ceaglex/VisualTTS_Chem · Datasets at Hugging Face

text stringlengths 19 206
Vzy2wd5oIKs-018\|The enthalpy is the energy change plus the pressure volume constant.
Vzy2wd5oIKs-022\|Again, for an ideal gas, the enthalpy changes will be directly proportional to temperature changes.
Vzy2wd5oIKs-025\|And that's proportional to an energy change for an ideal gas as well-- constant volume, constant pressure, energy changes.
kTJEs_u1fkQ-000\|Let's look at the relationship between the macroscopic properties for a sample of gas, the pressure, the volume, the temperature, the number of moles of gas.
kTJEs_u1fkQ-012\|This is the ideal gas law, the macroscopic properties of a gas, in one neat equation.
RffOkFeAkEE-001\|What's the entropy change?
RffOkFeAkEE-002\|Well, we know there's various ways to calculate entropy changes.
RffOkFeAkEE-011\|So I choose the middle equation here, the reversible heat over the temperature change.
RffOkFeAkEE-012\|So let's just do that.
RffOkFeAkEE-016\|So freezing requires a release of heat, so freezing is endothermic.
RffOkFeAkEE-018\|This is a positive heat.
RffOkFeAkEE-019\|And I find that that's 22 joules per Kelvin mole of water.
RffOkFeAkEE-020\|But we're not talking about a mole of water.
RffOkFeAkEE-021\|We're talking about just 10 grams of water.
uwL_j9yXkN8-000\|Let's look at protein synthesis and how it's coupled to the oxidation of glucose.
uwL_j9yXkN8-001\|So how many grams of glucose are required to make a mole of serum albumin?
uwL_j9yXkN8-002\|Now, serum albumin is around 100 amino acids.
uwL_j9yXkN8-003\|So I need about 100 peptide bonds per mole of serum albumin.
uwL_j9yXkN8-012\|We're talking about forming peptide bonds as we oxidize glucose.
uwL_j9yXkN8-014\|That's transferred via the conversion of ADP to ATP to the formation of peptide bonds.
uwL_j9yXkN8-015\|And you get about four moles of peptide bonds for every mole of glucose that you metabolize.
FmTLXnrPBPQ-000\|In oxidation reduction chemistry, oxidations and reductions always occur in pairs.
FmTLXnrPBPQ-001\|If you are going to be oxidized, that is you're going to give up some electrons, those electrons have to have a place to land.
FmTLXnrPBPQ-002\|They have to reduce some other compound.
FmTLXnrPBPQ-003\|So, let's look at a redox reaction and break it up into half cells, and determine the oxidation and reduction pairs.
FmTLXnrPBPQ-004\|And use our tables of standard reduction potentials to help us balance that reaction.
FmTLXnrPBPQ-007\|So, let's look at those two half cells.
FmTLXnrPBPQ-008\|We can go back to our table.
FmTLXnrPBPQ-009\|Here's our table of standard reduction potentials.
FmTLXnrPBPQ-014\|Here's the permanganate, and here's the bromine.
FmTLXnrPBPQ-018\|So we can balance the number of electrons, whenever oxidation and reduction occurs.
FmTLXnrPBPQ-019\|Of course, the number of electrons are conserved.
FmTLXnrPBPQ-020\|If I reduce you by two electrons, I must accept those two electrons.
FmTLXnrPBPQ-027\|Now, we can also calculate the relative cell potential for that.
FmTLXnrPBPQ-028\|We can go back to our table.
FmTLXnrPBPQ-032\|So, that overall cell potential, 0.43.
FmTLXnrPBPQ-033\|Now, notice again I didn't multiply my cell potentials, my standard reduction potentials, by two and five.
FmTLXnrPBPQ-034\|That's because cell potential is an intensive property, it's independent of the extent of the system.
FmTLXnrPBPQ-036\|That's downhill in free energy.
WvG46BwP-XY-001\|So if we take these three atomic orbitals, we can make three new atomic orbitals.
WvG46BwP-XY-004\|Now, I've drawn them off-center here.
WvG46BwP-XY-005\|So obviously, they're all about the same nucleus.
WvG46BwP-XY-006\|So let's collapse those together on the same nucleus.
WvG46BwP-XY-007\|And that will obscure the little negative lobes.
WvG46BwP-XY-008\|But the three positive lobes are appropriate for steric number three and a bond angle of 120 degrees.
WvG46BwP-XY-009\|Let's say we want to accommodate four, steric number four, or tetrahedral symmetry.
WvG46BwP-XY-010\|I think you can guess the pattern here.
WvG46BwP-XY-011\|It's actually possible to take the s and all three p orbitals.
WvG46BwP-XY-012\|Those four orbitals make four linear combinations and make four new atomic orbitals.
WvG46BwP-XY-015\|109 degrees, I'll bring them all to the same center here four sp3 orbitals.
WvG46BwP-XY-016\|Again, the negative lobes obscured.
WvG46BwP-XY-017\|But they'll accommodate a steric number of four and a bond angle of 109 degrees.
zsB5ZigPQHY-000\|Let's look at a situation where we're bracketing an oxidation-reduction reaction.
zsB5ZigPQHY-001\|So which of the following will be reduced by copper but not by bromide?
zsB5ZigPQHY-009\|We're looking for a compound that can oxidize copper metal, but not bromide.
zsB5ZigPQHY-010\|So if we look at our table of standard production potentials, we see nitrate is above the copper half cell.
zsB5ZigPQHY-011\|That means nitrate can force the oxidation of copper, but not the oxidation of bromide.
Sh42Wpvs42k-000\|Light has both the properties of a wave and the properties of a particle.
Sh42Wpvs42k-001\|It's a wave and a particle at the same time.
Sh42Wpvs42k-002\|There's a duality to the wave particle relationship.
Sh42Wpvs42k-004\|Particle properties that are moving, the most important property is the momentum, it's mass times its velocity.
Sh42Wpvs42k-005\|So how do we reconcile the two?
Sh42Wpvs42k-006\|Well light, photons of light, carry no mass.
Sh42Wpvs42k-007\|So how can they carry momentum?
Sh42Wpvs42k-012\|So using those two relationships we can derive that the momentum is Planck's constant divided by the wavelength.
Sh42Wpvs42k-016\|Wave, particle, acting together, sometimes the properties of the particle exert themselves.
Sh42Wpvs42k-017\|Sometimes the properties of the wave exert themselves.
Sh42Wpvs42k-018\|They both exist together all the time, particle and wave, nature of light.
DP4HnlNRR_4-000\|Nitrogen molecules, N2, have a triple bond.
DP4HnlNRR_4-001\|That's predicted both by molecular orbital theory and the Lewis dot structure.
DP4HnlNRR_4-002\|What I'd like to think about is the ionization N2 molecules-- that is, withdrawing an electron-- and adding an electron, electron capture by N2.
DP4HnlNRR_4-003\|That will form a positive and a negative ion.
DP4HnlNRR_4-004\|The question I have is, what's the relative bond strength?
DP4HnlNRR_4-010\|We're talking about ionization and electron capture in N2 molecules.
DP4HnlNRR_4-012\|And I filled them with the p electrons from nitrogen. Each nitrogen has 3 p electrons for a total of 6 in the molecule.
DP4HnlNRR_4-016\|Adding an electron to the next available orbital is an antibonding orbital.
DP4HnlNRR_4-017\|So electron capture into this antibonding orbital is going to reduce the bond order and the relative bond strength.
DP4HnlNRR_4-019\|So the bond order reduces to 2 and 1/2.
DP4HnlNRR_4-022\|And if you calculate the bond order here, again, 2 and 1/2.
DP4HnlNRR_4-023\|But interestingly, both electron capture and ionization reduce the bond order of nitrogen by about the same amount.
DP4HnlNRR_4-024\|So you'd predict that the bond strengths of N2+ and N2- are about the same.
s_xlDaR53v0-000\|Let's look at a broad range of temperatures on the absolute scale.
s_xlDaR53v0-001\|Here I've listed several.
s_xlDaR53v0-002\|The interior of the Sun for instance-- 10 to the 8th Kelvin, a very high temperature.
s_xlDaR53v0-003\|As we go down in temperature, we can go past the surface of the Sun, tens of thousands of degrees.
s_xlDaR53v0-004\|The melting point of tungsten-- about 1,000 degrees Kelvin.
s_xlDaR53v0-011\|That's a very ideal gas.
s_xlDaR53v0-013\|So that's a gas that behaves ideally over a broad range of temperatures.
s_xlDaR53v0-016\|If you slow down particles, you reduce their temperature.
s_xlDaR53v0-017\|We'll always associate the speed of molecules in a sample with their temperature.
s_xlDaR53v0-018\|Higher temperature-- more speed.
s_xlDaR53v0-019\|Some of the coldest temperatures ever achieved are in a Bose-Einstein condensation.
s_xlDaR53v0-020\|That's where you do laser cooling and magnetic trapping, and get molecules to virtually stop.
inLhIqLOouY-000\|When an acid reacts with water, it forms its conjugate base.
inLhIqLOouY-001\|Let's look at that.
inLhIqLOouY-010\|Again, products over reactants, water doesn't appear, pure water, in equilibrium constant expressions.
inLhIqLOouY-011\|Now, I know the stronger the conjugate acid, the weaker the conjugate base intuitively.
inLhIqLOouY-012\|The stronger this acid is, the more the equilibrium lies towards Ac-.
inLhIqLOouY-013\|So the more Ac- is likely to be in solution, here, the Ac- is favored.
inLhIqLOouY-014\|Here the Ac- is favored.

Vzy2wd5oIKs-018|The enthalpy is the energy change plus the pressure volume constant.

Vzy2wd5oIKs-022|Again, for an ideal gas, the enthalpy changes will be directly proportional to temperature changes.

Vzy2wd5oIKs-025|And that's proportional to an energy change for an ideal gas as well-- constant volume, constant pressure, energy changes.

kTJEs_u1fkQ-000|Let's look at the relationship between the macroscopic properties for a sample of gas, the pressure, the volume, the temperature, the number of moles of gas.

kTJEs_u1fkQ-012|This is the ideal gas law, the macroscopic properties of a gas, in one neat equation.

RffOkFeAkEE-001|What's the entropy change?

RffOkFeAkEE-002|Well, we know there's various ways to calculate entropy changes.

RffOkFeAkEE-011|So I choose the middle equation here, the reversible heat over the temperature change.

RffOkFeAkEE-012|So let's just do that.

RffOkFeAkEE-016|So freezing requires a release of heat, so freezing is endothermic.

RffOkFeAkEE-018|This is a positive heat.

RffOkFeAkEE-019|And I find that that's 22 joules per Kelvin mole of water.

RffOkFeAkEE-020|But we're not talking about a mole of water.

RffOkFeAkEE-021|We're talking about just 10 grams of water.

uwL_j9yXkN8-000|Let's look at protein synthesis and how it's coupled to the oxidation of glucose.

uwL_j9yXkN8-001|So how many grams of glucose are required to make a mole of serum albumin?

uwL_j9yXkN8-002|Now, serum albumin is around 100 amino acids.

uwL_j9yXkN8-003|So I need about 100 peptide bonds per mole of serum albumin.

uwL_j9yXkN8-012|We're talking about forming peptide bonds as we oxidize glucose.

uwL_j9yXkN8-014|That's transferred via the conversion of ADP to ATP to the formation of peptide bonds.

uwL_j9yXkN8-015|And you get about four moles of peptide bonds for every mole of glucose that you metabolize.

FmTLXnrPBPQ-000|In oxidation reduction chemistry, oxidations and reductions always occur in pairs.

FmTLXnrPBPQ-001|If you are going to be oxidized, that is you're going to give up some electrons, those electrons have to have a place to land.

FmTLXnrPBPQ-002|They have to reduce some other compound.

FmTLXnrPBPQ-003|So, let's look at a redox reaction and break it up into half cells, and determine the oxidation and reduction pairs.

FmTLXnrPBPQ-004|And use our tables of standard reduction potentials to help us balance that reaction.

FmTLXnrPBPQ-007|So, let's look at those two half cells.

FmTLXnrPBPQ-008|We can go back to our table.

FmTLXnrPBPQ-009|Here's our table of standard reduction potentials.

FmTLXnrPBPQ-014|Here's the permanganate, and here's the bromine.

FmTLXnrPBPQ-018|So we can balance the number of electrons, whenever oxidation and reduction occurs.

FmTLXnrPBPQ-019|Of course, the number of electrons are conserved.

FmTLXnrPBPQ-020|If I reduce you by two electrons, I must accept those two electrons.

FmTLXnrPBPQ-027|Now, we can also calculate the relative cell potential for that.

FmTLXnrPBPQ-028|We can go back to our table.

FmTLXnrPBPQ-032|So, that overall cell potential, 0.43.

FmTLXnrPBPQ-033|Now, notice again I didn't multiply my cell potentials, my standard reduction potentials, by two and five.

FmTLXnrPBPQ-034|That's because cell potential is an intensive property, it's independent of the extent of the system.

FmTLXnrPBPQ-036|That's downhill in free energy.

WvG46BwP-XY-001|So if we take these three atomic orbitals, we can make three new atomic orbitals.

WvG46BwP-XY-004|Now, I've drawn them off-center here.

WvG46BwP-XY-005|So obviously, they're all about the same nucleus.

WvG46BwP-XY-006|So let's collapse those together on the same nucleus.

WvG46BwP-XY-007|And that will obscure the little negative lobes.

WvG46BwP-XY-008|But the three positive lobes are appropriate for steric number three and a bond angle of 120 degrees.

WvG46BwP-XY-009|Let's say we want to accommodate four, steric number four, or tetrahedral symmetry.

WvG46BwP-XY-010|I think you can guess the pattern here.

WvG46BwP-XY-011|It's actually possible to take the s and all three p orbitals.

WvG46BwP-XY-012|Those four orbitals make four linear combinations and make four new atomic orbitals.

WvG46BwP-XY-015|109 degrees, I'll bring them all to the same center here four sp3 orbitals.

WvG46BwP-XY-016|Again, the negative lobes obscured.

WvG46BwP-XY-017|But they'll accommodate a steric number of four and a bond angle of 109 degrees.

zsB5ZigPQHY-000|Let's look at a situation where we're bracketing an oxidation-reduction reaction.

zsB5ZigPQHY-001|So which of the following will be reduced by copper but not by bromide?

zsB5ZigPQHY-009|We're looking for a compound that can oxidize copper metal, but not bromide.

zsB5ZigPQHY-010|So if we look at our table of standard production potentials, we see nitrate is above the copper half cell.

zsB5ZigPQHY-011|That means nitrate can force the oxidation of copper, but not the oxidation of bromide.

Sh42Wpvs42k-000|Light has both the properties of a wave and the properties of a particle.

Sh42Wpvs42k-001|It's a wave and a particle at the same time.

Sh42Wpvs42k-002|There's a duality to the wave particle relationship.

Sh42Wpvs42k-004|Particle properties that are moving, the most important property is the momentum, it's mass times its velocity.

Sh42Wpvs42k-005|So how do we reconcile the two?

Sh42Wpvs42k-006|Well light, photons of light, carry no mass.

Sh42Wpvs42k-007|So how can they carry momentum?

Sh42Wpvs42k-012|So using those two relationships we can derive that the momentum is Planck's constant divided by the wavelength.

Sh42Wpvs42k-016|Wave, particle, acting together, sometimes the properties of the particle exert themselves.

Sh42Wpvs42k-017|Sometimes the properties of the wave exert themselves.

Sh42Wpvs42k-018|They both exist together all the time, particle and wave, nature of light.

DP4HnlNRR_4-000|Nitrogen molecules, N2, have a triple bond.

DP4HnlNRR_4-001|That's predicted both by molecular orbital theory and the Lewis dot structure.

DP4HnlNRR_4-002|What I'd like to think about is the ionization N2 molecules-- that is, withdrawing an electron-- and adding an electron, electron capture by N2.

DP4HnlNRR_4-003|That will form a positive and a negative ion.

DP4HnlNRR_4-004|The question I have is, what's the relative bond strength?

DP4HnlNRR_4-010|We're talking about ionization and electron capture in N2 molecules.

DP4HnlNRR_4-012|And I filled them with the p electrons from nitrogen. Each nitrogen has 3 p electrons for a total of 6 in the molecule.

DP4HnlNRR_4-016|Adding an electron to the next available orbital is an antibonding orbital.

DP4HnlNRR_4-017|So electron capture into this antibonding orbital is going to reduce the bond order and the relative bond strength.

DP4HnlNRR_4-019|So the bond order reduces to 2 and 1/2.

DP4HnlNRR_4-022|And if you calculate the bond order here, again, 2 and 1/2.

DP4HnlNRR_4-023|But interestingly, both electron capture and ionization reduce the bond order of nitrogen by about the same amount.

DP4HnlNRR_4-024|So you'd predict that the bond strengths of N2+ and N2- are about the same.

s_xlDaR53v0-000|Let's look at a broad range of temperatures on the absolute scale.

s_xlDaR53v0-001|Here I've listed several.

s_xlDaR53v0-002|The interior of the Sun for instance-- 10 to the 8th Kelvin, a very high temperature.

s_xlDaR53v0-003|As we go down in temperature, we can go past the surface of the Sun, tens of thousands of degrees.

s_xlDaR53v0-004|The melting point of tungsten-- about 1,000 degrees Kelvin.

s_xlDaR53v0-011|That's a very ideal gas.

s_xlDaR53v0-013|So that's a gas that behaves ideally over a broad range of temperatures.

s_xlDaR53v0-016|If you slow down particles, you reduce their temperature.

s_xlDaR53v0-017|We'll always associate the speed of molecules in a sample with their temperature.

s_xlDaR53v0-018|Higher temperature-- more speed.

s_xlDaR53v0-019|Some of the coldest temperatures ever achieved are in a Bose-Einstein condensation.

s_xlDaR53v0-020|That's where you do laser cooling and magnetic trapping, and get molecules to virtually stop.

inLhIqLOouY-000|When an acid reacts with water, it forms its conjugate base.

inLhIqLOouY-001|Let's look at that.

inLhIqLOouY-010|Again, products over reactants, water doesn't appear, pure water, in equilibrium constant expressions.

inLhIqLOouY-011|Now, I know the stronger the conjugate acid, the weaker the conjugate base intuitively.

inLhIqLOouY-012|The stronger this acid is, the more the equilibrium lies towards Ac-.

inLhIqLOouY-013|So the more Ac- is likely to be in solution, here, the Ac- is favored.

inLhIqLOouY-014|Here the Ac- is favored.