text stringlengths 19 206 |
|---|
Hx9dFbQ2zhk-052|So the molecular orbital picture more powerful in determining a molecular property than the Lewis electron dot structure. |
VabAZE71dH0-000|Let's look at the chemical reaction hydrogen plus oxygen go into water under three different sets of circumstances. |
VabAZE71dH0-001|One, will ignite the hydrogen and oxygen. Two, will add platinum. |
VabAZE71dH0-002|And three, will just let the reaction go. |
VabAZE71dH0-011|We're looking at the chemical reaction hydrogen and oxygen forming water under three sets of circumstances. |
VabAZE71dH0-012|One, we ignite the reaction. |
VabAZE71dH0-013|Two, we add platinum. |
VabAZE71dH0-014|And three, we allow the reaction to proceed. |
VabAZE71dH0-015|Now, we've seen each of these in the demo lab. |
VabAZE71dH0-016|And when we ignite the chemical reaction it goes very rapidly. |
VabAZE71dH0-017|When you add platinum, the reaction goes at room temperature. |
VabAZE71dH0-018|Platinum is a catalyst for this chemical reaction and it lowers the activation energy. |
VabAZE71dH0-021|Reaction C and reaction A have the same activation energy. |
VabAZE71dH0-022|This one will go faster because we add energy, but their activation energies are the same. |
X0nCoOX0LAg-000|Let's look at a real gas that we're all familiar with, water vapor. |
X0nCoOX0LAg-006|Above that temperature, you'll always have a single phase. |
X0nCoOX0LAg-007|That is, you can compress the gas down. |
X0nCoOX0LAg-008|That phase might get very dense, but I won't see distinct liquid and gas phase. |
X0nCoOX0LAg-009|Below the critical temperature, I'll see distinct liquid and gas phases. |
X0nCoOX0LAg-013|That's 100 degrees C. |
X0nCoOX0LAg-014|Where does this flat line occur at 100 degrees C? |
X0nCoOX0LAg-015|Well, I think you can guess. |
X0nCoOX0LAg-016|It's going to be one atmosphere of pressure. |
X0nCoOX0LAg-020|So it's the partial pressure of the gas phase water above the liquid for that temperature. |
X0nCoOX0LAg-021|The normal boiling point is defined as where the vapor pressure equals atmospheric pressure. |
X0nCoOX0LAg-026|That's near room temperature. |
X0nCoOX0LAg-027|So the vapor pressure of water at room temperature is [? 0.31 ?] of an atmosphere. |
X0nCoOX0LAg-028|So that's the partial pressure of water above liquid water at room temperature. |
X0nCoOX0LAg-029|So the total pressure, if the sample is open to the atmosphere, you'll have a atmosphere of the atmospheric gases, nitrogen and oxygen. |
X0nCoOX0LAg-030|And [? 0.31 ?] of an atmosphere of water vapor pressure. |
X0nCoOX0LAg-031|So the vapor pressure can be defined as the pressure at which the gas spontaneously condenses for a given temperature. |
X0nCoOX0LAg-032|Or it can be defined as the partial pressure of the gas above a liquid sample. |
100n-E7o9Ug-000|When we make molecular orbitals from atomic orbitals, we take the atomic orbital wave functions and use linear combinations. |
100n-E7o9Ug-001|We add and subtract them with constant coefficients to make molecular orbitals. |
100n-E7o9Ug-005|So a helium molecule has four total electrons, and they need to occupy the molecular orbitals. |
100n-E7o9Ug-009|And I find that both the bonding and antibonding orbital for this molecule are completely full. |
100n-E7o9Ug-010|Now the bonding orbital is favorable for bonding. |
100n-E7o9Ug-011|The antibonding orbital is unfavorable for bonding. |
100n-E7o9Ug-012|A lower energy interaction and a higher energy interaction. |
100n-E7o9Ug-013|So those two cancel each other out. |
100n-E7o9Ug-017|Now I'm looking at a 2s electron. |
100n-E7o9Ug-018|A 2s electron I can take the 2s orbitals and use this same procedure. |
100n-E7o9Ug-019|Add and subtract them to take a linear combination, form a sigma 2s and a sigma 2s star antibonding orbital. |
100n-E7o9Ug-020|Those molecular orbitals are filled with the electrons from lithium. |
100n-E7o9Ug-021|Lithium has one electron in its 2s orbital, so two lithiums, two electrons, they fill up the bonding orbital for the lithium molecule. |
100n-E7o9Ug-022|Li2 we would predict to be a stable molecule. |
100n-E7o9Ug-023|It has a bond order of one. |
100n-E7o9Ug-024|Now the bond order will define as the sum of the bonding electrons minus the antibonding electrons. |
100n-E7o9Ug-027|Be2 has four electrons. |
100n-E7o9Ug-028|I'll put two electrons in the bonding of sigma and the two electrons in the antibonding sigma star orbital. |
100n-E7o9Ug-029|I'll calculate the bond order, two electrons here for bonding minus two antibonding electrons. |
100n-E7o9Ug-030|The bond order is zero. |
100n-E7o9Ug-031|So Be2, the beryllium 2 molecule, doesn't form. |
100n-E7o9Ug-032|So my molecular orbital theory has predictive power and it can predict the bond order and whether or not bonds form in molecules. |
c-yoRAiBQ1k-001|So to which energy level scheme-- A, B, or C-- does this emission spectrum correspond? |
c-yoRAiBQ1k-007|Let's look at the relationship between emission spectra and energy-level spacing in the matter. |
c-yoRAiBQ1k-008|So if you have an energy-level spacing that looks like A, what would the emission spectrum look like? |
c-yoRAiBQ1k-009|Well, we have to look at every possible transition in the system. |
c-yoRAiBQ1k-010|And here you can see three high energy level transitions-- those would give you high frequency lines-- and three low. |
c-yoRAiBQ1k-011|You could have this tiny transition, this tiny transition, and this tiny transition. |
c-yoRAiBQ1k-012|These two of equal energy, so they're degenerate. |
c-yoRAiBQ1k-013|They would fall right on top of each other and give you only one line even though there's two transitions. |
c-yoRAiBQ1k-014|But the two transitions have the same energy, so we can't resolve them in terms of energy. |
c-yoRAiBQ1k-015|So you get just two lines, one for this transition and one corresponding to both of these transitions. |
c-yoRAiBQ1k-016|So that doesn't look like the right answer. |
c-yoRAiBQ1k-018|They would fall right on top of each other. |
c-yoRAiBQ1k-019|So you'd have a very high energy level, a medium and a medium at the same energy, and a low in your high band. |
c-yoRAiBQ1k-020|That's a total of three from these four transitions since two are exactly the same. |
c-yoRAiBQ1k-021|And then two tiny low energy transitions, but again, they are of equal energy, so that would give you one line. |
c-yoRAiBQ1k-022|That looks like the spectrum we've seen, and if you look at C, that, of course, isn't anything like what we see. |
c-yoRAiBQ1k-024|So this pair gives you one line, this pair gives you one line, two intermediate lines, and then a small energy transition. |
c-yoRAiBQ1k-028|And remember, you can't look at tiny little matter in your microscope. |
c-yoRAiBQ1k-030|So here the correct answer is B. |
Ie1qr8XGi6g-000|When you form compounds from elements, there'll be an enthalpy associated with that. |
Ie1qr8XGi6g-001|So let's start with the elements in their standard states and form compounds. |
Ie1qr8XGi6g-002|What are the standard states of the elements? |
Ie1qr8XGi6g-003|Well, that's how you would find an element at 1 atmosphere of pressure and 25 degrees C. |
Ie1qr8XGi6g-004|So some elements from the periodic table will be a solid. |
Ie1qr8XGi6g-005|Some will be liquids. |
Ie1qr8XGi6g-006|Some will be gases. |
Ie1qr8XGi6g-007|Some will be diatomic. |
Ie1qr8XGi6g-012|Those are lower on enthalpy hill. |
Ie1qr8XGi6g-013|Overall, those compounds are a little more likely to be found because they're lower in energy than the elements. |
Ie1qr8XGi6g-014|So we have standard enthalpies of formation are the formation of compounds from their elements in their standard states. |
9sDdIaBhtgk-000|Let's look at weak acid, weak base, strong acid, and strong base solutions. |
9sDdIaBhtgk-004|So the pH is about one. |
9sDdIaBhtgk-006|The HAc weak acid dissociates to a lesser extent. |
9sDdIaBhtgk-007|So the pH is higher, indicating a lower concentration of H3O plus. |
9sDdIaBhtgk-008|Water, concentrations of H3O pluses at around 10 to the minus 7. |
9sDdIaBhtgk-009|NH3. |
9sDdIaBhtgk-010|Now we're getting to a base. |
9sDdIaBhtgk-011|So the H3O plus concentration drops even further, so I get a higher pH, 10 to the 10. |
9sDdIaBhtgk-013|The OH minus and the H3O plus concentrations, the product is always 10 to the minus 14 in water. |
9sDdIaBhtgk-014|So I can look at these, and we can put a little indicator in each. |
9sDdIaBhtgk-016|There's my strong acid solution. |
9sDdIaBhtgk-017|Here's my weak acid solution. |
9sDdIaBhtgk-018|Let's add a little more so that we get a little darker color. |
9sDdIaBhtgk-019|Here's my strong acid solution. |
9sDdIaBhtgk-020|Here's my weak acid solution. |
9sDdIaBhtgk-021|I already added a little indicator to the water, so we'll skip over that one. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.