ceaglex/VisualTTS_Chem · Datasets at Hugging Face

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vApxjo9sAmI-026\|E is 3/2 nRT.
vApxjo9sAmI-027\|For other systems like rubber, it's almost the same.
vApxjo9sAmI-028\|That is, energy changes and temperature changes are very often correlated, and in this case, they are.
TZuRh_epSyI-000\|Let's talk some more about bonding.
TZuRh_epSyI-001\|When two elements come together to bond, they can share electrons to fill their octets.
TZuRh_epSyI-002\|And when we write down elements and their electrons as dots to form bonds, we're writing down what's known as Lewis electron dot structures.
TZuRh_epSyI-003\|It was Gilbert Lewis who first proposed that eight electrons form a stable octet and the basis of much of bonding.
TZuRh_epSyI-004\|So let's look at a few examples.
TZuRh_epSyI-005\|Here's carbon bonded to four hydrogens.
TZuRh_epSyI-006\|And why does that occur?
TZuRh_epSyI-007\|Well, carbon has four valence electrons.
TZuRh_epSyI-008\|Hydrogen has one.
TZuRh_epSyI-009\|So if a carbon wants to fill its octet, it needs four more electrons.
TZuRh_epSyI-010\|Each hydrogen has one.
TZuRh_epSyI-011\|So if it bonds to four hydrogens it can get to its octet.
TZuRh_epSyI-012\|Carbon says I'll share one with you.
TZuRh_epSyI-013\|I'll share one with you.
TZuRh_epSyI-014\|I'll share one with you and one with you.
TZuRh_epSyI-015\|And what I'll end up with is a carbon with an octet.
TZuRh_epSyI-016\|And here, instead of the dots, I've drawn a single line to indicate the pair of electrons.
TZuRh_epSyI-017\|So carbon has two, four, six, eight electrons around it.
TZuRh_epSyI-018\|Each hydrogen has a pair of electrons.
TZuRh_epSyI-019\|And a pair is sufficient for hydrogen because hydrogen only goes up to two.
TZuRh_epSyI-020\|It fills up principal quantum level one.
TZuRh_epSyI-021\|And the maximum number of electrons there is two.
TZuRh_epSyI-022\|Now, there's some other examples.
TZuRh_epSyI-025\|Well, six on one and six on another is a total of 12.
TZuRh_epSyI-026\|How do these oxygens share 12 electrons so that each octet is fulfilled?
TZuRh_epSyI-027\|Well, here's the best way to do it.
TZuRh_epSyI-028\|I write the oxygens with a double bond.
TZuRh_epSyI-029\|Each of these lines represent two electrons that the oxygens are sharing.
TZuRh_epSyI-030\|So if I count up the electrons-- two, four, six, eight, 10, 12, that's the 12 valence electrons that I can use for bonding.
TZuRh_epSyI-031\|How did I share them?
TZuRh_epSyI-032\|Well, this oxygen can count two, four, six, eight in its octet.
TZuRh_epSyI-033\|And this oxygen can count two, four, six, eight in its octet.
TZuRh_epSyI-034\|The shared electrons can be counted on both atoms.
TZuRh_epSyI-035\|That's how nitrogen bonds as well.
TZuRh_epSyI-036\|Nitrogen, the molecule, is a 10 electron system.
TZuRh_epSyI-037\|Each nitrogen has five valence electrons.
TZuRh_epSyI-039\|We write the Lewis electron dot structure with a triple bond.
TZuRh_epSyI-040\|Here it is.
TZuRh_epSyI-041\|Nitrogen each sharing six electrons.
TZuRh_epSyI-042\|The total count is 10- two, four, six, eight, 10.
TZuRh_epSyI-043\|Those are all the valence electrons.
TZuRh_epSyI-044\|For each nitrogen it's two, four, six, eight, a stable octet.
TZuRh_epSyI-045\|Two, four, six, eight on the other nitrogen.
TZuRh_epSyI-046\|Now the octet rule is followed a lot of the time, but there are exceptions.
TZuRh_epSyI-048\|The boron trifluoride molecule is relatively stable, but boron only has six electrons around it.
TZuRh_epSyI-049\|But because it's not a stable octet, it turns out boron trifluoride as a molecule is very reactive.
TZuRh_epSyI-050\|Boron is looking to fill its octet.
TZuRh_epSyI-051\|In fact, boron trifluoride will react with ammonia.
TZuRh_epSyI-053\|When you mix boron trifluoride and ammonia, immediately they form a compound between the nitrogen and boron.
TZuRh_epSyI-054\|And a new molecule is formed.
TZuRh_epSyI-055\|And we'll see that in the demo lab.
TZuRh_epSyI-056\|Here's xenon with four florines bonded to it and two lone pairs.
TZuRh_epSyI-057\|That's two, four, six, eight 10, 12, 12 electrons around xenon.
TZuRh_epSyI-058\|That's an expanded octet, more than eight.
TZuRh_epSyI-059\|Elements that have larger nuclei have higher principle quantum levels they can access.
TZuRh_epSyI-060\|So xenon can go to d orbitals in higher principle quantum levels.
TZuRh_epSyI-061\|And it's not restricted to an octet it can expand its octet using extra orbitals for bonding.
TZuRh_epSyI-062\|The same thing with SF6.
TZuRh_epSyI-064\|So as you go below principle quantum level three, you get extra orbitals.
TZuRh_epSyI-065\|You don't have to satisfy yourself with just your s and p orbitals in an octet.
TZuRh_epSyI-067\|And that's stable.
TZuRh_epSyI-068\|But expanded octets are also explained by quantum mechanics.
TZuRh_epSyI-069\|It says as you go to higher principle quantum levels, you have more orbitals that you can use for bonding.
DVeicfdiyQk-003\|Here, we're not at that standard-state conditions.
DVeicfdiyQk-004\|We're at a different set of conditions.
DVeicfdiyQk-005\|We want to determine if that set of conditions is at equilibrium or will the reaction proceed towards products or reactants to reach equilibrium?
DVeicfdiyQk-006\|So we know the equilibrium constant is 41.
DVeicfdiyQk-011\|I can plug in the conditions I've been given.
DVeicfdiyQk-015\|So Q is less than K. That means-- less than k, that means the denominator must be too big.
DVeicfdiyQk-016\|The reactants are too big.
DVeicfdiyQk-017\|So I'm going to proceed toward products to get Q to equal k.
DVeicfdiyQk-019\|Now, I should anticipate that that's going to be negative.
DVeicfdiyQk-020\|Because I've already said the reaction is going to go towards products.
DVeicfdiyQk-021\|That means it has to proceed further downhill.
DVeicfdiyQk-022\|The reactant free energies are still too high compared to the product free energies.
DVeicfdiyQk-035\|So everything rings true.
DVeicfdiyQk-036\|I can calculate reaction quotients.
nEniUpvACdY-000\|Let's a get a chemical reaction and some data and see if we can determine the rate orders.
nEniUpvACdY-001\|So, what's the order with respect to the partial pressure of bromine gas in this chemical reaction?
nEniUpvACdY-002\|Hydrogen gas plus bromine gas makes hydrogen bromide gas.
nEniUpvACdY-003\|I have some data here, some possible rate orders.
nEniUpvACdY-010\|We're looking at a chemical reaction and some data and trying to determine rate orders.
nEniUpvACdY-011\|So our chemical reaction hydrogen gas bromine gas goes to hydrogen bromide, and some data for initial partial pressures and rates.
nEniUpvACdY-015\|When I quadruple the concentration of bromine, holding everything else constant, the rate doubles.
nEniUpvACdY-016\|So if I'm going to write down a rate law, it sounds like the rate should be proportional to Br2 to the 1/2 power, the square root.
nEniUpvACdY-017\|Quadruple the concentration gives you a doubling of the rate.
nEniUpvACdY-018\|Well, as long as we're here, what about hydrogen?
nEniUpvACdY-019\|Can we figure that out too, the power with respect to hydrogen or the order with respect to hydrogen?
nEniUpvACdY-021\|And when I double the H2 partial pressure, I double the rate holding the Br concentration constant.
nEniUpvACdY-022\|So that's constant, doesn't matter, folded into the rate constant.
nEniUpvACdY-023\|Doubling this doubles that, so that should be a power of one.
nEniUpvACdY-024\|So what it looks like is our overall power is 1 and 1/2, or 3/2, the overall order of the rate is 3/2.
nEniUpvACdY-025\|We just asked for the order with respect to Br2.
0JyN9PH2oNg-000\|Water can act as an acid or a base.
0JyN9PH2oNg-001\|Water can react with itself, one molecule acting like an acid, another acting like a base, to form H3O+.
0JyN9PH2oNg-002\|When this water molecule donates a proton to this water molecule, I'll form an H3O+ and leave behind an OH-.
0JyN9PH2oNg-007\|So the equilibrium expression for the autodissociation of water is the product of H3O+ and OH- concentrations.

vApxjo9sAmI-026|E is 3/2 nRT.

vApxjo9sAmI-027|For other systems like rubber, it's almost the same.

vApxjo9sAmI-028|That is, energy changes and temperature changes are very often correlated, and in this case, they are.

TZuRh_epSyI-000|Let's talk some more about bonding.

TZuRh_epSyI-001|When two elements come together to bond, they can share electrons to fill their octets.

TZuRh_epSyI-002|And when we write down elements and their electrons as dots to form bonds, we're writing down what's known as Lewis electron dot structures.

TZuRh_epSyI-003|It was Gilbert Lewis who first proposed that eight electrons form a stable octet and the basis of much of bonding.

TZuRh_epSyI-004|So let's look at a few examples.

TZuRh_epSyI-005|Here's carbon bonded to four hydrogens.

TZuRh_epSyI-006|And why does that occur?

TZuRh_epSyI-007|Well, carbon has four valence electrons.

TZuRh_epSyI-008|Hydrogen has one.

TZuRh_epSyI-009|So if a carbon wants to fill its octet, it needs four more electrons.

TZuRh_epSyI-010|Each hydrogen has one.

TZuRh_epSyI-011|So if it bonds to four hydrogens it can get to its octet.

TZuRh_epSyI-012|Carbon says I'll share one with you.

TZuRh_epSyI-013|I'll share one with you.

TZuRh_epSyI-014|I'll share one with you and one with you.

TZuRh_epSyI-015|And what I'll end up with is a carbon with an octet.

TZuRh_epSyI-016|And here, instead of the dots, I've drawn a single line to indicate the pair of electrons.

TZuRh_epSyI-017|So carbon has two, four, six, eight electrons around it.

TZuRh_epSyI-018|Each hydrogen has a pair of electrons.

TZuRh_epSyI-019|And a pair is sufficient for hydrogen because hydrogen only goes up to two.

TZuRh_epSyI-020|It fills up principal quantum level one.

TZuRh_epSyI-021|And the maximum number of electrons there is two.

TZuRh_epSyI-022|Now, there's some other examples.

TZuRh_epSyI-025|Well, six on one and six on another is a total of 12.

TZuRh_epSyI-026|How do these oxygens share 12 electrons so that each octet is fulfilled?

TZuRh_epSyI-027|Well, here's the best way to do it.

TZuRh_epSyI-028|I write the oxygens with a double bond.

TZuRh_epSyI-029|Each of these lines represent two electrons that the oxygens are sharing.

TZuRh_epSyI-030|So if I count up the electrons-- two, four, six, eight, 10, 12, that's the 12 valence electrons that I can use for bonding.

TZuRh_epSyI-031|How did I share them?

TZuRh_epSyI-032|Well, this oxygen can count two, four, six, eight in its octet.

TZuRh_epSyI-033|And this oxygen can count two, four, six, eight in its octet.

TZuRh_epSyI-034|The shared electrons can be counted on both atoms.

TZuRh_epSyI-035|That's how nitrogen bonds as well.

TZuRh_epSyI-036|Nitrogen, the molecule, is a 10 electron system.

TZuRh_epSyI-037|Each nitrogen has five valence electrons.

TZuRh_epSyI-039|We write the Lewis electron dot structure with a triple bond.

TZuRh_epSyI-040|Here it is.

TZuRh_epSyI-041|Nitrogen each sharing six electrons.

TZuRh_epSyI-042|The total count is 10- two, four, six, eight, 10.

TZuRh_epSyI-043|Those are all the valence electrons.

TZuRh_epSyI-044|For each nitrogen it's two, four, six, eight, a stable octet.

TZuRh_epSyI-045|Two, four, six, eight on the other nitrogen.

TZuRh_epSyI-046|Now the octet rule is followed a lot of the time, but there are exceptions.

TZuRh_epSyI-048|The boron trifluoride molecule is relatively stable, but boron only has six electrons around it.

TZuRh_epSyI-049|But because it's not a stable octet, it turns out boron trifluoride as a molecule is very reactive.

TZuRh_epSyI-050|Boron is looking to fill its octet.

TZuRh_epSyI-051|In fact, boron trifluoride will react with ammonia.

TZuRh_epSyI-053|When you mix boron trifluoride and ammonia, immediately they form a compound between the nitrogen and boron.

TZuRh_epSyI-054|And a new molecule is formed.

TZuRh_epSyI-055|And we'll see that in the demo lab.

TZuRh_epSyI-056|Here's xenon with four florines bonded to it and two lone pairs.

TZuRh_epSyI-057|That's two, four, six, eight 10, 12, 12 electrons around xenon.

TZuRh_epSyI-058|That's an expanded octet, more than eight.

TZuRh_epSyI-059|Elements that have larger nuclei have higher principle quantum levels they can access.

TZuRh_epSyI-060|So xenon can go to d orbitals in higher principle quantum levels.

TZuRh_epSyI-061|And it's not restricted to an octet it can expand its octet using extra orbitals for bonding.

TZuRh_epSyI-062|The same thing with SF6.

TZuRh_epSyI-064|So as you go below principle quantum level three, you get extra orbitals.

TZuRh_epSyI-065|You don't have to satisfy yourself with just your s and p orbitals in an octet.

TZuRh_epSyI-067|And that's stable.

TZuRh_epSyI-068|But expanded octets are also explained by quantum mechanics.

TZuRh_epSyI-069|It says as you go to higher principle quantum levels, you have more orbitals that you can use for bonding.

DVeicfdiyQk-003|Here, we're not at that standard-state conditions.

DVeicfdiyQk-004|We're at a different set of conditions.

DVeicfdiyQk-005|We want to determine if that set of conditions is at equilibrium or will the reaction proceed towards products or reactants to reach equilibrium?

DVeicfdiyQk-006|So we know the equilibrium constant is 41.

DVeicfdiyQk-011|I can plug in the conditions I've been given.

DVeicfdiyQk-015|So Q is less than K. That means-- less than k, that means the denominator must be too big.

DVeicfdiyQk-016|The reactants are too big.

DVeicfdiyQk-017|So I'm going to proceed toward products to get Q to equal k.

DVeicfdiyQk-019|Now, I should anticipate that that's going to be negative.

DVeicfdiyQk-020|Because I've already said the reaction is going to go towards products.

DVeicfdiyQk-021|That means it has to proceed further downhill.

DVeicfdiyQk-022|The reactant free energies are still too high compared to the product free energies.

DVeicfdiyQk-035|So everything rings true.

DVeicfdiyQk-036|I can calculate reaction quotients.

nEniUpvACdY-000|Let's a get a chemical reaction and some data and see if we can determine the rate orders.

nEniUpvACdY-001|So, what's the order with respect to the partial pressure of bromine gas in this chemical reaction?

nEniUpvACdY-002|Hydrogen gas plus bromine gas makes hydrogen bromide gas.

nEniUpvACdY-003|I have some data here, some possible rate orders.

nEniUpvACdY-010|We're looking at a chemical reaction and some data and trying to determine rate orders.

nEniUpvACdY-011|So our chemical reaction hydrogen gas bromine gas goes to hydrogen bromide, and some data for initial partial pressures and rates.

nEniUpvACdY-015|When I quadruple the concentration of bromine, holding everything else constant, the rate doubles.

nEniUpvACdY-016|So if I'm going to write down a rate law, it sounds like the rate should be proportional to Br2 to the 1/2 power, the square root.

nEniUpvACdY-017|Quadruple the concentration gives you a doubling of the rate.

nEniUpvACdY-018|Well, as long as we're here, what about hydrogen?

nEniUpvACdY-019|Can we figure that out too, the power with respect to hydrogen or the order with respect to hydrogen?

nEniUpvACdY-021|And when I double the H2 partial pressure, I double the rate holding the Br concentration constant.

nEniUpvACdY-022|So that's constant, doesn't matter, folded into the rate constant.

nEniUpvACdY-023|Doubling this doubles that, so that should be a power of one.

nEniUpvACdY-024|So what it looks like is our overall power is 1 and 1/2, or 3/2, the overall order of the rate is 3/2.

nEniUpvACdY-025|We just asked for the order with respect to Br2.

0JyN9PH2oNg-000|Water can act as an acid or a base.

0JyN9PH2oNg-001|Water can react with itself, one molecule acting like an acid, another acting like a base, to form H3O+.

0JyN9PH2oNg-002|When this water molecule donates a proton to this water molecule, I'll form an H3O+ and leave behind an OH-.

0JyN9PH2oNg-007|So the equilibrium expression for the autodissociation of water is the product of H3O+ and OH- concentrations.