ceaglex/VisualTTS_Chem · Datasets at Hugging Face

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9z0sMWVIm7A-020\|Well, in the reduction, the 2 chromium here, only 1 there, so I need a 2.
9z0sMWVIm7A-021\|And in fact, I'm just going to go through and apply the steps to the chromate first, and then come back and do it for the alcohol and aldehyde.
9z0sMWVIm7A-024\|Now I have to balance the hydrogens by adding H plus.
9z0sMWVIm7A-030\|So six pluses here and 2 times plus 3, 6 pluses there.
9z0sMWVIm7A-031\|Now, let's do the same thing for the alcohol to aldehyde, balance the elements.
9z0sMWVIm7A-032\|All the elements are balanced here, so nothing required there.
9z0sMWVIm7A-033\|Balance the oxygen with water.
9z0sMWVIm7A-034\|I have 1 oxygen on both sides, so nothing required there.
9z0sMWVIm7A-035\|Balance the hydrogens by adding H plus.
9z0sMWVIm7A-039\|Now I can bring my half reactions, now that they're balanced back together, and balance the electrons in both half cells.
9z0sMWVIm7A-040\|So this half cell, I noticed 2 electrons and I noticed that's an oxidation.
9z0sMWVIm7A-041\|6 electrons here and that's a reduction, that's good.
9z0sMWVIm7A-042\|That's what I thought was happening and my procedure has confirmed that.
9z0sMWVIm7A-043\|So, let's balance them out.
9z0sMWVIm7A-046\|I'll add the two reactions together with my factor of 3.
9z0sMWVIm7A-050\|And you can see now if you count up all the hydrogens, all the oxygens, all the chromium and carbons on this side, they exactly match up with that side.
9z0sMWVIm7A-052\|That's how we balance redox reactions.
11QPT4cbBO8-000\|Chemistry is the study of atoms, molecules, and their interactions.
11QPT4cbBO8-001\|Now, virtually every process that happens in the universe involves molecules, atoms, and their interchange.
11QPT4cbBO8-005\|So this reaction, the formation of water from hydrogen and oxygen, is favorable.
11QPT4cbBO8-006\|In fact, when that reaction occurs, energy is released.
11QPT4cbBO8-007\|It can be very dramatic.
11QPT4cbBO8-008\|So in a sense, the products, water, are lower energy than the reactants.
11QPT4cbBO8-009\|And we can plot that on a relative scale.
11QPT4cbBO8-010\|And we'll do this a lot in chemistry.
11QPT4cbBO8-011\|We'll plot energy on a vertical scale and the progress of a reaction along the horizontal scale.
11QPT4cbBO8-013\|Now, in a mechanical example, you'd say, well, if they were a ball on the top of this hill, it would naturally roll down to the bottom of the hill.
11QPT4cbBO8-014\|That's the natural way of things.
11QPT4cbBO8-015\|And it's the same for this chemical reaction.
11QPT4cbBO8-016\|The natural way is for hydrogen and oxygen to form water.
11QPT4cbBO8-017\|There's a natural favorable direction of this chemical reaction.
11QPT4cbBO8-018\|But there's one more interesting part.
11QPT4cbBO8-019\|And that is, if I take oxygen and hydrogen, and I mix them together in a balloon, I can wait quite a while and never find water.
11QPT4cbBO8-020\|I can come back and look at that balloon the next day, the next week, six months later, and it's still going to be a hydrogen oxygen balloon.
11QPT4cbBO8-021\|You won't find a balloon full of water.
11QPT4cbBO8-022\|And why is that?
11QPT4cbBO8-024\|That pulling and stretching and rearranging of bonds requires energy.
11QPT4cbBO8-025\|So there is a high energy intermediate state between the reactants and products.
11QPT4cbBO8-027\|And you might say, well, if there's that high road barrier, why do they ever react?
11QPT4cbBO8-028\|Well, let's go back to the mechanical example.
11QPT4cbBO8-029\|If I had a ball here, could I get it to the other side of that hill?
11QPT4cbBO8-031\|One way I could get it was to flick it-- give it some kinetic energy, and have it roll up over the hill.
11QPT4cbBO8-033\|So it's possible in the mechanical example.
11QPT4cbBO8-034\|Is it possible in the chemical example?
11QPT4cbBO8-035\|Well, it is.
11QPT4cbBO8-036\|I can't go in and flick individual atoms or molecules of hydrogen and oxygen, but I can get them to move faster, give them kinetic energy.
11QPT4cbBO8-037\|I can heat them up.
11QPT4cbBO8-038\|And throughout this course, we'll associate an increase in temperature with more molecular motion.
11QPT4cbBO8-039\|So I can get reactants to have enough energy that some of them will react and go over this hill.
11QPT4cbBO8-040\|And then it can become cooperative.
11QPT4cbBO8-041\|They'll release some energy.
11QPT4cbBO8-042\|they'll excite others in the mixture, and they'll all start to fall over that hill.
11QPT4cbBO8-049\|So both those would be interesting.
11QPT4cbBO8-050\|And in fact, a lot of these chemical reactions release energy, and they're fun to watch.
11QPT4cbBO8-051\|And we'll watch a lot of chemical reactions through this series.
11QPT4cbBO8-052\|The hydrogen oxygen reaction is one that I find most interesting, and I think you will, too.
XIJet3gIKRc-000\|Let's do a calculation involving a weak acid, its pKa, and determining the relative concentrations of the acid and base form.
XIJet3gIKRc-001\|So aspirin, acetylsalicylic acid has Ka, 3.2 times 10 to the minus 4.
XIJet3gIKRc-002\|A typical weak acid.
XIJet3gIKRc-003\|Now I can put it in solution and then adjust the pH to 4.13.
XIJet3gIKRc-008\|We'll just tell you what the acidic proton is unless it's very obvious.
XIJet3gIKRc-009\|So here's the acidic proton.
XIJet3gIKRc-010\|I know the acid dissociation constant for the reaction where this proton leaves the molecule is 3.2 times 10 to the minus 4.
XIJet3gIKRc-013\|I'm at a pH very close to that, really, within half a unit.
XIJet3gIKRc-017\|About equal concentrations.
XIJet3gIKRc-018\|I'm a little bit above, I'm a little bit to the basic side, of the pKa so that means the basic form will slightly predominate.
XIJet3gIKRc-019\|Not a factor of 10.
XIJet3gIKRc-021\|And we're not one full unit.
XIJet3gIKRc-022\|We'd have to go up to pH 4.49 to get a tenfold difference.
XIJet3gIKRc-023\|So we expect something less than tenfold difference.
XIJet3gIKRc-024\|Let's actually do the calculation.
XIJet3gIKRc-025\|So here's the acid and the base form, pKa 3.49.
XIJet3gIKRc-027\|In this case, we know these two, so we can solve for this ratio.
XIJet3gIKRc-028\|So let's just do that.
XIJet3gIKRc-033\|So the ratio of base form to acid form is 4.36.
XIJet3gIKRc-034\|There's about four times as much base form as there is acid form.
8y5KX4kzt0A-000\|Let's talk about the isothermal expansion and compression for an ideal gas.
8y5KX4kzt0A-001\|Isothermal means the temperature doesn't change.
8y5KX4kzt0A-002\|And if the temperature doesn't change for an ideal gas, the energy and the enthalpy don't change.
8y5KX4kzt0A-003\|Remember, the energy of an ideal gas, 3/2 nRT.
8y5KX4kzt0A-004\|So isothermal processes delta-E and delta-H are zero.
8y5KX4kzt0A-005\|So how can heat flow or work be done if the energy doesn't change?
8y5KX4kzt0A-006\|Well, the work and heat just have to balance each other with opposite sign.
8y5KX4kzt0A-007\|If you do a joule of work on the system, the system has to lose a joule of heat, joule for joule.
8y5KX4kzt0A-008\|Do a joule of work, lose a joule of heat.
8y5KX4kzt0A-009\|Do a joule of work, lose a joule of heat.
8y5KX4kzt0A-010\|If I expand, the system does work.
8y5KX4kzt0A-011\|So I do a joule of work, that would-- well, that's some of my energy, I need more energy to replace the energy I just used to do that work.
8y5KX4kzt0A-012\|Well, absorb a joule of heat.
8y5KX4kzt0A-013\|Do a joule of work, absorb a joule of heat.
8y5KX4kzt0A-020\|That's expansion and contraction isothermally for an ideal gas.
VcGOL6Uthmw-000\|Let's look at a couple of titrations and see if we can predict the equivalence point, or the end point, pH.
VcGOL6Uthmw-004\|So two acids and a base.
VcGOL6Uthmw-005\|We'll titrate them appropriately.
VcGOL6Uthmw-006\|If they're acids, we'll titrate them with a base.
VcGOL6Uthmw-007\|If it's a base, we'll titrate it with an acid.
VcGOL6Uthmw-008\|The question is at equivalence point, which has a pH less than 7?
VcGOL6Uthmw-018\|We're looking at the titration of some weak acid and base solutions.
VcGOL6Uthmw-019\|Now, when you titrate a weak acid or a weak base with a strong base or a strong acid, the titration curve looks something like this.
VcGOL6Uthmw-020\|We have the acid being converted to its conjugate base.

9z0sMWVIm7A-020|Well, in the reduction, the 2 chromium here, only 1 there, so I need a 2.

9z0sMWVIm7A-021|And in fact, I'm just going to go through and apply the steps to the chromate first, and then come back and do it for the alcohol and aldehyde.

9z0sMWVIm7A-024|Now I have to balance the hydrogens by adding H plus.

9z0sMWVIm7A-030|So six pluses here and 2 times plus 3, 6 pluses there.

9z0sMWVIm7A-031|Now, let's do the same thing for the alcohol to aldehyde, balance the elements.

9z0sMWVIm7A-032|All the elements are balanced here, so nothing required there.

9z0sMWVIm7A-033|Balance the oxygen with water.

9z0sMWVIm7A-034|I have 1 oxygen on both sides, so nothing required there.

9z0sMWVIm7A-035|Balance the hydrogens by adding H plus.

9z0sMWVIm7A-039|Now I can bring my half reactions, now that they're balanced back together, and balance the electrons in both half cells.

9z0sMWVIm7A-040|So this half cell, I noticed 2 electrons and I noticed that's an oxidation.

9z0sMWVIm7A-041|6 electrons here and that's a reduction, that's good.

9z0sMWVIm7A-042|That's what I thought was happening and my procedure has confirmed that.

9z0sMWVIm7A-043|So, let's balance them out.

9z0sMWVIm7A-046|I'll add the two reactions together with my factor of 3.

9z0sMWVIm7A-050|And you can see now if you count up all the hydrogens, all the oxygens, all the chromium and carbons on this side, they exactly match up with that side.

9z0sMWVIm7A-052|That's how we balance redox reactions.

11QPT4cbBO8-000|Chemistry is the study of atoms, molecules, and their interactions.

11QPT4cbBO8-001|Now, virtually every process that happens in the universe involves molecules, atoms, and their interchange.

11QPT4cbBO8-005|So this reaction, the formation of water from hydrogen and oxygen, is favorable.

11QPT4cbBO8-006|In fact, when that reaction occurs, energy is released.

11QPT4cbBO8-007|It can be very dramatic.

11QPT4cbBO8-008|So in a sense, the products, water, are lower energy than the reactants.

11QPT4cbBO8-009|And we can plot that on a relative scale.

11QPT4cbBO8-010|And we'll do this a lot in chemistry.

11QPT4cbBO8-011|We'll plot energy on a vertical scale and the progress of a reaction along the horizontal scale.

11QPT4cbBO8-013|Now, in a mechanical example, you'd say, well, if they were a ball on the top of this hill, it would naturally roll down to the bottom of the hill.

11QPT4cbBO8-014|That's the natural way of things.

11QPT4cbBO8-015|And it's the same for this chemical reaction.

11QPT4cbBO8-016|The natural way is for hydrogen and oxygen to form water.

11QPT4cbBO8-017|There's a natural favorable direction of this chemical reaction.

11QPT4cbBO8-018|But there's one more interesting part.

11QPT4cbBO8-019|And that is, if I take oxygen and hydrogen, and I mix them together in a balloon, I can wait quite a while and never find water.

11QPT4cbBO8-020|I can come back and look at that balloon the next day, the next week, six months later, and it's still going to be a hydrogen oxygen balloon.

11QPT4cbBO8-021|You won't find a balloon full of water.

11QPT4cbBO8-022|And why is that?

11QPT4cbBO8-024|That pulling and stretching and rearranging of bonds requires energy.

11QPT4cbBO8-025|So there is a high energy intermediate state between the reactants and products.

11QPT4cbBO8-027|And you might say, well, if there's that high road barrier, why do they ever react?

11QPT4cbBO8-028|Well, let's go back to the mechanical example.

11QPT4cbBO8-029|If I had a ball here, could I get it to the other side of that hill?

11QPT4cbBO8-031|One way I could get it was to flick it-- give it some kinetic energy, and have it roll up over the hill.

11QPT4cbBO8-033|So it's possible in the mechanical example.

11QPT4cbBO8-034|Is it possible in the chemical example?

11QPT4cbBO8-035|Well, it is.

11QPT4cbBO8-036|I can't go in and flick individual atoms or molecules of hydrogen and oxygen, but I can get them to move faster, give them kinetic energy.

11QPT4cbBO8-037|I can heat them up.

11QPT4cbBO8-038|And throughout this course, we'll associate an increase in temperature with more molecular motion.

11QPT4cbBO8-039|So I can get reactants to have enough energy that some of them will react and go over this hill.

11QPT4cbBO8-040|And then it can become cooperative.

11QPT4cbBO8-041|They'll release some energy.

11QPT4cbBO8-042|they'll excite others in the mixture, and they'll all start to fall over that hill.

11QPT4cbBO8-049|So both those would be interesting.

11QPT4cbBO8-050|And in fact, a lot of these chemical reactions release energy, and they're fun to watch.

11QPT4cbBO8-051|And we'll watch a lot of chemical reactions through this series.

11QPT4cbBO8-052|The hydrogen oxygen reaction is one that I find most interesting, and I think you will, too.

XIJet3gIKRc-000|Let's do a calculation involving a weak acid, its pKa, and determining the relative concentrations of the acid and base form.

XIJet3gIKRc-001|So aspirin, acetylsalicylic acid has Ka, 3.2 times 10 to the minus 4.

XIJet3gIKRc-002|A typical weak acid.

XIJet3gIKRc-003|Now I can put it in solution and then adjust the pH to 4.13.

XIJet3gIKRc-008|We'll just tell you what the acidic proton is unless it's very obvious.

XIJet3gIKRc-009|So here's the acidic proton.

XIJet3gIKRc-010|I know the acid dissociation constant for the reaction where this proton leaves the molecule is 3.2 times 10 to the minus 4.

XIJet3gIKRc-013|I'm at a pH very close to that, really, within half a unit.

XIJet3gIKRc-017|About equal concentrations.

XIJet3gIKRc-018|I'm a little bit above, I'm a little bit to the basic side, of the pKa so that means the basic form will slightly predominate.

XIJet3gIKRc-019|Not a factor of 10.

XIJet3gIKRc-021|And we're not one full unit.

XIJet3gIKRc-022|We'd have to go up to pH 4.49 to get a tenfold difference.

XIJet3gIKRc-023|So we expect something less than tenfold difference.

XIJet3gIKRc-024|Let's actually do the calculation.

XIJet3gIKRc-025|So here's the acid and the base form, pKa 3.49.

XIJet3gIKRc-027|In this case, we know these two, so we can solve for this ratio.

XIJet3gIKRc-028|So let's just do that.

XIJet3gIKRc-033|So the ratio of base form to acid form is 4.36.

XIJet3gIKRc-034|There's about four times as much base form as there is acid form.

8y5KX4kzt0A-000|Let's talk about the isothermal expansion and compression for an ideal gas.

8y5KX4kzt0A-001|Isothermal means the temperature doesn't change.

8y5KX4kzt0A-002|And if the temperature doesn't change for an ideal gas, the energy and the enthalpy don't change.

8y5KX4kzt0A-003|Remember, the energy of an ideal gas, 3/2 nRT.

8y5KX4kzt0A-004|So isothermal processes delta-E and delta-H are zero.

8y5KX4kzt0A-005|So how can heat flow or work be done if the energy doesn't change?

8y5KX4kzt0A-006|Well, the work and heat just have to balance each other with opposite sign.

8y5KX4kzt0A-007|If you do a joule of work on the system, the system has to lose a joule of heat, joule for joule.

8y5KX4kzt0A-008|Do a joule of work, lose a joule of heat.

8y5KX4kzt0A-009|Do a joule of work, lose a joule of heat.

8y5KX4kzt0A-010|If I expand, the system does work.

8y5KX4kzt0A-011|So I do a joule of work, that would-- well, that's some of my energy, I need more energy to replace the energy I just used to do that work.

8y5KX4kzt0A-012|Well, absorb a joule of heat.

8y5KX4kzt0A-013|Do a joule of work, absorb a joule of heat.

8y5KX4kzt0A-020|That's expansion and contraction isothermally for an ideal gas.

VcGOL6Uthmw-000|Let's look at a couple of titrations and see if we can predict the equivalence point, or the end point, pH.

VcGOL6Uthmw-004|So two acids and a base.

VcGOL6Uthmw-005|We'll titrate them appropriately.

VcGOL6Uthmw-006|If they're acids, we'll titrate them with a base.

VcGOL6Uthmw-007|If it's a base, we'll titrate it with an acid.

VcGOL6Uthmw-008|The question is at equivalence point, which has a pH less than 7?

VcGOL6Uthmw-018|We're looking at the titration of some weak acid and base solutions.

VcGOL6Uthmw-019|Now, when you titrate a weak acid or a weak base with a strong base or a strong acid, the titration curve looks something like this.

VcGOL6Uthmw-020|We have the acid being converted to its conjugate base.