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Phase Change of Water .txt | Okay? And melting according to this graph is endothermic. And that's because our final energy level is somewhere here. Our initial energy level is somewhere here. So a larger amount of energy minus a smaller amount of energy gives us a positive number. So the change in enthalpy of fusion is positive. |
Phase Change of Water .txt | Our initial energy level is somewhere here. So a larger amount of energy minus a smaller amount of energy gives us a positive number. So the change in enthalpy of fusion is positive. Melting is endothermic. Likewise, going backward or freezing is exothermic because we're taking this number and we're subtracting this number from this number. A smaller number minus a larger number gives you a negative number, and that means freezing. |
Phase Change of Water .txt | Melting is endothermic. Likewise, going backward or freezing is exothermic because we're taking this number and we're subtracting this number from this number. A smaller number minus a larger number gives you a negative number, and that means freezing. Going this way is exothermic. And so it releases energy into the environment, heating the environment and cooling our system. So let's look at vaporization. |
Phase Change of Water .txt | Going this way is exothermic. And so it releases energy into the environment, heating the environment and cooling our system. So let's look at vaporization. Vaporization is the process by which liquid molecules go into the gas state. So it's going from here to here. Once again, just like the change in enthalpy of fusion, change in enthalpy of vaporization is also positive. |
Phase Change of Water .txt | Vaporization is the process by which liquid molecules go into the gas state. So it's going from here to here. Once again, just like the change in enthalpy of fusion, change in enthalpy of vaporization is also positive. It's endothermic because we take a large number and we subtract a small number from the large number and we get a positive number. And once again, going from this stage to this stage, or from a gas to a liquid, which is called condensation, well, that's exothermic. That's negative, because we take this number, subtracted from this number, and we get a negative number. |
Acid Ionization Constant .txt | Remember, water molecules can act as both acids and bases. And in fact, if you add two water molecules together, one will act as an acid and the second one will act as a base, creating a conjugate acid and a conjugate base. In other words, if this is our acid, it will donate the H, creating an oh ion. And if this is our base, it will accept that H creating a hydronium ion. So now, if we wanted to, we can also write the equilibrium equation for this reaction. In other words, kw, our ionization constant for water is equal to hydronium concentration times the hydroxide concentration. |
Acid Ionization Constant .txt | And if this is our base, it will accept that H creating a hydronium ion. So now, if we wanted to, we can also write the equilibrium equation for this reaction. In other words, kw, our ionization constant for water is equal to hydronium concentration times the hydroxide concentration. And at 25 degrees Celsius, this equals 10 times ten to negative 14. Now, this process is called the autoimilization reaction. And if you want to learn more about this reaction, check out the link below. |
Acid Ionization Constant .txt | And at 25 degrees Celsius, this equals 10 times ten to negative 14. Now, this process is called the autoimilization reaction. And if you want to learn more about this reaction, check out the link below. Now, in the same way that we talk about ionization constants of water, we can also talk about ionization constants of acids. Except now they're not Kw, they're ka, where A is for our acid. So let's suppose we have a hypothetical acid Ha reacting with a water molecule in a liquid state. |
Acid Ionization Constant .txt | Now, in the same way that we talk about ionization constants of water, we can also talk about ionization constants of acids. Except now they're not Kw, they're ka, where A is for our acid. So let's suppose we have a hypothetical acid Ha reacting with a water molecule in a liquid state. Now, what will happen? Well, this acid will donate the H, releasing the H, while the base will accept that age, creating a hydronium ion and a conjugate base. So this is our conjugate acid, our conjugate base and our conjugate base and our conjugate acid. |
Acid Ionization Constant .txt | Now, what will happen? Well, this acid will donate the H, releasing the H, while the base will accept that age, creating a hydronium ion and a conjugate base. So this is our conjugate acid, our conjugate base and our conjugate base and our conjugate acid. Now, let's write the equilibrium constant stress in the same way we did for water for this acid. So ka, our acid annetzation constant is equal to our concentration of our hydronium times, the concentration of the conjugate base. Now, in this case, our conjugate base is simply this guy here. |
Acid Ionization Constant .txt | Now, let's write the equilibrium constant stress in the same way we did for water for this acid. So ka, our acid annetzation constant is equal to our concentration of our hydronium times, the concentration of the conjugate base. Now, in this case, our conjugate base is simply this guy here. Now, both guys are included in our numerator because both guys are in aqueous form. Remember, liquids and solids are not included. And that's why we didn't include these two water molecules. |
Acid Ionization Constant .txt | Now, both guys are included in our numerator because both guys are in aqueous form. Remember, liquids and solids are not included. And that's why we didn't include these two water molecules. Now, on the bottom, since we have an Ha in the Aqueous state, our conjugate acid, we must include the conjugate acid as well. So Ha gets incorporated into our equilibrium constant expression. Now, what is a k value or Ka value? |
Acid Ionization Constant .txt | Now, on the bottom, since we have an Ha in the Aqueous state, our conjugate acid, we must include the conjugate acid as well. So Ha gets incorporated into our equilibrium constant expression. Now, what is a k value or Ka value? Well, the ionization constant is simply a ratio from a mathematical perspective. And what the Ka is, it's the amount of product formed over the amount of reactants left over. So if this number is very large, then that means a lot of product was formed and very little reactants left over. |
Acid Ionization Constant .txt | Well, the ionization constant is simply a ratio from a mathematical perspective. And what the Ka is, it's the amount of product formed over the amount of reactants left over. So if this number is very large, then that means a lot of product was formed and very little reactants left over. So what does that tell us about the Ha? Well, if this reaction is favored this way, if a lot of product is formed and very little reactor is left over, that means this acid is very good at giving off that age. So it must be a very good acid by definition of an acid. |
Acid Ionization Constant .txt | So what does that tell us about the Ha? Well, if this reaction is favored this way, if a lot of product is formed and very little reactor is left over, that means this acid is very good at giving off that age. So it must be a very good acid by definition of an acid. So that means if our Ka is large, we have a good acid. Likewise, if our Ka is small, that means very little product is formed and a lot of reactors left over. That means this asset is very bad at releasing that age, so it's a bad asset. |
Acid Ionization Constant .txt | So that means if our Ka is large, we have a good acid. Likewise, if our Ka is small, that means very little product is formed and a lot of reactors left over. That means this asset is very bad at releasing that age, so it's a bad asset. Now, we can deduce that if the Ka is greater than one, then that means it's a strong acid. And if the Ka is less than one, that means it's a weak acid. Now let's look at one more thing. |
Acid Ionization Constant .txt | Now, we can deduce that if the Ka is greater than one, then that means it's a strong acid. And if the Ka is less than one, that means it's a weak acid. Now let's look at one more thing. So what happens if our Ka is less than one? Then this guy must be a weak acid. But that means our conjugate base must be a good base. |
Acid Ionization Constant .txt | So what happens if our Ka is less than one? Then this guy must be a weak acid. But that means our conjugate base must be a good base. Likewise, if this was a good acid and our Ka was above one, that means this was a bad conjugate base. Now let's look at a few examples. Nitric acid in aqueous state could act with water to produce hydronium ion plus nitrate ion. |
Acid Ionization Constant .txt | Likewise, if this was a good acid and our Ka was above one, that means this was a bad conjugate base. Now let's look at a few examples. Nitric acid in aqueous state could act with water to produce hydronium ion plus nitrate ion. Now, the ka for this reaction for this acid is 20. And that means, according to this theory, it's a very good acid. And indeed it is. |
Acid Ionization Constant .txt | Now, the ka for this reaction for this acid is 20. And that means, according to this theory, it's a very good acid. And indeed it is. Now let's look at hydrofluoric acid. So hydrofluoric acid in the equated states react with water in the liquid state to produce hydronium plus the fion. Now, the Ka for this reaction for this particular acid is very low. |
Acid Ionization Constant .txt | Now let's look at hydrofluoric acid. So hydrofluoric acid in the equated states react with water in the liquid state to produce hydronium plus the fion. Now, the Ka for this reaction for this particular acid is very low. It's 7.2 times ten to negative four. And that means, according to this theory, it's a bad acid. And in fact, it is a bad acid. |
Acid Ionization Constant .txt | It's 7.2 times ten to negative four. And that means, according to this theory, it's a bad acid. And in fact, it is a bad acid. It's a weak acid. So now we can use this ka value to determine whether or not an acid is a good acid or a bad asset. Now, before we have to look at the polarity of the bond, we have to look at the bond strength and we have to look at the conjugate base. |
Ion Pairing .txt | Now, ion Pairing is A Momentary aggregation of Electrically charged ions found In A Concentrated solution Of Two Or More compounds. Now to really grasp what we mean by Ion pairing, let's create a solution and see what happens with the natural solution. So let's mix liquid water and solid sodium chloride and see what happens. Well, once we mix them, the sodium chloride dissociates and forms ions. So we have a bunch of polar water molecules separated by ions. At any given time, two of these ions might be separated by a solvent water molecule, and at this point, they can't interact with one another because of this separation. |
Ion Pairing .txt | Well, once we mix them, the sodium chloride dissociates and forms ions. So we have a bunch of polar water molecules separated by ions. At any given time, two of these ions might be separated by a solvent water molecule, and at this point, they can't interact with one another because of this separation. However, at another place, the solid molecule, the water molecule, might not separate them. And at this point, because of their proximity, the ions will form an ion pair. And this is only for the moment. |
Ion Pairing .txt | However, at another place, the solid molecule, the water molecule, might not separate them. And at this point, because of their proximity, the ions will form an ion pair. And this is only for the moment. Imagine two ions floating or flying around in the liquid, right? Eventually they will get close to each other. And when they get close to each other, they will form that momentary bond. |
Ion Pairing .txt | Imagine two ions floating or flying around in the liquid, right? Eventually they will get close to each other. And when they get close to each other, they will form that momentary bond. But it's only for the moment. Because if they're traveling, they will attract and will continue to travel. So eventually they will break. |
Ion Pairing .txt | But it's only for the moment. Because if they're traveling, they will attract and will continue to travel. So eventually they will break. So they travel, attract and break. And it's only for the moment. And that's the difference between an Ion pair and an ionic bond. |
Ion Pairing .txt | So they travel, attract and break. And it's only for the moment. And that's the difference between an Ion pair and an ionic bond. An ionic bond is not for the moment. It Stays. It Exists. |
Ion Pairing .txt | An ionic bond is not for the moment. It Stays. It Exists. If this sodium chloride is left untouched, it will continue. The ionic bond will continue to exist. But in this case, the Ion pair only exists for a moment. |
Ion Pairing .txt | If this sodium chloride is left untouched, it will continue. The ionic bond will continue to exist. But in this case, the Ion pair only exists for a moment. Now Ion Pairing does not exist in an ideal solution. And that's because in an ideal solution, by definition, all the ions are separated from one another by solvent molecules. So no attraction between two ions exists. |
Ion Pairing .txt | Now Ion Pairing does not exist in an ideal solution. And that's because in an ideal solution, by definition, all the ions are separated from one another by solvent molecules. So no attraction between two ions exists. So Ion pairing only exists in nonideal conditions. Now, if you want to see why Ion Pairing is important and where it's applied, check out the video below. In that video, I talk about something called boiling point elevation and melting point depression. |
Introduction to Gibbs Free Energy .txt | Well, Gibbs free energy, just like enthalpy, is a manmade concept, which means it cannot be measured. So you cannot use an instrument to measure Gibbs free energy of some object. Gibbs free energy could only measure expansion fundamentally. Now, that's because Gibbs free energy is defined by a formula. And this formula only holds under certain conditions. If these conditions aren't met, Gibbs free energy breaks down. |
Introduction to Gibbs Free Energy .txt | Now, that's because Gibbs free energy is defined by a formula. And this formula only holds under certain conditions. If these conditions aren't met, Gibbs free energy breaks down. Now, let's look at these conditions. These conditions are constant temperature and pressure. The reaction must be reversible, and there is no mechanical work done. |
Introduction to Gibbs Free Energy .txt | Now, let's look at these conditions. These conditions are constant temperature and pressure. The reaction must be reversible, and there is no mechanical work done. Only TV work is allowed to be done. Now, let's see the result of constant temperature and pressure. In an isolated system, the number of moles stays the same. |
Introduction to Gibbs Free Energy .txt | Only TV work is allowed to be done. Now, let's see the result of constant temperature and pressure. In an isolated system, the number of moles stays the same. And that's because there is no exchange in mass. So, if the number of moles stay the same, if the temperature is constant and pressure is constant, and according to the ideal gas law, the volume remains constant, so there is no volume change. So, according to the formula for change in enthalpy, if there's no volume change, then this guy, the PV work done, is zero. |
Introduction to Gibbs Free Energy .txt | And that's because there is no exchange in mass. So, if the number of moles stay the same, if the temperature is constant and pressure is constant, and according to the ideal gas law, the volume remains constant, so there is no volume change. So, according to the formula for change in enthalpy, if there's no volume change, then this guy, the PV work done, is zero. And so we can approximate the change in enthalpy to simply equaling change in internal energy or change in energy or heat. Now, before we jump into the formula, let's look at entropy and entropy. So what is entropy? |
Introduction to Gibbs Free Energy .txt | And so we can approximate the change in enthalpy to simply equaling change in internal energy or change in energy or heat. Now, before we jump into the formula, let's look at entropy and entropy. So what is entropy? Well, entropy tells us if the reaction is exothermic or endothermic. It does not tell us if the reaction is spontaneous. Okay? |
Introduction to Gibbs Free Energy .txt | Well, entropy tells us if the reaction is exothermic or endothermic. It does not tell us if the reaction is spontaneous. Okay? That's important. Now let's look at entropy. Now, from another video, we learned that entropy is nature's tendency to create the most probable system, okay? |
Introduction to Gibbs Free Energy .txt | That's important. Now let's look at entropy. Now, from another video, we learned that entropy is nature's tendency to create the most probable system, okay? So, for example, if we have this isolated system, we have four molecules on this side. Entropy, by definition, will tell us that two of the molecules will want to go here and will want to create this system here. In other words, this system is more probable. |
Introduction to Gibbs Free Energy .txt | So, for example, if we have this isolated system, we have four molecules on this side. Entropy, by definition, will tell us that two of the molecules will want to go here and will want to create this system here. In other words, this system is more probable. Okay? So we see that entropy and not enthalpy tells us if the reaction is spontaneous. Therefore, entropy determines spontaneity and not enthalpy. |
Introduction to Gibbs Free Energy .txt | Okay? So we see that entropy and not enthalpy tells us if the reaction is spontaneous. Therefore, entropy determines spontaneity and not enthalpy. Okay? Now let's look at Gibbs free energy. So, gifts free energy combines entropy and enthalpy the same way that enthalpy combines internal energy and change in volume. |
Introduction to Gibbs Free Energy .txt | Okay? Now let's look at Gibbs free energy. So, gifts free energy combines entropy and enthalpy the same way that enthalpy combines internal energy and change in volume. The formula for Gibbs free energy is as follows. The change in Gibbs free energy is equal to the change in entropy minus temperature times change in entropy. Now, when the change in Gibbs free energy is zero, the reaction is said to be at equilibrium. |
Introduction to Gibbs Free Energy .txt | The formula for Gibbs free energy is as follows. The change in Gibbs free energy is equal to the change in entropy minus temperature times change in entropy. Now, when the change in Gibbs free energy is zero, the reaction is said to be at equilibrium. Okay? The rate of the full reaction is the same as the rate of the reverse reaction. Now, when GIBS free energy is negative, the reaction is said to be spontaneous. |
Introduction to Gibbs Free Energy .txt | Okay? The rate of the full reaction is the same as the rate of the reverse reaction. Now, when GIBS free energy is negative, the reaction is said to be spontaneous. If it's positive, it's said to be non spontaneous or non spontaneous. Okay? Now, what git speed energy basically tells us is that even an exothermic reaction can be non spontaneous if this portion, if the change in entropy is negative, enough. |
Half Reactions .txt | So recall that redox reactions are chemical reactions that involve the transfer of electrons from one atom to another atom. Now, this means one atom is oxidized and one atom is reduced. So that means any redox reaction can be broken down into two types of reactions an oxidation reactions, in which an atom loses electrons and a reduction reaction, in which atoms gain electrons. So let's look at a very simple redox reaction. So, zinc metal reacts with aqueous copper to produce aqueous zinc and metal copper. So let's break this reaction into an oxidation and a reduction reaction. |
Half Reactions .txt | So let's look at a very simple redox reaction. So, zinc metal reacts with aqueous copper to produce aqueous zinc and metal copper. So let's break this reaction into an oxidation and a reduction reaction. So, what gets oxidized? Well, our zinc solid goes from a neutral charge to a plus two charge. That means our zinc loses two electrons. |
Half Reactions .txt | So, what gets oxidized? Well, our zinc solid goes from a neutral charge to a plus two charge. That means our zinc loses two electrons. So let's write the equation for that. So, let's write the oxidation equation. So, zinc solid becomes zinc plus two because it loses two electrons. |
Half Reactions .txt | So let's write the equation for that. So, let's write the oxidation equation. So, zinc solid becomes zinc plus two because it loses two electrons. So our oxidation reaction involves the loss of electrons. Now let's write the reduction reaction. In our reduction reaction, our copper gets reduced because it gains those two electrons that are lost by the zinc. |
Half Reactions .txt | So our oxidation reaction involves the loss of electrons. Now let's write the reduction reaction. In our reduction reaction, our copper gets reduced because it gains those two electrons that are lost by the zinc. So our copper plus two gains two electrons plus two electrons producing a neutral copper solid molecule. So this is our oxidation and this is our reduction reactions. Now, these two reactions are each called the half reaction. |
Half Reactions .txt | So our copper plus two gains two electrons plus two electrons producing a neutral copper solid molecule. So this is our oxidation and this is our reduction reactions. Now, these two reactions are each called the half reaction. And the half reaction is simply a way for us to visualize more clearly the transfer of electrons from one atom to another. Because notice in this net reaction, there was no transfer of electrons. We couldn't visualize the transfer electrons. |
Half Reactions .txt | And the half reaction is simply a way for us to visualize more clearly the transfer of electrons from one atom to another. Because notice in this net reaction, there was no transfer of electrons. We couldn't visualize the transfer electrons. But in these half reactions, we have the plus two E and plus two e.
So this is simply a better way for us to see the movement of electrons. Now, to go back to our net reaction net reduce reaction. What we simply do is we add up the two half reactions. |
Half Reactions .txt | But in these half reactions, we have the plus two E and plus two e.
So this is simply a better way for us to see the movement of electrons. Now, to go back to our net reaction net reduce reaction. What we simply do is we add up the two half reactions. So we add up the two half reactions by first adding up all the molecules on this side and then add up all the molecules on this side. So zinc solid plus acreage copper plus two electrons gives us this produces everything on this side. So zinc or acreage zinc plus copper solid plus our two electrons. |
Half Reactions .txt | So we add up the two half reactions by first adding up all the molecules on this side and then add up all the molecules on this side. So zinc solid plus acreage copper plus two electrons gives us this produces everything on this side. So zinc or acreage zinc plus copper solid plus our two electrons. Now, notice one thing. That two electrons up here on this side and on this side. And that means by using simple algebra we just cross these guys out, we subtract two e, and what we get is our final net reaction. |
Half Reactions .txt | Now, notice one thing. That two electrons up here on this side and on this side. And that means by using simple algebra we just cross these guys out, we subtract two e, and what we get is our final net reaction. So that means no electrons appear in the net reaction ever. And that's because the number of electrons released by our oxidation reaction is equal to the number gained by our reduction reaction. So whatever is gained must be lost somewhere. |
Freezing Point depression Example .txt | Now, our initial freezing point is six degrees Celsius. Our final freezing point is 2.7 degrees Celsius. Our constant for freezing is 20 Celsius times kilograms per mole. Now, so we basically start with 100 grams of cyclohexane which freezes at six degrees Celsius. We add 2 grams of unknown compound to our beaker of 100 grams of cyclohexane and that solution freezing point drops to 2.7 degrees Celsius. Now, our goal is to use the freezing point depression formula to find the Molar mass. |
Freezing Point depression Example .txt | Now, so we basically start with 100 grams of cyclohexane which freezes at six degrees Celsius. We add 2 grams of unknown compound to our beaker of 100 grams of cyclohexane and that solution freezing point drops to 2.7 degrees Celsius. Now, our goal is to use the freezing point depression formula to find the Molar mass. In the first step we write our freezing depression formula and that basically state that's the change in temperature equal there constant times molarity times I I is a bond half factor and in this case it's one. Because our unknown compound does not associate into anything, it stays the way it is. So you plug in our values and we get six -20 or 2.7 degrees Celsius. |
Freezing Point depression Example .txt | In the first step we write our freezing depression formula and that basically state that's the change in temperature equal there constant times molarity times I I is a bond half factor and in this case it's one. Because our unknown compound does not associate into anything, it stays the way it is. So you plug in our values and we get six -20 or 2.7 degrees Celsius. Divided by 20 our constant Celsius cancels moles goes on top and we get 0.165
moles per kilogram. And this is our Molarity. Now, in the second step, let's look at the formula for molality. |
Freezing Point depression Example .txt | Divided by 20 our constant Celsius cancels moles goes on top and we get 0.165
moles per kilogram. And this is our Molarity. Now, in the second step, let's look at the formula for molality. Molarity equals moles of compound divided by a kilogram of solids. Now we're given kilogram of solids, so we know that we also know molality. Now let's change our moles of compound formula to something else. |
Freezing Point depression Example .txt | Molarity equals moles of compound divided by a kilogram of solids. Now we're given kilogram of solids, so we know that we also know molality. Now let's change our moles of compound formula to something else. Let's see, how else can we represent the moles of compound? Now, whenever you want to find the moles of compound, we basically take our amount in grams divided by the molecular weight of that same compound and we get our moles. So another way of finding or representing moles of compound is simply grams of compound divided by molecular weight. |
Freezing Point depression Example .txt | Let's see, how else can we represent the moles of compound? Now, whenever you want to find the moles of compound, we basically take our amount in grams divided by the molecular weight of that same compound and we get our moles. So another way of finding or representing moles of compound is simply grams of compound divided by molecular weight. But molecular weight and molar mass mean the same thing. So we simply write molar mass and now we have this, we have Molarity and all we need is to find the Molar mass. Remember, the grams of compound was given initially 2 grams of unknown compound and we can check using our units that this makes sense, grams divided by grams per mole. |
Freezing Point depression Example .txt | But molecular weight and molar mass mean the same thing. So we simply write molar mass and now we have this, we have Molarity and all we need is to find the Molar mass. Remember, the grams of compound was given initially 2 grams of unknown compound and we can check using our units that this makes sense, grams divided by grams per mole. The grams cancel moles goes on top and we're left with moles per kilogram. And that's exactly what molality is. Now, our third step. |
Freezing Point depression Example .txt | The grams cancel moles goes on top and we're left with moles per kilogram. And that's exactly what molality is. Now, our third step. In our last step, we basically plug in our values. So for molality, which we got from the first step is zero point 65 equals two divided by x. Our unknown. |
Freezing Point depression Example .txt | In our last step, we basically plug in our values. So for molality, which we got from the first step is zero point 65 equals two divided by x. Our unknown. The molar mass entire thing divided by 0.10.1
comes from 100 grams. Remember, we want to deal with kilograms of our solid and we're given 100 grams. So divide this by 1000, we get 0.1
kg, bring this over, multiply this out, then bring it back and we at two divided by 0.165 equals x. |
Balancing Redox reactions example .txt | We want to follow seven steps that I've outlined in another lecture and want to get a final net balanced redox reaction. So let's follow these seven steps. In the first step, we want to find or determine the oxidized atom and the reduced atom. So let's find a reduced atom first. So, which atom gains electrons? So I've written out the oxidation states for each atom. |
Balancing Redox reactions example .txt | So let's find a reduced atom first. So, which atom gains electrons? So I've written out the oxidation states for each atom. So let's see which atom becomes more negative. So, our Mm, or permanganate atom, goes from a plus seven to a plus two. That means it gains five electrons. |
Balancing Redox reactions example .txt | So let's see which atom becomes more negative. So, our Mm, or permanganate atom, goes from a plus seven to a plus two. That means it gains five electrons. So that means our Mm is reduced. Now, likewise, let's find an atom that loses those electrons. Well, our carbon atom is plus three on this side and plus four on that side. |
Balancing Redox reactions example .txt | So that means our Mm is reduced. Now, likewise, let's find an atom that loses those electrons. Well, our carbon atom is plus three on this side and plus four on that side. And that means when it goes from here to here, it is oxidized. It loses those electrons gained by this Mm. So our carbon is oxidized. |
Balancing Redox reactions example .txt | And that means when it goes from here to here, it is oxidized. It loses those electrons gained by this Mm. So our carbon is oxidized. We're double step one. Let's go to step two. And step two. |
Balancing Redox reactions example .txt | We're double step one. Let's go to step two. And step two. We want to write out the two half reactions for our unbalanced equations. So we want to write out the oxidation reaction and the reduction reaction. So, oxidation basically states that this guy goes from this guy to this guy in here. |
Balancing Redox reactions example .txt | We want to write out the two half reactions for our unbalanced equations. So we want to write out the oxidation reaction and the reduction reaction. So, oxidation basically states that this guy goes from this guy to this guy in here. And then the reduction reaction, our ion, goes to this ion. So this is our reduction reaction. Now, I combine steps three and four in steps. |
Balancing Redox reactions example .txt | And then the reduction reaction, our ion, goes to this ion. So this is our reduction reaction. Now, I combine steps three and four in steps. In step three, I basically want to balance out the atoms that are not oxygen atoms and not hydrogen atoms. So let's start with oxidation reaction. So, an oxidation health reaction. |
Balancing Redox reactions example .txt | In step three, I basically want to balance out the atoms that are not oxygen atoms and not hydrogen atoms. So let's start with oxidation reaction. So, an oxidation health reaction. Let's balance out the carbon atoms. So, to balance this carbon atom out, there are two atoms on this side and only one atom on this side. That means I want to multiply this guy by two. |
Balancing Redox reactions example .txt | Let's balance out the carbon atoms. So, to balance this carbon atom out, there are two atoms on this side and only one atom on this side. That means I want to multiply this guy by two. And that's exactly what I do here. Now, I have two carbon atoms on this side and two carbon atoms on this side. Next, I want to balance out the carbon or the MN atoms on this side in this reduction reaction. |
Balancing Redox reactions example .txt | And that's exactly what I do here. Now, I have two carbon atoms on this side and two carbon atoms on this side. Next, I want to balance out the carbon or the MN atoms on this side in this reduction reaction. So, since I have one MN here and one Amen here, it's already balanced. So, in the fourth step, let's balance out the oxygen molecules by adding water molecules. And let's balance the h atoms by adding H plus ions. |
Balancing Redox reactions example .txt | So, since I have one MN here and one Amen here, it's already balanced. So, in the fourth step, let's balance out the oxygen molecules by adding water molecules. And let's balance the h atoms by adding H plus ions. So let's see our oxidation reaction. So, we have four O atoms on this side and only two atoms on this side. But I already multiply this carbon dioxide by two. |
Balancing Redox reactions example .txt | So let's see our oxidation reaction. So, we have four O atoms on this side and only two atoms on this side. But I already multiply this carbon dioxide by two. That means our oxygens already bounced out. So I have four here and four here. Two times two. |
Balancing Redox reactions example .txt | That means our oxygens already bounced out. So I have four here and four here. Two times two. So the oxygen is bounced out. Let's balance out the h atoms. I have two atoms on this side and no h atoms on that side, right? |
Balancing Redox reactions example .txt | So the oxygen is bounced out. Let's balance out the h atoms. I have two atoms on this side and no h atoms on that side, right? So in order to balance this, I have to add two h atoms on this side. That's exactly what I do. So I'm done with my oxidation half reaction. |
Balancing Redox reactions example .txt | So in order to balance this, I have to add two h atoms on this side. That's exactly what I do. So I'm done with my oxidation half reaction. Let's look at the reduction half reaction. Remember, the MNS are already balanced out, so let's balance out oxygen. I have four oxygen this side, so that means you have to add four water molecules on this side. |
Balancing Redox reactions example .txt | Let's look at the reduction half reaction. Remember, the MNS are already balanced out, so let's balance out oxygen. I have four oxygen this side, so that means you have to add four water molecules on this side. If I add four water molecules on this side, I balance out the O. But now I have to balance out the H, because now I have four times two, eight H ions on this side or atoms on this side. That means I balance this guy out by adding eight H plus ions to this side, and I get this following reaction. |
Balancing Redox reactions example .txt | If I add four water molecules on this side, I balance out the O. But now I have to balance out the H, because now I have four times two, eight H ions on this side or atoms on this side. That means I balance this guy out by adding eight H plus ions to this side, and I get this following reaction. So I'm done with steps three and four. Now let's go to step five. In step five, you basically want to balance out the entire charge down on this side and this side, and then the entire charge down on this side and this side. |
Balancing Redox reactions example .txt | So I'm done with steps three and four. Now let's go to step five. In step five, you basically want to balance out the entire charge down on this side and this side, and then the entire charge down on this side and this side. In other words, on this side of the oxidation half reaction, our charge is neutral. On this side, it's not neutral because we have two positive charges. So this side is neutral. |
Balancing Redox reactions example .txt | In other words, on this side of the oxidation half reaction, our charge is neutral. On this side, it's not neutral because we have two positive charges. So this side is neutral. This side is plus two. So to make this side zero, I have to add two electrons. That's exactly what I do. |
Balancing Redox reactions example .txt | This side is plus two. So to make this side zero, I have to add two electrons. That's exactly what I do. And now my charge is neutral and neutral because these two pluses cancel these two minuses. Now let's follow the same steps for this reduction reaction. So on this side, I have a minus one and eight pluses. |
Balancing Redox reactions example .txt | And now my charge is neutral and neutral because these two pluses cancel these two minuses. Now let's follow the same steps for this reduction reaction. So on this side, I have a minus one and eight pluses. That means our charge on this side is plus seven. Now on this side, my charge is plus two. So I want to neutralize my charges. |
Balancing Redox reactions example .txt | That means our charge on this side is plus seven. Now on this side, my charge is plus two. So I want to neutralize my charges. That means if I have plus seven here, I have to add seven electrons, and that's exactly what I do. Now, on this side, to neutralize this charge, I have to add two electrons. That's exactly what I do. |
Balancing Redox reactions example .txt | That means if I have plus seven here, I have to add seven electrons, and that's exactly what I do. Now, on this side, to neutralize this charge, I have to add two electrons. That's exactly what I do. I add two electrons. But now notice that I have two electrons here and seven electrons here. That means I could subtract two electrons from both sides. |
Balancing Redox reactions example .txt | I add two electrons. But now notice that I have two electrons here and seven electrons here. That means I could subtract two electrons from both sides. And what happens is this guy cancels out and this guy becomes a five, because seven minus five is seven minus two is five. So I'm done. Step five. |
Balancing Redox reactions example .txt | And what happens is this guy cancels out and this guy becomes a five, because seven minus five is seven minus two is five. So I'm done. Step five. In step six, I want to consider the fact that energy or charge is concerned, like mass is concerned. And that means I have to whatever amount of electrons lost by one atom or by one side has to be gained by another side. So notice on this side, I have two electrons, on this side, I have five electrons. |
Balancing Redox reactions example .txt | In step six, I want to consider the fact that energy or charge is concerned, like mass is concerned. And that means I have to whatever amount of electrons lost by one atom or by one side has to be gained by another side. So notice on this side, I have two electrons, on this side, I have five electrons. That means I want to find a factor or a common factor of five and two. And this factor is ten, because two goes into ten five times, and five goes into ten two times. So I basically multiply this whole guy out by five, and I multiply this whole guy out by two, and I get the following. |
Balancing Redox reactions example .txt | That means I want to find a factor or a common factor of five and two. And this factor is ten, because two goes into ten five times, and five goes into ten two times. So I basically multiply this whole guy out by five, and I multiply this whole guy out by two, and I get the following. So five times this guy gives me this, five times this guy gives me this guy five times, two H gives me ten H, and five times this guy gives me ten electrons. I multiply this whole guy by two. I get two of these guys plus 16 of H pluses plus ten of east equals two of these guys and eight of these guys. |
Balancing Redox reactions example .txt | So five times this guy gives me this, five times this guy gives me this guy five times, two H gives me ten H, and five times this guy gives me ten electrons. I multiply this whole guy by two. I get two of these guys plus 16 of H pluses plus ten of east equals two of these guys and eight of these guys. Now, when I go from step six to seven, I basically want to add these equations up, right? I want to add my two half reactions. So I add all the guys on the left side, and then I add all the guys on the right side. |
Balancing Redox reactions example .txt | Now, when I go from step six to seven, I basically want to add these equations up, right? I want to add my two half reactions. So I add all the guys on the left side, and then I add all the guys on the right side. So I get this whole guy plus nothing more on this side. So I go down to my other reactions, plus two of these guys, plus 16 of these guys, plus ten electrons. And now I'm done with everything on the left side. |
Balancing Redox reactions example .txt | So I get this whole guy plus nothing more on this side. So I go down to my other reactions, plus two of these guys, plus 16 of these guys, plus ten electrons. And now I'm done with everything on the left side. Now I move on to the right side. This guy equals ten carbon dioxide molecules plus two Mm, two plus molecules, right? Plus ten H plus molecules, plus eight water molecules plus ten electrons. |
Balancing Redox reactions example .txt | Now I move on to the right side. This guy equals ten carbon dioxide molecules plus two Mm, two plus molecules, right? Plus ten H plus molecules, plus eight water molecules plus ten electrons. So this is my equation. Now, my next goal before I get to my finalized answer, I have to cross out some things. The first thing I crossed out is a ten electron. |
Balancing Redox reactions example .txt | So this is my equation. Now, my next goal before I get to my finalized answer, I have to cross out some things. The first thing I crossed out is a ten electron. So we have ten electrons here and seven electrons here. That cancels out. Next, notice that I have 16 H plus ions on this side and ten plus H ions on this side. |
Balancing Redox reactions example .txt | So we have ten electrons here and seven electrons here. That cancels out. Next, notice that I have 16 H plus ions on this side and ten plus H ions on this side. That means I could cancel out this guy by subtracting ten H plus from this side and ten H plus from this side. So I canceled out, and 16 minus ten is six. So my final equation, my final balanced redox equation, is five H, two C, two four plus two M and minus one four plus six of these guys. |
Raoul’s Law for Ideal Fluids .txt | Today we're going to talk about roll slaw for ideal fluids. So what is a pure liquid? A pure liquid is simply a liquid that is not contaminated by any other compound or molecule. For example, suppose we have a closed system, a closed container, and inside this container we have pure water. But that simply means that the only types of molecules found within our system or water molecules. Now, what will happen to our system if it's left untouched? |
Raoul’s Law for Ideal Fluids .txt | For example, suppose we have a closed system, a closed container, and inside this container we have pure water. But that simply means that the only types of molecules found within our system or water molecules. Now, what will happen to our system if it's left untouched? Well, eventually, some of the water molecules found on the surface of the liquid will escape into the gas state and will become gas molecules. And this is called evaporation. Now, on the rate of evaporation and condensation are equal. |
Raoul’s Law for Ideal Fluids .txt | Well, eventually, some of the water molecules found on the surface of the liquid will escape into the gas state and will become gas molecules. And this is called evaporation. Now, on the rate of evaporation and condensation are equal. The system is said to be in dynamic equilibrium. At this point, we could measure something called vapor pressure. Vapor pressure is the pressure exerted by evaporating molecules or gas molecules found in dynamic equilibrium with the pure liquid molecules. |