ceaglex/VisualTTS_Chem · Datasets at Hugging Face

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pxxhXa3TZAU-015\|The rates, if I cast it in terms of the zinc ion, should have the opposite sign, and there should be a factor of two.
pxxhXa3TZAU-016\|And that is, of course, C here.
pxxhXa3TZAU-018\|In this case, the correct answer is C.
7UFQPsGBreg-000\|Let's continue writing Lewis electron dot structures.
7UFQPsGBreg-001\|The cyanide ion, CN minus, is written down partially here.
7UFQPsGBreg-002\|Or is it partial?
7UFQPsGBreg-003\|Are there no bonds required to complete it?
7UFQPsGBreg-004\|One bond required to complete it?
7UFQPsGBreg-012\|We're writing the Lewis electron structure for CN minus, the cyanide ion.
7UFQPsGBreg-013\|First, we count up the valence electrons.
7UFQPsGBreg-014\|Carbon contributes 4, nitrogen contributes 5, and the ion has a negative charge, so there must be an extra electron.
7UFQPsGBreg-015\|That contributes 1.
7UFQPsGBreg-016\|So 10 total electrons.
7UFQPsGBreg-017\|That's actually the same number of electrons as the N2 molecule.
7UFQPsGBreg-018\|So CN minus requires two more bonds to get 10 total electrons.
7UFQPsGBreg-019\|I have 2, 4, 6, 8, 10 electrons in the cyanide ion.
7UFQPsGBreg-021\|CN minus, isoelectronic with N2, is a triple bonded structure.
K6LVoYlzJ08-000\|An acid is a compound that donates a proton.
K6LVoYlzJ08-001\|A base is a compound that accepts a proton.
K6LVoYlzJ08-002\|In water solutions, the acid donates its proton to water.
K6LVoYlzJ08-011\|So this Ka has the special label, but it behaves like a normal equilibrium constant.
K6LVoYlzJ08-012\|In particular, if it's large, then the equilibrium lies towards the products.
K6LVoYlzJ08-013\|A lot of H3O+ is produced, a lot of the HA, a high percentage, dissociates for strong acids.
K6LVoYlzJ08-014\|For weak acids, a smaller percentage dissociates.
K6LVoYlzJ08-019\|By accepting a proton, it forms OH- in water solution.
K6LVoYlzJ08-020\|So the presence of OH- is indicative of a base in water.
K6LVoYlzJ08-024\|Again, water, a pure liquid, doesn't appear in the expression.
K6LVoYlzJ08-025\|Again, the size of the K determines the strength of the base.
K6LVoYlzJ08-027\|So we have Ka and Kb, acid and base dissociation constants.
vC2v1n-bZQU-000\|Equilibrium is the condition where macroscopically, a reaction appears to have stopped-- the macroscopic concentrations or pressures aren't changing.
vC2v1n-bZQU-001\|But microscopically, there's a dynamic equilibrium.
vC2v1n-bZQU-002\|Products are changing into reactants and reactants are changing into products.
vC2v1n-bZQU-003\|You can approach equilibrium in a variety of ways.
vC2v1n-bZQU-010\|So this equilibrium constant K would be large, it would favor the products.
vC2v1n-bZQU-013\|Now, how can I approach this equilibrium state?
vC2v1n-bZQU-018\|I can demonstrate that.
vC2v1n-bZQU-020\|So let's see that happening.
vC2v1n-bZQU-022\|So equilibrium reached in a step.
vC2v1n-bZQU-023\|We can also reach equilibrium in an oscillatory fashion.
vC2v1n-bZQU-025\|So let's see that chemical reaction occur.
vC2v1n-bZQU-026\|So here's a chemical reaction, turns to yellow forming, but then to blue as it goes by the products.
vC2v1n-bZQU-031\|There's an overall envelope of reaching equilibrium, but it happens in an oscillatory fashion.
vC2v1n-bZQU-032\|There's another instance you could imagine, have an approach to equilibrium that's chaotic.
vC2v1n-bZQU-034\|All reactions if you wait long enough will approach the equilibrium.
vC2v1n-bZQU-035\|The thermodynamically favored state is the favored state at long times.
vC2v1n-bZQU-036\|So we expect favored states to occur if we wait long enough.
vC2v1n-bZQU-037\|So several ways to approach equilibrium.
vC2v1n-bZQU-038\|Either way, they're dynamic.
TUBGGXVJj3Q-000\|Let's look at adding a strong base solution to a strong acid solution.
TUBGGXVJj3Q-002\|Now, strong acids and strong bases totally dissociate in water-- that's the definition of strong acid and strong base.
TUBGGXVJj3Q-003\|So HCl completely dissociates into H3O+ and Cl-.
TUBGGXVJj3Q-004\|We can look at in steps, though.
TUBGGXVJj3Q-017\|I could, but it doesn't make much sense, the H3O+ plus concentration of 0.2 plus to the minus 7, 10 to the minus 7 is tiny by comparison, right?
TUBGGXVJj3Q-018\|This is 0.2000001.
TUBGGXVJj3Q-019\|So that's essentially 0.2.
TUBGGXVJj3Q-021\|Now, we want to add 10 mils of sodium hydroxide at one molar.
TUBGGXVJj3Q-022\|How many moles of OH- is that?
TUBGGXVJj3Q-026\|If I add 0.01 moles of sodium hydroxide, I'm actually adding 0.01 moles of hydroxide ion.
TUBGGXVJj3Q-027\|So let's look at that.
TUBGGXVJj3Q-033\|Do I need to do an equilibrium calculation here?
TUBGGXVJj3Q-034\|No I don't, because the K for this reaction, the reverse reaction is K W, So 1 over a K W is the K for this reaction.
TUBGGXVJj3Q-035\|K W is times 10 to the minus 14, so the K for this reaction 10 to the plus 14.
TUBGGXVJj3Q-036\|So this goes completely to the water.
TUBGGXVJj3Q-037\|So every mole of OH- finds a mole of H3O+ and forms water.
TUBGGXVJj3Q-038\|One for one.
TUBGGXVJj3Q-048\|And that 0.19 moles is in about a liter.
TUBGGXVJj3Q-049\|It's in a liter plus the 10 mils of any OH we added.
TUBGGXVJj3Q-050\|About a liter.
TUBGGXVJj3Q-051\|So it's about 0.19 molar.
TUBGGXVJj3Q-052\|So the H3O+ concentration now will be about 0.19 molar.
TUBGGXVJj3Q-053\|Minus log of 0.19 molar, 0.72.
TUBGGXVJj3Q-054\|So the pH has changed from 0.7 to 0.72.
TUBGGXVJj3Q-055\|Not very much.
TUBGGXVJj3Q-061\|Now, I can look at adding acid and base.
TUBGGXVJj3Q-062\|Here I have an acid solution and a base solution.
TUBGGXVJj3Q-063\|So this is concentrated, this is not as concentrated.
TUBGGXVJj3Q-064\|As I add base to acid, we can watch the pH change by the color of the solution.
TUBGGXVJj3Q-066\|So let's watch this acid-base titration.
TUBGGXVJj3Q-067\|Adding a strong base to a strong acid solution, let's watch that happen.
TUBGGXVJj3Q-068\|PH changes slowly at first, and then rapidly to blue-- the base solution.
TUBGGXVJj3Q-069\|So this is well down here on the titration curve.
TUBGGXVJj3Q-070\|I've gone past the equivalence point to the point where the base is determining the pH.
TUBGGXVJj3Q-071\|A basic solution here maybe around pH 10 to 13-- a blue color for my indicator.
TUBGGXVJj3Q-072\|So there's adding a strong base to a strong acid solution.
vApxjo9sAmI-000\|A rubber band is a system that could do work or have work done on it.
vApxjo9sAmI-001\|So let's take a rubber band and stretch it, hold it there so it comes to thermal equilibrium, and then suddenly release it.
vApxjo9sAmI-007\|We're talking about stretching and releasing a rubber band, or any piece of rubber.
vApxjo9sAmI-008\|I have a rubber glove here.
vApxjo9sAmI-009\|I can stretch it and hold it, let it come to thermal equilibrium.
vApxjo9sAmI-010\|I can even check that against my lip.
vApxjo9sAmI-011\|Yeah, that's room temperature.
vApxjo9sAmI-012\|And then suddenly release it and touch it to my lip again.
vApxjo9sAmI-013\|What happened to the temperature change?
vApxjo9sAmI-014\|Well you can do that experiment, I'll do the math.
vApxjo9sAmI-015\|It's an adiabatic change, and we arranged for it to be adiabatic simply by doing it quickly.
vApxjo9sAmI-020\|The rubber has to pull my hands back together.
vApxjo9sAmI-021\|So a system that does work uses its internal energy to do the work.
vApxjo9sAmI-022\|So the energy must go down.
vApxjo9sAmI-024\|The energy change and the temperature change are locked for an ideal gas.
vApxjo9sAmI-025\|That is, for an ideal gas, the energy and the temperature change are essentially the same parameter.

pxxhXa3TZAU-015|The rates, if I cast it in terms of the zinc ion, should have the opposite sign, and there should be a factor of two.

pxxhXa3TZAU-016|And that is, of course, C here.

pxxhXa3TZAU-018|In this case, the correct answer is C.

7UFQPsGBreg-000|Let's continue writing Lewis electron dot structures.

7UFQPsGBreg-001|The cyanide ion, CN minus, is written down partially here.

7UFQPsGBreg-002|Or is it partial?

7UFQPsGBreg-003|Are there no bonds required to complete it?

7UFQPsGBreg-004|One bond required to complete it?

7UFQPsGBreg-012|We're writing the Lewis electron structure for CN minus, the cyanide ion.

7UFQPsGBreg-013|First, we count up the valence electrons.

7UFQPsGBreg-014|Carbon contributes 4, nitrogen contributes 5, and the ion has a negative charge, so there must be an extra electron.

7UFQPsGBreg-015|That contributes 1.

7UFQPsGBreg-016|So 10 total electrons.

7UFQPsGBreg-017|That's actually the same number of electrons as the N2 molecule.

7UFQPsGBreg-018|So CN minus requires two more bonds to get 10 total electrons.

7UFQPsGBreg-019|I have 2, 4, 6, 8, 10 electrons in the cyanide ion.

7UFQPsGBreg-021|CN minus, isoelectronic with N2, is a triple bonded structure.

K6LVoYlzJ08-000|An acid is a compound that donates a proton.

K6LVoYlzJ08-001|A base is a compound that accepts a proton.

K6LVoYlzJ08-002|In water solutions, the acid donates its proton to water.

K6LVoYlzJ08-011|So this Ka has the special label, but it behaves like a normal equilibrium constant.

K6LVoYlzJ08-012|In particular, if it's large, then the equilibrium lies towards the products.

K6LVoYlzJ08-013|A lot of H3O+ is produced, a lot of the HA, a high percentage, dissociates for strong acids.

K6LVoYlzJ08-014|For weak acids, a smaller percentage dissociates.

K6LVoYlzJ08-019|By accepting a proton, it forms OH- in water solution.

K6LVoYlzJ08-020|So the presence of OH- is indicative of a base in water.

K6LVoYlzJ08-024|Again, water, a pure liquid, doesn't appear in the expression.

K6LVoYlzJ08-025|Again, the size of the K determines the strength of the base.

K6LVoYlzJ08-027|So we have Ka and Kb, acid and base dissociation constants.

vC2v1n-bZQU-000|Equilibrium is the condition where macroscopically, a reaction appears to have stopped-- the macroscopic concentrations or pressures aren't changing.

vC2v1n-bZQU-001|But microscopically, there's a dynamic equilibrium.

vC2v1n-bZQU-002|Products are changing into reactants and reactants are changing into products.

vC2v1n-bZQU-003|You can approach equilibrium in a variety of ways.

vC2v1n-bZQU-010|So this equilibrium constant K would be large, it would favor the products.

vC2v1n-bZQU-013|Now, how can I approach this equilibrium state?

vC2v1n-bZQU-018|I can demonstrate that.

vC2v1n-bZQU-020|So let's see that happening.

vC2v1n-bZQU-022|So equilibrium reached in a step.

vC2v1n-bZQU-023|We can also reach equilibrium in an oscillatory fashion.

vC2v1n-bZQU-025|So let's see that chemical reaction occur.

vC2v1n-bZQU-026|So here's a chemical reaction, turns to yellow forming, but then to blue as it goes by the products.

vC2v1n-bZQU-031|There's an overall envelope of reaching equilibrium, but it happens in an oscillatory fashion.

vC2v1n-bZQU-032|There's another instance you could imagine, have an approach to equilibrium that's chaotic.

vC2v1n-bZQU-034|All reactions if you wait long enough will approach the equilibrium.

vC2v1n-bZQU-035|The thermodynamically favored state is the favored state at long times.

vC2v1n-bZQU-036|So we expect favored states to occur if we wait long enough.

vC2v1n-bZQU-037|So several ways to approach equilibrium.

vC2v1n-bZQU-038|Either way, they're dynamic.

TUBGGXVJj3Q-000|Let's look at adding a strong base solution to a strong acid solution.

TUBGGXVJj3Q-002|Now, strong acids and strong bases totally dissociate in water-- that's the definition of strong acid and strong base.

TUBGGXVJj3Q-003|So HCl completely dissociates into H3O+ and Cl-.

TUBGGXVJj3Q-004|We can look at in steps, though.

TUBGGXVJj3Q-017|I could, but it doesn't make much sense, the H3O+ plus concentration of 0.2 plus to the minus 7, 10 to the minus 7 is tiny by comparison, right?

TUBGGXVJj3Q-018|This is 0.2000001.

TUBGGXVJj3Q-019|So that's essentially 0.2.

TUBGGXVJj3Q-021|Now, we want to add 10 mils of sodium hydroxide at one molar.

TUBGGXVJj3Q-022|How many moles of OH- is that?

TUBGGXVJj3Q-026|If I add 0.01 moles of sodium hydroxide, I'm actually adding 0.01 moles of hydroxide ion.

TUBGGXVJj3Q-027|So let's look at that.

TUBGGXVJj3Q-033|Do I need to do an equilibrium calculation here?

TUBGGXVJj3Q-034|No I don't, because the K for this reaction, the reverse reaction is K W, So 1 over a K W is the K for this reaction.

TUBGGXVJj3Q-035|K W is times 10 to the minus 14, so the K for this reaction 10 to the plus 14.

TUBGGXVJj3Q-036|So this goes completely to the water.

TUBGGXVJj3Q-037|So every mole of OH- finds a mole of H3O+ and forms water.

TUBGGXVJj3Q-038|One for one.

TUBGGXVJj3Q-048|And that 0.19 moles is in about a liter.

TUBGGXVJj3Q-049|It's in a liter plus the 10 mils of any OH we added.

TUBGGXVJj3Q-050|About a liter.

TUBGGXVJj3Q-051|So it's about 0.19 molar.

TUBGGXVJj3Q-052|So the H3O+ concentration now will be about 0.19 molar.

TUBGGXVJj3Q-053|Minus log of 0.19 molar, 0.72.

TUBGGXVJj3Q-054|So the pH has changed from 0.7 to 0.72.

TUBGGXVJj3Q-055|Not very much.

TUBGGXVJj3Q-061|Now, I can look at adding acid and base.

TUBGGXVJj3Q-062|Here I have an acid solution and a base solution.

TUBGGXVJj3Q-063|So this is concentrated, this is not as concentrated.

TUBGGXVJj3Q-064|As I add base to acid, we can watch the pH change by the color of the solution.

TUBGGXVJj3Q-066|So let's watch this acid-base titration.

TUBGGXVJj3Q-067|Adding a strong base to a strong acid solution, let's watch that happen.

TUBGGXVJj3Q-068|PH changes slowly at first, and then rapidly to blue-- the base solution.

TUBGGXVJj3Q-069|So this is well down here on the titration curve.

TUBGGXVJj3Q-070|I've gone past the equivalence point to the point where the base is determining the pH.

TUBGGXVJj3Q-071|A basic solution here maybe around pH 10 to 13-- a blue color for my indicator.

TUBGGXVJj3Q-072|So there's adding a strong base to a strong acid solution.

vApxjo9sAmI-000|A rubber band is a system that could do work or have work done on it.

vApxjo9sAmI-001|So let's take a rubber band and stretch it, hold it there so it comes to thermal equilibrium, and then suddenly release it.

vApxjo9sAmI-007|We're talking about stretching and releasing a rubber band, or any piece of rubber.

vApxjo9sAmI-008|I have a rubber glove here.

vApxjo9sAmI-009|I can stretch it and hold it, let it come to thermal equilibrium.

vApxjo9sAmI-010|I can even check that against my lip.

vApxjo9sAmI-011|Yeah, that's room temperature.

vApxjo9sAmI-012|And then suddenly release it and touch it to my lip again.

vApxjo9sAmI-013|What happened to the temperature change?

vApxjo9sAmI-014|Well you can do that experiment, I'll do the math.

vApxjo9sAmI-015|It's an adiabatic change, and we arranged for it to be adiabatic simply by doing it quickly.

vApxjo9sAmI-020|The rubber has to pull my hands back together.

vApxjo9sAmI-021|So a system that does work uses its internal energy to do the work.

vApxjo9sAmI-022|So the energy must go down.

vApxjo9sAmI-024|The energy change and the temperature change are locked for an ideal gas.

vApxjo9sAmI-025|That is, for an ideal gas, the energy and the temperature change are essentially the same parameter.