ceaglex/VisualTTS_Chem · Datasets at Hugging Face

text stringlengths 19 206
43YS58bxWuU-029\|So we can do that.
43YS58bxWuU-030\|We have a mini neon tube.
43YS58bxWuU-031\|And it's called neon generically, of course.
43YS58bxWuU-032\|In our tubes, there's several different gases.
43YS58bxWuU-033\|We have, in fact, all the gas of the atmosphere.
43YS58bxWuU-034\|We have nitrogen, oxygen, neon, helium, hydrogen. All the gases that are found, to a certain extent, in the atmosphere.
43YS58bxWuU-035\|When I excite them with an electric charge, the electrons will be promoted, and they will emit specific energies.
43YS58bxWuU-036\|They'll emit them all at once, so the colors will be a mixture of these, they'll also be emitting that ultraviolet radiation.
43YS58bxWuU-037\|We won't be able to see that, but it's emitted at the same time.
43YS58bxWuU-038\|So let's look at the emission spectrum of atoms and molecules.
43YS58bxWuU-040\|Let's look at that.
43YS58bxWuU-042\|You can see the various colors, a lot of them bluish, where there is high energy photons.
43YS58bxWuU-043\|And there's even photons, as we said, in the ultraviolet being emitted.
43YS58bxWuU-044\|So we're doing kind of a spectroscopy experiment here.
43YS58bxWuU-046\|If we know exactly the wavelength, we can calculate exactly the energy difference of the atomic orbitals in these atoms.
43YS58bxWuU-047\|So you can see how powerful spectroscopy is to understand the properties and the structure of atoms.
SUkBTHfKqf4-000\|Titration is the process of reacting acid and base together.
SUkBTHfKqf4-001\|And I do that in a controlled way.
SUkBTHfKqf4-002\|So I can take an acid solution and a base.
SUkBTHfKqf4-003\|The acid solution will originally have low pH, a high acid concentration.
SUkBTHfKqf4-004\|As you add base, that base will react with the acid and lower that acid concentration.
SUkBTHfKqf4-005\|Lowering the acid concentration raises the pH.
SUkBTHfKqf4-006\|The solution becomes more basic, which makes sense because you're adding a base.
SUkBTHfKqf4-007\|Eventually, you'll use up all of the acid that's there.
SUkBTHfKqf4-008\|And if you continue to add base, then that base will dominate the pH.
SUkBTHfKqf4-009\|And you'll have a basic solution.
SUkBTHfKqf4-010\|You can plot that in something that's called a titration curve.
SUkBTHfKqf4-013\|This totally dissociates, forms 0.1 or 10 to the minus 1 moles of H3O+ moles per liter.
SUkBTHfKqf4-014\|The pH is minus log of 10 to the minus 1 or 1.
SUkBTHfKqf4-015\|So I start with pH 1.
SUkBTHfKqf4-016\|I add some base.
SUkBTHfKqf4-017\|And as I add moles of base, those react with my moles of acid to form water.
SUkBTHfKqf4-018\|And that continues to happen until I use up all the original acid that is there.
SUkBTHfKqf4-019\|When I've added 1 mole of base for every mole of acid in my original solution, I've used up all my acid, and I've used up all the base that I've added.
SUkBTHfKqf4-020\|They've all completely reacted together.
SUkBTHfKqf4-022\|As you continue beyond that, then the base that you're adding dominates the pH, and the pH rapidly goes to a basic level.
SUkBTHfKqf4-023\|You can do the reverse.
SUkBTHfKqf4-024\|You can start with a strong base solution or any base solution and add an acid solution.
SUkBTHfKqf4-025\|Here the pH will start high because the H3O+ concentration is low.
SUkBTHfKqf4-026\|And as you add acid, that pH will drop.
SUkBTHfKqf4-028\|But a sodium chloride solution has pH 7.
SUkBTHfKqf4-029\|The ions sodium and chlorine don't contribute to the pH.
SUkBTHfKqf4-030\|We'll talk about that more later.
SUkBTHfKqf4-031\|So you have a pH of 7, a neutral salt water solution.
SUkBTHfKqf4-032\|As I continue to add acid, the acid will dominate the pH.
SUkBTHfKqf4-033\|And you'll go to a low pH solution.
SUkBTHfKqf4-034\|So this is a strong acid or strong base titration.
qvaEyckBI0g-000\|So let's look at electrons in their various orbitals in forms of a ChemQuiz.
qvaEyckBI0g-001\|When an electron spin flips, for which of the following is the energy not altered?
qvaEyckBI0g-009\|So we're talking about nitrogen, oxygen, and fluorine, and we're trying to decide can we flip an electron spin and not change the overall energy of the atom?
qvaEyckBI0g-010\|Nitrogen has the electronic configuration 1s2 2s2 2p3, three unpaired electrons in those 2p orbitals.
qvaEyckBI0g-011\|Oxygen, 2p4, fluorine, 2p5.
qvaEyckBI0g-014\|In oxygen and nitrogen, there's other electrons.
qvaEyckBI0g-015\|So these electrons will try to be parallel.
qvaEyckBI0g-016\|So the energy of this electron depends on the spin of this electron.
qvaEyckBI0g-017\|But fluorine, that's not the case, everybody is paired up except for this unpaired electron, so fluorine can either have spin up or spin down.
7TZXu6s0KA0-000\|Let's look at the Maxwell Boltzmann distribution for a pair of gasses.
7TZXu6s0KA0-001\|I've plotted them here, and I've indicated where the root mean square velocity would fall for gas X at 400 Kelvin, and gas Y at 800 Kelvin.
7TZXu6s0KA0-002\|The question is what are gases X and Y?
7TZXu6s0KA0-009\|We're looking at the Maxwell Boltzmann distribution plots for two different gases at two different temperatures.
7TZXu6s0KA0-012\|The temperature is doubling, but that wouldn't give us a factor of 2.
7TZXu6s0KA0-013\|That would give us a factor of square root 2 in the rms velocity.
X5KLAdXoDxc-000\|Let's look at the effect of temperature on a chemical reaction.
X5KLAdXoDxc-001\|Now, we know intuitively if you raise the temperature, chemical reactions proceed more rapidly.
X5KLAdXoDxc-002\|The reason is that at higher temperature, there are more molecules that have energies in excess of the activation energy for the chemical reaction.
X5KLAdXoDxc-003\|So they can get over the barrier that separates products and reactants.
X5KLAdXoDxc-006\|Now, that doesn't mean all those molecules will react.
X5KLAdXoDxc-007\|They still have to collide, and collide with the appropriate orientation.
X5KLAdXoDxc-010\|So now I have a higher percentage of molecules with energies in excess of the activation energy.
X5KLAdXoDxc-011\|So why do reactions go faster with the increasing temperature?
X5KLAdXoDxc-012\|More molecules, more particles with higher energies.
X5KLAdXoDxc-013\|Now if you think about kinetics, our kinetic rate laws look like the rate is the rate constant times some function of the concentrations.
X5KLAdXoDxc-014\|Now, the concentrations are not a function of temperature.
X5KLAdXoDxc-015\|So for rates to increase with temperature, our rate constants must be a function of temperature.
X5KLAdXoDxc-018\|So rate constants are a function of temperature.
X5KLAdXoDxc-023\|k, the rate constant for the chemical reaction, is a function of temperature.
vhlYDz16s0w-000\|Let's look at some examples of oxidation reduction reactions.
vhlYDz16s0w-001\|Here's a common one-- the rusting of iron.
vhlYDz16s0w-002\|Now, it's interesting to note, simply putting a nail made of iron in pure water, that won't rust.
vhlYDz16s0w-003\|You actually need the presence of dissolved oxygen.
vhlYDz16s0w-004\|So here's the actual reaction.
vhlYDz16s0w-007\|Iron is in its 0 oxidation state, it's oxidation number increases.
vhlYDz16s0w-008\|That's an oxidation to plus 2.
vhlYDz16s0w-009\|So to help us balance this, let's go to our table of standard production potentials.
vhlYDz16s0w-012\|We can use those to help balance this reaction, so let's do that.
vhlYDz16s0w-016\|And I'll change the sign of my reduction potential.
vhlYDz16s0w-021\|The cell potential for the oxidation of iron is 0.81 volts.
vhlYDz16s0w-023\|These are farther down the table.
vhlYDz16s0w-026\|They'll protect the iron from oxidation.
vhlYDz16s0w-027\|And you know this.
NLGrUdD5uDw-000\|To describe molecules quantum mechanically, we can take the atomic orbitals from the atoms and recombine them to form molecular orbitals.
NLGrUdD5uDw-001\|And that works well for diatomics.
NLGrUdD5uDw-002\|We can take s and p orbitals and overlap them and form molecular orbitals.
NLGrUdD5uDw-003\|The s and p orbitals have the appropriate arrangement and symmetry for a diatomic molecule.
NLGrUdD5uDw-004\|But what if we get more complicated?
NLGrUdD5uDw-005\|Even if you go to a very simple triatomic or if you go to something with steric number 2, how do we accommodate steric number 2?
NLGrUdD5uDw-006\|How do we have a carbon bond to something on this side and something on this side?
NLGrUdD5uDw-007\|If I use my p orbital over here to make a molecular orbital with this, I don't have a p orbital available to make a molecular orbital with this.
NLGrUdD5uDw-008\|I can't reuse my orbitals to make more bonds.
NLGrUdD5uDw-009\|So I have to have bonds and orbitals that have the right orientation.

43YS58bxWuU-029|So we can do that.

43YS58bxWuU-030|We have a mini neon tube.

43YS58bxWuU-031|And it's called neon generically, of course.

43YS58bxWuU-032|In our tubes, there's several different gases.

43YS58bxWuU-033|We have, in fact, all the gas of the atmosphere.

43YS58bxWuU-034|We have nitrogen, oxygen, neon, helium, hydrogen. All the gases that are found, to a certain extent, in the atmosphere.

43YS58bxWuU-035|When I excite them with an electric charge, the electrons will be promoted, and they will emit specific energies.

43YS58bxWuU-036|They'll emit them all at once, so the colors will be a mixture of these, they'll also be emitting that ultraviolet radiation.

43YS58bxWuU-037|We won't be able to see that, but it's emitted at the same time.

43YS58bxWuU-038|So let's look at the emission spectrum of atoms and molecules.

43YS58bxWuU-040|Let's look at that.

43YS58bxWuU-042|You can see the various colors, a lot of them bluish, where there is high energy photons.

43YS58bxWuU-043|And there's even photons, as we said, in the ultraviolet being emitted.

43YS58bxWuU-044|So we're doing kind of a spectroscopy experiment here.

43YS58bxWuU-046|If we know exactly the wavelength, we can calculate exactly the energy difference of the atomic orbitals in these atoms.

43YS58bxWuU-047|So you can see how powerful spectroscopy is to understand the properties and the structure of atoms.

SUkBTHfKqf4-000|Titration is the process of reacting acid and base together.

SUkBTHfKqf4-001|And I do that in a controlled way.

SUkBTHfKqf4-002|So I can take an acid solution and a base.

SUkBTHfKqf4-003|The acid solution will originally have low pH, a high acid concentration.

SUkBTHfKqf4-004|As you add base, that base will react with the acid and lower that acid concentration.

SUkBTHfKqf4-005|Lowering the acid concentration raises the pH.

SUkBTHfKqf4-006|The solution becomes more basic, which makes sense because you're adding a base.

SUkBTHfKqf4-007|Eventually, you'll use up all of the acid that's there.

SUkBTHfKqf4-008|And if you continue to add base, then that base will dominate the pH.

SUkBTHfKqf4-009|And you'll have a basic solution.

SUkBTHfKqf4-010|You can plot that in something that's called a titration curve.

SUkBTHfKqf4-013|This totally dissociates, forms 0.1 or 10 to the minus 1 moles of H3O+ moles per liter.

SUkBTHfKqf4-014|The pH is minus log of 10 to the minus 1 or 1.

SUkBTHfKqf4-015|So I start with pH 1.

SUkBTHfKqf4-016|I add some base.

SUkBTHfKqf4-017|And as I add moles of base, those react with my moles of acid to form water.

SUkBTHfKqf4-018|And that continues to happen until I use up all the original acid that is there.

SUkBTHfKqf4-019|When I've added 1 mole of base for every mole of acid in my original solution, I've used up all my acid, and I've used up all the base that I've added.

SUkBTHfKqf4-020|They've all completely reacted together.

SUkBTHfKqf4-022|As you continue beyond that, then the base that you're adding dominates the pH, and the pH rapidly goes to a basic level.

SUkBTHfKqf4-023|You can do the reverse.

SUkBTHfKqf4-024|You can start with a strong base solution or any base solution and add an acid solution.

SUkBTHfKqf4-025|Here the pH will start high because the H3O+ concentration is low.

SUkBTHfKqf4-026|And as you add acid, that pH will drop.

SUkBTHfKqf4-028|But a sodium chloride solution has pH 7.

SUkBTHfKqf4-029|The ions sodium and chlorine don't contribute to the pH.

SUkBTHfKqf4-030|We'll talk about that more later.

SUkBTHfKqf4-031|So you have a pH of 7, a neutral salt water solution.

SUkBTHfKqf4-032|As I continue to add acid, the acid will dominate the pH.

SUkBTHfKqf4-033|And you'll go to a low pH solution.

SUkBTHfKqf4-034|So this is a strong acid or strong base titration.

qvaEyckBI0g-000|So let's look at electrons in their various orbitals in forms of a ChemQuiz.

qvaEyckBI0g-001|When an electron spin flips, for which of the following is the energy not altered?

qvaEyckBI0g-009|So we're talking about nitrogen, oxygen, and fluorine, and we're trying to decide can we flip an electron spin and not change the overall energy of the atom?

qvaEyckBI0g-010|Nitrogen has the electronic configuration 1s2 2s2 2p3, three unpaired electrons in those 2p orbitals.

qvaEyckBI0g-011|Oxygen, 2p4, fluorine, 2p5.

qvaEyckBI0g-014|In oxygen and nitrogen, there's other electrons.

qvaEyckBI0g-015|So these electrons will try to be parallel.

qvaEyckBI0g-016|So the energy of this electron depends on the spin of this electron.

qvaEyckBI0g-017|But fluorine, that's not the case, everybody is paired up except for this unpaired electron, so fluorine can either have spin up or spin down.

7TZXu6s0KA0-000|Let's look at the Maxwell Boltzmann distribution for a pair of gasses.

7TZXu6s0KA0-001|I've plotted them here, and I've indicated where the root mean square velocity would fall for gas X at 400 Kelvin, and gas Y at 800 Kelvin.

7TZXu6s0KA0-002|The question is what are gases X and Y?

7TZXu6s0KA0-009|We're looking at the Maxwell Boltzmann distribution plots for two different gases at two different temperatures.

7TZXu6s0KA0-012|The temperature is doubling, but that wouldn't give us a factor of 2.

7TZXu6s0KA0-013|That would give us a factor of square root 2 in the rms velocity.

X5KLAdXoDxc-000|Let's look at the effect of temperature on a chemical reaction.

X5KLAdXoDxc-001|Now, we know intuitively if you raise the temperature, chemical reactions proceed more rapidly.

X5KLAdXoDxc-002|The reason is that at higher temperature, there are more molecules that have energies in excess of the activation energy for the chemical reaction.

X5KLAdXoDxc-003|So they can get over the barrier that separates products and reactants.

X5KLAdXoDxc-006|Now, that doesn't mean all those molecules will react.

X5KLAdXoDxc-007|They still have to collide, and collide with the appropriate orientation.

X5KLAdXoDxc-010|So now I have a higher percentage of molecules with energies in excess of the activation energy.

X5KLAdXoDxc-011|So why do reactions go faster with the increasing temperature?

X5KLAdXoDxc-012|More molecules, more particles with higher energies.

X5KLAdXoDxc-013|Now if you think about kinetics, our kinetic rate laws look like the rate is the rate constant times some function of the concentrations.

X5KLAdXoDxc-014|Now, the concentrations are not a function of temperature.

X5KLAdXoDxc-015|So for rates to increase with temperature, our rate constants must be a function of temperature.

X5KLAdXoDxc-018|So rate constants are a function of temperature.

X5KLAdXoDxc-023|k, the rate constant for the chemical reaction, is a function of temperature.

vhlYDz16s0w-000|Let's look at some examples of oxidation reduction reactions.

vhlYDz16s0w-001|Here's a common one-- the rusting of iron.

vhlYDz16s0w-002|Now, it's interesting to note, simply putting a nail made of iron in pure water, that won't rust.

vhlYDz16s0w-003|You actually need the presence of dissolved oxygen.

vhlYDz16s0w-004|So here's the actual reaction.

vhlYDz16s0w-007|Iron is in its 0 oxidation state, it's oxidation number increases.

vhlYDz16s0w-008|That's an oxidation to plus 2.

vhlYDz16s0w-009|So to help us balance this, let's go to our table of standard production potentials.

vhlYDz16s0w-012|We can use those to help balance this reaction, so let's do that.

vhlYDz16s0w-016|And I'll change the sign of my reduction potential.

vhlYDz16s0w-021|The cell potential for the oxidation of iron is 0.81 volts.

vhlYDz16s0w-023|These are farther down the table.

vhlYDz16s0w-026|They'll protect the iron from oxidation.

vhlYDz16s0w-027|And you know this.

NLGrUdD5uDw-000|To describe molecules quantum mechanically, we can take the atomic orbitals from the atoms and recombine them to form molecular orbitals.

NLGrUdD5uDw-001|And that works well for diatomics.

NLGrUdD5uDw-002|We can take s and p orbitals and overlap them and form molecular orbitals.

NLGrUdD5uDw-003|The s and p orbitals have the appropriate arrangement and symmetry for a diatomic molecule.

NLGrUdD5uDw-004|But what if we get more complicated?

NLGrUdD5uDw-005|Even if you go to a very simple triatomic or if you go to something with steric number 2, how do we accommodate steric number 2?

NLGrUdD5uDw-006|How do we have a carbon bond to something on this side and something on this side?

NLGrUdD5uDw-007|If I use my p orbital over here to make a molecular orbital with this, I don't have a p orbital available to make a molecular orbital with this.

NLGrUdD5uDw-008|I can't reuse my orbitals to make more bonds.

NLGrUdD5uDw-009|So I have to have bonds and orbitals that have the right orientation.