ceaglex/VisualTTS_Chem · Datasets at Hugging Face

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OlehBTSyPDA-107\|That's just coincidental.
OlehBTSyPDA-108\|That occurs at the midpoint of titrations, halfway between the start and the equivalence point.
OlehBTSyPDA-109\|So where am I on my titration curve?
OlehBTSyPDA-111\|My titration curve, I go for my weak acid to its conjugate base.
OlehBTSyPDA-112\|And in this calculation, I've added enough to convert half of my initial acid into its conjugate base.
OlehBTSyPDA-113\|I'm right smack in the middle here of this buffer region.
OlehBTSyPDA-114\|I'm half way towards the equivalence point.
OlehBTSyPDA-116\|That's just a feature of titration curves.
OlehBTSyPDA-117\|It's interesting numerical equivalence.
OlehBTSyPDA-119\|And that's numerically equal to the pKa.
BBqjKt0Qfd4-000\|Now we understand how multiple electrons can exist around an atom.
BBqjKt0Qfd4-001\|They go into the various orbitals, and they fill the various orbitals following some fairly specific rules.
BBqjKt0Qfd4-010\|Here's the first few elements on the periodic table.
BBqjKt0Qfd4-011\|Hydrogen has one electron.
BBqjKt0Qfd4-012\|It will enter the 1s, the lowest available orbital.
BBqjKt0Qfd4-014\|It's called a paramagnetic species.
BBqjKt0Qfd4-015\|Hydrogen is paramagnetic in its atomic state.
BBqjKt0Qfd4-016\|Helium has paired electrons, because the next electron will go in the 1s orbital paired up.
BBqjKt0Qfd4-017\|The 2s orbital is higher in energy, so it's cheaper to spend that little energy to pair than to put the energy in the higher 2s orbital.
BBqjKt0Qfd4-025\|So it's cheaper to put 2s paired than to put an electron in the higher energy 2p.
BBqjKt0Qfd4-026\|So we're going to have 1s 2, 2s 2 for beryllium.
BBqjKt0Qfd4-027\|And that will not be magnetic.
BBqjKt0Qfd4-029\|Now, we can continue.
BBqjKt0Qfd4-030\|Here's carbon and nitrogen, oxygen, fluorine.
BBqjKt0Qfd4-031\|And you'll see that carbon has maximum multiplicity.
BBqjKt0Qfd4-032\|The 2p orbitals are the same energy.
BBqjKt0Qfd4-033\|So the spins go in parallel.
BBqjKt0Qfd4-034\|Nitrogen, same thing-- three unpaired electrons.
BBqjKt0Qfd4-035\|So 1s 2, 2s 2, 2p 3-- 3 times as magnetic as carbon, which is more magnetic than boron.
BBqjKt0Qfd4-036\|In fact, that's how we can tell Hund's rule is being followed.
BBqjKt0Qfd4-037\|Nitrogen is more magnetic than carbon, which is more magnetic than boron.
BBqjKt0Qfd4-038\|So the spins are going in parallel.
BBqjKt0Qfd4-039\|We can continue with fluorine, neon.
BBqjKt0Qfd4-040\|And we see that the electrons start to pair in the 2s and 2p orbitals.
BBqjKt0Qfd4-041\|So neon now has 2p 6.
BBqjKt0Qfd4-042\|All the electrons are paired in the p shell.
BBqjKt0Qfd4-043\|Neon is not magnetic.
BBqjKt0Qfd4-044\|But sodium, the next element on the periodic table, will have one more electron.
BBqjKt0Qfd4-045\|And now we can start to use a shorthand in our electronic configurations.
BBqjKt0Qfd4-046\|We can say sodium has all the electrons that neon has plus a 3s 1.
BBqjKt0Qfd4-047\|And of course, an unpaired electron gives you a magnetic state.
BBqjKt0Qfd4-048\|So what we're doing is we're filling up the periodic table.
HBsY9ZOsWvY-000\|Multiple equilibria occur when there's steps in the chemical reaction or reactions can happen sequentially.
HBsY9ZOsWvY-002\|That has equilibrium constant K3 and delta H, delta H3.
HBsY9ZOsWvY-003\|So how are K1, K2, and K3 related?
HBsY9ZOsWvY-005\|Delta H3 is delta H1 plus delta H2.
HBsY9ZOsWvY-011\|That's the equilibrium constant K2.
HBsY9ZOsWvY-012\|And I have B over A. That's the equilibrium constant K1.
HBsY9ZOsWvY-013\|So K3 is actually K1 times K2.
HBsY9ZOsWvY-014\|So sequential reactions or sums of reactions, I multiply.
a2TpE4LATkc-000\|Let's look at a different kind of titration, a solubility titration.
a2TpE4LATkc-001\|I'm going to take 0.1 molar NaX solutions, where X is either chlorine, bromine, or iodine.
a2TpE4LATkc-002\|So sodium chloride, sodium bromide, and sodium iodide are all very soluble.
a2TpE4LATkc-003\|So I'll get a 100% dissociation.
a2TpE4LATkc-004\|And I'll have chloride ions, or bromide ions, or iodide ions at 0.1 molar in solution.
a2TpE4LATkc-005\|I'm going to add silver nitrate.
a2TpE4LATkc-006\|Now, silver chloride, silver iodide, and silver bromide are low soluble salts.
a2TpE4LATkc-007\|They have a very small Ksp value.
a2TpE4LATkc-008\|So silver chloride, silver bromide, and silver iodide will precipitate out of solution.
a2TpE4LATkc-009\|The question I have is, from this curve, can you tell which is the least soluble salt?
a2TpE4LATkc-019\|We're looking at the titration of sodium salt, sodium chloride, bromide, or iodide, with a silver solution.
a2TpE4LATkc-020\|And the silver salts, silver chloride, silver bromide, and silver iodide, are insoluble.
a2TpE4LATkc-021\|So they precipitate out of solution.
a2TpE4LATkc-023\|Well, the higher the pX, the lower the X concentration.
a2TpE4LATkc-024\|Remember, it's minus log of the concentration that gives you the pX, so minus log of X concentration.
a2TpE4LATkc-025\|So if the pX is high, then you have 10 to the minus high number versus 10 to the minus low number.
a2TpE4LATkc-026\|10 to the minus high number is a smaller number.
a2TpE4LATkc-027\|So this represents the smallest concentration of iodine.
DzmtbC7bYUI-000\|Let's look at some various kinetic rate laws and various plot to determine the characteristic plot for various rate laws.
DzmtbC7bYUI-001\|So for instance, zero order kinetics means the rate is independent of any concentration, so the rate is just a constant.
DzmtbC7bYUI-008\|Let's look at first order kinetics.
DzmtbC7bYUI-009\|First order kinetics, the rate is k times A to the first power.
DzmtbC7bYUI-010\|And if I write that versus time, that gives me a exponential.
DzmtbC7bYUI-011\|It's the concentration is e to the minus kt versus time.
DzmtbC7bYUI-018\|And I can write that versus time that concentration versus time goes as this expression here.
DzmtbC7bYUI-019\|And I have a half life that's dependent on 1 over the initial concentration.
DzmtbC7bYUI-024\|So when you find the linear plot, you determine the kinetics.
DzmtbC7bYUI-025\|So that's a summary of our characteristic kinetic plot and the parameters for zero order, first order, and second order kinetics.
1N6F0E1ReHk-000\|Let's look at the molecular orbitals in benzene.
1N6F0E1ReHk-008\|We're talking about the molecular orbitals in benzene.
1N6F0E1ReHk-009\|Now, benzene is one of these conjugated, alternating double-bond systems, where each carbon is SP2 hybridized.
1N6F0E1ReHk-010\|So it has an available p orbital to form extended molecular orbitals.
1N6F0E1ReHk-011\|The question is which combination is the highest energy.
1N6F0E1ReHk-012\|Again, we correlate the number of nodes with energy.
1N6F0E1ReHk-013\|As the number of nodes increase, the energy goes up.
1N6F0E1ReHk-014\|So A has a single node.
1N6F0E1ReHk-015\|B has one two three nodes-- and C, two nodes.
1N6F0E1ReHk-016\|So the highest energy orbital is the one with the most nodes-- in this case, B.
7A5UuW3EJPc-000\|Let's look at a gas phase equilibrium, the equilibrium between NO2, a brown gas, and N2O4, the dimer, which is a clear gas.
7A5UuW3EJPc-001\|They're in equilibrium here in this flask, and you can see the brown gas.
7A5UuW3EJPc-003\|So since the equilibrium constant is larger than 1, the products are slightly favored, in this case, over the reactants.
7A5UuW3EJPc-004\|So the products slightly favored.
7A5UuW3EJPc-005\|Even though those products are slightly favored, you can see there are reactants-- that's the brown gas that we can see-- are still present at equilibrium.
7A5UuW3EJPc-006\|There's an interchange between these two.
7A5UuW3EJPc-008\|The difference in free energy between N2O4 and NO2 at equilibrium is 0.
7A5UuW3EJPc-009\|There's no free energy penalty to switch-- interchange-- between products and reactants.
7A5UuW3EJPc-010\|And so they do so-- back and forth-- with no energy penalty.
7A5UuW3EJPc-011\|That's the nature of equilibrium.
7A5UuW3EJPc-012\|You can go between products and reactants because the free energy doesn't change as you do that.
7A5UuW3EJPc-013\|Delta G is 0 for equilibrium systems.

OlehBTSyPDA-107|That's just coincidental.

OlehBTSyPDA-108|That occurs at the midpoint of titrations, halfway between the start and the equivalence point.

OlehBTSyPDA-109|So where am I on my titration curve?

OlehBTSyPDA-111|My titration curve, I go for my weak acid to its conjugate base.

OlehBTSyPDA-112|And in this calculation, I've added enough to convert half of my initial acid into its conjugate base.

OlehBTSyPDA-113|I'm right smack in the middle here of this buffer region.

OlehBTSyPDA-114|I'm half way towards the equivalence point.

OlehBTSyPDA-116|That's just a feature of titration curves.

OlehBTSyPDA-117|It's interesting numerical equivalence.

OlehBTSyPDA-119|And that's numerically equal to the pKa.

BBqjKt0Qfd4-000|Now we understand how multiple electrons can exist around an atom.

BBqjKt0Qfd4-001|They go into the various orbitals, and they fill the various orbitals following some fairly specific rules.

BBqjKt0Qfd4-010|Here's the first few elements on the periodic table.

BBqjKt0Qfd4-011|Hydrogen has one electron.

BBqjKt0Qfd4-012|It will enter the 1s, the lowest available orbital.

BBqjKt0Qfd4-014|It's called a paramagnetic species.

BBqjKt0Qfd4-015|Hydrogen is paramagnetic in its atomic state.

BBqjKt0Qfd4-016|Helium has paired electrons, because the next electron will go in the 1s orbital paired up.

BBqjKt0Qfd4-017|The 2s orbital is higher in energy, so it's cheaper to spend that little energy to pair than to put the energy in the higher 2s orbital.

BBqjKt0Qfd4-025|So it's cheaper to put 2s paired than to put an electron in the higher energy 2p.

BBqjKt0Qfd4-026|So we're going to have 1s 2, 2s 2 for beryllium.

BBqjKt0Qfd4-027|And that will not be magnetic.

BBqjKt0Qfd4-029|Now, we can continue.

BBqjKt0Qfd4-030|Here's carbon and nitrogen, oxygen, fluorine.

BBqjKt0Qfd4-031|And you'll see that carbon has maximum multiplicity.

BBqjKt0Qfd4-032|The 2p orbitals are the same energy.

BBqjKt0Qfd4-033|So the spins go in parallel.

BBqjKt0Qfd4-034|Nitrogen, same thing-- three unpaired electrons.

BBqjKt0Qfd4-035|So 1s 2, 2s 2, 2p 3-- 3 times as magnetic as carbon, which is more magnetic than boron.

BBqjKt0Qfd4-036|In fact, that's how we can tell Hund's rule is being followed.

BBqjKt0Qfd4-037|Nitrogen is more magnetic than carbon, which is more magnetic than boron.

BBqjKt0Qfd4-038|So the spins are going in parallel.

BBqjKt0Qfd4-039|We can continue with fluorine, neon.

BBqjKt0Qfd4-040|And we see that the electrons start to pair in the 2s and 2p orbitals.

BBqjKt0Qfd4-041|So neon now has 2p 6.

BBqjKt0Qfd4-042|All the electrons are paired in the p shell.

BBqjKt0Qfd4-043|Neon is not magnetic.

BBqjKt0Qfd4-044|But sodium, the next element on the periodic table, will have one more electron.

BBqjKt0Qfd4-045|And now we can start to use a shorthand in our electronic configurations.

BBqjKt0Qfd4-046|We can say sodium has all the electrons that neon has plus a 3s 1.

BBqjKt0Qfd4-047|And of course, an unpaired electron gives you a magnetic state.

BBqjKt0Qfd4-048|So what we're doing is we're filling up the periodic table.

HBsY9ZOsWvY-000|Multiple equilibria occur when there's steps in the chemical reaction or reactions can happen sequentially.

HBsY9ZOsWvY-002|That has equilibrium constant K3 and delta H, delta H3.

HBsY9ZOsWvY-003|So how are K1, K2, and K3 related?

HBsY9ZOsWvY-005|Delta H3 is delta H1 plus delta H2.

HBsY9ZOsWvY-011|That's the equilibrium constant K2.

HBsY9ZOsWvY-012|And I have B over A. That's the equilibrium constant K1.

HBsY9ZOsWvY-013|So K3 is actually K1 times K2.

HBsY9ZOsWvY-014|So sequential reactions or sums of reactions, I multiply.

a2TpE4LATkc-000|Let's look at a different kind of titration, a solubility titration.

a2TpE4LATkc-001|I'm going to take 0.1 molar NaX solutions, where X is either chlorine, bromine, or iodine.

a2TpE4LATkc-002|So sodium chloride, sodium bromide, and sodium iodide are all very soluble.

a2TpE4LATkc-003|So I'll get a 100% dissociation.

a2TpE4LATkc-004|And I'll have chloride ions, or bromide ions, or iodide ions at 0.1 molar in solution.

a2TpE4LATkc-005|I'm going to add silver nitrate.

a2TpE4LATkc-006|Now, silver chloride, silver iodide, and silver bromide are low soluble salts.

a2TpE4LATkc-007|They have a very small Ksp value.

a2TpE4LATkc-008|So silver chloride, silver bromide, and silver iodide will precipitate out of solution.

a2TpE4LATkc-009|The question I have is, from this curve, can you tell which is the least soluble salt?

a2TpE4LATkc-019|We're looking at the titration of sodium salt, sodium chloride, bromide, or iodide, with a silver solution.

a2TpE4LATkc-020|And the silver salts, silver chloride, silver bromide, and silver iodide, are insoluble.

a2TpE4LATkc-021|So they precipitate out of solution.

a2TpE4LATkc-023|Well, the higher the pX, the lower the X concentration.

a2TpE4LATkc-024|Remember, it's minus log of the concentration that gives you the pX, so minus log of X concentration.

a2TpE4LATkc-025|So if the pX is high, then you have 10 to the minus high number versus 10 to the minus low number.

a2TpE4LATkc-026|10 to the minus high number is a smaller number.

a2TpE4LATkc-027|So this represents the smallest concentration of iodine.

DzmtbC7bYUI-000|Let's look at some various kinetic rate laws and various plot to determine the characteristic plot for various rate laws.

DzmtbC7bYUI-001|So for instance, zero order kinetics means the rate is independent of any concentration, so the rate is just a constant.

DzmtbC7bYUI-008|Let's look at first order kinetics.

DzmtbC7bYUI-009|First order kinetics, the rate is k times A to the first power.

DzmtbC7bYUI-010|And if I write that versus time, that gives me a exponential.

DzmtbC7bYUI-011|It's the concentration is e to the minus kt versus time.

DzmtbC7bYUI-018|And I can write that versus time that concentration versus time goes as this expression here.

DzmtbC7bYUI-019|And I have a half life that's dependent on 1 over the initial concentration.

DzmtbC7bYUI-024|So when you find the linear plot, you determine the kinetics.

DzmtbC7bYUI-025|So that's a summary of our characteristic kinetic plot and the parameters for zero order, first order, and second order kinetics.

1N6F0E1ReHk-000|Let's look at the molecular orbitals in benzene.

1N6F0E1ReHk-008|We're talking about the molecular orbitals in benzene.

1N6F0E1ReHk-009|Now, benzene is one of these conjugated, alternating double-bond systems, where each carbon is SP2 hybridized.

1N6F0E1ReHk-010|So it has an available p orbital to form extended molecular orbitals.

1N6F0E1ReHk-011|The question is which combination is the highest energy.

1N6F0E1ReHk-012|Again, we correlate the number of nodes with energy.

1N6F0E1ReHk-013|As the number of nodes increase, the energy goes up.

1N6F0E1ReHk-014|So A has a single node.

1N6F0E1ReHk-015|B has one two three nodes-- and C, two nodes.

1N6F0E1ReHk-016|So the highest energy orbital is the one with the most nodes-- in this case, B.

7A5UuW3EJPc-000|Let's look at a gas phase equilibrium, the equilibrium between NO2, a brown gas, and N2O4, the dimer, which is a clear gas.

7A5UuW3EJPc-001|They're in equilibrium here in this flask, and you can see the brown gas.

7A5UuW3EJPc-003|So since the equilibrium constant is larger than 1, the products are slightly favored, in this case, over the reactants.

7A5UuW3EJPc-004|So the products slightly favored.

7A5UuW3EJPc-005|Even though those products are slightly favored, you can see there are reactants-- that's the brown gas that we can see-- are still present at equilibrium.

7A5UuW3EJPc-006|There's an interchange between these two.

7A5UuW3EJPc-008|The difference in free energy between N2O4 and NO2 at equilibrium is 0.

7A5UuW3EJPc-009|There's no free energy penalty to switch-- interchange-- between products and reactants.

7A5UuW3EJPc-010|And so they do so-- back and forth-- with no energy penalty.

7A5UuW3EJPc-011|That's the nature of equilibrium.

7A5UuW3EJPc-012|You can go between products and reactants because the free energy doesn't change as you do that.

7A5UuW3EJPc-013|Delta G is 0 for equilibrium systems.