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Acidic Basic and Neutral Salts .txt | Now let's examine why we have an acidic salt. Remember, we begin with a weak acid that has a low KB value. And that means equilibrium will be far to the left. Our reactants will be favored, or at least our ammonia will be favored. So that means this weak base has a strong or relatively strong conjugate acid. So what happens when our conjugate acid is formed? |
Acidic Basic and Neutral Salts .txt | Our reactants will be favored, or at least our ammonia will be favored. So that means this weak base has a strong or relatively strong conjugate acid. So what happens when our conjugate acid is formed? What happens when Ammonium is formed, when this guy combines with this guy? Well, we get ammonia and Ammonium will not want to exist in this state because it's a relatively strong conjugate acid. So it will want to dissociate back into our ammonia. |
Acidic Basic and Neutral Salts .txt | What happens when Ammonium is formed, when this guy combines with this guy? Well, we get ammonia and Ammonium will not want to exist in this state because it's a relatively strong conjugate acid. So it will want to dissociate back into our ammonia. But now it dissociates back into ammonia in the presence of water. And when water is in the mixture, what happens? Well, this guy is an acid, so this guy must be a base. |
Acidic Basic and Neutral Salts .txt | But now it dissociates back into ammonia in the presence of water. And when water is in the mixture, what happens? Well, this guy is an acid, so this guy must be a base. And our base will accept the H ion forming back our ammonia and creating hydronium. So now, equilibrium, we're going to have more hydronium. And that means our acidity of our solution will increase. |
Acidic Basic and Neutral Salts .txt | And our base will accept the H ion forming back our ammonia and creating hydronium. So now, equilibrium, we're going to have more hydronium. And that means our acidity of our solution will increase. So our salt will be acidic. Therefore, we see that conjugate acid determines the PH of our solution. And the higher the K. A, the stronger our conjugate acid and the more acidic our salt or the more acidic our solution. |
Acidic Basic and Neutral Salts .txt | So our salt will be acidic. Therefore, we see that conjugate acid determines the PH of our solution. And the higher the K. A, the stronger our conjugate acid and the more acidic our salt or the more acidic our solution. Now, that means whenever we mix strong acids and weak bases, we create acidic salt. So let's look at what types of salts are created when weak bases and weak acids are reactive. So there is actually a competition between the conjugate acid and the conjugate base. |
Intro to Quantum Mechanics .txt | Now, in this lecture, I'm going to give a very, very brief introduction to quantum mechanics. The reason this branch of physics is important in chemistry is it becomes important when we talk about quantum numbers and the photoelectric effect. Now, quantum mechanics is the study of microscopic phenomena such as electromagnetic magnetic waves and the study of subatomic particles such as electrons, protons, neutrons, et cetera. Now, the main idea, the main conclusion that we get from quantum mechanics is the following tiny subatomic particles called electrons and protons and neutrons and other particles. Now, subatomic simply means smaller than an atom gain or lose energy in discrete amount called photons. Now, we're going to talk more about photons and what they are when we'll talk about the photoelectric effect. |
Intro to Quantum Mechanics .txt | Now, the main idea, the main conclusion that we get from quantum mechanics is the following tiny subatomic particles called electrons and protons and neutrons and other particles. Now, subatomic simply means smaller than an atom gain or lose energy in discrete amount called photons. Now, we're going to talk more about photons and what they are when we'll talk about the photoelectric effect. Now, this statement is analogous to the following everyday situation. Suppose we have a vending machine. And this vending machine only accepts coins that are $0.25. |
Intro to Quantum Mechanics .txt | Now, this statement is analogous to the following everyday situation. Suppose we have a vending machine. And this vending machine only accepts coins that are $0.25. So it won't accept two coins. It won't accept five cent coins. It won't accept dollar bills. |
Intro to Quantum Mechanics .txt | So it won't accept two coins. It won't accept five cent coins. It won't accept dollar bills. It only accepts a single coin known as the quarter. Now, in the same analogous way, electrons and protons and neutrons and other subatomic particles accept energy bursts this feet amount called a photon. Now, one other result from quantum mechanics is the following. |
Average Kinetic Energy and Root Mean Square Velocity .txt | So we already spoke about the root mean square velocity of any set of points. Now, we're going to look specifically at the root mean square velocity for gases. Now, in gases, root mean square velocity is given by the following formula, which can be derived using calculus. Now, I will spare you the calculus and simply give you the formula. But if you're curious about where this formula comes from, leave A comment and I'll Show you. So, VRMs is equal to the square root of three times R times T divided by M, where M is our molar mass of gas, t is our temperature in Kelvin, and R is the molar gas constant. |
Average Kinetic Energy and Root Mean Square Velocity .txt | Now, I will spare you the calculus and simply give you the formula. But if you're curious about where this formula comes from, leave A comment and I'll Show you. So, VRMs is equal to the square root of three times R times T divided by M, where M is our molar mass of gas, t is our temperature in Kelvin, and R is the molar gas constant. Now, my goal is I want to use this VRMs to find the average kinetic energy. Because remember, kinetic energy of anything is given by one half times mass times V squared. Now, to find the kinetic average, I basically plug in my Dr meh into my D, and that will give me the average, because remember, this guy is the average or the quadratic average of my sets of points. |
Average Kinetic Energy and Root Mean Square Velocity .txt | Now, my goal is I want to use this VRMs to find the average kinetic energy. Because remember, kinetic energy of anything is given by one half times mass times V squared. Now, to find the kinetic average, I basically plug in my Dr meh into my D, and that will give me the average, because remember, this guy is the average or the quadratic average of my sets of points. So average kinetic energy is equal to one half times and times VR mass squared equals one half times mass. Now, I take my formula for BRMs, for Dances, and plug into my equation. The radical will disappear, because the radical has an exponent of a half a half times two is one. |
Average Kinetic Energy and Root Mean Square Velocity .txt | So average kinetic energy is equal to one half times and times VR mass squared equals one half times mass. Now, I take my formula for BRMs, for Dances, and plug into my equation. The radical will disappear, because the radical has an exponent of a half a half times two is one. So I simply get one half times mass over molam mass times R and T. Now, remember, molar mass has units of mass divided by moles. So the mass will cancel, and moles will go on top, and we'll get one half the mass will cancel. So this M and this M will cancel. |
Average Kinetic Energy and Root Mean Square Velocity .txt | So I simply get one half times mass over molam mass times R and T. Now, remember, molar mass has units of mass divided by moles. So the mass will cancel, and moles will go on top, and we'll get one half the mass will cancel. So this M and this M will cancel. The moles will go on top, and we'll have R times T. So my file final representation of my kinetic or average kinetic energy is three times N times R times T divided by two, where N is our number of moles. So Suppose we have 1 mol of any gas. Well, since we have 1 Mol, our N will be one. |
Average Kinetic Energy and Root Mean Square Velocity .txt | The moles will go on top, and we'll have R times T. So my file final representation of my kinetic or average kinetic energy is three times N times R times T divided by two, where N is our number of moles. So Suppose we have 1 mol of any gas. Well, since we have 1 Mol, our N will be one. And for 1 Mol, our formula becomes average kinetic energy of 1 mol of gas. That's equal to three times R times T divided by two. So the only non constant in this equation is T. So we see kinetic energy depends strictly on temperature, or 1 mol of gas depends strictly on the temperature. |
Average Kinetic Energy and Root Mean Square Velocity .txt | And for 1 Mol, our formula becomes average kinetic energy of 1 mol of gas. That's equal to three times R times T divided by two. So the only non constant in this equation is T. So we see kinetic energy depends strictly on temperature, or 1 mol of gas depends strictly on the temperature. Now, suppose instead I want to find one molecule. What's the kinetic average of one molecule of gas? Not 1 mol. |
Average Kinetic Energy and Root Mean Square Velocity .txt | Now, suppose instead I want to find one molecule. What's the kinetic average of one molecule of gas? Not 1 mol. Well, remember, a mole has an avocado's number of molecules. So I simply take my formula and divide it by N. So three times R times two divided by two N. The reason I singled out the R and the N is because this guy is known as a Boltzmann constant. In other words, scientists use some value k that equals r divided by n. So we replace this R divided by n with k called the Bolton constant, and our equation becomes three times k times t over two. |
Average Kinetic Energy and Root Mean Square Velocity .txt | Well, remember, a mole has an avocado's number of molecules. So I simply take my formula and divide it by N. So three times R times two divided by two N. The reason I singled out the R and the N is because this guy is known as a Boltzmann constant. In other words, scientists use some value k that equals r divided by n. So we replace this R divided by n with k called the Bolton constant, and our equation becomes three times k times t over two. Which is the same thing as saying three times R times t divided by two times n.
So once again, for one molecule, a single molecule, to find the kinetic average of that molecule, we simply use this formula where k is at boltsman constant. Now, to find the average kinetic energy of a mole of molecules, we have to use this formula. Now note that at any given time, our molecule can have any kinetic energy. |
Colloids .txt | That means that they are nonhomogeneous solutions. Recall that a homogeneous solution is a solution in which the compounds that compose the mixture are in the same state while colloids are composed of compounds in different states. Now, they also contain larger solid particles. For example, blood, which is a colloid, is composed mostly of water and hemoglobin. Hemoglobin is the larger particle, the solute, while water is a solvent, okay? Now, the larger particles are also called colloidal particles. |
Colloids .txt | For example, blood, which is a colloid, is composed mostly of water and hemoglobin. Hemoglobin is the larger particle, the solute, while water is a solvent, okay? Now, the larger particles are also called colloidal particles. So hemoglobin is the colloidal particle. Now, other examples of colloids exist. Fog. |
Colloids .txt | So hemoglobin is the colloidal particle. Now, other examples of colloids exist. Fog. Fog is composed of liquid particles found in the gas state. Smoke is composed of solid particles or carbon, found in the gas state. Wood cream is composed of gas particles found in the liquid state. |
Colloids .txt | Fog is composed of liquid particles found in the gas state. Smoke is composed of solid particles or carbon, found in the gas state. Wood cream is composed of gas particles found in the liquid state. And paint is composed of solid particles found in the liquid state. Now, carbon systems also experience something called a Tyndall effect. The tyle effect is a scattering of light due to the presence of large particles. |
Colloids .txt | And paint is composed of solid particles found in the liquid state. Now, carbon systems also experience something called a Tyndall effect. The tyle effect is a scattering of light due to the presence of large particles. Now, if you look at this system here, pretend that this is a cup and inside is a colloidal system, okay? And these are the colloidal particles or the large particles. Now, when light will enter it's going to bounce back and forth between these particles and it's going to lose energy. |
Colloids .txt | Now, if you look at this system here, pretend that this is a cup and inside is a colloidal system, okay? And these are the colloidal particles or the large particles. Now, when light will enter it's going to bounce back and forth between these particles and it's going to lose energy. Eventually, when it comes out, it's going to come out with less energy because remember, light carries energy. Light is composed of photons. That means that the system, the colloidal system will be translucent so you won't be able to see through it very clearly. |
Colloids .txt | Eventually, when it comes out, it's going to come out with less energy because remember, light carries energy. Light is composed of photons. That means that the system, the colloidal system will be translucent so you won't be able to see through it very clearly. The solid portion of the colloid is called a continuous phase. The solid portion of the colloid is called dispersed phase. Now, when water is a solvent, classification occurs in two ways. |
Colloids .txt | The solid portion of the colloid is called a continuous phase. The solid portion of the colloid is called dispersed phase. Now, when water is a solvent, classification occurs in two ways. Lyophilicaloids are those colloids that have strong attractions between continuous and dispersed phases. Examples include protein water, which is found in blood. Protein is a dispersed phase. |
Colloids .txt | Lyophilicaloids are those colloids that have strong attractions between continuous and dispersed phases. Examples include protein water, which is found in blood. Protein is a dispersed phase. Water is a continuous phase and they mix very well. Biophobic colloids are those colloids that lack attraction between continuous and dispersed phases. And examples include fat and water mixtures. |
Colloids .txt | Water is a continuous phase and they mix very well. Biophobic colloids are those colloids that lack attraction between continuous and dispersed phases. And examples include fat and water mixtures. We know that fat, the dispersed phase and water continuous days don't mix very well. Now, there are two ways to filter colloids. The first way is by heating and then filtration. |
Colloids .txt | We know that fat, the dispersed phase and water continuous days don't mix very well. Now, there are two ways to filter colloids. The first way is by heating and then filtration. If you heat a colloid, this will cause the salute to coagulate and then you could filter it by simple filtration. The second way is via dialysis. Dialysis is a separation using a semi permeable membrane. |
Colloids .txt | If you heat a colloid, this will cause the salute to coagulate and then you could filter it by simple filtration. The second way is via dialysis. Dialysis is a separation using a semi permeable membrane. For example, you have a semi permeable membrane here. Red blood cells mixed with water. The red blood cells are the dispersed phase. |
Cell voltage example .txt | So we begin with a certain electrochemical cell that's composed of cadmium and zinc. Now, we are given the cell voltage or the electromotive force around cells, and it's given to be zero point 36 volts. Under standard conditions, that means 1 bar pressure and one molar concentration. Now, this guy is also at 25 degrees Celsius. So we are also given the cell voltage for one of the two half reactions, one of the two half cells. And the zinc half reaction is given negative zero point 76 volts. |
Cell voltage example .txt | Now, this guy is also at 25 degrees Celsius. So we are also given the cell voltage for one of the two half reactions, one of the two half cells. And the zinc half reaction is given negative zero point 76 volts. We want to find our goal is to find the cell voltage for the other half cell for the other half reaction. So, in the first step, we begin our problem by writing their reduction reaction for the entire equation for the entire process. So, solid zinc reacts with cadmium ions to produce aqueous zinc and solid cadmium. |
Cell voltage example .txt | We want to find our goal is to find the cell voltage for the other half cell for the other half reaction. So, in the first step, we begin our problem by writing their reduction reaction for the entire equation for the entire process. So, solid zinc reacts with cadmium ions to produce aqueous zinc and solid cadmium. So let's figure out which one is oxidized and which one is reduced. This will help us place the metal that belongs to the anode and the metal that belongs to the cathode. So, zinc solid goes from a neutral charge to a plus two charge. |
Cell voltage example .txt | So let's figure out which one is oxidized and which one is reduced. This will help us place the metal that belongs to the anode and the metal that belongs to the cathode. So, zinc solid goes from a neutral charge to a plus two charge. That means it loses electrons. So that means it is oxidized. It's a reducing agent. |
Cell voltage example .txt | That means it loses electrons. So that means it is oxidized. It's a reducing agent. Now, this guy cadmium ion goes from an ion to a neutral charge, and that means this guy is reduced. So it's the oxidizing agent. It gains those electrons that are released by this zinc solid metal. |
Cell voltage example .txt | Now, this guy cadmium ion goes from an ion to a neutral charge, and that means this guy is reduced. So it's the oxidizing agent. It gains those electrons that are released by this zinc solid metal. So that means in step two, when we draw our electrochemical cell, the first half cell will contain our zinc solid. And that's because zinc is oxidized, and the anode always contains the oxidation reaction. So this is our zinc metal. |
Cell voltage example .txt | So that means in step two, when we draw our electrochemical cell, the first half cell will contain our zinc solid. And that's because zinc is oxidized, and the anode always contains the oxidation reaction. So this is our zinc metal. That means this must be our cadmium metal. And so electrons will travel via the conductor, via this volt meter and into this cathode, into this electrode. The volt meter, by the way, is the thing that reads zero point 36 volts. |
Cell voltage example .txt | That means this must be our cadmium metal. And so electrons will travel via the conductor, via this volt meter and into this cathode, into this electrode. The volt meter, by the way, is the thing that reads zero point 36 volts. So this sold bridge is placed here because it plays the role of closing the circuit. Without the sole bridge, this guy would not function. Electrons would not flow, and it's very important. |
Cell voltage example .txt | So this sold bridge is placed here because it plays the role of closing the circuit. Without the sole bridge, this guy would not function. Electrons would not flow, and it's very important. So, once again, our anode, our place where oxidation occurs, and our cathode, the place where reduction occurs. So electrons travel from this way, from this electrode to this electrode. So when they leave this cell, this solid zinc releases electrons, right? |
Cell voltage example .txt | So, once again, our anode, our place where oxidation occurs, and our cathode, the place where reduction occurs. So electrons travel from this way, from this electrode to this electrode. So when they leave this cell, this solid zinc releases electrons, right? Electrons begin to flow here. It also releases a zinc ion into this solution. And when the electrons travel this way, they combine. |
Cell voltage example .txt | Electrons begin to flow here. It also releases a zinc ion into this solution. And when the electrons travel this way, they combine. When they reach this metal, they combine with the cadmium ions forming our cadmium solid. So, in the third step, we basically want to use this cell voltage formula to find our cell voltage for the cathode. Remember? |
Cell voltage example .txt | When they reach this metal, they combine with the cadmium ions forming our cadmium solid. So, in the third step, we basically want to use this cell voltage formula to find our cell voltage for the cathode. Remember? Now, we know that this guy, this value of negative zero point 76 represents the value for the anode. So we use this formula to solve for our cathode. So we know our 0.36 volts, which is told by the voltmeter. |
Cell voltage example .txt | Now, we know that this guy, this value of negative zero point 76 represents the value for the anode. So we use this formula to solve for our cathode. So we know our 0.36 volts, which is told by the voltmeter. Now, this we don't know. So we let a DX minus now in parentheses we have a negative 0.76 volts, right? So negative and negative becomes a positive. |
Charles Law .txt | And what this theory did for us is it helped us gain more intuition about how individual gas molecules interact on a very small microscopic level. Now we also spoke about a law called Boils Law. And what Boils Law did for us is it helps help us gain more intuition about the macroscopic behavior of gases. In other words, what happens at constant temperature when we take a balloon filled with air and squeeze it. Well, we said that, and Boyle's Law explained this, that when we squeeze the balloon, we decrease volume, increase pressure, and eventually our balloon will pop. Now we're going to look at another law called Charles Law which also helps us explain the macroscopic large scale behavior of many gas molecules and how they interact with one another and with our system. |
Charles Law .txt | In other words, what happens at constant temperature when we take a balloon filled with air and squeeze it. Well, we said that, and Boyle's Law explained this, that when we squeeze the balloon, we decrease volume, increase pressure, and eventually our balloon will pop. Now we're going to look at another law called Charles Law which also helps us explain the macroscopic large scale behavior of many gas molecules and how they interact with one another and with our system. So for Charles Law to work, two conditions must hold. We must have constant pressure and we must have constant number of molecules. So our N number of moles stays the same. |
Charles Law .txt | So for Charles Law to work, two conditions must hold. We must have constant pressure and we must have constant number of molecules. So our N number of moles stays the same. And what Charles Law does is it relates volume and temperature. Remember Boyle's Law related volume and pressure? Well, Charles Law relates temperature and volume when pressure is constant. |
Charles Law .txt | And what Charles Law does is it relates volume and temperature. Remember Boyle's Law related volume and pressure? Well, Charles Law relates temperature and volume when pressure is constant. And what it says is that volume is directly proportional to temperature. And this means if we bring the T over or if we multiply a constant by T and we bring T over, what we get is the following v divided by T is equal to a constant. Remember, in Boyle's Law we saw that P times V gave us a constant and that if we increase pressure, our volume must decrease while keeping the constant the same. |
Charles Law .txt | And what it says is that volume is directly proportional to temperature. And this means if we bring the T over or if we multiply a constant by T and we bring T over, what we get is the following v divided by T is equal to a constant. Remember, in Boyle's Law we saw that P times V gave us a constant and that if we increase pressure, our volume must decrease while keeping the constant the same. Well, in this case, we have the same kind of situation, except now we have volume divided by temperature. In other words. Now for this constant to remain a constant and not change if we increase volume, say by two, our temperature also must increase by the same amount by two. |
Charles Law .txt | Well, in this case, we have the same kind of situation, except now we have volume divided by temperature. In other words. Now for this constant to remain a constant and not change if we increase volume, say by two, our temperature also must increase by the same amount by two. So whenever we increase volume, we increase temperature. Or if we decrease volume, we must decrease the temperature for this guy to remain constant. Now notice this constant remains or depends on two things on the pressure and on the number of molecules. |
Charles Law .txt | So whenever we increase volume, we increase temperature. Or if we decrease volume, we must decrease the temperature for this guy to remain constant. Now notice this constant remains or depends on two things on the pressure and on the number of molecules. For example, if we have more molecules, our constant will be higher. And we'll learn more about that when we talk about the ideal gas law. For now, it's sufficient to say that our cost depends on both of these guys. |
Charles Law .txt | For example, if we have more molecules, our constant will be higher. And we'll learn more about that when we talk about the ideal gas law. For now, it's sufficient to say that our cost depends on both of these guys. So the same way we did for Boyle's Law, for Charles Law, we can rewrite this relation or this relation in the following manner. Now, suppose I have some system, some gas system and I have two sets of different conditions for this gas system. Well, I can relate them in the following manner. |
Charles Law .txt | So the same way we did for Boyle's Law, for Charles Law, we can rewrite this relation or this relation in the following manner. Now, suppose I have some system, some gas system and I have two sets of different conditions for this gas system. Well, I can relate them in the following manner. Assuming that these guys are both constant. In other words, note that the constant always stays the same when our pressure number of moles stays the same. That means if we have some conditions, v one and T one and a second condition of V two and T two. |
Charles Law .txt | Assuming that these guys are both constant. In other words, note that the constant always stays the same when our pressure number of moles stays the same. That means if we have some conditions, v one and T one and a second condition of V two and T two. And these two conditions are under the same pressure and number of moles, that means our constant will be the same. And so I can say V one over T one equals V two over t two equals that same constant. So once again, for a gas sample with two different sets of T's and Vs for two different conditions, this is our law. |
Charles Law .txt | And these two conditions are under the same pressure and number of moles, that means our constant will be the same. And so I can say V one over T one equals V two over t two equals that same constant. So once again, for a gas sample with two different sets of T's and Vs for two different conditions, this is our law. This is our Charles Law. Now, whenever we talk about gases, and we talk about these different laws that revolve around gases, our temperature is never in Celsius, it's only in Kelvin. And that means we have to convert Celsius to Kelvin. |
Charles Law .txt | This is our Charles Law. Now, whenever we talk about gases, and we talk about these different laws that revolve around gases, our temperature is never in Celsius, it's only in Kelvin. And that means we have to convert Celsius to Kelvin. And this is how you do it. To get the Kelvin temperature, you take the Celsius temperature and you add 273.15 to it. For example, if our Celsius temperature is ten degrees Celsius, I simply add 273.15 and I get 283.15 Kelvin. |
Charles Law .txt | And this is how you do it. To get the Kelvin temperature, you take the Celsius temperature and you add 273.15 to it. For example, if our Celsius temperature is ten degrees Celsius, I simply add 273.15 and I get 283.15 Kelvin. That's our temperature in Kelvin. So let's think of an example where Charles Law is evident. So suppose it's someone's birthday and it's winter. |
Charles Law .txt | That's our temperature in Kelvin. So let's think of an example where Charles Law is evident. So suppose it's someone's birthday and it's winter. So it's cold outside and you need to go and buy somebody a present. So he decides to buy a balloon. So you go inside the store, you fill up the balloon with some helium. |
Charles Law .txt | So it's cold outside and you need to go and buy somebody a present. So he decides to buy a balloon. So you go inside the store, you fill up the balloon with some helium. Now suppose once you fill the balloon up, there are no holes in the balloon. So the number of moles or number of molecules inside our balloon remains constant. Also, let's suppose that our pressure inside the store and outside is one ATM atmospheric pressure. |
Charles Law .txt | Now suppose once you fill the balloon up, there are no holes in the balloon. So the number of moles or number of molecules inside our balloon remains constant. Also, let's suppose that our pressure inside the store and outside is one ATM atmospheric pressure. So let's suppose that our pressure is also constant. Now, obviously, if it's winter, it's much colder outside than inside. So suppose I fill up my balloon with N number of moles of helium, and suppose now I go outside with that balloon. |
Charles Law .txt | So let's suppose that our pressure is also constant. Now, obviously, if it's winter, it's much colder outside than inside. So suppose I fill up my balloon with N number of moles of helium, and suppose now I go outside with that balloon. Well, on the inside, inside the store, we were at one condition. We had some D two and t two, right? Once we step outside, our temperature drops. |
Charles Law .txt | Well, on the inside, inside the store, we were at one condition. We had some D two and t two, right? Once we step outside, our temperature drops. So what happens to volume? Well, according to Charles Law, this guy states that if we go outside and my temperature drops, then my volume must drop as well to make sure that our constant stays the same. That means if I go outside with my balloon, my balloon will shrivel, it will become smaller, and that's because of Charles Law. |
Charles Law .txt | So what happens to volume? Well, according to Charles Law, this guy states that if we go outside and my temperature drops, then my volume must drop as well to make sure that our constant stays the same. That means if I go outside with my balloon, my balloon will shrivel, it will become smaller, and that's because of Charles Law. So once again, we see how macro scale events are explained by Charles Law, just like they were explained by Boyle's Law. So we have yet another law that helps us explain how gas molecules behave when we have a lot of them together, not simply individual gas molecules. Now, we can, of course, explain how Charles Law functions on an individual level using the kinetic theory. |
Charles Law .txt | So once again, we see how macro scale events are explained by Charles Law, just like they were explained by Boyle's Law. So we have yet another law that helps us explain how gas molecules behave when we have a lot of them together, not simply individual gas molecules. Now, we can, of course, explain how Charles Law functions on an individual level using the kinetic theory. Now, kinetic theory explains Charles Law on a microscopic level at constant pressure. So we have constant pressure, right? Our pressure doesn't change. |
Charles Law .txt | Now, kinetic theory explains Charles Law on a microscopic level at constant pressure. So we have constant pressure, right? Our pressure doesn't change. So if our temperature increases, what happens to the molecules? The molecules gain more kinetic energy. And if they gain more kinetic energy, they gain more speed. |
Charles Law .txt | So if our temperature increases, what happens to the molecules? The molecules gain more kinetic energy. And if they gain more kinetic energy, they gain more speed. They become quicker. And that means the only way that our pressure stays constant with higher kinetic energy is if the volume expands. So that's exactly what happens. |
Charles Law .txt | They become quicker. And that means the only way that our pressure stays constant with higher kinetic energy is if the volume expands. So that's exactly what happens. In order to ensure that there is constant pressure. An increase in temperature means there's an increase in kinetic energy. And this increase in kinetic energy means that there are more molecules, or the molecules are pushing against the wall of the container with greater force. |
Charles Law .txt | In order to ensure that there is constant pressure. An increase in temperature means there's an increase in kinetic energy. And this increase in kinetic energy means that there are more molecules, or the molecules are pushing against the wall of the container with greater force. And so they must expand container, increase the volume. So let's look at the graph of volume times our temperature. Now. |
Charles Law .txt | And so they must expand container, increase the volume. So let's look at the graph of volume times our temperature. Now. Temperature here is in Kelvin. So our origin is at zero. Zero? |
Charles Law .txt | Temperature here is in Kelvin. So our origin is at zero. Zero? Remember, we can't have negative volume. Our small volume of zero is zero, so can't go below this. Now our temperature in Kelvin is zero. |
Charles Law .txt | Remember, we can't have negative volume. Our small volume of zero is zero, so can't go below this. Now our temperature in Kelvin is zero. So we saw it as zero. Zero? At zero volume. |
Charles Law .txt | So we saw it as zero. Zero? At zero volume. We have zero Kelvin. But remember, zero Kelvin is unattainable. Everything has volume. |
Charles Law .txt | We have zero Kelvin. But remember, zero Kelvin is unattainable. Everything has volume. And that means our temperature must be somewhere above zero. Also notice that v one over t one means that our t one can't be zero. Otherwise, we get an undefined number, a number that has infinity as its answer. |
Charles Law .txt | And that means our temperature must be somewhere above zero. Also notice that v one over t one means that our t one can't be zero. Otherwise, we get an undefined number, a number that has infinity as its answer. Because any number over zero is infinity. So let's look at this law. Why is it that we have a linear function? |
Charles Law .txt | Because any number over zero is infinity. So let's look at this law. Why is it that we have a linear function? Linear line? Well, that's because this is our slope. A constant slope. |
Charles Law .txt | Linear line? Well, that's because this is our slope. A constant slope. So this guy's constant. Whenever D increases, team must increase by the same amount. If this is doubled, this is doubled. |
Sp Hybridization.txt | So let's begin with the following depiction. In our first case, let's suppose we have two different atoms. The first atom donates an S orbital, and the second atom donates a p orbital. So these two orbitals will combine in a way to form the following molecular orbitals. So, because we input two, we should get back to orbitals. And that's exactly what we get. |
Sp Hybridization.txt | So these two orbitals will combine in a way to form the following molecular orbitals. So, because we input two, we should get back to orbitals. And that's exactly what we get. The first molecular orbital is known as the bonding molecular orbital. It's lower in energy. And the second one is known as the anti bonding molecular orbital. |
Sp Hybridization.txt | The first molecular orbital is known as the bonding molecular orbital. It's lower in energy. And the second one is known as the anti bonding molecular orbital. It's the one higher in energy. So that's the first case. That's the normal case that we're used to seeing. |
Sp Hybridization.txt | It's the one higher in energy. So that's the first case. That's the normal case that we're used to seeing. So let's suppose we try a different thing. Now, let's suppose we have a single atom. And that single atom has both an S orbital as well as a p orbital. |
Sp Hybridization.txt | So let's suppose we try a different thing. Now, let's suppose we have a single atom. And that single atom has both an S orbital as well as a p orbital. What happens is, within that single atom, these two orbitals can combine in such a way to produce something that we know as hybridized orbitals. In other words, we have a single atom. Within that single atom, an S orbital interacts with a p orbital to produce two hybridized orbitals. |
Sp Hybridization.txt | What happens is, within that single atom, these two orbitals can combine in such a way to produce something that we know as hybridized orbitals. In other words, we have a single atom. Within that single atom, an S orbital interacts with a p orbital to produce two hybridized orbitals. Now, once again, we input two orbitals. So we should get back to hybridized orbitals. And that's exactly what we see happen here. |
Sp Hybridization.txt | Now, once again, we input two orbitals. So we should get back to hybridized orbitals. And that's exactly what we see happen here. Now, when this S combines with this positive P, we get the following hybridized orbitals. In other words, this positive region simply combines with this positive region, and this becomes smaller. So a positive S orbital combines with a positive p orbital. |
Sp Hybridization.txt | Now, when this S combines with this positive P, we get the following hybridized orbitals. In other words, this positive region simply combines with this positive region, and this becomes smaller. So a positive S orbital combines with a positive p orbital. The two greens combine, the blue becomes smaller to form an enlarged positive green lobe and a smaller or thinner negative blue lobe. And the same happens when this part is negative. We get the following because two negative lobes combine to form this enlarged negative section, enlarged negative lobe, and the smaller positive green lobe. |
Sp Hybridization.txt | The two greens combine, the blue becomes smaller to form an enlarged positive green lobe and a smaller or thinner negative blue lobe. And the same happens when this part is negative. We get the following because two negative lobes combine to form this enlarged negative section, enlarged negative lobe, and the smaller positive green lobe. Now, in this lecture, we're going to only talk about SP hypersized orbitals. In future lectures, we're also going to talk about SP two and SP three hyperze orbitals. So, what is an SP hyperized orbital? |
Sp Hybridization.txt | Now, in this lecture, we're going to only talk about SP hypersized orbitals. In future lectures, we're also going to talk about SP two and SP three hyperze orbitals. So, what is an SP hyperized orbital? Well, this is simply an orbital produced using 50% S orbitals and 50% P orbitals. In other words, when we're combining our orbitals within that given atom, 50% comes from S and 50% comes from P. And this is known as an SP hyper dice orbital. That's exactly what we have in this situation here. |
Sp Hybridization.txt | Well, this is simply an orbital produced using 50% S orbitals and 50% P orbitals. In other words, when we're combining our orbitals within that given atom, 50% comes from S and 50% comes from P. And this is known as an SP hyper dice orbital. That's exactly what we have in this situation here. So let's look at an example in nature. So where is this evidence? So let's look at one particular example in which a Beryllium atom combines with two H atoms. |
Sp Hybridization.txt | So let's look at an example in nature. So where is this evidence? So let's look at one particular example in which a Beryllium atom combines with two H atoms. So let's examine the electron configuration of Beryllium. So, Beryllium, in its neutral state, has four electrons, four protons, four neutrons. So the electron configuration goes like this. |
Sp Hybridization.txt | So let's examine the electron configuration of Beryllium. So, Beryllium, in its neutral state, has four electrons, four protons, four neutrons. So the electron configuration goes like this. We have two electrons that go into our one S, and we have two electrons that go into the two S. Now, we also have the two p orbitals. But since we have no more electrons, there are zero electrons in the two P orbital. So we can either represent it this way, or we can simply remove the two P. Now, for my purposes, I'm going to leave it in this way, and we'll see why. |
Sp Hybridization.txt | We have two electrons that go into our one S, and we have two electrons that go into the two S. Now, we also have the two p orbitals. But since we have no more electrons, there are zero electrons in the two P orbital. So we can either represent it this way, or we can simply remove the two P. Now, for my purposes, I'm going to leave it in this way, and we'll see why. So my question is the following will this Be donate a two P orbital to bind with the H, or will it donate a hybrid orbital? In other words, which situation is more stable? So let's examine it this way. |
Sp Hybridization.txt | So my question is the following will this Be donate a two P orbital to bind with the H, or will it donate a hybrid orbital? In other words, which situation is more stable? So let's examine it this way. Let's draw out our pictures. Let's suppose that Be forms this hybridized orbital, and then this hybridized orbital interacts with the H atom to form our covalent bond. And let's also suppose that we have a Beryllium atom that donates a simple two P orbital to interact with that H atom. |
Sp Hybridization.txt | Let's draw out our pictures. Let's suppose that Be forms this hybridized orbital, and then this hybridized orbital interacts with the H atom to form our covalent bond. And let's also suppose that we have a Beryllium atom that donates a simple two P orbital to interact with that H atom. Let's see which one is more stable. Well, recall that whenever bonds are formed, bonds or covalent bonds are formed by the overlap of atomic orbitals, as we see here. And we know that the better the overlap, the larger the lobes, the more stable our compound is, the more stable our bond is. |
Sp Hybridization.txt | Let's see which one is more stable. Well, recall that whenever bonds are formed, bonds or covalent bonds are formed by the overlap of atomic orbitals, as we see here. And we know that the better the overlap, the larger the lobes, the more stable our compound is, the more stable our bond is. So in which situation do we have a more stabilized overlap? A larger overlap? Well, clearly this case has a much bigger lobe. |
Sp Hybridization.txt | So in which situation do we have a more stabilized overlap? A larger overlap? Well, clearly this case has a much bigger lobe. And that means the interaction will be much better in this hybridized interaction. In other words, this hybridized lobe creates a larger lobe. And that means, because we have a larger lobe, we have a better overlap. |
Sp Hybridization.txt | And that means the interaction will be much better in this hybridized interaction. In other words, this hybridized lobe creates a larger lobe. And that means, because we have a larger lobe, we have a better overlap. And so that means this is much more stable. And so this will not occur. We're going to have this type of bond. |
Sp Hybridization.txt | And so that means this is much more stable. And so this will not occur. We're going to have this type of bond. In other words, within this benh, when Be bonds to H, it creates a hybridized orbital, which then bonds to the one S of the H. And let's see exactly that in this energy diagram. So we can imagine this being the energy diagram. So the higher up we go, the more energy we have. |
Sp Hybridization.txt | In other words, within this benh, when Be bonds to H, it creates a hybridized orbital, which then bonds to the one S of the H. And let's see exactly that in this energy diagram. So we can imagine this being the energy diagram. So the higher up we go, the more energy we have. The lower we go, the less energy we have. What happens is the following. The Beryllium creates this hybridized orbital SD hybridized which comes from one S and one P an St hybridized orbital. |
Sp Hybridization.txt | The lower we go, the less energy we have. What happens is the following. The Beryllium creates this hybridized orbital SD hybridized which comes from one S and one P an St hybridized orbital. And that orbital, which is a bit slightly higher in energy than our one S of the H atom. So this is the H atom, and this is the one S orbital. And this is the SP hybridized orbital of our Beryllium. |
Sp Hybridization.txt | And that orbital, which is a bit slightly higher in energy than our one S of the H atom. So this is the H atom, and this is the one S orbital. And this is the SP hybridized orbital of our Beryllium. They will interact. So we input two atomic orbitals, and we get two molecular orbitals. So we get the one lower in energy and the one higher in energy. |
Sp Hybridization.txt | They will interact. So we input two atomic orbitals, and we get two molecular orbitals. So we get the one lower in energy and the one higher in energy. So, once again, let's recap. So hybridization is simply a process that occurs within an atom. Within an atom, the orbitals can interact in a way to produce these hybridized orbitals that contain larger sections and smaller sections. |
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