ceaglex/VisualTTS_Chem · Datasets at Hugging Face

text stringlengths 19 206
2hikObHZrXg-077\|Well, pH 8 is between 6 and 10.
2hikObHZrXg-078\|It's well above this initial pH.
2hikObHZrXg-079\|So pKa1, that proton is certainly fully removed by the time we're at pH 8.
2hikObHZrXg-080\|So this is a negative ion right now.
2hikObHZrXg-081\|The other two species, we have to worry about.
2hikObHZrXg-082\|Now this proton here, this won't be removed until we're above pH 9.9.
2hikObHZrXg-083\|So that one is certainly going to be attached, so we don't have to worry about this pKa.
2hikObHZrXg-084\|We just have to worry about this pKa6.
2hikObHZrXg-085\|We're about two pH units above that.
2hikObHZrXg-086\|So at two pH units above this pKa, what species is present in solution?
2hikObHZrXg-090\|This proton either attached or removed.
2hikObHZrXg-091\|The pKa is 6 for that, but we're at pH 8, so the difference between pH and pKa is 2.
2hikObHZrXg-092\|So the difference is 2, so 2 has to be log of this concentration ratio.
2hikObHZrXg-095\|So these two forms, about a 100 to 1.
2hikObHZrXg-096\|So clearly, mostly the base form of this equilibrium is present at pH 8.
2hikObHZrXg-097\|Now there's also some sodium ions from the sodium hydroxide that we've been adding.
2hikObHZrXg-098\|But there are no hydroxide ions, per se.
2hikObHZrXg-099\|Very, very low concentration of hydroxide ions because all the hydroxide ions have been used up in converting the acid form to the base form.
2hikObHZrXg-101\|Those are the ions present in highest concentration at pH 8.
hOwQQryixJM-000\|Let's look at the pi bonding orbitals in butadiene.
hOwQQryixJM-001\|Can you determine which is the lowest energy pi bonding orbital?
hOwQQryixJM-002\|A, B, or C?
hOwQQryixJM-013\|We're looking for the lowest energy pi molecular orbital in butadiene, and we have three choices.
hOwQQryixJM-014\|This first choice actually isn't a pi orbital.
hOwQQryixJM-015\|It's a low energy orbital, if this were to form.
hOwQQryixJM-016\|Because it's along the internuclear axis though, we would call it a sigma orbital.
hOwQQryixJM-017\|This orbital has a node along the internuclear axis.
hOwQQryixJM-018\|And that's common for pi orbitals.
hOwQQryixJM-019\|They have internuclear axis nodes, and they have electron density above and below the axis.
hOwQQryixJM-020\|So this is a pi orbital and this is a pi orbital.
hOwQQryixJM-021\|And among these two, this has the lower number of nodes.
hOwQQryixJM-022\|So it's the lower energy.
hOwQQryixJM-023\|So in this case, B is the lowest energy pi molecular orbital among those choices.
nE480ADhZsY-000\|The free energy change, or delta G, is what determines whether a process is favored.
nE480ADhZsY-001\|Negative delta Gs indicate the favored direction for a chemical reaction.
nE480ADhZsY-002\|A reaction is spontaneous if delta G is negative.
nE480ADhZsY-003\|Delta G is composed of delta H and delta S and the temperature.
nE480ADhZsY-004\|So we need to look at three factors to determine where our negative delta G, or favored reactions, occur.
nE480ADhZsY-010\|Because a negative delta S makes the minus T delta S term positive.
nE480ADhZsY-011\|Remember, temperatures are always positive.
nE480ADhZsY-012\|So minus T delta S will be positive for negative delta S.
nE480ADhZsY-013\|So if we look at the overall space and say, well, what about the regions where delta S is negative and delta H is positive?
nE480ADhZsY-017\|No conditions of temperature will make that spontaneous.
nE480ADhZsY-018\|Let's look at a couple other conditions.
nE480ADhZsY-020\|Both of these favor delta G being negative.
nE480ADhZsY-021\|Of course, exothermic, delta H is negative.
nE480ADhZsY-022\|And an increase in entropy, that's a positive delta H times a positive temperature.
nE480ADhZsY-023\|With this minus sign, this term will always be negative, and this term will always be negative.
nE480ADhZsY-024\|So when delta S is positive and delta H is negative, that's favorable for all temperatures.
nE480ADhZsY-031\|But you can overcome it as long as you have a high enough temperature and a positive delta S.
u2xEJdSJZOs-000\|Now we can look at that trend in atomic radius in a little more detail.
u2xEJdSJZOs-003\|And then I've gone from sodium to argon across the next row of the periodic table.
u2xEJdSJZOs-005\|So lithium at 152 picometers, fluorine at 71.
u2xEJdSJZOs-006\|We have a decrease as I go across.
u2xEJdSJZOs-007\|And we understand that now.
u2xEJdSJZOs-008\|We understand that we're going into the same principle quantum level, but we're adding more nuclear charge.
u2xEJdSJZOs-010\|So those 2p orbitals that start out rather large at boron are contracting and getting smaller and smaller by the time I get over to fluorine.
u2xEJdSJZOs-011\|That's because a bigger charge on the nucleus.
u2xEJdSJZOs-013\|So we understand why ionization energy and atomic radii mirror each other in the trend because of the quantum mechanical structure of the atoms.
u2xEJdSJZOs-014\|We can understand trends in the periodic table because of our understanding of quantum mechanics.
z24IYbHtphM-000\|Matter absorbs or emits light based on its fine electronic structure.
z24IYbHtphM-001\|That is, you have electrons behaving like waves, they're bounded in the matter.
z24IYbHtphM-003\|Now let's take this.
z24IYbHtphM-004\|Let's say a certain piece of matter has this energy level scheme that's shown here, and it has an emission spectrum.
z24IYbHtphM-005\|Which of these three emission lines arises from the transition of high energy to low energy 3 to 1 emission in this system?
z24IYbHtphM-018\|We're trying to get from the energy level spacings in some matter to the actual emission spectrum that we observe.
z24IYbHtphM-019\|So if you look at the possible energy transitions, a transition between level 3 and 2 is possible.
z24IYbHtphM-020\|That would be the lowest energy of the possible.
z24IYbHtphM-021\|That's the smallest spacing here.
z24IYbHtphM-022\|So that's the smallest energy.
z24IYbHtphM-025\|The other transition from 3 to 1 is the highest possible for this system.
Yzz2eluxRYc-000\|The natural direction for a process is determined by the overall increase in entropy.
Yzz2eluxRYc-001\|If there's an overall increase in entropy between the system and the surroundings, then that's the favored direction for that process.
Yzz2eluxRYc-003\|So what's the difference between a process that has a natural direction and a process that has equilibrium?
Yzz2eluxRYc-004\|Both directions are equally favored.
Yzz2eluxRYc-005\|Well, it's a balance between the entropy change.
Yzz2eluxRYc-008\|I know if I go from the liquid to the gas, that's an increase in entropy.
Yzz2eluxRYc-009\|I go from the relatively constrained liquid to the many microstates of the gas.
Yzz2eluxRYc-010\|And of course, that's correct.
Yzz2eluxRYc-011\|But the error there is we're only thinking about the system.
Yzz2eluxRYc-012\|We need to think about the system and the surroundings to calculate the total entropy.
Yzz2eluxRYc-016\|You know as you evaporate the gas, as you go from the liquid to the gas, that you have to absorb energy.
Yzz2eluxRYc-017\|So the surroundings gives up some heat.
Yzz2eluxRYc-018\|As that surrounding gives up heat to the system, the entropy of the surroundings decrease.
Yzz2eluxRYc-019\|So system entropy increasing, surroundings entropy decreasing, the overall balance is 0.
Yzz2eluxRYc-020\|And there's no net tendency for the reaction to go one direction or the other.
Yzz2eluxRYc-025\|So a careful balance between the system and surrounding entropies can give you equilibrium situations.
Yzz2eluxRYc-026\|And if you're very careful, that equilibrium can exist between the three phases.
Yzz2eluxRYc-027\|And that's what happens at the triple point of water.
FiAyWh9kb9w-000\|Our understanding of gases and expansions and contractions help us understand something about the atmosphere, which is a big layer of gas.
FiAyWh9kb9w-002\|As you go up in altitude, the pressure decreases.
FiAyWh9kb9w-003\|There is less atmosphere above you, fewer molecules, lower pressure.
FiAyWh9kb9w-006\|Now warm gases near the surface of the Earth tend to rise.
FiAyWh9kb9w-007\|And as they rise through that decreasing pressure they expand against that pressure.
FiAyWh9kb9w-008\|That expansion against that pressure is an adiabatic process.
FiAyWh9kb9w-009\|It's adiabatic because there's no heat.
FiAyWh9kb9w-010\|There's no source of heat to return that energy it took to expand.
FiAyWh9kb9w-011\|So that adiabatic process, an adiabatic expansion is a cooling process.
FiAyWh9kb9w-012\|I use some of my energy to expand.
FiAyWh9kb9w-013\|I can't get it back as heat, so the gas cools off.

2hikObHZrXg-077|Well, pH 8 is between 6 and 10.

2hikObHZrXg-078|It's well above this initial pH.

2hikObHZrXg-079|So pKa1, that proton is certainly fully removed by the time we're at pH 8.

2hikObHZrXg-080|So this is a negative ion right now.

2hikObHZrXg-081|The other two species, we have to worry about.

2hikObHZrXg-082|Now this proton here, this won't be removed until we're above pH 9.9.

2hikObHZrXg-083|So that one is certainly going to be attached, so we don't have to worry about this pKa.

2hikObHZrXg-084|We just have to worry about this pKa6.

2hikObHZrXg-085|We're about two pH units above that.

2hikObHZrXg-086|So at two pH units above this pKa, what species is present in solution?

2hikObHZrXg-090|This proton either attached or removed.

2hikObHZrXg-091|The pKa is 6 for that, but we're at pH 8, so the difference between pH and pKa is 2.

2hikObHZrXg-092|So the difference is 2, so 2 has to be log of this concentration ratio.

2hikObHZrXg-095|So these two forms, about a 100 to 1.

2hikObHZrXg-096|So clearly, mostly the base form of this equilibrium is present at pH 8.

2hikObHZrXg-097|Now there's also some sodium ions from the sodium hydroxide that we've been adding.

2hikObHZrXg-098|But there are no hydroxide ions, per se.

2hikObHZrXg-099|Very, very low concentration of hydroxide ions because all the hydroxide ions have been used up in converting the acid form to the base form.

2hikObHZrXg-101|Those are the ions present in highest concentration at pH 8.

hOwQQryixJM-000|Let's look at the pi bonding orbitals in butadiene.

hOwQQryixJM-001|Can you determine which is the lowest energy pi bonding orbital?

hOwQQryixJM-002|A, B, or C?

hOwQQryixJM-013|We're looking for the lowest energy pi molecular orbital in butadiene, and we have three choices.

hOwQQryixJM-014|This first choice actually isn't a pi orbital.

hOwQQryixJM-015|It's a low energy orbital, if this were to form.

hOwQQryixJM-016|Because it's along the internuclear axis though, we would call it a sigma orbital.

hOwQQryixJM-017|This orbital has a node along the internuclear axis.

hOwQQryixJM-018|And that's common for pi orbitals.

hOwQQryixJM-019|They have internuclear axis nodes, and they have electron density above and below the axis.

hOwQQryixJM-020|So this is a pi orbital and this is a pi orbital.

hOwQQryixJM-021|And among these two, this has the lower number of nodes.

hOwQQryixJM-022|So it's the lower energy.

hOwQQryixJM-023|So in this case, B is the lowest energy pi molecular orbital among those choices.

nE480ADhZsY-000|The free energy change, or delta G, is what determines whether a process is favored.

nE480ADhZsY-001|Negative delta Gs indicate the favored direction for a chemical reaction.

nE480ADhZsY-002|A reaction is spontaneous if delta G is negative.

nE480ADhZsY-003|Delta G is composed of delta H and delta S and the temperature.

nE480ADhZsY-004|So we need to look at three factors to determine where our negative delta G, or favored reactions, occur.

nE480ADhZsY-010|Because a negative delta S makes the minus T delta S term positive.

nE480ADhZsY-011|Remember, temperatures are always positive.

nE480ADhZsY-012|So minus T delta S will be positive for negative delta S.

nE480ADhZsY-013|So if we look at the overall space and say, well, what about the regions where delta S is negative and delta H is positive?

nE480ADhZsY-017|No conditions of temperature will make that spontaneous.

nE480ADhZsY-018|Let's look at a couple other conditions.

nE480ADhZsY-020|Both of these favor delta G being negative.

nE480ADhZsY-021|Of course, exothermic, delta H is negative.

nE480ADhZsY-022|And an increase in entropy, that's a positive delta H times a positive temperature.

nE480ADhZsY-023|With this minus sign, this term will always be negative, and this term will always be negative.

nE480ADhZsY-024|So when delta S is positive and delta H is negative, that's favorable for all temperatures.

nE480ADhZsY-031|But you can overcome it as long as you have a high enough temperature and a positive delta S.

u2xEJdSJZOs-000|Now we can look at that trend in atomic radius in a little more detail.

u2xEJdSJZOs-003|And then I've gone from sodium to argon across the next row of the periodic table.

u2xEJdSJZOs-005|So lithium at 152 picometers, fluorine at 71.

u2xEJdSJZOs-006|We have a decrease as I go across.

u2xEJdSJZOs-007|And we understand that now.

u2xEJdSJZOs-008|We understand that we're going into the same principle quantum level, but we're adding more nuclear charge.

u2xEJdSJZOs-010|So those 2p orbitals that start out rather large at boron are contracting and getting smaller and smaller by the time I get over to fluorine.

u2xEJdSJZOs-011|That's because a bigger charge on the nucleus.

u2xEJdSJZOs-013|So we understand why ionization energy and atomic radii mirror each other in the trend because of the quantum mechanical structure of the atoms.

u2xEJdSJZOs-014|We can understand trends in the periodic table because of our understanding of quantum mechanics.

z24IYbHtphM-000|Matter absorbs or emits light based on its fine electronic structure.

z24IYbHtphM-001|That is, you have electrons behaving like waves, they're bounded in the matter.

z24IYbHtphM-003|Now let's take this.

z24IYbHtphM-004|Let's say a certain piece of matter has this energy level scheme that's shown here, and it has an emission spectrum.

z24IYbHtphM-005|Which of these three emission lines arises from the transition of high energy to low energy 3 to 1 emission in this system?

z24IYbHtphM-018|We're trying to get from the energy level spacings in some matter to the actual emission spectrum that we observe.

z24IYbHtphM-019|So if you look at the possible energy transitions, a transition between level 3 and 2 is possible.

z24IYbHtphM-020|That would be the lowest energy of the possible.

z24IYbHtphM-021|That's the smallest spacing here.

z24IYbHtphM-022|So that's the smallest energy.

z24IYbHtphM-025|The other transition from 3 to 1 is the highest possible for this system.

Yzz2eluxRYc-000|The natural direction for a process is determined by the overall increase in entropy.

Yzz2eluxRYc-001|If there's an overall increase in entropy between the system and the surroundings, then that's the favored direction for that process.

Yzz2eluxRYc-003|So what's the difference between a process that has a natural direction and a process that has equilibrium?

Yzz2eluxRYc-004|Both directions are equally favored.

Yzz2eluxRYc-005|Well, it's a balance between the entropy change.

Yzz2eluxRYc-008|I know if I go from the liquid to the gas, that's an increase in entropy.

Yzz2eluxRYc-009|I go from the relatively constrained liquid to the many microstates of the gas.

Yzz2eluxRYc-010|And of course, that's correct.

Yzz2eluxRYc-011|But the error there is we're only thinking about the system.

Yzz2eluxRYc-012|We need to think about the system and the surroundings to calculate the total entropy.

Yzz2eluxRYc-016|You know as you evaporate the gas, as you go from the liquid to the gas, that you have to absorb energy.

Yzz2eluxRYc-017|So the surroundings gives up some heat.

Yzz2eluxRYc-018|As that surrounding gives up heat to the system, the entropy of the surroundings decrease.

Yzz2eluxRYc-019|So system entropy increasing, surroundings entropy decreasing, the overall balance is 0.

Yzz2eluxRYc-020|And there's no net tendency for the reaction to go one direction or the other.

Yzz2eluxRYc-025|So a careful balance between the system and surrounding entropies can give you equilibrium situations.

Yzz2eluxRYc-026|And if you're very careful, that equilibrium can exist between the three phases.

Yzz2eluxRYc-027|And that's what happens at the triple point of water.

FiAyWh9kb9w-000|Our understanding of gases and expansions and contractions help us understand something about the atmosphere, which is a big layer of gas.

FiAyWh9kb9w-002|As you go up in altitude, the pressure decreases.

FiAyWh9kb9w-003|There is less atmosphere above you, fewer molecules, lower pressure.

FiAyWh9kb9w-006|Now warm gases near the surface of the Earth tend to rise.

FiAyWh9kb9w-007|And as they rise through that decreasing pressure they expand against that pressure.

FiAyWh9kb9w-008|That expansion against that pressure is an adiabatic process.

FiAyWh9kb9w-009|It's adiabatic because there's no heat.

FiAyWh9kb9w-010|There's no source of heat to return that energy it took to expand.

FiAyWh9kb9w-011|So that adiabatic process, an adiabatic expansion is a cooling process.

FiAyWh9kb9w-012|I use some of my energy to expand.

FiAyWh9kb9w-013|I can't get it back as heat, so the gas cools off.