ceaglex/VisualTTS_Chem · Datasets at Hugging Face

text stringlengths 19 206
Zwt_x-8hM54-000\|Let's look at the titration curve of a weak base with a strong acid.
Zwt_x-8hM54-002\|The formation of water is favored.
Zwt_x-8hM54-003\|So this reaction lies far towards products.
Zwt_x-8hM54-004\|And basically, every mole of acid you have will consume a mole of your weak base.
Zwt_x-8hM54-006\|So let's look at the various points along this curve.
Zwt_x-8hM54-013\|When I get to point, the equivalence point here, now I've converted all my base to the conjugate acid.
Zwt_x-8hM54-014\|I've added a mole of strong acid for every mole of base that I originally had, converting it all to its conjugate base.
Zwt_x-8hM54-015\|So you have a solution here of a conjugate acid.
Zwt_x-8hM54-016\|You can calculate the pH of a solution of a weak acid.
Zwt_x-8hM54-017\|As you can continue out, the pH will start to be dominated by the fact that you're just adding strong acid.
Zwt_x-8hM54-018\|So these are the regions of the titration curve for a weak base by a strong acid.
yoq9Vx6FvSE-000\|Among the most common equilibria that we'll look, and the most important, are acid base equilibria.
yoq9Vx6FvSE-001\|Acid base equilibria involve transferring a proton from one species to another.
yoq9Vx6FvSE-003\|Now I'm saying proton, by that I mean when this hydrogen leaves the HA compound, it leaves its electron behind.
yoq9Vx6FvSE-004\|So it's only the nucleus of the hydrogen that leaves.
yoq9Vx6FvSE-005\|And the nucleus of a hydrogen atom is a proton.
yoq9Vx6FvSE-006\|So we call this proton transfer.
yoq9Vx6FvSE-011\|Because this is an equilibrium.
yoq9Vx6FvSE-013\|So that'll be our definition of acid and base.
yoq9Vx6FvSE-017\|So I've abbreviated significantly here.
yoq9Vx6FvSE-018\|And of those four hydrogens only one is donated as a proton.
yoq9Vx6FvSE-019\|One is the acidic proton on acetic acid.
yoq9Vx6FvSE-020\|H3O plus, that's water with a proton added.
yoq9Vx6FvSE-021\|So it has an extra proton, and it will very readily donate that one.
yoq9Vx6FvSE-025\|So water can act as an acid or a base.
yoq9Vx6FvSE-027\|So here would be water donating a proton, you'd have OH minus leftover.
yoq9Vx6FvSE-031\|NH3, it can accept the proton, act as a base, and OH minus, as we've already discussed, the strongest of these three bases.
yoq9Vx6FvSE-032\|The most able to take a proton.
yoq9Vx6FvSE-035\|So our definition, an acid gives up a proton, a base accepts a proton.
AHUBz-4aAbE-000\|Let's look at some more orbital shapes and orientation.
AHUBz-4aAbE-008\|We're talking about orbitals in terms of their shape, and the number of nodes, and their orientation.
AHUBz-4aAbE-009\|Orbitals are described by three quantum numbers, n, l, and m sub l.
AHUBz-4aAbE-010\|n gives you the overall size and energy, l gives you the shape, and m sub l an orientation.
AHUBz-4aAbE-011\|Now what we want to know for this case is where and how are the nodes oriented.
AHUBz-4aAbE-012\|Well, let's look at nodes.
AHUBz-4aAbE-013\|Remember, for angular nodes, they go as the value of l.
AHUBz-4aAbE-014\|So if l is 0, there are 0 angular nodes.
AHUBz-4aAbE-015\|An s orbital, l equals 0, has 0 angular nodes.
AHUBz-4aAbE-018\|If you understand that there are angular nodes slicing up space, that pretty much tells you about the shape and orientation of the orbital.
AHUBz-4aAbE-020\|Or 2, 3 minus 1, or 1.
AHUBz-4aAbE-021\|So there's three nodes here, two nodes here, one node here.
AHUBz-4aAbE-025\|When l is 2, that's a d, there are two angular nodes.
AHUBz-4aAbE-026\|When l is 3, that's an f orbital, there are three angular nodes.
AHUBz-4aAbE-027\|So now we can easily tell what's going on, remember there's three nodes in the 4f, but they're all angular.
AHUBz-4aAbE-028\|So there are no radial nodes in a 4f orbital.
AHUBz-4aAbE-029\|The 3d, there's two total nodes, n minus 1, but they're both angular.
AHUBz-4aAbE-030\|Again, no radial nodes in the 3d.
AHUBz-4aAbE-031\|The 2s, there's one node, s equals 0, so the value of l is 0.
AHUBz-4aAbE-032\|So there are zero angular nodes, so that one node must be a radial node.
AHUBz-4aAbE-033\|So no radial nodes, no radial nodes, 1 radial node.
9JM7zZk1j_g-000\|Let's look at the isothermal compression of a gas.
9JM7zZk1j_g-001\|I'll take a gas, compress it, and I'd like to know-- what is the heat flow?
9JM7zZk1j_g-010\|We're talking about the isothermal compression of an ideal gas.
9JM7zZk1j_g-011\|Now isothermal is the key word because if the temperature doesn't change for an ideal gas, its energy didn't change.
9JM7zZk1j_g-012\|So no energy change in this system.
9JM7zZk1j_g-013\|Now it was a compression, so work is being done on the system.
9JM7zZk1j_g-014\|Delta-V will be negative, but we have a negative sign minus the external delta-V-- positive work.
9JM7zZk1j_g-019\|Now let's think about this practically.
9JM7zZk1j_g-020\|I compress the gas, I'm doing work on the gas.
9JM7zZk1j_g-021\|I'm giving energy to the gas.
9JM7zZk1j_g-022\|And you say, wait, but if I give energy to the gas, its temperature has to change.
9JM7zZk1j_g-023\|How can I give it energy without its temperature changing?
9JM7zZk1j_g-024\|Well, that energy has to flow out at the same time I give it.
9JM7zZk1j_g-025\|So how can that energy flow out?
9JM7zZk1j_g-026\|Well, I'll have work done on the system, but every joule of work will flow out as a joule of heat.
9JM7zZk1j_g-027\|And we call that negative heat or an exothermic process when heat leaves the system.
9JM7zZk1j_g-028\|Practically, it means you probably do this process very slowly.
9JM7zZk1j_g-029\|You compress it a little bit, wait for some heat to flow out.
9JM7zZk1j_g-030\|You compress a little more, wait for heat to flow out.
9JM7zZk1j_g-031\|Or you compress it and wait a while until all the heat flows out.
9JM7zZk1j_g-032\|In general, you can do work quickly on a system, but heat flow always takes time.
9JM7zZk1j_g-033\|So when you see isothermal, you often think slow, because I have to wait for the system to come back to the same temperature or become isothermal.
9JM7zZk1j_g-034\|That's an isothermal compression of an ideal gas.
jzHAfhGPQxg-003\|Well, this chemical reaction releases energy, delta H is negative it's exothermic.
jzHAfhGPQxg-004\|Delta acid is positive and you might have guessed that because there's a gas released from a solid and a liquid.
jzHAfhGPQxg-005\|So that's going to have many more microstate.
jzHAfhGPQxg-006\|You're going to have a dispersion of energy when a gas is created.
jzHAfhGPQxg-007\|So delta H negative, delta S positive.
jzHAfhGPQxg-008\|This reaction will have a negative delta G for all temperatures.
jzHAfhGPQxg-013\|We've actually seen this chemical reaction, but in reverse.
jzHAfhGPQxg-014\|We saw peroxide going to water and oxygen.
jzHAfhGPQxg-015\|And we know energy is released in this direction.
jzHAfhGPQxg-016\|So it has to be absorbed in this direction.
jzHAfhGPQxg-017\|Delta S, in this case, is negative.
jzHAfhGPQxg-018\|Again you might guess that.
jzHAfhGPQxg-019\|Gases forming liquids.
jzHAfhGPQxg-020\|So this reaction in this direction is never favorable, at any temperature.
jzHAfhGPQxg-024\|You're forming a liquid from a solid, so you'd expect an increase in entropy.
jzHAfhGPQxg-028\|This one it's a little less clear, everything's a solid.
jzHAfhGPQxg-029\|So entropy wise it's hard to predict.
jzHAfhGPQxg-030\|Entropy wise this is an exothermic chemical reaction, and entropically there there's a decrease in entropy as you go.
jzHAfhGPQxg-031\|So this reaction is favored at low temperatures.
jzHAfhGPQxg-032\|So we need, since we have this negative delta S contribution, we need T to be small.
jzHAfhGPQxg-033\|So that the overall and felt that contribution will have a contribution to delta G that is negative.
jzHAfhGPQxg-034\|So this reaction favored at low temperature, unfavorable at high temperature.
pxxhXa3TZAU-001\|Now, the change in concentration over time is the slope of the concentration versus time curve.
pxxhXa3TZAU-002\|And we could start at any initial concentrations and let this reaction go.
pxxhXa3TZAU-011\|We're talking about writing an initial rate for a chemical reaction, and we can write initial rates in terms of changes in concentration over a change in time.
pxxhXa3TZAU-013\|That is, two hydrogen ions will be consumed when a zinc ion is produced.
pxxhXa3TZAU-014\|So two things will happen.

Zwt_x-8hM54-000|Let's look at the titration curve of a weak base with a strong acid.

Zwt_x-8hM54-002|The formation of water is favored.

Zwt_x-8hM54-003|So this reaction lies far towards products.

Zwt_x-8hM54-004|And basically, every mole of acid you have will consume a mole of your weak base.

Zwt_x-8hM54-006|So let's look at the various points along this curve.

Zwt_x-8hM54-013|When I get to point, the equivalence point here, now I've converted all my base to the conjugate acid.

Zwt_x-8hM54-014|I've added a mole of strong acid for every mole of base that I originally had, converting it all to its conjugate base.

Zwt_x-8hM54-015|So you have a solution here of a conjugate acid.

Zwt_x-8hM54-016|You can calculate the pH of a solution of a weak acid.

Zwt_x-8hM54-017|As you can continue out, the pH will start to be dominated by the fact that you're just adding strong acid.

Zwt_x-8hM54-018|So these are the regions of the titration curve for a weak base by a strong acid.

yoq9Vx6FvSE-000|Among the most common equilibria that we'll look, and the most important, are acid base equilibria.

yoq9Vx6FvSE-001|Acid base equilibria involve transferring a proton from one species to another.

yoq9Vx6FvSE-003|Now I'm saying proton, by that I mean when this hydrogen leaves the HA compound, it leaves its electron behind.

yoq9Vx6FvSE-004|So it's only the nucleus of the hydrogen that leaves.

yoq9Vx6FvSE-005|And the nucleus of a hydrogen atom is a proton.

yoq9Vx6FvSE-006|So we call this proton transfer.

yoq9Vx6FvSE-011|Because this is an equilibrium.

yoq9Vx6FvSE-013|So that'll be our definition of acid and base.

yoq9Vx6FvSE-017|So I've abbreviated significantly here.

yoq9Vx6FvSE-018|And of those four hydrogens only one is donated as a proton.

yoq9Vx6FvSE-019|One is the acidic proton on acetic acid.

yoq9Vx6FvSE-020|H3O plus, that's water with a proton added.

yoq9Vx6FvSE-021|So it has an extra proton, and it will very readily donate that one.

yoq9Vx6FvSE-025|So water can act as an acid or a base.

yoq9Vx6FvSE-027|So here would be water donating a proton, you'd have OH minus leftover.

yoq9Vx6FvSE-031|NH3, it can accept the proton, act as a base, and OH minus, as we've already discussed, the strongest of these three bases.

yoq9Vx6FvSE-032|The most able to take a proton.

yoq9Vx6FvSE-035|So our definition, an acid gives up a proton, a base accepts a proton.

AHUBz-4aAbE-000|Let's look at some more orbital shapes and orientation.

AHUBz-4aAbE-008|We're talking about orbitals in terms of their shape, and the number of nodes, and their orientation.

AHUBz-4aAbE-009|Orbitals are described by three quantum numbers, n, l, and m sub l.

AHUBz-4aAbE-010|n gives you the overall size and energy, l gives you the shape, and m sub l an orientation.

AHUBz-4aAbE-011|Now what we want to know for this case is where and how are the nodes oriented.

AHUBz-4aAbE-012|Well, let's look at nodes.

AHUBz-4aAbE-013|Remember, for angular nodes, they go as the value of l.

AHUBz-4aAbE-014|So if l is 0, there are 0 angular nodes.

AHUBz-4aAbE-015|An s orbital, l equals 0, has 0 angular nodes.

AHUBz-4aAbE-018|If you understand that there are angular nodes slicing up space, that pretty much tells you about the shape and orientation of the orbital.

AHUBz-4aAbE-020|Or 2, 3 minus 1, or 1.

AHUBz-4aAbE-021|So there's three nodes here, two nodes here, one node here.

AHUBz-4aAbE-025|When l is 2, that's a d, there are two angular nodes.

AHUBz-4aAbE-026|When l is 3, that's an f orbital, there are three angular nodes.

AHUBz-4aAbE-027|So now we can easily tell what's going on, remember there's three nodes in the 4f, but they're all angular.

AHUBz-4aAbE-028|So there are no radial nodes in a 4f orbital.

AHUBz-4aAbE-029|The 3d, there's two total nodes, n minus 1, but they're both angular.

AHUBz-4aAbE-030|Again, no radial nodes in the 3d.

AHUBz-4aAbE-031|The 2s, there's one node, s equals 0, so the value of l is 0.

AHUBz-4aAbE-032|So there are zero angular nodes, so that one node must be a radial node.

AHUBz-4aAbE-033|So no radial nodes, no radial nodes, 1 radial node.

9JM7zZk1j_g-000|Let's look at the isothermal compression of a gas.

9JM7zZk1j_g-001|I'll take a gas, compress it, and I'd like to know-- what is the heat flow?

9JM7zZk1j_g-010|We're talking about the isothermal compression of an ideal gas.

9JM7zZk1j_g-011|Now isothermal is the key word because if the temperature doesn't change for an ideal gas, its energy didn't change.

9JM7zZk1j_g-012|So no energy change in this system.

9JM7zZk1j_g-013|Now it was a compression, so work is being done on the system.

9JM7zZk1j_g-014|Delta-V will be negative, but we have a negative sign minus the external delta-V-- positive work.

9JM7zZk1j_g-019|Now let's think about this practically.

9JM7zZk1j_g-020|I compress the gas, I'm doing work on the gas.

9JM7zZk1j_g-021|I'm giving energy to the gas.

9JM7zZk1j_g-022|And you say, wait, but if I give energy to the gas, its temperature has to change.

9JM7zZk1j_g-023|How can I give it energy without its temperature changing?

9JM7zZk1j_g-024|Well, that energy has to flow out at the same time I give it.

9JM7zZk1j_g-025|So how can that energy flow out?

9JM7zZk1j_g-026|Well, I'll have work done on the system, but every joule of work will flow out as a joule of heat.

9JM7zZk1j_g-027|And we call that negative heat or an exothermic process when heat leaves the system.

9JM7zZk1j_g-028|Practically, it means you probably do this process very slowly.

9JM7zZk1j_g-029|You compress it a little bit, wait for some heat to flow out.

9JM7zZk1j_g-030|You compress a little more, wait for heat to flow out.

9JM7zZk1j_g-031|Or you compress it and wait a while until all the heat flows out.

9JM7zZk1j_g-032|In general, you can do work quickly on a system, but heat flow always takes time.

9JM7zZk1j_g-033|So when you see isothermal, you often think slow, because I have to wait for the system to come back to the same temperature or become isothermal.

9JM7zZk1j_g-034|That's an isothermal compression of an ideal gas.

jzHAfhGPQxg-003|Well, this chemical reaction releases energy, delta H is negative it's exothermic.

jzHAfhGPQxg-004|Delta acid is positive and you might have guessed that because there's a gas released from a solid and a liquid.

jzHAfhGPQxg-005|So that's going to have many more microstate.

jzHAfhGPQxg-006|You're going to have a dispersion of energy when a gas is created.

jzHAfhGPQxg-007|So delta H negative, delta S positive.

jzHAfhGPQxg-008|This reaction will have a negative delta G for all temperatures.

jzHAfhGPQxg-013|We've actually seen this chemical reaction, but in reverse.

jzHAfhGPQxg-014|We saw peroxide going to water and oxygen.

jzHAfhGPQxg-015|And we know energy is released in this direction.

jzHAfhGPQxg-016|So it has to be absorbed in this direction.

jzHAfhGPQxg-017|Delta S, in this case, is negative.

jzHAfhGPQxg-018|Again you might guess that.

jzHAfhGPQxg-019|Gases forming liquids.

jzHAfhGPQxg-020|So this reaction in this direction is never favorable, at any temperature.

jzHAfhGPQxg-024|You're forming a liquid from a solid, so you'd expect an increase in entropy.

jzHAfhGPQxg-028|This one it's a little less clear, everything's a solid.

jzHAfhGPQxg-029|So entropy wise it's hard to predict.

jzHAfhGPQxg-030|Entropy wise this is an exothermic chemical reaction, and entropically there there's a decrease in entropy as you go.

jzHAfhGPQxg-031|So this reaction is favored at low temperatures.

jzHAfhGPQxg-032|So we need, since we have this negative delta S contribution, we need T to be small.

jzHAfhGPQxg-033|So that the overall and felt that contribution will have a contribution to delta G that is negative.

jzHAfhGPQxg-034|So this reaction favored at low temperature, unfavorable at high temperature.

pxxhXa3TZAU-001|Now, the change in concentration over time is the slope of the concentration versus time curve.

pxxhXa3TZAU-002|And we could start at any initial concentrations and let this reaction go.

pxxhXa3TZAU-011|We're talking about writing an initial rate for a chemical reaction, and we can write initial rates in terms of changes in concentration over a change in time.

pxxhXa3TZAU-013|That is, two hydrogen ions will be consumed when a zinc ion is produced.

pxxhXa3TZAU-014|So two things will happen.