ceaglex/VisualTTS_Chem · Datasets at Hugging Face

text stringlengths 19 206
P_e4n9XSjUw-033\|The experimenters who did this were awarded a Nobel Prize for what's called laser cooling.
P_e4n9XSjUw-035\|And then using this additional method, using lasers to trap the atoms, and bounce photons off them 'til they virtually come to a stop.
P_e4n9XSjUw-036\|The lowest temperatures ever achieved by what's called laser cooling.
P_e4n9XSjUw-037\|The correct answer for our laser cooling experiment is C, about 10,000 yellow photons to stop that sodium atom.
cy4vUW9x62s-003\|But don't be fooled.
cy4vUW9x62s-014\|I have those two here.
cy4vUW9x62s-015\|Here's the tetrahedral arrangement and the trigonal bipyramidal arrangement.
cy4vUW9x62s-016\|But remember, when we're talking about the actual molecular geometry, we ignore the lone pairs.
cy4vUW9x62s-017\|So for these two molecules here, there are two lone pairs.
cy4vUW9x62s-018\|So you have the tetrahedral geometry, but it's as if you're going to ignore two of the arms.
cy4vUW9x62s-019\|If you ignore two of the arms, you get a molecule that just looks bent.
cy4vUW9x62s-020\|And that's what we say here.
cy4vUW9x62s-022\|That angle is going to be something like 104 degrees, 105.
cy4vUW9x62s-023\|104 and 1/2, two lone pairs, and two bonding pairs in a tetrahedral configuration generally gives you about 105 degrees.
cy4vUW9x62s-024\|For our ICl2, we have five things.
cy4vUW9x62s-025\|Three of them along the equatorial positions are lone pairs, so we can ignore those.
cy4vUW9x62s-026\|Taking off the three lone pairs in the equatorial position just leaves us the two axial chlorines.
cy4vUW9x62s-028\|So of these three, ICl2- is the linear ion.
3WNdFxNR9tY-001\|So I'll mix solutions 1 and 5, equal volumes, equal concentration.
3WNdFxNR9tY-002\|So 0.1 molar HCl with an equal volume of 0.1 molar NaOH.
3WNdFxNR9tY-003\|Can you predict will the solution be acidic, neutral, or basic?
3WNdFxNR9tY-015\|We're going to mix a strong acid and a strong base and try to predict will it be acidic, neutral, or basic and the ionic strength.
3WNdFxNR9tY-016\|So let's do the experiment.
3WNdFxNR9tY-017\|Here I have my strong acid or my strong base, and I'm going to mix it with my strong acid.
3WNdFxNR9tY-018\|Now I think we can tell already from the color that the pH is going to be about neutral.
3WNdFxNR9tY-019\|It's very similar to my neutral water.
3WNdFxNR9tY-020\|And that's true.
3WNdFxNR9tY-022\|And a salt solution, sodium chloride solution, I think you understand is neutral.
3WNdFxNR9tY-023\|Neither the sodium nor chloride ions contribute H3O pluses or OH minuses.
3WNdFxNR9tY-024\|Neither of them is a good acid or a base.
3WNdFxNR9tY-025\|So the pH is simply the pH of neutral water.
3WNdFxNR9tY-026\|What about the ionic strength?
3WNdFxNR9tY-027\|Well, to do that, we can insert our ionic tester, our light bulb, and see the strength of the light intensity.
3WNdFxNR9tY-028\|I'll put this in and bright light.
3WNdFxNR9tY-029\|So that's interesting.
3WNdFxNR9tY-030\|We mix the strong acid and the strong base, and we got a bright light.
3WNdFxNR9tY-031\|Why is that?
3WNdFxNR9tY-032\|Because this salt that's produced, sodium chloride, completely ionizes to form sodium and chlorine ions.
3WNdFxNR9tY-033\|They stay in solution at a neutral pH but give you a bright light.
3WNdFxNR9tY-034\|So in this case, the answer neutral pH, relatively strong ionic strength-- lots of ions in solution, sodium and chloride.
3WNdFxNR9tY-035\|The answer's B and A.
M6AdsBPkJSU-000\|Let's look at making a buffer solution.
M6AdsBPkJSU-001\|Which of the following when added to NH3, ammonia, the base, forms a basic buffer at pH greater than 7?
M6AdsBPkJSU-002\|Is it the strong base, sodium hydroxide, sodium chloride, the salt, or the strong acid, hydrochloric acid, HCl.
M6AdsBPkJSU-009\|We're talking about making a buffer from NH3, ammonia, the weak base.
M6AdsBPkJSU-010\|Now to make a buffer, you need a base and it's conjugate acid or an acid in its conjugate base.
M6AdsBPkJSU-011\|They need to be in about equal concentrations.
M6AdsBPkJSU-012\|So if I'm talking about NH3, I need it's conjugate acid.
M6AdsBPkJSU-013\|It's a base already.
M6AdsBPkJSU-014\|I need it's conjugate acid.
M6AdsBPkJSU-015\|So to form it's conjugate acid, I would add a strong acid and convert some of my NH3 to NH4 plus.
M6AdsBPkJSU-016\|That would form the buffer.
M6AdsBPkJSU-018\|So the pKa of the acid NH4 plus is 9.3.
M6AdsBPkJSU-019\|So I'll form a buffer of NH4 plus and NH3, the ammonium ion and ammonia.
M6AdsBPkJSU-020\|The pH will be around 9.3.
M6AdsBPkJSU-021\|And in order to form it, I actually have to add a strong acid to convert some of my base to the acid form.
M6AdsBPkJSU-022\|So in this case, the correct answer is I add some HCl to get equal concentrations of an acid and it's conjugate base.
DCcIg-TtH7U-000\|Let's look at the ionization of helium plus.
DCcIg-TtH7U-001\|Helium plus is a one-electron system like hydrogen.
DCcIg-TtH7U-002\|The question I have is, can a 30 nanometer photon ionize He+ from its ground state?
DCcIg-TtH7U-003\|And I give you this information about hydrogen that we've been talking about.
DCcIg-TtH7U-004\|The 1 to 2 transition in hydrogen is in the ultraviolet at 120 nanometers.
DCcIg-TtH7U-005\|The question is, given that, can we deduce whether helium plus, also a one-electron system, will be ionized?
DCcIg-TtH7U-010\|So we're kind of doing a little spectroscopy here.
DCcIg-TtH7U-011\|We have a transition in hydrogen. It's 120 nanometers.
DCcIg-TtH7U-012\|And we do the calculation, and we say, well, that transition energy is about 3/4 of a Rydberg.
DCcIg-TtH7U-013\|That's the final state minus the initial state-- R minus 1/4R is 3/4R.
DCcIg-TtH7U-014\|So 120 nanometer photon corresponds to about 3/4 of a Rydberg.
DCcIg-TtH7U-015\|We have a 30 nanometer photon, which is four times as energetic.
DCcIg-TtH7U-018\|It's z squared over n squared.
DCcIg-TtH7U-019\|So z is 2 for helium.
DCcIg-TtH7U-020\|So it's minus 4 Rydbergs.
DCcIg-TtH7U-021\|So a photon with an energy of about 3 Rydbergs is not sufficient to ionize He+, which requires an energy of about 4 Rydbergs.
DCcIg-TtH7U-022\|So not enough energy in a 30 nanometer photon to ionize helium plus.
VL_TnSkqNYY-000\|Let's look at the equilibrium between NO2, a brown gas, and N2O4, a clear gas.
VL_TnSkqNYY-001\|Let's say I start, as I have it here, with a bulb of the equilibrium mixture.
VL_TnSkqNYY-002\|I open a valve and let it expand to twice its volume.
VL_TnSkqNYY-003\|Now, initially that I do that, obviously it'll get lighter because it'll expand and be less dense but then the new equilibrium will be established.
VL_TnSkqNYY-011\|What I'm going to do here is write the expression for Q for that chemical reaction, that dimerization.
VL_TnSkqNYY-012\|The products, N2O4, over the reactants squared, because the stoichiometric coefficient of the NO2 was two.
VL_TnSkqNYY-013\|So if I double the volume for this, what that will do is reduce the pressures instantaneously by a factor of two.
VL_TnSkqNYY-018\|So Q will be larger than K, it will be 2 times bigger than the equilibrium constant and I want to go back towards equilibrium.
VL_TnSkqNYY-019\|So I need to make Q smaller.
VL_TnSkqNYY-020\|It's two times too big, I need to make it smaller.
VL_TnSkqNYY-021\|How do I make it smaller?
VL_TnSkqNYY-022\|Well, I have to shift back to reactants.
VL_TnSkqNYY-023\|The products are on top, making the products bigger would make Q larger.
VL_TnSkqNYY-024\|Making products smaller and reactants bigger would make Q smaller.
VL_TnSkqNYY-025\|And I want Q to get smaller to go back towards K so I'll shift back toward the reactants and the reactants or the brown gas.
VL_TnSkqNYY-026\|So I'll get darker as I shift towards equilibrium, in this case.
VL_TnSkqNYY-027\|The correct answer here is increase.
D4R7ho3YnSo-000\|Let's calculate the pH at a specific point on the titration curve.
D4R7ho3YnSo-002\|And we've actually done this before, but let's go through the calculation.
D4R7ho3YnSo-009\|And solve for x.
D4R7ho3YnSo-010\|Now, to solve for x the easiest thing to do is assume that x is small with respect to 0.1.
D4R7ho3YnSo-011\|And I do that because I see that k is a relatively small number.
D4R7ho3YnSo-012\|And if k is small, then this reaction doesn't go very far an x's should be small.
D4R7ho3YnSo-013\|I should favor this side, the 0.1 side.
D4R7ho3YnSo-014\|So if only a small fraction of those 0.1 initial moles react, x will be small.
D4R7ho3YnSo-016\|x is the H3O plus concentration.

P_e4n9XSjUw-033|The experimenters who did this were awarded a Nobel Prize for what's called laser cooling.

P_e4n9XSjUw-035|And then using this additional method, using lasers to trap the atoms, and bounce photons off them 'til they virtually come to a stop.

P_e4n9XSjUw-036|The lowest temperatures ever achieved by what's called laser cooling.

P_e4n9XSjUw-037|The correct answer for our laser cooling experiment is C, about 10,000 yellow photons to stop that sodium atom.

cy4vUW9x62s-003|But don't be fooled.

cy4vUW9x62s-014|I have those two here.

cy4vUW9x62s-015|Here's the tetrahedral arrangement and the trigonal bipyramidal arrangement.

cy4vUW9x62s-016|But remember, when we're talking about the actual molecular geometry, we ignore the lone pairs.

cy4vUW9x62s-017|So for these two molecules here, there are two lone pairs.

cy4vUW9x62s-018|So you have the tetrahedral geometry, but it's as if you're going to ignore two of the arms.

cy4vUW9x62s-019|If you ignore two of the arms, you get a molecule that just looks bent.

cy4vUW9x62s-020|And that's what we say here.

cy4vUW9x62s-022|That angle is going to be something like 104 degrees, 105.

cy4vUW9x62s-023|104 and 1/2, two lone pairs, and two bonding pairs in a tetrahedral configuration generally gives you about 105 degrees.

cy4vUW9x62s-024|For our ICl2, we have five things.

cy4vUW9x62s-025|Three of them along the equatorial positions are lone pairs, so we can ignore those.

cy4vUW9x62s-026|Taking off the three lone pairs in the equatorial position just leaves us the two axial chlorines.

cy4vUW9x62s-028|So of these three, ICl2- is the linear ion.

3WNdFxNR9tY-001|So I'll mix solutions 1 and 5, equal volumes, equal concentration.

3WNdFxNR9tY-002|So 0.1 molar HCl with an equal volume of 0.1 molar NaOH.

3WNdFxNR9tY-003|Can you predict will the solution be acidic, neutral, or basic?

3WNdFxNR9tY-015|We're going to mix a strong acid and a strong base and try to predict will it be acidic, neutral, or basic and the ionic strength.

3WNdFxNR9tY-016|So let's do the experiment.

3WNdFxNR9tY-017|Here I have my strong acid or my strong base, and I'm going to mix it with my strong acid.

3WNdFxNR9tY-018|Now I think we can tell already from the color that the pH is going to be about neutral.

3WNdFxNR9tY-019|It's very similar to my neutral water.

3WNdFxNR9tY-020|And that's true.

3WNdFxNR9tY-022|And a salt solution, sodium chloride solution, I think you understand is neutral.

3WNdFxNR9tY-023|Neither the sodium nor chloride ions contribute H3O pluses or OH minuses.

3WNdFxNR9tY-024|Neither of them is a good acid or a base.

3WNdFxNR9tY-025|So the pH is simply the pH of neutral water.

3WNdFxNR9tY-026|What about the ionic strength?

3WNdFxNR9tY-027|Well, to do that, we can insert our ionic tester, our light bulb, and see the strength of the light intensity.

3WNdFxNR9tY-028|I'll put this in and bright light.

3WNdFxNR9tY-029|So that's interesting.

3WNdFxNR9tY-030|We mix the strong acid and the strong base, and we got a bright light.

3WNdFxNR9tY-031|Why is that?

3WNdFxNR9tY-032|Because this salt that's produced, sodium chloride, completely ionizes to form sodium and chlorine ions.

3WNdFxNR9tY-033|They stay in solution at a neutral pH but give you a bright light.

3WNdFxNR9tY-034|So in this case, the answer neutral pH, relatively strong ionic strength-- lots of ions in solution, sodium and chloride.

3WNdFxNR9tY-035|The answer's B and A.

M6AdsBPkJSU-000|Let's look at making a buffer solution.

M6AdsBPkJSU-001|Which of the following when added to NH3, ammonia, the base, forms a basic buffer at pH greater than 7?

M6AdsBPkJSU-002|Is it the strong base, sodium hydroxide, sodium chloride, the salt, or the strong acid, hydrochloric acid, HCl.

M6AdsBPkJSU-009|We're talking about making a buffer from NH3, ammonia, the weak base.

M6AdsBPkJSU-010|Now to make a buffer, you need a base and it's conjugate acid or an acid in its conjugate base.

M6AdsBPkJSU-011|They need to be in about equal concentrations.

M6AdsBPkJSU-012|So if I'm talking about NH3, I need it's conjugate acid.

M6AdsBPkJSU-013|It's a base already.

M6AdsBPkJSU-014|I need it's conjugate acid.

M6AdsBPkJSU-015|So to form it's conjugate acid, I would add a strong acid and convert some of my NH3 to NH4 plus.

M6AdsBPkJSU-016|That would form the buffer.

M6AdsBPkJSU-018|So the pKa of the acid NH4 plus is 9.3.

M6AdsBPkJSU-019|So I'll form a buffer of NH4 plus and NH3, the ammonium ion and ammonia.

M6AdsBPkJSU-020|The pH will be around 9.3.

M6AdsBPkJSU-021|And in order to form it, I actually have to add a strong acid to convert some of my base to the acid form.

M6AdsBPkJSU-022|So in this case, the correct answer is I add some HCl to get equal concentrations of an acid and it's conjugate base.

DCcIg-TtH7U-000|Let's look at the ionization of helium plus.

DCcIg-TtH7U-001|Helium plus is a one-electron system like hydrogen.

DCcIg-TtH7U-002|The question I have is, can a 30 nanometer photon ionize He+ from its ground state?

DCcIg-TtH7U-003|And I give you this information about hydrogen that we've been talking about.

DCcIg-TtH7U-004|The 1 to 2 transition in hydrogen is in the ultraviolet at 120 nanometers.

DCcIg-TtH7U-005|The question is, given that, can we deduce whether helium plus, also a one-electron system, will be ionized?

DCcIg-TtH7U-010|So we're kind of doing a little spectroscopy here.

DCcIg-TtH7U-011|We have a transition in hydrogen. It's 120 nanometers.

DCcIg-TtH7U-012|And we do the calculation, and we say, well, that transition energy is about 3/4 of a Rydberg.

DCcIg-TtH7U-013|That's the final state minus the initial state-- R minus 1/4R is 3/4R.

DCcIg-TtH7U-014|So 120 nanometer photon corresponds to about 3/4 of a Rydberg.

DCcIg-TtH7U-015|We have a 30 nanometer photon, which is four times as energetic.

DCcIg-TtH7U-018|It's z squared over n squared.

DCcIg-TtH7U-019|So z is 2 for helium.

DCcIg-TtH7U-020|So it's minus 4 Rydbergs.

DCcIg-TtH7U-021|So a photon with an energy of about 3 Rydbergs is not sufficient to ionize He+, which requires an energy of about 4 Rydbergs.

DCcIg-TtH7U-022|So not enough energy in a 30 nanometer photon to ionize helium plus.

VL_TnSkqNYY-000|Let's look at the equilibrium between NO2, a brown gas, and N2O4, a clear gas.

VL_TnSkqNYY-001|Let's say I start, as I have it here, with a bulb of the equilibrium mixture.

VL_TnSkqNYY-002|I open a valve and let it expand to twice its volume.

VL_TnSkqNYY-003|Now, initially that I do that, obviously it'll get lighter because it'll expand and be less dense but then the new equilibrium will be established.

VL_TnSkqNYY-011|What I'm going to do here is write the expression for Q for that chemical reaction, that dimerization.

VL_TnSkqNYY-012|The products, N2O4, over the reactants squared, because the stoichiometric coefficient of the NO2 was two.

VL_TnSkqNYY-013|So if I double the volume for this, what that will do is reduce the pressures instantaneously by a factor of two.

VL_TnSkqNYY-018|So Q will be larger than K, it will be 2 times bigger than the equilibrium constant and I want to go back towards equilibrium.

VL_TnSkqNYY-019|So I need to make Q smaller.

VL_TnSkqNYY-020|It's two times too big, I need to make it smaller.

VL_TnSkqNYY-021|How do I make it smaller?

VL_TnSkqNYY-022|Well, I have to shift back to reactants.

VL_TnSkqNYY-023|The products are on top, making the products bigger would make Q larger.

VL_TnSkqNYY-024|Making products smaller and reactants bigger would make Q smaller.

VL_TnSkqNYY-025|And I want Q to get smaller to go back towards K so I'll shift back toward the reactants and the reactants or the brown gas.

VL_TnSkqNYY-026|So I'll get darker as I shift towards equilibrium, in this case.

VL_TnSkqNYY-027|The correct answer here is increase.

D4R7ho3YnSo-000|Let's calculate the pH at a specific point on the titration curve.

D4R7ho3YnSo-002|And we've actually done this before, but let's go through the calculation.

D4R7ho3YnSo-009|And solve for x.

D4R7ho3YnSo-010|Now, to solve for x the easiest thing to do is assume that x is small with respect to 0.1.

D4R7ho3YnSo-011|And I do that because I see that k is a relatively small number.

D4R7ho3YnSo-012|And if k is small, then this reaction doesn't go very far an x's should be small.

D4R7ho3YnSo-013|I should favor this side, the 0.1 side.

D4R7ho3YnSo-014|So if only a small fraction of those 0.1 initial moles react, x will be small.

D4R7ho3YnSo-016|x is the H3O plus concentration.