This video provides a comprehensive walkthrough of the Physical Sciences Paper 2 Grade 12 June Exam 2026, covering key topics including organic chemistry (hydrocarbons, functional groups, isomerism, reaction types), chemical equilibrium (Le Chatelier's principle, equilibrium constants), reaction kinetics (rate calculations, percentage yield), and acid-base chemistry (titration, pH calculations). The exam requires understanding of fundamental concepts such as unsaturated hydrocarbons (CnH2n), esterification reactions, combustion equations, and the effects of catalysts on activation energy without changing enthalpy.
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PHYSICAL SCIENCES PAPER 2 GRADE 12 JUNE EXAM MEMO 2026 PHYSICAL SCIENCES PAPER 2 MEMO THUNDEREDUC
Added:[music] [music] [music] >> Good take a toss today. We are going to look at a June exam 2026.
Uh paper two. So scrolling down there, we are going to have our instructions. And then starting with the multiple choice questions, we are going to start with 1.1. Okay, so here the section reads It reads here, "Which of the following is an example of an unsaturated hydrocarbon?" Right, so we're looking for a hydrocarbon that is unsaturated. It means it has got multiple bond between carbon to carbon what atoms. Right, so if you look at that one, option A, it has got a Cl3 there. And then this other one also has got an oxygen there, which makes them no longer hydrocarbons. Remember hydrocarbons are carbon containing compounds that actually contain carbon and hydrogen atoms only. So in terms of this, the one that is unsaturated, looking from the general formula, we must have a CnH2n like that. Okay, I'm going to start with that one, the first one there, which is our option B. It means I'm going to have a C4 where n represents the number of carbons. The 1.2 H there, then 2 multiplied by How many carbons do I have there? Four. Therefore, it gives me a general formula the formula of what? C4H 8 there. So it means our correct option there is going to be option B. Then moving on to the next question there, it says here 1.2, "A compound A compound with the molecular formula 6 C6H12O could either be of the following. Could be either of the following, okay? So, what we are having here, we can see. So, we're looking for something that has got only one oxygen there, right? So, if you look at an ester, remember it's going to have COO C there. Therefore, this one is incorrect. And then, for a ketone there, what you're going to have is something like this, where you're going to have a double bond oxygen like that, okay? So, this one fits the criteria. And also, for for an aldehyde, we'll find that we are going to have something like this, like that, right? So, it means here, the correct options there that you're having here is going to be option two and option three, right? So, if we look at our options there, we will find that our option C is the one that is correct. Then, moving on to the next question, it reads here, 1.3, the equation below represents the cracking of a long hydrocarbon chain, right? So, we are cracking a long hydrocarbon chain, right? So, it says here, which of the following represents compound X, right? So, if we are to look at here, we are going to look at the number of carbons that we actually used here. We used We started with seven.
Now, we are having five of them. So, it means now, if we are to subtract it, means we are left with how many carbons?
Two other carbons, right? So, I'm going to say C2 there. Then, after that, we We started with 16 hydrogens. Now, we are having 10 of them. It means the ones that we are left there is six, right?
So, I'm going to say C2H6. C2H6, right?
So, if I'm to look at this, this one is going to give me something like an alkane. So, if I'm to draw the structure, you'll see I'm having two carbons, and then the rest here are going to be our hydrogens like that.
Hydrogens, hydrogens like that. So, if we are to look at this, a two-carbon chain is means is F, and then it's an alkane. Therefore, it's going to be our ethane there. So, our correct option there is going to be option D. Then, moving on to 1.4, 1.4 reads, "Given the IUPAC name of the organic compounds below are CH3 We are given CH3 like that like that like that. So, this is the the compound that is given there, right? So, now we are asked to give the IUPAC name of this one, right? So, we need to know where we start like our our numbering and all, right? So, if I can just like maybe highlight the to show that's what we are going to have there. So, I'm going to highlight this part here, which is our C2H5. There is going to be a branch and then another branch that we are going to have here is going to be a butyl again and then another methyl there. So, those are the three branches that you are going to have there. So, it means here if I'm to number here, I'm going to number this one as carbon number one, number two, then number three, and then this remember is going to be a branch, so carbon number four, then another branch there, then carbon number five, another branch there, and then six and then seven, right? So, we are clear the longest chain is going to be hept. So, we have ten because this is an an alkane, right? So, it means here if we can look here on carbon number three, right? What do we have there? You will find out that we are having an ethyl group, right? So, it means it has got two carbons. So, in terms of that, so I'm going to say on carbon number three, what do I have there?
I'm going to say on carbon number three, I'm having an ethyl, right? And then also on carbon number four, what do I have there? It means on carbon number four, I'm having a a methyl and on carbon number five, another methyl. So, it's going to be 4, 5 then di ethyl, right? Then after that now, what you are supposed to do is to look for the longest chain, which is hept, right? So, if you can see here, what do you have? Option B gives us three ethyl and then 4, 5 di methyl heptane. So, this is our correct answer there. Then moving on to 1.5. 1.5 reads, which of the following compounds has the strongest van der waals forces, right?
So, we are looking at the strongest van der waals forces, right? So, if we are to look at the options there, you will find that this one has got an OH, then COO, CN, ester, and then this one might be a ketone, and then uh the other one falls under what? Hydrocarbons there, right? So, the one that is going to have the strongest van der Waals forces is going to be our option D there. Then moving on to 1.6, 1.6 reads here, uh the diagram below shows the energy profile of an endothermic reaction, right? So, the type of reaction that is given there is an endothermic reaction, which is type of reaction that actually absorbs heat from the surroundings, right? You can see. Then after that now we are given uh the question reads, "Which of the following correctly indicates the effect of a catalyst, right?" So, we're looking for the effect of a of a catalyst uh on the activation energy and the enthalpy change for the reverse reaction, right? So, it means here if we are to add a catalyst, remember a catalyst just like uh lowers the activation energy and then doesn't change the amount of what? Of products, right? So, it means our catalyst is going to go like that like that like that like that and then go up somewhere there and then here it may be turns somewhere there and then also continue to join what? Where we are having our what? Our products there, right? So, you can see here what is it that actually changed there is that our our EA here is actually going to what? To change there.
So, our EA there is going to what? To change there. So, it means our EA in terms of our option there, I'm going to say our EA it decreases, right? So, I'm going to say our EA decreases. So, what does that mean? It means now we are looking for these two options where we are going to have our option uh B and what? And and and C there, right? So, these ones are incorrect. Then after that now if you look at the enthalpy change the enthalpy change, which is your delta H there, it remains what?
Unchanged. So, it means our correct option there for this one is going to be our option C, right? So, it just remember speeds up the rate of what? Of reaction by using an alternative route that actually uh uses less energy. Then moving on to 1.7, so under the here we're given the graph below shows the concentration of N2, H2, and NH3 against time for the reaction, right?
So, now, leaving that in mind now, this is the equation that you're given here. So, NH3, then after that you're given delta H is -93 kJ/mol, right? So, it means this is an exothermic reaction. It gives out what?
Heat energy there, right? Then after that now, they're asking us which is the correct representation for each time in the graph there, right? So, we're given T1 there, right? So, in terms of our T1, you can see here what is it that happened, right? So, you can see here in terms of our N2, there is a sharp increase, right? There is a sharp increase here in terms of our N2, right?
So, this is an increase in what? In concentration of what? Of N2, right? So, as a response, you can see what is going to happen, right? So, you can see after that it decreases, but if you look at the amount of products there, right? So, we are counteracting what are the the the the disturbance there. You can see your product there. Your NH3 is a product, right? It goes up, therefore it what? It increases, right? Then after that now, the first one there is going to be increase in concentration of what?
Of N2 there, right? Then after that now, at T2, let's look at T2 there. What happens there is that all of them, they go up a bit, they go up a bit, right?
So, now, in terms of this, what I'm going to say here is that uh it's going to be what? Our pressure there, right? So, what does pressure do, right? So, increase in pressure, so what is it that is going to happen if you increase pressure, right? So, increase in pressure favors the side with the least number of moles, right? So, if we had to look here, what happens to your NH3 again, right? You can see it goes up again, it increases, right? Increase in pressure favors the side with the least number of moles, right? So, I'm just say going to say kPa there. Increase in pressure favors the side with the least number of moles. It means if we had to increase uh within this what? A sealed or container, it means the forward direction is going to be favored, right?
We're going to have more of what? Of products, right? So, if we're having more of products, this is what [snorts] we're going to get there. Then, moving on after that, now I'm going to have my T3 here. So, T3 there, what is it that is going to happen there? We can see here we say the type of reaction there is going to be an exothermic reaction, right? So, if we are to increase our temperature here, if we increase the temperature, what is going to happen there? It means now we are going to favor the reverse what? Reaction, which is endothermic, right? So, if we are to look at this, what happens to our products there? Our products are going down, right? So, if our products are going down there, it means now we increased our what? Our temperature. As you can see, the reactants there the reactants they are increasing uh they are increasing the concentration of the reactants or the amount of product of reactants is actually what? Increasing.
So, it's increasing temperature, the temp T3, and then increasing uh pressure, and then also increasing concentration, right? So, this actually uh corresponds with option B, which is N2 increased concentration of N2 was increased, and then also pressure was also increased, and then the last one that says what? Temperature was increased there. Then, moving on to 1.8, 1.8 says here the reaction between sodium and oxygen is So, we're given the reaction of this there, and then also a delta H of -418.8 kJ/mol. It's also an exothermic what?
Reaction there. If the equilibrium concentration of O2 gas at 25° is equal to what? 2Y mol/dm³ right? So, if the equilibrium concentration we need to check that of the statement there, right? So, remember your equilibrium constant concentration refers to what?
To your KC there, right? So, it means the value of the equilibrium constant your KC there at this temperature will be, right? So, if I'm to look at this, is that uh remember we're going to say the amount of what? Of product all over if we are to do our KC expression we're going to say concentration of what? Of products over the concentration of what?
Of reactants, right? But also take note here what are we having? We're having a solid, right? So it means this one won't take part and also our product there is a solid which won't take part in terms of this, right? So it means now what we are going to have there is going to be 1/O2, right? So which gives us corresponds to what? 1/Y like that, right? So oxygen there remember if it has to be squared it means we're going to be having a coefficient somewhere there before what?
Our oxygen but in terms of this we do not have one so it means now it's going to be 1/Y concentration of what? Of Y.
So it means our correct option there is going to be option A. Moving on to 1.9 it says here learners conduct a titration experiment using the apparatus below, right? Then after that now it says uh correct the correct labels for X, Y, and Z are okay. So in terms of this what you're going to have there is our X, okay? Our X here I'm going to say my X there is a burette, right? So X is a burette and then also we have got our Y there, right? So our Y refers to this stand there so I'm going to say this is my retort stand, right? So this is a retort stand like that, okay? Then after that now what do you have there as Z which is our conical flask, right? So this is a a conical flask, okay? Like that, okay? Then after that now let's look at the option that actually corresponds. So the first one there we said X is a burette so that one takes and then the second one we said our Y is our okay is our retort stand so it means the only option there that we are remaining with is option D. This is not a what? A retort what? Uh This is not a a tripod stand, right? So this is not a tripod stand. Remember try means three so it means this is something like that is three legged like that, right? So whenever you are burning something there, so you can either have like a a tripod or a a bipod, right? So, then after that now we're moving to 1.10. It says, "According to the Lowry-Brønsted theory, a base is" right?
So, we know that a base is a proton what? Acceptor. So, is a base is a a proton acceptor like that, okay? So, if we are to look at our options there, that one that corresponds to this one, it says what? B, the proton is a proton acceptor, right? So, according to Lowry- Brønsted theory, an acid is a proton donor, but a base is a what? A proton what? Acceptor there. So, you should take note there. The main focus was for a base there, okay? Then after that now I'm going to like to proceed to the next question. So, our next question there is our next question reads uh uh question two, which is now our our structured questions, right? So, now I'm going to proceed with the structured question. It says here, "We are given letters A to H in the table below." Uh then they represent what? Organic what? Compounds there. Then it says here, "Option A, we're given a propanoic acid. B, like that like that." Okay, until option what? Uh letter H there. Then the first question reads as there, "We are being tasked to define a what? A homologous series." So, what's a homologous series, right? So, a homologous series is a series of organic compounds uh that can be described by the same general formula. So, I can just like write it down. A series of organic compounds compounds uh that can be can be described described by the same general formula.
The same general formula.
Like that, okay? Then after the 2.2 says, "What is the name of the functional group? What is the name of the functional group of the homologous series to which compound A belong?"
Right? So, now we're looking for the name of the functional group, okay? Take note of that. Let me just like erase that so that you can easily see. So, the question says, "What is the name of the functional group?" Right? So, we're looking for the name of the functional group, not the homologous series, right?
So, it means here this is a what? A propanoic acid, right? So, it's propanoic acid. It's uh falls under the carboxylic acid, but the name of the functional group there is going to be referred to as a what? A carboxyl group, right? So, this one is going to be a carboxyl group, like that, okay? So, that's what you're actually having there. Then moving on to 2.3 says, "Use the table above to write down the letter that represents each of the following."
Right? So, option 2.3.1, we're looking for an aldehyde, right? So, an aldehyde, remember your oxygen hides at the end there, right? So, it means our option there is going to be F there, which you are having our our aldehyde group. So, I'm going to say automatically here we're having our our F. Then an ester, remember ester there is uh easily seen by what? By a COO C there, right? So, we're looking for something that has got a COO C, right? So, if you can see, this is COOH, which refers to our our propanoic acid. And then here um carboxylic acid, sorry. This is a carboxylic acid. And then here what you're having here, you can see our COO C there. So, this is the part that you're looking for there. So, this one formulates our what? Our ester there.
So, it means there we are going to have our option what? Option C. So, this is the correct option, option C. Then 2.3.3, an unsaturated hydrocarbon, right? So, we're looking for something that is unsaturated and it is a hydrocarbon. So, remember hydrocarbons are carbon atom compounds that contain hydrogen and carbon atoms only. And then unsaturated there, right? So, it means now it has got what? Multiple bonds. So, we're looking for something that is contains hydrogen and carbon atoms only, and then also it has got what?
Uh multiple bonds, right? So, 2.3.3, what are we having there? I'm going to look at our options there. What are we going to have there?
Uh let's look at this one, yeah. This is our H, right? So, you can see there is a double bond there, right? So, that double bond there shows that this is a what? An unsaturated what? Hydrocarbon.
So, our option there for this one is going to be an H. And then 2.3.4, it says a chain isomer of 2.2 uh dimethyl butane, right? So, a chain isomer of uh 2.2 uh dimethyl butane, right? So, 2.2 dimethyl butane, right? So, it means how many carbons do we have there? It's one and two plus the bute there, it means a total of what? Six carbons, right? So, if you can look at here, we're having one, and then here plus four here, it means five, and then plus another one, which is six, right? So, our option there is going to be G. So, this one is going to be our our correct option there. So, a chain isomer of that one is going to be what? A chain. Remember, we're looking for the same molecular formula, but different one chains there.
Uh then 2.4, it says, "Write down the IUPAC name of compound C," right? So, compound C, let's go to compound C, okay? So, our compound C, remember, is an ester, and an ester consists of how many parts? Two parts. The alcohol part, and then the the the carboxylic part.
And then whenever we're naming, we always start with the alcohol part, right? So, now the first thing that you're supposed to do is to identify the the alcohol part, and then after that, the acid part. So, I'm going to just like indicate. So, this side there is going to represent our what? Our alcohol part, right? So, in terms of that, I'm going to now count how many do we have carbons? 2 3 4, right? So, it's going to be butyl, right? So, since it's an ester, it's going to be a butyl there. Then, after that now, I can identify the carboxylic acid, which is 1 2 3. So, it's a three-carbon chain there. I'm going to say prop, right? So, it's going to be butyl propanoate.
Remember, -oate is going to be our our suffix for what? For an ester. So, that's the IUPAC name of that one there.
Then, moving on to compound E. Again, we are requested to uh find the IUPAC name of compound E.
So, where is E? So, this is our E there.
So, if I'm to look at our E, what we're going to have here is that uh we need to see where we are going to have our our side chains and all, right?
So, if you can look here, our bromo there. Uh maybe if I can just like find some space somewhere there, so that I can rewrite this one. So, I'm going to like maybe like uh redo this one in terms of like uh the structure. Maybe if I can just like write it here. So, we're having a CH3. I'm going to say like uh write down the structure, CH3 there.
Then, after that, what am I having? I'm having a CH, so I'm going to put an H there. Also, we're having a bromo there.
And then, after that now, I'm going to have a CH2 again, CH2 like that. And then, you stretch it out again, another CH2.
That's the second CH2. And then, uh we are having a CH and then a CH like that, okay? Then, we're having this CH3 there, right? So, it means this CH3 acts as what? As a side chain there, like that like that like that. And then, after that now, we're having another CH3, right? So, CH3 like that, okay? So, in terms of the counting here, what I'm going to do is to count from my left uh going to the to the right. So, it's going to be 1 2 3 4 5 and 6. So, the longest chain is going to be six carbons, right? So, one might and why are we starting from this side instead of this side right? So you can see here what you are having there is a promo right? So the promo there remember we're going to like prioritize this one because here promo and methyl right? So which one comes first is going to be our promo then after that now I'm going to say our answer there is going to be two dash promo then after that what are we having there on carbon number five here? What are we having?
We're having a methyl right? So I'm going to say dash five dash methyl remember your letters uh and uh numerals or numbers they are separated using a what? A hyphen right? Uh or a dash there. So the longest chain is six carbons there after that now I'm going to say hexane like that okay? So that's our answer there or IUPAC name for compound uh which is compound this is compound E like that okay? Then uh moving on to 2.5. So 2.5 reads there uh compound D reacts so we're looking at compound D reacts with a primary alcohol containing four carbon atoms right? So it reacts with a a primary alcohol containing four carbon atoms in the presence of a a catalyst right? So write down the name of what?
Of a catalyst right? So write down the name of a catalyst right? So now we need to identify what compound D is right? So we need to go back and identify compound D right? So if you look here so this is a compound D C O O H tells us this is a a carboxylic acid right? So uh I'm going to say a carboxylic acid so it's going to be a carbo- xylic acid and then what are we adding there? A primary alcohol.
So I'm going to say plus an alcohol. So if we're having something like this there after that I'm going to say concentrated H2SO4 uh then after that now we're actually going to have an ester plus our water right as our product. So the whole this of this is going to what? To be referred to as esterification right? So name of the catalyst there, right? So, the name of this one is our sulfuric acid, right?
So, take note, instead of like H2SO4, you're supposed to write sulfuric acid because they said the the name, not the formula, right? So, the name of this one is going to what? Sulfuric acid.
Usually, it's what? Concentrated there, right? And then the type of reaction there, I've already mentioned it, it's what? Esterification, where we are is a mixture of what? Reacting of a carboxylic acid and an alcohol to produce an ester and water. And then they say name and formula of the inorganic product that is formed, right?
So, we need an inorganic product there is our what? Our water. So, I'm going to say the name there is water, and then the formula, I'm going to say our formula for water is what? Is H2O, right? So, this is what we are having in terms of our 2.5.3. Now, moving on to 2.6, it says here, compounds B and F are isomers, right? So, isomers, then after that, they're asking us to define what a structural isomer is, right?
So, these are organic compounds with the same molecular formula but different structures, right? So, organic compounds, I'm going to say organic compounds with the same molecular formula molecular formula but different structural formula, but different structural formula. Okay. So, these are what we refer to as our what? Structural isomers. Then 2.6.2 says, identify the type of isomers, right? Only chain, positional, or functional isomers, right? So, if we are looking at our F there, B and F B and F, maybe we can just like scroll a bit up, we're looking at our compound B. So, we're looking at this B here right and our F there, right? So, what do we have there? So, you can see the other one here is going to be an aldehyde and the other one here is going to be what? A ketone, right? So, if you had to look at their general formula CnH2nO, all of these they're going to have what?
The same what? General formula there.
So, it means these ones they formulate what we refer to as what? Functional isomers. So, they are functional isomers. So, the correct answer there, I'm just going to highlight it. It's referred to as what? Functional isomers.
Then, write down the structural formula of compound F.
So, structural formula of compound F, okay? So, structural formula of compound F, allow me to just like write it here so that it's easier to follow, right?
So, we're looking at this F there. So, I'll just like erase there. So, our compound there, which is our structure F. So, I'm going to have seven carbons, that's where I'm going to start. 1 2 3 4 5 6 and then the seventh carbon there, right? So, these are the carbons that you're having. Then, after that now, what am I going to have here? Here, because it's an aldehyde, it hides way at the end. So, I'm going to have a double O, our oxygen there hides at the end and then the rest here are going to be my hydrogens. Please, always remember these hydrogens are important. You are always supposed to have to include all these what?
Hydrogens like that, okay? So, this is what you're having in terms of this, right? Okay? So, yeah. Like that, then we're good to go. Then, moving on to the next question, which is our question 2.7 there. It says, "Write down the empirical formula of compound G, right?"
So, we are looking at the empirical formula of compound G, right? Okay? So, compound G, what are we having there? I'm going to say here, what am I having in terms of this? So, it means here basically I'm going to be having 1 2 and 3 carbons. So, it's going to be C3 here in terms of my empirical formula. Then after that, I'm going to count the number of hydrogens that I'm going to have is 3 4 5 6 7. All right.
So, it's going to be C H7 there like that. Okay. So, this is my general uh my empirical uh formula of compound G. And then 2.8 says here give the general formula of compound H. So, compound H we need to identify it falls under which homologous series. Okay. So, you can see here we are having a double bond. So, it means here these are alkenes, right? So, alkenes have got a general formula of CnH2n where n represents our number of of carbons. So, CnH2n there. CnH2n Maybe I can just rewrite that one. CnH2n and then 2.9 says compound G undergoes complete combustion, right? So, it's a combustion reaction. Always take note your products are always going to be uh carbon dioxide and water. And then using molecular formula, write down the balanced equation for this reaction, right? So, we need to look at compound G there. So, it's our compound G there.
What do we have there? I'm going to have say the total number of carbons that I'm having here for G is 3. It's 1 plus uh 4 which is a total of 5 and then 6. So, it's it's going to be C6 and then H how many? I'm having a total of 14 like that. C6H14, right? So, it's 6 C6H14.
So, it's going to be C C6H14, right? So, for anything to combust, they it should actually be in the presence of what? Of excess oxygen, right? And then our products, I said they are always going to be carbon dioxide plus our H2O, right? So, this is what we are having, right? So, take note now we need to balance our equation. So, our balancing it always start with carbons followed by hydrogens and then lastly you're going to end with your oxygen there. That's how I'm going to start with my carbons there. So, if I'm to look at things here, on my left I'm having a number of what of carbons which is I mean is six there, right? So, I'm having six carbons there. So, I'm going to put a six there.
Then after that the carbons are now okay, right? So, the number of carbons are now what? Balanced for now there.
Then after that I'm going to proceed to say how many hydrogens do I have in that order there. How many hydrogens do I have? In terms of this I'm going to look at this. How many do I have? I'm having 12 in this side. I'm having only two.
Therefore, I'm going to put a what? A coefficient of six there. Then after that now I'm going to look at my oxygens. The oxygens here I'm going to say 6 * 2 is 12 and then plus here what do I have there? What am I going to have there? I'm having a total of how many six plus six there which is a total of what?
Of 18, right? So, if I'm having an 18 there, what am I going to do with this?
Okay, I'm going to say uh Okay, okay.
Let [snorts] me check here. C6H12 something.
Oh, H14 here is 14 not 12. Sorry about that one. Let me just like correct it quickly rectify it. Here we are having H14 not 12. So, here is 14 hydrogens, right? So, I'm going to correct that one there. It means here we are no longer going to have I mean six of them there.
It means now we are going to have a total number of what?
14 here this side. So, we should multiply by what? By seven. So, it's 7 * 2 which is 14. So, yeah, okay. So, it means my hydrogens are now fine for now.
Then moving on to oxygens, I'm going to say 6 * 2 is 12 and then plus here seven there. It means I'm going to get a total of what? Of 19 there, right? So, if I'm to put my 19 here, what is actually going to happen there? You'll find that uh on this side here, I'm now having 19 * 2, which is more of like a 38, right?
So, it no longer balances. So, for me to balance everything now, I must multiply everything this side by what? By two.
So, I'm going to put a coefficient of two there. Then after that here, this 19 is now okay like that. Then here, since I was having six there, I'm supposed to multiply everything by what? By two, right? So, if I'm to multiply everything by two, this one now changes to to a total of what? 6 * 2, which is going to be a 12 there. So, it's an error there, remember? And then 7 * 2 now becomes what? 14, right? So, this is our now what? Newly balanced what? Equation. So, that's how we go about it. Hope you did enjoy question two. Now, we are now moving to question number three, right?
So, question number three, which is our physical properties, right? So, it says here, we're given the melting point of five organic compounds with uh known molar mass. So, the molar masses are known. We determined using uh during a practical investigation. And then after that now, we are saying the results were recorded in the table below, right? So, the first question there that you raised there, it says uh define the term melting point, right? What is it that is melting point, right? So, it's the temperature. Okay, so, it's important if you can go through your exam guide, you'll see this part here is usually underlined, right? So, it's an emphasis there. So, you start with the temperature at which a solid at which a solid and liquid phases and liquid phases and liquid phases of a substance are at equilibrium.
Uh at equilibri um.
Okay. So, that's how we attempt uh that first one there. Then moving on to the second one, right? They are saying for the compounds P, Q, and R, right?
So, it means now we are focusing on these three, right? P, Q, and as well as what is those three, right? Then now, write down the dependent variable, right? The dependent variable, right?
So, here we already given the dependent variable at the top. The melting points, right? So, this is the dependent variable, right? So, it depends on so many things. So, it can be like here in terms of this you can see your chain length increases there, right? So, this one becomes our dependent variable, which is going to be our our melting point, right? So, remember your dependent melting point your dependent dependent variable here most of the time there are three of them. It's either going to be melting point, uh boiling point, or vapor pressure, right? So, what is it that was controlled in terms of this? So, 3.2.2 says the controlled variable, right? So, between P, Q, and R, what is it that was controlled, right? So, you can see all these uh what are our alkanes, right? So, it means now they belong to the same homologous series, right? So, it's homologous series. They belong to the same homologous series. Then an investigative question, right? So, an investigative question, remember it's your ID, right?
How does the independent variable affect the dependent variable, right? So, it means here we already identified the independent variable. How does the independent variable affect the dependent variable, right? So, we already the dependent variable, right?
The independent variable here is going to be our our chain length, right? So, you can see your chain length is the one that increases, right? So, now I'm going to say how does Here's my question. How does uh the chain length How does chain length, okay? Chain length, right? So, chain length there represents my independent variable. How does chain length affect melting point? Affect I'm going to say affect a melting point melting point of organic compounds like of organic maybe I will just say co- compounds like that. Okay. So, all you take note.
Now, this is a question, right? So, it means all the time if you're asked in terms of a what? An investigative question there, right? So, you use your ID, start with the independent variable and then the dependent variable. Then the other thing that you should always take note of is that now this should be always be a closed question like that.
And then 3.2.4 it says here an explanation for the trend of the melting point of these compounds, right? Okay.
So, we need to explain the trend in this what? A melting point, right? So, now what I'm going to say here is that look at our melting points there, right?
So, from P going downwards, right? So, I can either say from P to R there, but is it that is happening there? You can see the boiling point or the melting point what uh goes down like that, right? So, why is that so? It means here chain length, right? So, increase in chain length increases what?
Uh the the the the the melting point, right? So, the longer the chain the higher the the melting point, right? So, the one with the highest melting point here is going to be what? This one there, right? So, after that now I'm going to say here in terms of this I'm actually going to say from R to P, right? So, I will also to P, right? So, I'm going to take a trend going what? Upwards instead of me going what? Downwards. So, I'll just like say from R to P. What is it that is happening there? I'm going to say from R to P.
Uh maybe from P to R. Okay. Let me just use from P to R so that it makes sense, okay? From P to R from P going to R.
What is it that is happening there?
According to our trend chain trend there I'm going to say our chain length chain length increases, right? So, our chain length increases.
Chain length increases, okay? So, this is going to be my first bullet there.
I'm going to say chain length increases.
And then after that, I'm going to say bullet number two. What is it that is actually what happening then? I'm going to say strength of intermolecular forces, strength. I'm going to talk about the strength of intermolecular forces, IMF. Please do write it in full.
Intermolecular forces what increases, right? So, we can see from what increases there from P to Q to R there.
Then after that now, as a conclusion there, I'm going to say more energy more energy is needed more energy is needed needed to overcome to overcome like intermolecular forces in R than in P in R than in what? In P. Like that, okay? So, that's how we explain that 3.2.4.
Then moving on to 3.3 there.
Question 3.3 says here, consider compound R. So, we've been asked to consider compound R as well as compound S. So, R and S there R and S, okay? So, these are the two there. Maybe let me just like remove that highlighting there. Then I like maybe using a different so we're looking at that R and S like that, okay? So, those are the two that we're looking at. Then they're saying here, consider compound R and S, right? Then after that, which compound will have a higher a boiling point, right? So, if we were to look at that R and S there R and S there, so the one that is going to have a higher boiling point there, you can see is -150 there. So, it means it's our R there.
So, our first answer there, I'm going to say it's going to be our R like that, okay? So, this is R and then the second one there question says here, which compound will have uh the lowest vapor pressure? Right, so remember the one that has got a high boiling point is also going to have what? A lower vapor pressure. So, I'm going to say I again. And then here, 3.3 says here, explain uh the answer in question 3.3.1, right? So, what you're going to do here is to maybe find some space. Okay, let me just like maybe write it here. So, I'm going to say in terms of this, what do we have there as our first bullet, right? So, I'm going to say if we're to look at uh both R and S, right? Both R and S there, you'd find that uh all these, right? Uh the dominant force there is uh London forces, right? So, I'm going to say both they have got London forces, so I'm going to start with the type of intermolecular forces that is dominant.
Both of them have London forces, right? So, London forces like that, okay? Then after that now, my second bullet is going to focus on uh uh the the the the chain lengths, right? So, in terms of my chain length there, what is it that I'm having, right? We can see here, this one is actually having what? A branch there, right? Our CH3 there. And then this one is is is is is a straight chain, right?
So, I'm going to say compound I is the longest straight chain. Compound R Compound R Compound R has the longest straight chain, but okay.
Straight chain. The other one has got what? Uh branches there. So, it means S has got what?
Branches there. Uh maybe I can just like say compound S. So, I'm going to include it here. So, I'm going to say compound S is branched, okay? Then after that now, I'm going to say uh uh because of that, right? So, I has got stronger IMF maybe uh has stronger, okay?
Intermolecular forces IMF than S there. Then, as a conclusion there, I'm going to say more energy is needed more energy is needed to overcome needed to overcome intermolecular forces overcome IMF in what? In R than in S than in S like that. Okay, that's how we conclude it there. Then, uh moving on to the next question which is 3.4 there.
Uh sorry about uh the writing there. It says here to which homologous series so we're looking at the homologous series does this compound belong, right? So, consider compound T, okay? So, we're now considering compound T and then they're asking us to which homologous series does it belong to, right? So, if you can look at that, there is a methyl of what?
Looking at that, what do you have there?
CH3 Okay, CH3 like that like that. Okay, it's actually going to fall under what?
Our alkanes there.
Uh I'm going to say an alkane.
Basically, this is what you're having there. I'm going to say CH3. So, this is our CH3 the first one. Then, after that you're having a carbon and then here we're having a CH3 two of them, right?
So, it means the other one is going to go at the top like that uh CH3 like that and then at the bottom again I'm having we're going to have our second CH3 like that. Okay? Then, after that now what do I have? I'm going to have a CH3 like that, right? So, it means this falls under our alkanes, right? So, it means now it belongs to uh a group or the homologous series there it falls under what? Alkanes there. I'm going to say this is an alkane, right? Then, after that now they're saying write down the IUPAC name of this what? Compound there, right? So, we're looking at the IUPAC name of this compound there. So, if I can go back there and try and number this, you'll see that what we're having here, I'm going to say carbon number one, carbon number one, carbon number two, and carbon number three, right? So, on carbon number two there, what are we having? We are going to have what? Our branches there. So, it means here I'm going to say two {comma} two because I'm having this methyl at the top and this methyl at the bottom. That's why I'm going to say two {comma} two. And then after that, uh because there are two, I'm no longer going to say methyl, but I'm going to say dimethyl because there are two of them. Therefore, I'm going to say two {comma} two {dash} dimethyl.
And then I'm going to look at the longest chain there. It's three carbons, right? So, it's three carbons, I'm going to say what? Propane like that. Pro pane like that. Okay. So, that's what we have there as the IUPAC name for that compound T, right? Then moving on to question number four, we are now done with question three, right?
Question number four reads Um the flow diagram below represents six different organic reactions. So, we're having six different organic reactions then undergone by a virus virus what?
Organic compounds, right? Then after that now they're saying name the type, right? So, name the type of addition reaction that takes place at a reaction number one, right? Okay. So, basically what you need to understand is that anything that uh points away from an alkene, right? So, I'm going to focus on this alkene. Anything that points away from this alkene there is going to be what? Uh sort of a bit. Let me just say I'm doing there. Okay. So, anything that points away there from this alkene there, I'm going to say this arrow there and that arrow there. And then this arrow there, all of these right the ones that are pointing away there, they are going to be our what? Our addition reactions, right? Then the ones that point towards, right? The ones that point towards this one, they are going to be our what? Our elimination reactions, right? Okay. So, having that in mind, I'm going to look at reaction number one, right? So, the type of reaction number one there is going to be So, I'm started with started with the pentene, it means I was having a double bond, right? So, now what is it that we added for it to be what? To be pentane, right? So, it means now we added our hydrogen there. I'm going to say then the type of reaction there, addition. So, it's addition reaction, yes. So, but what type of addition there? I'm going to say hydrogenation, right? Hydrogenation, right? Okay. Then after that now, reaction three, we're looking at reaction three. So, it's pentene then reaction three plus what? H2O there, right? So, it's an addition reaction again, but it What is it that we're adding there? We're adding water. So, if you're adding water, you're hydrating, right? So, I'm going to say hydration.
Then reaction four, so our reaction four here also points away from the pentene going to that side. So, what is it that we're now having? We're having a bromo, right? So, a bromo. So, it means now a bromo there from this double bond here.
Can you see you're having a bromo? So, for it to be an alkane, so it means now we need something like an HBR like that, right? So, I'm going to say here what did we have there? It's hydrohalogenation, right? So, I'm going to say hydro- hydrohalogenation like that, okay? That's what we're having there. Then moving on to question 4.2, it says for reaction two, so we are focusing on reaction number two. Write down the type of reaction that takes place, right? So, 4.21, for reaction number two, write type of reaction that takes place, right? So, our reaction two there, right? So, it means here it points towards the pentene there, right?
Which is our alkane. So, the type of reaction there is to be what?
Elimination, right? So, I'm going to say this one here elimination, right? So, I'm going to say elimination elimination reaction, right? So, but if I wanted to be specifically specifically I'm going to say dehydrohalogenation dehydrohalogenation, right? So, it means the opposite here if we are going this side here is a halogenation, right? So, it's going to be hydrohalogenation. So, going this side here for reaction four is hydrohalogenation then going back here is dehydrohalogenation.
We are eliminating the what? Uh uh is an elimination what? Reaction. Then moving on to 4.2.2 it says two reaction conditions, right? So, we are looking for the reaction conditions there for the reaction two, right? So, for reaction two for reaction two uh from two bromopentane to pentene there what do we need there? It means now we are need in need of what? A strong a concentrated strong base, right? So, I'm going to say here conc. strong base, right? So, usually the strong base here it can be sodium hydroxide or like uh potassium hydroxide, right? So, KOH, right? So, this is the condition that we actually need there. We need a strong concentrated uh uh concentrated strong what? A base, right? So, that's the condition that we need there, right? Then moving on to the next question there facing uh 4.2.3 uh balance uh four reaction two write down the balanced chemical equation using structural what? A formula, right? So, now we need to work out uh that there, right? So, for reaction two So, in terms of our reaction two I'm going to go back at the top there. So, what we are having there in terms of reaction two, right?
So, I'm going to start with this uh two bromo pentane, right? So, what do we have there? I'm going to say you have two bromo pentane. So, allow me to use this two bromo pentane. So, pent means how many carbons? I'm going to say five carbons, right? So, I'm going to say one, two, three, four, and five. Like that. So, these are the carbons that we are having. Remember, it's two bromo, so I'm going to put my BR there. And then the rest there are going to be my hydrogens like that. Hydrogens hydrogen another hydrogen there. So, I know they can seem like they are many, but you need to include all of them, right? Otherwise, they it will be not considered, right? So, here what did we say here? We said here what do we have there?
Plus okay. So, what I'm going to get there?
I'm going to say plus for me to move from here to two bromo pentane to for me to have something like this. What am I going to have there? So, it means now I need to add maybe an NABR.
I'm going to say plus NABR.
And then I'm going to get my product, right? So, my product here you can see is a is a pentene, right? So, pent means five. And then a one, two, three, four, and and five, right? So, now in terms of this my bromo is on carbon number two.
So, I'm going to put a double bond there, right? So, I'm going to put at the hydrogens again at the hydrogens. So, it's three and then four.
Three. Maybe I put this one at the bottom. And then after that like that.
Like that.
So, these are going to be my hydrogens.
This is a hydrogen also. So, then after that now the other product that we are going to have there is going to be our NABR, right? So, it's going to be our NABR. So, plus H2O, right? I'm going to say NABR.
NABR plus our our H2O there, right? So, this is what we are going to have there for 4.2.3. Then moving on to the next question there it says uh 4.3 uh consider reaction three. So we are going back to reaction number three. So if we look at reaction three, where is reaction three? Okay, so this is our reaction three there. So it says here reaction three. Then uh To which homologous series does compound A belong? But so at compound A there, right? So this is going to be our compound A. So we started with pentene and then after that now we know anything that points out there, what is it? It's addition reaction, right? So then after that now from pentene to compound A, therefore what do we have there? It means you are going to formulate what?
An alcohol, right? So our A there is going to be an alcohol, right? So I'm going to say in terms of this to which homologous series does A belong? It belongs to what? To alcohol. What is that? Alcohol. Alcohols. Okay. Then write down the structural formula of compound A, right? So in terms of this Remember what is going to happen here?
We said pent, right? So it means it's five carbons. So the number of carbons there they are not going to change, right? So they're going to remain the same. So what I'm going to like just write there is going to be like 1 2 3 4 5 in terms of that. Then after that now what I'm going to say on carbon number two where we are having our double bond there, that's where I'm going to have my put my OH there and then the rest there I'm going to put my hydrogens throughout. Hydrogens, hydrogens like that. Hydrogens, hydrogens, hydrogens.
Hydrogens then yeah, we are good to go.
So that's the structure that we're looking for there in terms of 4.2.3. Now moving on to 4.3.3. 4.3.3 says write down the IUPAC name of the compound A, right? So we're now naming it, right? So in terms of this, remember we name we number from the side closest to the functional group. 1 2 3 4 and carbon number five, right? So, it means now on carbon number five, I'm going to have carbon number two. That's where I'm having an alcohol, right? So, it's pentan- 2-ol, right? So, where do we find our OH there? It's going to be on carbon number what? Number two like that. So, this is the carbon where we are having that. Then, moving on to the next question, which is 4.4, it says, "Write down a balanced chemical equation."
Write down a balanced chemical equation for the reaction four using condensed structural what? Formula, right? Okay.
So, there we are going to look at equation number four, reaction number four. Okay, in terms of reaction four, where is this reaction number four?
Okay, so the reaction number four. So, it means now what did we start with? We started with a a pentene, right? So, this is pentene. Okay. So, maybe if I can just like maybe write it here as the two pentene, what I'm going to have there?
So, I'm going to say five carbons, it's going to be CH3.
Okay, [clears throat] and then a CH then another CH, okay? Then, uh CH2CH3.
CH2 CH3. Okay, so in between these two here, that's where there's going to be what? A double bond there, right? Okay. Then, after that now, I'm going to say uh plus what did we add there? We add uh for reaction four, we are going to add an HBR there like that, right? So, I'm going to say H BR. Then now, what are we going to have as our product there? I'm going to say it's going to give us something like this, right? So, it's going to be So, we are going to like overcome that double bond there. So, I'm going to retain the number of carbons there. It's going to be CH3, then uh we're now having a CHBR, CHBR like that. And then after that, uh it's going to be CH2CH2, CH2, CH2, and then with a CH3, like that. Okay, so if you can check the number of carbons, 1 2 3 4 5. They still what? Uh 5. We started with 5. Also, we ended with what? With 5 carbons there.
Okay, now moving on to question 4.5. So, it says 4.5, they name the type of reaction represented by reaction number five. Okay, so we go to reaction five. Where is our reaction five? Our reaction five goes way up past there, right? So, this is our reaction number number five there, right? So, it means now I'm going to say reaction number five. So, it starts from a pentene going to a 2-bromopentane, right? So, anything that points from away from a pentene going to there there, uh which is our our our haloalkane, the type of reaction is going to what? Is substitution. So, I'm going to say here, what do you have there? I'm going to say 4.5 there. We are going to say our answer there is going to what? Is substitution.
Substitu- tion reaction, right? Then after that now, we are looking at reaction number six. They say, "Consider reaction six, then identify the type of what? Of sub- substitution." So, they told us it's already what? Substitution. And then we are supposed to what? To identify the type of what?
Substitution. So, 60, right? Okay. So, we are going to have our products there as our what? As our alcohol, right? So, it means there what you are going to have there is going to be our hydro- lysis, right? So, this is going to be hydrolysis, like that, okay? Then moving on to the next uh question there, which is going to be We said here hydro- lysis, like that, okay? Then here, they are saying here, "Suggest the alternative Suggest an alternative reactant if water is not available, right? So, we're considering reaction six. Okay, so now in terms of this, we can make use of a what?
A strong base, right? Okay, strong base, strong base, right? Uh it should be what?
Diluted, right? So, I'm going to say dilute a strong base. They are supposed to say dilute strong base, right? So, it can be what? Uh your sodium hydroxide there. I'm going to say sodium hydroxide there, right? So, having but uh said we are done with question four. I do believe you did also enjoy it, okay?
Then, question number five, it says when chromium metal So, we're looking at this chromium metal there is added to hydrochloric acid, it undergoes a reaction that produces a characteristic green solution of of chromium three chloride as well as hydrogen gas, right?
Okay. So, then the balanced equation can be represented as follows. Therefore, this is what you having there's our equation, and then delta H is less than zero. Then, I'm going to proceed to define the first uh question to attempt the first question there. It says uh define the term reaction rate, right? So, this is the change in concentration. I'm going to say the change in concentration. Let me just change this color to blue. The change in concentration, concentration change in concentration of reactants or products of reactants or products or products per unit time.
Okay, take note of this, right? So, this part here, maybe I can just like circle heat. This or here is important, right?
Some learners are up to say the change in concentration of reactants uh per unit time, right? This or here is very important. It should always be what? Be included there. Then 5.3 says name two experimental methods that could be used to measure the rate of this reaction in a school laboratory. Right, so so in terms of this what you're going to have there, what you're going to need here is that uh, measure the loss of what? Cr mass right. So the mass of Cr is going to have a decrease there right. So measure I'm going to say number one there. Measure measure the loss of the loss of Cr mass or the mass of Cr right. And then after that now I'm also going to say maybe measure the volume of hydrogen gas that is formed right. Measure the volume of hydrogen gas hydro gen gas uh, that is formed right.
Measure the volume of the hydrogen gas formed by using making use of what? Of a syringe there right. Then 5.3 5.3 is right. In the experiment uh, so we're given a specific value there. 8.67 grams of powdered chromium metal was added to excess hydrochloric acid right. Okay. So then after that the reaction produced hydrogen gas the same direction produced hydrogen gas and then after that now they're saying uh, which was collected as at STP right.
Standard temperature and pressure. Then after that the volume of the hydrogen gas formed was measured at regular intervals and the results were used to plot the graph as shown below right. So this is the graph that is given here.
Take note here. This is the volume of the hydrogen gas in cubic centimeters and then this is our timing what? In seconds like that, okay? So, now uh, calculate the average 5.31 rate here.
Calculate the average rate rate in dm cubed. So, we are given the units of measurement there, right? So, dm cubed per second, right? So, it means here we are dividing this. So, it means automatically we are supposed to convert this cubic centimeters to dm cubed per second. It means we are dividing by time of the above reaction for the first one, 60 seconds. So, these are 60 seconds. We can see it's there, right? So, this is where we are having there. So, I'm going to say in terms of this year, what I'm going to do is like uh rate.
I'm going to just say rate here. I'll just say our rate there is equivalent to the change in volume, right? So, this is I'm going to say delta V, which is the volume there, all over our change in what? In time, right? So, which is equivalent to So, my change in volume there I'm going to say where did you start? You started from zero, and then we ended with what? 5,000, right? So, remember this to convert to dm cubed, I'm going to say 5,000 divided by 1,000, which is going to give me a five there.
So, I'm going to say five all over five maybe minus zero change in Y all over change in X, right? All over Let me just like erase that one there.
It's not quite visible. So, 5 minus zero all over So, the time that I'm having there is 60 seconds, right? So, we started 60 and then zero there. So, it is equivalent to So, if I do my math there, I'm going to get a value of 0.08 three. I can say 83 dm cubed dm cubed per second, like that, okay? So, the most important part there also take note. You are supposed to convert this what? Cubic to to dm cubed by dividing by a by a thousand there, right? So, after that now I'm going to proceed to 5.3.2. It says calculate the mass of hydrogen gas produced, right? So, we are looking for the mass of this hydrogen gas that was what produced there, right? So, for this to to to work out, what do we need there? We need to start by calculating the number of what of moles, right? So, now, how do you go ahead and calculate the number of moles? What you're going to have here is to say the number of moles, right?
So, remember, if you're given this I'm going to make use of what of this gas volume molar gas volume, right? So, the number of moles I denote here, but the volume that we are having here having here is what is five, right?
According to this data, this is the five that we're looking for. 5,000 the one that we converted to what to dm cubed there, right? So, I'm going to say five, and then after that, I'm going to say molar gas volume there is 22.4, right?
So, it's 22.4. You can refer to the data sheet for that one. Therefore, if I do the math there, like I'm going to say get a value of what 0. 2 2 2 3 2 moles, right? So, these are the moles that I'm having, right? Then after that now, I'm going to go back to the equation to say here, remember we asked to calculate what the mass of what of hydrogen gas. So, we're going to look for the mass of hydrogen gas, and then after that now, I'm going to say the number of moles here is equivalent to maybe number of moles, let me just be specific in terms of this one. The number of moles [clears throat] of what of hydrogen gas there is equivalent to the mass all over the molar mass, right?
Then after that now, I'm going to say here what do I have in terms of the number of moles? 0.2232 is equivalent to m all over the molar mass for hydrogen, which is two, right?
So, if I cross multiply there, I'm going to get a value of what 0.45 g, right? 0.45 g like that, okay? That's my final answer, okay? So, maybe someone might wonder how did I get to that there? So, if we can look here, what you're going to have there is that you'll find that you have a the same as what molar volume for what for this hydrogen case the one that you having there. So it means the number of moles that I'm having here are going to be the same here. So I'm going That's why I'm going to use what 0.2232 then after that I multiply. So number of moles is equivalent to M all over M then you're going to get what 0.45 what grams. Then I'll proceed to the next question which is 5.3.3. It says calculate the percentage yield of hydrogen gas in this reaction, right? So we are looking at the percentage yield, right? So they calculate the percentage yield, right?
Of the hydrogen gas in this what reaction, right? So in terms of this what you're going to have there we are going to start with the number of moles of Cr, right? Okay. So let me just like maybe underline that one, okay? So the number of moles of what of this chromium, right? Which is Cr.
Just erase there. So bracket there Cr is equivalent to M all over M, okay? So one might ask why did I call this what, right? So you can see here what is it that I'm given there? I'm given this experiment. So what did we use there? We are using 8.67 grams of what of chromium, right? So it means here if you are given this it means now I can calculate what the number of moles, right? So I'm going to proceed and say here N is equivalent to therefore the mass that I'm given there for chromium is 8.67 8.67 grams all over and then if I look at my data sheet there you'll find the molar mass of Cr is going to be about 52 there, right? So if I divide there I'm going to get an answer of 0.16 673.
I'm going to say moles like that, okay?
So now if I'm having those number of moles now I'm going to now proceed to the mole ratio between what Uh Cr and what H2. So, this is my Cr there, right?
And then the hydrogen gas that we're having there is going to what? Three, right? So, the mole ratio is 2 is to what? 2:3. So, I'm going to just say uh Cr here and then actually put my hydrogen gas there, right? So, I'm saying here what did you get here is what like two the mole ratio was 2:3, right? [clears throat] So, if I'm to compare these two here, what am I having? I'm going to say if I'm having this amount here, which is 0, 16 673, how much am I going to get this side, right? So, for me to get how much of each here, what I'm supposed to do is say 3 over 2 multiplied by 0, 1 6673, right? So, if I do my math correct there, I'm going to get a value of what?
0, 25. So, it's 0, 25. So, it means here these are the moles that you're going to get there, 0, 25, right? Then after that now I can now proceed to calculate the number of moles of what?
Uh the the mass of what? Of hydrogen gas that is going to what? Produce there.
I'm going to say H2 is equivalent to M all over M, right? Okay. So, now the number of moles I already have is 0, 25.
Therefore, I'm going to say 0, 25 is equivalent to M that I'm looking for and then the molar mass of hydrogen gas there is going to be two. Therefore, if I multiply that one there, it's going to give me an answer of 0, 5 g, right?
Okay. So, now having said that, we are done with question uh 5.3.3.
Now, moving on to the next question, which is 5.3.4. It says if a lump of uh chromium is used, how will this affect the rate of reaction, right? So, choose from increase, decrease, or remain the same, right? So, if a lump of chromium is used, how will this affect the rate of the reaction? Choose from increase, decrease, or remain the same, okay? Okay.
Oh.
It's actually going to to decrease. Oh.
Yeah. Yeah. Yeah, I left out something there. I didn't finish because here I was looking for uh sort of about that one. We're looking for the percentage yield, right? So, it means I didn't finish calculating there.
Therefore, now the mass that I'm going to get here, which is 0.5 g there, now I'm supposed to proceed and make use of it so that I can get the percentage yield, right? So, oh, percentage yield I'm going to say percentage yield. How do you calculate percentage yield? Is equivalent to the actual yield, right?
All over uh actual yield all over the theoretical yield, right?
Theoretical yield. Theoretical yield like that, okay? So, having that in mind now, it means the actual yield that we got there was I mean 0.45 g and then the the one that was at the theoretical uh is going to be 0.5, right? Then multiplied by 0.5 and then multiplied by 100. Therefore, we are going to get a value of what? How many percent? Which is 90%, okay?
There. That's how we go about that one there.
Uh then after that now, we are going to say, "If a lump of chromium is used, how will this affect the rate of the reaction?" Choose from increase, decrease, or remain the same, right? So, if we're to use a lump there, it means the rate of reaction there is going to what? To decrease there, right? So, how do you explain there? It's a matter of explaining in terms of what? A surface area, right? So, surface area decreases.
There's a decrease in surface area.
Decrease in surface area.
Okay? Then after that now, it means now we are going to have what? Uh less uh particles uh uh having sufficient EK, so less particles having uh, sufficient sufficient EK for effective effective collisions and then after that you're going to say reaction rate decreases.
Okay, so that's how we go about 5.3.5 there. Then we are now done with question five. Now moving on to question number six.
So question six reads here, the equilibrium mixture of nitrogen dioxide, which is our NO2, a brown gas with a sharp, uh, pungent, uh, smell and a colorless dinitrogen tetraoxide, which is our N2O4, is investigated using a sealed industrial gas syringe there.
This color change allows the progress of the equilibrium to be observed visually, right? So here we're having a pressure gauge and also then, uh, the balanced equation there is given. So this, uh, balanced equation there we're having brown to to colorless there, right? Then after that now they're saying state Le Chatelier's principle.
Le Chatelier's principle is stated as saying, uh, I'll just write it maybe at the top there. I'm going to say when the equilibrium the equilibrium So equilibrium in a closed system in a closed system So if an equilibrium in a closed system is disturbed, right? Is disturbed, okay?
So maybe if I can talk about the disturbance there, we are going to have three disturbances, which is our concentration, uh, pressure as well as what?
Temperature. Then, after that disturbance, the system the system uh will reinstate, will reinstate, reinstate in state.
A new equilibrium a new equilibrium, equilibrium, okay? So, the system will reinstate a new equilibrium uh by favoring by favoring direction direction that will oppose that will oppose the disturbance.
Okay, so that's how we define uh we state Le Chatelier's principle. Then, 6.2 says, "Is the system above open or closed and give a reason for for the answer." So, here we are going to say here we used what? A closed system, right? Uh and then uh because the reason why it is uh isolated there, right? So, it's isolated from it's what? From its surroundings. Then, 6.3 says, "How will each of the following changes affect the concentration of what? N2O4 gas at equilibrium, right?" So, choose from increase, decrease, or remain the same.
Then, 6.31 says, "The plunger is pressed to decrease the volume, right?" So, it means now we are increasing pressure.
So, if you increase pressure, what is it going to happen there? Uh how will each of the following affect the concentration? Right, so if we do that, like it means now our concentration is going to what? To increase.
It increases. And then, a catalyst is added has been added. So, remember this one favors both the forward and the reverse reaction at the same time, right? So, I'm going to say remains the same, right? Okay, so it's just like speeds up the the of reaction, this one.
Like but uh nothing is going to change then. Then uh the amount of products that are still going to remain the same, only that you're going to get them what?
At a faster rate. Then 6.4 it says the rate This syringe is now placed into the water bath filled with what? Ice cubes, right? So, there we should take note of these ice cubes. So, it means now we're cooling down everything. And then the gas in the syringe starts to turn what?
Colorless, right? So, is the forward reaction endothermic or exothermic, right? Then I'm going to say the forward reaction is going to be our exothermic.
Then we are asked to use Le Chatelier's principle to explain the answer to question 6.4.1, okay?
Uh so, now I'm going to say placing if we were to place the syringe in what? In the ice cubes, right? Syringe putting, okay?
Putting or placing, putting or placing uh the syringe Yeah, like that, okay? Uh in ice cubes, in ice cubes, ice cubes decreases the temperature.
Decreases the temperature. Then bullet number two there, I'm going to say endothermic reaction is favored.
The endothermic reaction is favored, right?
So, I'm going to say here the forward reaction is favored.
Okay? The forward reaction is is favored because the because the syringe turns colorless.
Okay. So, that's our answer there to 6.4.2.
Then 6.5 b it is here the salt in the straw uh gas syringe is removed from the water bath and is allowed to reach equilibrium at 298 K there.
Uh 298 K, then at equilibrium the mixture contains Okay, so it contains 0.52 moles of NO2 and 0.08 moles of N2O4, right? Uh the volume of the syringe is 1,000 cubic centimeters there, right? Okay. Then after that now, we are told the syringe is then placed in hot water, okay?
Uh but at 37 3 K, right? Then a new equilibrium is established at which point uh the mixture contains 8.28 g of what? N2O4.
Then calculate the number of moles of N2O4, right? So we're looking at the number of what? Of moles there of N2O4 at what? At equilibrium there, right?
So this one I'll just like maybe quickly uh work it out here. So I'm going to say the number of moles of N2O4 is equivalent to remember we're going to say the mass all over the the molar mass, right? So which is equivalent to the mass that is given here is 8.28 g there. So I'm going to say 8.28 all over the molar mass of N2O4, right?
So if you can look there, I'm going to say nitrogen is 14 * 2, then plus my oxygen is 16 by 4, right? So if I put this into my calculator, it gives me a value of 0.09 moles per bit, okay? So we are done with that one. Then moving on to 6.5.2, it says calculate the value of the equilibrium constant, which is your KC, at this what? Temperature, right? Okay.
So our equilibrium constant there at this temperature, what we need to do here is to formulate our what? Our ICE table, right? So, I'm going to say here in terms of this, maybe let's space there.
Okay, so this is what I'm going to do here. I'm going to have like my reactants and then my products and all, right? So, in terms of my reactants there, I'm having NO2.
Then uh in terms of my products here, what I'm having here is uh my N2O4, N2O4, right? Okay. Then after that now, I'm going to start by saying, "Okay, this is what I'm having there." I'm going to start with the with the uh which is going to be my ratio, right? So, okay.
So, in terms of the ratio here, we can see from the balanced equation there, uh is 2 is 2 1, right? So, you can say is 2 is 2 is 2 1 there. So, I'm going to just like maybe change color. So, I'm going to say this side I'm having two here, I'm having one, right? And then here, I'm going to say my I.
So, my I for ICE is the initial moles, right? So, initial number of moles that you're having there. Then after that now, I'm going to refer again, right?
So, here in terms of this, we are given the initial number of moles, right?
Okay.
Where is it? Okay. There. These are the initial number of moles at equilibrium, uh the mole contains 0.52, okay? So, I'm going to say initial moles 0.52 there.
And then also, what do we have there?
We're also having 0.8 moles of N2O4, right? So, I'm going to put there as uh my 0.52 here, 0.52.
And then here, I'm having 0.08 like that, okay? Then after that now, I'm going to say the change in moles, C which is change. I'm going to say change in moles like that, okay? So, what was the change in terms of our moles there, right? Okay.
Let me just like proceed with this line there going downwards like that, okay? So, that I can have more space.
And then also like continue like this, okay? So, here I'm going to say change in moles, then here I'm going to say equilib moles or moles at equilibrium, right?
And then here is going to be our concentration what? At equilib, okay?
That's what you have in there. After that now I'm going to like proceed to say, "Okay." So, uh having that in mind now, I'm going to look at this, right? So, I'm having 8.28 there. Uh it says a new equilibrium is established at which the point which point the mixture contains 8.28 g of N2O4, right? So, you can see the number of moles at equilibrium for this one. We already kept what them is what?
0.09, right? So, I'm going to put them here. So, these are my equilibrium moles 0. 09, right? So, as a matter of fact here, you should take note here. These are my products, right? So, my products are being formulated what? Whilst here, I'm going to get rid of making use of what? Of NO2. NO2 is being used up, whilst N2O4 is being what? Formulated, right? So, that's what I'm going to do here. So, if I started with 0.08 here, and then now I ended with 0.09, right?
So, what does that mean? It means I added what? 0.01 moles, right? Then after that now I'm going to use the ratio, the mole ratio to get this one, right? So, I'm going to say here the ratio is 2:1, right? So, if I change what? 0.01, so what does that mean this side? It means here I'm going to multiply by 2. So, I'm going to say 0.01 multiplied by 2, which will give me a 0.02 like that, okay? Then after that now I'm going to say moles at equilibrium, right?
So, the moles at equilibrium I'm going to just like subtract there. So, it means here I'm going to get a 2 - 2 is 0, 5 - 0, I'm going to have a 5. Then the 0 - 0 is going to be 0.50. Then after that now I'm going to calculate the the concentration at what at equilibrium, right? So, remember concentration at equilibrium concentration is given to number of moles all over the the volume, right?
So, I'm going to say this one for NO2.
For NO2, maybe let me just like say concentration of what? Of C brackets NO 2 like that, right? Is equivalent to that. And then the number of moles that I'm having there is 0.5, right? All over Then now I'm going to say volume, right?
So, if we are to look here, we are given the volume there is what? 8,000 cubic centimeters. So, this one converts to one if you convert it to what? To dm cubed, right? So, if I divide that one there, you'll find that I'm still going to retain the same value, which is 0.50, right? So, it's 0.05, 0.50 there. And then after that now I'm also going to do the same. The concentration of what?
N2O4 there is equivalent to N all over V. So, my number of moles that I'm having is 0.9, right? All over the volume, which is one there. So, still going to retain the same value, which is 0.09, right? Then after that I'm going to say my KC there is the concentration of products, which is N2O4 all over the concentration of my reactants there, which is all my NO2 like that, right?
So, if I look at that, the values that I'm going to have there is what? My product What do I have? I'm going to have 0.9 and then like that. And then all over then 0.50.
Then I'm supposed to have to square that one there. So, if I do that, I'm going to get a value of what?
>> [gasps] >> 0.36, right? So, remember KC does not have what? Units like that, okay? So, that's how we go about in terms of calculating what? Our KC expression, right? Okay?
Then moving on to the seventh question there.
Question seven reads, hydrochloric acid, which is our HCL, is a strong acid.
It dissolves in water as shown in the equation below, right? Okay, so this is the equation that is being given there.
Then they say explain what is meant by a strong acid. Okay. So, in strong acid, what we know it ionizes completely. So, I'm going to say ionizes completely.
Okay, completely.
Completely.
Okay, in water. In water.
To form to form a high concentration a high concentration concentration of H3O+ H3O+ ions like that. Okay. So, they're asking us 7.2 to identify one conjugate acid-base pair, right? So, one conjugate acid-base pair, right? So, if you are to look at this, right? So, how do you pair these ones? Automatically, the difference between this one here and that one there it should be a hydrogen, right? So, if you can look at that the difference is going to be what? A hydrogen. So, either here we can go for the first option which is going to be our H3O+ H3O+ Okay.
Maybe let me start with my reactants then the products, right? Okay, so what I'm going to have there is going to say H2O H2O versus my what? My H3O+ H3O+ Then also what do I have there? I also have my HCL versus my Cl- there. Okay, so one of the two there is going to what? To be correct. So, you either goes for this one or go for that one, right? Then hydrochloric acid uh 7.3 hydrochloric acid is monoprotic, right? Uh give a reason for this uh for this, right? So, it's more of us defining the term monoprotic, right? So, it means here it can donate uh one proton, right?
Can donate one proton.
Maybe if I can be specific, I just can say only can donate one proton there, right? That's the reason why this monoprotic, right? Then moving on to the next question, which is our 7.4. 7.4 is here. So, we're given a volume of 150 cubic centimeter hydrochloric acid your HCl there uh solution with a concentration of what? So, this is the concentration available for the lab uh experiments. Calculate the number of moles, right? So, we're given concentration volume, then we're asked to calculate the the number of what? Of moles there, right? Okay. So, basically what I can say here is that uh I can just say here uh concentration of uh HCl there is equivalent to number of moles.
Concentration is given to number of moles over volume. Therefore, here we're asked in terms of number of moles, which is same as CV. Therefore, I'm going to like quickly multiply this one. So, it's 0.8, which is our mole concentration there, and then the volume there, I'm going to say 0.1 by 5 0.15, right? So, what did I do? I converted this to dm cube by dividing by what? By 1,000 there. So, you divide by 1,000, then it gives you that 0.15. So, if I multiply that one there, it gives me a value of 0.12 moles like that, okay? So, we're done with that one there. Then moving on to 7.4.1.
7.4.1, so uh it's here. uh >> [clears throat] >> Oh, wait. Let me save it for now. Then after that now it says all the all of the 150 cubic centimeter hydrochloric acid solution which is your HCl is allowed to react with x grams of calcium carbonate right? So which is your CaCO3 solid according to the balanced chemical equation. So this is the balanced equation that we are given there. Then after that now they are saying the hydrochloric solution is found to be in what? In excess. So we are having excess HCl. Then after that the excess HCl is neutralized by 60 cubic centimeters of sodium hydroxide which is NaOH solution with a concentration of that, right? Then uh mole dm cubed there. Then the neutralization reaction is given like this, okay? So this is what you are having there. Then after that we are asked to calculate the value of what? Of x there.
Okay? So x is in grams, right? Which is our mass there. So we're looking for the mass there, right? So what I'm going to say here is that I'm going to start with the number of moles. Maybe let me start with the number of moles of what? Of sodium hydroxide, right? Okay? So I'm going to say the number of moles of sodium hydroxide, okay? So the number of moles of NaOH is equivalent to >> [clears throat] >> okay? So if I go back there, what am I given there? So it says here 0.5 right? So uh the excess hydrochloric acid is neutralized by so I'm given the volume there and then I'm also given what? The concentration of what? NaOH. So it means now I can calculate the number of moles there. So I'm going to say 0.06 multiplied by number of moles there. Remember concentration we started from c is equivalent to n over v. Then here it means is c times what? Times volume there, right? So, it means here the volume concentration is 0.5 and then the volume is 60 cubic centimeters.
Therefore, I'm supposed to convert that one then it's going to give me 0.5. So, if I multiply there, it's going to give me 0.03 moles, right? So, this is CV the concentration and volume. Or maybe let me just say VC so that everyone understand which one is the concentration and which one is the the volume there, right? Then after that now, look for excess HCl, right? So, excess HCl is going to be equivalent to 0.03 moles like that, okay? So, these are the moles in excess of what? Of HCl there, right? So, now how much of this HCl was used by calcium carbonate? So, I'm going to say your CaCO3 there, how much of it was used there? I'm going to say 0.012 0.012 from this one there at the top, right?
So, that's what I'm going to get there.
I'm going to say 0.012 and then what am I going to subtract there? I'm going to subtract these ones, right? So, it's going to be 0.03, right?
So, what am I going to get there? I'm going to get a value of 0.09 moles, right? So, these are the moles that I'm going to have there. So, of the ones that actually used, right? So, then for me to calculate the number of moles of calcium carbonate there, I'm going to say is equivalent to Remember, number of moles is equivalent to concentration over volume, so I'm going to say 0.09 all over concentration is equivalent to number of moles over volume, but so I'm going to say all over what? Two there, right? So, which is going to give me 0.045, right? So, how did I get to 0.09 divided [clears throat] by two there, right? Okay, so if you can go back there in terms of this, you'll find that the mole ratio here is 1:2, right? So, this is the mole ratio there 1:1:2 there, so that's why dividing this one by what? By 2, because the mole ratio of for HCl versus CaCO3 there is 1 is to 2, right?
So, it's more of saying if I'm having 0,09 this side, uh how much am I going to have this side, right? So, it's going to be like uh Uh this is 2, sorry, this is 1, right?
So, this side I'm going to have a half of that, so which is 0,045 like that, okay? Then after that now, I'm going to proceed to say to say here, uh after getting that, I'm going to say the mass of X now, I'm going to proceed to get the mass of X, right? So, the mass of X is equivalent to I'm going to say mass of X, this is the one that you're having there, right?
Is equivalent to uh 0, of X there is equivalent to 0, 0.045, and then I'm going to multiply it by what? By 100 there. So, which is going to give me an answer of what? 4.5 g there, okay? Then after that now, I'm going to proceed to the next question, which is our 4 7.4.3.
So, 7.4.3 it says, "Which will the salt that is formed in a neutralization reaction undergo hydrolysis?" Give a reason for your answer. So, I'm going to say no. So, the reason behind is that sodium chloride is a salt of a what? Of a strong base and a strong what? Acid. It is formed by a strong base and a strong uh acid there, right? Then after that now, so it's going to be actually what? In neutral. So, after that 7.4.4, bromothymol blue is used as an indicator. Explain why it is a most suitable indicator for this titration by referring to the pH of the equivalent what? Point there, right? Okay, so it means now, what I'm going to say here, uh is to say uh maybe if you can go ahead and calculate, explain why it is the most suitable indicator for this titration, right? So, this is more of like a titration between what? A strong acid and what? A strong base there. That's the reason it is there. Then after that now, moving on to the next question there, which is going to be our 7.5. It says in an aqueous solution there, the concentration of hydronium ion and hydroxide ions determine the acidity or alkalinity of a solution, right? Then we are asked to calculate the pH there of potassium hydroxide, which is KOH solution with a concentration of what? 2.5 * 10 -7 mol/dm³ there. Then at 298 what? Kelvin's like that, okay? So, now for me to calculate there, I'm going to say, okay, basically what am I given there? I'm given the concentration of hydroxide ions, which is equivalent to 2.5 * 10 -7 there, right? So, in terms of my pH, remember I can say pH pH plus my pOH is equivalent to 14 like that, okay? So, if I have that in mind now, I can proceed and say pOH is equivalent to negative log then bracket 2.5 * 10 -7 like that, right? So, if I do the math there, it's going to give me a value of 6.60.
Then after that now, I can say my pH is equivalent to 14, right? So, if I can transpose there, it's going to be 14 and then I'm just going to subtract the 6.60, right? So, it means my pH there is going to give me a pH of what? 7.40, okay? So, with that being said, uh we've come to an end of our paper. I hope you did actually enjoy the discussion. Thank you for watching.
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