chemical reaction it goes very rapidly.

VabAZE71dH0-017|When you add platinum, the reaction goes at room temperature.

VabAZE71dH0-018|Platinum is a catalyst for this chemical reaction and it lowers the activation energy.

VabAZE71dH0-021|Reaction C and reaction A have the same activation energy.

VabAZE71dH0-022|This one will go faster because we add energy, but their activation energies are the same.

X0nCoOX0LAg-000|Let's look at a real gas that we're all familiar with, water vapor.

X0nCoOX0LAg-006|Above that temperature, you'll always have a single phase.

X0nCoOX0LAg-007|That is, you can compress the gas down.

X0nCoOX0LAg-008|That phase might get very dense, but I won't see distinct liquid and gas phase.

X0nCoOX0LAg-009|Below the critical temperature, I'll see distinct liquid and gas phases.

X0nCoOX0LAg-013|That's 100 degrees C.

X0nCoOX0LAg-014|Where does this flat line occur at 100 degrees C?

X0nCoOX0LAg-015|Well, I think you can guess.

X0nCoOX0LAg-016|It's going to be one atmosphere of pressure.

X0nCoOX0LAg-020|So it's the partial pressure of the gas phase water above the liquid for that temperature.

X0nCoOX0LAg-021|The normal boiling point is defined as where the vapor pressure equals atmospheric pressure.

X0nCoOX0LAg-026|That's near room temperature.

X0nCoOX0LAg-027|So the vapor pressure of water at room temperature is [? 0.31 ?] of an atmosphere.

X0nCoOX0LAg-028|So that's the partial pressure of water above liquid water at room temperature.

X0nCoOX0LAg-029|So the total pressure, if the sample is open to the atmosphere, you'll have a atmosphere of the atmospheric gases, nitrogen and oxygen.

X0nCoOX0LAg-030|And [? 0.31 ?] of an atmosphere of water vapor pressure.

X0nCoOX0LAg-031|So the vapor pressure can be defined as the pressure at which the gas spontaneously condenses for a given temperature.

X0nCoOX0LAg-032|Or it can be defined as the partial pressure of the gas above a liquid sample.

100n-E7o9Ug-000|When we make molecular orbitals from atomic orbitals, we take the atomic orbital wave functions and use linear combinations.

100n-E7o9Ug-001|We add and subtract them with constant coefficients to make molecular orbitals.

100n-E7o9Ug-005|So a helium molecule has four total electrons, and they need to occupy the molecular orbitals.

100n-E7o9Ug-009|And I find that both the bonding and antibonding orbital for this molecule are completely full.

100n-E7o9Ug-010|Now the bonding orbital is favorable for bonding.

100n-E7o9Ug-011|The antibonding orbital is unfavorable for bonding.

100n-E7o9Ug-012|A lower energy interaction and a higher energy interaction.

100n-E7o9Ug-013|So those two cancel each other out.

100n-E7o9Ug-017|Now I'm looking at a 2s electron.

100n-E7o9Ug-018|A 2s electron I can take the 2s orbitals and use this same procedure.

100n-E7o9Ug-019|Add and subtract them to take a linear combination, form a sigma 2s and a sigma 2s star antibonding orbital.

100n-E7o9Ug-020|Those molecular orbitals are filled with the electrons from lithium.

100n-E7o9Ug-021|Lithium has one electron in its 2s orbital, so two lithiums, two electrons, they fill up the bonding orbital for the lithium molecule.

100n-E7o9Ug-022|Li2 we would predict to be a stable molecule.

100n-E7o9Ug-023|It has a bond order of one.

100n-E7o9Ug-024|Now the bond order will define as the sum of the bonding electrons minus the antibonding electrons.

100n-E7o9Ug-027|Be2 has four electrons.

100n-E7o9Ug-028|I'll put two electrons in the bonding of sigma and the two electrons in the antibonding sigma star orbital.

100n-E7o9Ug-029|I'll calculate the bond order, two electrons here for bonding minus two antibonding electrons.

100n-E7o9Ug-030|The bond order is zero.

100n-E7o9Ug-031|So Be2, the beryllium 2 molecule, doesn't form.

100n-E7o9Ug-032|So my molecular orbital theory has predictive power and it can predict the bond order and whether or not bonds form in molecules.

c-yoRAiBQ1k-001|So to which energy level scheme-- A, B, or C-- does this emission spectrum correspond?

c-yoRAiBQ1k-007|Let's look at the relationship between emission spectra and energy-level spacing in the matter.

c-yoRAiBQ1k-008|So if you have an energy-level spacing that looks like A, what would the emission spectrum look like?

c-yoRAiBQ1k-009|Well, we have to look at every possible transition in the system.

c-yoRAiBQ1k-010|And here you can see three high energy level transitions-- those would give you high frequency lines-- and three low.

c-yoRAiBQ1k-011|You could have this tiny transition, this tiny transition, and this tiny transition.

c-yoRAiBQ1k-012|These two of equal energy, so they're degenerate.

c-yoRAiBQ1k-013|They would fall right on top of each other and give you only one line even though there's two transitions.

c-yoRAiBQ1k-014|But the two transitions have the same energy, so we can't resolve them in terms of energy.

c-yoRAiBQ1k-015|So you get just two lines, one for this transition and one corresponding to both of these transitions.

c-yoRAiBQ1k-016|So that doesn't look like the right answer.

c-yoRAiBQ1k-018|They would fall right on top of each other.

c-yoRAiBQ1k-019|So you'd have a very high energy level, a medium and a medium at the same energy, and a low in your high band.

c-yoRAiBQ1k-020|That's a total of three from these four transitions since two are exactly the same.

c-yoRAiBQ1k-021|And then two tiny low energy transitions, but again, they are of equal energy, so that would give you one line.

c-yoRAiBQ1k-022|That looks like the spectrum we've seen, and if you look at C, that, of course, isn't anything like what we see.

c-yoRAiBQ1k-024|So this pair gives you one line, this pair gives you one line, two intermediate lines, and then a small energy transition.

c-yoRAiBQ1k-028|And remember, you can't look at tiny little matter in your microscope.

c-yoRAiBQ1k-030|So here the correct answer is B.

Ie1qr8XGi6g-000|When you form compounds from elements, there'll be an enthalpy associated with that.

Ie1qr8XGi6g-001|So let's start with the elements in their standard states and form compounds.

Ie1qr8XGi6g-002|What are the standard states of the elements?

Ie1qr8XGi6g-003|Well, that's how you would find an element at 1 atmosphere of pressure and 25 degrees C.

Ie1qr8XGi6g-004|So some elements from the periodic table will be a solid.

Ie1qr8XGi6g-005|Some will be liquids.

Ie1qr8XGi6g-006|Some will be gases.

Ie1qr8XGi6g-007|Some will be diatomic.

Ie1qr8XGi6g-012|Those are lower on enthalpy hill.

Ie1qr8XGi6g-013|Overall, those compounds are a little more likely to be found because they're lower in energy than the elements.

Ie1qr8XGi6g-014|So we have standard enthalpies of formation are the formation of compounds from their elements in their standard states.

9sDdIaBhtgk-000|Let's look at weak acid, weak base, strong acid, and strong base solutions.

9sDdIaBhtgk-004|So the pH is about one.

9sDdIaBhtgk-006|The HAc weak acid dissociates to a lesser extent.

9sDdIaBhtgk-007|So the pH is higher, indicating a lower concentration of H3O plus.

9sDdIaBhtgk-008|Water, concentrations of H3O pluses at around 10 to the minus 7.

9sDdIaBhtgk-009|NH3.

9sDdIaBhtgk-010|Now we're getting to a base.

9sDdIaBhtgk-011|So the H3O plus concentration drops even further, so I get a higher pH, 10 to the 10.

9sDdIaBhtgk-013|The OH minus and the H3O plus concentrations, the product is always 10 to the minus 14 in water.

9sDdIaBhtgk-014|So I can look at these, and we can put a little indicator in each.

9sDdIaBhtgk-016|There's my strong acid solution.

9sDdIaBhtgk-017|Here's my weak acid solution.

9sDdIaBhtgk-018|Let's add a little more so that we get a little darker color.

9sDdIaBhtgk-019|Here's my strong acid solution.

9sDdIaBhtgk-020|Here's my weak acid solution.

9sDdIaBhtgk-021|I already added a little indicator to the water, so we'll skip over that one.