alexignite/chemset1 · Datasets at Hugging Face

title stringlengths 8 78	text stringlengths 3 643
Rate Law .txt	Let's plug in 0.5 for this guy, 0.5 for this guy, and 0.2 for our rate of reaction, the forward direction.
Rate Law .txt	So we plug these guys in and we solve for KF and we get 0.002 divided by zero.
Rate Law .txt	Five 0.5 times 0.05 gives us 0.08.
Rate Law .txt	So that means our KF, our rate constant for this reaction going this way is 0.8.
Rate Law .txt	So now we plug this into our reaction and we find the final rate law to be the rate of our four reaction is equal to 0.8 times the concentration of our methyl acetate to the first power.
Rate Law .txt	Times the concentration of our hydroxide to the first power.
Rate Law .txt	Now, let's think about it.
Rate Law .txt	So our initial concentration increases and that increase increases our initial rates.
Rate Law .txt	Why is that?
Rate Law .txt	Well, think about it.
Rate Law .txt	The only way that these guys react is if they collide.
Rate Law .txt	And if you increase the concentration of either guy, we have more collisions happening.
Rate Law .txt	And if there are more collisions happening, that means our rate should increase.
Rate Law .txt	In other words, these guys will convert quicker to our products from the reactants.
Rate Law .txt	Now, likewise, if you decrease either concentration, our rate should decrease because we have less collisions occurring.
Rate Law .txt	So our rate constant will be less and therefore our rate will be less.
Molarity Example .txt	One of the different ways that people could measure the concentration of a solution is using molarity.
Molarity Example .txt	What is molarity?
Molarity Example .txt	Molarity is represented by the capital letter M, and it has the unit moles of solid over volume of solution.
Molarity Example .txt	And this includes the volume of both the solvent and the solid.
Molarity Example .txt	Now let's do an example using molar.
Molarity Example .txt	The question tells us that we have 0.6 liters of a three molar solution.
Molarity Example .txt	We need to find the number of liters that we need to add to go from three molar to a one molar solution.
Molarity Example .txt	So we're diluting.
Molarity Example .txt	Whenever we dilute, we want to keep the number of solid the same.
Molarity Example .txt	We want to increase the number of solids.
Molarity Example .txt	So this is our current situation.
Molarity Example .txt	We have a three molar solution where the red dots are the solid, blue dots are the solid.
Molarity Example .txt	We want to go from a three molar to a one molar solution.
Molarity Example .txt	So we need to ask ourselves, how many more blue dots do we need to add to go from a three molar to a one molar?
Molarity Example .txt	So since the number of red dots stays the same, we need to find the constant.
Molarity Example .txt	The constant here is the number of moles of solute.
Molarity Example .txt	To find the number of moles of solute, we take 0.6 liters and we multiply it by three molar.
Molarity Example .txt	0.6 liters times three moles of solute over liters.
Molarity Example .txt	The L's cross out 0.6 times three.
Molarity Example .txt	We get 1.8 moles of solution.
Molarity Example .txt	Now, we found the number of red dots or the moles of red dots.
Molarity Example .txt	Now, our goal is a one molar solution.
Molarity Example .txt	So we set up an equation.
Molarity Example .txt	One molar is equal to the thing that stays constant, one moles of solution.
Molarity Example .txt	Over what the amount?
Molarity Example .txt	We already have 0.6 liters plus the amount we need to add the mount of blue dots that we need to add to the system to get one molar solution.
Molarity Example .txt	Okay, now we do a little bit of algebra, and we get x equals 1.2 liters of solvent.
Molarity Example .txt	In other words, we need to add 1.2 liters worth of blue dots to get a one molar solution.
Sp2 Hybridization.txt	So in the previous lecture, we began our discussion on hybridization and we develop sphybidized orbitals.
Sp2 Hybridization.txt	So we essentially took one S orbital, we took one P orbital, we combine them, and we formed two different sphypedized orbitals.
Sp2 Hybridization.txt	Now we're going to look at SP two hybridized orbitals.
Sp2 Hybridization.txt	So let's suppose we want to construct a BH three molecule.
Sp2 Hybridization.txt	Now, in order to construct this molecule, we need three H atoms and one boron atom.
Sp2 Hybridization.txt	Now, boron has five protons, so has five electrons.
Sp2 Hybridization.txt	Two go into the one S, two go into the two S, and one goes into the two piece.
Sp2 Hybridization.txt	So we have three balanced electrons.
Sp2 Hybridization.txt	Now, the H atom each has one electron.
Sp2 Hybridization.txt	So the one electron goes into the one S orbital.
Sp2 Hybridization.txt	Now, before these atoms can combine to form, our BH three molecule hybridization of boron must take place.
Sp2 Hybridization.txt	The question is, how many hybrid orbitals should boron form before this molecule can be created?
Sp2 Hybridization.txt	The answer lies in this molecule itself.
Sp2 Hybridization.txt	How many times does boron bond two H?
Sp2 Hybridization.txt	Well, since there's one boron and three H atoms, that means there are three different orbitals.
Sp2 Hybridization.txt	So that means we must develop three hybrid orbitals.
Sp2 Hybridization.txt	So that means we can no longer use SP hybridization, because SP hybridization produces only two hybrid orbitals.
Sp2 Hybridization.txt	And we need three, as we see in this case here.
Sp2 Hybridization.txt	So that means we're not combining S and P, but we're combining three orbitals.
Sp2 Hybridization.txt	And these three orbitals are the two S orbital, the two PX orbital, and the two PY orbital.
Sp2 Hybridization.txt	So we combine these atomic orbitals of the boron atom, and we form three identical SP, two orbitals or hybrid orbitals.
Sp2 Hybridization.txt	And these guys look like this hybrid orbital here.
Sp2 Hybridization.txt	The only difference is they all lie in different directions.
Sp2 Hybridization.txt	So they point in different directions.
Sp2 Hybridization.txt	So that means because we're combining two P orbitals and one two S orbital, we're going to have 66% P character and 33% S character.
Sp2 Hybridization.txt	So now that we formed the three different hybrid orbitals, we are ready for these orbitals to interact with the one S orbitals of the H. So we take three one S orbitals, we combine them with three SP two orbitals, and we form the following picture.
Sp2 Hybridization.txt	So here we have our boron atom.
Sp2 Hybridization.txt	The nucleus is in the middle here.
Sp2 Hybridization.txt	It's not shown.
Sp2 Hybridization.txt	We have one SP hybridized orbital pointing into the page into the board, and it's bonding to one of the S's.
Sp2 Hybridization.txt	One is coming out of the board, and that is bonding to a second one S orbital.
Sp2 Hybridization.txt	And a third is lying on the board, and it's bonding with a third one S orbital.
Sp2 Hybridization.txt	And here we have our BH three molecule that also looks like this.
Sp2 Hybridization.txt	So once again, the boron nucleus, one bond is coming out of the page.
Sp2 Hybridization.txt	The second bond is going into the page, and the third is going this way.
Sp2 Hybridization.txt	Now, if we grab the XYZ axis, we see that this molecule lies on the same plane.
Sp2 Hybridization.txt	All the bonds lie on the same plane.
Sp2 Hybridization.txt	So that means since we have three angles, each angle must be 120 degrees.
Sp2 Hybridization.txt	So, once again, let's review.
Sp2 Hybridization.txt	So what is an SP two hybridized orbital?
Sp2 Hybridization.txt	Well, an SP two hybridized orbital is simply the combination of three different atomic orbitals found within an atom.
Sp2 Hybridization.txt	In this case, it was the Boron atom.
Sp2 Hybridization.txt	When these three different atomic orbitals combined, they form identical hybrid orbitals.
Sp2 Hybridization.txt	Remember, the amount of orbitals that go into the combination must equal to the amount of hybrid orbitals we form.
Sp2 Hybridization.txt	So if we include three if we input three we you must produce three orbitals.
Polyprotic Acids .txt	Now, as of now, we've only really spoken about monoprotic acids.
Polyprotic Acids .txt	Now, monoprotic acid is an acid that can donate a single H plus ion.
Polyprotic Acids .txt	Let's look at a few examples.
Polyprotic Acids .txt	Hydrochloric acid, acetic acid, nitric acid, hydrobromic acid, hydrochloric acid, chloro acid, and perchloric acid are all examples of monopolic acids because they can all donate a single H plus ion.
Polyprotic Acids .txt	Now let's look at the ionization of these acids in water.
Polyprotic Acids .txt	Let's choose the hypothetical monopolic acid.
Polyprotic Acids .txt	So let's choose Ha to be our acid.
Polyprotic Acids .txt	And Ha will react in water to produce hydronium and a conjugate base.
Polyprotic Acids .txt	Now, we've spoken about something called Ka, or the acid ionization concept.
Polyprotic Acids .txt	If you don't know what K is, check out the link above.
Polyprotic Acids .txt	But Ka is basically the product of the concentrations of the product.

Rate Law .txt

Let's plug in 0.5 for this guy, 0.5 for this guy, and 0.2 for our rate of reaction, the forward direction.

Rate Law .txt

So we plug these guys in and we solve for KF and we get 0.002 divided by zero.

Rate Law .txt

Five 0.5 times 0.05 gives us 0.08.

Rate Law .txt

So that means our KF, our rate constant for this reaction going this way is 0.8.

Rate Law .txt

So now we plug this into our reaction and we find the final rate law to be the rate of our four reaction is equal to 0.8 times the concentration of our methyl acetate to the first power.

Rate Law .txt

Times the concentration of our hydroxide to the first power.

Rate Law .txt

Now, let's think about it.

Rate Law .txt

So our initial concentration increases and that increase increases our initial rates.

Rate Law .txt

Why is that?

Rate Law .txt

Well, think about it.

Rate Law .txt

The only way that these guys react is if they collide.

Rate Law .txt

And if you increase the concentration of either guy, we have more collisions happening.

Rate Law .txt

And if there are more collisions happening, that means our rate should increase.

Rate Law .txt

In other words, these guys will convert quicker to our products from the reactants.

Rate Law .txt

Now, likewise, if you decrease either concentration, our rate should decrease because we have less collisions occurring.

Rate Law .txt

So our rate constant will be less and therefore our rate will be less.

Molarity Example .txt

One of the different ways that people could measure the concentration of a solution is using molarity.

Molarity Example .txt

What is molarity?

Molarity Example .txt

Molarity is represented by the capital letter M, and it has the unit moles of solid over volume of solution.

Molarity Example .txt

And this includes the volume of both the solvent and the solid.

Molarity Example .txt

Now let's do an example using molar.

Molarity Example .txt

The question tells us that we have 0.6 liters of a three molar solution.

Molarity Example .txt

We need to find the number of liters that we need to add to go from three molar to a one molar solution.

Molarity Example .txt

So we're diluting.

Molarity Example .txt

Whenever we dilute, we want to keep the number of solid the same.

Molarity Example .txt

We want to increase the number of solids.

Molarity Example .txt

So this is our current situation.

Molarity Example .txt

We have a three molar solution where the red dots are the solid, blue dots are the solid.

Molarity Example .txt

We want to go from a three molar to a one molar solution.

Molarity Example .txt

So we need to ask ourselves, how many more blue dots do we need to add to go from a three molar to a one molar?

Molarity Example .txt

So since the number of red dots stays the same, we need to find the constant.

Molarity Example .txt

The constant here is the number of moles of solute.

Molarity Example .txt

To find the number of moles of solute, we take 0.6 liters and we multiply it by three molar.

Molarity Example .txt

0.6 liters times three moles of solute over liters.

Molarity Example .txt

The L's cross out 0.6 times three.

Molarity Example .txt

We get 1.8 moles of solution.

Molarity Example .txt

Now, we found the number of red dots or the moles of red dots.

Molarity Example .txt

Now, our goal is a one molar solution.

Molarity Example .txt

So we set up an equation.

Molarity Example .txt

One molar is equal to the thing that stays constant, one moles of solution.

Molarity Example .txt

Over what the amount?

Molarity Example .txt

We already have 0.6 liters plus the amount we need to add the mount of blue dots that we need to add to the system to get one molar solution.

Molarity Example .txt

Okay, now we do a little bit of algebra, and we get x equals 1.2 liters of solvent.

Molarity Example .txt

In other words, we need to add 1.2 liters worth of blue dots to get a one molar solution.

Sp2 Hybridization.txt

So in the previous lecture, we began our discussion on hybridization and we develop sphybidized orbitals.

Sp2 Hybridization.txt

So we essentially took one S orbital, we took one P orbital, we combine them, and we formed two different sphypedized orbitals.

Sp2 Hybridization.txt

Now we're going to look at SP two hybridized orbitals.

Sp2 Hybridization.txt

So let's suppose we want to construct a BH three molecule.

Sp2 Hybridization.txt

Now, in order to construct this molecule, we need three H atoms and one boron atom.

Sp2 Hybridization.txt

Now, boron has five protons, so has five electrons.

Sp2 Hybridization.txt

Two go into the one S, two go into the two S, and one goes into the two piece.

Sp2 Hybridization.txt

So we have three balanced electrons.

Sp2 Hybridization.txt

Now, the H atom each has one electron.

Sp2 Hybridization.txt

So the one electron goes into the one S orbital.

Sp2 Hybridization.txt

Now, before these atoms can combine to form, our BH three molecule hybridization of boron must take place.

Sp2 Hybridization.txt

The question is, how many hybrid orbitals should boron form before this molecule can be created?

Sp2 Hybridization.txt

The answer lies in this molecule itself.

Sp2 Hybridization.txt

How many times does boron bond two H?

Sp2 Hybridization.txt

Well, since there's one boron and three H atoms, that means there are three different orbitals.

Sp2 Hybridization.txt

So that means we must develop three hybrid orbitals.

Sp2 Hybridization.txt

So that means we can no longer use SP hybridization, because SP hybridization produces only two hybrid orbitals.

Sp2 Hybridization.txt

And we need three, as we see in this case here.

Sp2 Hybridization.txt

So that means we're not combining S and P, but we're combining three orbitals.

Sp2 Hybridization.txt

And these three orbitals are the two S orbital, the two PX orbital, and the two PY orbital.

Sp2 Hybridization.txt

So we combine these atomic orbitals of the boron atom, and we form three identical SP, two orbitals or hybrid orbitals.

Sp2 Hybridization.txt

And these guys look like this hybrid orbital here.

Sp2 Hybridization.txt

The only difference is they all lie in different directions.

Sp2 Hybridization.txt

So they point in different directions.

Sp2 Hybridization.txt

So that means because we're combining two P orbitals and one two S orbital, we're going to have 66% P character and 33% S character.

Sp2 Hybridization.txt

So now that we formed the three different hybrid orbitals, we are ready for these orbitals to interact with the one S orbitals of the H. So we take three one S orbitals, we combine them with three SP two orbitals, and we form the following picture.

Sp2 Hybridization.txt

So here we have our boron atom.

Sp2 Hybridization.txt

The nucleus is in the middle here.

Sp2 Hybridization.txt

It's not shown.

Sp2 Hybridization.txt

We have one SP hybridized orbital pointing into the page into the board, and it's bonding to one of the S's.

Sp2 Hybridization.txt

One is coming out of the board, and that is bonding to a second one S orbital.

Sp2 Hybridization.txt

And a third is lying on the board, and it's bonding with a third one S orbital.

Sp2 Hybridization.txt

And here we have our BH three molecule that also looks like this.

Sp2 Hybridization.txt

So once again, the boron nucleus, one bond is coming out of the page.

Sp2 Hybridization.txt

The second bond is going into the page, and the third is going this way.

Sp2 Hybridization.txt

Now, if we grab the XYZ axis, we see that this molecule lies on the same plane.

Sp2 Hybridization.txt

All the bonds lie on the same plane.

Sp2 Hybridization.txt

So that means since we have three angles, each angle must be 120 degrees.

Sp2 Hybridization.txt

So, once again, let's review.

Sp2 Hybridization.txt

So what is an SP two hybridized orbital?

Sp2 Hybridization.txt

Well, an SP two hybridized orbital is simply the combination of three different atomic orbitals found within an atom.

Sp2 Hybridization.txt

In this case, it was the Boron atom.

Sp2 Hybridization.txt

When these three different atomic orbitals combined, they form identical hybrid orbitals.

Sp2 Hybridization.txt

Remember, the amount of orbitals that go into the combination must equal to the amount of hybrid orbitals we form.

Sp2 Hybridization.txt

So if we include three if we input three we you must produce three orbitals.

Polyprotic Acids .txt

Now, as of now, we've only really spoken about monoprotic acids.

Polyprotic Acids .txt

Now, monoprotic acid is an acid that can donate a single H plus ion.

Polyprotic Acids .txt

Let's look at a few examples.

Polyprotic Acids .txt

Hydrochloric acid, acetic acid, nitric acid, hydrobromic acid, hydrochloric acid, chloro acid, and perchloric acid are all examples of monopolic acids because they can all donate a single H plus ion.

Polyprotic Acids .txt

Now let's look at the ionization of these acids in water.

Polyprotic Acids .txt

Let's choose the hypothetical monopolic acid.

Polyprotic Acids .txt

So let's choose Ha to be our acid.

Polyprotic Acids .txt

And Ha will react in water to produce hydronium and a conjugate base.

Polyprotic Acids .txt

Now, we've spoken about something called Ka, or the acid ionization concept.

Polyprotic Acids .txt

If you don't know what K is, check out the link above.

Polyprotic Acids .txt

But Ka is basically the product of the concentrations of the product.