ceaglex/VisualTTS_Chem · Datasets at Hugging Face

text stringlengths 19 206
pB0yR5kXQjI-058\|But that gives me many more 90 degree lone pair-bond pair interactions.
pB0yR5kXQjI-059\|And those are very sterically bad.
pB0yR5kXQjI-060\|The 90-degree lone pair-bond pair arrangements we want to minimize in our structures.
pB0yR5kXQjI-061\|So this is a better arrangement of five things-- two lone pairs, three atoms.
pB0yR5kXQjI-062\|We can go to steric number 6.
pB0yR5kXQjI-063\|Steric number 6 will have an octahedron configuration.
pB0yR5kXQjI-064\|Now remember, it's octahedral because this shape is an octahedron, which has eight sides but six vertices.
pB0yR5kXQjI-065\|So six vertices, each position identical, 90-degree bond angles between all of them.
pB0yR5kXQjI-066\|Some examples-- SF6.
pB0yR5kXQjI-067\|Sulfur hexafluoride, that's steric number 6 around the fluorine.
pB0yR5kXQjI-068\|Each fluorine identical, steric number 6 around the sulfur, each fluorine identical.
pB0yR5kXQjI-069\|Or xenon tetrafluoride.
pB0yR5kXQjI-070\|Again, steric number 6, but this case, two of the items are lone pairs.
pB0yR5kXQjI-071\|And in this case, they are arranged axially, and they give an overall shape to xenon tetrafluoride of square planar.
pB0yR5kXQjI-074\|So we have bonded atoms that determine the overall shape.
pB0yR5kXQjI-075\|When we have bonded atoms and lone pairs, we ignore the lone pairs and name the shape based on bonded atoms alone.
pB0yR5kXQjI-076\|That's how we do shapes of molecules and molecular geometry.
81v3NxDRdLk-000\|Let's look at a few stepwise equilibria and look at manipulations with the equilibrium constants.
81v3NxDRdLk-001\|So here I have nickel hydrate, a complex ion, reacting with ammonia to form a nickel ammonia complex ion.
81v3NxDRdLk-006\|If I take my green solution of the hydrate and add ethylene diamine I'll get a purple ethylene diamine complex.
81v3NxDRdLk-008\|Well, let's look at these equilibrium constants more carefully and see if we can make a determination.
81v3NxDRdLk-011\|And here the nickel ammonia complex with the ethylene diamine as a reactant, displacing that.
81v3NxDRdLk-012\|This is the K we want to determine.
81v3NxDRdLk-015\|And when you add reactions, you multiply equilibrium constants.
81v3NxDRdLk-016\|And we can prove that.
81v3NxDRdLk-017\|Let's do this even a little more explicitly.
81v3NxDRdLk-018\|We'll show K2 is K3 over K1.
81v3NxDRdLk-019\|Or K3 is K1 times K2.
81v3NxDRdLk-020\|We'll write out.
81v3NxDRdLk-036\|So now we're in a position to say, well, what is the reaction, K2?
81v3NxDRdLk-037\|How big is K2?
81v3NxDRdLk-038\|We know it's K3 over K1.
81v3NxDRdLk-039\|We can just do the math here.
81v3NxDRdLk-040\|So if K3 is larger than K1, then K2 will be bigger than 1.
81v3NxDRdLk-041\|And I'll go towards the ethylene diamine.
81v3NxDRdLk-042\|If K3 is less than K1, then this K will be less than 1.
81v3NxDRdLk-045\|So in order to figure this out, we're just going to have to do the experiment.
81v3NxDRdLk-046\|So let's go ahead and do that.
81v3NxDRdLk-047\|I have these down here.
81v3NxDRdLk-048\|And I have the nickel hydrate, two vials.
81v3NxDRdLk-049\|And I have some ammonia here.
81v3NxDRdLk-052\|And indeed, I produced my blue nickel ammonia complex.
81v3NxDRdLk-053\|Now here's the nickel hydrate.
81v3NxDRdLk-058\|Which of these Ks is larger, K3 or K1?
81v3NxDRdLk-060\|So here's ammonia.
81v3NxDRdLk-064\|It looks like this K is going to be bigger than 1.
81v3NxDRdLk-065\|It favors the ethylene diamine Let's prove that.
81v3NxDRdLk-066\|Here is their exact reaction.
81v3NxDRdLk-070\|It will be this violet color if K3 is bigger.
81v3NxDRdLk-071\|It would be a blue color if K3 was smaller.
81v3NxDRdLk-072\|But what we've shown is K3 is bigger.
81v3NxDRdLk-073\|The ethylene diamine is the stronger binding ligand.
81v3NxDRdLk-074\|And what we have is an analysis of equilibrium constants in multiple equilibria.
6BZb96mqmbg-000\|Let's look at some second order chemical kinetics.
6BZb96mqmbg-004\|Or the rate is the product of two concentrations.
6BZb96mqmbg-005\|So the sum of the powers 1 plus 1 gives you second order kinetics.
6BZb96mqmbg-007\|So a plot of 1 over concentration versus time is linear.
6BZb96mqmbg-008\|And that's how you determine if you have a second order chemical reaction.
6BZb96mqmbg-012\|Now, we can look at the half life for second order reactions as well.
6BZb96mqmbg-014\|So the initial amount-- the initial concentration matters in terms of the half life.
6BZb96mqmbg-015\|The amount of time it takes to go from initial concentration to molar down to initial concentration one molar depends on the fact that it's two molar.
6BZb96mqmbg-016\|That is it's going to take longer for four molar to go to two molar than one molar down to half molar.
6BZb96mqmbg-017\|And that tells you something about the kinetics.
6BZb96mqmbg-018\|A second order reaction means there's some collision somewhere in there.
6BZb96mqmbg-020\|Those have to come together.
6BZb96mqmbg-021\|If the concentration of iodine increases, the rate increases, because the number of collisions increase.
6BZb96mqmbg-022\|More iodine there, more collisions.
6BZb96mqmbg-023\|So the frequency of collisions is reflected in this half life.
6BZb96mqmbg-024\|If you have a high initial concentration, you'll have high initial concentrations and high initial collisions.
6BZb96mqmbg-025\|And that will give you a low half life.
6BZb96mqmbg-026\|But if you're a dilute solution, low initial concentration, then you'll have a longer half life.
6BZb96mqmbg-027\|Collisions aren't as frequent.
fiJ6UDSt8hU-000\|Let's talk about the p orbitals in terms of a ChemQuiz.
fiJ6UDSt8hU-001\|Remember, an orbital is a group of three quantum numbers that define a wave function.
fiJ6UDSt8hU-002\|So we need a value of n, a value of l, and a value of m sub l.
fiJ6UDSt8hU-003\|If the value of n is 2, the value of l is 1, and the value of m sub l is 0, that's a p orbital.
fiJ6UDSt8hU-004\|So for p orbitals, I've written a plot of psi and psi squared versus phi, that angle from the positive x-axis.
fiJ6UDSt8hU-009\|We're talking about the 2p orbitals and we're plotting the wave function psi versus the angle phi from the positive x-axis.
fiJ6UDSt8hU-010\|Now, there's three possible 2p orbitals.
fiJ6UDSt8hU-014\|I've chosen to plot the 2py orbital here.
fiJ6UDSt8hU-015\|And I know it's the 2py, because the maximum occurs along the y-axis.
fiJ6UDSt8hU-016\|So let's look at the wave function, which the square is kind of shown here.
fiJ6UDSt8hU-019\|The angle phi gets bigger as you go away from the positive x-axis.
fiJ6UDSt8hU-020\|So this angle phi equals 0 is the x-axis.
fiJ6UDSt8hU-022\|So anywhere along the x-axis or the z-axis, the wave function has to be 0.
fiJ6UDSt8hU-023\|And that's what we have in this case.
fiJ6UDSt8hU-024\|The wave function is 0 for values of phi that are equal to 0.
fiJ6UDSt8hU-025\|If I let phi get bigger, what I'll find is, well, I'll start to sweep into these positive values of the wave function.
fiJ6UDSt8hU-026\|The wave function should go positive.
fiJ6UDSt8hU-029\|That's as big as it's going to get along the positive y-axis.
fiJ6UDSt8hU-030\|So that should be a maximum in the wave function and the square of the wave function.
fiJ6UDSt8hU-031\|And I can continue around.
fiJ6UDSt8hU-032\|I can let phi go to pi.
fiJ6UDSt8hU-033\|I'll come along to the negative x-axis, and of course, anywhere along the x-axis the function has to go to 0, and of course, the square of the wave function goes to 0.
oqN1aemQnu0-000\|The three common phases of matter are solid, liquid, and gas.
oqN1aemQnu0-001\|And we can plot where they occur on a pressure-temperature plot.
oqN1aemQnu0-002\|And when we do, we call that a phase diagram.
oqN1aemQnu0-003\|On a phase diagram, at low pressures, you'd expect the gas phase to exist.
oqN1aemQnu0-008\|This is a plot for carbon dioxide.
oqN1aemQnu0-009\|And here's the liquid-gas equilibrium line.

pB0yR5kXQjI-058|But that gives me many more 90 degree lone pair-bond pair interactions.

pB0yR5kXQjI-059|And those are very sterically bad.

pB0yR5kXQjI-060|The 90-degree lone pair-bond pair arrangements we want to minimize in our structures.

pB0yR5kXQjI-061|So this is a better arrangement of five things-- two lone pairs, three atoms.

pB0yR5kXQjI-062|We can go to steric number 6.

pB0yR5kXQjI-063|Steric number 6 will have an octahedron configuration.

pB0yR5kXQjI-064|Now remember, it's octahedral because this shape is an octahedron, which has eight sides but six vertices.

pB0yR5kXQjI-065|So six vertices, each position identical, 90-degree bond angles between all of them.

pB0yR5kXQjI-066|Some examples-- SF6.

pB0yR5kXQjI-067|Sulfur hexafluoride, that's steric number 6 around the fluorine.

pB0yR5kXQjI-068|Each fluorine identical, steric number 6 around the sulfur, each fluorine identical.

pB0yR5kXQjI-069|Or xenon tetrafluoride.

pB0yR5kXQjI-070|Again, steric number 6, but this case, two of the items are lone pairs.

pB0yR5kXQjI-071|And in this case, they are arranged axially, and they give an overall shape to xenon tetrafluoride of square planar.

pB0yR5kXQjI-074|So we have bonded atoms that determine the overall shape.

pB0yR5kXQjI-075|When we have bonded atoms and lone pairs, we ignore the lone pairs and name the shape based on bonded atoms alone.

pB0yR5kXQjI-076|That's how we do shapes of molecules and molecular geometry.

81v3NxDRdLk-000|Let's look at a few stepwise equilibria and look at manipulations with the equilibrium constants.

81v3NxDRdLk-001|So here I have nickel hydrate, a complex ion, reacting with ammonia to form a nickel ammonia complex ion.

81v3NxDRdLk-006|If I take my green solution of the hydrate and add ethylene diamine I'll get a purple ethylene diamine complex.

81v3NxDRdLk-008|Well, let's look at these equilibrium constants more carefully and see if we can make a determination.

81v3NxDRdLk-011|And here the nickel ammonia complex with the ethylene diamine as a reactant, displacing that.

81v3NxDRdLk-012|This is the K we want to determine.

81v3NxDRdLk-015|And when you add reactions, you multiply equilibrium constants.

81v3NxDRdLk-016|And we can prove that.

81v3NxDRdLk-017|Let's do this even a little more explicitly.

81v3NxDRdLk-018|We'll show K2 is K3 over K1.

81v3NxDRdLk-019|Or K3 is K1 times K2.

81v3NxDRdLk-020|We'll write out.

81v3NxDRdLk-036|So now we're in a position to say, well, what is the reaction, K2?

81v3NxDRdLk-037|How big is K2?

81v3NxDRdLk-038|We know it's K3 over K1.

81v3NxDRdLk-039|We can just do the math here.

81v3NxDRdLk-040|So if K3 is larger than K1, then K2 will be bigger than 1.

81v3NxDRdLk-041|And I'll go towards the ethylene diamine.

81v3NxDRdLk-042|If K3 is less than K1, then this K will be less than 1.

81v3NxDRdLk-045|So in order to figure this out, we're just going to have to do the experiment.

81v3NxDRdLk-046|So let's go ahead and do that.

81v3NxDRdLk-047|I have these down here.

81v3NxDRdLk-048|And I have the nickel hydrate, two vials.

81v3NxDRdLk-049|And I have some ammonia here.

81v3NxDRdLk-052|And indeed, I produced my blue nickel ammonia complex.

81v3NxDRdLk-053|Now here's the nickel hydrate.

81v3NxDRdLk-058|Which of these Ks is larger, K3 or K1?

81v3NxDRdLk-060|So here's ammonia.

81v3NxDRdLk-064|It looks like this K is going to be bigger than 1.

81v3NxDRdLk-065|It favors the ethylene diamine Let's prove that.

81v3NxDRdLk-066|Here is their exact reaction.

81v3NxDRdLk-070|It will be this violet color if K3 is bigger.

81v3NxDRdLk-071|It would be a blue color if K3 was smaller.

81v3NxDRdLk-072|But what we've shown is K3 is bigger.

81v3NxDRdLk-073|The ethylene diamine is the stronger binding ligand.

81v3NxDRdLk-074|And what we have is an analysis of equilibrium constants in multiple equilibria.

6BZb96mqmbg-000|Let's look at some second order chemical kinetics.

6BZb96mqmbg-004|Or the rate is the product of two concentrations.

6BZb96mqmbg-005|So the sum of the powers 1 plus 1 gives you second order kinetics.

6BZb96mqmbg-007|So a plot of 1 over concentration versus time is linear.

6BZb96mqmbg-008|And that's how you determine if you have a second order chemical reaction.

6BZb96mqmbg-012|Now, we can look at the half life for second order reactions as well.

6BZb96mqmbg-014|So the initial amount-- the initial concentration matters in terms of the half life.

6BZb96mqmbg-015|The amount of time it takes to go from initial concentration to molar down to initial concentration one molar depends on the fact that it's two molar.

6BZb96mqmbg-016|That is it's going to take longer for four molar to go to two molar than one molar down to half molar.

6BZb96mqmbg-017|And that tells you something about the kinetics.

6BZb96mqmbg-018|A second order reaction means there's some collision somewhere in there.

6BZb96mqmbg-020|Those have to come together.

6BZb96mqmbg-021|If the concentration of iodine increases, the rate increases, because the number of collisions increase.

6BZb96mqmbg-022|More iodine there, more collisions.

6BZb96mqmbg-023|So the frequency of collisions is reflected in this half life.

6BZb96mqmbg-024|If you have a high initial concentration, you'll have high initial concentrations and high initial collisions.

6BZb96mqmbg-025|And that will give you a low half life.

6BZb96mqmbg-026|But if you're a dilute solution, low initial concentration, then you'll have a longer half life.

6BZb96mqmbg-027|Collisions aren't as frequent.

fiJ6UDSt8hU-000|Let's talk about the p orbitals in terms of a ChemQuiz.

fiJ6UDSt8hU-001|Remember, an orbital is a group of three quantum numbers that define a wave function.

fiJ6UDSt8hU-002|So we need a value of n, a value of l, and a value of m sub l.

fiJ6UDSt8hU-003|If the value of n is 2, the value of l is 1, and the value of m sub l is 0, that's a p orbital.

fiJ6UDSt8hU-004|So for p orbitals, I've written a plot of psi and psi squared versus phi, that angle from the positive x-axis.

fiJ6UDSt8hU-009|We're talking about the 2p orbitals and we're plotting the wave function psi versus the angle phi from the positive x-axis.

fiJ6UDSt8hU-010|Now, there's three possible 2p orbitals.

fiJ6UDSt8hU-014|I've chosen to plot the 2py orbital here.

fiJ6UDSt8hU-015|And I know it's the 2py, because the maximum occurs along the y-axis.

fiJ6UDSt8hU-016|So let's look at the wave function, which the square is kind of shown here.

fiJ6UDSt8hU-019|The angle phi gets bigger as you go away from the positive x-axis.

fiJ6UDSt8hU-020|So this angle phi equals 0 is the x-axis.

fiJ6UDSt8hU-022|So anywhere along the x-axis or the z-axis, the wave function has to be 0.

fiJ6UDSt8hU-023|And that's what we have in this case.

fiJ6UDSt8hU-024|The wave function is 0 for values of phi that are equal to 0.

fiJ6UDSt8hU-025|If I let phi get bigger, what I'll find is, well, I'll start to sweep into these positive values of the wave function.

fiJ6UDSt8hU-026|The wave function should go positive.

fiJ6UDSt8hU-029|That's as big as it's going to get along the positive y-axis.

fiJ6UDSt8hU-030|So that should be a maximum in the wave function and the square of the wave function.

fiJ6UDSt8hU-031|And I can continue around.

fiJ6UDSt8hU-032|I can let phi go to pi.

fiJ6UDSt8hU-033|I'll come along to the negative x-axis, and of course, anywhere along the x-axis the function has to go to 0, and of course, the square of the wave function goes to 0.

oqN1aemQnu0-000|The three common phases of matter are solid, liquid, and gas.

oqN1aemQnu0-001|And we can plot where they occur on a pressure-temperature plot.

oqN1aemQnu0-002|And when we do, we call that a phase diagram.

oqN1aemQnu0-003|On a phase diagram, at low pressures, you'd expect the gas phase to exist.

oqN1aemQnu0-008|This is a plot for carbon dioxide.

oqN1aemQnu0-009|And here's the liquid-gas equilibrium line.