ceaglex/VisualTTS_Chem · Datasets at Hugging Face

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OCK36oYeAFA-006\|Nitrogen in its standard state is N2 gas.
OCK36oYeAFA-007\|Oxygen in its standard state is O2 gas.
OCK36oYeAFA-008\|So I can write N2 gas plus O2 gas goes to two NO, just balancing that chemical equation.
OCK36oYeAFA-009\|Of course, the enthalpy information is for a single mole.
OCK36oYeAFA-010\|I've formed two moles here, so this enthalpy is the enthalpy for 1/2 this chemical reaction.
OCK36oYeAFA-011\|I've got 2 here, so I'll just double this enthalpy.
OCK36oYeAFA-013\|181 kilojoules.
OCK36oYeAFA-014\|Now I want to find the enthalpy of formation for N2O5.
OCK36oYeAFA-015\|I can write that because it's the enthalpy of formation.
OCK36oYeAFA-018\|So I'd like to sum these to form this one.
OCK36oYeAFA-019\|And I noticed right off, well, this reaction here, that would give me two moles of N2O5 on the product side.
OCK36oYeAFA-020\|I only want one mole, so I'm going to divide this one through by 2.
OCK36oYeAFA-021\|If I divide this reaction through by 2, I have to divide the enthalpy by 2.
OCK36oYeAFA-022\|So let's do that.
OCK36oYeAFA-023\|And then, when we add down, you'll see that these three reactions add to give the reaction we're looking for.
OCK36oYeAFA-024\|So it's a Hess's law situation.
OCK36oYeAFA-025\|I can add these three.
OCK36oYeAFA-033\|It's slightly endothermic.
2hohw6e19_s-000\|We can calculate the pH in the buffer region using the Henderson-Hasselbalch expression.
2hohw6e19_s-002\|So here, I've expanded the buffer region.
2hohw6e19_s-003\|And we know that, at half equivalence, halfway to the equivalence point, the concentrations of the acid and base form are equal.
2hohw6e19_s-004\|So this ratio is 1, the log term is 0, and the pH is numerically equal to the pKa at half equivalence.
2hohw6e19_s-007\|This term would be negative.
2hohw6e19_s-008\|And of course, if the base form predominates, you'd expect this term to be positive.
2hohw6e19_s-010\|So the Henderson-Hasselbalch expression allows us to calculate the pH throughout this region.
2hohw6e19_s-013\|The buffer region would be from 3.75 to 5.75, one unit around the pKa.
2hohw6e19_s-014\|So let's look at that in more detail.
2hohw6e19_s-023\|So a buffer resists change in pH from adding acid, from adding base, and from dilution.
-zXxTQt2slA-000\|Let's look at hybridization in benzene.
-zXxTQt2slA-001\|Benzene, C6H6, I've drawn it like this with this circle representing the alternating double bonds.
-zXxTQt2slA-007\|We're talking about the hybridization of carbons in benzene.
-zXxTQt2slA-008\|Now, we drew benzene schematically, but that's enough to determine the steric number of each carbon, and that's what you need to determine hybridization.
-zXxTQt2slA-009\|Carbon in benzene has to accommodate three things-- two other carbons and a hydrogen.
-zXxTQt2slA-010\|So to accommodate three things with steric number 3, I have hybridize three atomic orbitals-- an s and two p's.
-zXxTQt2slA-011\|That leads to sp2 hybridization, 3 equivalent orbitals that have a 120 degree bond angle.
-zXxTQt2slA-013\|All the carbons are equivalent, so these carbons here will also be sp2 hybridized.
-zXxTQt2slA-016\|So the correct answer here-- sp2 hybridisation for the carbons in benzene.
y97cegGFECA-000\|When salts dissolve in water there's a wide variation in the solubility.
y97cegGFECA-001\|Some salts like sodium chloride dissolve quite well and they form high concentrations of sodium ions and chloride ions in solution.
y97cegGFECA-002\|Others like barium sulfate are not very soluble.
y97cegGFECA-004\|So a small K means I favor the reactant side, the solid as opposed to the ions.
y97cegGFECA-007\|It kind of means that barium and sulfate ions don't like to exist together in solution.
y97cegGFECA-008\|The more stable state is the barium sulfate solid.
y97cegGFECA-009\|So if barium comes from one source and sulfate from another, those two ions seek each other out.
y97cegGFECA-017\|So the K for this reaction, the reverse, is 1 over 10 to the minus 10, or 10 to the plus 10.
y97cegGFECA-018\|A large K means I strongly favor the barium sulfate solid.
y97cegGFECA-020\|So both these reactions strongly favor the solid and water.
y97cegGFECA-022\|So the ion concentration drops if you mix these two solutions.
y97cegGFECA-024\|The barium and hydroxide ions would conduct electricity across this gap and allow this light to come on.
y97cegGFECA-025\|When there's high ionic concentration, that light will be bright.
y95SN6mDh0Y-000\|Now the intensity of light, or electromagnetic radiation, may be more of an intuitive property.
y95SN6mDh0Y-001\|Let's look at it this way.
y95SN6mDh0Y-003\|So I0 going through a pair of identical filters.
y95SN6mDh0Y-004\|Would that reduce the intensity down to a quarter, an eighth, or 0 of the original intensity?
y95SN6mDh0Y-011\|So we've been talking about intensity, the brightness of light, in terms of attenuating it, or reducing it, with a filter.
y95SN6mDh0Y-013\|Well if the first filter reduces it by one-half, so it's half as bright here.
y95SN6mDh0Y-014\|A second filter would make this half as bright.
y95SN6mDh0Y-015\|So you'd go from half the original intensity to a quarter of the original intensity after two filters.
aoAdE8L2ASw-000\|Let's look at our liquid-to-gas transition for water and ask, what is the standard state free energy difference?
aoAdE8L2ASw-002\|What is the standard state free energy difference for this physical change at 100 degrees C?
aoAdE8L2ASw-010\|We're talking about the liquid-to-gas phase transition.
aoAdE8L2ASw-011\|And at 100 degrees C, what's the standard state free energy difference?
aoAdE8L2ASw-014\|So we can talk about those states of matter at different temperatures.
aoAdE8L2ASw-015\|And in this case, we're talking about it at 100 degrees C.
aoAdE8L2ASw-017\|And I think you could say, well, that's what boiling is.
aoAdE8L2ASw-018\|Boiling occurs when there's a vapor pressure of one atmosphere above the liquid.
aoAdE8L2ASw-019\|So that is the equilibrium point.
aoAdE8L2ASw-021\|So what we have here is, where is the free energy difference between the one atmosphere of gas and liquid exactly 0?
aoAdE8L2ASw-025\|If you went to a different temperature and you were talking about the standard state, you would still have one atmosphere of gas.
aoAdE8L2ASw-026\|So that's why you say, well, if I go to a lower temperature and I still have one atmosphere of gas, well, then of course, I favor the liquid.
aoAdE8L2ASw-027\|That's a higher amount of gas, one atmosphere, than the vapor pressure at these lower temperatures.
aoAdE8L2ASw-028\|And as I go to higher temperatures, then I have the one atmosphere of gas is the favored state of the system.
aoAdE8L2ASw-029\|So delta G standard is a function of temperature.
aoAdE8L2ASw-030\|In this case, I'm right at the equilibrium where the gas and the liquid are equally favored.
aoAdE8L2ASw-031\|In this case, the correct answer is B.
Hx9dFbQ2zhk-000\|Let's look at both the quantum mechanical description and the Lewis dot structure description of a molecule.
Hx9dFbQ2zhk-001\|We'll pick oxygen, O2.
Hx9dFbQ2zhk-002\|Oxygen, each atom has 6 valence electrons in principle quantum level 2.
Hx9dFbQ2zhk-004\|Each oxygen obeys the octet, it has 2, 4, 6, 8 electrons around it.
Hx9dFbQ2zhk-005\|So that's our good Lewis electron dot structure.
Hx9dFbQ2zhk-006\|Let's look at the quantum mechanical description.
Hx9dFbQ2zhk-016\|So those are the s orbitals, let's take the p orbitals.
Hx9dFbQ2zhk-017\|There are 6 p orbitals, 3 on each atom.
Hx9dFbQ2zhk-024\|Again, sigma bonding, sigma antibonding, energetically distributed.
Hx9dFbQ2zhk-025\|Now, let's look at the px and the py.
Hx9dFbQ2zhk-026\|The px and the py lie in front and behind and above and below the internuclear axis.
Hx9dFbQ2zhk-027\|And you can see that on a model.
Hx9dFbQ2zhk-030\|It doesn't really matter which one I look at, they're both identical in energy.
Hx9dFbQ2zhk-035\|And the minus combination gives me antibonding, pi orbitals, two of those.
Hx9dFbQ2zhk-036\|One from the minus combination of px and one from the minus combination of py.
Hx9dFbQ2zhk-038\|Each oxygen had an s and three p's.
Hx9dFbQ2zhk-039\|So 4 orbitals on each oxygen, 8 total, make 8 molecular orbitals.
Hx9dFbQ2zhk-040\|I need to fill those with the 12 valence electrons from oxygen.
Hx9dFbQ2zhk-043\|So the bond order, I have a sigma bonding and antibonding orbital to start.
Hx9dFbQ2zhk-044\|Two electrons in each, those cancel each other out, zero bond order from that.
Hx9dFbQ2zhk-046\|So 6 minus 2 is 4 bonding electrons divided by 2-- 2 formal bonds.
Hx9dFbQ2zhk-047\|So I have a formal bond order of 2 in oxygen, and that is consistent with the double bond that my Lewis dot structure predicted.
Hx9dFbQ2zhk-048\|But here's something that the Lewis dot structure didn't predict-- oxygen has two unpaired electrons.
Hx9dFbQ2zhk-049\|Oxygen is paramagnetic.
Hx9dFbQ2zhk-051\|And if you do the experiment, you find that oxygen is actually paramagnetic and you'll see that in the demonstration lab.

OCK36oYeAFA-006|Nitrogen in its standard state is N2 gas.

OCK36oYeAFA-007|Oxygen in its standard state is O2 gas.

OCK36oYeAFA-008|So I can write N2 gas plus O2 gas goes to two NO, just balancing that chemical equation.

OCK36oYeAFA-009|Of course, the enthalpy information is for a single mole.

OCK36oYeAFA-010|I've formed two moles here, so this enthalpy is the enthalpy for 1/2 this chemical reaction.

OCK36oYeAFA-011|I've got 2 here, so I'll just double this enthalpy.

OCK36oYeAFA-013|181 kilojoules.

OCK36oYeAFA-014|Now I want to find the enthalpy of formation for N2O5.

OCK36oYeAFA-015|I can write that because it's the enthalpy of formation.

OCK36oYeAFA-018|So I'd like to sum these to form this one.

OCK36oYeAFA-019|And I noticed right off, well, this reaction here, that would give me two moles of N2O5 on the product side.

OCK36oYeAFA-020|I only want one mole, so I'm going to divide this one through by 2.

OCK36oYeAFA-021|If I divide this reaction through by 2, I have to divide the enthalpy by 2.

OCK36oYeAFA-022|So let's do that.

OCK36oYeAFA-023|And then, when we add down, you'll see that these three reactions add to give the reaction we're looking for.

OCK36oYeAFA-024|So it's a Hess's law situation.

OCK36oYeAFA-025|I can add these three.

OCK36oYeAFA-033|It's slightly endothermic.

2hohw6e19_s-000|We can calculate the pH in the buffer region using the Henderson-Hasselbalch expression.

2hohw6e19_s-002|So here, I've expanded the buffer region.

2hohw6e19_s-003|And we know that, at half equivalence, halfway to the equivalence point, the concentrations of the acid and base form are equal.

2hohw6e19_s-004|So this ratio is 1, the log term is 0, and the pH is numerically equal to the pKa at half equivalence.

2hohw6e19_s-007|This term would be negative.

2hohw6e19_s-008|And of course, if the base form predominates, you'd expect this term to be positive.

2hohw6e19_s-010|So the Henderson-Hasselbalch expression allows us to calculate the pH throughout this region.

2hohw6e19_s-013|The buffer region would be from 3.75 to 5.75, one unit around the pKa.

2hohw6e19_s-014|So let's look at that in more detail.

2hohw6e19_s-023|So a buffer resists change in pH from adding acid, from adding base, and from dilution.

-zXxTQt2slA-000|Let's look at hybridization in benzene.

-zXxTQt2slA-001|Benzene, C6H6, I've drawn it like this with this circle representing the alternating double bonds.

-zXxTQt2slA-007|We're talking about the hybridization of carbons in benzene.

-zXxTQt2slA-008|Now, we drew benzene schematically, but that's enough to determine the steric number of each carbon, and that's what you need to determine hybridization.

-zXxTQt2slA-009|Carbon in benzene has to accommodate three things-- two other carbons and a hydrogen.

-zXxTQt2slA-010|So to accommodate three things with steric number 3, I have hybridize three atomic orbitals-- an s and two p's.

-zXxTQt2slA-011|That leads to sp2 hybridization, 3 equivalent orbitals that have a 120 degree bond angle.

-zXxTQt2slA-013|All the carbons are equivalent, so these carbons here will also be sp2 hybridized.

-zXxTQt2slA-016|So the correct answer here-- sp2 hybridisation for the carbons in benzene.

y97cegGFECA-000|When salts dissolve in water there's a wide variation in the solubility.

y97cegGFECA-001|Some salts like sodium chloride dissolve quite well and they form high concentrations of sodium ions and chloride ions in solution.

y97cegGFECA-002|Others like barium sulfate are not very soluble.

y97cegGFECA-004|So a small K means I favor the reactant side, the solid as opposed to the ions.

y97cegGFECA-007|It kind of means that barium and sulfate ions don't like to exist together in solution.

y97cegGFECA-008|The more stable state is the barium sulfate solid.

y97cegGFECA-009|So if barium comes from one source and sulfate from another, those two ions seek each other out.

y97cegGFECA-017|So the K for this reaction, the reverse, is 1 over 10 to the minus 10, or 10 to the plus 10.

y97cegGFECA-018|A large K means I strongly favor the barium sulfate solid.

y97cegGFECA-020|So both these reactions strongly favor the solid and water.

y97cegGFECA-022|So the ion concentration drops if you mix these two solutions.

y97cegGFECA-024|The barium and hydroxide ions would conduct electricity across this gap and allow this light to come on.

y97cegGFECA-025|When there's high ionic concentration, that light will be bright.

y95SN6mDh0Y-000|Now the intensity of light, or electromagnetic radiation, may be more of an intuitive property.

y95SN6mDh0Y-001|Let's look at it this way.

y95SN6mDh0Y-003|So I0 going through a pair of identical filters.

y95SN6mDh0Y-004|Would that reduce the intensity down to a quarter, an eighth, or 0 of the original intensity?

y95SN6mDh0Y-011|So we've been talking about intensity, the brightness of light, in terms of attenuating it, or reducing it, with a filter.

y95SN6mDh0Y-013|Well if the first filter reduces it by one-half, so it's half as bright here.

y95SN6mDh0Y-014|A second filter would make this half as bright.

y95SN6mDh0Y-015|So you'd go from half the original intensity to a quarter of the original intensity after two filters.

aoAdE8L2ASw-000|Let's look at our liquid-to-gas transition for water and ask, what is the standard state free energy difference?

aoAdE8L2ASw-002|What is the standard state free energy difference for this physical change at 100 degrees C?

aoAdE8L2ASw-010|We're talking about the liquid-to-gas phase transition.

aoAdE8L2ASw-011|And at 100 degrees C, what's the standard state free energy difference?

aoAdE8L2ASw-014|So we can talk about those states of matter at different temperatures.

aoAdE8L2ASw-015|And in this case, we're talking about it at 100 degrees C.

aoAdE8L2ASw-017|And I think you could say, well, that's what boiling is.

aoAdE8L2ASw-018|Boiling occurs when there's a vapor pressure of one atmosphere above the liquid.

aoAdE8L2ASw-019|So that is the equilibrium point.

aoAdE8L2ASw-021|So what we have here is, where is the free energy difference between the one atmosphere of gas and liquid exactly 0?

aoAdE8L2ASw-025|If you went to a different temperature and you were talking about the standard state, you would still have one atmosphere of gas.

aoAdE8L2ASw-026|So that's why you say, well, if I go to a lower temperature and I still have one atmosphere of gas, well, then of course, I favor the liquid.

aoAdE8L2ASw-027|That's a higher amount of gas, one atmosphere, than the vapor pressure at these lower temperatures.

aoAdE8L2ASw-028|And as I go to higher temperatures, then I have the one atmosphere of gas is the favored state of the system.

aoAdE8L2ASw-029|So delta G standard is a function of temperature.

aoAdE8L2ASw-030|In this case, I'm right at the equilibrium where the gas and the liquid are equally favored.

aoAdE8L2ASw-031|In this case, the correct answer is B.

Hx9dFbQ2zhk-000|Let's look at both the quantum mechanical description and the Lewis dot structure description of a molecule.

Hx9dFbQ2zhk-001|We'll pick oxygen, O2.

Hx9dFbQ2zhk-002|Oxygen, each atom has 6 valence electrons in principle quantum level 2.

Hx9dFbQ2zhk-004|Each oxygen obeys the octet, it has 2, 4, 6, 8 electrons around it.

Hx9dFbQ2zhk-005|So that's our good Lewis electron dot structure.

Hx9dFbQ2zhk-006|Let's look at the quantum mechanical description.

Hx9dFbQ2zhk-016|So those are the s orbitals, let's take the p orbitals.

Hx9dFbQ2zhk-017|There are 6 p orbitals, 3 on each atom.

Hx9dFbQ2zhk-024|Again, sigma bonding, sigma antibonding, energetically distributed.

Hx9dFbQ2zhk-025|Now, let's look at the px and the py.

Hx9dFbQ2zhk-026|The px and the py lie in front and behind and above and below the internuclear axis.

Hx9dFbQ2zhk-027|And you can see that on a model.

Hx9dFbQ2zhk-030|It doesn't really matter which one I look at, they're both identical in energy.

Hx9dFbQ2zhk-035|And the minus combination gives me antibonding, pi orbitals, two of those.

Hx9dFbQ2zhk-036|One from the minus combination of px and one from the minus combination of py.

Hx9dFbQ2zhk-038|Each oxygen had an s and three p's.

Hx9dFbQ2zhk-039|So 4 orbitals on each oxygen, 8 total, make 8 molecular orbitals.

Hx9dFbQ2zhk-040|I need to fill those with the 12 valence electrons from oxygen.

Hx9dFbQ2zhk-043|So the bond order, I have a sigma bonding and antibonding orbital to start.

Hx9dFbQ2zhk-044|Two electrons in each, those cancel each other out, zero bond order from that.

Hx9dFbQ2zhk-046|So 6 minus 2 is 4 bonding electrons divided by 2-- 2 formal bonds.

Hx9dFbQ2zhk-047|So I have a formal bond order of 2 in oxygen, and that is consistent with the double bond that my Lewis dot structure predicted.

Hx9dFbQ2zhk-048|But here's something that the Lewis dot structure didn't predict-- oxygen has two unpaired electrons.

Hx9dFbQ2zhk-049|Oxygen is paramagnetic.

Hx9dFbQ2zhk-051|And if you do the experiment, you find that oxygen is actually paramagnetic and you'll see that in the demonstration lab.