This lesson provides a masterfully systematic breakdown of chemical separation, turning complex physical properties into a logical and accessible workflow. It is a highly efficient educational tool that prioritizes procedural clarity for student success.
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MARANDA Mock Exams 2026 Chemistry Paper
Added:Okay guys, welcome to our next revision lesson. So in our lesson today, we are going to discuss the Miranda High School Primoke exam chemistry paper 1 2026 and I would just like to say this this paper is on fire and the revision is going to be intense. Okay, so buckle up and let us begin. Now in our first question we are being told describe how the following mixture can be separated to obtain a crystal of the salt in bold.
Okay we have sodium chloride, ammonium chloride, iron fillings and lead to chloride. Do you know when I first went through this question I was like the salt in bold what is that? I I kept on tracking you know I didn't even proceed to the next part where they were listing the examples of the salt I was just like salt in bold salt in bold what does that even mean only to go further and realize that lead to chloride has actually been written in bold okay so this simply means that we are having a mixture containing the following four compounds okay not really compounds but the following four components we having sodium chloride ammonium ium chloride ion fillings. Okay, just tiny pieces of iron metal and then lead 2 chloride. So what the question requires from us is that by the end of all of this we should end up having lead to chloride. So we should be removing the other three components okay in order until we remain with lead to chloride. Okay. Now what I'm going to do is this. I am first going to explain and then we will go through the answer or how you're supposed to answer together.
Okay. Now the very first thing that we are going to use is uh we need to get rid of the iron fillings because I believe this is the you know the easiest part whereby we are simply going to take you know the mixture and then using a magnet you know use it to attract the iron feelings with the magnet because the iron feelings are magnetic. So we can just easily you know pick those up using using the magnet. Ignore my drawing.
I am very poor in drawing here. So that is the mixture and the block above that is our magnet. Okay. Yeah. You know this is something that I will actually like to improve. Uh I'm planning on taking tutorials of drawing at least so that when I'm illustrating you know whatever point it is during our lessons I don't make your eyes hurt. Okay. Like in this case anyways. So where was I? Yes. So I was saying that we are going to use a magnet in order to pick off the ion feelings. Okay? Because they are magnetic. Sorry, they are magnetic and will be easily attracted by the magnet.
But I need to point out something. If we were to place them in a beaker and they are, you know, piled up on top of one another, it's not going to be effective because what about the other pieces that are going to be below? Will they be attracted to the magnet? uh probably not because you know they're not going to be exposed to it. So what you're going to need is that we need to place them in uh you know the mixture in a container or maybe in a flat surface whereby as we are moving the magnet okay from one point to another it becomes very easy to pick up on the iron fillings. Okay. So essentially lay them on a flat surface so that they can be all exposed to the magnet that you're going to pass above them. Now by doing this we will essentially be removing ion feelings. So ion feelings done. We have removed them.
So that means we are left with sodium chloride and ammonium chloride. Now our next bit will involve ammonium chloride.
Now the reason for this is because ammonium chloride is a supplement. Okay, that means that on heating ammonium chloride it supplies. Okay, changes directly from a solid to a liquid, right? So if for example whatever mixture we are remaining with at this point uh not whatever mixture we we we do know the components that are present. Okay. So the mixture containing sodium chloride, ammonium chloride and lead 2 chloride was heated at this point. What will happen is that the ammonium chloride is simply going to sublime. Okay. Yeah. And the reason because it's a supplement. So when it sublimes you know in a toka in form of vapor. Okay. Now this can be collected by a watch class or not. But at the end of the day ammonium chloride yeah has been removed. So ammonium chloride game over your tomato. Okay. So what we are now going to be left with is sodium chloride and of course the lead 2 chloride just a minute.
So we have manage uh managed to eliminate ammonium chloride ion fillings. Okay ion fillings through the use of a magnet. Ammonium chloride by heating it sublimes and so you know it's simply released in form of a gas. So our mixture at this point is going to contain two things, right? It's going to have of course u we're going to have the lead 2 chloride and we are also going to have sodium chloride. Now we need to separate these two. Now we can do so with the introduction of water. So if we have let's say our beaker at this point okay with the mixture h and we add water and star what will happen is that sodium chloride which is very soluble in water is going to dissolve. Okay it's going to dissolve in water to form a solution.
Lead 2 chloride is insoluble in cold water. When we talk about cold water we simply mean you know water at room temperature. Lead 2 chloride is not soluble in cold water. It does dissolve but in warm water. So if you were to heat it, it would dissolve. But since we are not doing that, okay, our goal is to separate these two. We are going to use cold water stuff. Sodium chloride will end up dissolving. Lead 2 chloride will not. Okay. So at this point, we going to have a mixture. We're going to have a solution and a precipitate lead to chloride. So we can easily separate this using filtration. Okay. So if we were to filter, what would happen is that we are going to end up collecting the lead chloride as our residue. Okay. Because it's insoluble in water. And the sodium chloride is going to be our solution.
And there we have it.
Wow. Hey, those diagrams. Okay, bear with me.
Ah yeah guys let us put this into words.
So number one spread the mixture on a fl uh sorry on a flat surface and pass a magnet to attract the iron fillings to the reason for the flat surface you know is so as we can capture all we can attract all the ion feelings. Heat the remaining mixture to allow ammonium chloride to sublime.
Add cold water and stir to dissolve sodium chloride and then filter the mixture to obtain lead 2 chloride as the residue. Now, because the lead 2 chloride was coming from sodium chloride solution, we are going to wash it with distilled water to remove any traces of the sodium chloride solution. And then we will dry between filter papers. Now, I want to say this, and I always say this during my lessons. If you feel like you need more time, okay, to look at the questions or maybe to try them out by yourselves, pause the video, take as much time as you need. And when you're ready, we are going to be here. Hi guys, to our next question.
Now in our next question we are being told that chemical reactions occur as a result of collisions of particles give a reason for effective stroke fruitful or successful collisions of reactants.
Okay, let me explain.
Whenever we talk about particles reacting with one another to form products, we usually think of it as a simple uh process, right? For example, hydrogen reacting with chlorine to give us hydrogen chloride. In reality, what will be happening is that the reactants, okay, which is hydrogen and chlorine will need to collide with one another.
Okay, they will need to collide with one another with enough energy in order to bring about the formation of the product which is hydrogen chloride. Now, if that happens, we call those collisions successful collisions, okay? Or effective collisions because they have brought about the formation of a product. Now, sometimes particles do collide with one another but they don't have enough energy. Okay, this is what we call activation energy. they don't have enough energy to bring about the formation of a product. So they collide with one another and then they simply bounce away from one another. No product formed. Okay. Now in this question we are being told that why or how. Okay, not why. What is the criteria for forming effective collisions? For successful collisions to occur, the reactant particles need to have activation energy. So this is the minimum energy that is required by particles to bring about successful collisions. You know the formation of a product. So if they don't have activation energy or whatever energy they have is less than the activation energy then yeah no product is going to be formed. Okay. Part B. Explain the effect of increase in pressure on the rate of the following reaction. Now if we increase the pressure okay for the following reaction what will happen is that we are going to bring the reacting particles closer to one another. Okay let us pause there. Whenever we mention pressure it's usually tied to volume.
Okay volume and pressure have an inverse relationship. If one increases the other decreases. So for example in this case okay let's imagine initially this is going to be like uh okay my container okay this is going to be my container where I am going to have the hydrogen and the chlorine atoms inside the particles inside right so it's going to have a specific volume and a specific pressure now if I wanted to increase the pressure what I would need to do is I will need to reduce the volume Okay. So inverse relation relationship by reducing the volume I will be increasing the pressure of uh the particles. So in this case as you can see the volume is less and therefore the pressure will increase. How will the pressure increase? Because the volume has been reduced. What will happen is that the particles are now going to collide with one another more frequently. Okay.
particles, especially gaseous particles, are in continuous random motion. But if you have a container that has a small volume, the particles are going to collide with one another more frequently. So this will cause an increase in the pressure. Now by colliding with one another, we are also increasing the probability of the collisions becoming successful.
we are having a higher chance of formation of a product successful collision product formed. So if we reduce the volume okay we bring the reacting particles closer towards one another therefore increasing the chances of successful collisions happening and if we have successful collisions won't that increase the rate of the reaction yes it would. So by increasing the pressure in this case we are going to increase the rate of the reaction. Now I want to say this someone might be thinking in terms of equilibrium. Those are two different things. Pressure can affect the rate of a reaction. It can also affect okay the position of the equilibrium. In this case we are just focused on the rate of the reaction. If for example we had been asked okay material something similar to this notin you know the following are at equilibrium uh tell us how the reaction will be affected you know if we increase the pressure then yes we could talk in terms of that now if we were talking about how it would affect the equilibrium okay whether it will favor the forward or the backward reaction then yeah it will have no effect on the equilibrium because at the end of the day when it comes to equilibri We are simply looking for the side that has a larger volume or more moles. In this case, we are having one mole of hydrogen gas reacting with one mole of chlorine gas to form two moles of hydrogen chloride gas. So at the end of the day, yeah, we have an equal volume of gases on either side. So an increase in pressure will have no effect on the equilibrium, meaning that it will not favor the forward or the backward reaction.
But what are we discussing in this question? We are being asked for the rate of the reaction is a forward backward. It's simply just one reaction.
Okay, forward reaction and how it will affect. And so that is our explanation.
Okay, let us proceed to our next question.
So define a half life. Okay. Now, how do we define a halflife? This is simply the time taken for the mass of a radioactive substance to decay by half. Now, radioactive substances tend to break down. Okay? H they tend to break down over time. Now if you have an original mass and it breaks down to half the mass for example 20 g and then it breaks down it decays to 10 g the time taken for that decay is what we call the half life. Now I want to say this. I have a playlist dedicated to radioactivity.
Okay. I tackle the particles of radioactive. You know radioactive particles alpha, gamma rays. What am I talking about? Alpha, beta, and gamma rays. I also tackle a few application questions specifically those in KCS. So if you feel like this is uh something that you will need more clarity on, check it out. Part B. The table below gives the rate of decay for a sample of radioactive element P. Okay, let's look at the table. So we have mass of P in g.
We started out with 48 g. 48 g took how many days? It took okay at 0 days. So at 0 days this is our starting time. Okay, the first day literally.
So the mass of P decayed to 18 g and that took a period of 90 days. So after 90 days we it decayed from 48 to 18 and then finally 6 g and 6 g took another 90 days. So from initial to 6 g we took 180 days. So determine the halflife of the radioactive element P. Okay. Now if you look at the information provided it's not in order to say half the original mass. So if we had started with 48 the next one if it had been going in order should have been 24 and then 12 and then and six but in order. So what we going to do is we are going to condense this information but in order taking into account the half lives. So we are going to start with the first one. 48 g. Okay.
So, 48 g to half of it which is going to 24 g. Okay. That is going to be our first half life. 48 g to 24 g. Okay. So, okay. Uh the halfife we are supposed to determine the half life 24 g to half of it which is going to be 12 g. Right. And then the last one which is going to be 12 g again to 6 g.
Dingdong. We are going to stop there.
And why are we going to stop there?
Because we can clearly see six grams equal when table. Now if we condense this, that means that from 48 g up to 6 g, it took a total of 180 days. That is what we are told, right? From 48 to 60 to 6 g, uh it took 180 days. Now let's look at our half uh our half lives.
Sorry. How many half lives do we have there? We can clearly see we have one two and three. So we have three half lives. These three half lives have been spread across 18 days. Sing the way it's easy. So all we need to do is take 180 divide by 3 to get the equivalent of one half life. And that ladies and gentlemen is going to be 60 days. So that is going to be our half life. So it takes 60 days for a specific mass of P to decay by half.
Okay. Proceeding to our next question.
The table below shows the results obtained when a current of 2 ampers is passed through copper 2 sulfate solution for 15 minutes. Okay, Farah's law.
Calculate the quantity of electricity required to deposit one mole of copper.
Okay, now we are going to break down our workout. Okay, into three. Now we're going to start with the first one. Okay.
Now in the first one we are going to calculate the total charge that was used. Now when it comes to charge, charge is given by it. Right?
So I is going to be our current which is going to be right. So our current is 2 amp and then we are going to multiply that with the time. Now as we can see our time to power 15 minutes but time needs to be in seconds. So we're going to take 15 multiplied by 60. This will convert the time total time into seconds and then we will multiply it with the two in order to get our charge. And our charge is going to be 1,800 columns. Okay. So C simply refers to the unit of charge. So this is the total electrical charge that was utilized or that was passed through the copper 2 sulfate solution.
Now in our next step, what we are going to do is we are going to determine the moles of copper. Okay, the moles of copper that were used. Now when it comes to this information, let's go let's look at the table.
Initial mass of the cathode is 1 g.
Final mass of the cathode is 1.6 g. The change in mass at the cathode is 0.6 g.
Now when it comes to the cathode, the cathode is our negative terminal. Okay.
In electrolysis, we have the positive terminal and the negative terminal. The positive terminal is the anode. The negative terminal is the cathode. Now at the cathode is where we going to have reduction taking place. Okay. What is reduction? Reduction is gaining of electrons. So oil rig oxidation is the loss of electrons. Reduction is the gaining of electrons. Okay. So at the cathode we are going to have reduction and reduction is gaining of electrons.
So that means whatever ions end up at the cathode which in this case copper 2 ions. So the copper 2 ions that end up at the cathode are going to gain electrons.
Now they're going to gain electrons and they will be deposited as atoms. So the change in mass that we are seeing the 0.6 g is because copper atoms are going to be deposited at the cathode.
Specifically a mass of 0.6 g. Now we need to find out the number of moles that are present in 0.6 g and we can easily do that. By the way when we find out the number of moles we simply take the mass okay sorry the mass over the mass. Now the mass in this case is 0.6.
Okay even the change in mass is 0.6.
What is our marass? 63.5 g. Right. So the mar mass is simply the mass of one mole of a substance. RM of copper is 63.5 g. Okay. So that means one mole of copper will have a mass of 63.5 g.
They're just the same especi mass has unit. It's mass. So in a unit in a gram. So of course these cancel one another and we end up having an answer.
Okay, let me use my calc because if I remember any answer in many decimal digits. Hey, I was right. Okay, so we're going to have 0.
Okay, so those are going to be the number of moves. Now, someone was asking in the previous lesson, uh, how many decimal digits should we round it off?
If it's in the middle of a calculation and you are having an answer like this that has so many digits in there, just write as many as you can. Okay? Don't write all of them. This answer has so many of them, but as many as you can. Never ever. Okay? round it off to a value that has less than four decimal digits decimal places 0.09.
So we have now gotten our moles. Okay.
Now since we have now gotten our moles, we can proceed with the last bit. Okay.
Calculate the quantity of electricity required to deposit one mole. Now we have our charge. Okay. which is going to be just the quantity of electricity. We have the moles of copper that were deposited by the current by the charge.
So we can equate it. So how are we going to do so? Simple. So if for example just a minute if 0.94488 okay were deposited using a charge of 1,800 columns, right? Those were the number of moles of copper that were deposited by 1,800 columns. What about if we needed one mole? Okay, how much charge would we need? So, at this point, you simply cross multiply and you end up having your answer and you're done.
So, the first thing you calculate the charge. Now I will say this as a student and you know the first step but you don't know the second or the third step do the first step.
So when it comes to the charge it's very easy to calculate it. Okay mark try to do your best later on it's okay but you'll still get that mark and then we calculate the number of moles and then simply equate this. So our answer is going to be as such.
We give you a moment to go through it.
Okay, let us proceed now in our next question to Nambia. Using an energy cycle diagram, calculate the enthalpy change of formation of carbon dulfide.
Okay, so we need to calculate okay the enalpy of formation of carbon dulfide.
So carbon dulfide is simply as such.
So we've been provided with three equation and all of the equations by the way if you look at them they're all about combustion. The first one sulfur with oxygen giving us sulfur oxide. The second one carbon dulfide with oxygen giving us carbon 4 oxide and sulfur oxide. And the last one carbon and oxygen giving us carbon 4 oxide. Okay.
Now we are supposed to use these equations to get the enthalpy change of formation of carbon dulfide. Now when we talk about enalpy of formation what do we mean by this? Okay enalpy of formation.
Enthalpy of formation refers to the heat change that occurs okay when you have a substance being formed from its constituent elements. Now in this case if we are forming carbon dulfide we need to have carbon reacting with sulfur to give us carbon dulfide. So let us write down that equation. So carbon reacting with sulfur to give us carbon dulfide as such. Now if you look at the equation it's not balanced. Okay. So let us balance it. It's a simple equation.
Two in front of the sulfur and boom to.
So carbon of course is a solid sulfur is a solid and carbon dulfide is a gas the second equation it's a gas now this is going to be our first equation dulfide okay I am looking at the second equation carbon dulfide is going to react with oxygen to form carbon 4 oxide and sulfur oxide carbon dulfide.
Okay. Yes. As such. Now what is the arrow? What is going to be present at the end of the arrow? Simple. The products that are formed when carbon dulfide reacts with oxygen. So this is going to lead to the formation of carbon for oxide and sulfur for oxide. Okay.
As such, let us not forget oxygen.
Guys, let's balance the equation, right? We have at the beginning. Okay, we have CO2. What about Okay. Yeah.
We only have one sulfur. So, we are going to introduce two in front of the sulfa oxide to balance the number of sulfur atoms as such.
Now let's look at the last bit oxygen.
oxygen sour of course not let's count the total number of oxygen okay that are present in carbon 4 oxide and sulfur oxide to now one molecule 2 three we have three molecules okay yeah so that means we need to come over here where the oxygen atom the oxygen molecule is three so as to balance the number of oxygen molecules remember oxygen if we had written it in the form of an equation reactants like you can see above the equation provided we are now done with those two equations. Now we are going to complete the third and fourth bit. Okay. If we go back to the equations provided reacting with oxygen to form carbon 4 oxide sulfa reacting with oxygen to form sulfa 4 oxide. So all we need to do is simply draw arrows from carbon and sulfur to the products carbon 4 oxide and sulfur oxide and boom our energy cycle diagram is now complete actually not let's not forget oxygen and of course they need to be balanced right so oxygen carbon reacting with oxygen to form carbon for oxide everything is perfect there. What about sulfur? Sulfur reacting with oxygen to form sulfa for oxide to make up or two in front of the oxygen. And now yes, everything is balanced the way it should be. Okay guys, now this is our energy cycle diagram.
Okay.
He does not care whether the reaction will take place in a single step or multiple steps. As long as you start with the same reactants and end up with the same products, the enthalpy change will be the same. So what does say oxide? Okay, that is one path. Okay, now in the other path the direct way we oxidize them they reacted with oxygen and boom to capa carbon 4 oxide and sulfa 4 oxide. So according to H this will be the direct route to me carbon sulfa in one step to meala with carbon 4 oxide and sulfa 4 oxide. Indirect step is the blue one to meanza carbon sulfa.
We reacted them to get carbon 2 sulfide and then carbon 2 sulfide okay was burnt in oxygen to finally give us carbon 4 oxide and sulfur oxide. So according to H. Okay. Uh because both of these okay.
Okay. According to H because both of these occurred using the same pro uh reactants ended up with the same products. Yeah. Their enthalpy changes will be the same. So let me assign the enthalpy changes and then we create an equation from it.
So those are going to be enthalpy 1 and enthalpy 2 okay for the direct route and those will be enthalpy 3 and enthalpy 4 but it doesn't matter what you assign them who you come assign one who comes assign three if you plan it out correctly you should end up having the same value okay now according to hes and in this is enthalpy 1 plus enthalpy 2 should be equivalent to enthalpy 3 plus enthalpy 4. Using these we should end up having the same value according to H and H is right all you need to do is you need to fill in the correct values. So enthalpy.
Okay, let's look at enalp1. Where am I?
Where are we? Enalp1h.
It's carbon with carbon 4 oxide. That is going to be I need more space.
I need more space.
Okay. En carbon with carbon 4 oxide that is going to be 1072 kJ. Okay. Yeah. So let me try to fit it in here in the way I have big handwriting. Hey, so 10 72. Okay. So -1072 that is enthalpy 1. Okay. For carbon reacting with oxygen.
Now this we will add it to enthalpy 2.
Enthalpy 2. Enthalpy 2 sulfa sulfa 4 oxide that is -294.
But let us not be quick. Is it going to be -294 by itself? Of course not. To Kangalia, the enthalpy provided in the equation is for one mole of sulfur. Okay. Burning in oxygen to form one mole of sulfur oxide.
What are we having in the balanced equation in our energy cycle diagram? We are having two moles of sulfur burning in oxygen. So that means we will need to take -294 multiplied by 2 to make everything you know the way it's supposed to be multiplied by two.
Now on the other side we are going to have let's see uh enthalpy H3 enthalpy 3 ah that is what we after this is whom we are after and then we are going to have the last one enthalpy 4 now enp 4 is for combustion of CS2 Huh?
Oh my god. I have just realized something. I have realized something guys.
Why am I repeating 1072 again? Okay.
I used a wrong value. So let us just backtrack. Okay. Enthalpy one was supposed to be for combustion of carbon with oxygen. It's supposed to be -392.
I mistakenly took that of conversion of carbon dulfide. I have corrected it. There we go.
So we are going to add enthalpy 3 to enalpy 4 and enalpy 4 is that of the combustion of carbon dulfide which is going to be 10.
I'm trying to fit these things which is going to be where 1072 oh my lord do you know the reason I'm struggling I forgot to include the marking scheme for this particular question and so I have to do it and unfortunately with the constraints of space I am a bit challenged Hi guys. So I would like you to do me a favor, right? I want you okay to complete this. I have filled out all the values. So kindly complete it for me.
Okay. And uh I will provide the answer in the comment section in uh under the video. Right. Yeah. Thank you for your understanding.
So if you need to pause the video to undertake this and maybe look at it from another angle, uh, pause the video.
Okay. Yeah.
Okay. Question number six. The apparatus below was set up to prepare carbon 2 oxide gas. Use it to answer the questions that follow. Okay. Let's look at the diagram. So we are having a flask containing sodium methano. Okay. And above it we have a this funnel containing concentrated sulfuric 6 acid.
Okay. So essentially a reaction between sodium methano and con sulfuric acid.
Now we can see that our product at least the final product is a gas and that is carbon gas.
Okay. Identify one mistake in the setup above and the mistake is regarding the thistle funnel. Now when it comes to the thistle funnel, it has no tap. Okay, it has no tap and whenever you're using a thistle funnel, what you need to do is you need to have it immersed into the solution.
So in this case it was supposed to proceed into the solution and as we can see from our diagram that is not really the the case right it's hanging above and this is not correct because we are having a gas being produced carbon to oxide. So if it's hanging above the solution what will happen is that any gas produced will simply you know flow out. set up together concentrated sulfuric acid almost all of actually not almost all of it will flow into the flask below. So it will serve as an exit point for any gas produced and that is not our goal. We need to collect the gas produced. So essentially it beats the purpose. Okay. So the thistle funnel is not immersed in the solution. Now I want to say this. Okay. The thistle funnel is different from the dropping funnel. That is the dropping funnel. The one that you're seeing at the corner. Okay, it's our dropping funnel. Now, a dropping funnel has a tap as you can see. And the tap is very important because you twist the tap to open up the the opening so that the liquid can flow downwards.
Immediately you're done, you close the the tap and the exit is closed. So, okay. With a this funnel, we don't have a tap. So, you need to immerse it into the solution solution. What happens is that the solution comes up a bit in into the stem. Okay? The stem is just this part, okay? Comes a bit into the stem and it serves as a seal. Okay? And any gas that wants to pass through cannot pass through. Okay? Part B. Write an equation for the reaction that occurred in the flask. Okay.
I that is going to be our equation. We are going to have sodium methanoid reacting with con sulfuric 6 acid to form sodium hydrogen sulfate and we are also going to have methanoic acid. Okay.
So the last bit that is methaninoic acid HCO.
Now at this point I would assume that students are going to have questions.
Okay. So uh I'm assuming there are two questions because those are the only ones that I could think of that students will be wondering about. So number one we are having a sodium salt. Okay, this is sodium ethanoid. It's a salt of sodium reacting with sulfuric 6 acid and giving us sodium hydrogen sulfate. Why sodium hydrogen sulfate? Why not sodium sulfate? In most of the cases whenever we talk about you know a sodium salt reacting with sulfuric acid we end up having sodium sulfate. So in this case why are we having sodium hydrogen sulfate. Okay. Now when it comes to acids we know that acids usually you know dissociate in water to give us hydrogen ions. Okay. For example, hydrogen uh uh sorry, hydrochloric acid will dissociate in water to give us hydrogen ions and chloride ions. Right?
Now, in the case of sulfuric acid, sulfuric 6 acid, we are going to have a similar case. You're going to have sulfuric 6 acid dissociating in water to give us, you know, hydrogen ions and sulfate ions. This is what we are expecting. Sulfuric 6 acid is a dibasic acid similar two hydrogen ions. So a diabasic acid means that when it dissociates in water it releases two replaceable hydrogen ions diabetic di basic acid. Hydrochloric acid is a monobasic acid because there's only one replaceable hydrogen ion. Now in this case okay this will be what will be happening right? M so we expecting a case whereby sulfuric 6 acid will dissociate to form two hydrogen ions.
But what happens is that because the sulfuric 6 acid we are using is highly concentrated highly concentrated in that solution you are going to have more acid molecules and very few water molecules. Okay.
Yeah. Concentrated means that. So because of that, instead of us having sulfuric acid dissociating to give us two hydrogen ions, it's actually going to dissociate to give us okay one just a minute.
It's actually going to dissociate to give us one hydrogen ion and a hydrogen sulfate ion. Okay. So partial dissociation because there are few molecules water molecules that are present. For us to get complete dissociation we need to have a more dilute solution like what you use normally in the lab. But since we are using concentrated sulfuric acid we end up having um it dissociate to form hydrogen ions and hydrogen sulfate ions.
Okay. And that is the reason why we end up having sodium hydrogen sulfate. Boom.
I've answered your question. You haven't asked me about it, but I can read my telepathy. Okay. So, that is answer number one. Answer number two. Question number two will be you are telling us that the products are going to be sodium hydrogen sulfate and methaninoic acid.
Why then are we getting carbon 2 oxide?
Amma why are we collecting carbon 2 oxide as our product? Very good question. I might say the reason is because in this particular setup we are not only having one reaction but we are actually having two consecutive reactions in a funny and we get methaninoic acid present. Okay. Yeah.
That is the first step. Now the methaninoic acid that is produced is going to be in the presence of what? of whom? Of concentrated sulfuric 6 acid.
So yes, we do get methaninoic acid but because methaninoic acid is going to be present okay in concentrated sulfuric 6 acid con sulfuric 6 acid is a strong dehydrating agent. Okay, it's capable of removing water or elements of water from a compound. So what will it do? it will remove the elements of water from methaninoic acid such that we end up having water and carbon 2 oxide and that is where the carbon 2 oxide is actually coming from. Okay. Yeah. From the dehydration of methaninoic acid.
Part C. State two other methods used to prepare carbon 2 oxide in the lab. Okay.
So we are going oh I made that too fast.
Okay. Uh I did not include this. So let me just state it. So the other two methods that are used for preparing carbon to oxide is number one reduction by carbon. Okay. So if you have a setup for example where you are going to have um carbon 4 oxide it can it can be reduced by carbon and you end up having carbon 2 oxide. Okay. Yeah. So the hot carbon can reduce carbon 4 oxide to give you carbon 2 oxide. And number two is through dehydration of oxylic acid.
Okay.
two and our next question. Bromine reacted with compound Q to form a compound with the structural formula as such. Okay. So to require a compound Q it reacted with bromine and the product that is formed is whatever we are seeing over there. So write the structural formula of Q. Okay, we need to go back to what Q is. Now, how are we going to do that? Simple. By removing the bromine atoms because is in that you know that is what was added. So what was there initially? So what will happen is that we are then going to draw the structure the way it is. Okay, the carbon atoms that are present you know the hydrogen atoms that were there is up everything actually with the exception of the bromine atoms.
Now once we have that we are going to look okay at whatever is missing. For carbon to be stable it needs to have four bonds. Okay remember the electron configuration of carbon is 24. Okay. So it needs four additional electrons in order to become stable. So that means it needs to form four coalent bonds. Okay?
In order for it to become stable with each sharing an electron. Now in the first carbon atom the first to our left on this side if you look at the carbon on that side. Okay. How many bonds does it have? Four. A. Okay. What about the middle carbon? It has three bonds. one with the hydrogen atom above, one with each carbon atom on either side. That means it's missing a bond. Carbon atom we show on the right again similar case.
It's having three bonds. So what we need to do is we are going to introduce a double bond between the two carbon atoms. Now if we do that, everyone is happy and stable and balanced. So this is going to be the structural formula of Q.
Now if for example you were to draw it as such would that also be okay? Yes.
You know hydrogen atomishi it's okay as long as you placed the double bond where it's supposed to be between the two carbon atoms starting from the right.
Okay, let us proceed to our next question. In the lab, hydrogen sulfide gas is prepared by the action of dilute hydrochloric acid on a suitable metal sulfide. Okay, so we are having a metal sulfide reacting with dilute hydrochloric acid to give us hydrogen sulfide. So we need to write an equation for a reaction that produces hydrogen sulfide. So you can do it as such. Now if you're not I have written two equations. Okay. So the metal sulfide used can be zinc sulfide which will give you zinc chloride and hydrogen sulfide.
It's okay. You can also utilize ion 2 sulfide which will also give you hydrogen sulfide. Now I want to say this. You cannot use the sulfides of the more reactive metals. Can you imagine using sodium sulfide? Yes, you're still going to end up having hydrogen sulfide, but because sodium by nature is very very reactive, the reaction is going to progress very very quickly. You are going to end up collecting, okay?
Getting so much hydrogen sulfide gas that uh it's going to be challenging for you to collect and contain. Remember hydrogen sulfide is a toxic gas. So, danger danger. Now if you use other metals like uh lead sulfide or copper sulfide you know for the less reactive metals no reaction will take place because these are metals that are less reactive than hydrogen. So under normal conditions they will not be able to react with dilute hydrochloric acid. So you will not be able to get your hydrogen sulfide.
Okay. Part B. Write an equation to illustrate the reason why conch sulfuric 6 acid cannot be used to to dry the gas.
Okay. Now there is our equation. Okay.
When you use a substance as a drying agent, it needs to be inert towards the substance being dried. Okay. What do I mean by this? If I'm using kong sulfuric 6 acid as a drying agent, it shouldn't react with whatever I'm drying. Okay.
dry. I want to remove moisture from hydrogen sulfide. Do I want to end up having something else apart from the gas? No, I don't. So, I need to have a drying agent that will be inert, not in general. No, just inert to what I'm collecting, you know, just unreactive to what I'm collecting. So, because of this, I cannot use con sulfuric 6 acid.
It will end up reacting with hydrogen sulfide to form sulfur and water. A a big no. I want to end up having hydrogen sulfide but dry. Okay. Not something else entirely.
I part C. Give one chemical test for hydrogen sulfide gas. Okay. So hydrogen sulfide gas forms a black precipitate with lead to solution. Okay. So if you have lead 2 solution, okay, and you bubble hydrogen sulfide gas through it, it will react with the lead 2 ions to form lead 2 sulfide.
So lead 2 sulfide is the black precipitate. So this is our test for hydrogen sulfide gas.
Okay, proceeding to our next question.
The table below shows the tests carried out in a sample of water and the results obtained. Okay, let us look at the table together.
So sample A, we are adding sodium hydroxide dropwise until in excess.
Okay, initially we get a white precipitate which dissolves in excess.
Okay, what does that tell us? Zap is present, right? hair. So that means zinc, aluminium or lead to ions are present in the sample of water.
Sample B addition of excess aquous ammonia solution a white precipitate was formed. So precipitate that means the white precipitate persisted. Okay, it did not precipitate.
Okay. Now when we look at excess ammonia solution out of the three ions only zinc ions will dissolve in excess ammonia solution. Okay they will dissolve in excess to form a colorless solution. So that means that story because we got a white precipitate that did not dissolve in excess. So that means the ones that are present could either be aluminium or lead 2 ions iod addition of berium chloride followed by dilute nitric 5 acid. Okay. Now when we added berium chloride what did we get a white precipitate. So that means whatever salt was formed is actually an insoluble salt of bium. Okay. A white precipitate. So which salts are insoluble? Which berium salts are insoluble? We have berium carbonate, berium sulfate and berium sulfite. Okay.
Yeah. So the white precipitate that was initially formed tells us that it could be one of these three.
We added dilute nitric 5 acid. What happened?
White precipitate.
So insoluble in the p. Now if for example we had had a sulfite okayh what we would have observed is that we would have observed effvesence okay yeah bubbling would have been seen okay just a minute bubbling would have been seen with the evolution of a colorless gas okay reason being that sulfites okay sulfite ions would dissolve in an acid such as nitric 5 acid And we will end up having sulfur oxide being formed. So you will get the white precipitate dissolving and evolution of ais gas. Did we have that? No. So that means it's not bium sulfite. What about carbonate? If we had berium car berium carbonate what would have happened is that again it will also dissolve to form a solution with the evolution of a gas. What gas will we be having? Carbon 4 oxide. Now we didn't get any gas. We got a white precipitate insoluble in the acid. So that means sulfate ions are present.
Number two. Step number two. It could either be aluminium or lead ions. Right?
But I am here to tell you it cannot be lead to ions. It cannot be led to ions.
We just Oh, we made a but it's okay. So it's not going to be lead to ions. Why is it not going to be lead to ions?
Because when it comes to lead lead sulfate, our annion is sulfate ions.
When it comes to lead sulfate, lead sulfate is insoluble. If we had our sample of water and it had lead sulfate, it would, you know, it would actually have a precipitate before we even started the experiments already.
If lead sulfate had been present. Yeah.
A white precipitate has been formed when we already had a white precipitate to start with. So it cannot be lead sulfate. The only other alternative is that we are going to have aluminium ions. Okay. Yeah. So that if we have aluminium sulfate, aluminium sulfates remember is soluble z. Yes. All sulfates are soluble except for not berium. All sulfates are soluble except for oh my god I have forgotten lead sulfate.
Yeah, actually berium sulfate and calcium sulfate which is slightly solid.
So in this case my point is this it cannot be lead sulfate. Okay. Yeah.
So the only other viable alternative is aluminium ions are the ones that are present. So our salt the substance that was present is aluminium.
Write the ionic equation for the reaction in C. What is happening in C?
We are having berium ions reacting with the sulfate ions to form berium sulfate.
There we go. So that is our ionic equation. So the white precipitate we are seeing in part C is berium sulfate.
Roman number three. Write the formula of the complex ion in a. Okay. So the ion that was present or the cation that was present was aluminium. Right. Yeah. So aluminium if you add sodium hydroxide in excess will give you tetra hydroxo aluminum uh sorry tetra hydroxo aluminium ion. And there we have it. I I am a bit tired today. Don't know. Maybe I'm shrubbing a lot but if I don't do this video this week I will end up just postponing it to another week and another week. Yeah. So pardon me. Okay. Understand my situation. So that is the complex ion that is going to be present in A.
Okay. Question number 10. Study the scheme below and answer the questions that follow. Okay. So we are going to look at it together. So we start with ion feelings and steam. Okay. Now iron, you know, just iron. ion can react with steam uh and this reaction will end up giving us two products. We will end up having triion tetra oxide. Okay, the way it is as such and we will also end up having hydrogen gas. Okay, so those are going to be the two components that we will form. So to compound A and gas Q compound A is simply going to be triion tetra because tetra oxide because oxygen atoms are full and then gas Q of course is going to be hydrogen gas. Now in step two gas Q is going to be passed across hot lead to oxide. Now if you have lead to oxide okay and it is reacted with hydrogen we are going to end up having a classic redux reaction okay hydrogen is more reactive than lead. So what will happen is that it will react with oxygen okay to form water and lead to oxide is going to be reduced to lead metal okay so this is what we call the redux reaction. So lead oxide is going to be reduced and hydrogen is going to be oxidized to water.
Step number three. Oh, okay. That is that. Okay. Step number three. We are going to end up having solid B. Solid B2 will be the lead metal. Liquid C is just going to be water.
Guys, so part A, give the condition under which the reaction in step one occur. Okay. For this to take place, we need heat. Okay. Yeah. Because how are we going to form steam? By heating of water. Okay. until it evaporates to form steam.
Part B. Identify liquid C. It will liquid C is going to be water. Name the type of reaction taking place between gas Q and hot lead to oxide. The reaction is going to be redux reaction.
Do not write reduction. That will be wrong. You cannot have reduction taking place without oxidation occurring simultaneously.
So it's a redux reaction the reaction taking place. Okay.
Yeah. No no I can't think of something right now. Okay guys that was that.
Moving on to our next question. Describe how a sample of sodium sulfate crystals can be prepared starting with 20 cm of 0.2 molar sodium hydroxide. Okay. So we are starting with sodium hydroxide. Now if we are having sodium hydroxide, right? And we want to end up having sodium sulfate. It's simple what we need to do, right? We need to use sulfuric acid. So we going to react sodium hydroxide with sulfuric 6 acid and this will give us sodium sulfate plus water.
Remember a base and an acid gives you an a salt plus water.
Okay. So we need to balance our equation. It's not balanced as of yet.
And there we go. The mole ratio is very important. If you get a question and it's telling you how can you prepare okay uh a sample of a salt and then it goes ahead and lists a certain reagent but it gives you okay the specific volume or the specific concentration you need to do your calculations. Okay.
Yeah. So you need to do your calculations to ensure that at the end of the day the amount of uh reagent that you use is just enough not in excess and not you know not limited. Okay. So in this case we need to find out the specific okay concentration and volume of sulfuric 6 acid we need to use to get the correct number of moles. Okay let's pause there.
Our mole ratio is 2 is to one. So that means two moles of sodium hydroxide are going to be reacting with one moles one mole of sulfuric 6 acid. So let us calculate the moles of sodium hydroxide.
We can easily do that right considering we have the volume and we have the marity. So we going to take the volume multiplied by the marity over 1,00 and this will give us the number of moles of sodium hydroxide.
Okay. So the number of moles that we are going to get is 0.004 moles. Let me just confirm and I am unaware. [clears throat] Okay. It seems to be okay. So the moles of sodium hydroxide are 0.004.
What are going to be the moles of sulfuric acid? Half of those. Still no 2 is to one. So the moles of sulfuric acid that we need to utilize are going to be 0.02 0.002 moles half of those moles of sodium hydroxide. We need to ensure that complete neutralization takes place.
Okay.
Now whatever values we are going to end up using should ensure that at the end of the day the moles of sulfuric acid are going to be those. Okay. So for example if I used okay um the volume of sulfuric acid as 20 cm but I utilized a marity of 0.1 will that be okay yes it will be okay because at the end of the day this will give me the number of moles as 02 now if for example I utilized a volume of 10 cm m but a marity of 0.2 molar again this will also be accurate because it will give me the same number of moles of sulfuric acid. So what is my point?
My point is you can decide okay the volume and the concentration that you are planning on using as long as the number of moles you end up having are the same.
Okay.
So let us it add 10 cm of 0.2 molar sulfuric 6 acid to 20 cm of sodium hydroxide provided and stir. Eat the solution to saturation and allow it to cool to form crystals of sodium sulfate.
Wash the crystals and dry between filter papers.
Okay. I What did I do?
Okay.
Hi. Let us proceed. Question 13 part A.
define the term electrolysis. Okay, so electrolysis is the process by which a compound under goes chemical decomposition. Okay, it breaks down chemically uh through the passage of an electric current. So if you pass an electric current, what will happen is that a compound okay if it's an electrolyte will will carry out chemical decomposition.
Part B.
Aluminium oxide is electrolyed using graphite rods. Write the equation of the reaction at the anode and at the cathode. Okay. There is going to be our equation. Okay. Now I want to do let us do this. Okay. Let us do this and let it serve as you know just revision. So what I'm going to end up starting with is number one. The ions that are going to be present are simply going to be two.
Okay, we're going to have aluminium ions and oxygen ions. Okay, it's a binary electrolyte. It only has two ions, a single cation and a single anion. So graphite very good. It won't interfere with the product. Okay. Yeah. So at the anode it's our positive terminal. The cathode is our negative terminal. So at the positive terminal we are going to have oxygen ions proceeding to this. Why did they even draw a line in between these two? Okay. So we're going to have oxide ions proceeding towards the anode. Now to mema what will happen at the anode is we are going to have oxidation taking place. Do you remember the pneumonics an ox and then we have the other one?
I for that red cat guys one like. Hey by the way if you have a friend a classmate or whatever if you are in a group where there are other students please share the videos or share the link to the YouTube channel. I have found so many students that uh felt you know disappointed that they were not aware of the channel in due time. Okay. Yeah. And the reason we are doing this is so as to help the students and if we don't get to our target audience. Yeah. In short if you know of someone who could benefit and there are so many students out there share the link with them. Okay. Let us share knowledge with one another. Let us not restrict it. If you have something that could help a fellow candidate or a fellow student share it. Okay. So we are saying that at the anode we are going to have oxidation taking place. At the cathode we are going to have reduction taking place.
Now this is for students who sometimes get confused whether oxidation or reduction occurs at the cathode and the anode. These pneumonics are for you. So oxide ions are going to undergo oxidation and oxidation is simply the loss of electrons. So we are going to have it as such. Okay.
[snorts] Okay just a bit.
It's going to be as such. A single oxygen ion is going to lose two electrons in order to form an oxygen atom. But oxygen oxygen tends to be diatomic. You have two oxygen atoms.
They bond with one another to form a molecule. So what will happen is that we are going to have a molecule O2. So that means we need to change a bit regarding the electrons. Okay. So for us to form an oxygen molecule, we need to have two oxygen ions to start out with. Okay. Each ion will lose two electrons. So that at the end we are going to have four electrons being lost.
Cath at the cathode we are going to have the aluminium ions. Okay remember cathode negatively charged aluminium ions positively charged. So they are going to be attracted towards the cathode. Cation and aqua attracted towards the cathode. So what will happen at the cathode? at the cathode is where we are going to have reduction and reduction is where we're going to have gaining of electrons right so the electrons are going to be gained by the aluminium ions and we are going to have aluminium atoms being formed now in this case what will happen is that if you look at our equation our half equations they are not balanced but if you write them as such it's usually acceptable if you balance them Even better kubuka the electrons that are lost at the anode are the ones that are gained at the cathode. So it could be better if you balanced the total number of electrons. How are you going to do?
Just look at the electrons that we have.
4 3. So you look for the LCM which is going to be 12. Okay. Yeah. So 12 will give us the equivalent number of electrons. So if we take the first one multiplied by 3, we will end up having 12 electrons. Okay. So we are going to have a case where 2* 3 we are going to have six oxygen ions and three oxygen molecules and we have 12 electrons as such.
Now similar case we are now going to multiply by 12. Okay. So this is going to be our half equation.
Now my point is this. You can write it simply without balancing the equation or you can balance. I find balancing the equation to be the more accurate version. Yeah. Because 12 electrons being lost at the 12 electrons being gained at the cathode. Boom. But writing without balancing such as in this case is also okay.
Okay. Proceeding to our next question.
Study the flowchart below and answer the question that follow that follows. Okay.
We are having copper pyite. Okay. That is the chief O of copper. So this is extraction of copper. Now I am going to say this I will just pass through this.
Okay. Yeah.
And uh the reason being that I have playlist dedicated to metals.
It's metals. I have talked about metals the six individual metals that are tested in our syllabus. I've also tackled past questions on metals. So feel free to go through them. Just a minute.
Okay. So, write an equation for the reaction in step two. What are we having in step two? We are having the copper pyite. Okay. And it's being oxidized.
Okay. Fortnite are roasting. and it's being roasted in air to give us copper 1 sulfide and ion 2 oxide and sulfur 4 oxide. Okay. Now when it comes to copper pyate actually when it comes to all OS they tend to come with impurities. Okay.
Now a way in which we use to minimize the impur in the case of copper is a process that is known as froth rotation.
Okay. In froth notation, what simply happens is that we are going to have air bubbled into copper pyates. Okay, you're going to have uh what will happen is that you are going to have a setup similar to this and then you're going to have water and then your O. Okay, the O has been crushed into tiny tiny bits. Okay, in order to increase its surface area. So, we're going to have uh water plus uh some special oils like pine oil added onto it. And then what happens is that we are going to have air bubble into it.
So, you're going to have air bubbled into it and then the air because it's being bubbled under high pressure creates a froth for you. Now that froth is going to contain a high concentration of the pilate of the minerals we are after takataka soil clay other impurities are going to settle at the bottom in form of gang okay ear waste and then what we want the minerals that we are after are going to be in the okay hi sasa this process is known as froth flotation right I Now after froth rotation you are going to have the pyrite which is the or being roasted. It's going to be heated under very high temperatures in the presence of oxygen and it will form the following three products. So product number one is going to be sulfur oxide. Similar sulfa 4 oxide. Yeah. Okay. Product number two is going to be ion 2 oxide. Now solid X is actually going to be ion 2 oxide in this case. And then lastly we are going to have copper one sulfide. Okay that is our main compound the one we are after.
Now usually when we talk about copper compounds we usually focus on them having an oxidation number of two. So copper 2 sulfate copper 2 nitrate copper 2 chloride. But the copper that is present here is having an oxidation number of one. Copper can have both. It can have oxidation number of one or oxidation number of two. In this case it has an oxidation number of one copper 1 sulfide copper one oxide as such.
So that is our first equation. Okay.
Part B. Identify solid X and process W.
Okay. I process X. Where is process X?
Process X is over here. No, that is process W. Where is Oh, solid X. Sorry.
solid X to be identified that was supposed to be ion 2 oxide and process W that is going to be electrolysis. So the copper produced is not going to be completely free of impurities. So in order to purify it further it needs to be electrolyed. Okay. And then we end up having pure copper metal.
Okay. Proceeding to question number 17.
Aluminium chloride is dissolved in small quantities of water in a test tube.
White fumes are observed or were observed. Roman number one explain the observation above. Okay. Now when it comes to aluminium chloride okay essentially what happens is that in the presence of water it hydrayes. Okay. To form hydrogen chloride gas. Now I want to say something unique about aluminium chloride. Whenever we talk about ionic compounds, they are mostly formed when a metal reacts with a nonmetal, right?
Because metals tend to prefer to lose electrons in order to become stable.
Non-metals tend to prefer to gain electrons in order to become stable. So between these two, you end up having one losing one gaining the electrons that are lost and you have an ionic component. Now in the case of aluminium chloride one would expect to have the same thing happening. Okay, aluminium is a metal. Uh, chlorine is a nonmetal. And between these two, we why not why not have um what why not have an ionic component?
But in the case of aluminium ion, it's a bit unique because when it comes to the aluminium ion, number one, it's quite small. Okay, it's smaller than magnesium. It's smaller than sodium, I mean the atom itself. And then number two, it has a very high charge. Okay, a high positive charge and that is 3+ the ion. Okay, so it's going to be attracting the electrons to itself. So what happens is that instead yeah instead of us ending up having an ionic bond yes you're going to have aluminium ion you know initially electrons ina because chlorine atoms are highly electrogative but is electron they are still going to be attracted towards the aluminium ion because of its you know small size and because of its high positive charge. Remember electrons are negatively charged. So instead of having the electrons ending up in the chlorine atom, they will just be shared in between the two. So aluminium chloride is going to be largely coalent instead of actually being ionic. Okay?
Yeah. Electrons are going to end up being shared between the two. Now when aluminium chloride is dissolved in water, it hydrayes to form hydrogen chloride gas. Okay? Now in this question we have been specifically told that the amount of water that was used was quite small. Okay and this is important to note. So if you have aluminium chloride dissolved in enough water okay it will form hydrogen chloride gas which will dissolve in the water and we will end up having hydrochloric solution a fumes okay because enough water is there hydrochloric acid is going to be formed because hydrogen chloride gas will dissolve in the water present now in this case what we having is we are having a small quantity of water being used so aluminium chlor chloride is is going to hydrayze in water to form hydrogen chloride fuels. But because the amount of water present is not enough to fully dissolve all of the gas, it's going to escape into the atmosphere.
Now once in the atmosphere, it's going to dissolve in the moisture present to form hydrochloric acid. So the fumes that you're actually going to see are going to be, you know, hydrogen chloride gas. But yeah, and hydrochloric acid forming droplets of that was quite a lot. So aluminium chloride hydrayes in water to give hydrogen chloride gas fields. Roman number two, write a chemical equation.
Oh, by the way, let me pause there. If it was not clear, we are getting white fumes. Okay, okay, white fumes. The white fumes are the hydrogen chloride gas fumes. Okay, the reason we are having fumes is because the quantity of water used was not enough to dissolve all the hydrogen chloride gas that was produced. So, some of it will escape.
Roman number two, write a chemical equation to show the reaction taking place in the test tube. Okay. So, aluminium chloride will hydraulize in water to form aluminium hydroxide and hydrogen chloride gas.
So, the hydrogen chloride gas okay is the one we are seeing as fumes.
Hi, Roman number three. State the observation which can be made if little if a little [clears throat] sodium carbonate is added to the resultant solution. So if sodium carbonate was added, we are going to get bubbles of a catalyst gas. Now the solution is going to contain aluminium hydroxide but it will also contain hydrogen chloride gas that will dissolve to form hydrochloric acid. Okay. Yeah. So to add sodium carbonate, a carbonate will react with an acidic solution to form carbon oxide.
So bubbles of a colorless gas are going to be seen.
Okay, proceeding to the next question. A student reacted 0.5 g of zinc granules with 2 m hydrochloric acid and the volume of hydrogen gas produced was measured at various intervals. Okay. So if we have zinc, of course it's going to react with hydrochloric acid. Zinc is a metal. hydrochloric acid. This will form a salt which is going to be zinc chloride plus of course hydrogen gas.
Right? So we can measure the rate of the reaction by the volume of hydrogen gas produced. The more hydrogen gas produced, the faster the rate of the reaction. Okay. A sketch graph of volume against time is as shown below. Explain why the graph is steepest at the beginning. A steep graph simply means you know eco you know steep you know steep I I don't even know how to explain it but yeah a steep curve as such okay now the reason why we are having a steep curve okay as we are seeing here is simple it's at the beginning of the reaction and at the beginning of the reaction we are going to have the con you know the highest concentration of the reactants which reactants are these zinc and hydrochloric acid. So they're not yet used up. So more of them will be reacting with one another producing a larger volume of hydrogen gas. So as the reaction proceeds, the reactants are going to be used up and therefore less hydrogen gas will be produced until finally it stabilizes.
So why the gas is steepest at the beginning? concentration of the reacting particles is highest at the beginning leading to more successful collisions.
The more reactant particles we have, the more chance we have of them colliding with one another leading to the formation of hydrogen gas. Therefore, the rate of the reaction is going to be highest.
Question number 19. The grid below is part of the periodic table. Use it to answer the questions that follow. Okay.
So, take a moment to go through the grid. Identify the elements that have been labeled. So, number which letters represent alkali metals? Now, alkaline metals are found in the first column.
Okay, these are the group one elements.
So, that means H and P. But it's not going to be H because H is actually hydrogen. Hydrogen is also placed in group one because it has one valence electron. It literally has one electron.
So it's placed there but it's not an alkali metal. So that means the only correct answer is N or P. By the way, the answer should be N and P. Which letter? Which letters? So you have to include both of them. Okay. Part two.
Select a letter which represents the element with the highest ionization energy and then give a reason why. Now when it comes to ionization energy, ionization energy refers to the energy that is required to remove an electron from an atom in its gaseous state. Now normally an electron is going to be highly attracted by its nucleus, right?
So we need energy in order to forcefully pull out the electron away from uh you know away from away from uh the atom. I'm just experimenting with the diagram.
Looks crooked though.
Anyways, so coming back, all of these, by the way, all of the elements here are going to have ionization energies. It's just a matter of which has more and which has less. Okay. Now, how do we know which has a higher ionization energy? Simple. Look at the one that is going to be small. And in this case, that is stable. The only stable one that we have here is you know a group eight.
That means it has a stable or octed configuration. So if it's stable, hey the amount of ionization that you will require energy to pull that electron will be the highest of any that is present there. Okay. Now I want to give an example. Let's imagine U was not present. Okay. Hu. We were left with the other that are unstable. What would our answer be? The correct answer would be S. Okay.
Now when it comes to the ionization energies they are highest in the atoms that are smallest as you are moving from the right to the left side of the periodic table the atoms tend to become smaller okay so that means the atoms that are present on the right side oh that is the right side I've switched them actually switch okay so as you're moving from the left to the right side of the periodic table, the atoms become smaller and smaller. So that means S is going to have the smallest atoms out of all of that.
So the highest ionization energy if you was not present but the answer should only be you.
Okay.
Roman number three. The product formed when potassium reacts with excess oxygen is known as potassium peroxide. Write an equation of the reaction that occurs when the product is dissolved in water.
Okay, let's stop there. Potassium can react with oxygen to actually form three oxides. Okay, the first oxide is that that is the one we're used to, right?
Potassium oxide, right? Potassium has a valency of one. Oxygen has a valency of two. When these interchange, you end up having K2 potassium oxide.
And then you're going to have potassium peroxide. Okay, potassium peroxide is going to be written as such. K2 O2. We are usually more familiar with sodium peroxide. So, sodium peroxide min2.
potassium peroxide and then now we have potassium superoxide. Now potassium superoxide is written as such KO2 KO2 KO2. So if we need to write an equation between potassium peroxide and water it's going to be simple.
Remember all oxides of group one actually most oxides of metals tend to be basic. Okay. So they will react with water to form potassium sorry to form hydroxides. So in this case it will react with water to form potassium hydroxide and of course we are also going to have oxygen. Yeah oxygen.
H to a delay.
Oh, just a minute. I want to check on something.
Okay, I'll check on it later. Okay, to our next question. Now in our question in our next question to Nambia when a stream of carbon 2 oxide is passed over heated oxide of ion until no further change in mass the following data was recorded. So we have the mass of the crucible 30.29.
Okay. So the crucible is simply our container. So that by itself has a mass of 30.296 higher. If you add the oxide inside of it, you are going to end up having a total mass of 33.70.
So yeah, so the difference between these two should give us the mass of the iron oxide.
So these two the mass of the iron oxide and then the last value the mass of the crucible plus the residue. Okay. So, if you have a setup such as this and you heat the iron oxide, what will happen is that uh it's going to decompose and you're going to end up having just the iron metal. Okay? So, oxygen yeah in a layer. Oxygen in a layer.
Okay. So if we want to get the values uh sorry not the values the mass of the ion and the mass of oxygen we can simply do so using the values provided to so the mass of the ion 2 oxide how do we get it we simply take 33.709 minus the mass of the crucible the difference will give us the mass of the ion to oxide. Although to be honest, we do not need this. Okay, what we need is the mass of the ion itself and the mass of oxygen in order for us to get the empirical formula. So the mass of the ion, what is the mass of the ion?
Okay, the oxide, it's not really oxide because we have not yet determined conclusively that it is actually iron 2 oxide. It's actually notion.
Okay, I don't want to say to give the hint away, but the mass of the oxideion [snorts] oxideion oxide, we don't know it. We just know it's an iron oxide. So, the mass of let's look at the where are we? So, how do we get the mass of the ion? Simple, right? A ha ha. Last one. This is the mass of the crucible plus the residue after oxygen has been lost. So if we take this okay subtract the mass of the crucible that should give us the mass of the ion. Okay. Yeah. That should give us the mass of the ion. So this is the mass of the ion. Now the mass of the oxide.
How are we going to get the mass of the oxygen that was present in the oxide?
Simple. We are going to take the mass of the crucible plus the oxide and then minus the mass of the crucible okay with the residue. So the difference between these two should give us the mass of the oxygen.
Okay guys so those two the masses of iron the masses of oxygen and we can now start.
So there we have it.
So you'll take the mass divide by the relative atomic mass. The relative atomic mass has been provided for over here. That of ion is 56. That of oxygen is 60. So the mass of ion divided by the mass of oxygen divided by the we get our moles. We're going to divide by the smaller value to get the simplest ratio.
Our ratio is going to be 1 is to 1.5.
We cannot have whole numbers when atoms are concerned. So we are going to multiply by two. Okay. To get rid of the decimal place. So at the end that will give us 23. So our oxide is actually going to be ion 3 oxide. Fe23.
Okay, proceeding to question number 21.
In an experiment to determine the molar heat of reaction when magnesium displaces copper, 0.15 g of magnesium metal was added to 25 cm of 2 m copper 2 chloride solution. The temperature of the solution rose from 25 to 43°.
Part A. What is meant by the molar heat of displacement? So this is the enalpy change. Enalpy change simply refers to the heat change that occurs. Okay?
Whether the heat the amount of heat energy that has either been lost or gained in a particular reaction. So this is the enthalpy change that occurs when one mole of a substance is displaced from a solution of its ions. Okay. So we have to specify because heat of displacement. No remember the molar heat of displacement. So this is actually going to be the heat of displacement of copper. Okay. It's the copper ions that are being displaced from their solution.
By whom? by magnesium because magnesium is the more reactive metal.
Part B, determine the molar heat of displacement of copper by magnesium. Now for this we utilize our equation. Okay.
So, so the molar heat over displacement the enthalpy change sorry is given by the mass of the solution multiplied by the specific heat capacity multiplied by the change in temperature. What is the mass of our solution? Which solution? By the way, we are not using magnesium. Okay?
Magnesium is not being used at this point. When it comes to the mass, it's referring to the mass of the solution.
Now there's only one solution and that is copper 2 chloride solution. So the mass of the solution can be given by taking the volume multiplied by the density density, right? The volume is 25. 25 * 1 will give us the mass over the solution as 25 g. But we are not going to stop there because I want us to make a note of the specific heat capacity and this is very very important. Look at the specific heat capacity specific heat capacity has been given in 4.2 kg per kilogram per kelvin.
You need to make a note of this. Okay.
The value 4.2 never changes but the units. So sometimes you can be given as 4.2 2 J per g per but it can either be J or KJ g or kilog so you need to make a note this and why is it important because if it's given in kilogram you need to convert the g or the mass into kilog how do you do that simple by dividing by a th00and so our mass over here divide by a th00and to convert the g into kg we will then multiply by the specific heat capacity and lastly by our change in temperature.
So 43 minus 25. Now sometimes students we are used or are used to having the initial minus the final. So 25 minus 43 and then they end up having a negative value and it confuses them. It should we are just talking about the change in temperature. We do not care whether initial initial we just want the change just the change. Okay. So just take the larger value subtract the smaller value from it. So okay and if you take all of that what will happen is that what value am I getting? Uh 1890. Okay. Is it 1890?
No it should.
Okay. Allow me to do this.
>> [snorts] >> time 4.2 * 80 Oh my lord. Working under pressure.
Oh my lord.
Uh where am I?
You're probably thinking, "Oh my H okay I finally quoted the value do not judge do not judge [laughter] so we are going to end up having our value as 18.9 kilogjles okay 18.9 kJ is that actually correct let me just do it one last time yeah just to ease myself.
You see, you see my Lord.
Yeah, there we go. So, let me just rub this off.
So it's going to be 1.89 kJ. There we go. Now we write as such no matter the size that will be considered as a capital letter. Okay. So our answer is going to be 1.89 kiloj. Okay. This is the amount of heat energy that was evoked. But you're not going to stop there, right?
This is an exothermic reaction. How do we know it's an exothermic reaction?
Because we can clearly tell that temperature rose from 25 to 43. So an increase in temperature translates to an exothermic reaction and we assign it a negative sign. Now we are still having the second part. Okay. Now in the second part that is the easier one. We will simply be calculating. Okay. Let me just check. Yeah, it's there.
So what we will do is we are going to determine the number of moles that are present in magnesium. Okay. So once we have determined the number of moles that are present in magnesium, we will then equate them to the heat energy. Right.
Question number one. Determine the molar heat of displacement of copper by magnesium. So there we go. So the moles of magnesium. Okay, we are going to take the mass over the mass over the okay over the mass. So our mass is 0.15. The molar mass is 24.
Divide this and we get the number of moles as 0.625.
Our last step, what do we need to do now? 0.00625 moles of magnesium are the ones that are responsible for the evolution, okay, of -1.89 kJ.
What about one mole? Okay, what about one mole? What would one mole get us?
So, you'll simply take 1.89 over 0.625 625 and that will give you your final answer. So the final answer is going to be -32.4 KJ per moj because aren't we after the heat of displacement? Yes, we are now because the reaction is exotic.
So that is the question that is the answer.
Okay. Question number 22. Figure six shows a setup of apparatus used to prepare a sample of nitrogen gas in the lab. Study it and answer the questions that follow. Okay. So we have solution X and sodium nitrite. Okay.
Now part A describe how nitrogen gas is formed in the flask. So what will happen is that you are going to have sodium nitrate reacting with ammonium chloride.
Okay. Yeah. Solution of X is simply ammonium chloride. So it's going to react with ammonium chloride to form ammonium nitrate. Now ammonium nitrate because we have heating taking place is then going to decompose okay to give us our nitrogen gas. So essentially we have two reactions taking place. The first one sodium nitrate reacting with ammonium chloride. This will give us ammonium nitrate. The second reaction ammonium nitrate undergoing thermal decomposition. Okay. It's going to break down in the presence of heat to give us nitrogen gas.
What property of nitrogen gas makes it suitable to be collected as shown? Okay, it's being collected over water. The reason why this is possible is because nitrogen gas is only slightly soluble in water.
Okay, our next question. Use the information below and answer the questions that follow. The letters are not the actual symbols of the elements.
Okay.
So, let's look at the first one.
So, we have element E, F, and G. Now, if I was to take a guess, okay, not a guess.
It's actually is it I guess it's not a guess. It's based on the uh the information provided. These are going to be metals. When it comes to the standard electro potentials, if they are assigned a negative value, that means that they have a lower tendency to gain electrons compared to hydrogen. Now, when it comes to metals, okay, metals prefer losing electrons, right? They prefer losing electrons in order to become stable.
Non-metals prefer to gain electrons in order for them to become stable. Now, if you were to compare hydrogen, which is a nonmetal, to a metal like sodium, okay, and you want to know between these two, which is more likely to gain electrons, it's definitely going to be hydrogen.
Okay, hydrogen will have a higher tendency to gain electrons than sodium.
So because sodium's tendency to gain electrons is lower than that of hydrogen, it's assigned a negative value. Now in most cases, metals toma lose electrons to become stable. So compared to hydrogen, they will have a lower tendency of gaining electrons. So they will be assigned a negative value.
In fact, the only exception to this rule is copper. And the reason is because copper is a metal, but it's reactivity is very low. So copper compared to hydrogen will actually be will have a higher tendency of gaining than hydrogen. But as for the rest negative negative negative now we need to calculate okay the uh the electromotive value.
What is that? Oh the standard electro potential of x in the electrochemical series. Okay in the electrochemical cell. Now if we look at this we have two electrochemical cells right we have that one yeah F and then we have the second one yeah G and the arrangement is very important okay whichever is usually on the left okay represents the anode literally the left hand side whichever is on the right represents the cathode.
Now this is important because it lets us know which will undergo oxidation and which will undergo reduction. So our formula is going to be as such uh the electro the emf value of a cell is given by the standard electron potential of the cell that under goes reduction minus the one that under goes oxidation.
So we've already identified the one that goes reduction and the one that goes oxidation the emf of the whole cell will be the one at the cathode at the anode is where oxidation takes place at the cathode is where reduction takes place. So reduction.
Okay.
Now F will be the one okay at the anode where oxidation takes place. So now once we've reached that point it's simply just a matter of making X the subject of our formula and we will end up having the correct answer. So let us do so.
There we have it.
So if we if we make x the subject of our formula x plus 1.22 you simplify it and you end up having a value that is1.66 vol. Yeah.
So fu and a negative 1.66 66.
[music] I am Okay, let us proceed to the next bit. Part B. Arrange the elements in order of reactivity starting with the least reactive. Okay.
1.66 volma negative value means that they have a lower tendency compared to hydrogen of gain electrons. So what you'll not reactive a metal is the higher the negative valuery.
So the higher the negative value that means the more reactive the metal is. So out of these three more reactive that is obvious it's F reactive followed by E and then last one this is going to be the least reactive question arrange the elements in order of reactivity starting with the least reactive we need to look for the one that is the least reactive followed by E and Then finally F F being the most reactive.
Part explain if it will be advisable to store element G in a solution containing E ions. Okay. So let us imagine we are having a container as such and in this container we are having a solution containing E ions. Okay that is what we're being told.
Okay, element G. Would it be safe for us if we stored element G? Okay, in a solution containing E ions. Now, is it safe? They don't mean is it okay? They mean that will there be a reaction taking place at the end of it? Are we going to have G as G remaining as G. So to what we have okay uh let us look at the reactivity between G and E which is more reactive definitely E is more reactive is the least reactive less reactive more reactive will be any reaction taking place. No, they will not. If I was to have a case whereby I took magnesium metal, place it in a solution of copper 2 ions.
Magnesium is more reactive than copper 2 ions. Oh, actually it it should have been the reverse. It should have it should have been the reverse. here. So let us reverse it. Okay. So we are having a solution containing magnesium ions and these have been and we have placed copper metal inside. Copper is less reactive than magnesium. So if we place copper inside the solution of magnesium, no reaction will take place. Copper does not have the power to displace magnesium. So everything will remain the way it is. So G is less reactive than E.
So it will not be able to displace E from its solution. So that means no reaction will take place and therefore yes it is safe to store G in a solution containing E ions.
Okay guys, that brings us to the end of our lesson today and I will be seeing you next time when we revise a new paper and Yeah. Nagaza. Okay. Yeah. Because I don't want to disappoint my students.
Yeah. I need you to pass. I need you to pass in biology and chemistry. So hopefully next week I will have another video ready for you.
Yeah, guys. Bye-bye.
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