Datasets:
ceaglex
/
VisualTTS_Chem

Dataset card Data Studio Files
xet
 Community
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.