ceaglex/VisualTTS_Chem · Datasets at Hugging Face

text stringlengths 19 206
ZuwQf_7yXrg-020\|That means going back in this direction, we'd have 1 over that, something like K to the plus 5.
ZuwQf_7yXrg-021\|Same thing here.
ZuwQf_7yXrg-022\|This is the reaction of the weak base with water going in reverse.
ZuwQf_7yXrg-023\|So for acetic acid, this has something like 10 to the minus 10.
ZuwQf_7yXrg-024\|So going in this direction, 10 to the plus 10.
ZuwQf_7yXrg-025\|So both of these equilibria lie strongly towards products.
ZuwQf_7yXrg-026\|Both of them consume acid and base.
ZuwQf_7yXrg-027\|Strong acid and base consumed, converted into weak acid and base.
ZuwQf_7yXrg-028\|So whether I add a little acid or a little base, the pH doesn't change very much.
ZuwQf_7yXrg-029\|And we can watch that happen.
ZuwQf_7yXrg-031\|I'm going to add acid to all three of these, and I'm going to do it in a unique way.
ZuwQf_7yXrg-032\|I'm going to add solid carbon dioxide.
ZuwQf_7yXrg-033\|When I add solid carbon dioxide, that dissolves and forms carbonic acid.
ZuwQf_7yXrg-034\|So that will acidify the solutions.
ZuwQf_7yXrg-035\|I'll acidify the solutions, and let's watch how fast the pH changes.
ZuwQf_7yXrg-036\|So here to the buffered solution, and here to the unbuffered solution.
ZuwQf_7yXrg-038\|As the solutions are acidified, the unbuffered solution, pH changes rapidly.
ZuwQf_7yXrg-039\|And you can see that by the color change in the indicator.
ZuwQf_7yXrg-040\|In the buffered solution, a slow change in pH.
ZuwQf_7yXrg-041\|This buffered solution resists change in pH.
ZuwQf_7yXrg-042\|As acid is added, it's converted to a weak base, and the pH changes slowly.
ZuwQf_7yXrg-043\|That's the effect of a buffer in solution.
Zghfd6Nz5Rk-000\|Let's do some calculations with the ideal gas law.
Zghfd6Nz5Rk-001\|I'm going to take a flask, one liter, at room temperature, put a mole of a metal in it.
Zghfd6Nz5Rk-002\|Then I'm going to evacuate the flask, so there's no gas particles.
Zghfd6Nz5Rk-003\|And add in chlorine gas, Cl2 gas, until the pressure comes up to 48.6 atmospheres.
Zghfd6Nz5Rk-004\|Now that 48.6 atmospheres refers to a specific number of particles of chlorine gas.
Zghfd6Nz5Rk-005\|I'm going to let the reaction occur.
Zghfd6Nz5Rk-006\|After the reaction occurs, the pressure of chlorine gas has gone down to 24.3 atmospheres.
Zghfd6Nz5Rk-007\|So some of the chlorine has been consumed.
Zghfd6Nz5Rk-011\|And it's the ratio of the metal to the chlorine, this y to x ratio, that'll allow us to predict what metal that is.
Zghfd6Nz5Rk-012\|So let's see if we can figure that out.
Zghfd6Nz5Rk-013\|We know a in this equation, the coefficient of the metal is 1.
Zghfd6Nz5Rk-014\|B is the number of moles of chlorine gas, which I can figure out, because I know the pressure went from 48.6 to 24.3.
Zghfd6Nz5Rk-015\|The pressure dropped by a factor of two.
Zghfd6Nz5Rk-016\|That means half of the original chlorine molecules were used up in that chemical reaction.
Zghfd6Nz5Rk-026\|And I know the pressure, volume and the temperature.
Zghfd6Nz5Rk-027\|So I know how many atmospheres of chlorine were consumed, the volume and their temperature.
Zghfd6Nz5Rk-030\|Excuse me, one mole.
Zghfd6Nz5Rk-031\|So one mole of chlorine is consumed.
Zghfd6Nz5Rk-033\|So the actual relationship is C, 1.
Zghfd6Nz5Rk-034\|That is, this coefficient is 1.
Zghfd6Nz5Rk-035\|I have 1 mole of metal with 2 moles of chlorine atoms.
Zghfd6Nz5Rk-036\|So y is 1, x is 2.
Zghfd6Nz5Rk-039\|Well, it's all the metals in row two of the periodic table.
Zghfd6Nz5Rk-040\|Like magnesium and calcium, for instance.
Zghfd6Nz5Rk-041\|So we can tell about which metal that's going to be or predict which metal that is based on how it reacts with chlorine.
Zghfd6Nz5Rk-042\|And the number of moles of chlorine it reacts with, we figure from the ideal gas law.
Zghfd6Nz5Rk-043\|How the pressure changed in a chemical reaction.
_JgKgx7Jn-g-000\|The reaction of alkali metals with water is a really good way to demonstrate the trend in ionization energies in the alkali metals.
_JgKgx7Jn-g-006\|And this reaction, the formation of the hydroxide ion, releases quite a bit of energy.
_JgKgx7Jn-g-007\|But the reaction of ionization we know absorbs energy.
_JgKgx7Jn-g-008\|You have to put energy in to ionize, but then you get some back out when you form the hydroxide.
_JgKgx7Jn-g-010\|So if you have a lower ionization energy, you'll have a more vigorous reaction overall because more energy is allowed to be released.
_JgKgx7Jn-g-012\|We have metal being ionized, water forming hydroxide ion, more vigorous reaction for lower ionization energies.
XBua9WZkTeU-000\|Let's look at a galvanic cell where we use concentration difference instead of two different cell components.
XBua9WZkTeU-001\|So on each side we have a copper ion, copper metal half cell.
XBua9WZkTeU-002\|The only difference is the concentration on each side.
XBua9WZkTeU-012\|We're looking at a cell where the difference is the concentration of the ions on either side.
XBua9WZkTeU-013\|So the overall cell reaction is just high concentration copper going to low concentration copper.
XBua9WZkTeU-014\|And we know intuitively if we were to mix these two, that's what would happen, the concentration would equalize.
XBua9WZkTeU-015\|So intuitively, we know the concentration here of copper ions will decrease and the concentration here will increase.
XBua9WZkTeU-016\|But how does that correspond to a cell potential?
XBua9WZkTeU-017\|Well, let's look at that.
XBua9WZkTeU-019\|That would be 0 potential.
XBua9WZkTeU-020\|We don't have one molar on both sides.
XBua9WZkTeU-021\|We have 0.1 molar and one molar.
XBua9WZkTeU-022\|So we have a situation where we have reactants over products, the Q inserted here.
XBua9WZkTeU-023\|And that Q is less than 1.
XBua9WZkTeU-024\|That means this natural log term is negative.
XBua9WZkTeU-025\|So the overall potential difference is going to be positive.
XBua9WZkTeU-026\|So a positive cell potential difference in our convention means current flows from left to right, and that we understand.
XBua9WZkTeU-027\|Current flowing from left to right would have an oxidation occurring here.
m6SHjX_QcwE-000\|Let's look at chemical reactions and we'll compare them by plotting the initial rate versus the initial concentration of a reactant.
m6SHjX_QcwE-001\|Now reaction I plotted here is first order.
m6SHjX_QcwE-002\|The question is, what is the order of reaction II?
m6SHjX_QcwE-003\|Is that first order, second order, or is it indeterminant?
m6SHjX_QcwE-010\|We're looking at initial rates versus concentration for two reactions.
m6SHjX_QcwE-011\|Now, reaction I, we've plotted it here.
m6SHjX_QcwE-012\|And we've measured initial rates versus initial concentrations.
m6SHjX_QcwE-013\|So how would you do that experiment?
m6SHjX_QcwE-014\|You'd set up the reaction, you'd measure the rate.
m6SHjX_QcwE-015\|Then you'd set up the reaction again at a different initial concentration, and you'd measure the initial rate.
m6SHjX_QcwE-016\|Then, you'd set up the reaction again at a different initial concentration, measure the rate, and you would plot those out.
m6SHjX_QcwE-017\|And the plots you get for reaction I is linear, and that makes sense for a first order rate law.
m6SHjX_QcwE-018\|We said it was first order.
m6SHjX_QcwE-019\|The rate is proportional to the concentration.
m6SHjX_QcwE-020\|The proportionality constant is k, so the slope of this line would be the rate constant k.
m6SHjX_QcwE-021\|If it were a second order reaction, a second order reaction has overall powers add 2, in this case, there's just a single power of 2.
m6SHjX_QcwE-024\|So it can't be second order.
m6SHjX_QcwE-025\|It must be first orders well the quadratic relationship doesn't appear.
m6SHjX_QcwE-027\|The correct answer here is A, first order.
dVdQsZ8kmWg-000\|Let's compare three carbo-bases.
dVdQsZ8kmWg-001\|Which is the strongest base, acetate, ethanoate, or carbonate?
dVdQsZ8kmWg-002\|And here I've written their structures, acetate, ethanoate, carbonate.
dVdQsZ8kmWg-010\|We're talking about three bases, ethanoate, acetate, and carbonate.
dVdQsZ8kmWg-011\|We want to know which is the strongest.
dVdQsZ8kmWg-015\|Well, what's going on here?
dVdQsZ8kmWg-016\|This is a sequence where the oxidation number, the oxidation state of the carbon is changing.
dVdQsZ8kmWg-017\|I'm adding more electron withdrawing oxygens.

ZuwQf_7yXrg-020|That means going back in this direction, we'd have 1 over that, something like K to the plus 5.

ZuwQf_7yXrg-021|Same thing here.

ZuwQf_7yXrg-022|This is the reaction of the weak base with water going in reverse.

ZuwQf_7yXrg-023|So for acetic acid, this has something like 10 to the minus 10.

ZuwQf_7yXrg-024|So going in this direction, 10 to the plus 10.

ZuwQf_7yXrg-025|So both of these equilibria lie strongly towards products.

ZuwQf_7yXrg-026|Both of them consume acid and base.

ZuwQf_7yXrg-027|Strong acid and base consumed, converted into weak acid and base.

ZuwQf_7yXrg-028|So whether I add a little acid or a little base, the pH doesn't change very much.

ZuwQf_7yXrg-029|And we can watch that happen.

ZuwQf_7yXrg-031|I'm going to add acid to all three of these, and I'm going to do it in a unique way.

ZuwQf_7yXrg-032|I'm going to add solid carbon dioxide.

ZuwQf_7yXrg-033|When I add solid carbon dioxide, that dissolves and forms carbonic acid.

ZuwQf_7yXrg-034|So that will acidify the solutions.

ZuwQf_7yXrg-035|I'll acidify the solutions, and let's watch how fast the pH changes.

ZuwQf_7yXrg-036|So here to the buffered solution, and here to the unbuffered solution.

ZuwQf_7yXrg-038|As the solutions are acidified, the unbuffered solution, pH changes rapidly.

ZuwQf_7yXrg-039|And you can see that by the color change in the indicator.

ZuwQf_7yXrg-040|In the buffered solution, a slow change in pH.

ZuwQf_7yXrg-041|This buffered solution resists change in pH.

ZuwQf_7yXrg-042|As acid is added, it's converted to a weak base, and the pH changes slowly.

ZuwQf_7yXrg-043|That's the effect of a buffer in solution.

Zghfd6Nz5Rk-000|Let's do some calculations with the ideal gas law.

Zghfd6Nz5Rk-001|I'm going to take a flask, one liter, at room temperature, put a mole of a metal in it.

Zghfd6Nz5Rk-002|Then I'm going to evacuate the flask, so there's no gas particles.

Zghfd6Nz5Rk-003|And add in chlorine gas, Cl2 gas, until the pressure comes up to 48.6 atmospheres.

Zghfd6Nz5Rk-004|Now that 48.6 atmospheres refers to a specific number of particles of chlorine gas.

Zghfd6Nz5Rk-005|I'm going to let the reaction occur.

Zghfd6Nz5Rk-006|After the reaction occurs, the pressure of chlorine gas has gone down to 24.3 atmospheres.

Zghfd6Nz5Rk-007|So some of the chlorine has been consumed.

Zghfd6Nz5Rk-011|And it's the ratio of the metal to the chlorine, this y to x ratio, that'll allow us to predict what metal that is.

Zghfd6Nz5Rk-012|So let's see if we can figure that out.

Zghfd6Nz5Rk-013|We know a in this equation, the coefficient of the metal is 1.

Zghfd6Nz5Rk-014|B is the number of moles of chlorine gas, which I can figure out, because I know the pressure went from 48.6 to 24.3.

Zghfd6Nz5Rk-015|The pressure dropped by a factor of two.

Zghfd6Nz5Rk-016|That means half of the original chlorine molecules were used up in that chemical reaction.

Zghfd6Nz5Rk-026|And I know the pressure, volume and the temperature.

Zghfd6Nz5Rk-027|So I know how many atmospheres of chlorine were consumed, the volume and their temperature.

Zghfd6Nz5Rk-030|Excuse me, one mole.

Zghfd6Nz5Rk-031|So one mole of chlorine is consumed.

Zghfd6Nz5Rk-033|So the actual relationship is C, 1.

Zghfd6Nz5Rk-034|That is, this coefficient is 1.

Zghfd6Nz5Rk-035|I have 1 mole of metal with 2 moles of chlorine atoms.

Zghfd6Nz5Rk-036|So y is 1, x is 2.

Zghfd6Nz5Rk-039|Well, it's all the metals in row two of the periodic table.

Zghfd6Nz5Rk-040|Like magnesium and calcium, for instance.

Zghfd6Nz5Rk-041|So we can tell about which metal that's going to be or predict which metal that is based on how it reacts with chlorine.

Zghfd6Nz5Rk-042|And the number of moles of chlorine it reacts with, we figure from the ideal gas law.

Zghfd6Nz5Rk-043|How the pressure changed in a chemical reaction.

_JgKgx7Jn-g-000|The reaction of alkali metals with water is a really good way to demonstrate the trend in ionization energies in the alkali metals.

_JgKgx7Jn-g-006|And this reaction, the formation of the hydroxide ion, releases quite a bit of energy.

_JgKgx7Jn-g-007|But the reaction of ionization we know absorbs energy.

_JgKgx7Jn-g-008|You have to put energy in to ionize, but then you get some back out when you form the hydroxide.

_JgKgx7Jn-g-010|So if you have a lower ionization energy, you'll have a more vigorous reaction overall because more energy is allowed to be released.

_JgKgx7Jn-g-012|We have metal being ionized, water forming hydroxide ion, more vigorous reaction for lower ionization energies.

XBua9WZkTeU-000|Let's look at a galvanic cell where we use concentration difference instead of two different cell components.

XBua9WZkTeU-001|So on each side we have a copper ion, copper metal half cell.

XBua9WZkTeU-002|The only difference is the concentration on each side.

XBua9WZkTeU-012|We're looking at a cell where the difference is the concentration of the ions on either side.

XBua9WZkTeU-013|So the overall cell reaction is just high concentration copper going to low concentration copper.

XBua9WZkTeU-014|And we know intuitively if we were to mix these two, that's what would happen, the concentration would equalize.

XBua9WZkTeU-015|So intuitively, we know the concentration here of copper ions will decrease and the concentration here will increase.

XBua9WZkTeU-016|But how does that correspond to a cell potential?

XBua9WZkTeU-017|Well, let's look at that.

XBua9WZkTeU-019|That would be 0 potential.

XBua9WZkTeU-020|We don't have one molar on both sides.

XBua9WZkTeU-021|We have 0.1 molar and one molar.

XBua9WZkTeU-022|So we have a situation where we have reactants over products, the Q inserted here.

XBua9WZkTeU-023|And that Q is less than 1.

XBua9WZkTeU-024|That means this natural log term is negative.

XBua9WZkTeU-025|So the overall potential difference is going to be positive.

XBua9WZkTeU-026|So a positive cell potential difference in our convention means current flows from left to right, and that we understand.

XBua9WZkTeU-027|Current flowing from left to right would have an oxidation occurring here.

m6SHjX_QcwE-000|Let's look at chemical reactions and we'll compare them by plotting the initial rate versus the initial concentration of a reactant.

m6SHjX_QcwE-001|Now reaction I plotted here is first order.

m6SHjX_QcwE-002|The question is, what is the order of reaction II?

m6SHjX_QcwE-003|Is that first order, second order, or is it indeterminant?

m6SHjX_QcwE-010|We're looking at initial rates versus concentration for two reactions.

m6SHjX_QcwE-011|Now, reaction I, we've plotted it here.

m6SHjX_QcwE-012|And we've measured initial rates versus initial concentrations.

m6SHjX_QcwE-013|So how would you do that experiment?

m6SHjX_QcwE-014|You'd set up the reaction, you'd measure the rate.

m6SHjX_QcwE-015|Then you'd set up the reaction again at a different initial concentration, and you'd measure the initial rate.

m6SHjX_QcwE-016|Then, you'd set up the reaction again at a different initial concentration, measure the rate, and you would plot those out.

m6SHjX_QcwE-017|And the plots you get for reaction I is linear, and that makes sense for a first order rate law.

m6SHjX_QcwE-018|We said it was first order.

m6SHjX_QcwE-019|The rate is proportional to the concentration.

m6SHjX_QcwE-020|The proportionality constant is k, so the slope of this line would be the rate constant k.

m6SHjX_QcwE-021|If it were a second order reaction, a second order reaction has overall powers add 2, in this case, there's just a single power of 2.

m6SHjX_QcwE-024|So it can't be second order.

m6SHjX_QcwE-025|It must be first orders well the quadratic relationship doesn't appear.

m6SHjX_QcwE-027|The correct answer here is A, first order.

dVdQsZ8kmWg-000|Let's compare three carbo-bases.

dVdQsZ8kmWg-001|Which is the strongest base, acetate, ethanoate, or carbonate?

dVdQsZ8kmWg-002|And here I've written their structures, acetate, ethanoate, carbonate.

dVdQsZ8kmWg-010|We're talking about three bases, ethanoate, acetate, and carbonate.

dVdQsZ8kmWg-011|We want to know which is the strongest.

dVdQsZ8kmWg-015|Well, what's going on here?

dVdQsZ8kmWg-016|This is a sequence where the oxidation number, the oxidation state of the carbon is changing.

dVdQsZ8kmWg-017|I'm adding more electron withdrawing oxygens.