Datasets:
ceaglex
/
VisualTTS_Chem

Dataset card Data Studio Files
xet
 Community
text
stringlengths
19
206
Vzy2wd5oIKs-018|The enthalpy is the energy change plus the pressure volume constant.
Vzy2wd5oIKs-022|Again, for an ideal gas, the enthalpy changes will be directly proportional to temperature changes.
Vzy2wd5oIKs-025|And that's proportional to an energy change for an ideal gas as well-- constant volume, constant pressure, energy changes.
kTJEs_u1fkQ-000|Let's look at the relationship between the macroscopic properties for a sample of gas, the pressure, the volume, the temperature, the number of moles of gas.
kTJEs_u1fkQ-012|This is the ideal gas law, the macroscopic properties of a gas, in one neat equation.
RffOkFeAkEE-001|What's the entropy change?
RffOkFeAkEE-002|Well, we know there's various ways to calculate entropy changes.
RffOkFeAkEE-011|So I choose the middle equation here, the reversible heat over the temperature change.
RffOkFeAkEE-012|So let's just do that.
RffOkFeAkEE-016|So freezing requires a release of heat, so freezing is endothermic.
RffOkFeAkEE-018|This is a positive heat.
RffOkFeAkEE-019|And I find that that's 22 joules per Kelvin mole of water.
RffOkFeAkEE-020|But we're not talking about a mole of water.
RffOkFeAkEE-021|We're talking about just 10 grams of water.
uwL_j9yXkN8-000|Let's look at protein synthesis and how it's coupled to the oxidation of glucose.
uwL_j9yXkN8-001|So how many grams of glucose are required to make a mole of serum albumin?
uwL_j9yXkN8-002|Now, serum albumin is around 100 amino acids.
uwL_j9yXkN8-003|So I need about 100 peptide bonds per mole of serum albumin.
uwL_j9yXkN8-012|We're talking about forming peptide bonds as we oxidize glucose.
uwL_j9yXkN8-014|That's transferred via the conversion of ADP to ATP to the formation of peptide bonds.
uwL_j9yXkN8-015|And you get about four moles of peptide bonds for every mole of glucose that you metabolize.
FmTLXnrPBPQ-000|In oxidation reduction chemistry, oxidations and reductions always occur in pairs.
FmTLXnrPBPQ-001|If you are going to be oxidized, that is you're going to give up some electrons, those electrons have to have a place to land.
FmTLXnrPBPQ-002|They have to reduce some other compound.
FmTLXnrPBPQ-003|So, let's look at a redox reaction and break it up into half cells, and determine the oxidation and reduction pairs.
FmTLXnrPBPQ-004|And use our tables of standard reduction potentials to help us balance that reaction.
FmTLXnrPBPQ-007|So, let's look at those two half cells.
FmTLXnrPBPQ-008|We can go back to our table.
FmTLXnrPBPQ-009|Here's our table of standard reduction potentials.
FmTLXnrPBPQ-014|Here's the permanganate, and here's the bromine.
FmTLXnrPBPQ-018|So we can balance the number of electrons, whenever oxidation and reduction occurs.
FmTLXnrPBPQ-019|Of course, the number of electrons are conserved.
FmTLXnrPBPQ-020|If I reduce you by two electrons, I must accept those two electrons.
FmTLXnrPBPQ-027|Now, we can also calculate the relative cell potential for that.
FmTLXnrPBPQ-028|We can go back to our table.
FmTLXnrPBPQ-032|So, that overall cell potential, 0.43.
FmTLXnrPBPQ-033|Now, notice again I didn't multiply my cell potentials, my standard reduction potentials, by two and five.
FmTLXnrPBPQ-034|That's because cell potential is an intensive property, it's independent of the extent of the system.
FmTLXnrPBPQ-036|That's downhill in free energy.
WvG46BwP-XY-001|So if we take these three atomic orbitals, we can make three new atomic orbitals.
WvG46BwP-XY-004|Now, I've drawn them off-center here.
WvG46BwP-XY-005|So obviously, they're all about the same nucleus.
WvG46BwP-XY-006|So let's collapse those together on the same nucleus.
WvG46BwP-XY-007|And that will obscure the little negative lobes.
WvG46BwP-XY-008|But the three positive lobes are appropriate for steric number three and a bond angle of 120 degrees.
WvG46BwP-XY-009|Let's say we want to accommodate four, steric number four, or tetrahedral symmetry.
WvG46BwP-XY-010|I think you can guess the pattern here.
WvG46BwP-XY-011|It's actually possible to take the s and all three p orbitals.
WvG46BwP-XY-012|Those four orbitals make four linear combinations and make four new atomic orbitals.
WvG46BwP-XY-015|109 degrees, I'll bring them all to the same center here four sp3 orbitals.
WvG46BwP-XY-016|Again, the negative lobes obscured.
WvG46BwP-XY-017|But they'll accommodate a steric number of four and a bond angle of 109 degrees.
zsB5ZigPQHY-000|Let's look at a situation where we're bracketing an oxidation-reduction reaction.
zsB5ZigPQHY-001|So which of the following will be reduced by copper but not by bromide?
zsB5ZigPQHY-009|We're looking for a compound that can oxidize copper metal, but not bromide.
zsB5ZigPQHY-010|So if we look at our table of standard production potentials, we see nitrate is above the copper half cell.
zsB5ZigPQHY-011|That means nitrate can force the oxidation of copper, but not the oxidation of bromide.
Sh42Wpvs42k-000|Light has both the properties of a wave and the properties of a particle.
Sh42Wpvs42k-001|It's a wave and a particle at the same time.
Sh42Wpvs42k-002|There's a duality to the wave particle relationship.
Sh42Wpvs42k-004|Particle properties that are moving, the most important property is the momentum, it's mass times its velocity.
Sh42Wpvs42k-005|So how do we reconcile the two?
Sh42Wpvs42k-006|Well light, photons of light, carry no mass.
Sh42Wpvs42k-007|So how can they carry momentum?
Sh42Wpvs42k-012|So using those two relationships we can derive that the momentum is Planck's constant divided by the wavelength.
Sh42Wpvs42k-016|Wave, particle, acting together, sometimes the properties of the particle exert themselves.
Sh42Wpvs42k-017|Sometimes the properties of the wave exert themselves.
Sh42Wpvs42k-018|They both exist together all the time, particle and wave, nature of light.
DP4HnlNRR_4-000|Nitrogen molecules, N2, have a triple bond.
DP4HnlNRR_4-001|That's predicted both by molecular orbital theory and the Lewis dot structure.
DP4HnlNRR_4-002|What I'd like to think about is the ionization N2 molecules-- that is, withdrawing an electron-- and adding an electron, electron capture by N2.
DP4HnlNRR_4-003|That will form a positive and a negative ion.
DP4HnlNRR_4-004|The question I have is, what's the relative bond strength?
DP4HnlNRR_4-010|We're talking about ionization and electron capture in N2 molecules.
DP4HnlNRR_4-012|And I filled them with the p electrons from nitrogen. Each nitrogen has 3 p electrons for a total of 6 in the molecule.
DP4HnlNRR_4-016|Adding an electron to the next available orbital is an antibonding orbital.
DP4HnlNRR_4-017|So electron capture into this antibonding orbital is going to reduce the bond order and the relative bond strength.
DP4HnlNRR_4-019|So the bond order reduces to 2 and 1/2.
DP4HnlNRR_4-022|And if you calculate the bond order here, again, 2 and 1/2.
DP4HnlNRR_4-023|But interestingly, both electron capture and ionization reduce the bond order of nitrogen by about the same amount.
DP4HnlNRR_4-024|So you'd predict that the bond strengths of N2+ and N2- are about the same.
s_xlDaR53v0-000|Let's look at a broad range of temperatures on the absolute scale.
s_xlDaR53v0-001|Here I've listed several.
s_xlDaR53v0-002|The interior of the Sun for instance-- 10 to the 8th Kelvin, a very high temperature.
s_xlDaR53v0-003|As we go down in temperature, we can go past the surface of the Sun, tens of thousands of degrees.
s_xlDaR53v0-004|The melting point of tungsten-- about 1,000 degrees Kelvin.
s_xlDaR53v0-011|That's a very ideal gas.
s_xlDaR53v0-013|So that's a gas that behaves ideally over a broad range of temperatures.
s_xlDaR53v0-016|If you slow down particles, you reduce their temperature.
s_xlDaR53v0-017|We'll always associate the speed of molecules in a sample with their temperature.
s_xlDaR53v0-018|Higher temperature-- more speed.
s_xlDaR53v0-019|Some of the coldest temperatures ever achieved are in a Bose-Einstein condensation.
s_xlDaR53v0-020|That's where you do laser cooling and magnetic trapping, and get molecules to virtually stop.
inLhIqLOouY-000|When an acid reacts with water, it forms its conjugate base.
inLhIqLOouY-001|Let's look at that.
inLhIqLOouY-010|Again, products over reactants, water doesn't appear, pure water, in equilibrium constant expressions.
inLhIqLOouY-011|Now, I know the stronger the conjugate acid, the weaker the conjugate base intuitively.
inLhIqLOouY-012|The stronger this acid is, the more the equilibrium lies towards Ac-.
inLhIqLOouY-013|So the more Ac- is likely to be in solution, here, the Ac- is favored.
inLhIqLOouY-014|Here the Ac- is favored.