ceaglex/VisualTTS_Chem · Datasets at Hugging Face

text stringlengths 19 206
nbeE9b2JV5Q-017\|The pure solid doesn't appear in equilibrium expressions.
nbeE9b2JV5Q-018\|So we have a Ksp, solubility product, for magnesium hydroxide.
nbeE9b2JV5Q-020\|Well, you can think about this in a couple of ways.
nbeE9b2JV5Q-021\|This A is a strong acid.
nbeE9b2JV5Q-022\|Adding a strong acid will react with the OH minus quite strongly.
nbeE9b2JV5Q-023\|They'll form water, H3O pluses and OH minuses form water, that will lower this concentration.
nbeE9b2JV5Q-024\|If I lower that concentration, the reaction will shift, Le Chatelier's principle, shift towards the products to raise these concentrations.
nbeE9b2JV5Q-025\|So the magnesium ion concentration will go up, and you'll dissolve more solid by adding some strong acid.
nbeE9b2JV5Q-026\|Actually, the reverse occurs if you go with B. Adding NaOH adds sodium hydroxide, adds hydroxide ions.
nbeE9b2JV5Q-027\|That will shift the reaction back towards product.
nbeE9b2JV5Q-028\|Excuse me, back towards reactant, the solid.
nbeE9b2JV5Q-029\|And the solid will increase.
nbeE9b2JV5Q-030\|So this will actually make these guys come out of solution and precipitate to form the solid.
nbeE9b2JV5Q-031\|So this is actually worst case scenario.
nbeE9b2JV5Q-032\|If I add plain water, that will dilute the solution.
nbeE9b2JV5Q-034\|The question is which goes faster?
nbeE9b2JV5Q-035\|Adding some acid or adding the straight water?
nbeE9b2JV5Q-041\|So this is just some math, but what it shows you, it shows you, well, I understand how equilibrias hold.
nbeE9b2JV5Q-042\|They all hold at the same time.
nbeE9b2JV5Q-044\|So Kw is H3O plus times OH minus, so I can use that equilibrium and this equilibrium simultaneously.
nbeE9b2JV5Q-045\|When I do, I get an analytical expression for the magnesium ion concentration versus the acid concentration.
nbeE9b2JV5Q-046\|And it's a strong function.
nbeE9b2JV5Q-047\|It goes as the square of the acid concentration.
nbeE9b2JV5Q-048\|So as H3O plus goes up, the magnesium goes up as well in a squared fashion.
nbeE9b2JV5Q-049\|So a quadratic increase in my magnesium iron concentration by adding strong acid.
nbeE9b2JV5Q-050\|In this case, the best answer is A.
jW15zq0_78Q-000\|Let's do a calculation where we look at formal charges to determine the best Lewis electron dot structures.
jW15zq0_78Q-003\|Now, in order to do that, we need Lewis electron dot structures.
jW15zq0_78Q-004\|And to do that, we need to draw them all, all the various resonance structures and determine the best ones based on formal charges.
jW15zq0_78Q-005\|So let's look at a few.
jW15zq0_78Q-006\|Here's Lewis electron dot structure number one for our sulfate ion.
jW15zq0_78Q-007\|You can see, I've distributed the electrons around the atoms and bonded this all together.
jW15zq0_78Q-008\|All the rules for Lewis electron dot structures are followed for this molecule.
jW15zq0_78Q-009\|Now, let me assign formal charges.
jW15zq0_78Q-011\|So 1, 2, 3, 4, 5, 6, 7 electrons around the oxygen.
jW15zq0_78Q-012\|Oxygen normally has 6 as an atom, has one extra, that's a minus 1 formal charge.
jW15zq0_78Q-013\|All those oxygens are the same.
jW15zq0_78Q-014\|So they all have minus 1 formal charge.
jW15zq0_78Q-017\|So there, we've determined the formal charges for that Lewis electron dot structure.
jW15zq0_78Q-021\|Is this one better than this one is our question.
jW15zq0_78Q-022\|Well, let's, again, calculate formal charges.
jW15zq0_78Q-023\|Here, oxygen-- 1, 2, 3, 4, 5, 6.
jW15zq0_78Q-024\|Again, these two lone pairs are assigned to that oxygen, and this double bond has 4 electrons and I'm going to share them.
jW15zq0_78Q-025\|So 2 electrons are assigned to the oxygen, plus these two plus these two-- 6 electrons are assigned to oxygen.
jW15zq0_78Q-026\|It normally has 6 as an atom, so formal charge of 0 there.
jW15zq0_78Q-027\|This oxygen-- identical, so it'll have a 0.
jW15zq0_78Q-028\|And these oxygens look just like these oxygens, they have a formal charge of minus 1 again.
jW15zq0_78Q-029\|The sulfur, let's calculate its formal charge.
jW15zq0_78Q-030\|1, 2, 3, 4, 5, 6 electrons shared in the bonds around sulfur.
jW15zq0_78Q-031\|6 electrons-- as an atom it has 6, so it has a formal charge of 0.
jW15zq0_78Q-035\|We know these oxygens are similar to these oxygens, so they'll have formal charge 0 again.
jW15zq0_78Q-036\|The sulfur shares 1 electron from each of these 8 bonds, that gives it 8 electrons around it.
jW15zq0_78Q-038\|Now here, I have a lot of 0 formal charges, that's nice, but I have a minus 2 on a lesser electronegative element.
jW15zq0_78Q-039\|When I compare sulfur and oxygen, oxygen is the more electronegative.
jW15zq0_78Q-040\|I'd rather have the formal charges negative on the more electronegative elements.
jW15zq0_78Q-041\|So this Lewis dot structure I don't like a lot because of the numbers and distribution of formal charges.
jW15zq0_78Q-042\|This Lewis dot structure, I've got a negative charge on the less electronegative element.
jW15zq0_78Q-043\|So those two I don't like as much as I like this one, which has low formal charges and the negative formal charges are on the more electronegative element.
jW15zq0_78Q-044\|So if I calculate the bond order here, I think it's pretty obvious, the bond order is 1 in an S-O bond here.
jW15zq0_78Q-045\|An S-O bond here, I can see two single bonds, two double bonds, I'll get a bond order of 1 and 1/2.
jW15zq0_78Q-046\|Here, it's obvious that the bond order is going to be 2.
jW15zq0_78Q-047\|All of them are identical and double bonds.
jW15zq0_78Q-048\|So I'll have a bond order of 2 here.
jW15zq0_78Q-049\|But I'm going to reject these two Lewis electron dot structures based on their formal charges.
eCc24p1bLig-000\|Another way to form bonds and fill octets is to share the electrons rather than to transfer them from one element to another.
eCc24p1bLig-001\|If I share electrons to fill octets that's called covalent bonding.
eCc24p1bLig-005\|Same thing with chlorine.
eCc24p1bLig-006\|Chlorine can come together and fill its octet of eight by sharing one valence electron, forming a covalent bond.
eCc24p1bLig-007\|And of course, hydrogen chloride could form.
eCc24p1bLig-008\|Hydrogen can share its valence electron with chlorine.
eCc24p1bLig-009\|Chlorine gets its full octet by sharing.
eCc24p1bLig-010\|Hydrogen gets its full outer shell of two by sharing.
eCc24p1bLig-011\|And they form a stable bond.
eCc24p1bLig-012\|So covalent bonding is sharing electrons to fill octets.
43YS58bxWuU-000\|When an electronic transition occurs, a photon is either absorbed or emitted.
43YS58bxWuU-001\|So if a photon is absorbed, an electron goes from a low energy state to a high energy state.
43YS58bxWuU-002\|When a photon is emitted, that's an electron in an excited state dropping down to a lower energy state.
43YS58bxWuU-003\|When we observe those things, we say, oh, there's blue light, or there's an ultraviolet photon being absorbed or emitted.
43YS58bxWuU-004\|We can infer something about the electronic structure of that atom.
43YS58bxWuU-006\|So let's look at that.
43YS58bxWuU-007\|In the hydrogen atom, you have various energy levels that we're now familiar with.
43YS58bxWuU-008\|And if you have transitions that end or start in principle quantum level 1, those transitions will all be in the ultraviolet.
43YS58bxWuU-009\|Our eyes won't be sensitive to them.
43YS58bxWuU-010\|You can design a detector that detects ultraviolet, but our eyes don't work to detect these.
43YS58bxWuU-011\|They're too high energy of photons.
43YS58bxWuU-012\|If transitions start or end-- that is they end in 2 if it was an emission.
43YS58bxWuU-013\|They would start in 2 if there was an absorption of a photon.
43YS58bxWuU-014\|We can have the first, which is the 3 to 2 transition, actually is in a visible region.
43YS58bxWuU-015\|So it's about 657 nanometers.
43YS58bxWuU-016\|That's a red photon.
43YS58bxWuU-017\|And we can detect that with our eyes.
43YS58bxWuU-018\|The 4 to 2 is in the green, about 487 nanometers.
43YS58bxWuU-021\|The ultraviolet Lyman series of lines, or the Balmer series of lines-- those ending or starting in an equal 2.
43YS58bxWuU-022\|Now, since there is these visible emission lines, we can detect those.
43YS58bxWuU-023\|If we excite hydrogen, or any number of gases, there'll be emissions that are in the visible spectrum, and we can see them.
43YS58bxWuU-024\|And if you do excitation in particular-- if you excite gases, and let the photons be emitted, you can actually see them.
43YS58bxWuU-025\|And that's how neon signs work.
43YS58bxWuU-026\|They excite gases.
43YS58bxWuU-027\|And then those gases emit photons.
43YS58bxWuU-028\|And those photons are visible to us.

nbeE9b2JV5Q-017|The pure solid doesn't appear in equilibrium expressions.

nbeE9b2JV5Q-018|So we have a Ksp, solubility product, for magnesium hydroxide.

nbeE9b2JV5Q-020|Well, you can think about this in a couple of ways.

nbeE9b2JV5Q-021|This A is a strong acid.

nbeE9b2JV5Q-022|Adding a strong acid will react with the OH minus quite strongly.

nbeE9b2JV5Q-023|They'll form water, H3O pluses and OH minuses form water, that will lower this concentration.

nbeE9b2JV5Q-024|If I lower that concentration, the reaction will shift, Le Chatelier's principle, shift towards the products to raise these concentrations.

nbeE9b2JV5Q-025|So the magnesium ion concentration will go up, and you'll dissolve more solid by adding some strong acid.

nbeE9b2JV5Q-026|Actually, the reverse occurs if you go with B. Adding NaOH adds sodium hydroxide, adds hydroxide ions.

nbeE9b2JV5Q-027|That will shift the reaction back towards product.

nbeE9b2JV5Q-028|Excuse me, back towards reactant, the solid.

nbeE9b2JV5Q-029|And the solid will increase.

nbeE9b2JV5Q-030|So this will actually make these guys come out of solution and precipitate to form the solid.

nbeE9b2JV5Q-031|So this is actually worst case scenario.

nbeE9b2JV5Q-032|If I add plain water, that will dilute the solution.

nbeE9b2JV5Q-034|The question is which goes faster?

nbeE9b2JV5Q-035|Adding some acid or adding the straight water?

nbeE9b2JV5Q-041|So this is just some math, but what it shows you, it shows you, well, I understand how equilibrias hold.

nbeE9b2JV5Q-042|They all hold at the same time.

nbeE9b2JV5Q-044|So Kw is H3O plus times OH minus, so I can use that equilibrium and this equilibrium simultaneously.

nbeE9b2JV5Q-045|When I do, I get an analytical expression for the magnesium ion concentration versus the acid concentration.

nbeE9b2JV5Q-046|And it's a strong function.

nbeE9b2JV5Q-047|It goes as the square of the acid concentration.

nbeE9b2JV5Q-048|So as H3O plus goes up, the magnesium goes up as well in a squared fashion.

nbeE9b2JV5Q-049|So a quadratic increase in my magnesium iron concentration by adding strong acid.

nbeE9b2JV5Q-050|In this case, the best answer is A.

jW15zq0_78Q-000|Let's do a calculation where we look at formal charges to determine the best Lewis electron dot structures.

jW15zq0_78Q-003|Now, in order to do that, we need Lewis electron dot structures.

jW15zq0_78Q-004|And to do that, we need to draw them all, all the various resonance structures and determine the best ones based on formal charges.

jW15zq0_78Q-005|So let's look at a few.

jW15zq0_78Q-006|Here's Lewis electron dot structure number one for our sulfate ion.

jW15zq0_78Q-007|You can see, I've distributed the electrons around the atoms and bonded this all together.

jW15zq0_78Q-008|All the rules for Lewis electron dot structures are followed for this molecule.

jW15zq0_78Q-009|Now, let me assign formal charges.

jW15zq0_78Q-011|So 1, 2, 3, 4, 5, 6, 7 electrons around the oxygen.

jW15zq0_78Q-012|Oxygen normally has 6 as an atom, has one extra, that's a minus 1 formal charge.

jW15zq0_78Q-013|All those oxygens are the same.

jW15zq0_78Q-014|So they all have minus 1 formal charge.

jW15zq0_78Q-017|So there, we've determined the formal charges for that Lewis electron dot structure.

jW15zq0_78Q-021|Is this one better than this one is our question.

jW15zq0_78Q-022|Well, let's, again, calculate formal charges.

jW15zq0_78Q-023|Here, oxygen-- 1, 2, 3, 4, 5, 6.

jW15zq0_78Q-024|Again, these two lone pairs are assigned to that oxygen, and this double bond has 4 electrons and I'm going to share them.

jW15zq0_78Q-025|So 2 electrons are assigned to the oxygen, plus these two plus these two-- 6 electrons are assigned to oxygen.

jW15zq0_78Q-026|It normally has 6 as an atom, so formal charge of 0 there.

jW15zq0_78Q-027|This oxygen-- identical, so it'll have a 0.

jW15zq0_78Q-028|And these oxygens look just like these oxygens, they have a formal charge of minus 1 again.

jW15zq0_78Q-029|The sulfur, let's calculate its formal charge.

jW15zq0_78Q-030|1, 2, 3, 4, 5, 6 electrons shared in the bonds around sulfur.

jW15zq0_78Q-031|6 electrons-- as an atom it has 6, so it has a formal charge of 0.

jW15zq0_78Q-035|We know these oxygens are similar to these oxygens, so they'll have formal charge 0 again.

jW15zq0_78Q-036|The sulfur shares 1 electron from each of these 8 bonds, that gives it 8 electrons around it.

jW15zq0_78Q-038|Now here, I have a lot of 0 formal charges, that's nice, but I have a minus 2 on a lesser electronegative element.

jW15zq0_78Q-039|When I compare sulfur and oxygen, oxygen is the more electronegative.

jW15zq0_78Q-040|I'd rather have the formal charges negative on the more electronegative elements.

jW15zq0_78Q-041|So this Lewis dot structure I don't like a lot because of the numbers and distribution of formal charges.

jW15zq0_78Q-042|This Lewis dot structure, I've got a negative charge on the less electronegative element.

jW15zq0_78Q-043|So those two I don't like as much as I like this one, which has low formal charges and the negative formal charges are on the more electronegative element.

jW15zq0_78Q-044|So if I calculate the bond order here, I think it's pretty obvious, the bond order is 1 in an S-O bond here.

jW15zq0_78Q-045|An S-O bond here, I can see two single bonds, two double bonds, I'll get a bond order of 1 and 1/2.

jW15zq0_78Q-046|Here, it's obvious that the bond order is going to be 2.

jW15zq0_78Q-047|All of them are identical and double bonds.

jW15zq0_78Q-048|So I'll have a bond order of 2 here.

jW15zq0_78Q-049|But I'm going to reject these two Lewis electron dot structures based on their formal charges.

eCc24p1bLig-000|Another way to form bonds and fill octets is to share the electrons rather than to transfer them from one element to another.

eCc24p1bLig-001|If I share electrons to fill octets that's called covalent bonding.

eCc24p1bLig-005|Same thing with chlorine.

eCc24p1bLig-006|Chlorine can come together and fill its octet of eight by sharing one valence electron, forming a covalent bond.

eCc24p1bLig-007|And of course, hydrogen chloride could form.

eCc24p1bLig-008|Hydrogen can share its valence electron with chlorine.

eCc24p1bLig-009|Chlorine gets its full octet by sharing.

eCc24p1bLig-010|Hydrogen gets its full outer shell of two by sharing.

eCc24p1bLig-011|And they form a stable bond.

eCc24p1bLig-012|So covalent bonding is sharing electrons to fill octets.

43YS58bxWuU-000|When an electronic transition occurs, a photon is either absorbed or emitted.

43YS58bxWuU-001|So if a photon is absorbed, an electron goes from a low energy state to a high energy state.

43YS58bxWuU-002|When a photon is emitted, that's an electron in an excited state dropping down to a lower energy state.

43YS58bxWuU-003|When we observe those things, we say, oh, there's blue light, or there's an ultraviolet photon being absorbed or emitted.

43YS58bxWuU-004|We can infer something about the electronic structure of that atom.

43YS58bxWuU-006|So let's look at that.

43YS58bxWuU-007|In the hydrogen atom, you have various energy levels that we're now familiar with.

43YS58bxWuU-008|And if you have transitions that end or start in principle quantum level 1, those transitions will all be in the ultraviolet.

43YS58bxWuU-009|Our eyes won't be sensitive to them.

43YS58bxWuU-010|You can design a detector that detects ultraviolet, but our eyes don't work to detect these.

43YS58bxWuU-011|They're too high energy of photons.

43YS58bxWuU-012|If transitions start or end-- that is they end in 2 if it was an emission.

43YS58bxWuU-013|They would start in 2 if there was an absorption of a photon.

43YS58bxWuU-014|We can have the first, which is the 3 to 2 transition, actually is in a visible region.

43YS58bxWuU-015|So it's about 657 nanometers.

43YS58bxWuU-016|That's a red photon.

43YS58bxWuU-017|And we can detect that with our eyes.

43YS58bxWuU-018|The 4 to 2 is in the green, about 487 nanometers.

43YS58bxWuU-021|The ultraviolet Lyman series of lines, or the Balmer series of lines-- those ending or starting in an equal 2.

43YS58bxWuU-022|Now, since there is these visible emission lines, we can detect those.

43YS58bxWuU-023|If we excite hydrogen, or any number of gases, there'll be emissions that are in the visible spectrum, and we can see them.

43YS58bxWuU-024|And if you do excitation in particular-- if you excite gases, and let the photons be emitted, you can actually see them.

43YS58bxWuU-025|And that's how neon signs work.

43YS58bxWuU-026|They excite gases.

43YS58bxWuU-027|And then those gases emit photons.

43YS58bxWuU-028|And those photons are visible to us.