ceaglex/VisualTTS_Chem · Datasets at Hugging Face

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inLhIqLOouY-015\|Here it's a product.
inLhIqLOouY-016\|Here it's a reactant.
inLhIqLOouY-017\|But the stronger the acid, the more that ion should be favored.
inLhIqLOouY-018\|Now, let's look at the two.
inLhIqLOouY-019\|If I look at Ka and Kb, they're not the reciprocals of each other.
inLhIqLOouY-020\|That is Ka is not 1/Kb.
inLhIqLOouY-026\|So now, I have analytically that the product Ka times Kb is a constant.
inLhIqLOouY-027\|So if Ka gets larger, a stronger acid, Kb must get smaller.
inLhIqLOouY-028\|So I have an analytical expression for strong acid means a weaker conjugate base.
ZpzjU4NfAhw-000\|So let's look at the various orbitals of a hydrogen atom.
ZpzjU4NfAhw-001\|Now, remember each orbital is a specific wave function, and that wave function will be described by three quantum numbers-- n, l, and m sub l.
ZpzjU4NfAhw-002\|Once I give you three quantum numbers, you should be able to tell me the orbital.
ZpzjU4NfAhw-003\|So let's look at them in sequence.
ZpzjU4NfAhw-004\|For n equal 1-- that's the ground state, the lowest possible energy level-- we'll have one value of l.
ZpzjU4NfAhw-005\|Now, remember l goes from 0 to n minus 1.
ZpzjU4NfAhw-006\|So if n is 1, the only possible value of l is 0.
ZpzjU4NfAhw-007\|So for n equal 1, we can have l equal 0.
ZpzjU4NfAhw-008\|And we're also going to use the designation s for the quantum number l.
ZpzjU4NfAhw-009\|So if I say l is s or if I say l is 0, I'm saying the same thing.
ZpzjU4NfAhw-010\|You have to memorize that little connection.
ZpzjU4NfAhw-011\|Now, for l equals 0, the only possible values of m sub l are 0.
ZpzjU4NfAhw-012\|Remember, m sub l goes in integer steps from minus l to l.
ZpzjU4NfAhw-014\|So now we have three quantum numbers that describes an orbital, and we're going to call this orbital the 1s orbital.
ZpzjU4NfAhw-015\|We're going to use the designations of the quantum numbers to describe the orbitals.
ZpzjU4NfAhw-016\|What does that look like?
ZpzjU4NfAhw-017\|Well, it turns out it's spherically symmetric.
ZpzjU4NfAhw-018\|The 1s orbital-- in fact, all s orbitals are spherical symmetric.
ZpzjU4NfAhw-019\|Remember, the l quantum number-- in this case, 0 equal to s-- that gives you something to understand the shape of the orbital.
ZpzjU4NfAhw-020\|In this case, the shape is spherical.
ZpzjU4NfAhw-021\|In that case, the s might be a little bit helpful.
ZpzjU4NfAhw-022\|s spherical l equals 0 is spherical.
ZpzjU4NfAhw-023\|So the 1s orbital is spherical, and I've written this plus sign to indicate there's no nodes.
ZpzjU4NfAhw-024\|The wave function does not change sign.
ZpzjU4NfAhw-027\|And of course, the number of nodes is n minus 1, so 1 minus 1 is 0.
ZpzjU4NfAhw-028\|There's no nodes in the 1s orbital.
ZpzjU4NfAhw-029\|Well, can we go to n equal 2?
ZpzjU4NfAhw-030\|We're done with n equal 1.
ZpzjU4NfAhw-031\|There's no other possibilities of the quantum numbers.
ZpzjU4NfAhw-032\|We go to n equal 2.
ZpzjU4NfAhw-033\|For n equal 2, one possible value of l now is 0.
ZpzjU4NfAhw-034\|So l can be 0 or 1.
ZpzjU4NfAhw-035\|But let's look at the 0 value first.
ZpzjU4NfAhw-036\|So that's s.
ZpzjU4NfAhw-037\|m sub l is still 0.
ZpzjU4NfAhw-038\|So what we have here is another spherically symmetric orbital, but in this case, it's a 2s orbital, but n equals 1.
ZpzjU4NfAhw-039\|So the total number of nodes and minus 1 is 1.
ZpzjU4NfAhw-042\|It had a positive sign in this region, a negative sign in that region.
ZpzjU4NfAhw-043\|The square of the wave function is always positive, but the wave function changed sign, giving you this radial node.
7Nv8QLeF3yY-000\|Let's look at that NO2 to NO3- reaction, and this time think about the oxidation number change.
7Nv8QLeF3yY-001\|So how does the oxidation number of nitrogen change in the transition NO2- to NO3-?
7Nv8QLeF3yY-010\|If we assign the electrons in the bonds to the more electronegative element, this nitrogen has only two electrons around it in a lone pair.
7Nv8QLeF3yY-014\|So in NO2-, it has three fewer than it normally would for a plus 3 oxidation number.
7Nv8QLeF3yY-015\|In this case, NO3-, nitrogen has five fewer electrons than it would as a neutral atom.
7Nv8QLeF3yY-016\|So an oxidation number of plus 5.
-vJXlNkMqYE-003\|Now, this reaction profile doesn't tell us anything about how fast that energy will be released.
-vJXlNkMqYE-007\|But intuitively, we know that time is a parameter that we can look at.
-vJXlNkMqYE-008\|Some reactions go quickly and some go slowly.
-vJXlNkMqYE-009\|Here's a reaction of zinc and acid.
-vJXlNkMqYE-010\|Now, the zinc and acid, same amount of zinc in both of these flasks.
-vJXlNkMqYE-012\|But since there's the same amount of zinc, zinc limits the reaction.
-vJXlNkMqYE-013\|When the zinc is consumed, the reaction is over.
-vJXlNkMqYE-014\|So the same amount of energy is released in both cases.
-vJXlNkMqYE-017\|So this time parameter is the realm of chemical kinetics.
-vJXlNkMqYE-018\|We know that when we mix H2 and O2 together, they can sit there almost indefinitely.
-vJXlNkMqYE-019\|And the reason is the collisions between the hydrogen and the oxygen often don't have sufficient energy to go from products to reactants.
-vJXlNkMqYE-021\|That barrier says you need energy to go from reactants to products.
-vJXlNkMqYE-023\|Everything else being equal-- concentrations, temperatures, pressures.
-vJXlNkMqYE-025\|So you can see along a reaction profile you have two separate domains-- the domain of thermodynamics and the domain of kinetics.
myr7i1nYj-4-000\|Let's do a chem quiz involving isotopes.
myr7i1nYj-4-003\|I'm going to let them react to form oxygen molecules.
myr7i1nYj-4-011\|Our problem was oxygen atoms in a flask.
myr7i1nYj-4-012\|Two different isotopes, oxygen 16, and oxygen 18.
myr7i1nYj-4-013\|They're going to react to form oxygen molecules.
myr7i1nYj-4-014\|So I'll get oxygen molecules with three different masses.
myr7i1nYj-4-015\|If an oxygen 16 reacts with an oxygen 16, that gives me an oxygen molecule of mass 32.
myr7i1nYj-4-016\|If an auction 16 reacts with an auction 18, an oxygen molecule of mass 34.
myr7i1nYj-4-017\|And if an 18 reacts with an 18, an oxygen molecule of mass 36.
myr7i1nYj-4-018\|So what's the distribution of masses?
myr7i1nYj-4-025\|It turns out it's actually twice as likely that you get the 16-18 combination.
myr7i1nYj-4-026\|It's the same thing in this physical example, where I have blue and red chocolate-covered candies.
myr7i1nYj-4-027\|If I mix them together-- equal numbers of each-- and then pick them out at random, what kind of combinations would I get?
myr7i1nYj-4-028\|Well, let's pick out one with each hand.
myr7i1nYj-4-029\|If I reach in and I pick out a blue, and I reach in with the other hand, there's two possible combinations.
myr7i1nYj-4-030\|I could pick out a red, or I could reach in and pick out a blue.
myr7i1nYj-4-031\|But what if this hand first picked out a red?
myr7i1nYj-4-032\|Well, the same thing could occur.
myr7i1nYj-4-033\|I could pick out a blue, or I could pick out a red.
myr7i1nYj-4-034\|So the combination of red-blue, occurred both times, regardless of what was in this hand.
myr7i1nYj-4-035\|It's twice as likely that the red-blue combination occurs, just like in an equimolar mixture of oxygen atoms.
myr7i1nYj-4-038\|And it's always likely that chocolate-covered candies are incredibly delicious.
ZFZ2J8gwIkE-000\|Equilibrium can occur in a variety of situations.
ZFZ2J8gwIkE-001\|A heterogeneous equilibrium would be one between a gas and a liquid, two different phases, or a solid and a liquid.
ZFZ2J8gwIkE-002\|If the phases are different, we call it a heterogeneous equilibrium.
ZFZ2J8gwIkE-006\|If you wait long enough these two values will come to equilibrium, the quotient will stop changing.
ZFZ2J8gwIkE-007\|It will become a constant.
ZFZ2J8gwIkE-011\|When it's sealed, there's a high pressure of carbon dioxide gas.
ZFZ2J8gwIkE-012\|A high pressure because by Le Chatelier's principle that drives the reaction towards the dissolved carbon dioxide.
ZFZ2J8gwIkE-013\|So you can get more carbon dioxide gas dissolved at high pressure.
ZFZ2J8gwIkE-014\|When you open it, you reduce the pressure, lowering the pressure here causes the equilibrium to shift back this way.
ZFZ2J8gwIkE-015\|So aqueous carbon dioxide comes out of solution, and it spontaneously comes out of solution all through the solution.

inLhIqLOouY-015|Here it's a product.

inLhIqLOouY-016|Here it's a reactant.

inLhIqLOouY-017|But the stronger the acid, the more that ion should be favored.

inLhIqLOouY-018|Now, let's look at the two.

inLhIqLOouY-019|If I look at Ka and Kb, they're not the reciprocals of each other.

inLhIqLOouY-020|That is Ka is not 1/Kb.

inLhIqLOouY-026|So now, I have analytically that the product Ka times Kb is a constant.

inLhIqLOouY-027|So if Ka gets larger, a stronger acid, Kb must get smaller.

inLhIqLOouY-028|So I have an analytical expression for strong acid means a weaker conjugate base.

ZpzjU4NfAhw-000|So let's look at the various orbitals of a hydrogen atom.

ZpzjU4NfAhw-001|Now, remember each orbital is a specific wave function, and that wave function will be described by three quantum numbers-- n, l, and m sub l.

ZpzjU4NfAhw-002|Once I give you three quantum numbers, you should be able to tell me the orbital.

ZpzjU4NfAhw-003|So let's look at them in sequence.

ZpzjU4NfAhw-004|For n equal 1-- that's the ground state, the lowest possible energy level-- we'll have one value of l.

ZpzjU4NfAhw-005|Now, remember l goes from 0 to n minus 1.

ZpzjU4NfAhw-006|So if n is 1, the only possible value of l is 0.

ZpzjU4NfAhw-007|So for n equal 1, we can have l equal 0.

ZpzjU4NfAhw-008|And we're also going to use the designation s for the quantum number l.

ZpzjU4NfAhw-009|So if I say l is s or if I say l is 0, I'm saying the same thing.

ZpzjU4NfAhw-010|You have to memorize that little connection.

ZpzjU4NfAhw-011|Now, for l equals 0, the only possible values of m sub l are 0.

ZpzjU4NfAhw-012|Remember, m sub l goes in integer steps from minus l to l.

ZpzjU4NfAhw-014|So now we have three quantum numbers that describes an orbital, and we're going to call this orbital the 1s orbital.

ZpzjU4NfAhw-015|We're going to use the designations of the quantum numbers to describe the orbitals.

ZpzjU4NfAhw-016|What does that look like?

ZpzjU4NfAhw-017|Well, it turns out it's spherically symmetric.

ZpzjU4NfAhw-018|The 1s orbital-- in fact, all s orbitals are spherical symmetric.

ZpzjU4NfAhw-019|Remember, the l quantum number-- in this case, 0 equal to s-- that gives you something to understand the shape of the orbital.

ZpzjU4NfAhw-020|In this case, the shape is spherical.

ZpzjU4NfAhw-021|In that case, the s might be a little bit helpful.

ZpzjU4NfAhw-022|s spherical l equals 0 is spherical.

ZpzjU4NfAhw-023|So the 1s orbital is spherical, and I've written this plus sign to indicate there's no nodes.

ZpzjU4NfAhw-024|The wave function does not change sign.

ZpzjU4NfAhw-027|And of course, the number of nodes is n minus 1, so 1 minus 1 is 0.

ZpzjU4NfAhw-028|There's no nodes in the 1s orbital.

ZpzjU4NfAhw-029|Well, can we go to n equal 2?

ZpzjU4NfAhw-030|We're done with n equal 1.

ZpzjU4NfAhw-031|There's no other possibilities of the quantum numbers.

ZpzjU4NfAhw-032|We go to n equal 2.

ZpzjU4NfAhw-033|For n equal 2, one possible value of l now is 0.

ZpzjU4NfAhw-034|So l can be 0 or 1.

ZpzjU4NfAhw-035|But let's look at the 0 value first.

ZpzjU4NfAhw-036|So that's s.

ZpzjU4NfAhw-037|m sub l is still 0.

ZpzjU4NfAhw-038|So what we have here is another spherically symmetric orbital, but in this case, it's a 2s orbital, but n equals 1.

ZpzjU4NfAhw-039|So the total number of nodes and minus 1 is 1.

ZpzjU4NfAhw-042|It had a positive sign in this region, a negative sign in that region.

ZpzjU4NfAhw-043|The square of the wave function is always positive, but the wave function changed sign, giving you this radial node.

7Nv8QLeF3yY-000|Let's look at that NO2 to NO3- reaction, and this time think about the oxidation number change.

7Nv8QLeF3yY-001|So how does the oxidation number of nitrogen change in the transition NO2- to NO3-?

7Nv8QLeF3yY-010|If we assign the electrons in the bonds to the more electronegative element, this nitrogen has only two electrons around it in a lone pair.

7Nv8QLeF3yY-014|So in NO2-, it has three fewer than it normally would for a plus 3 oxidation number.

7Nv8QLeF3yY-015|In this case, NO3-, nitrogen has five fewer electrons than it would as a neutral atom.

7Nv8QLeF3yY-016|So an oxidation number of plus 5.

-vJXlNkMqYE-003|Now, this reaction profile doesn't tell us anything about how fast that energy will be released.

-vJXlNkMqYE-007|But intuitively, we know that time is a parameter that we can look at.

-vJXlNkMqYE-008|Some reactions go quickly and some go slowly.

-vJXlNkMqYE-009|Here's a reaction of zinc and acid.

-vJXlNkMqYE-010|Now, the zinc and acid, same amount of zinc in both of these flasks.

-vJXlNkMqYE-012|But since there's the same amount of zinc, zinc limits the reaction.

-vJXlNkMqYE-013|When the zinc is consumed, the reaction is over.

-vJXlNkMqYE-014|So the same amount of energy is released in both cases.

-vJXlNkMqYE-017|So this time parameter is the realm of chemical kinetics.

-vJXlNkMqYE-018|We know that when we mix H2 and O2 together, they can sit there almost indefinitely.

-vJXlNkMqYE-019|And the reason is the collisions between the hydrogen and the oxygen often don't have sufficient energy to go from products to reactants.

-vJXlNkMqYE-021|That barrier says you need energy to go from reactants to products.

-vJXlNkMqYE-023|Everything else being equal-- concentrations, temperatures, pressures.

-vJXlNkMqYE-025|So you can see along a reaction profile you have two separate domains-- the domain of thermodynamics and the domain of kinetics.

myr7i1nYj-4-000|Let's do a chem quiz involving isotopes.

myr7i1nYj-4-003|I'm going to let them react to form oxygen molecules.

myr7i1nYj-4-011|Our problem was oxygen atoms in a flask.

myr7i1nYj-4-012|Two different isotopes, oxygen 16, and oxygen 18.

myr7i1nYj-4-013|They're going to react to form oxygen molecules.

myr7i1nYj-4-014|So I'll get oxygen molecules with three different masses.

myr7i1nYj-4-015|If an oxygen 16 reacts with an oxygen 16, that gives me an oxygen molecule of mass 32.

myr7i1nYj-4-016|If an auction 16 reacts with an auction 18, an oxygen molecule of mass 34.

myr7i1nYj-4-017|And if an 18 reacts with an 18, an oxygen molecule of mass 36.

myr7i1nYj-4-018|So what's the distribution of masses?

myr7i1nYj-4-025|It turns out it's actually twice as likely that you get the 16-18 combination.

myr7i1nYj-4-026|It's the same thing in this physical example, where I have blue and red chocolate-covered candies.

myr7i1nYj-4-027|If I mix them together-- equal numbers of each-- and then pick them out at random, what kind of combinations would I get?

myr7i1nYj-4-028|Well, let's pick out one with each hand.

myr7i1nYj-4-029|If I reach in and I pick out a blue, and I reach in with the other hand, there's two possible combinations.

myr7i1nYj-4-030|I could pick out a red, or I could reach in and pick out a blue.

myr7i1nYj-4-031|But what if this hand first picked out a red?

myr7i1nYj-4-032|Well, the same thing could occur.

myr7i1nYj-4-033|I could pick out a blue, or I could pick out a red.

myr7i1nYj-4-034|So the combination of red-blue, occurred both times, regardless of what was in this hand.

myr7i1nYj-4-035|It's twice as likely that the red-blue combination occurs, just like in an equimolar mixture of oxygen atoms.

myr7i1nYj-4-038|And it's always likely that chocolate-covered candies are incredibly delicious.

ZFZ2J8gwIkE-000|Equilibrium can occur in a variety of situations.

ZFZ2J8gwIkE-001|A heterogeneous equilibrium would be one between a gas and a liquid, two different phases, or a solid and a liquid.

ZFZ2J8gwIkE-002|If the phases are different, we call it a heterogeneous equilibrium.

ZFZ2J8gwIkE-006|If you wait long enough these two values will come to equilibrium, the quotient will stop changing.

ZFZ2J8gwIkE-007|It will become a constant.

ZFZ2J8gwIkE-011|When it's sealed, there's a high pressure of carbon dioxide gas.

ZFZ2J8gwIkE-012|A high pressure because by Le Chatelier's principle that drives the reaction towards the dissolved carbon dioxide.

ZFZ2J8gwIkE-013|So you can get more carbon dioxide gas dissolved at high pressure.

ZFZ2J8gwIkE-014|When you open it, you reduce the pressure, lowering the pressure here causes the equilibrium to shift back this way.

ZFZ2J8gwIkE-015|So aqueous carbon dioxide comes out of solution, and it spontaneously comes out of solution all through the solution.