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Introduction to Gas State .txt | And that leads straight to the third point. Gas molecules exert a force on whatever they hit. And this force can be calculated in terms of pressure. So let's go back to this example, right? So when I take my ball and I push this ball, I can only push it to a certain extent at some point. I can't push it any further unless I exert a stronger force. |
Introduction to Gas State .txt | So let's go back to this example, right? So when I take my ball and I push this ball, I can only push it to a certain extent at some point. I can't push it any further unless I exert a stronger force. Well, why is that? Well, the reason for that is because all these molecules are moving at tremendous velocities and every time they hit the container, hit the walls, they exert a force or a pressure. Pressure is simply force per some given area. |
Introduction to Gas State .txt | Well, why is that? Well, the reason for that is because all these molecules are moving at tremendous velocities and every time they hit the container, hit the walls, they exert a force or a pressure. Pressure is simply force per some given area. And that means the added effect of all these molecules hitting the wall will be tremendous. And that's exactly why I can't push this ball any further than, say this, right? So whenever we talk about force of molecules, force of gas molecules, we really want to talk about the pressure. |
Introduction to Gas State .txt | And that means the added effect of all these molecules hitting the wall will be tremendous. And that's exactly why I can't push this ball any further than, say this, right? So whenever we talk about force of molecules, force of gas molecules, we really want to talk about the pressure. And so when we talk about things like vapor pressure of a gas, vapor pressure is exactly this. It's the force that these molecules exert on the walls of the container at equilibrium that's vapor pressure. So let's jump into number or part D.
So gas molecules expand into container. |
Introduction to Gas State .txt | And so when we talk about things like vapor pressure of a gas, vapor pressure is exactly this. It's the force that these molecules exert on the walls of the container at equilibrium that's vapor pressure. So let's jump into number or part D.
So gas molecules expand into container. And that's because whenever I have this container and does this container out of air, if I open this up, what will happen to gas molecules? Well, remember, gas molecules are moving at high speeds. So if I open something up, they will escape. |
Introduction to Gas State .txt | And that's because whenever I have this container and does this container out of air, if I open this up, what will happen to gas molecules? Well, remember, gas molecules are moving at high speeds. So if I open something up, they will escape. And also, because they're moving at high speeds, they don't feel intermocular attractions, so there's nothing stopping them from exiting into the container, so they can completely expand the container. And that means, on a similar note, different gases will mix completely because there is no inter molecular interactions between any two gas molecules. Now, compare this to liquids and solids, right? |
Introduction to Gas State .txt | And also, because they're moving at high speeds, they don't feel intermocular attractions, so there's nothing stopping them from exiting into the container, so they can completely expand the container. And that means, on a similar note, different gases will mix completely because there is no inter molecular interactions between any two gas molecules. Now, compare this to liquids and solids, right? If I take the same container of liquid and I spill it into the room, what will happen? Well, they won't escape. They will just create a small puddle and they will stay there, right? |
Introduction to Gas State .txt | If I take the same container of liquid and I spill it into the room, what will happen? Well, they won't escape. They will just create a small puddle and they will stay there, right? That's because they do experience intermolecular attractions and they don't have high velocities or high translational speeds. And that means these intermolecular attractions and low velocities will keep the molecules together. Now, of course, some of these guys will have enough kinetic energy to escape this environment, but that's a different idea. |
Introduction to Gas State .txt | That's because they do experience intermolecular attractions and they don't have high velocities or high translational speeds. And that means these intermolecular attractions and low velocities will keep the molecules together. Now, of course, some of these guys will have enough kinetic energy to escape this environment, but that's a different idea. That's called evaporation. And when you evaporate from a liquid to a gas, the gas gains that high velocity escapes. And now it no longer experiences any intermolecular attractions. |
Introduction to Gas State .txt | That's called evaporation. And when you evaporate from a liquid to a gas, the gas gains that high velocity escapes. And now it no longer experiences any intermolecular attractions. And the final thing we want to look at is the following. Now, we normally describe gas as using a few things temperature. And later we'll see how temperature and kinetic energy speed of our molecules are related. |
Osmotic Pressure Example .txt | So before we begin, I want to illustrate exactly, exactly what's meant by asthmaic pressure. So let's look at this system. Suppose we have two sides, one side and a second side. And the two sides are blocked off by a membrane. This membrane allows one to pass through, but does not allow our solid molecules to pass through. Now this side simply contains 20 solvent. |
Osmotic Pressure Example .txt | And the two sides are blocked off by a membrane. This membrane allows one to pass through, but does not allow our solid molecules to pass through. Now this side simply contains 20 solvent. This side contains a solution of 20 ML that contains 3 grams of our polymer. So the red dots are our polymer dots. Now what will happen? |
Osmotic Pressure Example .txt | This side contains a solution of 20 ML that contains 3 grams of our polymer. So the red dots are our polymer dots. Now what will happen? Well, since the concentration of our solute is greater in this side, then water will tend to move from this side to this side, right? So osmotic pressure basically says well, I need to apply this much pressure on this side of the membrane to stop the movement of that water, to stop osmosis from occurring. So that's what is meant by asthmatic pressure. |
Osmotic Pressure Example .txt | Well, since the concentration of our solute is greater in this side, then water will tend to move from this side to this side, right? So osmotic pressure basically says well, I need to apply this much pressure on this side of the membrane to stop the movement of that water, to stop osmosis from occurring. So that's what is meant by asthmatic pressure. So now we basically want to use all the given info and use our formula for asthmatic pressure and find our molar mass of polymer. Now, first step is let's write down our formula and let's see which ones we have and which ones we don't. So I we have I just simply won. |
Osmotic Pressure Example .txt | So now we basically want to use all the given info and use our formula for asthmatic pressure and find our molar mass of polymer. Now, first step is let's write down our formula and let's see which ones we have and which ones we don't. So I we have I just simply won. And that's because our polymer doesn't associate into anything. So I is one. Our key in Kelvin is 25 plus 273 gives us 298 Kelvin. |
Osmotic Pressure Example .txt | And that's because our polymer doesn't associate into anything. So I is one. Our key in Kelvin is 25 plus 273 gives us 298 Kelvin. Our R is a constant. So it's zero point 81 litres times ATM over moles times Kelvin. And our molarity, well, we don't have our molarity, but we do have our asthma pressure. |
Osmotic Pressure Example .txt | Our R is a constant. So it's zero point 81 litres times ATM over moles times Kelvin. And our molarity, well, we don't have our molarity, but we do have our asthma pressure. So this is what we want to find. And we want to find this guy and use that to find our grams per mole or molar mass. Because remember, molar mass is the amount of mass per 1 mol. |
Osmotic Pressure Example .txt | So this is what we want to find. And we want to find this guy and use that to find our grams per mole or molar mass. Because remember, molar mass is the amount of mass per 1 mol. So let's plug these guys in into this equation and we got this by rearranging this, by bringing all these guys on this side. So we got that. So we plug in our atmosphere pressure zero point 13 ATMs over our constant times, our temperature times I or one. |
Osmotic Pressure Example .txt | So let's plug these guys in into this equation and we got this by rearranging this, by bringing all these guys on this side. So we got that. So we plug in our atmosphere pressure zero point 13 ATMs over our constant times, our temperature times I or one. So ATMs cancel, the Kelvins cancel and we get 5.31 times ten to negative four moles per liter. So this is our molarity. Now let's find the number of moles of polymer. |
Osmotic Pressure Example .txt | So ATMs cancel, the Kelvins cancel and we get 5.31 times ten to negative four moles per liter. So this is our molarity. Now let's find the number of moles of polymer. If we find the number of moles of polymer, we could take our amount in grams, divide that by our moles and we get mol and mass or mass per mole. So 5.31 times ten to negative four molar which we got from here, times the amount of solution we have, which is 20 ML or 0.2 liters because we have to convert milliliters to liters. We divide this by thousand and we get to zero two the liters cancel and we get 1.6 times ten to negative four moles of polymer. |
Raoul’s Law for Non-Ideal Fluids .txt | Let's remember what a non ideal fluid is. A non ideal fluid is a fluid in which the volume of molecules is not neglected and intermolecular forces connecting the molecules exist. And these forces will play a role in decreasing the vapor repression of the fluid. And we'll see that in a second. Now, before we talk about that, let's talk about the heat of solution. Now, we have already spoken about heat of solution in another video, so let's briefly discuss it. |
Raoul’s Law for Non-Ideal Fluids .txt | And we'll see that in a second. Now, before we talk about that, let's talk about the heat of solution. Now, we have already spoken about heat of solution in another video, so let's briefly discuss it. So the heat of solution is the sum of energy required to break the intermolecular forces of compound one. Plus the sum of the energy is required to break the intermolecular forces of compound two plus the energy release when you form the new solution. And that gives you the enthalpy of solution or heat of solution. |
Raoul’s Law for Non-Ideal Fluids .txt | So the heat of solution is the sum of energy required to break the intermolecular forces of compound one. Plus the sum of the energy is required to break the intermolecular forces of compound two plus the energy release when you form the new solution. And that gives you the enthalpy of solution or heat of solution. Now, when this guy is negative, when he's exothermic the bonds form are stronger than the bonds broken, so they're more stable. Now, when it's endothermic, when it's positive, the bonds form are weaker and less stable than the bonds that were broken. Now let's compare ideal and non ideal situations. |
Raoul’s Law for Non-Ideal Fluids .txt | Now, when this guy is negative, when he's exothermic the bonds form are stronger than the bonds broken, so they're more stable. Now, when it's endothermic, when it's positive, the bonds form are weaker and less stable than the bonds that were broken. Now let's compare ideal and non ideal situations. Now, when we talk about ideal situations, we can graph Roth's law. And the Y axis is vapor pressure and the x axis is mole fraction. Now, we have two compounds here, compound A and compound B. |
Raoul’s Law for Non-Ideal Fluids .txt | Now, when we talk about ideal situations, we can graph Roth's law. And the Y axis is vapor pressure and the x axis is mole fraction. Now, we have two compounds here, compound A and compound B. Now for compound A when we go from here to here along the x axis the mole fraction increases. For compound B, when we go from here to here on the x axis the mole fraction decreases. The brown line represents the pressure. |
Raoul’s Law for Non-Ideal Fluids .txt | Now for compound A when we go from here to here along the x axis the mole fraction increases. For compound B, when we go from here to here on the x axis the mole fraction decreases. The brown line represents the pressure. It's the slope for compound A and the green line represents the pressure of the slope for compound B. Now, as we go as we increase the mole concentration in compound one or compound A, the pressure of it increases. And as we go this way, the mole fraction of this guy compound B decreases. |
Raoul’s Law for Non-Ideal Fluids .txt | It's the slope for compound A and the green line represents the pressure of the slope for compound B. Now, as we go as we increase the mole concentration in compound one or compound A, the pressure of it increases. And as we go this way, the mole fraction of this guy compound B decreases. So as we go this way, the vapor pressure of compound B decreases. Now, at any given time, we could find a final vapor pressure and we find the final vapor pressure by summing the two pressures. So for example, say at this point the total pressure is this guy plus this guy gives you this plus this. |
Raoul’s Law for Non-Ideal Fluids .txt | So as we go this way, the vapor pressure of compound B decreases. Now, at any given time, we could find a final vapor pressure and we find the final vapor pressure by summing the two pressures. So for example, say at this point the total pressure is this guy plus this guy gives you this plus this. So it's somewhere over here and that's what the red line is. Now this is for an ideal fluid. Now, in an ideal fluid the surface molecules are not connected by any intermolecular forces because remember, in an ideal fluid we're neglecting those intermolecular forces. |
Raoul’s Law for Non-Ideal Fluids .txt | So it's somewhere over here and that's what the red line is. Now this is for an ideal fluid. Now, in an ideal fluid the surface molecules are not connected by any intermolecular forces because remember, in an ideal fluid we're neglecting those intermolecular forces. So nothing holds the molecules together so they could freely escape into the environment if they have the kinetic energy. However, in a non ideal fluid there are intermolecular forces and these forces hold the molecules together. They inhibit the molecules from escaping into the environment. |
Raoul’s Law for Non-Ideal Fluids .txt | So nothing holds the molecules together so they could freely escape into the environment if they have the kinetic energy. However, in a non ideal fluid there are intermolecular forces and these forces hold the molecules together. They inhibit the molecules from escaping into the environment. So at any given time, less molecules will be able to escape in a non ideal fluid than in an ideal fluid. And because there are less molecules present in the states above the pressure will be less. Therefore, the stronger the intermolecular forces, the less likely that they will evaporate. |
Raoul’s Law for Non-Ideal Fluids .txt | So at any given time, less molecules will be able to escape in a non ideal fluid than in an ideal fluid. And because there are less molecules present in the states above the pressure will be less. Therefore, the stronger the intermolecular forces, the less likely that they will evaporate. Likewise, the weaker the intermolecular forces holding the molecules together, the more likely that they will evaporate. Now, let's look at some graphs for non ideal fluids. Now in this graph we're going to compare two situations. |
Raoul’s Law for Non-Ideal Fluids .txt | Likewise, the weaker the intermolecular forces holding the molecules together, the more likely that they will evaporate. Now, let's look at some graphs for non ideal fluids. Now in this graph we're going to compare two situations. We're going to compare positive or endothermic cube solution and negative or exothermicial solution. We're also going to compare non ideal situations we're presented to by dashed lines and ideal situations they're presented by solid lines. Now, in this graph, the y axis is vapor pressure and the x axis is the mole fraction. |
Raoul’s Law for Non-Ideal Fluids .txt | We're going to compare positive or endothermic cube solution and negative or exothermicial solution. We're also going to compare non ideal situations we're presented to by dashed lines and ideal situations they're presented by solid lines. Now, in this graph, the y axis is vapor pressure and the x axis is the mole fraction. We're dealing with two compounds, compounds A, compound B. Compound B is the green line. Compound A is the brown line. |
Raoul’s Law for Non-Ideal Fluids .txt | We're dealing with two compounds, compounds A, compound B. Compound B is the green line. Compound A is the brown line. Now, in an endothermic reaction the bonds form are weaker than the bonds broken. And that means the bonds formed are going to be less likely to hold the molecules on the surface. That means more molecules will evaporate. |
Raoul’s Law for Non-Ideal Fluids .txt | Now, in an endothermic reaction the bonds form are weaker than the bonds broken. And that means the bonds formed are going to be less likely to hold the molecules on the surface. That means more molecules will evaporate. And if more molecules evaporate, more molecules will be present in the space above, more gas molecules. And if more gas molecules are present in the space above, the vapor pressure is higher. Therefore, compared to ideal situations for each case in a non ideal situation that's endothermic the pressure for each will be higher and that's why they curve upward. |
Raoul’s Law for Non-Ideal Fluids .txt | And if more molecules evaporate, more molecules will be present in the space above, more gas molecules. And if more gas molecules are present in the space above, the vapor pressure is higher. Therefore, compared to ideal situations for each case in a non ideal situation that's endothermic the pressure for each will be higher and that's why they curve upward. Now, in an exothermic, the opposite holds. In an exothermic reaction, the bonds formed are stronger than the bonds broken. Therefore, the final solution will hold its molecules very tightly and will not let them go. |
Raoul’s Law for Non-Ideal Fluids .txt | Now, in an exothermic, the opposite holds. In an exothermic reaction, the bonds formed are stronger than the bonds broken. Therefore, the final solution will hold its molecules very tightly and will not let them go. They will not be able to escape, will be less likely to go into the gas state. And that's why less gas molecules will be present in the space above and the deeper pressure will be lower. And that's why in each situation they're curved downward. |
Gibbs Free Energy Derivation .txt | Firstly, whenever we talk about gives free energy, we talk about closed systems or reactions occurring under closed systems. So what is a closed system? A closed system simply means that no mass is exchanged, only energy is exchanged. And since mass is matter, and matter are molecules, that means no molecules go into the system and no molecules lead the system. So the number of moles remains constant. Secondly, reactions are under constant temperature and pressure. |
Gibbs Free Energy Derivation .txt | And since mass is matter, and matter are molecules, that means no molecules go into the system and no molecules lead the system. So the number of moles remains constant. Secondly, reactions are under constant temperature and pressure. If this wasn't true, the equation would break down. It would not work. So according to the ideal gas law, when number of moles is held constant, temperature is held constant, and pressure is held constant, then volume must remain constant. |
Gibbs Free Energy Derivation .txt | If this wasn't true, the equation would break down. It would not work. So according to the ideal gas law, when number of moles is held constant, temperature is held constant, and pressure is held constant, then volume must remain constant. If volume is constant, then change in volume is zero, because the final and the initial are the same. So the TV work done is zero. And that means that we could approximate change in enthalpy. |
Gibbs Free Energy Derivation .txt | If volume is constant, then change in volume is zero, because the final and the initial are the same. So the TV work done is zero. And that means that we could approximate change in enthalpy. And simply change in enthalpy of the system is equal to change in internal energy or simply change in energy. The PV term disappears because it's zero. Now thirdly, reactions are reversible, which means that if this wasn't true, the equation would break down as well. |
Gibbs Free Energy Derivation .txt | And simply change in enthalpy of the system is equal to change in internal energy or simply change in energy. The PV term disappears because it's zero. Now thirdly, reactions are reversible, which means that if this wasn't true, the equation would break down as well. Now, since reactions are reversible and enthalpy is a state function, hess's law tells us the change in enthalpy of the four reaction is equal to the negative change in enthalpy of the reverse reaction. So if we look at this reversible reaction here, a plus B is equal to C plus D, where the change in enthalpy is negative ten. And the reverse reaction, according to Hess's Law, is C plus D equal A plus B, and this is negative of this. |
Gibbs Free Energy Derivation .txt | Now, since reactions are reversible and enthalpy is a state function, hess's law tells us the change in enthalpy of the four reaction is equal to the negative change in enthalpy of the reverse reaction. So if we look at this reversible reaction here, a plus B is equal to C plus D, where the change in enthalpy is negative ten. And the reverse reaction, according to Hess's Law, is C plus D equal A plus B, and this is negative of this. So it's positive ten kilojoules per mole. The last thing we should mention is the mathematical formula for entropy, which basically states that change in entropy is the heat or change in energy over time, over temperature, and temperature is in Kelvin. Step one. |
Gibbs Free Energy Derivation .txt | So it's positive ten kilojoules per mole. The last thing we should mention is the mathematical formula for entropy, which basically states that change in entropy is the heat or change in energy over time, over temperature, and temperature is in Kelvin. Step one. In step one, I apply the formula for a change in entropy. That is, change in entropy surroundings is equal to change in energy over temperature, as seen here. Step two remember when we talk about free energy, or gifts free energy, we talk about a closed system, constant pressure and constant temperature. |
Gibbs Free Energy Derivation .txt | In step one, I apply the formula for a change in entropy. That is, change in entropy surroundings is equal to change in energy over temperature, as seen here. Step two remember when we talk about free energy, or gifts free energy, we talk about a closed system, constant pressure and constant temperature. And what this basically means, according to the ideal gas law, is that volume is constant. So change in volume is zero. So the PV work done is zero. |
Gibbs Free Energy Derivation .txt | And what this basically means, according to the ideal gas law, is that volume is constant. So change in volume is zero. So the PV work done is zero. So when we look at the equation for change in enthalpy, the PV work term disappears. And that means that our equation becomes change in enthalpy. The surroundings is equal to change in energy or heat. |
Gibbs Free Energy Derivation .txt | So when we look at the equation for change in enthalpy, the PV work term disappears. And that means that our equation becomes change in enthalpy. The surroundings is equal to change in energy or heat. So this becomes this. Now, step three remember when we talk about Gibbs free energy, we talk about a reversible reaction. And because enthalpy is a state function, according to Hess's law, hess's law states that the entropy change of the forward reaction is equal to the negative entropy change of the reverse reaction. |
Gibbs Free Energy Derivation .txt | So this becomes this. Now, step three remember when we talk about Gibbs free energy, we talk about a reversible reaction. And because enthalpy is a state function, according to Hess's law, hess's law states that the entropy change of the forward reaction is equal to the negative entropy change of the reverse reaction. So the amount of energy that leaves the system is the same amount of energy, except negative that into the system. Okay? So the magnitude remains the same, but the signs change. |
Gibbs Free Energy Derivation .txt | So the amount of energy that leaves the system is the same amount of energy, except negative that into the system. Okay? So the magnitude remains the same, but the signs change. Therefore, the change in entropy of the surroundings is equal to negative change in entropy of the system. Step four. Remember, change in entropy of the universe is equal to change in entropy the surroundings plus change in entropy of the system. |
Gibbs Free Energy Derivation .txt | Therefore, the change in entropy of the surroundings is equal to negative change in entropy of the system. Step four. Remember, change in entropy of the universe is equal to change in entropy the surroundings plus change in entropy of the system. Now, this guy equals this guy. Our next step will be to see that this guy can be replaced by this guy, okay? And that's exactly what we do in step five. |
Gibbs Free Energy Derivation .txt | Now, this guy equals this guy. Our next step will be to see that this guy can be replaced by this guy, okay? And that's exactly what we do in step five. We simply take this and plug it into here. And that's what we get. Here. |
Gibbs Free Energy Derivation .txt | We simply take this and plug it into here. And that's what we get. Here. The next step we multiply out by negative t.
And that's because we want to get rid of this t, okay? So let's do that. Negative t times this gives you this negative t times this gives you that negative t times a negative whatever gives us a positive. |
Gibbs Free Energy Derivation .txt | The next step we multiply out by negative t.
And that's because we want to get rid of this t, okay? So let's do that. Negative t times this gives you this negative t times this gives you that negative t times a negative whatever gives us a positive. So plus this whole term, step seven, is basically the TS cancel out. So we get this whole thing minus these t's. Next step, what I do is I switch these guys to make it look more like the actual equation. |
Gibbs Free Energy Derivation .txt | So plus this whole term, step seven, is basically the TS cancel out. So we get this whole thing minus these t's. Next step, what I do is I switch these guys to make it look more like the actual equation. And now I equate this to changing Gibbs energy. And now I get our final equation change. In Gibbs, the energy is equal to change in entropy of the system minus temperature times change in entropy of the system. |
Carbon and Nitrogen Groups .txt | In this lecture we're going to look at and compare two more important groups found in our periodic table. We're going to look at group 14 or four A and group 15 and five A elements. Now let's begin with the group 14 or four A elements. Now this group consists of five elements or at least five elements. And the important elements are are listed below. We have carbon, which is a nonmetal. |
Carbon and Nitrogen Groups .txt | Now this group consists of five elements or at least five elements. And the important elements are are listed below. We have carbon, which is a nonmetal. We have silicon, which is a metalloid. We have germanium, which is also a metalloid. And we have two metals, tin and lead. |
Carbon and Nitrogen Groups .txt | We have silicon, which is a metalloid. We have germanium, which is also a metalloid. And we have two metals, tin and lead. Now notice that in our group we have at least one of each. We have at least one nonmetal, at least one metalloid and at least one metal. Now every single element within this group can form four Covalent bonds with other nonmetals. |
Carbon and Nitrogen Groups .txt | Now notice that in our group we have at least one of each. We have at least one nonmetal, at least one metalloid and at least one metal. Now every single element within this group can form four Covalent bonds with other nonmetals. And every atom except the sea atom, except our carbon can form two more additional bonds with lowest bases. Now remember what a lewis bases. It's simply an atom or a molecule that has an extra lone pair of electrons that it can donate to some other lewis acid. |
Carbon and Nitrogen Groups .txt | And every atom except the sea atom, except our carbon can form two more additional bonds with lowest bases. Now remember what a lewis bases. It's simply an atom or a molecule that has an extra lone pair of electrons that it can donate to some other lewis acid. Now let's look at D. Now, only carbon is capable of forming strong pi bonds. In other words, it can form a strong double bond or a strong triple bond. No other atom found in our group 14 or four A has that capability. |
Carbon and Nitrogen Groups .txt | Now let's look at D. Now, only carbon is capable of forming strong pi bonds. In other words, it can form a strong double bond or a strong triple bond. No other atom found in our group 14 or four A has that capability. Now let's jump to group 15. In group 15 or five A, we have also at least five atoms. These are the important atoms listed. |
Carbon and Nitrogen Groups .txt | Now let's jump to group 15. In group 15 or five A, we have also at least five atoms. These are the important atoms listed. So we have nitrogen and phosphorus, which is a nonmetal or both nonmetals. We have arsenic and antimony, which are both metalloids, and bismuth, which is a metal. Now once again, just like group 14 or four A, we have at least one of each in our group 15 or five day. |
Carbon and Nitrogen Groups .txt | So we have nitrogen and phosphorus, which is a nonmetal or both nonmetals. We have arsenic and antimony, which are both metalloids, and bismuth, which is a metal. Now once again, just like group 14 or four A, we have at least one of each in our group 15 or five day. Now, unlike this group which forms four Covalent bonds with other nonmetals, group 15 or five day forms three Covalent bonds and in addition, every single atom, except that nitrogen can form two more bonds using their higher D orbitals. Now we're going to talk more about D orbitals and P orbitals and S orbitals in another lecture. Let's look at C. Now nitrogen, this atom can form pi bonds just like carbon can form pi bonds in group 14. |
Carbon and Nitrogen Groups .txt | Now, unlike this group which forms four Covalent bonds with other nonmetals, group 15 or five day forms three Covalent bonds and in addition, every single atom, except that nitrogen can form two more bonds using their higher D orbitals. Now we're going to talk more about D orbitals and P orbitals and S orbitals in another lecture. Let's look at C. Now nitrogen, this atom can form pi bonds just like carbon can form pi bonds in group 14. And these pi bonds can be triple bonds or double bonds. And they're relatively strong. But unlike this group, we have one more atom here, our phosphorus, that can form a double bond, but it's a weak double bond. |
Carbon and Nitrogen Groups .txt | And these pi bonds can be triple bonds or double bonds. And they're relatively strong. But unlike this group, we have one more atom here, our phosphorus, that can form a double bond, but it's a weak double bond. But regardless, it's still a double bond, a pi bond. Now let's look at one more thing. Let's look at our N. Now normally when we have nitrogen in a molecule or compound that nitrogen has a lone pair of electrons. |
Carbon and Nitrogen Groups .txt | But regardless, it's still a double bond, a pi bond. Now let's look at one more thing. Let's look at our N. Now normally when we have nitrogen in a molecule or compound that nitrogen has a lone pair of electrons. For example, in ammonia. And ammonia can take one more H because it has a lone pair of electrons. In other words, our nitrogen has the capability of forming not three Covalent bonds, but four Covalent bonds. |
Henry’s Law Example .txt | Now, as an example, let's take a section of our capillary found within our lung. And let's assume this section contains exactly one liter of blood or one liter of the red molecules. Now, a capillary is simply a very, very small vessel that carries blood. And within a capillary, oxygen and nitrogen can be exchanged with the environment. And we're using lungs because gases are exchanged within our lungs. So we want to find a number of blue molecules and a number of green molecules in terms of grounds found within one liter of this blood. |
Henry’s Law Example .txt | And within a capillary, oxygen and nitrogen can be exchanged with the environment. And we're using lungs because gases are exchanged within our lungs. So we want to find a number of blue molecules and a number of green molecules in terms of grounds found within one liter of this blood. So the first step is to realize that air is composed approximately 79% nitrogen and approximately 21% oxygen. Now, this translates to zero 79 mole fraction nitrogen and 0.2 1 mol fraction of oxygen. So the first step is to use rolled slow. |
Henry’s Law Example .txt | So the first step is to realize that air is composed approximately 79% nitrogen and approximately 21% oxygen. Now, this translates to zero 79 mole fraction nitrogen and 0.2 1 mol fraction of oxygen. So the first step is to use rolled slow. But before we use a roll slow, we must assume that there is dynamic equilibriums between the gas molecules in the space above and the gas molecules dissolved within our blood. So let's make that assumption. Now, we can find the partial pressures of each gas by simply using the formula. |
Henry’s Law Example .txt | But before we use a roll slow, we must assume that there is dynamic equilibriums between the gas molecules in the space above and the gas molecules dissolved within our blood. So let's make that assumption. Now, we can find the partial pressures of each gas by simply using the formula. Now, before we use the formula, let's let's realize why we're using 760 mmhg. So the total pressure above our system is the air that we're breathing in is the pressure of the air that we're breathing in. And the air we're breathing in is at atmospheric pressure. |
Henry’s Law Example .txt | Now, before we use the formula, let's let's realize why we're using 760 mmhg. So the total pressure above our system is the air that we're breathing in is the pressure of the air that we're breathing in. And the air we're breathing in is at atmospheric pressure. And atmospheric pressure is one ATM or 760 mercury. So defined partial pressure of oxygen, we simply multiply mole fraction by our total pressure and we get 159.6 mmhg for oxygen and 600.4
mmhg for nitrogen. So second step is to find the Molarity. |
Henry’s Law Example .txt | And atmospheric pressure is one ATM or 760 mercury. So defined partial pressure of oxygen, we simply multiply mole fraction by our total pressure and we get 159.6 mmhg for oxygen and 600.4
mmhg for nitrogen. So second step is to find the Molarity. We can use these guys, the partial pressures to find the Molarity. Now, Henry's loss states. So we can find the Molarity by simply taking our constant and multiplying it by the partial pressure of the gas. |
Henry’s Law Example .txt | We can use these guys, the partial pressures to find the Molarity. Now, Henry's loss states. So we can find the Molarity by simply taking our constant and multiplying it by the partial pressure of the gas. Therefore, we get 1.66
times ten to negative six times 1.59.6 and we get 2.64 times ten to negative four molar for two or moles per liter. Now, we do the same thing, except we change the partial pressure for nitrogen and the concept of nitrogen and we get 5.55 times ten to negative four molar or moles per liter. The third step is to find the moles of oxygen and the moles of nitrogen dissolved within our blood. |
Henry’s Law Example .txt | Therefore, we get 1.66
times ten to negative six times 1.59.6 and we get 2.64 times ten to negative four molar for two or moles per liter. Now, we do the same thing, except we change the partial pressure for nitrogen and the concept of nitrogen and we get 5.55 times ten to negative four molar or moles per liter. The third step is to find the moles of oxygen and the moles of nitrogen dissolved within our blood. To find the moles of oxygen, we simply take our Molar concentration, multiply that by one liter of total solution and we get 0.000,264 moles of O two and 0.0055 moles for M two. Finally, we can use the moles, multiplying that by the molecular weight of each respective compound. That will give us the number in grams of each compound resolved within our system. |
Henry’s Law Example .txt | To find the moles of oxygen, we simply take our Molar concentration, multiply that by one liter of total solution and we get 0.000,264 moles of O two and 0.0055 moles for M two. Finally, we can use the moles, multiplying that by the molecular weight of each respective compound. That will give us the number in grams of each compound resolved within our system. So molecular weight of oxygen, which is 32 grams/mol multiplied by the moles, which is 0.00,134
moles, gets us oh, I'm sorry. I made a mistake here. This should be 0.00,264
moles. |
Henry’s Law Example .txt | So molecular weight of oxygen, which is 32 grams/mol multiplied by the moles, which is 0.00,134
moles, gets us oh, I'm sorry. I made a mistake here. This should be 0.00,264
moles. So it's 32 centrist is right. 32 times .0,006,264. And that is correct. |
Mole Fraction Example .txt | Molality is simply another way of measuring the concentration of a solution. The symbol for molality is lowercase letter M and the formula is moles of solute over kilograms of solvent. Now let's do an example. Using molality. The question tells us that we have 90 grams of glucose and 200 grams of H 20. And we want to find the molality of the solution. |
Mole Fraction Example .txt | Using molality. The question tells us that we have 90 grams of glucose and 200 grams of H 20. And we want to find the molality of the solution. We're basically taking glucose, the salute and water, the solvent. We're mixing it and we want to find the concentration. In terms of molality, the first step is to find the molecular weight of glucose. |
Mole Fraction Example .txt | We're basically taking glucose, the salute and water, the solvent. We're mixing it and we want to find the concentration. In terms of molality, the first step is to find the molecular weight of glucose. This will help us find the number of moles of glucose within our solution. The first step is to find the atomic weight of each atom and multiplied by each subscript. So the atomic weight of carbon is 12 grams/mol, the atomic weight of H is 1 gram/mol, and the atomic weight of oxygen is 16 grams/mol. |
Mole Fraction Example .txt | This will help us find the number of moles of glucose within our solution. The first step is to find the atomic weight of each atom and multiplied by each subscript. So the atomic weight of carbon is 12 grams/mol, the atomic weight of H is 1 gram/mol, and the atomic weight of oxygen is 16 grams/mol. So we multiply six by twelve, add that to twelve times one and add that to six times 16 and we get approximately 180 grams/mol. Now, using this number, we can take our 90 grams of glucose and find the number of moles. 90 grams of glucose divided by 180 grams of glucose per mole gets us 0.5
or one half moles of glucose. |
Mole Fraction Example .txt | So we multiply six by twelve, add that to twelve times one and add that to six times 16 and we get approximately 180 grams/mol. Now, using this number, we can take our 90 grams of glucose and find the number of moles. 90 grams of glucose divided by 180 grams of glucose per mole gets us 0.5
or one half moles of glucose. That's our number of moles within this solution. The final step is to use the formula for molality. Lowercase M for molality is equal to one half number of moles of salute divided by now notice we're getting 200 grams and molality ditosomality are moles per kilogram. |
Henderson-Hasselbach Equation .txt | So in this lecture we're going to look at something called the Hamilton Hasselbach equation. Now this equation has two uses. First, we can use it to find the PH of the buffet system. And second, we can use it to find the ratio of conjugate base to conjugate acid or the ratio of conjugate acid to conjugate base of our buffet system. Now let's see where this equation comes from. Let's derive it. |
Henderson-Hasselbach Equation .txt | And second, we can use it to find the ratio of conjugate base to conjugate acid or the ratio of conjugate acid to conjugate base of our buffet system. Now let's see where this equation comes from. Let's derive it. But first, let's look at the reaction of an acid in an aqueous state with the water molecule in the liquid state. Well, what we get is a conjugate base in the acreage state and a conjugate acid in the aqueous states. So let's write the equilibrium constant expression for this reaction. |
Henderson-Hasselbach Equation .txt | But first, let's look at the reaction of an acid in an aqueous state with the water molecule in the liquid state. Well, what we get is a conjugate base in the acreage state and a conjugate acid in the aqueous states. So let's write the equilibrium constant expression for this reaction. So we get the acid ionization constant for this acid is equal to the concentration of hydronium times the concentration of the conjugate base. These two guys divided by the concentration of the conjugate acid for this guy. Now, if you are confused where this expression comes from, check out the link below. |
Henderson-Hasselbach Equation .txt | So we get the acid ionization constant for this acid is equal to the concentration of hydronium times the concentration of the conjugate base. These two guys divided by the concentration of the conjugate acid for this guy. Now, if you are confused where this expression comes from, check out the link below. Now remember, our goal is to find an expression that we can use to solve for the PH. So which one of these guys determines PH? Well, it's this guy. |
Henderson-Hasselbach Equation .txt | Now remember, our goal is to find an expression that we can use to solve for the PH. So which one of these guys determines PH? Well, it's this guy. Remember, PH is equal to negative log of the hydronium concentration. So our goal will be to isolate this guy first. So let's isolate him. |
Henderson-Hasselbach Equation .txt | Remember, PH is equal to negative log of the hydronium concentration. So our goal will be to isolate this guy first. So let's isolate him. We get ka times the concentration of conjugate acid. This guy divided by the concentration of conjugate base for this guy equals the hydronium concentration. Now our next step is to convert these guys into logs because our PH involves logs. |
Henderson-Hasselbach Equation .txt | We get ka times the concentration of conjugate acid. This guy divided by the concentration of conjugate base for this guy equals the hydronium concentration. Now our next step is to convert these guys into logs because our PH involves logs. So we take the logs of both sides and we get log of base ten of the hydronium concentration of this guy equals log of base ten of this whole guy. Now our next goal is to use the laws of logs to convert this guy from this more compact state to a less compact state. Now, what we get is the following log of the hydronium concentration. |
Henderson-Hasselbach Equation .txt | So we take the logs of both sides and we get log of base ten of the hydronium concentration of this guy equals log of base ten of this whole guy. Now our next goal is to use the laws of logs to convert this guy from this more compact state to a less compact state. Now, what we get is the following log of the hydronium concentration. This guy is equal to notice inside this log we're multiplying ka times this whole thing, this ratio. So another way of expressing this is to use the logs of logs and we get log of ka plus log of this guy. So the concentration of the conjugate axis divided by the concentration of conjugate base. |
Henderson-Hasselbach Equation .txt | This guy is equal to notice inside this log we're multiplying ka times this whole thing, this ratio. So another way of expressing this is to use the logs of logs and we get log of ka plus log of this guy. So the concentration of the conjugate axis divided by the concentration of conjugate base. Now our next step is to basically rewrite this guy in the form of the PH formula. We can't say this guy is equal to PH because remember, PH is negative log of this guy. So let's take the negative of the whole expression. |
Henderson-Hasselbach Equation .txt | Now our next step is to basically rewrite this guy in the form of the PH formula. We can't say this guy is equal to PH because remember, PH is negative log of this guy. So let's take the negative of the whole expression. This way we can get our PH. So we multiply two by negative one and we get negative this whole thing. So negative this equals negative this guy minus this guy. |
Henderson-Hasselbach Equation .txt | This way we can get our PH. So we multiply two by negative one and we get negative this whole thing. So negative this equals negative this guy minus this guy. So next step is to convert this guy into PH. And we get PH is equal to negative of log ka. Now notice what happens here whenever we have a negative on the outside of our log. |
Henderson-Hasselbach Equation .txt | So next step is to convert this guy into PH. And we get PH is equal to negative of log ka. Now notice what happens here whenever we have a negative on the outside of our log. That simply means we're taking this guy and raising it to the negative one. This is equivalent to saying plus log of this guy to the negative one. But this guy to the negative one is simply reciprocating these guys. |
Henderson-Hasselbach Equation .txt | That simply means we're taking this guy and raising it to the negative one. This is equivalent to saying plus log of this guy to the negative one. But this guy to the negative one is simply reciprocating these guys. So negative log of this guy is the same as saying positive log of the reciprocal of this guy. So the concentration of conjugate base divided by the concentration of conjugate acid. So Now, A final formula states we find that PH is equal to. |
Henderson-Hasselbach Equation .txt | So negative log of this guy is the same as saying positive log of the reciprocal of this guy. So the concentration of conjugate base divided by the concentration of conjugate acid. So Now, A final formula states we find that PH is equal to. Remember, negative log of Ka is simply PKA. And we get PH is equal to PKA plus log of the ratio of conjugate base over conjugate acid. And this is our Henderson half of block formula. |
Henderson-Hasselbach Equation .txt | Remember, negative log of Ka is simply PKA. And we get PH is equal to PKA plus log of the ratio of conjugate base over conjugate acid. And this is our Henderson half of block formula. Now, if you want to do an example using this formula, check out the link below. But remember, this formula can Be used to find Two things. So, if You Know the PKA and you know our ratio, you can find the PH. |
Lewis Acids and Bases .txt | So let's begin by defining what a lowest acid and lowest base is. Lewis acids are molecules containing an empty, unfilled atomic orbital, and lowest bases are molecules containing a pair of electrons, bonds, or a filled orbital. So let's look at a Lewis acid. So, this is one example of a Lewis acid. It's a methylcation. This methylcation has three SP two hybridized covalent bonds between the C and the HS. |
Lewis Acids and Bases .txt | So, this is one example of a Lewis acid. It's a methylcation. This methylcation has three SP two hybridized covalent bonds between the C and the HS. And it has an empty pur two P orbital given here. So it has a positive charge. So here's an example of a Lewis base. |
Lewis Acids and Bases .txt | And it has an empty pur two P orbital given here. So it has a positive charge. So here's an example of a Lewis base. Once again, a Lewis base is a molecule containing a pair of electrons, a pair of non bonding electrons in an orbital. So hydrant is an example of a Lewis base because it has two electrons, a pair of non bonding electrons within the one S orbital. Now, when the Lewis acid react with the Lewis base, that basically means their atomic orbitals interact. |
Lewis Acids and Bases .txt | Once again, a Lewis base is a molecule containing a pair of electrons, a pair of non bonding electrons in an orbital. So hydrant is an example of a Lewis base because it has two electrons, a pair of non bonding electrons within the one S orbital. Now, when the Lewis acid react with the Lewis base, that basically means their atomic orbitals interact. They overlap, producing a bond. So in this case, when this Lewis acid interacts with this Lewis base, this one S interacts with this two p.
These two electrons are donated to this two P orbital. Now, as these come closer, this load becomes smaller, this becomes larger so that the overlap is better. |
Lewis Acids and Bases .txt | They overlap, producing a bond. So in this case, when this Lewis acid interacts with this Lewis base, this one S interacts with this two p.
These two electrons are donated to this two P orbital. Now, as these come closer, this load becomes smaller, this becomes larger so that the overlap is better. And we form SP three hybridized bonds. So 1234 SP three hybridized bonds. And now this carbon and this H share a pair of electrons. |
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