ceaglex/VisualTTS_Chem · Datasets at Hugging Face

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ZFZ2J8gwIkE-016\|That's what the bubbles are.
ZFZ2J8gwIkE-017\|They're coming out of solution and then they bubble to the surface.
0Ps7ux68RMU-000\|Let's look at a phase diagram for water and try to predict where the 373 kelvin isotherm is.
0Ps7ux68RMU-006\|We're trying to position the 373 kelvin isotherm on the phase diagram for water.
0Ps7ux68RMU-007\|Now, the 373 isotherm is a 100 degrees Celsius, so that's the normal boiling point.
0Ps7ux68RMU-008\|The normal boiling point is the transition between the liquid and the gas water.
0Ps7ux68RMU-009\|So if there are two phases present, I must be below the critical temperature.
0Ps7ux68RMU-010\|Remember the critical temperature, the definition is, above the critical temperature, I have a single phase regardless of the pressure.
0Ps7ux68RMU-011\|If I have below the critical point, this critical temperature, then I can have phase transitions.
0Ps7ux68RMU-012\|So the phase transition at 100 degrees C implies that I'm below the critical temperature.
0Ps7ux68RMU-014\|If you decrease the pressure, you go from liquid to gas, or if you increase the pressure, you go from the gas to the liquid.
_7s29Q76st0-000\|Let's look at three molecules and see if we can determine if they have a permanent dipole moment.
_7s29Q76st0-001\|Three molecules, carbon monoxide, carbon dioxide, and acetone.
_7s29Q76st0-002\|Which has no permanent dipole moment?
_7s29Q76st0-003\|It's interesting to note that no permanent dipole moment means a molecule will be transparent to microwaves.
_7s29Q76st0-004\|It turns out the permanent dipole moment is what allows molecules to absorb microwaves.
_7s29Q76st0-005\|It's what allows microwave spectroscopy to work, and even your microwave oven.
_7s29Q76st0-006\|So while you're thinking about which of these has no permanent dipole moment, you can think about which of these would be transparent to microwaves at the same time.
_7s29Q76st0-013\|We're looking at three molecules and trying to determine their electric dipole moments.
_7s29Q76st0-016\|Carbon monoxide.
_7s29Q76st0-017\|There's a diatomic molecule, two atoms, unequal electronegativities between oxygen and carbon.
_7s29Q76st0-018\|So there'll be an electric dipole moment in carbon monoxide.
_7s29Q76st0-019\|Carbon dioxide.
_7s29Q76st0-020\|Now here we have two bonds to consider.
_7s29Q76st0-021\|Remember, we need to consider each bond in the molecule.
_7s29Q76st0-022\|Each bond will be polar.
_7s29Q76st0-023\|Each will have a tiny dipole moment that will add to give the total dipole moment for the molecule.
_7s29Q76st0-024\|In the case of carbon dioxide, the dipole moments are equal and opposite, so they'll identically cancel out.
_7s29Q76st0-025\|How about acetone?
_7s29Q76st0-027\|Now those are not equal and opposite.
_7s29Q76st0-028\|In fact, they'll point in different directions and have different magnitudes, and they will not cancel out.
_7s29Q76st0-030\|So the answer here, carbon dioxide has no permanent dipole moment.
_7s29Q76st0-031\|Now those of you that are looking at this very closely might say, wait a minute, isn't oxygen more electronegative than carbon?
_7s29Q76st0-032\|Shouldn't this electric dipole moment point the other way?
_7s29Q76st0-033\|Shouldn't there be a higher probability of finding electrons around the more electronegative element, oxygen, in this diatomic molecule?
_7s29Q76st0-034\|Well, your initial impression would be correct.
_7s29Q76st0-035\|Your feeling from electronegativities is that this dipole moment should point in the other direction.
_7s29Q76st0-036\|And in fact, it does point from the oxygen to the carbon.
zdupTfmI3ME-000\|Let's look at the catalyzed reaction of hydrogen and oxygen to form water.
zdupTfmI3ME-013\|Now, we weren't given the reagents hydrogen and oxygen in that ratio.
zdupTfmI3ME-014\|In fact, you hardly ever are.
zdupTfmI3ME-015\|You don't provide things in exact chemical ratio and when you don't one thing will run out before the other thing.
zdupTfmI3ME-016\|One thing will be in excess.
zdupTfmI3ME-017\|And in this case, I think you can see we have much more oxygen than we need to react with one mole of hydrogen.
zdupTfmI3ME-018\|To react with one mole of hydrogen we'd only need half a mole of oxygen and we have two.
zdupTfmI3ME-022\|Hydrogen and water will produce in one to one ratios.
zdupTfmI3ME-023\|They have the same stoichiometric coefficient.
zdupTfmI3ME-024\|So one mole of hydrogen, if there's enough oxygen, which there is, will produce one mole of water.
zdupTfmI3ME-025\|But there's oxygen left over.
zdupTfmI3ME-026\|We call hydrogen here the limiting reagent because when it ran out it limited how much product I could form.
WH4YuPwH9l8-003\|So which of these transitions in thymol blue has the greatest absorbance at pH 10?
WH4YuPwH9l8-011\|We're looking at the indicator thymol blue, which is a polyprotic acid, and it has several color variations.
WH4YuPwH9l8-015\|So we're talking about thymol blue in the region where we have a pH of 10.
WH4YuPwH9l8-016\|So at pH 10, the blue form predominates.
WH4YuPwH9l8-017\|Now, if a solution appears blue, where does it absorb?
WH4YuPwH9l8-018\|Remember, solutions that appear blue absorb other regions.
WH4YuPwH9l8-019\|And something that absorbs strongly in the red will pass blue colors and appear blue.
5_H82ywrQIY-000\|Let's look at dissolving barium sulfate, a sparingly soluble solid, in water.
5_H82ywrQIY-001\|If we try to dissolve it and a little, tiny speck remains, say about a milligram remains in the bottom, what's the best way to dissolve that additional milligram?
5_H82ywrQIY-009\|We're looking at barium sulfate in water, and dissolving a tiny, milligram speck, and the most efficient way to do that.
5_H82ywrQIY-010\|Should we add some acid?
5_H82ywrQIY-011\|H2SO4?
5_H82ywrQIY-012\|Should we add barium chloride, BaCl2, or water?
5_H82ywrQIY-013\|Well, if you look at the reaction for barium sulfate solid dissolving in water, barium sulfate forms barium ions and sulfate ions.
5_H82ywrQIY-018\|What you want is more ions to get this to dissolve.
5_H82ywrQIY-019\|Sulfuric acid, H2SO4, the same thing, you're going to add more sulfate ions to solution.
5_H82ywrQIY-021\|So in this case, the best strategy is simply add more water, increase the concentration.
5_H82ywrQIY-022\|That gives you diluting these concentrations, making them smaller.
5_H82ywrQIY-023\|By Le Chatelier's principle, I'll shift to make these concentrations higher again.
5_H82ywrQIY-024\|So in this case, add more water to dissolve a one milligram speck of barium sulfate.
aNEDU6EL8jc-000\|Let's look at the course of chemical reactions.
aNEDU6EL8jc-006\|And I can calculate this value Q at any time.
aNEDU6EL8jc-013\|So I can calculate this number Q at any time.
aNEDU6EL8jc-016\|You can see they flatline here.
aNEDU6EL8jc-017\|There is no change with time for the concentrations.
aNEDU6EL8jc-018\|So if you wait long enough, these values of concentration stop changing and that means Q stops changing.
aNEDU6EL8jc-019\|It becomes a constant after some time.
aNEDU6EL8jc-021\|And your reaction quotient stops changing.
aNEDU6EL8jc-022\|It becomes a constant.
aNEDU6EL8jc-024\|That is, A and B are still changing into C and D. It's just at the same time, C and D are changing back into A and B.
aNEDU6EL8jc-025\|So macroscopically, things don't change, but the reverse and forward rates have equalized.
aNEDU6EL8jc-026\|So it's a dynamic equilibrium.
aNEDU6EL8jc-029\|So it's indeed a constant, and it's a characteristic of the chemical reaction.
aNEDU6EL8jc-030\|So if you measure lots of these K's you can predict where the reaction's going to go.
aNEDU6EL8jc-032\|The numerator is the products.
aNEDU6EL8jc-033\|So if these products are too small-- because Q is too small-- I should go make more of them.
aNEDU6EL8jc-034\|So I should go towards products.
aNEDU6EL8jc-035\|So comparing Q that you measure at any time with your known constant gives you predictive power.
aNEDU6EL8jc-036\|Which way is that reaction going to go?
aNEDU6EL8jc-037\|Same thing, if you measure Q and its bigger than K, well, that says, this numerator's too large.
aNEDU6EL8jc-038\|Too many products, I should go back to reactants, because I know Q's always go back to the K.
aNEDU6EL8jc-039\|If I wait long enough, reactions will proceed.
aNEDU6EL8jc-040\|So Q, the instantaneous concentrations, stops changing at this value of K.
aNEDU6EL8jc-041\|And it will progress towards K in a relatively straightforward monotonic fashion.
aNEDU6EL8jc-042\|So here we have Q equal K at equilibrium.
aNEDU6EL8jc-043\|Now, the approach from Q to K can be monotonic, as I've written it, but it actually could also oscillate.
aNEDU6EL8jc-044\|Whatever it is, it'll get to K eventually.
aNEDU6EL8jc-045\|So Q equals K at equilibrium.
aNEDU6EL8jc-046\|And what I have is a comparison between Q and K will tell me which direction I have to go to get equilibrium.
aNEDU6EL8jc-047\|I can measure this value of K. What I need to do is measure it at different temperatures because K will change with temperature.

ZFZ2J8gwIkE-016|That's what the bubbles are.

ZFZ2J8gwIkE-017|They're coming out of solution and then they bubble to the surface.

0Ps7ux68RMU-000|Let's look at a phase diagram for water and try to predict where the 373 kelvin isotherm is.

0Ps7ux68RMU-006|We're trying to position the 373 kelvin isotherm on the phase diagram for water.

0Ps7ux68RMU-007|Now, the 373 isotherm is a 100 degrees Celsius, so that's the normal boiling point.

0Ps7ux68RMU-008|The normal boiling point is the transition between the liquid and the gas water.

0Ps7ux68RMU-009|So if there are two phases present, I must be below the critical temperature.

0Ps7ux68RMU-010|Remember the critical temperature, the definition is, above the critical temperature, I have a single phase regardless of the pressure.

0Ps7ux68RMU-011|If I have below the critical point, this critical temperature, then I can have phase transitions.

0Ps7ux68RMU-012|So the phase transition at 100 degrees C implies that I'm below the critical temperature.

0Ps7ux68RMU-014|If you decrease the pressure, you go from liquid to gas, or if you increase the pressure, you go from the gas to the liquid.

_7s29Q76st0-000|Let's look at three molecules and see if we can determine if they have a permanent dipole moment.

_7s29Q76st0-001|Three molecules, carbon monoxide, carbon dioxide, and acetone.

_7s29Q76st0-002|Which has no permanent dipole moment?

_7s29Q76st0-003|It's interesting to note that no permanent dipole moment means a molecule will be transparent to microwaves.

_7s29Q76st0-004|It turns out the permanent dipole moment is what allows molecules to absorb microwaves.

_7s29Q76st0-005|It's what allows microwave spectroscopy to work, and even your microwave oven.

_7s29Q76st0-006|So while you're thinking about which of these has no permanent dipole moment, you can think about which of these would be transparent to microwaves at the same time.

_7s29Q76st0-013|We're looking at three molecules and trying to determine their electric dipole moments.

_7s29Q76st0-016|Carbon monoxide.

_7s29Q76st0-017|There's a diatomic molecule, two atoms, unequal electronegativities between oxygen and carbon.

_7s29Q76st0-018|So there'll be an electric dipole moment in carbon monoxide.

_7s29Q76st0-019|Carbon dioxide.

_7s29Q76st0-020|Now here we have two bonds to consider.

_7s29Q76st0-021|Remember, we need to consider each bond in the molecule.

_7s29Q76st0-022|Each bond will be polar.

_7s29Q76st0-023|Each will have a tiny dipole moment that will add to give the total dipole moment for the molecule.

_7s29Q76st0-024|In the case of carbon dioxide, the dipole moments are equal and opposite, so they'll identically cancel out.

_7s29Q76st0-025|How about acetone?

_7s29Q76st0-027|Now those are not equal and opposite.

_7s29Q76st0-028|In fact, they'll point in different directions and have different magnitudes, and they will not cancel out.

_7s29Q76st0-030|So the answer here, carbon dioxide has no permanent dipole moment.

_7s29Q76st0-031|Now those of you that are looking at this very closely might say, wait a minute, isn't oxygen more electronegative than carbon?

_7s29Q76st0-032|Shouldn't this electric dipole moment point the other way?

_7s29Q76st0-033|Shouldn't there be a higher probability of finding electrons around the more electronegative element, oxygen, in this diatomic molecule?

_7s29Q76st0-034|Well, your initial impression would be correct.

_7s29Q76st0-035|Your feeling from electronegativities is that this dipole moment should point in the other direction.

_7s29Q76st0-036|And in fact, it does point from the oxygen to the carbon.

zdupTfmI3ME-000|Let's look at the catalyzed reaction of hydrogen and oxygen to form water.

zdupTfmI3ME-013|Now, we weren't given the reagents hydrogen and oxygen in that ratio.

zdupTfmI3ME-014|In fact, you hardly ever are.

zdupTfmI3ME-015|You don't provide things in exact chemical ratio and when you don't one thing will run out before the other thing.

zdupTfmI3ME-016|One thing will be in excess.

zdupTfmI3ME-017|And in this case, I think you can see we have much more oxygen than we need to react with one mole of hydrogen.

zdupTfmI3ME-018|To react with one mole of hydrogen we'd only need half a mole of oxygen and we have two.

zdupTfmI3ME-022|Hydrogen and water will produce in one to one ratios.

zdupTfmI3ME-023|They have the same stoichiometric coefficient.

zdupTfmI3ME-024|So one mole of hydrogen, if there's enough oxygen, which there is, will produce one mole of water.

zdupTfmI3ME-025|But there's oxygen left over.

zdupTfmI3ME-026|We call hydrogen here the limiting reagent because when it ran out it limited how much product I could form.

WH4YuPwH9l8-003|So which of these transitions in thymol blue has the greatest absorbance at pH 10?

WH4YuPwH9l8-011|We're looking at the indicator thymol blue, which is a polyprotic acid, and it has several color variations.

WH4YuPwH9l8-015|So we're talking about thymol blue in the region where we have a pH of 10.

WH4YuPwH9l8-016|So at pH 10, the blue form predominates.

WH4YuPwH9l8-017|Now, if a solution appears blue, where does it absorb?

WH4YuPwH9l8-018|Remember, solutions that appear blue absorb other regions.

WH4YuPwH9l8-019|And something that absorbs strongly in the red will pass blue colors and appear blue.

5_H82ywrQIY-000|Let's look at dissolving barium sulfate, a sparingly soluble solid, in water.

5_H82ywrQIY-001|If we try to dissolve it and a little, tiny speck remains, say about a milligram remains in the bottom, what's the best way to dissolve that additional milligram?

5_H82ywrQIY-009|We're looking at barium sulfate in water, and dissolving a tiny, milligram speck, and the most efficient way to do that.

5_H82ywrQIY-010|Should we add some acid?

5_H82ywrQIY-011|H2SO4?

5_H82ywrQIY-012|Should we add barium chloride, BaCl2, or water?

5_H82ywrQIY-013|Well, if you look at the reaction for barium sulfate solid dissolving in water, barium sulfate forms barium ions and sulfate ions.

5_H82ywrQIY-018|What you want is more ions to get this to dissolve.

5_H82ywrQIY-019|Sulfuric acid, H2SO4, the same thing, you're going to add more sulfate ions to solution.

5_H82ywrQIY-021|So in this case, the best strategy is simply add more water, increase the concentration.

5_H82ywrQIY-022|That gives you diluting these concentrations, making them smaller.

5_H82ywrQIY-023|By Le Chatelier's principle, I'll shift to make these concentrations higher again.

5_H82ywrQIY-024|So in this case, add more water to dissolve a one milligram speck of barium sulfate.

aNEDU6EL8jc-000|Let's look at the course of chemical reactions.

aNEDU6EL8jc-006|And I can calculate this value Q at any time.

aNEDU6EL8jc-013|So I can calculate this number Q at any time.

aNEDU6EL8jc-016|You can see they flatline here.

aNEDU6EL8jc-017|There is no change with time for the concentrations.

aNEDU6EL8jc-018|So if you wait long enough, these values of concentration stop changing and that means Q stops changing.

aNEDU6EL8jc-019|It becomes a constant after some time.

aNEDU6EL8jc-021|And your reaction quotient stops changing.

aNEDU6EL8jc-022|It becomes a constant.

aNEDU6EL8jc-024|That is, A and B are still changing into C and D. It's just at the same time, C and D are changing back into A and B.

aNEDU6EL8jc-025|So macroscopically, things don't change, but the reverse and forward rates have equalized.

aNEDU6EL8jc-026|So it's a dynamic equilibrium.

aNEDU6EL8jc-029|So it's indeed a constant, and it's a characteristic of the chemical reaction.

aNEDU6EL8jc-030|So if you measure lots of these K's you can predict where the reaction's going to go.

aNEDU6EL8jc-032|The numerator is the products.

aNEDU6EL8jc-033|So if these products are too small-- because Q is too small-- I should go make more of them.

aNEDU6EL8jc-034|So I should go towards products.

aNEDU6EL8jc-035|So comparing Q that you measure at any time with your known constant gives you predictive power.

aNEDU6EL8jc-036|Which way is that reaction going to go?

aNEDU6EL8jc-037|Same thing, if you measure Q and its bigger than K, well, that says, this numerator's too large.

aNEDU6EL8jc-038|Too many products, I should go back to reactants, because I know Q's always go back to the K.

aNEDU6EL8jc-039|If I wait long enough, reactions will proceed.

aNEDU6EL8jc-040|So Q, the instantaneous concentrations, stops changing at this value of K.

aNEDU6EL8jc-041|And it will progress towards K in a relatively straightforward monotonic fashion.

aNEDU6EL8jc-042|So here we have Q equal K at equilibrium.

aNEDU6EL8jc-043|Now, the approach from Q to K can be monotonic, as I've written it, but it actually could also oscillate.

aNEDU6EL8jc-044|Whatever it is, it'll get to K eventually.

aNEDU6EL8jc-045|So Q equals K at equilibrium.

aNEDU6EL8jc-046|And what I have is a comparison between Q and K will tell me which direction I have to go to get equilibrium.

aNEDU6EL8jc-047|I can measure this value of K. What I need to do is measure it at different temperatures because K will change with temperature.