ceaglex/VisualTTS_Chem · Datasets at Hugging Face

text stringlengths 19 206
9sDdIaBhtgk-022\|And my weak base solution.
9sDdIaBhtgk-023\|So the base turns this indicator blue.
9sDdIaBhtgk-024\|That's a nice little mnemonic to remember, because the bases are blue, both starting with the same letter.
9sDdIaBhtgk-026\|So that indicator in the strong acid, weak acid, water, weak base, and strong base solution.
9sDdIaBhtgk-027\|Now, the number of ions in each of these solutions is indicative of the acid and base strength, as well.
9sDdIaBhtgk-028\|HCl, remember, completely disassociates.
9sDdIaBhtgk-029\|So 0.1 molar HCl makes 0.1 molar H plus ions and 0.1 molar Cl minus ions.
9sDdIaBhtgk-030\|So I have 0.2 molar total ion concentration.
9sDdIaBhtgk-031\|And that ion concentration will make the solution conductive.
9sDdIaBhtgk-032\|So if I test the conductivity, just kind of qualitatively with this light bulb, I'll see that HCl, there's ions in there.
9sDdIaBhtgk-033\|Ions conduct, and I'll have a strongly conductive solution.
9sDdIaBhtgk-035\|And in general, water, pure, is not a conductive solution.
9sDdIaBhtgk-036\|So let's look at the weak acid.
9sDdIaBhtgk-037\|Now, the weak acid dissociates much less strongly than the strong acid.
9sDdIaBhtgk-039\|We should see the same correlation between the weak base and the strong base.
9sDdIaBhtgk-045\|A correlation between pH, color, and ionic strength for acid and base solutions.
hTimaCA_A88-000\|Let's look at the phase diagram of carbon dioxide and see if we can predict where the sublimation of carbon dioxide occurs at 1 atmosphere.
hTimaCA_A88-004\|We're trying to find the sublimation line, a sublimation at constant pressure, transition for carbon dioxide.
hTimaCA_A88-005\|Well, if it's a constant pressure transition, it has to be a horizontal line.
hTimaCA_A88-006\|And if it's a sublimation, it must cross the solid/gas transition line.
hTimaCA_A88-007\|Sublimation is the conversion from solid to gas.
hTimaCA_A88-009\|The correct answer here is A.
--imKPteSwQ-000\|When I burn the hydrocarbon butane, heat is released.
--imKPteSwQ-009\|We're talking about burning butane in oxygen, and in each case oxygen is in excess.
--imKPteSwQ-010\|So butane limits the amount of energy released.
--imKPteSwQ-014\|But the speed of the reaction and the overall energy release are independent.
--imKPteSwQ-015\|I can release a lot of energy over a long time, or release a small amount of energy quickly.
-dpZipe4Hm0-000\|Let's look at some of the molecular orbitals involved in forming multiple bonds.
-dpZipe4Hm0-001\|So when we have a carbon that's sp2 hybridized-- and I've drawn one here-- I have three equivalent sp2 orbitals.
-dpZipe4Hm0-002\|And I have a model that looks like that here.
-dpZipe4Hm0-003\|The three bonds here at 120 degrees from each other are sp2 hybrid orbitals.
-dpZipe4Hm0-004\|Now, sp2 hybrids leave behind one p.
-dpZipe4Hm0-005\|Sp2 hybrids are formed from an s and two p's.
-dpZipe4Hm0-006\|So there's an unchanged p orbital that I represented here by this plus minus lobe.
-dpZipe4Hm0-012\|I'll form a sigma bond between the two carbons.
-dpZipe4Hm0-014\|Those two atomic hybrid orbitals will form a sigma bonding and antibonding molecular orbital.
-dpZipe4Hm0-015\|So two atomic orbitals, both of them sp2, form two molecular orbitals, a sigma bonding and antibonding orbital.
-dpZipe4Hm0-016\|Now, the p orbital left over can overlap to form a pi bond, a second bond.
-dpZipe4Hm0-017\|And remember, pi bonds are bonds that are above and below the internuclear axis, usually formed by p orbitals.
-dpZipe4Hm0-019\|There.
-dpZipe4Hm0-020\|There is the sp sigma bond.
-dpZipe4Hm0-021\|And now the p orbitals can overlap and form a bond above and below the axis, a second bond.
-dpZipe4Hm0-022\|So I have a sigma and a pi bond.
-dpZipe4Hm0-023\|The first bond formed between atoms is always sigma.
-dpZipe4Hm0-024\|Other bonds are pi.
-dpZipe4Hm0-030\|So multiple bonds, a sigma and a pi bond, formed from a combination of hybrid orbitals and leftover p orbitals on carbon.
oTCKWNiWdus-000\|Just like the entropy and the enthalpy, the free energy is a state function.
oTCKWNiWdus-002\|Now, here's a table of standard free energies of formations.
oTCKWNiWdus-004\|That's because it wouldn't be fair to take the reactants at, say, two or three atmospheres of pressure and compare that to the products at half an atmosphere of pressure.
oTCKWNiWdus-005\|We compare everything across the board in a standard state of conditions.
oTCKWNiWdus-006\|So in a sense, it makes it a little artificial.
oTCKWNiWdus-008\|And that's not necessarily how you set it up in the laboratory, but it'll still help you determine whether overall the products or the reactants are favored.
oTCKWNiWdus-009\|If delta g is overall negative, then the products are favored.
oTCKWNiWdus-010\|So here we have the tables now of enthalpy, entropy, and free energy.
oTCKWNiWdus-011\|So we can use these to calculate free energies, standard state enthalpies, and standard entropy differences.
ma-MKQ2TAGk-000\|Our bodies use the metabolism of glucose as an energy source.
ma-MKQ2TAGk-002\|Now, that oxidation that's down hill in free energy, and that's coupled to an uphill in free energy reaction.
ma-MKQ2TAGk-003\|The phosphorylation, the addition of a phosphate group, to adenosine diphosphate to form adenosine triphosphate.
ma-MKQ2TAGk-007\|So ATP hydrolysis from ATP to ADP, that releases energy.
ma-MKQ2TAGk-008\|And that can be coupled with building something in your cells.
ma-MKQ2TAGk-009\|So for instance, taking free amino acids and making them into proteins.
ma-MKQ2TAGk-010\|So free amino acids are small individual molecules.
ma-MKQ2TAGk-011\|They're linked together to form proteins.
ma-MKQ2TAGk-012\|And proteins are molecules that are catalysts in our body and they're structural in your body.
ma-MKQ2TAGk-013\|They're a very important molecule in your body.
ma-MKQ2TAGk-015\|For protein synthesis, it turns out that about four moles of ATP are hydrolyzed for every peptide bond that's formed.
ma-MKQ2TAGk-016\|Now, the peptide bond is the bond between the amino acids.
ma-MKQ2TAGk-017\|So as you build up a protein, you add amino acids.
ma-MKQ2TAGk-018\|For each amino acid you add, you need to form a peptide bond.
ma-MKQ2TAGk-019\|And for each peptide bond you form, you hydrolyze about four moles of ATP.
ma-MKQ2TAGk-020\|So that's an overall accounting of how energy is used in your cells.
cLXnZlwrU04-000\|Let's look at a gas phase equilibrium.
cLXnZlwrU04-001\|Here's the reaction of nitrogen and hydrogen to form ammonia.
cLXnZlwrU04-002\|Now, this reaction is important industrially, because nitrogen, although it's abundant-- there's 70% of it in the air we breathe-- it's unreactive.
cLXnZlwrU04-003\|There's a triple bond in N2.
cLXnZlwrU04-004\|It's an unreactive molecule.
cLXnZlwrU04-005\|But nitrogen is required for all life.
cLXnZlwrU04-006\|It's in our proteins and DNA.
cLXnZlwrU04-007\|So getting nitrogen from this unreactive form to a reactive form is important in both nature and industry.
cLXnZlwrU04-008\|In industry, to make fertilizers, and in nature, to fix the nitrogen so plants and animals can grow.
cLXnZlwrU04-009\|This nitrogen fixation reaction is exothermic.
cLXnZlwrU04-010\|It releases 92 kilojoules per mole of nitrogen that reacts.
cLXnZlwrU04-014\|Raising the temperature increases the motion of the molecules and accelerates the kinetics of the reaction.
cLXnZlwrU04-015\|But you're fighting thermodynamics, because as you raise the temperature, an exothermic chemical reaction will tend to favor the reactants.
cLXnZlwrU04-016\|So K decreases.
cLXnZlwrU04-017\|So you start to favor reactants as the temperature increases.
cLXnZlwrU04-018\|So you want to fight that, and one way to do it is to raise the pressure.
cLXnZlwrU04-019\|Le Chatelier's principle tells you if you raise the pressure, you're going to favor the ammonia side.
cLXnZlwrU04-020\|And why is that?
cLXnZlwrU04-021\|Well, raising the pressure, you have 4 moles of gas on this side and 2 moles of gas on this side.
cLXnZlwrU04-022\|So the product side is favored.
cLXnZlwrU04-023\|It would shift towards products to relieve that high pressure situation, shift towards the fewer molecules.
cLXnZlwrU04-026\|That will efficiently balance the kinetics, the speed of the reaction, and the thermodynamics that are tending to make it favor reactants.
6DGPuhFoiJI-000\|Let's take a week acid, acetic acid, originally at pH 3, dilute it by a factor of 10, and ask what is the new pH?
6DGPuhFoiJI-001\|So acetic acid, pH 3, dilute with water a factor of 10.
6DGPuhFoiJI-012\|Well, the pH is 3, so the H3O plus concentration is 10 to the minus 3.
6DGPuhFoiJI-016\|This reaction favors the reactants.
6DGPuhFoiJI-017\|So to get a product concentration of 10 to the minus 3, we have to have a much higher reactant concentration.
6DGPuhFoiJI-018\|We think that's to be the case.
6DGPuhFoiJI-019\|Now we dilute it.

9sDdIaBhtgk-022|And my weak base solution.

9sDdIaBhtgk-023|So the base turns this indicator blue.

9sDdIaBhtgk-024|That's a nice little mnemonic to remember, because the bases are blue, both starting with the same letter.

9sDdIaBhtgk-026|So that indicator in the strong acid, weak acid, water, weak base, and strong base solution.

9sDdIaBhtgk-027|Now, the number of ions in each of these solutions is indicative of the acid and base strength, as well.

9sDdIaBhtgk-028|HCl, remember, completely disassociates.

9sDdIaBhtgk-029|So 0.1 molar HCl makes 0.1 molar H plus ions and 0.1 molar Cl minus ions.

9sDdIaBhtgk-030|So I have 0.2 molar total ion concentration.

9sDdIaBhtgk-031|And that ion concentration will make the solution conductive.

9sDdIaBhtgk-032|So if I test the conductivity, just kind of qualitatively with this light bulb, I'll see that HCl, there's ions in there.

9sDdIaBhtgk-033|Ions conduct, and I'll have a strongly conductive solution.

9sDdIaBhtgk-035|And in general, water, pure, is not a conductive solution.

9sDdIaBhtgk-036|So let's look at the weak acid.

9sDdIaBhtgk-037|Now, the weak acid dissociates much less strongly than the strong acid.

9sDdIaBhtgk-039|We should see the same correlation between the weak base and the strong base.

9sDdIaBhtgk-045|A correlation between pH, color, and ionic strength for acid and base solutions.

hTimaCA_A88-000|Let's look at the phase diagram of carbon dioxide and see if we can predict where the sublimation of carbon dioxide occurs at 1 atmosphere.

hTimaCA_A88-004|We're trying to find the sublimation line, a sublimation at constant pressure, transition for carbon dioxide.

hTimaCA_A88-005|Well, if it's a constant pressure transition, it has to be a horizontal line.

hTimaCA_A88-006|And if it's a sublimation, it must cross the solid/gas transition line.

hTimaCA_A88-007|Sublimation is the conversion from solid to gas.

hTimaCA_A88-009|The correct answer here is A.

--imKPteSwQ-000|When I burn the hydrocarbon butane, heat is released.

--imKPteSwQ-009|We're talking about burning butane in oxygen, and in each case oxygen is in excess.

--imKPteSwQ-010|So butane limits the amount of energy released.

--imKPteSwQ-014|But the speed of the reaction and the overall energy release are independent.

--imKPteSwQ-015|I can release a lot of energy over a long time, or release a small amount of energy quickly.

-dpZipe4Hm0-000|Let's look at some of the molecular orbitals involved in forming multiple bonds.

-dpZipe4Hm0-001|So when we have a carbon that's sp2 hybridized-- and I've drawn one here-- I have three equivalent sp2 orbitals.

-dpZipe4Hm0-002|And I have a model that looks like that here.

-dpZipe4Hm0-003|The three bonds here at 120 degrees from each other are sp2 hybrid orbitals.

-dpZipe4Hm0-004|Now, sp2 hybrids leave behind one p.

-dpZipe4Hm0-005|Sp2 hybrids are formed from an s and two p's.

-dpZipe4Hm0-006|So there's an unchanged p orbital that I represented here by this plus minus lobe.

-dpZipe4Hm0-012|I'll form a sigma bond between the two carbons.

-dpZipe4Hm0-014|Those two atomic hybrid orbitals will form a sigma bonding and antibonding molecular orbital.

-dpZipe4Hm0-015|So two atomic orbitals, both of them sp2, form two molecular orbitals, a sigma bonding and antibonding orbital.

-dpZipe4Hm0-016|Now, the p orbital left over can overlap to form a pi bond, a second bond.

-dpZipe4Hm0-017|And remember, pi bonds are bonds that are above and below the internuclear axis, usually formed by p orbitals.

-dpZipe4Hm0-019|There.

-dpZipe4Hm0-020|There is the sp sigma bond.

-dpZipe4Hm0-021|And now the p orbitals can overlap and form a bond above and below the axis, a second bond.

-dpZipe4Hm0-022|So I have a sigma and a pi bond.

-dpZipe4Hm0-023|The first bond formed between atoms is always sigma.

-dpZipe4Hm0-024|Other bonds are pi.

-dpZipe4Hm0-030|So multiple bonds, a sigma and a pi bond, formed from a combination of hybrid orbitals and leftover p orbitals on carbon.

oTCKWNiWdus-000|Just like the entropy and the enthalpy, the free energy is a state function.

oTCKWNiWdus-002|Now, here's a table of standard free energies of formations.

oTCKWNiWdus-004|That's because it wouldn't be fair to take the reactants at, say, two or three atmospheres of pressure and compare that to the products at half an atmosphere of pressure.

oTCKWNiWdus-005|We compare everything across the board in a standard state of conditions.

oTCKWNiWdus-006|So in a sense, it makes it a little artificial.

oTCKWNiWdus-008|And that's not necessarily how you set it up in the laboratory, but it'll still help you determine whether overall the products or the reactants are favored.

oTCKWNiWdus-009|If delta g is overall negative, then the products are favored.

oTCKWNiWdus-010|So here we have the tables now of enthalpy, entropy, and free energy.

oTCKWNiWdus-011|So we can use these to calculate free energies, standard state enthalpies, and standard entropy differences.

ma-MKQ2TAGk-000|Our bodies use the metabolism of glucose as an energy source.

ma-MKQ2TAGk-002|Now, that oxidation that's down hill in free energy, and that's coupled to an uphill in free energy reaction.

ma-MKQ2TAGk-003|The phosphorylation, the addition of a phosphate group, to adenosine diphosphate to form adenosine triphosphate.

ma-MKQ2TAGk-007|So ATP hydrolysis from ATP to ADP, that releases energy.

ma-MKQ2TAGk-008|And that can be coupled with building something in your cells.

ma-MKQ2TAGk-009|So for instance, taking free amino acids and making them into proteins.

ma-MKQ2TAGk-010|So free amino acids are small individual molecules.

ma-MKQ2TAGk-011|They're linked together to form proteins.

ma-MKQ2TAGk-012|And proteins are molecules that are catalysts in our body and they're structural in your body.

ma-MKQ2TAGk-013|They're a very important molecule in your body.

ma-MKQ2TAGk-015|For protein synthesis, it turns out that about four moles of ATP are hydrolyzed for every peptide bond that's formed.

ma-MKQ2TAGk-016|Now, the peptide bond is the bond between the amino acids.

ma-MKQ2TAGk-017|So as you build up a protein, you add amino acids.

ma-MKQ2TAGk-018|For each amino acid you add, you need to form a peptide bond.

ma-MKQ2TAGk-019|And for each peptide bond you form, you hydrolyze about four moles of ATP.

ma-MKQ2TAGk-020|So that's an overall accounting of how energy is used in your cells.

cLXnZlwrU04-000|Let's look at a gas phase equilibrium.

cLXnZlwrU04-001|Here's the reaction of nitrogen and hydrogen to form ammonia.

cLXnZlwrU04-002|Now, this reaction is important industrially, because nitrogen, although it's abundant-- there's 70% of it in the air we breathe-- it's unreactive.

cLXnZlwrU04-003|There's a triple bond in N2.

cLXnZlwrU04-004|It's an unreactive molecule.

cLXnZlwrU04-005|But nitrogen is required for all life.

cLXnZlwrU04-006|It's in our proteins and DNA.

cLXnZlwrU04-007|So getting nitrogen from this unreactive form to a reactive form is important in both nature and industry.

cLXnZlwrU04-008|In industry, to make fertilizers, and in nature, to fix the nitrogen so plants and animals can grow.

cLXnZlwrU04-009|This nitrogen fixation reaction is exothermic.

cLXnZlwrU04-010|It releases 92 kilojoules per mole of nitrogen that reacts.

cLXnZlwrU04-014|Raising the temperature increases the motion of the molecules and accelerates the kinetics of the reaction.

cLXnZlwrU04-015|But you're fighting thermodynamics, because as you raise the temperature, an exothermic chemical reaction will tend to favor the reactants.

cLXnZlwrU04-016|So K decreases.

cLXnZlwrU04-017|So you start to favor reactants as the temperature increases.

cLXnZlwrU04-018|So you want to fight that, and one way to do it is to raise the pressure.

cLXnZlwrU04-019|Le Chatelier's principle tells you if you raise the pressure, you're going to favor the ammonia side.

cLXnZlwrU04-020|And why is that?

cLXnZlwrU04-021|Well, raising the pressure, you have 4 moles of gas on this side and 2 moles of gas on this side.

cLXnZlwrU04-022|So the product side is favored.

cLXnZlwrU04-023|It would shift towards products to relieve that high pressure situation, shift towards the fewer molecules.

cLXnZlwrU04-026|That will efficiently balance the kinetics, the speed of the reaction, and the thermodynamics that are tending to make it favor reactants.

6DGPuhFoiJI-000|Let's take a week acid, acetic acid, originally at pH 3, dilute it by a factor of 10, and ask what is the new pH?

6DGPuhFoiJI-001|So acetic acid, pH 3, dilute with water a factor of 10.

6DGPuhFoiJI-012|Well, the pH is 3, so the H3O plus concentration is 10 to the minus 3.

6DGPuhFoiJI-016|This reaction favors the reactants.

6DGPuhFoiJI-017|So to get a product concentration of 10 to the minus 3, we have to have a much higher reactant concentration.

6DGPuhFoiJI-018|We think that's to be the case.

6DGPuhFoiJI-019|Now we dilute it.