ceaglex/VisualTTS_Chem · Datasets at Hugging Face

text stringlengths 19 206
yHIDWAYbYBs-015\|This second balloon has only oxygen.
yHIDWAYbYBs-016\|Now, many people think that oxygen is a very dangerous gas, very explosive.
yHIDWAYbYBs-017\|But it turns out oxygen is an explosive gas in the presence of other molecules to react with.
yHIDWAYbYBs-018\|Oxygen by itself is not an explosive gas.
yHIDWAYbYBs-019\|It will react rapidly with other molecules.
yHIDWAYbYBs-020\|In this case, we haven't given it any other molecules, except maybe the material in the balloon.
yHIDWAYbYBs-021\|But in general, this is going to give almost no explosion.
yHIDWAYbYBs-022\|The third case, I have hydrogen and oxygen intimately mixed together.
yHIDWAYbYBs-023\|Now in this case when the flame comes through the balloon, hydrogen and oxygen are primed to react.
yHIDWAYbYBs-024\|They're very close to each other.
yHIDWAYbYBs-025\|A hydrogen doesn't have to go very far to find an oxygen.
yHIDWAYbYBs-028\|Now Chemical Quizzes that we're going to do-- ChemQuizzes-- throughout this series are going to require you to think outside the box or outside the balloon.
yHIDWAYbYBs-029\|This one had a lot of chemical principles, especially for early on in the course, but I wanted to get you a feel for what you have to think about to do ChemQuizzes.
yHIDWAYbYBs-030\|You might not have got this one because of the complicated nature of the quick reaction versus the slow reaction, but keep trying.
yHIDWAYbYBs-031\|We're going to give you more principles and more tools to be able to perform on these ChemQuizzes.
eqHY6DH8QRI-000\|For an ideal gas, the product of the pressure and the volume is a constant for a constant temperature.
eqHY6DH8QRI-001\|If you change the temperature, the value of the constant changes, but the product of pressure and volume is still a constant.
eqHY6DH8QRI-002\|Let's look at how changing the temperature affects our curves.
eqHY6DH8QRI-007\|Now the product of pressure and volume being a constant, means pressure and volume are inversely proportional.
eqHY6DH8QRI-008\|As I increase pressure, volume decreases.
eqHY6DH8QRI-009\|Decrease pressure volume increases.
eqHY6DH8QRI-010\|So their product is always a constant.
eqHY6DH8QRI-011\|It turns out, that's how your lungs work.
eqHY6DH8QRI-012\|In order for your lungs to draw air in, your chest cavity expands higher value, that means lower pressure.
eqHY6DH8QRI-013\|And high pressure air from the atmosphere can rush into your lungs.
eqHY6DH8QRI-016\|There's a model chest cavity.
eqHY6DH8QRI-017\|The model chest cavity has that muscle in your chest, the diaphragm.
eqHY6DH8QRI-018\|And when that contracts, the volume of your chest cavity increases.
eqHY6DH8QRI-019\|Higher volume, lower pressure, air rushes in.
eqHY6DH8QRI-022\|Decrease volume, higher pressure, air rushes out.
eqHY6DH8QRI-025\|Taking advantage of the relationship p times v is the constant.
9mTP53W2_RY-000\|Glucose is an energy source for your body.
9mTP53W2_RY-001\|That is glucose is oxidized to release energy.
9mTP53W2_RY-002\|Now, that oxidation you could do on the benchtop.
9mTP53W2_RY-003\|You could burn glucose in oxygen and it would release energy.
9mTP53W2_RY-004\|In your body, it's done via enzymes, and the reaction occurs at body temperatures.
9mTP53W2_RY-006\|So if the initial and final states are the same, we know that free energy and enthalpy and entropy changes are the same.
9mTP53W2_RY-007\|They're a state function.
9mTP53W2_RY-008\|So the same amount of energy that you release on the benchtop is available to your body when you oxidize glucose in your body.
9mTP53W2_RY-013\|So rather than drawing it as a ring, I just bent this bond all the way around to show that this carbon is attached to that carbon.
9mTP53W2_RY-014\|Here's oxygen, carbon dioxide, and water.
9mTP53W2_RY-015\|And we know how to assign oxidation numbers from Lewis structures.
9mTP53W2_RY-019\|So oxygen gets these two.
9mTP53W2_RY-020\|Carbon and carbon here, that's the same elements.
9mTP53W2_RY-021\|So I'll share these electrons and assign one to this carbon.
9mTP53W2_RY-022\|And carbon is slightly more electronegative than hydrogen, so I'll assign both of them to the carbon here.
9mTP53W2_RY-023\|So I'm assigning one, two, three, four electrons to the carbon based on relative electronegativities.
9mTP53W2_RY-024\|And carbon normally has four electrons around it in its valence shell.
9mTP53W2_RY-025\|So it has oxidation number 0.
9mTP53W2_RY-027\|Then I see that it has a plus 4 oxidation number here.
9mTP53W2_RY-028\|So it goes from 0 to plus 4, that's an increase in oxidation number-- formerly, an oxidation.
9mTP53W2_RY-029\|So something has to be reduced.
9mTP53W2_RY-030\|If there's an oxidation, there must be a reduction as well.
9mTP53W2_RY-031\|Here's elemental oxygen, where the oxidation number is 0.
9mTP53W2_RY-032\|And what about the oxygen over here?
9mTP53W2_RY-033\|The oxygen in water, I assign all of the electrons to the oxygen. It's the more electronegative element.
9mTP53W2_RY-034\|So it has 1, 2, 3, 4, 5, 6, 7, 8 electrons around it.
9mTP53W2_RY-035\|The atomic oxygen would have 6, so it has two more than it normally has as an atom.
9mTP53W2_RY-036\|So that's an oxidation number of minus 2.
9mTP53W2_RY-037\|Indeed, that's a reduction.
9mTP53W2_RY-038\|I go from 0 to minus 2.
9mTP53W2_RY-039\|A reduction in oxidation number is a reduction.
g0nvNTN4XHc-000\|Let's look at the valence electrons of some atoms and see how they participate in bonding.
g0nvNTN4XHc-001\|Now, we've talked about several kinds of bonding-- ionic bonding and covalent bonding.
g0nvNTN4XHc-002\|Let's look at ionic bonding first.
g0nvNTN4XHc-003\|If you look at lithium and fluorine, sodium and chlorine and their valence electrons.
g0nvNTN4XHc-004\|You'll notice that fluorine and chlorine has seven valence electrons.
g0nvNTN4XHc-005\|If they could pick up one more, they'd get to eight and that stable octet.
g0nvNTN4XHc-006\|Sodium and lithium, should they lose one, then they'll have a closed shell underneath that primary shell.
g0nvNTN4XHc-007\|So lithium and sodium are kind of primed to give up an electron and go to the closed shell underneath.
g0nvNTN4XHc-008\|Fluorine and chlorine are primed to accept electrons and close their existing shells at that stable octet.
g0nvNTN4XHc-009\|So one way that you can describe the bonding is an electron transfer.
g0nvNTN4XHc-010\|One element gives up an electron to another element.
g0nvNTN4XHc-011\|You form a positive ion and a negative ion.
g0nvNTN4XHc-012\|And then Coulombic attraction, the attraction between plus and minus charges, holds those elements together in a bond.
g0nvNTN4XHc-013\|So for lithium and fluorine, an electron goes from the lithium to the fluorine.
g0nvNTN4XHc-014\|You get a Li plus and F minus holding each other together.
g0nvNTN4XHc-016\|And this can happen with multiple atoms as well.
g0nvNTN4XHc-018\|It can give up one to each of two chlorines.
g0nvNTN4XHc-019\|It becomes plus 2 charge.
g0nvNTN4XHc-020\|Each chlorine is negative 1 charge.
g0nvNTN4XHc-021\|And all three of those are held together by Coulombic interaction.
g0nvNTN4XHc-022\|So ionic bonding is the transfer of electrons from one species to another species.
g0nvNTN4XHc-023\|And then the plus-minus interaction holds those two together in a bond.
-WxXRDp-IuU-000\|Let's look at the sublimation of two different solids, iodine and carbon dioxide.
-WxXRDp-IuU-002\|Would it look like A, B, or C?
-WxXRDp-IuU-003\|We can start these two subliming.
-WxXRDp-IuU-012\|We're talking about the sublimation of iodine and carbon dioxide.
-WxXRDp-IuU-015\|So with that information, can we determine how the plot of lnK versus 1/T look relative to each other?
-WxXRDp-IuU-016\|Well, lnK versus 1/T, the slope is determined by the enthalpy, the intercept by the entropy.
-WxXRDp-IuU-017\|So the entropy of sublimation probably about the same.
-WxXRDp-IuU-019\|So we expect the entropy of sublimation is determined mainly by the fact that it's a phase change from solid to gas.
-WxXRDp-IuU-020\|Entropy of sublimation about the same means the intercept of these two plots should be about the same.
-WxXRDp-IuU-023\|That sublimes at a very low temperature.
-WxXRDp-IuU-024\|And the bonds of carbon dioxide, not held together in the solid as strongly as iodine.
-WxXRDp-IuU-025\|Because we see the iodine at room temperature and we don't see the solid carbon dioxide at room temperature.
-WxXRDp-IuU-026\|So we expect the enthalpy of sublimation to be lower.
-WxXRDp-IuU-027\|The slope determines the enthalpy.
-WxXRDp-IuU-028\|Enthalpy is determined by the slope for lnK versus 1/T.
-WxXRDp-IuU-029\|So we expect a less dramatic slope for the carbon dioxide, a lower enthalpy of sublimation, so a less negative slope.

yHIDWAYbYBs-015|This second balloon has only oxygen.

yHIDWAYbYBs-016|Now, many people think that oxygen is a very dangerous gas, very explosive.

yHIDWAYbYBs-017|But it turns out oxygen is an explosive gas in the presence of other molecules to react with.

yHIDWAYbYBs-018|Oxygen by itself is not an explosive gas.

yHIDWAYbYBs-019|It will react rapidly with other molecules.

yHIDWAYbYBs-020|In this case, we haven't given it any other molecules, except maybe the material in the balloon.

yHIDWAYbYBs-021|But in general, this is going to give almost no explosion.

yHIDWAYbYBs-022|The third case, I have hydrogen and oxygen intimately mixed together.

yHIDWAYbYBs-023|Now in this case when the flame comes through the balloon, hydrogen and oxygen are primed to react.

yHIDWAYbYBs-024|They're very close to each other.

yHIDWAYbYBs-025|A hydrogen doesn't have to go very far to find an oxygen.

yHIDWAYbYBs-028|Now Chemical Quizzes that we're going to do-- ChemQuizzes-- throughout this series are going to require you to think outside the box or outside the balloon.

yHIDWAYbYBs-029|This one had a lot of chemical principles, especially for early on in the course, but I wanted to get you a feel for what you have to think about to do ChemQuizzes.

yHIDWAYbYBs-030|You might not have got this one because of the complicated nature of the quick reaction versus the slow reaction, but keep trying.

yHIDWAYbYBs-031|We're going to give you more principles and more tools to be able to perform on these ChemQuizzes.

eqHY6DH8QRI-000|For an ideal gas, the product of the pressure and the volume is a constant for a constant temperature.

eqHY6DH8QRI-001|If you change the temperature, the value of the constant changes, but the product of pressure and volume is still a constant.

eqHY6DH8QRI-002|Let's look at how changing the temperature affects our curves.

eqHY6DH8QRI-007|Now the product of pressure and volume being a constant, means pressure and volume are inversely proportional.

eqHY6DH8QRI-008|As I increase pressure, volume decreases.

eqHY6DH8QRI-009|Decrease pressure volume increases.

eqHY6DH8QRI-010|So their product is always a constant.

eqHY6DH8QRI-011|It turns out, that's how your lungs work.

eqHY6DH8QRI-012|In order for your lungs to draw air in, your chest cavity expands higher value, that means lower pressure.

eqHY6DH8QRI-013|And high pressure air from the atmosphere can rush into your lungs.

eqHY6DH8QRI-016|There's a model chest cavity.

eqHY6DH8QRI-017|The model chest cavity has that muscle in your chest, the diaphragm.

eqHY6DH8QRI-018|And when that contracts, the volume of your chest cavity increases.

eqHY6DH8QRI-019|Higher volume, lower pressure, air rushes in.

eqHY6DH8QRI-022|Decrease volume, higher pressure, air rushes out.

eqHY6DH8QRI-025|Taking advantage of the relationship p times v is the constant.

9mTP53W2_RY-000|Glucose is an energy source for your body.

9mTP53W2_RY-001|That is glucose is oxidized to release energy.

9mTP53W2_RY-002|Now, that oxidation you could do on the benchtop.

9mTP53W2_RY-003|You could burn glucose in oxygen and it would release energy.

9mTP53W2_RY-004|In your body, it's done via enzymes, and the reaction occurs at body temperatures.

9mTP53W2_RY-006|So if the initial and final states are the same, we know that free energy and enthalpy and entropy changes are the same.

9mTP53W2_RY-007|They're a state function.

9mTP53W2_RY-008|So the same amount of energy that you release on the benchtop is available to your body when you oxidize glucose in your body.

9mTP53W2_RY-013|So rather than drawing it as a ring, I just bent this bond all the way around to show that this carbon is attached to that carbon.

9mTP53W2_RY-014|Here's oxygen, carbon dioxide, and water.

9mTP53W2_RY-015|And we know how to assign oxidation numbers from Lewis structures.

9mTP53W2_RY-019|So oxygen gets these two.

9mTP53W2_RY-020|Carbon and carbon here, that's the same elements.

9mTP53W2_RY-021|So I'll share these electrons and assign one to this carbon.

9mTP53W2_RY-022|And carbon is slightly more electronegative than hydrogen, so I'll assign both of them to the carbon here.

9mTP53W2_RY-023|So I'm assigning one, two, three, four electrons to the carbon based on relative electronegativities.

9mTP53W2_RY-024|And carbon normally has four electrons around it in its valence shell.

9mTP53W2_RY-025|So it has oxidation number 0.

9mTP53W2_RY-027|Then I see that it has a plus 4 oxidation number here.

9mTP53W2_RY-028|So it goes from 0 to plus 4, that's an increase in oxidation number-- formerly, an oxidation.

9mTP53W2_RY-029|So something has to be reduced.

9mTP53W2_RY-030|If there's an oxidation, there must be a reduction as well.

9mTP53W2_RY-031|Here's elemental oxygen, where the oxidation number is 0.

9mTP53W2_RY-032|And what about the oxygen over here?

9mTP53W2_RY-033|The oxygen in water, I assign all of the electrons to the oxygen. It's the more electronegative element.

9mTP53W2_RY-034|So it has 1, 2, 3, 4, 5, 6, 7, 8 electrons around it.

9mTP53W2_RY-035|The atomic oxygen would have 6, so it has two more than it normally has as an atom.

9mTP53W2_RY-036|So that's an oxidation number of minus 2.

9mTP53W2_RY-037|Indeed, that's a reduction.

9mTP53W2_RY-038|I go from 0 to minus 2.

9mTP53W2_RY-039|A reduction in oxidation number is a reduction.

g0nvNTN4XHc-000|Let's look at the valence electrons of some atoms and see how they participate in bonding.

g0nvNTN4XHc-001|Now, we've talked about several kinds of bonding-- ionic bonding and covalent bonding.

g0nvNTN4XHc-002|Let's look at ionic bonding first.

g0nvNTN4XHc-003|If you look at lithium and fluorine, sodium and chlorine and their valence electrons.

g0nvNTN4XHc-004|You'll notice that fluorine and chlorine has seven valence electrons.

g0nvNTN4XHc-005|If they could pick up one more, they'd get to eight and that stable octet.

g0nvNTN4XHc-006|Sodium and lithium, should they lose one, then they'll have a closed shell underneath that primary shell.

g0nvNTN4XHc-007|So lithium and sodium are kind of primed to give up an electron and go to the closed shell underneath.

g0nvNTN4XHc-008|Fluorine and chlorine are primed to accept electrons and close their existing shells at that stable octet.

g0nvNTN4XHc-009|So one way that you can describe the bonding is an electron transfer.

g0nvNTN4XHc-010|One element gives up an electron to another element.

g0nvNTN4XHc-011|You form a positive ion and a negative ion.

g0nvNTN4XHc-012|And then Coulombic attraction, the attraction between plus and minus charges, holds those elements together in a bond.

g0nvNTN4XHc-013|So for lithium and fluorine, an electron goes from the lithium to the fluorine.

g0nvNTN4XHc-014|You get a Li plus and F minus holding each other together.

g0nvNTN4XHc-016|And this can happen with multiple atoms as well.

g0nvNTN4XHc-018|It can give up one to each of two chlorines.

g0nvNTN4XHc-019|It becomes plus 2 charge.

g0nvNTN4XHc-020|Each chlorine is negative 1 charge.

g0nvNTN4XHc-021|And all three of those are held together by Coulombic interaction.

g0nvNTN4XHc-022|So ionic bonding is the transfer of electrons from one species to another species.

g0nvNTN4XHc-023|And then the plus-minus interaction holds those two together in a bond.

-WxXRDp-IuU-000|Let's look at the sublimation of two different solids, iodine and carbon dioxide.

-WxXRDp-IuU-002|Would it look like A, B, or C?

-WxXRDp-IuU-003|We can start these two subliming.

-WxXRDp-IuU-012|We're talking about the sublimation of iodine and carbon dioxide.

-WxXRDp-IuU-015|So with that information, can we determine how the plot of lnK versus 1/T look relative to each other?

-WxXRDp-IuU-016|Well, lnK versus 1/T, the slope is determined by the enthalpy, the intercept by the entropy.

-WxXRDp-IuU-017|So the entropy of sublimation probably about the same.

-WxXRDp-IuU-019|So we expect the entropy of sublimation is determined mainly by the fact that it's a phase change from solid to gas.

-WxXRDp-IuU-020|Entropy of sublimation about the same means the intercept of these two plots should be about the same.

-WxXRDp-IuU-023|That sublimes at a very low temperature.

-WxXRDp-IuU-024|And the bonds of carbon dioxide, not held together in the solid as strongly as iodine.

-WxXRDp-IuU-025|Because we see the iodine at room temperature and we don't see the solid carbon dioxide at room temperature.

-WxXRDp-IuU-026|So we expect the enthalpy of sublimation to be lower.

-WxXRDp-IuU-027|The slope determines the enthalpy.

-WxXRDp-IuU-028|Enthalpy is determined by the slope for lnK versus 1/T.

-WxXRDp-IuU-029|So we expect a less dramatic slope for the carbon dioxide, a lower enthalpy of sublimation, so a less negative slope.