ceaglex/VisualTTS_Chem · Datasets at Hugging Face

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AlwmlHuZRUU-026\|I can take the difference of those two.
AlwmlHuZRUU-027\|And now I'll have high amplitude on this side.
AlwmlHuZRUU-028\|This side will become green, this side will become red.
AlwmlHuZRUU-029\|And I'll have high amplitude on this side, 180 degrees from this high amplitude.
AlwmlHuZRUU-030\|This is a combination of s minus the pz.
AlwmlHuZRUU-036\|I still have my remaining p orbitals that I didn't use.
AlwmlHuZRUU-037\|I used the pz, so the px and the py are still available.
AlwmlHuZRUU-038\|But I have a new set, two sps, and a px, and a py.
AlwmlHuZRUU-039\|Those are the atomic orbitals that are appropriate for a 180 degree bond angle.
3CHic39yxjs-000\|Let's look at a couple of filters and how they'll affect a blue object.
3CHic39yxjs-001\|So three possible filters put in front of a blue object.
3CHic39yxjs-009\|So we've got a blue object.
3CHic39yxjs-010\|We're going to put filters between the blue object and us.
3CHic39yxjs-011\|And we're going to see what we see.
3CHic39yxjs-012\|So a blue object, wavelengths of blue light are being emitted from that object.
3CHic39yxjs-013\|That's why it appears blue.
3CHic39yxjs-014\|How do we make it appear black?
3CHic39yxjs-016\|So we need to find a filter that absorbs strongly in the blue.
3CHic39yxjs-017\|And that is filter number two.
SKDF7sz_wgI-000\|Let's see if we can deduce an atomic radius from the information we have so.
SKDF7sz_wgI-001\|Far we haven't really talked about the trend in atomic radius.
SKDF7sz_wgI-002\|We've talked about ionization energy and electron affinity.
SKDF7sz_wgI-003\|But you might be able to surmise how the atoms go based on our other atomic trends.
SKDF7sz_wgI-004\|So let's take sulfur, chlorine, and potassium and ask, which has the lowest atomic radius?
SKDF7sz_wgI-008\|We're trying to deduce atomic radius from other properties that have periodic trends that we understand.
SKDF7sz_wgI-009\|So we're taking sulfur, chlorine, and potassium-- three atoms.
SKDF7sz_wgI-026\|So for smallest atomic radii, your best bet, chlorine.
JIKXNx5TK3Q-000\|Let's look at two titrations-- the same acid, but tenfold diluted.
JIKXNx5TK3Q-001\|So I'm going to take the weak acid, acetic acid, and titrate it with a strong base, potassium hydroxide.
JIKXNx5TK3Q-002\|That is plotted here in the dotted yellow line.
JIKXNx5TK3Q-003\|Then I'm going to dilute by a factor of 10 and reperform the titration.
JIKXNx5TK3Q-011\|We're looking at titrating HAc, acetic acid, two times.
JIKXNx5TK3Q-012\|One titration, then dilute by a factor of 10, and titrate again.
JIKXNx5TK3Q-013\|Now, if you dilute by a factor of 10, the weak acid, initially, the pH will change by one unit because the H3O+ concentration changes by a factor of 10.
JIKXNx5TK3Q-014\|But remember, in a weak acid, there's plenty of undissociated HAc around that will dissociate after that dilution and lower the pH.
JIKXNx5TK3Q-017\|So the total amount of acid is the same, so the total amount of base required to reach equivalence point is the same.
JIKXNx5TK3Q-018\|In the buffer region, remember, buffers are also resistant to changes in dilution.
JIKXNx5TK3Q-019\|You dilute a buffer by a factor of 10, but the ratio of acid to base is what's important in a buffer.
DGTtekW_2J8-000\|Let's look at a point on the titration curve.
DGTtekW_2J8-002\|At point C, I've added one mole of my strong base for every mole of weak acid that I originally had.
DGTtekW_2J8-004\|So what you have here is a solution of weak base.
DGTtekW_2J8-005\|So you can calculate the pH here at point C by using a calculation of a weak base.
DGTtekW_2J8-006\|Let's do that.
DGTtekW_2J8-007\|Here's the weak base, in this case the conjugate base of acetic acid, reacting with water, because these are the two species left at the endpoint of the titration.
DGTtekW_2J8-008\|They form the conjugate acid and OH minus.
DGTtekW_2J8-009\|So at this point in the titration, we've titrated a weak acid with a strong base.
DGTtekW_2J8-010\|At the endpoint, where I've used up all the acid, the solution is actually slightly basic.
DGTtekW_2J8-011\|And it's slightly basic because I've converted all of the weak acid into its conjugate base.
DGTtekW_2J8-014\|And I say, well, I have approximately 0.1 moles per volume.
DGTtekW_2J8-015\|And the volume doesn't change much during a titration, because we can say we still have about 0.1 molar Ac ions.
DGTtekW_2J8-016\|A little of that reacts to go back and form HAc and some OH minus.
DGTtekW_2J8-022\|If I know HAc, I know Ka times Kb is Kw.
DGTtekW_2J8-023\|So I can calculate a Kb.
DGTtekW_2J8-024\|Kb is small.
DGTtekW_2J8-025\|Kb is small, so I'll assume that 0.1 minus x is essentially 0.1, that x is small compared to 0.1.
DGTtekW_2J8-026\|When I do that, the math gets relatively easy.
DGTtekW_2J8-027\|x squared is 0.1Kb.
DGTtekW_2J8-028\|So x is 7.5 times 10 to the minus 6.
DGTtekW_2J8-029\|Now, x is the OH minus concentration.
DGTtekW_2J8-030\|If I want the pH, I need the H3O plus concentration.
DGTtekW_2J8-031\|And of course, I can get that because H3O plus and OH minus in water are always a product to 10 to the minus 14.
DGTtekW_2J8-032\|So H3O plus times OH minus is 10 to the minus 14th all the time.
DGTtekW_2J8-033\|That's an equilibrium that's holding.
DGTtekW_2J8-034\|At the same time, this equilibrium is holding.
DGTtekW_2J8-035\|So we satisfy them both at the same time.
DGTtekW_2J8-037\|And in this case, the pH, 8.88.
DGTtekW_2J8-039\|And if you have just weak base, the pH will be slightly basic.
w_9DNwm1DFI-000\|Let's look at the amino acid cysteine at a specific pH.
w_9DNwm1DFI-001\|So here I've drawn cysteine, and I've shown you the pKa's of each group.
w_9DNwm1DFI-002\|So the pKa of this proton, the pKa of one of these protons, and the pKa of this proton.
w_9DNwm1DFI-003\|The question is at pH 5, what's the overall charge?
w_9DNwm1DFI-004\|So what are the protonation states?
w_9DNwm1DFI-005\|Is the overall charge plus 1, 0, or minus 1?
w_9DNwm1DFI-012\|We're looking at the amino acid cysteine at pH 5.
w_9DNwm1DFI-014\|So, for instance, here the pKa is 9, so the pH is less than the pKa.
w_9DNwm1DFI-015\|If the pH is less than the pKa, you're on the acidic side, so the acidic form predominates.
w_9DNwm1DFI-016\|In fact, you're four units away from this.
w_9DNwm1DFI-017\|So you're four units to the acidic side, so this is all in its acidic form.
w_9DNwm1DFI-019\|Now here, pKa 8.
w_9DNwm1DFI-020\|Again, the pH is 5, well to the acidic side.
w_9DNwm1DFI-024\|Here, pKa 2.
w_9DNwm1DFI-025\|This is the carboxylic acid part of the molecule.
w_9DNwm1DFI-026\|The pKa is 2.
w_9DNwm1DFI-027\|Now, I am on the basic side.
w_9DNwm1DFI-028\|The pH is 5.
w_9DNwm1DFI-029\|The pKa is 2.
w_9DNwm1DFI-030\|So I'm at a pH above the pKa.
w_9DNwm1DFI-031\|On the basic side, so the basic form predominates.
w_9DNwm1DFI-032\|The basic form is the form without the proton.
w_9DNwm1DFI-036\|So in this case, the overall charge is 0.
w_9DNwm1DFI-037\|Minus 1, plus 1, and 0 all sum to 0.
nbeE9b2JV5Q-000\|Let's look at dissolving a solid.
nbeE9b2JV5Q-001\|We'll have magnesium hydroxide, and we'll put it in water.
nbeE9b2JV5Q-003\|Now, I'd like that to dissolve.
nbeE9b2JV5Q-004\|What's the best strategy for dissolving that additional speck of magnesium hydroxide?
nbeE9b2JV5Q-005\|Should I add 0.1 molar H2SO4, sulfuric acid?
nbeE9b2JV5Q-006\|0.01 molar sodium hydroxide?
nbeE9b2JV5Q-014\|We're dissolving magnesium hydroxide in water to form magnesium and hydroxide ions.
nbeE9b2JV5Q-015\|Magnesium hydroxide, sparingly soluble, so the reaction favors the solid, and you can write the equilibrium expression.
nbeE9b2JV5Q-016\|Magnesium ions and hydroxide ions squared, in this case, stoichiometric coefficient of 2.

AlwmlHuZRUU-026|I can take the difference of those two.

AlwmlHuZRUU-027|And now I'll have high amplitude on this side.

AlwmlHuZRUU-028|This side will become green, this side will become red.

AlwmlHuZRUU-029|And I'll have high amplitude on this side, 180 degrees from this high amplitude.

AlwmlHuZRUU-030|This is a combination of s minus the pz.

AlwmlHuZRUU-036|I still have my remaining p orbitals that I didn't use.

AlwmlHuZRUU-037|I used the pz, so the px and the py are still available.

AlwmlHuZRUU-038|But I have a new set, two sps, and a px, and a py.

AlwmlHuZRUU-039|Those are the atomic orbitals that are appropriate for a 180 degree bond angle.

3CHic39yxjs-000|Let's look at a couple of filters and how they'll affect a blue object.

3CHic39yxjs-001|So three possible filters put in front of a blue object.

3CHic39yxjs-009|So we've got a blue object.

3CHic39yxjs-010|We're going to put filters between the blue object and us.

3CHic39yxjs-011|And we're going to see what we see.

3CHic39yxjs-012|So a blue object, wavelengths of blue light are being emitted from that object.

3CHic39yxjs-013|That's why it appears blue.

3CHic39yxjs-014|How do we make it appear black?

3CHic39yxjs-016|So we need to find a filter that absorbs strongly in the blue.

3CHic39yxjs-017|And that is filter number two.

SKDF7sz_wgI-000|Let's see if we can deduce an atomic radius from the information we have so.

SKDF7sz_wgI-001|Far we haven't really talked about the trend in atomic radius.

SKDF7sz_wgI-002|We've talked about ionization energy and electron affinity.

SKDF7sz_wgI-003|But you might be able to surmise how the atoms go based on our other atomic trends.

SKDF7sz_wgI-004|So let's take sulfur, chlorine, and potassium and ask, which has the lowest atomic radius?

SKDF7sz_wgI-008|We're trying to deduce atomic radius from other properties that have periodic trends that we understand.

SKDF7sz_wgI-009|So we're taking sulfur, chlorine, and potassium-- three atoms.

SKDF7sz_wgI-026|So for smallest atomic radii, your best bet, chlorine.

JIKXNx5TK3Q-000|Let's look at two titrations-- the same acid, but tenfold diluted.

JIKXNx5TK3Q-001|So I'm going to take the weak acid, acetic acid, and titrate it with a strong base, potassium hydroxide.

JIKXNx5TK3Q-002|That is plotted here in the dotted yellow line.

JIKXNx5TK3Q-003|Then I'm going to dilute by a factor of 10 and reperform the titration.

JIKXNx5TK3Q-011|We're looking at titrating HAc, acetic acid, two times.

JIKXNx5TK3Q-012|One titration, then dilute by a factor of 10, and titrate again.

JIKXNx5TK3Q-013|Now, if you dilute by a factor of 10, the weak acid, initially, the pH will change by one unit because the H3O+ concentration changes by a factor of 10.

JIKXNx5TK3Q-014|But remember, in a weak acid, there's plenty of undissociated HAc around that will dissociate after that dilution and lower the pH.

JIKXNx5TK3Q-017|So the total amount of acid is the same, so the total amount of base required to reach equivalence point is the same.

JIKXNx5TK3Q-018|In the buffer region, remember, buffers are also resistant to changes in dilution.

JIKXNx5TK3Q-019|You dilute a buffer by a factor of 10, but the ratio of acid to base is what's important in a buffer.

DGTtekW_2J8-000|Let's look at a point on the titration curve.

DGTtekW_2J8-002|At point C, I've added one mole of my strong base for every mole of weak acid that I originally had.

DGTtekW_2J8-004|So what you have here is a solution of weak base.

DGTtekW_2J8-005|So you can calculate the pH here at point C by using a calculation of a weak base.

DGTtekW_2J8-006|Let's do that.

DGTtekW_2J8-007|Here's the weak base, in this case the conjugate base of acetic acid, reacting with water, because these are the two species left at the endpoint of the titration.

DGTtekW_2J8-008|They form the conjugate acid and OH minus.

DGTtekW_2J8-009|So at this point in the titration, we've titrated a weak acid with a strong base.

DGTtekW_2J8-010|At the endpoint, where I've used up all the acid, the solution is actually slightly basic.

DGTtekW_2J8-011|And it's slightly basic because I've converted all of the weak acid into its conjugate base.

DGTtekW_2J8-014|And I say, well, I have approximately 0.1 moles per volume.

DGTtekW_2J8-015|And the volume doesn't change much during a titration, because we can say we still have about 0.1 molar Ac ions.

DGTtekW_2J8-016|A little of that reacts to go back and form HAc and some OH minus.

DGTtekW_2J8-022|If I know HAc, I know Ka times Kb is Kw.

DGTtekW_2J8-023|So I can calculate a Kb.

DGTtekW_2J8-024|Kb is small.

DGTtekW_2J8-025|Kb is small, so I'll assume that 0.1 minus x is essentially 0.1, that x is small compared to 0.1.

DGTtekW_2J8-026|When I do that, the math gets relatively easy.

DGTtekW_2J8-027|x squared is 0.1Kb.

DGTtekW_2J8-028|So x is 7.5 times 10 to the minus 6.

DGTtekW_2J8-029|Now, x is the OH minus concentration.

DGTtekW_2J8-030|If I want the pH, I need the H3O plus concentration.

DGTtekW_2J8-031|And of course, I can get that because H3O plus and OH minus in water are always a product to 10 to the minus 14.

DGTtekW_2J8-032|So H3O plus times OH minus is 10 to the minus 14th all the time.

DGTtekW_2J8-033|That's an equilibrium that's holding.

DGTtekW_2J8-034|At the same time, this equilibrium is holding.

DGTtekW_2J8-035|So we satisfy them both at the same time.

DGTtekW_2J8-037|And in this case, the pH, 8.88.

DGTtekW_2J8-039|And if you have just weak base, the pH will be slightly basic.

w_9DNwm1DFI-000|Let's look at the amino acid cysteine at a specific pH.

w_9DNwm1DFI-001|So here I've drawn cysteine, and I've shown you the pKa's of each group.

w_9DNwm1DFI-002|So the pKa of this proton, the pKa of one of these protons, and the pKa of this proton.

w_9DNwm1DFI-003|The question is at pH 5, what's the overall charge?

w_9DNwm1DFI-004|So what are the protonation states?

w_9DNwm1DFI-005|Is the overall charge plus 1, 0, or minus 1?

w_9DNwm1DFI-012|We're looking at the amino acid cysteine at pH 5.

w_9DNwm1DFI-014|So, for instance, here the pKa is 9, so the pH is less than the pKa.

w_9DNwm1DFI-015|If the pH is less than the pKa, you're on the acidic side, so the acidic form predominates.

w_9DNwm1DFI-016|In fact, you're four units away from this.

w_9DNwm1DFI-017|So you're four units to the acidic side, so this is all in its acidic form.

w_9DNwm1DFI-019|Now here, pKa 8.

w_9DNwm1DFI-020|Again, the pH is 5, well to the acidic side.

w_9DNwm1DFI-024|Here, pKa 2.

w_9DNwm1DFI-025|This is the carboxylic acid part of the molecule.

w_9DNwm1DFI-026|The pKa is 2.

w_9DNwm1DFI-027|Now, I am on the basic side.

w_9DNwm1DFI-028|The pH is 5.

w_9DNwm1DFI-029|The pKa is 2.

w_9DNwm1DFI-030|So I'm at a pH above the pKa.

w_9DNwm1DFI-031|On the basic side, so the basic form predominates.

w_9DNwm1DFI-032|The basic form is the form without the proton.

w_9DNwm1DFI-036|So in this case, the overall charge is 0.

w_9DNwm1DFI-037|Minus 1, plus 1, and 0 all sum to 0.

nbeE9b2JV5Q-000|Let's look at dissolving a solid.

nbeE9b2JV5Q-001|We'll have magnesium hydroxide, and we'll put it in water.

nbeE9b2JV5Q-003|Now, I'd like that to dissolve.

nbeE9b2JV5Q-004|What's the best strategy for dissolving that additional speck of magnesium hydroxide?

nbeE9b2JV5Q-005|Should I add 0.1 molar H2SO4, sulfuric acid?

nbeE9b2JV5Q-006|0.01 molar sodium hydroxide?

nbeE9b2JV5Q-014|We're dissolving magnesium hydroxide in water to form magnesium and hydroxide ions.

nbeE9b2JV5Q-015|Magnesium hydroxide, sparingly soluble, so the reaction favors the solid, and you can write the equilibrium expression.

nbeE9b2JV5Q-016|Magnesium ions and hydroxide ions squared, in this case, stoichiometric coefficient of 2.