ceaglex/VisualTTS_Chem · Datasets at Hugging Face

text stringlengths 19 206
NLGrUdD5uDw-010\|The atomic orbitals forming bond angle 180 can't quite do that.
NLGrUdD5uDw-011\|So forming acetylene isn't really possible.
NLGrUdD5uDw-012\|And it gets a little more complicated when I go to steric number 3.
NLGrUdD5uDw-013\|So if I have a bond angle of 120 degrees, there aren't even atomic orbitals that have 120 degree orientation to each other.
NLGrUdD5uDw-017\|So what I need are a set of atomic orbitals where I can combine them and get the various angles for the various steric numbers.
NLGrUdD5uDw-018\|And that's what we'll look at next.
aLwZdRbk8_4-000\|Let's look at the molecular orbital description of several atoms going across the periodic table.
aLwZdRbk8_4-001\|We'll go from boron to neon and form the diatomic molecules, and look at how they look in a molecular orbital description.
aLwZdRbk8_4-002\|When you start with boron, WE'LL take boron and we'll take the p orbital contributions only to the molecular orbitals.
aLwZdRbk8_4-004\|So those don't contribute to the bond order.
aLwZdRbk8_4-005\|Sigma s and sigma s star cancel each other out.
aLwZdRbk8_4-006\|So we'll just look at the p contributions to the molecular orbitals.
aLwZdRbk8_4-008\|It has a 2p electron, we can find that 2p electron, one from each boron and put it into the molecular orbitals.
aLwZdRbk8_4-009\|So two electrons in the molecular orbitals.
aLwZdRbk8_4-010\|Here's boron, the lowest energy molecular orbitals are the pi bonding orbitals.
aLwZdRbk8_4-012\|Two bonding electrons, no anti-bonding electrons, so 2 divided by 2 is 1.
aLwZdRbk8_4-013\|We can do the same thing with carbon.
aLwZdRbk8_4-015\|So I have 1p electron, or 2p electrons from each carbon for a total of four going into my molecular orbitals.
aLwZdRbk8_4-016\|And when I put in four electrons to my molecular orbitals I get a diamagnetic carbon molecule with a double bond.
aLwZdRbk8_4-017\|Four bonding electrons divided by 2, bond order two.
aLwZdRbk8_4-019\|Oxygen we would predict would be paramagnetic from this.
aLwZdRbk8_4-020\|Now as we go across the periodic table, the lower energy orbitals are going to switch here.
aLwZdRbk8_4-021\|The sigma bonding orbital becomes the lower energy as I go across the periodic table.
aLwZdRbk8_4-022\|And you might predict that these molecular orbitals are formed from linear combinations of atomic orbitals.
aLwZdRbk8_4-023\|And you have to do the mathematical calculation and calculate the energies, and you know the orbitals change as you go across the periodic table.
aLwZdRbk8_4-024\|Atomic radii decrease.
aLwZdRbk8_4-025\|So the character of the individual atomic orbital changes, and the character of the molecular orbitals that result changes.
aLwZdRbk8_4-026\|When we give these orbital configurations, and it's necessary for you to know them, we'll make it very clear which orbital designation you should use.
aLwZdRbk8_4-027\|So oxygen has two unpaired electrons, oxygen is paramagnetic.
aLwZdRbk8_4-028\|We can go ahead to fluorine for the Lewis electron dot structure for fluorine, we'd predict a single bond.
aLwZdRbk8_4-029\|Does the molecular orbital theory predict that?
aLwZdRbk8_4-033\|So what we have is our molecular orbital theory predicting neon would not form a bond.
aLwZdRbk8_4-034\|It would have a bond order zero.
wEwYuEk6dzs-000\|Let's do a calculation involving both the ideal gas law and the van der Waals expression for gases and compare the two.
wEwYuEk6dzs-004\|Well these last two we can answer right away.
wEwYuEk6dzs-005\|Which guess has more collisions per second with the walls?
wEwYuEk6dzs-006\|Well that's the gas that's moving at the greater average velocity.
wEwYuEk6dzs-007\|And the greater average velocity goes as the temperature-- but they're both at the same temperature-- or the mass.
wEwYuEk6dzs-008\|The lower the mass, the higher the average velocity.
wEwYuEk6dzs-009\|So the one with the higher rms velocity, in this case, it's helium.
wEwYuEk6dzs-010\|So helium will have more collisions with the wall.
wEwYuEk6dzs-011\|Remember, they both can have the same pressure.
wEwYuEk6dzs-012\|Just because helium is hitting the wall more often to impart its momentum, carbon dioxide hits with a bigger punch.
wEwYuEk6dzs-013\|So they both exert the same pressure on the wall.
wEwYuEk6dzs-014\|Now, what about the greater attractions?
wEwYuEk6dzs-015\|Carbon dioxide will have the greater intermolecular attractions.
wEwYuEk6dzs-016\|And we would see that reflected in the size of the A parameter in the van der Waals expression.
wEwYuEk6dzs-017\|So we can go directly to that.
wEwYuEk6dzs-019\|Now, they could be mixed or they could be in separate flasks, it doesn't matter.
wEwYuEk6dzs-020\|If they're mixed, we'll calculate a partial pressure for each.
wEwYuEk6dzs-021\|If they're separate, the partial pressure and the total pressure will be the same.
wEwYuEk6dzs-022\|If they're mixed, the total pressure will be the sum of both pressures.
wEwYuEk6dzs-023\|But either way, we can calculate the pressure for helium and carbon dioxide.
wEwYuEk6dzs-024\|So first we'll do it for a real gas situation.
wEwYuEk6dzs-025\|So we'll assume they behave like ideal gases.
wEwYuEk6dzs-030\|The ideal gas law doesn't include the nature of the particle.
wEwYuEk6dzs-031\|So both gases would have a pressure 1.63 atmospheres at 15 liters.
wEwYuEk6dzs-033\|So the ideal gas calculation is easy because we don't have to include which gas we're talking about.
RI6HdlWwDtw-000\|Let's do a calculation where we look at the standard state free energy difference for the combustion of glucose.
RI6HdlWwDtw-001\|And we'll also look at what temperature range is that reaction spontaneous.
RI6HdlWwDtw-002\|Now, it's the standard state free energy difference.
RI6HdlWwDtw-004\|How can I calculate that?
RI6HdlWwDtw-006\|So here it is.
RI6HdlWwDtw-013\|What temperature range is this reaction spontaneous for?
RI6HdlWwDtw-014\|Well, here's the chemical reaction in the free energy difference.
RI6HdlWwDtw-015\|Is it spontaneous over a broad temperature range?
RI6HdlWwDtw-016\|In order to do that, we need to know the enthalpy and the entropy.
RI6HdlWwDtw-017\|Those are relatively independent of temperature.
RI6HdlWwDtw-018\|And they tell us how delta G varies with temperature.
RI6HdlWwDtw-019\|So delta S for this reaction-- I think we can intuitively say, without doing a calculation-- delta S is greater than zero.
RI6HdlWwDtw-020\|We're producing a liquid and a gas from a solid and this gas.
RI6HdlWwDtw-022\|The enthalpy-- you may know.
RI6HdlWwDtw-023\|You've probably seen this chemical reaction, glucose burning.
RI6HdlWwDtw-024\|It is exothermic.
RI6HdlWwDtw-025\|But we don't have to use intuition for either one.
RI6HdlWwDtw-033\|As temperature increases, delta S is positive.
RI6HdlWwDtw-034\|That means the slope is negative.
njRitLAU3Ik-007\|Now, four electrons from the o orbitals will fill these molecular orbitals.
njRitLAU3Ik-008\|So we'll put those in-- one, two, three, four.
njRitLAU3Ik-010\|The Lowest Unoccupied Molecular Orbital or LUMO is a pi-star or antibonding orbital.
njRitLAU3Ik-011\|And it's common the transitions in these conjugated systems are a pi to pi-star transition.
njRitLAU3Ik-016\|They go from a pi to pi-star.
njRitLAU3Ik-017\|So the bond order here is reduced.
njRitLAU3Ik-019\|Those are two names for the same compound.
njRitLAU3Ik-020\|This compound that's in your eye acts as a light receptor.
njRitLAU3Ik-021\|And when this transition occurs, this isomerism, that triggers a chain of chemical reactions that allow you to detect light in your eye.
njRitLAU3Ik-022\|This is actually the light recepting molecule in your eye.
njRitLAU3Ik-023\|So a pi to pi-star transition, very important in human vision.
9z0sMWVIm7A-000\|Let's balance a redox reaction.
9z0sMWVIm7A-001\|Here's a redox reaction here.
9z0sMWVIm7A-002\|And when you balance redox reactions, you should follow a series of steps, and they're outlined in many textbooks.
9z0sMWVIm7A-003\|I'll go through them right now.
9z0sMWVIm7A-007\|Well, let's go through and see which is which.
9z0sMWVIm7A-008\|So here's chromium.
9z0sMWVIm7A-010\|And that turns out to be plus 6.
9z0sMWVIm7A-011\|And the oxidation number here is obviously plus 3, so chromium is being reduced.
9z0sMWVIm7A-012\|It's oxidation number is reduced.
9z0sMWVIm7A-013\|So hopefully that means an oxidation is occurring over here.
9z0sMWVIm7A-014\|And if we analyze this using the same rules, oxygen is always minus 2, hydrogen is always a plus 1.
9z0sMWVIm7A-019\|So the next step, balance the elements in your two skeletal half cells.

NLGrUdD5uDw-010|The atomic orbitals forming bond angle 180 can't quite do that.

NLGrUdD5uDw-011|So forming acetylene isn't really possible.

NLGrUdD5uDw-012|And it gets a little more complicated when I go to steric number 3.

NLGrUdD5uDw-013|So if I have a bond angle of 120 degrees, there aren't even atomic orbitals that have 120 degree orientation to each other.

NLGrUdD5uDw-017|So what I need are a set of atomic orbitals where I can combine them and get the various angles for the various steric numbers.

NLGrUdD5uDw-018|And that's what we'll look at next.

aLwZdRbk8_4-000|Let's look at the molecular orbital description of several atoms going across the periodic table.

aLwZdRbk8_4-001|We'll go from boron to neon and form the diatomic molecules, and look at how they look in a molecular orbital description.

aLwZdRbk8_4-002|When you start with boron, WE'LL take boron and we'll take the p orbital contributions only to the molecular orbitals.

aLwZdRbk8_4-004|So those don't contribute to the bond order.

aLwZdRbk8_4-005|Sigma s and sigma s star cancel each other out.

aLwZdRbk8_4-006|So we'll just look at the p contributions to the molecular orbitals.

aLwZdRbk8_4-008|It has a 2p electron, we can find that 2p electron, one from each boron and put it into the molecular orbitals.

aLwZdRbk8_4-009|So two electrons in the molecular orbitals.

aLwZdRbk8_4-010|Here's boron, the lowest energy molecular orbitals are the pi bonding orbitals.

aLwZdRbk8_4-012|Two bonding electrons, no anti-bonding electrons, so 2 divided by 2 is 1.

aLwZdRbk8_4-013|We can do the same thing with carbon.

aLwZdRbk8_4-015|So I have 1p electron, or 2p electrons from each carbon for a total of four going into my molecular orbitals.

aLwZdRbk8_4-016|And when I put in four electrons to my molecular orbitals I get a diamagnetic carbon molecule with a double bond.

aLwZdRbk8_4-017|Four bonding electrons divided by 2, bond order two.

aLwZdRbk8_4-019|Oxygen we would predict would be paramagnetic from this.

aLwZdRbk8_4-020|Now as we go across the periodic table, the lower energy orbitals are going to switch here.

aLwZdRbk8_4-021|The sigma bonding orbital becomes the lower energy as I go across the periodic table.

aLwZdRbk8_4-022|And you might predict that these molecular orbitals are formed from linear combinations of atomic orbitals.

aLwZdRbk8_4-023|And you have to do the mathematical calculation and calculate the energies, and you know the orbitals change as you go across the periodic table.

aLwZdRbk8_4-024|Atomic radii decrease.

aLwZdRbk8_4-025|So the character of the individual atomic orbital changes, and the character of the molecular orbitals that result changes.

aLwZdRbk8_4-026|When we give these orbital configurations, and it's necessary for you to know them, we'll make it very clear which orbital designation you should use.

aLwZdRbk8_4-027|So oxygen has two unpaired electrons, oxygen is paramagnetic.

aLwZdRbk8_4-028|We can go ahead to fluorine for the Lewis electron dot structure for fluorine, we'd predict a single bond.

aLwZdRbk8_4-029|Does the molecular orbital theory predict that?

aLwZdRbk8_4-033|So what we have is our molecular orbital theory predicting neon would not form a bond.

aLwZdRbk8_4-034|It would have a bond order zero.

wEwYuEk6dzs-000|Let's do a calculation involving both the ideal gas law and the van der Waals expression for gases and compare the two.

wEwYuEk6dzs-004|Well these last two we can answer right away.

wEwYuEk6dzs-005|Which guess has more collisions per second with the walls?

wEwYuEk6dzs-006|Well that's the gas that's moving at the greater average velocity.

wEwYuEk6dzs-007|And the greater average velocity goes as the temperature-- but they're both at the same temperature-- or the mass.

wEwYuEk6dzs-008|The lower the mass, the higher the average velocity.

wEwYuEk6dzs-009|So the one with the higher rms velocity, in this case, it's helium.

wEwYuEk6dzs-010|So helium will have more collisions with the wall.

wEwYuEk6dzs-011|Remember, they both can have the same pressure.

wEwYuEk6dzs-012|Just because helium is hitting the wall more often to impart its momentum, carbon dioxide hits with a bigger punch.

wEwYuEk6dzs-013|So they both exert the same pressure on the wall.

wEwYuEk6dzs-014|Now, what about the greater attractions?

wEwYuEk6dzs-015|Carbon dioxide will have the greater intermolecular attractions.

wEwYuEk6dzs-016|And we would see that reflected in the size of the A parameter in the van der Waals expression.

wEwYuEk6dzs-017|So we can go directly to that.

wEwYuEk6dzs-019|Now, they could be mixed or they could be in separate flasks, it doesn't matter.

wEwYuEk6dzs-020|If they're mixed, we'll calculate a partial pressure for each.

wEwYuEk6dzs-021|If they're separate, the partial pressure and the total pressure will be the same.

wEwYuEk6dzs-022|If they're mixed, the total pressure will be the sum of both pressures.

wEwYuEk6dzs-023|But either way, we can calculate the pressure for helium and carbon dioxide.

wEwYuEk6dzs-024|So first we'll do it for a real gas situation.

wEwYuEk6dzs-025|So we'll assume they behave like ideal gases.

wEwYuEk6dzs-030|The ideal gas law doesn't include the nature of the particle.

wEwYuEk6dzs-031|So both gases would have a pressure 1.63 atmospheres at 15 liters.

wEwYuEk6dzs-033|So the ideal gas calculation is easy because we don't have to include which gas we're talking about.

RI6HdlWwDtw-000|Let's do a calculation where we look at the standard state free energy difference for the combustion of glucose.

RI6HdlWwDtw-001|And we'll also look at what temperature range is that reaction spontaneous.

RI6HdlWwDtw-002|Now, it's the standard state free energy difference.

RI6HdlWwDtw-004|How can I calculate that?

RI6HdlWwDtw-006|So here it is.

RI6HdlWwDtw-013|What temperature range is this reaction spontaneous for?

RI6HdlWwDtw-014|Well, here's the chemical reaction in the free energy difference.

RI6HdlWwDtw-015|Is it spontaneous over a broad temperature range?

RI6HdlWwDtw-016|In order to do that, we need to know the enthalpy and the entropy.

RI6HdlWwDtw-017|Those are relatively independent of temperature.

RI6HdlWwDtw-018|And they tell us how delta G varies with temperature.

RI6HdlWwDtw-019|So delta S for this reaction-- I think we can intuitively say, without doing a calculation-- delta S is greater than zero.

RI6HdlWwDtw-020|We're producing a liquid and a gas from a solid and this gas.

RI6HdlWwDtw-022|The enthalpy-- you may know.

RI6HdlWwDtw-023|You've probably seen this chemical reaction, glucose burning.

RI6HdlWwDtw-024|It is exothermic.

RI6HdlWwDtw-025|But we don't have to use intuition for either one.

RI6HdlWwDtw-033|As temperature increases, delta S is positive.

RI6HdlWwDtw-034|That means the slope is negative.

njRitLAU3Ik-007|Now, four electrons from the o orbitals will fill these molecular orbitals.

njRitLAU3Ik-008|So we'll put those in-- one, two, three, four.

njRitLAU3Ik-010|The Lowest Unoccupied Molecular Orbital or LUMO is a pi-star or antibonding orbital.

njRitLAU3Ik-011|And it's common the transitions in these conjugated systems are a pi to pi-star transition.

njRitLAU3Ik-016|They go from a pi to pi-star.

njRitLAU3Ik-017|So the bond order here is reduced.

njRitLAU3Ik-019|Those are two names for the same compound.

njRitLAU3Ik-020|This compound that's in your eye acts as a light receptor.

njRitLAU3Ik-021|And when this transition occurs, this isomerism, that triggers a chain of chemical reactions that allow you to detect light in your eye.

njRitLAU3Ik-022|This is actually the light recepting molecule in your eye.

njRitLAU3Ik-023|So a pi to pi-star transition, very important in human vision.

9z0sMWVIm7A-000|Let's balance a redox reaction.

9z0sMWVIm7A-001|Here's a redox reaction here.

9z0sMWVIm7A-002|And when you balance redox reactions, you should follow a series of steps, and they're outlined in many textbooks.

9z0sMWVIm7A-003|I'll go through them right now.

9z0sMWVIm7A-007|Well, let's go through and see which is which.

9z0sMWVIm7A-008|So here's chromium.

9z0sMWVIm7A-010|And that turns out to be plus 6.

9z0sMWVIm7A-011|And the oxidation number here is obviously plus 3, so chromium is being reduced.

9z0sMWVIm7A-012|It's oxidation number is reduced.

9z0sMWVIm7A-013|So hopefully that means an oxidation is occurring over here.

9z0sMWVIm7A-014|And if we analyze this using the same rules, oxygen is always minus 2, hydrogen is always a plus 1.

9z0sMWVIm7A-019|So the next step, balance the elements in your two skeletal half cells.