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VisualTTS_Chem

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jv9_JbddC7Y-017|So you have 0.1x equals 10 to the minus 10, or x equals 10 to the minus 9.
jv9_JbddC7Y-018|That's a very small concentration.
jv9_JbddC7Y-019|If you had barium sulfate in pure water, then the concentration, you can figure that would be x squared equals Ksp.
jv9_JbddC7Y-020|The concentration would be 10 to the minus 5th.
jv9_JbddC7Y-021|So adding just a little sulfate reduces the barium sulfate concentration by a factor of 10 to the 4.
SpwSZfpYOLY-000|Let's talk more about the strength of ionic interactions.
SpwSZfpYOLY-001|There's two components to an ionic interaction.
SpwSZfpYOLY-002|One is the charge-- in this case, plus 1 and minus 1.
SpwSZfpYOLY-003|The other is the distance between them.
SpwSZfpYOLY-004|As the distance gets smaller, that ionic interaction gets stronger and that forms a stronger bond.
SpwSZfpYOLY-006|When chlorine approaches a lithium or a sodium or potassium.
SpwSZfpYOLY-007|The chlorine wants to pull an electron.
SpwSZfpYOLY-008|The sodium and lithium and potassium want to give up an electron.
SpwSZfpYOLY-009|The question is at what distance does it make sense for them to do so?
SpwSZfpYOLY-010|When is it energetically efficient for the electron to move from the potassium or the sodium or the lithium over to the chlorine?
SpwSZfpYOLY-011|Let me ask you this question.
SpwSZfpYOLY-012|The R, the distance between nuclear centers, where an electron jumps from one element to the other varies.
SpwSZfpYOLY-020|We're looking at the formation of chlorides and we're trying to decide when the electron jumps from either sodium, potassium, or lithium onto the chlorine.
SpwSZfpYOLY-021|We know when it does, there'll be an ionic bond formation.
SpwSZfpYOLY-022|So we're balancing two things.
SpwSZfpYOLY-023|We get a bunch of energy out when we form the ionic bond.
SpwSZfpYOLY-024|But it takes a bunch of energy to transfer that electron from the potassium, sodium, or lithium over to the chlorine.
SpwSZfpYOLY-025|The question is, how do I balance that distance?
SpwSZfpYOLY-026|The closer the distance, the stronger the bond.
SpwSZfpYOLY-027|So I get more and more energy released the closer I get.
SpwSZfpYOLY-028|If I'm far apart, I don't get as much energy released.
SpwSZfpYOLY-029|So if I'm far apart, I need it to be easy to peel the electron off the sodium, potassium, or lithium.
SpwSZfpYOLY-030|Whereas if I get close together, I can peel the electron off something that's less willing to give it up.
SpwSZfpYOLY-031|So it's the ionization energy of the sodium or lithium or potassium that I have to look at.
SpwSZfpYOLY-036|I can do so at a greater distance.
SpwSZfpYOLY-037|Potassium's the right answer in this case.
SpwSZfpYOLY-038|Now there's a lot going on in this chem quiz.
SpwSZfpYOLY-039|We had to talk about ionization.
SpwSZfpYOLY-040|We had to talk about coulombic interaction.
SpwSZfpYOLY-041|We had to talk about ionization energy.
SpwSZfpYOLY-042|If you didn't get everything here, don't worry about that.
SpwSZfpYOLY-043|We'll have chem quizzes like this where we bring everything together, and it might be really tough.
SpwSZfpYOLY-044|The point is, stop and think about it.
SpwSZfpYOLY-045|Talk about it with your friends.
SpwSZfpYOLY-046|If you get it, great.
SpwSZfpYOLY-047|If you don't, listen to the explanation again, and see if you can.
BceZjaeLCsw-000|Here's a chemical reaction-- the reaction of iron with oxygen to form iron oxide.
BceZjaeLCsw-001|The question is, what's the enthalpy change for that chemical reaction?
BceZjaeLCsw-007|So we can find the enthalpy for this reaction by looking up on a table the enthalpy of formation of iron oxide.
VYDCI3l22Q0-000|Let's do a calculation involving oxidation numbers determined from Lewis dot structures.
VYDCI3l22Q0-004|Well, in order to do that, we need a Lewis electron dot structure for each molecule.
VYDCI3l22Q0-005|I can draw those here.
VYDCI3l22Q0-006|And I've simplified them a little, because we're interested only in this carbon.
VYDCI3l22Q0-007|So I haven't drawn the full Lewis electron dot structure, but the relevant parts around that carbon.
VYDCI3l22Q0-008|Now, to do an oxidation number, I have to assign the electrons based on relative electronegativities.
VYDCI3l22Q0-009|I'm going to assign electrons to this carbon based on electronegativities on the things it's bonded to.
VYDCI3l22Q0-010|This carbon is more electronegative than the hydrogen. So it will get all of these electrons.
VYDCI3l22Q0-011|It's less electronegative than this oxygen.
VYDCI3l22Q0-012|So it won't get any of those.
VYDCI3l22Q0-013|It's equally electronegative with this carbon.
VYDCI3l22Q0-014|So to count them up, the carbon gets two electrons from that bond.
VYDCI3l22Q0-015|It gets two electrons from this bond, being more electronegative than the hydrogen.
VYDCI3l22Q0-017|So two, four, five electrons around carbon in that molecule.
VYDCI3l22Q0-018|Carbon, as an atom, has four electrons around it.
VYDCI3l22Q0-019|This carbon has one more than that.
VYDCI3l22Q0-020|So it has an oxidation number of minus one.
VYDCI3l22Q0-021|How about this molecule?
VYDCI3l22Q0-022|Again, we'll do the same thing.
VYDCI3l22Q0-023|It'll get one electron here, because it's equally electronegative with the carbon next to it.
VYDCI3l22Q0-024|And it will get two electrons from that hydrogen, because that bond, both electrons will be assigned to the more electronegative carbon.
VYDCI3l22Q0-025|And none in this double bond, because those electrons will be assigned to the oxygen.
VYDCI3l22Q0-026|So one, two, three is the number of electrons here.
VYDCI3l22Q0-027|Normally, it has four.
VYDCI3l22Q0-030|So the oxidation number changes two steps in this oxidation reaction.
hXLiubyd58Y-001|Now, the reaction goes because the reaction of the weak acid and the strong base forms water.
hXLiubyd58Y-002|And the formation of the water is very favored.
hXLiubyd58Y-003|So this reaction lies very strongly towards the products.
hXLiubyd58Y-005|Let's see what that reaction looks like.
hXLiubyd58Y-006|We have a typical titration curve plotted out here.
hXLiubyd58Y-007|And at several points in the titration curve, the pH is relatively easy to calculate.
hXLiubyd58Y-011|But we can also talk about here point A, the beginning of the titration, the initial point.
hXLiubyd58Y-012|That's when I have simply a solution of the weak acid.
hXLiubyd58Y-013|Haven't added any base yet.
hXLiubyd58Y-015|If you think about what's happening at point A, it really is just a solution of a weak acid.
hXLiubyd58Y-017|So we've converted half of the HA to A minus.
hXLiubyd58Y-018|And these concentrations are about equal.
hXLiubyd58Y-019|In fact, you could find a point directly in the center where they are equal, the HA concentration is equal to the A minus concentration.
hXLiubyd58Y-020|This region is called the buffer region.
hXLiubyd58Y-021|When HA concentrations and A minus concentrations are about equal, the solution resists changes in pH.
hXLiubyd58Y-022|And you can see the line is relatively flat in this region.
hXLiubyd58Y-023|That resistance to change in pH is called buffering.
hXLiubyd58Y-024|And the property of weak acids to act like buffers is very important.
hXLiubyd58Y-028|And I get to what's called the end point or the equivalence point or the stoichiometric point.
hXLiubyd58Y-029|All different names for the same thing, where you've added a mole of base for every mole of acid that you originally had.
hXLiubyd58Y-030|So these are the major points along the titration curve.
hXLiubyd58Y-031|Out here at point D, it's simply really a solution of just a strong base.
hXLiubyd58Y-032|You've used up all your weak acid, and now the pH rapidly changes because you continue to add strong base.
hXLiubyd58Y-033|And then the strong base dominates the pH.
hXLiubyd58Y-034|So these are the major points along a titration curve of a weak acid by a strong base.
BwPxen6eBV8-006|We're trying to determine which of these structural isomers is chiral.
BwPxen6eBV8-007|And remember, chirality involves non-superimposable mirror images, and that's a very subtle thing.
BwPxen6eBV8-008|But the basis of chirality is an asymmetric center, so a carbon with four different things attached, and that's the key, that's what we have to look for in our molecules.
BwPxen6eBV8-009|It's much too difficult to draw the molecules and try to imagine them superimposing in their mirror images.
BwPxen6eBV8-010|The thing that we can really latch onto is a carbon with four things attached.
BwPxen6eBV8-011|So in molecule A, if you look through all the carbons, each one has at least two substituents the same.