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VisualTTS_Chem

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6DGPuhFoiJI-020|When you dilute by a factor of 10, initially, this H3O plus concentration drops from 10 to the minus 3 to 10 to the minus 4.
6DGPuhFoiJI-021|If you look at this reaction, what you're doing is you're adding water.
6DGPuhFoiJI-022|If you're adding water, that shifts the reaction towards the product side.
6DGPuhFoiJI-023|If you have a greater volume, you make more ions to fill that volume.
6DGPuhFoiJI-024|So a shift towards the product side.
6DGPuhFoiJI-029|So the answer here is less than 4.
6DGPuhFoiJI-030|Now that's interesting.
6DGPuhFoiJI-031|If you did the strong acid, and this was HCl, that totally dissociates.
6DGPuhFoiJI-032|So you have H3O plus, 10 to the minus 3 originally, you dilute by a factor of 10.
6DGPuhFoiJI-033|It goes to 10 to the minus 4.
6DGPuhFoiJI-034|And there's no more H's to be had.
6DGPuhFoiJI-035|HCl completely dissociates in one shot.
6DGPuhFoiJI-036|So the pH can't change after the initial dilution.
6DGPuhFoiJI-038|In this case the pH less than 4.
jUhvGL5Jwl4-000|Let's look at a reaction with iron oxide that produces molten iron.
jUhvGL5Jwl4-001|So, which of the following-- carbon, iron, or aluminum-- should I react with iron oxide-- and I'll give you a hint, it will be an exchange reaction.
jUhvGL5Jwl4-010|Well, let's look at the possible oxides of carbon, iron, and aluminum with respect to the elements in their standard states.
jUhvGL5Jwl4-019|So, let's look at that.
jUhvGL5Jwl4-020|Here is aluminum oxide on our table relative to the elements in their standard state.
jUhvGL5Jwl4-023|So if I convert between one and another, I can tell what the entropy change would be.
jUhvGL5Jwl4-026|So the correct answer here-- aluminum to form aluminum oxide.
--pzKBW13FE-000|PolyProtic acids are compounds with more than one acidic proton.
--pzKBW13FE-001|Some examples are carbonic acid with two acidic protons, phosphoric with three, and sulfuric acid with two acidic protons.
--pzKBW13FE-002|When you titrate a PolyProtic acid, in general, the protons come off stepwise.
--pzKBW13FE-003|Another way to say that is the pKas are often separated by many units.
--pzKBW13FE-004|So you'll have an independent equilibrium for each of the acidic protons.
--pzKBW13FE-008|At pH 6.1 there would be an equal mixture of both of those.
--pzKBW13FE-013|And notice I use the terminology, the proton comes off.
--pzKBW13FE-015|So I can sketch out a titration curve.
--pzKBW13FE-016|Here's a titration curve with two buffer regions and two equivalence points.
--pzKBW13FE-019|PHs below that, the acid form here dominates, doubly protonated.
--pzKBW13FE-020|Above that you have the singly protonated species, one proton removed.
--pzKBW13FE-021|And that species, they'll be 100% that species at the first equivalence point.
--pzKBW13FE-023|As you pass through pH 10.2, then the doubly deprotonated, the doubly basic form predominates.
--pzKBW13FE-024|And then you'll get to an equivalence point where it's 100% essentially of the doubly deprotonated form.
--pzKBW13FE-025|So by looking at the pH and the various pKas, you can predict which forms are present in solution.
--pzKBW13FE-026|That's the titration curve for a PolyProtic acid.
upr1m4DR00M-000|Let's look at the pressure-volume relationship for a real gas.
upr1m4DR00M-001|Now, if the temperature is high enough, real gases behave like ideal gases.
upr1m4DR00M-002|That is, they have a PV plot that looks like PV is a constant.
upr1m4DR00M-005|The critical temperature defines the relationship between ideal and real gas behavior.
upr1m4DR00M-006|Above the critical temperature gases behave like ideal gases, and below the critical temperature they behave like real gases.
upr1m4DR00M-007|So what is real gas behavior?
upr1m4DR00M-013|Well, how do I decrease volume and not increase the pressure?
upr1m4DR00M-014|I can't change the temperature.
upr1m4DR00M-015|We're on an isotherm.
upr1m4DR00M-016|So the only other parameter left is the number of particles, but this is a fixed cylinder.
upr1m4DR00M-017|How do I decrease the number of particles in that cylinder?
upr1m4DR00M-018|Well, I can't decrease the number of particles, but I can change their phase.
upr1m4DR00M-019|I can take some particles from the gas phase to the liquid phase, where they occupy very, very small volume by comparison.
upr1m4DR00M-020|So right here, what's happening is particles are moving from the gas phase to the liquid phase as I decrease the volume.
upr1m4DR00M-021|Since the number of particles decrease, the pressure can stay the same.
upr1m4DR00M-023|So ideal gases, as you compress them, the compressibility decreases.
upr1m4DR00M-024|Real gases, as you compress them below their critical temperature, will undergo a phase transition.
upr1m4DR00M-025|Phase transition is the critical separating point between real and ideal gases.
U6uWvuiO1rA-000|Let's look at lactic acid and try to decide which atomic orbitals make up the molecular orbitals in a bond.
U6uWvuiO1rA-001|We'll look at the carbon-oxygen bond shown here.
U6uWvuiO1rA-011|We're talking about the molecular orbitals that are formed between the carbon and the oxygen in lactic acid.
U6uWvuiO1rA-012|So the carbon has to accommodate this oxygen, this oxygen, and this carbon.
U6uWvuiO1rA-013|It'll be sp2 hybridized, accommodate three things with three equivalent orbitals.
U6uWvuiO1rA-014|The oxygen has to accommodate the carbon, the hydrogen, and two lone pairs, four things.
U6uWvuiO1rA-015|So it'll have four equivalent orbitals or sp3's.
U6uWvuiO1rA-016|So the molecular orbital is going to be an overlap of the sp3 on the oxygen with the sp2 on the carbon.
U6uWvuiO1rA-017|Those will form a sigma bonding, an antibonding orbital, to make up this sigma bond.
U6uWvuiO1rA-018|So the correct answer here sp2 on carbon to sp3 on oxygen.
KtxKNXYhpy4-000|We can use the fact that enthalpy is a state function, and it only depends on the initial and final state to track enthalpies and add enthalpies.
KtxKNXYhpy4-002|That process is known as an application of Hess's law.
KtxKNXYhpy4-003|So let's just do that.
KtxKNXYhpy4-004|Here's a process, methane burning and oxygen to give carbon dioxide and liquid water.
KtxKNXYhpy4-005|The enthalpy for that is 890 kilojoules.
KtxKNXYhpy4-014|We can look at that a little differently.
KtxKNXYhpy4-015|We can look at the relative enthalpy scale.
KtxKNXYhpy4-019|So the enthalpy of vaporization is added here and takes this final state to a higher state by the enthalpy of vaporization of water.
KtxKNXYhpy4-022|So I can add this enthalpy and this enthalpy to get to this enthalpy-- two known steps to get to an unknown step.
KtxKNXYhpy4-023|That's an application of Hess's law.
9j22V8j_x0M-004|This rate will react to a temperature change more dramatically than the forward rate.
9j22V8j_x0M-007|Now, let's look at the effect of a catalyst.
9j22V8j_x0M-008|Here's the both reactions under the effect of a catalyst.
9j22V8j_x0M-009|A catalyst lowers activation energies.
9j22V8j_x0M-013|So you can run a catalyzed chemical reaction and the same amount of energy will be released or absorbed when that reaction proceeds.
9j22V8j_x0M-014|So this is a summary of the thermodynamic and kinetic variables in a chemical reaction.
-AOyf8eA1ic-000|We understand the relationship between free energy and temperature.
-AOyf8eA1ic-001|The standard state free energy is delta H minus T delta S.
-AOyf8eA1ic-007|Now, I have a linear function of natural log k and 1 over T.
-AOyf8eA1ic-010|So this slope I expect to be positive.
-AOyf8eA1ic-013|Heat is a reactant, so if I raise the temperature, I start to favor products.
-AOyf8eA1ic-014|For exothermic chemical reactions, heat is a product.
-AOyf8eA1ic-015|So as I add heat, I start to favor reactants.
-AOyf8eA1ic-016|k gets smaller.
-AOyf8eA1ic-017|Now, this helps me in another way.
-AOyf8eA1ic-018|This k is something that's relatively easy to measure.
-AOyf8eA1ic-019|I can mix reactants together, let them go to equilibrium-- essentially, wait some time.
-AOyf8eA1ic-020|When the macroscopic properties stop changing, I can measure the concentrations or pressures.
-AOyf8eA1ic-021|When the macroscopic pressures and concentration stop changing, I'm at equilibrium.
-AOyf8eA1ic-022|So I can measure equilibrium constants.
-AOyf8eA1ic-023|I can also measure temperatures.
-AOyf8eA1ic-026|Measure k versus T, plot it out, and you can measure entropies, which are very difficult to measure in the laboratory.
-AOyf8eA1ic-027|So the relationship between k and temperature, linear in lnK versus 1 over T, is displayed here.
fCGDtkmkMgU-000|Let's look at a plating reaction, where an ion from solution plates out onto a metal.
fCGDtkmkMgU-008|If I look at my standard reduction potentials, I can find that silver, as a reduction, has a higher potential than copper.