ceaglex/VisualTTS_Chem · Datasets at Hugging Face

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eoFw8xEy-m4-003\|So we're up to n equal 2, and one possible value for n equal 2 is l equal 1.
eoFw8xEy-m4-007\|So the value of l equal 1, we give the letter designation p.
eoFw8xEy-m4-008\|So these are going to be p orbitals.
eoFw8xEy-m4-009\|How about the values of m sub l?
eoFw8xEy-m4-014\|An angular node is a node that has a planar shape rather than a spherical shape for the radial nodes.
eoFw8xEy-m4-017\|The square of the wave function is large around the x-axis for the x.
eoFw8xEy-m4-019\|That helps you find the angular node.
eoFw8xEy-m4-020\|Now, there's one other value of m sub l, equal 0.
eoFw8xEy-m4-021\|And that will give you the pz orbital, where the square of the wave function has its maximum along the z-axis.
eoFw8xEy-m4-024\|And the m sub l equals 0 already neatly lies along the z-axis.
eoFw8xEy-m4-027\|And we'll call them the 2px, 2py, and 2pz orbitals.
eoFw8xEy-m4-029\|These are the p orbitals.
eoFw8xEy-m4-030\|Now, we can continue.
eoFw8xEy-m4-032\|It'll have two nodes now, two radial nodes, because the total number of nodes is n minus 1.
eoFw8xEy-m4-033\|We'll have some 3p orbitals-- that is, n equal 3, l equals 1.
eT6IA89Vt4Y-000\|We're comparing two acids, HF, a weak acid and HBr, a strong acid.
eT6IA89Vt4Y-001\|We know the relative bond enthalpies go like this.
eT6IA89Vt4Y-004\|Now, that's really the indicator of what we're looking for here because if K is less than 1, then the standard state free energy difference must be positive somewhere.
eT6IA89Vt4Y-005\|It must have some positive values.
eT6IA89Vt4Y-007\|If you dissolve HBr in water, that's an exothermic reaction.
U0N3Vqy1QtQ-001\|As energy is dispersed over more equal energy microstates, the entropy increases.
U0N3Vqy1QtQ-002\|Now there's another definition, the thermodynamic definition, that says as heat is transferred that entropy increases.
U0N3Vqy1QtQ-003\|So can I rationalize those two and bring them together as the same definition?
U0N3Vqy1QtQ-007\|Let's look at those both in terms of a system, say like a particle in a box.
U0N3Vqy1QtQ-008\|A very simple system where there are several microstates available to the system.
U0N3Vqy1QtQ-010\|Now more microstates are available to the system.
U0N3Vqy1QtQ-011\|That is, I can distribute the energy over many more microstates.
h-7G6XgIPGY-000\|Let's look at a specific equilibrium-- water liquid and water gas.
h-7G6XgIPGY-001\|Now, here I have water liquid and water gas at about 25 degrees C, and they're in equilibrium.
h-7G6XgIPGY-002\|There's a little bit of vapor above this liquid.
h-7G6XgIPGY-003\|We could write a reaction quotient for this.
h-7G6XgIPGY-004\|And when we do, that reaction quotient is the pressure of H2O.
h-7G6XgIPGY-005\|That's because the liquid water is pure liquid water, and pure liquids and solids do not appear in equilibrium expressions.
h-7G6XgIPGY-006\|So products over reactants, products is the partial pressure of the gas, the reactants are liquid water.
h-7G6XgIPGY-007\|But it's a pure liquid, so that doesn't appear in the equilibrium expression.
h-7G6XgIPGY-008\|So this reaction quotient is just equivalent to the partial pressure of the gaseous water.
h-7G6XgIPGY-009\|Now that's the vapor pressure of water.
h-7G6XgIPGY-010\|When this is in equilibrium, that will become the vapor pressure.
h-7G6XgIPGY-011\|So the vapor pressure is the equilibrium constant.
h-7G6XgIPGY-013\|And we could find those equilibrium vapor pressures.
h-7G6XgIPGY-015\|Notice here I've plotted several different temperatures for this gas.
h-7G6XgIPGY-016\|So as the temperature increases, the vapor pressure increases.
h-7G6XgIPGY-017\|And you can see that.
h-7G6XgIPGY-028\|So in this case, we'd have the equilibrium constant K prime is 1 over K for this forward reaction that we've been talking about.
h-7G6XgIPGY-029\|So equilibrium expressions-- when we write Qs and Ks, we omit pure liquids and solids.
h-7G6XgIPGY-030\|In these physical equilibria, vapor pressure and K are equivalent, and they are functions of temperature that we already stand.
h-7G6XgIPGY-031\|If we reverse a reaction, we invert the equilibrium constant.
h-7G6XgIPGY-032\|Those are the properties of Q and K.
XwD8___Mm1I-000\|When an electron is promoted from one energy level to another energy level in an atom, an electronic transition occurs, and there's many orbitals in an atom that an electron can occupy.
XwD8___Mm1I-001\|Some are low energy some are high energy, but they have specific energies between them.
XwD8___Mm1I-002\|They are quantized energy levels and an electron will absorb a photon of a specific energy to go from one energy level to another.
XwD8___Mm1I-003\|Let's look at those photon energies.
XwD8___Mm1I-004\|So the energy levels in the atom go as Z squared over n squared R infinity.
XwD8___Mm1I-005\|This quantity tells us the energy level based on n, the principle quantum level.
XwD8___Mm1I-008\|So if I start here and I emit a photon, the energy difference can be calculated by this expression.
XwD8___Mm1I-009\|This will allow us to take photons that we see being emitted from an atom and correlate those to transitions within the electronic structure.
JAMNm0dG7jk-000\|Let's look at the pH gradient across the mitochondrial membrane.
JAMNm0dG7jk-010\|We're talking about the pH gradient across the mitochondrial membrane during ATP synthesis.
JAMNm0dG7jk-012\|And that's accomplished by the oxidation of glucose.
JAMNm0dG7jk-013\|This high concentration of H plus on the outside means the pH is lower on the outside than the inside.
JAMNm0dG7jk-014\|It's more basic on the inside and more acidic on the outside of the mitochondria.
ss4JSV-YRec-000\|Let's try to predict which species is present in the titration of carboning.
ss4JSV-YRec-001\|So carbonic acid all three is titrated.
ss4JSV-YRec-002\|And when we get to pH 7, which species predominates?
ss4JSV-YRec-014\|We go from this form to this form, at around pH equal to pKa 1.
ss4JSV-YRec-015\|At pH 6, there's an equal mixture of these two.
ss4JSV-YRec-018\|We're at pH 7, which is between those two-- one greater than the first pKa and 3 less than the second pKa.
ss4JSV-YRec-019\|So we have titrated along the titration curve.
ss4JSV-YRec-020\|We've passed this point where there's an equal mixture where the pH is equal to the pKa.
s7PIqFpYXJA-001\|Standard conditions and standard states.
s7PIqFpYXJA-002\|So 25 degrees C, that will give us a standard condition.
s7PIqFpYXJA-003\|And standard states, one atmosphere of pressure, pure liquids and solids, 1 molar concentrations.
s7PIqFpYXJA-004\|So I have pure liquid water.
s7PIqFpYXJA-010\|We're talking about water liquid, water gas, in equilibrium.
s7PIqFpYXJA-011\|But we've made specific conditions.
s7PIqFpYXJA-012\|We've said standard conditions.
s7PIqFpYXJA-013\|25 degrees C and standard states.
s7PIqFpYXJA-014\|So 1 atmosphere of pressure of gas and pure liquids.
s7PIqFpYXJA-015\|So that means I have 1 atmosphere of gas and pure liquid water.
s7PIqFpYXJA-016\|What's the situation with this equilibrium?
s7PIqFpYXJA-018\|And at equilibrium, that's the vapor pressure of water for that temperature.
s7PIqFpYXJA-022\|So if Q is bigger than K, I'm going to go back towards reactants.
s7PIqFpYXJA-023\|And I think you'd predict that what's happened.
s7PIqFpYXJA-024\|If you had a whole atmosphere of pressure of water at 25 degrees C, that's too high a pressure.
s7PIqFpYXJA-026\|So the answer here, Q is larger than K.
Vzy2wd5oIKs-000\|Let's look at some energy changes under very specific situations.
Vzy2wd5oIKs-001\|The first is I'll fix the volume.
Vzy2wd5oIKs-002\|So if I fix the volume of the system, what can happen?
Vzy2wd5oIKs-003\|Well, energy is still heat plus work.
Vzy2wd5oIKs-004\|But in a fixed situation, the system can't expand and do any work, or contract and have work done on it.
Vzy2wd5oIKs-005\|So work is zero in this case.
Vzy2wd5oIKs-006\|That means all the energy exchanges have to come from heat flow.
Vzy2wd5oIKs-007\|That means, in this specific case, heat kind of behaves like a state function.
Vzy2wd5oIKs-008\|In general, heat depends on the path.
Vzy2wd5oIKs-009\|But we're saying this specific path, a fixed volume pathway, heat behaves like a state function.
Vzy2wd5oIKs-010\|It's equal to the energy change.
Vzy2wd5oIKs-011\|So that energy change will be reflected in a temperature change.
Vzy2wd5oIKs-015\|Then the heat flow is not a good measure of the energy change.
Vzy2wd5oIKs-016\|And that's actually convenient to just measure one parameter to get the energy change.
Vzy2wd5oIKs-017\|So what we do is we define a new state function called the enthalpy.

eoFw8xEy-m4-003|So we're up to n equal 2, and one possible value for n equal 2 is l equal 1.

eoFw8xEy-m4-007|So the value of l equal 1, we give the letter designation p.

eoFw8xEy-m4-008|So these are going to be p orbitals.

eoFw8xEy-m4-009|How about the values of m sub l?

eoFw8xEy-m4-014|An angular node is a node that has a planar shape rather than a spherical shape for the radial nodes.

eoFw8xEy-m4-017|The square of the wave function is large around the x-axis for the x.

eoFw8xEy-m4-019|That helps you find the angular node.

eoFw8xEy-m4-020|Now, there's one other value of m sub l, equal 0.

eoFw8xEy-m4-021|And that will give you the pz orbital, where the square of the wave function has its maximum along the z-axis.

eoFw8xEy-m4-024|And the m sub l equals 0 already neatly lies along the z-axis.

eoFw8xEy-m4-027|And we'll call them the 2px, 2py, and 2pz orbitals.

eoFw8xEy-m4-029|These are the p orbitals.

eoFw8xEy-m4-030|Now, we can continue.

eoFw8xEy-m4-032|It'll have two nodes now, two radial nodes, because the total number of nodes is n minus 1.

eoFw8xEy-m4-033|We'll have some 3p orbitals-- that is, n equal 3, l equals 1.

eT6IA89Vt4Y-000|We're comparing two acids, HF, a weak acid and HBr, a strong acid.

eT6IA89Vt4Y-001|We know the relative bond enthalpies go like this.

eT6IA89Vt4Y-004|Now, that's really the indicator of what we're looking for here because if K is less than 1, then the standard state free energy difference must be positive somewhere.

eT6IA89Vt4Y-005|It must have some positive values.

eT6IA89Vt4Y-007|If you dissolve HBr in water, that's an exothermic reaction.

U0N3Vqy1QtQ-001|As energy is dispersed over more equal energy microstates, the entropy increases.

U0N3Vqy1QtQ-002|Now there's another definition, the thermodynamic definition, that says as heat is transferred that entropy increases.

U0N3Vqy1QtQ-003|So can I rationalize those two and bring them together as the same definition?

U0N3Vqy1QtQ-007|Let's look at those both in terms of a system, say like a particle in a box.

U0N3Vqy1QtQ-008|A very simple system where there are several microstates available to the system.

U0N3Vqy1QtQ-010|Now more microstates are available to the system.

U0N3Vqy1QtQ-011|That is, I can distribute the energy over many more microstates.

h-7G6XgIPGY-000|Let's look at a specific equilibrium-- water liquid and water gas.

h-7G6XgIPGY-001|Now, here I have water liquid and water gas at about 25 degrees C, and they're in equilibrium.

h-7G6XgIPGY-002|There's a little bit of vapor above this liquid.

h-7G6XgIPGY-003|We could write a reaction quotient for this.

h-7G6XgIPGY-004|And when we do, that reaction quotient is the pressure of H2O.

h-7G6XgIPGY-005|That's because the liquid water is pure liquid water, and pure liquids and solids do not appear in equilibrium expressions.

h-7G6XgIPGY-006|So products over reactants, products is the partial pressure of the gas, the reactants are liquid water.

h-7G6XgIPGY-007|But it's a pure liquid, so that doesn't appear in the equilibrium expression.

h-7G6XgIPGY-008|So this reaction quotient is just equivalent to the partial pressure of the gaseous water.

h-7G6XgIPGY-009|Now that's the vapor pressure of water.

h-7G6XgIPGY-010|When this is in equilibrium, that will become the vapor pressure.

h-7G6XgIPGY-011|So the vapor pressure is the equilibrium constant.

h-7G6XgIPGY-013|And we could find those equilibrium vapor pressures.

h-7G6XgIPGY-015|Notice here I've plotted several different temperatures for this gas.

h-7G6XgIPGY-016|So as the temperature increases, the vapor pressure increases.

h-7G6XgIPGY-017|And you can see that.

h-7G6XgIPGY-028|So in this case, we'd have the equilibrium constant K prime is 1 over K for this forward reaction that we've been talking about.

h-7G6XgIPGY-029|So equilibrium expressions-- when we write Qs and Ks, we omit pure liquids and solids.

h-7G6XgIPGY-030|In these physical equilibria, vapor pressure and K are equivalent, and they are functions of temperature that we already stand.

h-7G6XgIPGY-031|If we reverse a reaction, we invert the equilibrium constant.

h-7G6XgIPGY-032|Those are the properties of Q and K.

XwD8___Mm1I-000|When an electron is promoted from one energy level to another energy level in an atom, an electronic transition occurs, and there's many orbitals in an atom that an electron can occupy.

XwD8___Mm1I-001|Some are low energy some are high energy, but they have specific energies between them.

XwD8___Mm1I-002|They are quantized energy levels and an electron will absorb a photon of a specific energy to go from one energy level to another.

XwD8___Mm1I-003|Let's look at those photon energies.

XwD8___Mm1I-004|So the energy levels in the atom go as Z squared over n squared R infinity.

XwD8___Mm1I-005|This quantity tells us the energy level based on n, the principle quantum level.

XwD8___Mm1I-008|So if I start here and I emit a photon, the energy difference can be calculated by this expression.

XwD8___Mm1I-009|This will allow us to take photons that we see being emitted from an atom and correlate those to transitions within the electronic structure.

JAMNm0dG7jk-000|Let's look at the pH gradient across the mitochondrial membrane.

JAMNm0dG7jk-010|We're talking about the pH gradient across the mitochondrial membrane during ATP synthesis.

JAMNm0dG7jk-012|And that's accomplished by the oxidation of glucose.

JAMNm0dG7jk-013|This high concentration of H plus on the outside means the pH is lower on the outside than the inside.

JAMNm0dG7jk-014|It's more basic on the inside and more acidic on the outside of the mitochondria.

ss4JSV-YRec-000|Let's try to predict which species is present in the titration of carboning.

ss4JSV-YRec-001|So carbonic acid all three is titrated.

ss4JSV-YRec-002|And when we get to pH 7, which species predominates?

ss4JSV-YRec-014|We go from this form to this form, at around pH equal to pKa 1.

ss4JSV-YRec-015|At pH 6, there's an equal mixture of these two.

ss4JSV-YRec-018|We're at pH 7, which is between those two-- one greater than the first pKa and 3 less than the second pKa.

ss4JSV-YRec-019|So we have titrated along the titration curve.

ss4JSV-YRec-020|We've passed this point where there's an equal mixture where the pH is equal to the pKa.

s7PIqFpYXJA-001|Standard conditions and standard states.

s7PIqFpYXJA-002|So 25 degrees C, that will give us a standard condition.

s7PIqFpYXJA-003|And standard states, one atmosphere of pressure, pure liquids and solids, 1 molar concentrations.

s7PIqFpYXJA-004|So I have pure liquid water.

s7PIqFpYXJA-010|We're talking about water liquid, water gas, in equilibrium.

s7PIqFpYXJA-011|But we've made specific conditions.

s7PIqFpYXJA-012|We've said standard conditions.

s7PIqFpYXJA-013|25 degrees C and standard states.

s7PIqFpYXJA-014|So 1 atmosphere of pressure of gas and pure liquids.

s7PIqFpYXJA-015|So that means I have 1 atmosphere of gas and pure liquid water.

s7PIqFpYXJA-016|What's the situation with this equilibrium?

s7PIqFpYXJA-018|And at equilibrium, that's the vapor pressure of water for that temperature.

s7PIqFpYXJA-022|So if Q is bigger than K, I'm going to go back towards reactants.

s7PIqFpYXJA-023|And I think you'd predict that what's happened.

s7PIqFpYXJA-024|If you had a whole atmosphere of pressure of water at 25 degrees C, that's too high a pressure.

s7PIqFpYXJA-026|So the answer here, Q is larger than K.

Vzy2wd5oIKs-000|Let's look at some energy changes under very specific situations.

Vzy2wd5oIKs-001|The first is I'll fix the volume.

Vzy2wd5oIKs-002|So if I fix the volume of the system, what can happen?

Vzy2wd5oIKs-003|Well, energy is still heat plus work.

Vzy2wd5oIKs-004|But in a fixed situation, the system can't expand and do any work, or contract and have work done on it.

Vzy2wd5oIKs-005|So work is zero in this case.

Vzy2wd5oIKs-006|That means all the energy exchanges have to come from heat flow.

Vzy2wd5oIKs-007|That means, in this specific case, heat kind of behaves like a state function.

Vzy2wd5oIKs-008|In general, heat depends on the path.

Vzy2wd5oIKs-009|But we're saying this specific path, a fixed volume pathway, heat behaves like a state function.

Vzy2wd5oIKs-010|It's equal to the energy change.

Vzy2wd5oIKs-011|So that energy change will be reflected in a temperature change.

Vzy2wd5oIKs-015|Then the heat flow is not a good measure of the energy change.

Vzy2wd5oIKs-016|And that's actually convenient to just measure one parameter to get the energy change.

Vzy2wd5oIKs-017|So what we do is we define a new state function called the enthalpy.