ceaglex/VisualTTS_Chem · Datasets at Hugging Face

text stringlengths 19 206
Nb76my9vHLY-016\|So that's a total of 12 valence electrons.
Nb76my9vHLY-017\|So our structures each must have 12 valence electrons to hold them together.
Nb76my9vHLY-020\|There aren't 14 valence electrons among those atoms, so that's not a proper Lewis electron dot structure.
Nb76my9vHLY-022\|If you look at B, there is a subtle error here.
Nb76my9vHLY-023\|This has 12 electrons two, four, six, eight, 10, 12.
Nb76my9vHLY-024\|But carbon has too many.
Nb76my9vHLY-025\|Carbon wants an octet, but it's sharing two, four, six, eight, 10 electrons.
Nb76my9vHLY-026\|So there's no octet on carbon here.
m9XOfqalSgs-000\|Let's look at the correlation between intermolecular forces and Van der Waals a parameters.
m9XOfqalSgs-004\|And in this case, it's obvious that water has the largest intermolecular forces.
Dc69O-XLKtc-000\|Let's construct an electrolytic cell with gold and zinc.
Dc69O-XLKtc-005\|So if gold is reduced, electrons flow to the gold electrode and at the gold electrode, gold medal is plated out from ions.
Dc69O-XLKtc-006\|So gold ions accept electrons-- they're reduced to gold metal.
Dc69O-XLKtc-007\|At the same time, you'd have zinc metal being oxidized to zinc ions.
Dc69O-XLKtc-011\|The mass of the gold electrode would increase, the mass of the zinc electrode would decrease.
Dc69O-XLKtc-012\|We're going to make this go in the electrolytic direction though by providing a external voltage.
Dc69O-XLKtc-013\|So the external voltage will force electrons in this direction.
Dc69O-XLKtc-014\|So electrons will flow from the gold electrode to the zinc electrode.
Dc69O-XLKtc-015\|To get electrons to flow in this direction, that means that gold metal has to be oxidized to gold ions.
Dc69O-XLKtc-019\|So that's theoretically what's going to happen.
Dc69O-XLKtc-020\|Let's do some mathematics-- a little algebra and arithmetic to figure out the actual voltages.
Dc69O-XLKtc-021\|So here's gold and zinc half cells.
Dc69O-XLKtc-022\|Here are their voltages.
Dc69O-XLKtc-023\|Of course, in the electrolytic reaction, zinc will be oxidized while gold is reduced.
Dc69O-XLKtc-024\|So I'll reverse the zinc reaction and change the sign of the potential.
Dc69O-XLKtc-025\|I can then add these if I keep track of the number of electrons-- so two electrons here and two electrons here.
Dc69O-XLKtc-026\|That gives me a two electron transfer process with a total voltage of the sum of 1.69 and 0.76 volts.
Dc69O-XLKtc-027\|So the overall voltage of the standard galvanic direction would be 2.54.
Dc69O-XLKtc-028\|I need to overcome that to go in the electrolytic direction.
Dc69O-XLKtc-029\|So the minimum extra voltage I need is something greater than 2.45.
Dc69O-XLKtc-030\|How much greater, than what kind of overvoltage I need, depends on the configuration of the system.
Dc69O-XLKtc-032\|Minimum voltage of 2.54 volts is the minimum overvoltage we need.
Dc69O-XLKtc-034\|So that's one coulomb per second for 3,600 seconds is 36,000 coulombs.
Dc69O-XLKtc-035\|So 36,000 coulombs of charge are going to be transferred.
Dc69O-XLKtc-036\|Well, how many electrons is that?
Dc69O-XLKtc-037\|Well we can convert 36,000 coulombs of charge into moles of electrons using Faraday's constant.
Dc69O-XLKtc-038\|Faraday's constant is the charge in coulombs on a mole of electrons.
Dc69O-XLKtc-040\|Well, how much zinc metal does that 0.37 moles of electrons produce?
Dc69O-XLKtc-041\|Let's look back at our chemical reaction-- it's gold ions and zinc producing zinc ions and gold metal.
Dc69O-XLKtc-042\|It's a two electron process that I'm going to make go in reverse to make zinc metal.
Dc69O-XLKtc-044\|So overall, my 0.37 moles of electrons produces 12 grams of zinc.
Dc69O-XLKtc-045\|So this electrolytic process, over 10 hours, produces 12 grams of zinc metal.
NI0c8hjRWc0-000\|The equilibrium of NO2 gas and N2O4 is dynamic.
NI0c8hjRWc0-002\|N2O4 dimers are breaking down to form NO2.
NI0c8hjRWc0-004\|The dynamic nature of that equilibrium allows the equilibrium to adjust if there's a change in the system.
NI0c8hjRWc0-005\|And it does so in a logical way.
NI0c8hjRWc0-007\|So let's look at that.
NI0c8hjRWc0-008\|Here's the system in equilibrium, as I have it here.
NI0c8hjRWc0-010\|But now, it's not at equilibrium anymore.
NI0c8hjRWc0-011\|In fact, Q will be greater than K.
NI0c8hjRWc0-012\|And you can prove that to yourself by putting in one half the partial pressures for the equilibrium quotient expression.
NI0c8hjRWc0-015\|Now, what you'll have is a different partial pressure of NO2 and a different partial pressure of N2O4 then you started here.
NI0c8hjRWc0-017\|What you'll notice here is you could have thought of this another way.
NI0c8hjRWc0-018\|The system expanded and went to a larger volume.
NI0c8hjRWc0-019\|So the system wants to expand to fill that volume.
NI0c8hjRWc0-020\|Well, how does the system expand to fill that volume?
NI0c8hjRWc0-021\|It goes to the side where there are more particles.
NI0c8hjRWc0-022\|If some of these particles dissociate, it can fill the volume more effectively.
NI0c8hjRWc0-023\|So I shift towards the sides with more particles when I go to larger volume.
NI0c8hjRWc0-024\|So this is an example of Le Chatelier's principle, where a system moves in response to stress and re-establishes equilibrium.
1ZTXQhi9aW4-005\|But I can also form all of the reactants from elements in their standard states.
1ZTXQhi9aW4-010\|Now what if I want to find this enthalpy of reactants going to products?
1ZTXQhi9aW4-011\|Well, the thing about enthalpy is it's a state function.
1ZTXQhi9aW4-012\|So it doesn't matter the path that I go from reactants to products, it only matters that I start at reactants and end at products.
1ZTXQhi9aW4-013\|The initial and final state are the only thing that determine the enthalpy change.
1ZTXQhi9aW4-015\|This path.
1ZTXQhi9aW4-016\|Now this path from reactants to products has the same enthalpy as this path.
1ZTXQhi9aW4-018\|The negative sign here, because I've gone from reactants to elements in their standard state.
1ZTXQhi9aW4-019\|I've taken the reverse reaction of heats of formation.
1ZTXQhi9aW4-021\|A very powerful tool.
3L5bi-9jO74-000\|Let's look at two gases mixed in a balloon.
3L5bi-9jO74-001\|I'll take bromine and hydrogen gas, their relative masses shown here.
3L5bi-9jO74-002\|And I'd like to know what are their average molar kinetic energies?
3L5bi-9jO74-003\|Does bromine gas have the higher kinetic energy?
3L5bi-9jO74-004\|Are they about equal?
3L5bi-9jO74-012\|We're talking about hydrogen and bromine gas mixed together in a balloon, one mole of each.
3L5bi-9jO74-013\|Which has the higher kinetic energy?
-LFgLSytl1U-000\|We can apply our notions of heat flow to applications in the home.
-LFgLSytl1U-001\|Let's say I'm going to bake a cake in my oven, and I know it takes half hour to bring my cake to 325 degrees in my oven.
-LFgLSytl1U-002\|Now, that's the minimum time it takes.
-LFgLSytl1U-003\|I can't heat it any faster than that.
-LFgLSytl1U-004\|What if two cakes are placed in that same oven?
-LFgLSytl1U-005\|After that same half an hour, what will the relative temperatures be?
-LFgLSytl1U-011\|Let's look at baking cakes in an oven.
-LFgLSytl1U-012\|Now, when you bake the first cake in an oven, the heat flow is similar to liquid flow.
-LFgLSytl1U-013\|So what you have is a reservoir of heat in the oven that you deliver to the cakes.
-LFgLSytl1U-014\|It's like delivering water through a spigot and the flow is limited by the spigot.
-LFgLSytl1U-021\|And you may have heard this.
-LFgLSytl1U-023\|If you put two cakes or five cakes in that same oven, the heat flow will be distributed between all the cakes, you'll effectively overload the oven.
-LFgLSytl1U-025\|So thermodynamics has its place in the home when you're baking.
sihnvASQFSw-000\|Let's look at radioactive decay in terms of kinetics.
sihnvASQFSw-002\|Now, millicurie is just a measure of the amount of tritium, so it's like a concentration.
sihnvASQFSw-003\|So if you plot the natural log of tritium concentration versus time, it decays away like this.
sihnvASQFSw-004\|Natural log concentration versus time is linear.
sihnvASQFSw-012\|We're talking about the decay of tritium over time.
sihnvASQFSw-013\|And we're actually talking about the biological half-life.
sihnvASQFSw-014\|If you ingest some tritium, the tritium acts just like hydrogen, and that's expelled from your body relatively quickly.
sihnvASQFSw-015\|So if you plot the tritium concentration in your body over time, the natural log versus time, it's linear.
sihnvASQFSw-019\|So in five half-lives, I'm down to less than a millicurie in my body.
sihnvASQFSw-020\|So what if I had a 40 millicurie exposure?

Nb76my9vHLY-016|So that's a total of 12 valence electrons.

Nb76my9vHLY-017|So our structures each must have 12 valence electrons to hold them together.

Nb76my9vHLY-020|There aren't 14 valence electrons among those atoms, so that's not a proper Lewis electron dot structure.

Nb76my9vHLY-022|If you look at B, there is a subtle error here.

Nb76my9vHLY-023|This has 12 electrons two, four, six, eight, 10, 12.

Nb76my9vHLY-024|But carbon has too many.

Nb76my9vHLY-025|Carbon wants an octet, but it's sharing two, four, six, eight, 10 electrons.

Nb76my9vHLY-026|So there's no octet on carbon here.

m9XOfqalSgs-000|Let's look at the correlation between intermolecular forces and Van der Waals a parameters.

m9XOfqalSgs-004|And in this case, it's obvious that water has the largest intermolecular forces.

Dc69O-XLKtc-000|Let's construct an electrolytic cell with gold and zinc.

Dc69O-XLKtc-005|So if gold is reduced, electrons flow to the gold electrode and at the gold electrode, gold medal is plated out from ions.

Dc69O-XLKtc-006|So gold ions accept electrons-- they're reduced to gold metal.

Dc69O-XLKtc-007|At the same time, you'd have zinc metal being oxidized to zinc ions.

Dc69O-XLKtc-011|The mass of the gold electrode would increase, the mass of the zinc electrode would decrease.

Dc69O-XLKtc-012|We're going to make this go in the electrolytic direction though by providing a external voltage.

Dc69O-XLKtc-013|So the external voltage will force electrons in this direction.

Dc69O-XLKtc-014|So electrons will flow from the gold electrode to the zinc electrode.

Dc69O-XLKtc-015|To get electrons to flow in this direction, that means that gold metal has to be oxidized to gold ions.

Dc69O-XLKtc-019|So that's theoretically what's going to happen.

Dc69O-XLKtc-020|Let's do some mathematics-- a little algebra and arithmetic to figure out the actual voltages.

Dc69O-XLKtc-021|So here's gold and zinc half cells.

Dc69O-XLKtc-022|Here are their voltages.

Dc69O-XLKtc-023|Of course, in the electrolytic reaction, zinc will be oxidized while gold is reduced.

Dc69O-XLKtc-024|So I'll reverse the zinc reaction and change the sign of the potential.

Dc69O-XLKtc-025|I can then add these if I keep track of the number of electrons-- so two electrons here and two electrons here.

Dc69O-XLKtc-026|That gives me a two electron transfer process with a total voltage of the sum of 1.69 and 0.76 volts.

Dc69O-XLKtc-027|So the overall voltage of the standard galvanic direction would be 2.54.

Dc69O-XLKtc-028|I need to overcome that to go in the electrolytic direction.

Dc69O-XLKtc-029|So the minimum extra voltage I need is something greater than 2.45.

Dc69O-XLKtc-030|How much greater, than what kind of overvoltage I need, depends on the configuration of the system.

Dc69O-XLKtc-032|Minimum voltage of 2.54 volts is the minimum overvoltage we need.

Dc69O-XLKtc-034|So that's one coulomb per second for 3,600 seconds is 36,000 coulombs.

Dc69O-XLKtc-035|So 36,000 coulombs of charge are going to be transferred.

Dc69O-XLKtc-036|Well, how many electrons is that?

Dc69O-XLKtc-037|Well we can convert 36,000 coulombs of charge into moles of electrons using Faraday's constant.

Dc69O-XLKtc-038|Faraday's constant is the charge in coulombs on a mole of electrons.

Dc69O-XLKtc-040|Well, how much zinc metal does that 0.37 moles of electrons produce?

Dc69O-XLKtc-041|Let's look back at our chemical reaction-- it's gold ions and zinc producing zinc ions and gold metal.

Dc69O-XLKtc-042|It's a two electron process that I'm going to make go in reverse to make zinc metal.

Dc69O-XLKtc-044|So overall, my 0.37 moles of electrons produces 12 grams of zinc.

Dc69O-XLKtc-045|So this electrolytic process, over 10 hours, produces 12 grams of zinc metal.

NI0c8hjRWc0-000|The equilibrium of NO2 gas and N2O4 is dynamic.

NI0c8hjRWc0-002|N2O4 dimers are breaking down to form NO2.

NI0c8hjRWc0-004|The dynamic nature of that equilibrium allows the equilibrium to adjust if there's a change in the system.

NI0c8hjRWc0-005|And it does so in a logical way.

NI0c8hjRWc0-007|So let's look at that.

NI0c8hjRWc0-008|Here's the system in equilibrium, as I have it here.

NI0c8hjRWc0-010|But now, it's not at equilibrium anymore.

NI0c8hjRWc0-011|In fact, Q will be greater than K.

NI0c8hjRWc0-012|And you can prove that to yourself by putting in one half the partial pressures for the equilibrium quotient expression.

NI0c8hjRWc0-015|Now, what you'll have is a different partial pressure of NO2 and a different partial pressure of N2O4 then you started here.

NI0c8hjRWc0-017|What you'll notice here is you could have thought of this another way.

NI0c8hjRWc0-018|The system expanded and went to a larger volume.

NI0c8hjRWc0-019|So the system wants to expand to fill that volume.

NI0c8hjRWc0-020|Well, how does the system expand to fill that volume?

NI0c8hjRWc0-021|It goes to the side where there are more particles.

NI0c8hjRWc0-022|If some of these particles dissociate, it can fill the volume more effectively.

NI0c8hjRWc0-023|So I shift towards the sides with more particles when I go to larger volume.

NI0c8hjRWc0-024|So this is an example of Le Chatelier's principle, where a system moves in response to stress and re-establishes equilibrium.

1ZTXQhi9aW4-005|But I can also form all of the reactants from elements in their standard states.

1ZTXQhi9aW4-010|Now what if I want to find this enthalpy of reactants going to products?

1ZTXQhi9aW4-011|Well, the thing about enthalpy is it's a state function.

1ZTXQhi9aW4-012|So it doesn't matter the path that I go from reactants to products, it only matters that I start at reactants and end at products.

1ZTXQhi9aW4-013|The initial and final state are the only thing that determine the enthalpy change.

1ZTXQhi9aW4-015|This path.

1ZTXQhi9aW4-016|Now this path from reactants to products has the same enthalpy as this path.

1ZTXQhi9aW4-018|The negative sign here, because I've gone from reactants to elements in their standard state.

1ZTXQhi9aW4-019|I've taken the reverse reaction of heats of formation.

1ZTXQhi9aW4-021|A very powerful tool.

3L5bi-9jO74-000|Let's look at two gases mixed in a balloon.

3L5bi-9jO74-001|I'll take bromine and hydrogen gas, their relative masses shown here.

3L5bi-9jO74-002|And I'd like to know what are their average molar kinetic energies?

3L5bi-9jO74-003|Does bromine gas have the higher kinetic energy?

3L5bi-9jO74-004|Are they about equal?

3L5bi-9jO74-012|We're talking about hydrogen and bromine gas mixed together in a balloon, one mole of each.

3L5bi-9jO74-013|Which has the higher kinetic energy?

-LFgLSytl1U-000|We can apply our notions of heat flow to applications in the home.

-LFgLSytl1U-001|Let's say I'm going to bake a cake in my oven, and I know it takes half hour to bring my cake to 325 degrees in my oven.

-LFgLSytl1U-002|Now, that's the minimum time it takes.

-LFgLSytl1U-003|I can't heat it any faster than that.

-LFgLSytl1U-004|What if two cakes are placed in that same oven?

-LFgLSytl1U-005|After that same half an hour, what will the relative temperatures be?

-LFgLSytl1U-011|Let's look at baking cakes in an oven.

-LFgLSytl1U-012|Now, when you bake the first cake in an oven, the heat flow is similar to liquid flow.

-LFgLSytl1U-013|So what you have is a reservoir of heat in the oven that you deliver to the cakes.

-LFgLSytl1U-014|It's like delivering water through a spigot and the flow is limited by the spigot.

-LFgLSytl1U-021|And you may have heard this.

-LFgLSytl1U-023|If you put two cakes or five cakes in that same oven, the heat flow will be distributed between all the cakes, you'll effectively overload the oven.

-LFgLSytl1U-025|So thermodynamics has its place in the home when you're baking.

sihnvASQFSw-000|Let's look at radioactive decay in terms of kinetics.

sihnvASQFSw-002|Now, millicurie is just a measure of the amount of tritium, so it's like a concentration.

sihnvASQFSw-003|So if you plot the natural log of tritium concentration versus time, it decays away like this.

sihnvASQFSw-004|Natural log concentration versus time is linear.

sihnvASQFSw-012|We're talking about the decay of tritium over time.

sihnvASQFSw-013|And we're actually talking about the biological half-life.

sihnvASQFSw-014|If you ingest some tritium, the tritium acts just like hydrogen, and that's expelled from your body relatively quickly.

sihnvASQFSw-015|So if you plot the tritium concentration in your body over time, the natural log versus time, it's linear.

sihnvASQFSw-019|So in five half-lives, I'm down to less than a millicurie in my body.

sihnvASQFSw-020|So what if I had a 40 millicurie exposure?