CXC CSET Chemistry Exam Insights

This is the CXC CSET Chemistry Paper 2 from May June 2023. Right, so we're just going to get right into it. Proceed with caution. I should give the disclaimer that these are not the answers from CXC. I'm giving you a breakdown of what the answer should be, what CXC accepts, you know, it might. vary but we um as best as possible will aim to give the correct answers right so so number one is your good old data analysis um question we can always depend on that um number one here we're being asked not to spend more than 30 minutes on question one no number one hydrogen peroxide h2o2 decomposes to produce oxygen and water in the presence of potassium iodide as a catalyst. The effect of the concentration of the hydrogen peroxide on the rate of reaction was investigated. For each experiment, the concentration of the hydrogen peroxide was changed and the time taken for the hydrogen peroxide to decompose was measured using a stop clock as shown in figure 1. 1 of A. Define the term rate of reaction. Change in concentration of reactant or product per unit time. Anything having to do with rate, time has to be measured. Define the term catalyst. So it's a substance that is used to speed up the rate of a reaction but remains chemically unchanged in the process. So the catalyst does not take part in the reaction. We could put that, we could state that as well. A substance used to speed up the rate of a reaction but does not take part in the reaction. All right, so that will give us three marks. All right, so part B. For each experiment shown on the stop clocks in figure one on page four, record in table one the time taken for the hydrogen peroxide to decompose. Write down these values to be used in our table a little later on before we get into anything big. So at first we always look to see what the scale is on whatever the instrument is. each of those stroke there is representing one second so for the first one in experiment one this is 59 in experiment two we have this at 42 in experiment three this is at 31 in experiment four this is at 25 in experiment five this is at 21. so we're going to record those we're going to keep those close by for a little later on. So we're going to do that. We made a note of those times. So we just need to fill in that in the time column. And then after that, in part C, we're just going to do everything in one. In part C, we're to complete table one by calculating the rate of reaction. This time, the rate of reaction is one over time, and they want it calculated to three decimal places for each of the experiments. So, the first one, we had mentioned that the first one, for the first one, the time is 59 seconds. Second one is 42 seconds. The third one, 31 seconds. The fourth one, 25 seconds. The fifth one, 21 seconds. And we're just going to take the inverse of those to say 1 divided by 59, and that will give us 0. 0.017 that's three decimal places then 1 divided by 42 0.024 1 divided by 31 0.032 1 divided by 25 0.040 three decimal places and then 1 divided by 21 which would be 0.048 So we're going to keep these in mind we have to bear these in mind because we're going to Use these to actually plot our graph. Using the axes provided in figure 2 on page 7, plot a graph of rate of reaction versus concentration of hydrogen peroxide from the information in table 1. Draw the line of best fit through the points and this is for five marks. So we're going to head on over to our graph sheet and we're going to plot a graph of Rate of reaction versus concentration. So rate of reaction here is our dependent variable. The rate of reaction will get its value from the concentration of the hydrogen peroxide. So the hydrogen peroxide then is the independent. The concentration of the hydrogen peroxide is the independent variable that goes on the x-axis. And then our responding variable. or our dependent variable is our rate of reaction that goes on the y-axis. So let's go, let's figure out, let's go, let's figure out the scales on each axis and then let's plot and then let's draw our line of this fit and get five marks. All right so we have our values that we're going to plot, we wrote them down a little while ago so we just go right into plotting but before we plot we have to establish we have to figure out the scale in each case all right so on the let's look on the x-axis we find that um 2cm they use 2cm 2cm to represent 0.05 units right so that means that each of those little milli what's that each of those millimeter going up by 0.005 each of those little millimeter is going up by 0.005 and on the y-axis we're seeing where 2 cm 2 centimeters there they're being used here to represent 0.00 So it means that each of those, each of those millimeter, that's the tiny box, each of those right here, that's the one I can actually touch. Each of those would be representing 0.0005. 0.1 mole per dm cube the rate there is 0.017 so we need to find 0.017 so this is 0.015 so this would be zero okay don't need to mark that so we have 0.015 there so 0.016 0.017 and we would circle this and then the next one 0.15 is 0.024 so we find 0.15 then we go up to 0.024 that's 0.025 here so two two behind let's see 0.020 0.021 0.022 0.023 and 024 all right and then following that we have 0.2 0.2 and that would be 0.032 0.032 0.032 would be here we're circling we can use n circle dots or we can use x's all right and then the next one 0.0 sorry 0.25 0.25 on the x and for that one we have 0.04 0.25 and 0.04 so that would be here and then after that we now have 0.30 and zero point corresponding with that 0.30 we have 0.048 0.048 would be it would be here, right? So the next thing that we need to do is to know. So we've plotted all our points, all five points correctly. The next thing would be to now draw our line of best fit. Now our line of best fit does not always have to go through the origin. But it appears that this one would actually go through the origin. So let's do that. We need a ruler to do it. I'm exempted based on what I'm using. But you would use your ruler. You try to get all the points to fall on this. on a straight line if it doesn't if it doesn't you try if they don't then you try to get as many of them falling on the line as possible with an equal number falling off but this this seems like a good trend line or that will get um them falling falling on a straight line here Even if it did not go to the origin, that would have been fine. So we're now going to move on to the questions using this very graph that we have plotted. All right. So we're at part E. Using your graph, well, part one of E, describe the relationship between the rate of reaction and the concentration of the hydrogen peroxide for two marks. So we need to go back to our graph and look to see what. what the relationship is we look to see how the rate is being affected by changing the concentration so let us see what's happening here so from this we're seeing that what are we seeing we're seeing that as as they so follow on the x-axis as the concentration increases so increases taking place here let's use green So as the concentration is increasing, as we're going over here, we're seeing that, hey, the rate is also increasing. So we're seeing that as the concentration of the peroxide increases, the rate of reaction increases as well. So we would need to state that. So we just need to toggle between. So we're flipping pages now. So we're going between pages. So we need to write that down as the concentration of. peroxide increases the rate of reaction increases right and then the next question they want us to determine the concentration of the hydrogen peroxide given that the rate of reaction is 0.045 per second. All right. So we would need to go back and we'd find 0.045 per second on the y-axis. So we need to find 0.045 seconds per second rather on the y-axis. so let's use a different color to represent this and once we find it on the y-axis we're going to draw a line from it going across and wherever that line cuts our best fit line our curve we're going to draw a line or a straight line from that coming down to cut the x-axis wherever it cuts wherever it cuts the x-axis right did it come all the way we need to come all the way right and from what we see here this looks like it is zero point zero point two sorry zero point two five six so that we'll write it boldly so that is zero point two five six mole per dm cube that would be the concentration okay so we make a note of that and then so in part f we are to calculate the mass of the hydrogen peroxide present in the concentration identified in E2. which is 0.25 moles per day cubed. So we're pretty much converting here from moles to, from moles to mass. Molar concentration, it's, it's the same as saying, so mass concentration here is the same as saying moles being converted to grams. but the number of moles is in a set volume so we say moles over the m cube or moles per dmq now in order for us to move from moles to the mass we'd have to multiply by the molar mass which is in grams per mole and if we look at this we see that mole will cancel out mole and will be left with grams per dm cube so this will give us this will be 0.2 0.256 moles per dm cubed times 34.01 grams per mole and this looks like a lot of working out for one we're not just working we're not just looking for the answer we're looking at how we arrive at the answer so we're teaching you doing everything all in one so that's 0.256 times 34.01 and that will give us everything so far has been to three decimal places so this is and then we have three decimal places here so we'll just be consistent so this is 8.707 grams per bmq and they want us to know write a balanced chemical equation including state symbols show the decomposition of hydrogen peroxide all right so hydrogen peroxide you know pure hydrogen peroxide is a liquid but they were varying the concentration here so it means that it would be a a solution of a liquid in a liquid peroxide is the solute and the water is the solvent so we would write h2o2 very important it's not the same as water that second two makes a big difference don't take my word for it or you should take my word for it don't want you to try drinking this now all right so this will give us h2o which is a liquid plus we'll get o2 gas so to balance this would need to put a two oxygens are off we have three on the right side two on the on the left side now if you put a two here and a two here that would fix pretty much everything for everyone so that's the balance equation for three more marks and 25 marks full marks let's check our time did we spend more than 30 minutes on this question all right let's move on to the second question okay number two part one of a or a at atmospheric pressure water exists in three states of matter while carbon dioxide exists in only two states one list the three states of matter in which water can exist and this is for for one mark and as usual we're thankful for small mercies so those three states that water can exist in at atmospheric pressure would be solid liquid and gas. Then they want us to describe the energy of the particles in each of the three states listed in part one of a above. So let's go with the solid state. So in the solid state the energy that we're describing here is the kinetic energy that the particles would have as a result of their motion. So in a solid the particles are arranged in a fixed way held by very strong forces of attraction so they have nowhere to move to they can only vibrate. So in a solid there is very little there is little. very little to no kinetic energy. Alright? In a liquid we could say there is a moderate amount and then in a gas the forces of attraction are very it's almost non-existent so these particles can move in any direction all about the place with a large amount of space between them so we say that there's large or enormous amount of kinetic energy present part three named the process which of course when carbon dioxide changes from one state to another. We have to appreciate the two states that it's in first. It can be in the solid state, it can be in the gaseous state, so that would be sublimation. Part four, describe how the arrangement of the carbon dioxide particles changes as carbon dioxide undergoes the process named in the a3 above. its sublimation we'd have to decide which state we're moving from whether we're going from the gas to the solid or the solid to the gas all right so we're gonna start with the gas here so let's just describe what happens first when we're in the gaseous state well let's describe what happens first when we're in the solid state and then as we move from the solid to the gaseous state heat would have to be applied so let's go Alright, so we could mention here that it actually skips or they skip the liquid phase. So in the solid state the particles in dry ice are very close to each other in a regular way. as heat is added or as heat is applied the particles begin to vibrate and quickly move into the state where they have large amounts of space between them that's into the into the gaseous state so that is the gaseous state and here they want us to use appropriate diagrams to illustrate the lattice structure of sodium chloride crystals and the giant molecular structure of diamond. I believe there's a neater way to do it than I'm doing it right now but as always this is not technical drawing we just want to show we just want to be able to draw something that looks like a square. then we get our parallel lines so we can get it to look like a cube then we can take it from there all right so we're going to look at the lines at the back which are going down well that one that will join that it's not a perfect cube but we can actually work with this so all right so in this one i'm gonna put chloride at each vertex or what we call corner so chlorides are the vertices so i'm using the big one to represent chloride so we have alternating chloride and sodium so this would be let's do the key so this is chloride ion and then this little one will be the sodium ion na plus ion so sodium would be here Sodium would be here as well. Sodium here. Sodium here. And in the middle there would be a chloride and would draw a line to show that there there's you know electrostatic attraction between them. There's the bond right there, the ionic bond. All right so sodium would be here. Here as well. Here. Chloride would be here. And we could have put sodium at each vertex. What's important to note is that each chloride ion is surrounded by six sodium ion and each Sodium ion is surrounded by 6 chloride ions, whichever way we sample. All right, so let's continue. So here we have sodium, sodium, sodium, chloride. Sodium would be here. Sodium would be here. Chloride would be there at the back. we can line that up much better all right so line would be coming down through there one would be going across there there's a chloride at the one at the back line coming down line going across it probably should be blowing your mind now there is a chloride at the front i didn't focus on the front one so then right there and then there's one more so them would be in the middle a small one and then from the top going through at the bottom so that should give us our full mark then a much simpler one for diamond we're going to use carbon and we're going to show that one carbon is bonded to four more so this is our carbon in the center so each c represents a carbon atom and the line here represents a strong or we don't have to say strong we could say a covalent bond so that would give us three marks each and that would make it six so we're down to the last two in number two to make it up to 15 we have 13 so far part c in solution two metals a and b form a plus and b2 plus ions respectively so it means the a forms A plus and B forms B2 plus. Metal A displaces Fe from a solution containing Fe2 plus ions but metal B does not. Write a balanced ionic equation to show the reaction between metal A and Fe2 plus. All right, kind of expecting something more like a reactivity series. based on this but this is where they stop it so we'll just take it and be thankful for small mercies again so all right so a is going to displace fe2 plus a is in is being treated as something in group one it has a valency of one so it will be it will be a and a here would be a solid so Fe2 plus would be an aqueous medium and from this we would end up with A plus plus Fe solid. But each time for every atom or every mole of A that reacts, we get one mole of A plus or we get one mole of electrons actually being liberated. for each mole of the Fe2 plus we need two moles of electrons so it means then even though we're not seeing the electrons we would actually we'd actually have this is what we have happening A is actually reacting to give A plus plus an electron all right and then on so that is um oxidation right and then on the other side we have Fe2 plus actually gaining two electrons to be liberated as Fe the solid. So if we have two electrons on the left hand side we need two electrons on the right hand side. So to get that we would need two electrons but we can't go in and multiply the electrons by two without multiplying the A plus by two and anything we do to one side we have to do it to the other side. So we'll end up with that. Then we remove the electrons and then we write back what we have. So we need 2A Right here. And we would need 2A plus. And that would do it. And this would give us the additional two marks to make it up to 15. How many have you gotten so far? All right. Just like that, we've come to the end of this question. Continue with the others and see how many marks you've missed so far. Couple later. Okay, so we're at number three. And this is an organic chemistry question. Figure three shows the fully displayed structures of three compounds, A, B, and C, which are from different homologous series. All right, so let's examine compound A. like a saturated hydrocarbon all we have is a carbon carbon single bond and carbon hydrogen single bond all right so in compound b we have a functional group and that's the c double bond o o h and then in compound c we have a functional group that's the o h functional group so okay so for a saturated hydrocarbon only made of carbon and hydrogen so it's like uh hydrocarbon is in the true truest sense of the word so this is coming from the alkane series so that belongs to the alkanes compound b cooh that is our carboxylic group so this is coming from the carboxylic acid group or the carboxylic it's just called carboxylic acids or the alkanoic acids now they're saying compound c burns with a blue flame in oxygen yes it does that's some ethanol and they want us to write a balanced equation for this reaction so then let's just get right ahead and do that so ethanol here c2h5oh that's the liquid burns in oxygen whenever we write oxygen as an element like this we write it as o2 anything that ends in gen r in is diatomic so this will give us just like when any other organic compound burns this will give us carbon dioxide and water water here in the form of a vapor okay so straight up let's count we have two carbons on the left we have one on the right so to fix that and we're always balancing the order so we go with the c's first so we put that two there that fixes the two carbons for the right one the left so that's good when i look at the hydrogen cho so h comes next we have five here on the left plus one that's six over that side and over the right side we have you 2 so to make that equal to 6 we need to put a 3 right here right in front of the water and that makes 3 2 6 so the carbons the hydrogens are good so it's now a matter of the oxygen so this would be two twos four from the carbon dioxide and then three from the water that's seven a little oddish so we have one oxygen over here on the left from the ethanol and we have two here if we put a three here that will give us three to six plus the one over here in the ethanol that would make it seven so this would be our balance equation and then for part c state which of the two compounds a or c is more soluble in water give a reason for your response all right a or c well for one you straight up off the bat we're gonna go with compound c compound c is more soluble in water for one water is water is polar and compound c compound c so water is polar and we know that light dissolves light now compound c has an oh group you So the presence of the OH group means that this compound is polar, unlike compound A, where we have just carbon and hydrogen. So this is a non-polar compound. So compound A is polar. So we just need to state, OK, compound C has the OH group, so is polar. and dissolves in Water, which is also polar. We could also state that, not to over-answer, but we could state that compound A is non-polar. And we know that light dissolves light. Okay. Part D. State whether compound B or compound C would react more vigorously with sodium metal and give a reason for your choice. No. Both of them would react with... sodium here now it comes down to a matter of which of them would let go of that hydrogen more readily that hydrogen at the end well for one compound b is a weak acid compound c you know it can be as an acid but it's even weaker than um compound b so the one that would be more reactive in this case the one that would react more vigorously would be compound a B and so we should give we should give a reason for our answer now the hydrogen in compound B is more readily lost or is more readily replaced by the metal than that hydrogen in compound C so the hydrogen in compound B is more readily lost than the hydrogen in compound C. And, you know, this is due to the fact that we have the presence of a carbon carbon, sorry, not a carbon carbon, carbon oxygen double bond, which adds to the polarity of the CO. Well, it adds to the polarity of the OH group in compound B. All right. But that would be, that's probably even over answering. So we're just going to state exactly why that the We'll state that B is more polar with the C double bond O. So the H in the OH is more readily given up than the H. in the OH. So the hydrogen in the OH of COOH in compound B is more readily lost than the hydrogen in the OH of compound C. And we'll just leave it right there. So, so far for this question, that's 2, 4, 6, 8 marks. So let's go for the other 7 marks. Okay, in E, write a balanced equation for the reaction of compound C with sodium metal. Okay, so compound C is ethanol. So we're going to go again that C2H5OH. And this is a liquid which is going to react with sodium, which is a solid. In this case, that H at the end is going to be acting as a replaceable hydrogen with a very... metal like sodium so we'd end up with C 2 H 5 O and E and this is actually can put a LC this is an alcoholic you call it alcoholic alcoholic medium it's not aqueous no water is here we're talking about liquid ethanol and then this would liberate hydrogen but we cannot write h by itself it's diatomic so it would have to be h2 gas and then of course if we're going to do that we have to just finish off and ensure that we balance so for us to have h2 there we would need two right there and then that would make everything here two on the left you the ethanol so we just need to match that off with a two just ensure that this is visible two and then the sodium would need to be two and then that is it now part f describe one test that could be used to identify the gas that is produced in the reaction of compound c with sodium metal of course we'll have to tell them what we expect not just the test but what we expect to observe so we could to just hold a lighted splint at the mouth of the test tube and so all right so the gas present will put out the lighted splint with a pop it's hydrogen use the right terms so the pop that we hear is a mini explosion sometimes it's not really poppy it's squeaky sometimes And it just depends on the volume of the gas that is present. And that little pop or that squeak is just a mini explosion because hydrogen is explosive. Yes, that's right. Part G. Compound B and compound C react together in the presence of a catalyst to form compound E. State the name of the catalyst. All right, just a reminder of what compound E is. Compound E. propanoic acid that's compound b and compound c we said earlier is ethanol so they're asking for the name of the catalyst the name of the catalyst well would use um sulfuric acid sulfuric acid is our catalyst and they want the fully displayed structure of compound d now whenever compound B, this is propanoic acid. Whenever we're drawing or we're showing the product, the ester formed when an organic acid reacts with an alcohol, we draw the acid part first. Okay, so probably what we should do, perhaps what we could do, we could redraw this in blue. We're actually seeing where everything is coming from. Retract the red, retract the blue. So we're going to pinch off. Let's use black to show or green to show what's happening. We're going to join these together. We're going to pinch off the OH from the acid and the H from the alcohol, which will give us water. We're not so concerned about the water. We want the fully displayed structure of compound D. All right. So we're going to go with red first. So we have a CH3 from the propanoic acid. And we're just going to put. on the H's as we go along. Then we have a CH2. Then we have C with the double bond O. And then from that now we're going to go on to the blue compound. So we're going to have now C. We have C double bond O from the red. So we're going to have the blue. Then we're going to have a CH2. Then we're going to have a pH 3. They didn't ask us to name this. So this is our compound. And whenever I draw, whenever I draw an ester, it's just second nature for me to highlight the carb, highlight the, for me to highlight the ester linkage. So that is our compound. They did not ask us to name it. It's ethyl propanoate nonetheless. All right. just like that we've come to the end of the section how many months number four sulfur and magnesium are two elements in the same period of the periodic table the different properties of the oxides of these elements are presented in table two all right so we have the oxide of sulfur it's a gas that's the state the melting point is negative 72 degrees celsius and for the oxide of magnesium. It's a solid and the melting point is 2,852. Whoa, that's a lot. That requires a lot of energy. Now, with reference to bonding, explain the difference in melting point between the oxides of sulfur and magnesium. So we're going to be talking about the bonding that's there. Don't want to get too much into the structure. It's hard to talk about the bonding without talking about the structure, but we want to be very disciplined so we can get our full marks. So let's just go. So for the oxides of sulfur, these are covalent compounds and they have covalent bonds. You could say the intramolecular bond is covalent. Each molecule though is joined to another by weak intermolecular forces of attraction. As a result of this, because we have weak intermolecular forces of attraction, very little energy, very little energy. is required to break these molecules from these forces of attraction, holding them together in the solid state. So they have very low melting point. Low melting point of negative 72 degrees Celsius. Now, for the magnesium oxide, this is an ionic compound. They have ionic bonds present, and these arise from very strong electrostatic forces of attraction between the cations and the anions. The forces require lots of energy to break, well, to break them apart. To break them apart when they're in the solid state. So, magnesium has a melting point of 2852, or you could say a high melting point. That should give us our six marks. Part 2 of 4A. Explain whether the oxides of sulfur and the magnesium will conduct electricity, and if so, Under what conditions? All right. So let's just tell them. So the oxides of sulfur. So the oxides of sulfur will not conduct electricity under any condition. Now, these are covalent compounds. So that's that. straight up no ions are present no ions are present there nothing nothing to carry a current if we were to apply a voltage no magnesium the oxide of magnesium so mgo mgo all right so we need to say here that the magnesium oxide it will conduct electricity when molten as that time the ions are set free and can move and carry a current for voltage. is applied now we need not say if it's molten because magnesium oxide is not soluble in water all right now we need not to go into the details here you probably need to look back over your solubility of um your oxides so magnesium oxide is insoluble in water so we need not to need not talk about um when it's dissolved in aqueous medium this is enough right here for the four marks Okay, so part B for the win. Figure 4 is a partial diagram of the apparatus a group of students intend or propose to use to investigate whether ethanol, aqueous ammonia, and aqueous lead to nitrate would conduct electricity. So part 1, complete the diagram in figure 4 in order to make it a circuit that is suitable for achieving the aim of this experiment. So we have the battery. and they're showing us the positive terminal. They're showing us electrode A connected to the positive terminal, B connected to the negative terminal. So what we didn't know is some sources will say it's a cell, or some will just say it's a container. So we just would need that to be present, and we'd need to have our electrolytes, so our electrodes must be immersed. into the electrolyte or not the electrolyte whatever the liquid we're testing to see if it's an electrolyte if it's an electrolyte then it will conduct so we would put our ethanol in the container then at another time or aqueous ammonia and at another time or aqueous lead to nitrate so we've done that so part two now that should give us the two mark so we could state that we're putting the liquid here liquid to be tested and what we would need as well would need since this is all we have we could put in a a lamp that would glow of course if we um get if we get um current passing that's if the liquid is if the liquid conducts electricity so they want us to know classify the three liquids given as conductors or non-conductors know our knowledge of you know we we call it snow weak acid weak bases should chip in but they said conductors are non-conductors so even if it's a weak base or a weak acid we don't have any weak acid here once we'll get conduction occurring we'd have to classify the substance as a conductor we can't say weak conductor it's just a conductor or a non-conductor so we'll start with the most well obvious one conductors aqueous lead nitrate would be a conductor so we would put that here The other thing that will conduct aqueous ammonia. And then our non-conductor will be ethanol. That would give us the other three marks to top it up to 15. How are you faring so far? Leave a comment. Did you get this right? What did you put for this? What do you have for your answer here? Okay. all right so this is number five part well 5a anaerobic fermentation occurs when yeast is used in the production of wine yeast very important all right part one define the term anaerobic fermentation all right so fermentation we know once um when something ferments it's forming an alcohol and if they say anaerobic we'll have to tell them what anaerobic means aerobic air and means without so we're looking at the formation of alcohol in the absence of oxygen so if we put that we'll get our full marks so we could say it's the conversion of glucose to alcohol in the absence of oxygen all right and they said here you That's it. when yeast is used. So when they ask us now why high temperatures are not suitable for anaerobic fermentation in part two, we have to bear in mind that in yeast would have an enzyme and high temperatures would actually destroy the enzyme that is in yeast. That would actually cause them to become, you could say, denatured. But if you say high temperatures will kill the enzymes in yeast or will destroy the enzyme in yeast, then that will give you the marks. So the high temperatures will destroy the yeast, making the enzyme inactive. And they want us to write a balanced equation for the anaerobic fermentation of glucose. That's part three. So that's C6H12O6. aqueous medium. This we're not using any air, it's anaerobic so we need to just write this form in ethanol C2H5. Perhaps ethanol is a theme for this paper because this is the third time we're encountering ethanol or we have to do something with ethanol. So this will be ethanol here in aqueous medium. I guess we'd have to distill it, we'd have to distill it to actually get the ethanol. from the water it's not like our equation earlier where we saw sodium a very reactive method um a very active metal reacting with um ethanol so we'd get this would get co2 and to balance we have six carbons on the left we would need a change of color here we would need two here and that would make four and then you know you can look at it to put two right here that would fix it because we need six twelve hydrogen so two five ten plus two times one there two so that would give us our twelve and then for the oxygen we'd have two times this one here which is two and then two two's four so everything is checked so that will give us our full mark so there we go we have five already let's go for the other ten part b of five Soaps are formed from the alkaline hydrolysis or saponification of natural oils and fats, esters. Compound E, shown below, is an ester which is hydrolyzed by acrosodium hydroxide. Draw the fully displayed structures of the hydrolysis products. So what you could do if you're at this point and I don't want to say clueless, I'm recommending that you check the link in the description below that outlines well it that outlines reactions of esters all right so if you if you had covered that before the exam then you should be you should be good all right so this is an ester so whenever we're going to um hydrolyze we're going to have hydrolysis of an ester we always cut right here right so cut right there so we can draw our pair of scissors right there that's where we cut And when we cut there, we're splitting water into OH and H, and we're putting the OH back onto the acid portion, which is the part normally drawn first. That's the part with the C double bond, O. And then we're putting the H back onto our alcohol. So in this case, we're going to get the salt of the acid here. Okay, so if we're putting on the OH, the OH would have to now react. with the sodium hydroxide even though i'm getting too much into the mechanism and that would give us the salt of the acid so we would have ch3 then there's a c double bond o and then we would have o n a even if the oh goes back on it will not have to react with the sodium hydroxide to give us the salt and we would end up with the corresponding alcohol and the alcohol here would have the oh group again It's ethanol. So it's the fourth time we are seeing ethanol on this paper overall. All right. So that is that. That should give us four marks. So let's keep racking them up. Okay. Part one of C. Name the byproduct of the saponification of fats and oils. Now, earlier they told us that the saponification of fats and oils give us a soap. So. We cannot say about soap here. We'd have to give them the other thing that we'd get. all right so we'd actually get um would actually get a special alcohol which is called glycerol and as i said before yet be sure to check out the link in the description below to the video that actually breaks down reactions of esters and now they're asking us so i guess since we're talking about soap they're looking at applications of soap and sopless detergents all right state one difference part two one difference between the effect of using soaps and sopless detergents on hard water okay and it's just one mark so we're just gonna hit it right on the head so so when we use soap in um hard water soaps do not lather well In hard water, we get the formation of a scum. So, it's a deformed scum. Alright, and then sopless detergents on the other hand. So, we would not get any scum with our sopless detergents. Part D. Figure 5 shows the structure. of an amino acid and propene right amino amino we have the NH group there and then acid portion we have the COOH and you can check out another link in the description below that will look at amino acids and how they're joined to actually give us polyamides you can look at that addition versus condensation perhaps this question is leading us there but you can check out that video for more details on how the processes are work state the type of polymerization that amina acid in figure the amina acid in figure five would undergo all right so amina acids would undergo condensation polymerization state the general name for the type of polymer formed from the amino acid in figure five so it would be a polyamide all right be sure to check out that link to be on the safe side for the last two marks state two chemical tests that can be used to distinguish between propene and its polymer all right two marks so the polymer of propene would be polypropene you so test one test both with acidified amn o4 propene decolorizes it changes it from purple to colorless all right polymer does not all right the next test we could use we could test both with Both of them with Br2 liquid in CCl4. So bromine liquid in tetrachloromethane. Again, propene would decolorize this solution. Propene would change. Propene changes this solution from reddish brown to brown. colorless and we could have used aqueous bromine that's br2 aqueous bromine in aqueous medium that would be yellow and then the propene would would decolorize that as well and the polymer would have no change on it so of the three tests um the propene would decolorize all of the substances and then the you polymer would have no effect on the testing reagents all right and that would give us our full marks so we're at number six part a farmer brown reaped a crop of vegetables and reported that the yield was far below what was expected an analysis of the soil revealed that there was a deficiency of magnesium oh no outline the importance of magnesium on plant health. now we have to read our question carefully now plant health in general why is magnesium importance of magnesium and plant health so we'd say all right so magnesium forms part of chlorophyll and chlorophyll is needed to me is needed to make um food so that's very important for the plant and then the magnesium it aids in protein formation all right and that's that's two they're more like it's used in energy transfer to get a little more technical but those will give us the marks in part two now we're supposed to explain our magnesium or magnesium deficiency can result in the low yield of vegetables all right so we said earlier that magnesium forms a part of chlorophyll let's make the link now If there's a deficiency in magnesium, then there will be a low amount of chlorophyll. If there is a small amount of chlorophyll, then there's going to be a reduced ability, you know, in the plant making food. And then if there's little food being manufactured, then there just won't be much to store. And so there will be low yield. All right, let's put that in words. All right, so if a lot of food is manufactured, then there will be a lot of food to be stored. The yield will be high. So magnesium forms a part of chlorophyll. Plants use chlorophyll to trap sunlight and make food. If there's a deficiency in magnesium, then it means that we'll have very little chlorophyll. If there's little chlorophyll, then we'll have little food being manufactured. If little food is being manufactured. then there would be little food available to be stored, and so we'll have a low yield. That should give us our marks. Part 3 of A of 6A. State two other metal ions which are important to plant growth and the result of each deficiency. So straight up, just like in us, yes, we're similar to plants, calcium ions. the deficiency stunted growth and then we could give potassium ion and then if we have a deficiency we'll end up with weak spindly leaves first finely leaves also we can get undersized seeds or poorly developed seeds all right so that's um that's two i think we could have put iron this would give us many yellow yellow leaves but we just want um we just want two so we'll work with those two and that would give us four months okay part b um this one speaks about um the pollution plastic pollution and its effect on the environment um be sure to check out um a video that i did on this in celebration of world ocean day june 8th please check the link in the description for that one we're just going to get straight to the point here but if you're really interested in looking after the environment just having an idea having you know more insight into how pervasive this problem of plastic pollution of the marine waters or of our oceans is then you can just check that out let's just get straight to the point the caribbean sea and many oceans around the world have been found to be polluted by solid waste. mainly in the form of plastics. Part 1 states two examples of plastic waste that are commonly found in the sea and oceans. So we could plastic bags, plastic bottles, all right, that will do it. We could do fishing gears. probably not as common but common still discuss two harmful effects of plastics on marine life and the possibilities here the possibilities are endless we're just going to give um two so all right so harmful effects um harmful is you know it doesn't have to necessarily mean death but it actually happens that that can be one of the effects So sometimes these animals, they can become trapped or become entangled, you know, in these plastic materials that are left. And they can actually be cut. They can get, you know, bad wounds, lacerations. So they can cause deep lacerations. So you can end up with some organisms being wounded. We're going to just give like two, for example. You can get wounding of the whales and turtles. Not only those, but entanglement can cause de-placerations. And then sometimes when these animals eat the plastic, for example, or turtles, they will mistake a plastic bag for a jellyfish. And they'll eat that, and then their guts become filled with plastic. It gives them the sense of them being full. And so they end up... dying of hunger later on. And I mean, some of them know when they become trapped, it reduces their ability to actually swim. And so they now become, you know, easy pickings for predators. They can't escape danger. or they can't swim and find food and eventually they die so we'll just leave it there remember to check out the link in the description below or you can check out the card um above just to actually look at that presentation in honor or in celebration of world ocean day remember use what's in your hand take that bottle recycle it reduce reuse recycle and just like that we've come to the end of this session i hope this was informative i hope you actually were able to work the questions with me and you actually messed up the marks just like um i did here i hope you actually did very well right remember to like share and consider subscribing if you find value in what i do here on couple later

vary but we um as best as possible will aim to give the correct answers right so so number one is your good old data analysis um question we can always depend on that um number one here we're being asked not to spend more than 30 minutes on question one no number one hydrogen peroxide h2o2 decomposes to produce oxygen and water in the presence of potassium iodide as a catalyst. The effect of the concentration of the hydrogen peroxide on the rate of reaction was investigated. For each experiment, the concentration of the hydrogen peroxide was changed and the time taken for the hydrogen peroxide to decompose was measured using a stop clock as shown in figure 1. 1 of A.

Define the term rate of reaction. Change in concentration of reactant or product per unit time. Anything having to do with rate, time has to be measured. Define the term catalyst.

So it's a substance that is used to speed up the rate of a reaction but remains chemically unchanged in the process. So the catalyst does not take part in the reaction. We could put that, we could state that as well.

A substance used to speed up the rate of a reaction but does not take part in the reaction. All right, so that will give us three marks. All right, so part B. For each experiment shown on the stop clocks in figure one on page four, record in table one the time taken for the hydrogen peroxide to decompose. Write down these values to be used in our table a little later on before we get into anything big.

So at first we always look to see what the scale is on whatever the instrument is. each of those stroke there is representing one second so for the first one in experiment one this is 59 in experiment two we have this at 42 in experiment three this is at 31 in experiment four this is at 25 in experiment five this is at 21. so we're going to record those we're going to keep those close by for a little later on. So we're going to do that. We made a note of those times. So we just need to fill in that in the time column. And then after that, in part C, we're just going to do everything in one.

In part C, we're to complete table one by calculating the rate of reaction. This time, the rate of reaction is one over time, and they want it calculated to three decimal places for each of the experiments. So, the first one, we had mentioned that the first one, for the first one, the time is 59 seconds.

Second one is 42 seconds. The third one, 31 seconds. The fourth one, 25 seconds. The fifth one, 21 seconds. And we're just going to take the inverse of those to say 1 divided by 59, and that will give us 0. 0.017 that's three decimal places then 1 divided by 42 0.024 1 divided by 31 0.032 1 divided by 25 0.040 three decimal places and then 1 divided by 21 which would be 0.048 So we're going to keep these in mind we have to bear these in mind because we're going to Use these to actually plot our graph.

Using the axes provided in figure 2 on page 7, plot a graph of rate of reaction versus concentration of hydrogen peroxide from the information in table 1. Draw the line of best fit through the points and this is for five marks. So we're going to head on over to our graph sheet and we're going to plot a graph of Rate of reaction versus concentration. So rate of reaction here is our dependent variable.

The rate of reaction will get its value from the concentration of the hydrogen peroxide. So the hydrogen peroxide then is the independent. The concentration of the hydrogen peroxide is the independent variable that goes on the x-axis.

And then our responding variable. or our dependent variable is our rate of reaction that goes on the y-axis. So let's go, let's figure out, let's go, let's figure out the scales on each axis and then let's plot and then let's draw our line of this fit and get five marks. All right so we have our values that we're going to plot, we wrote them down a little while ago so we just go right into plotting but before we plot we have to establish we have to figure out the scale in each case all right so on the let's look on the x-axis we find that um 2cm they use 2cm 2cm to represent 0.05 units right so that means that each of those little milli what's that each of those millimeter going up by 0.005 each of those little millimeter is going up by 0.005 and on the y-axis we're seeing where 2 cm 2 centimeters there they're being used here to represent 0.00 So it means that each of those, each of those millimeter, that's the tiny box, each of those right here, that's the one I can actually touch.

Each of those would be representing 0.0005. 0.1 mole per dm cube the rate there is 0.017 so we need to find 0.017 so this is 0.015 so this would be zero okay don't need to mark that so we have 0.015 there so 0.016 0.017 and we would circle this and then the next one 0.15 is 0.024 so we find 0.15 then we go up to 0.024 that's 0.025 here so two two behind let's see 0.020 0.021 0.022 0.023 and 024 all right and then following that we have 0.2 0.2 and that would be 0.032 0.032 0.032 would be here we're circling we can use n circle dots or we can use x's all right and then the next one 0.0 sorry 0.25 0.25 on the x and for that one we have 0.04 0.25 and 0.04 so that would be here and then after that we now have 0.30 and zero point corresponding with that 0.30 we have 0.048 0.048 would be it would be here, right? So the next thing that we need to do is to know. So we've plotted all our points, all five points correctly. The next thing would be to now draw our line of best fit.

Now our line of best fit does not always have to go through the origin. But it appears that this one would actually go through the origin. So let's do that. We need a ruler to do it. I'm exempted based on what I'm using.

But you would use your ruler. You try to get all the points to fall on this. on a straight line if it doesn't if it doesn't you try if they don't then you try to get as many of them falling on the line as possible with an equal number falling off but this this seems like a good trend line or that will get um them falling falling on a straight line here Even if it did not go to the origin, that would have been fine.

So we're now going to move on to the questions using this very graph that we have plotted. All right. So we're at part E. Using your graph, well, part one of E, describe the relationship between the rate of reaction and the concentration of the hydrogen peroxide for two marks.

So we need to go back to our graph and look to see what. what the relationship is we look to see how the rate is being affected by changing the concentration so let us see what's happening here so from this we're seeing that what are we seeing we're seeing that as as they so follow on the x-axis as the concentration increases so increases taking place here let's use green So as the concentration is increasing, as we're going over here, we're seeing that, hey, the rate is also increasing. So we're seeing that as the concentration of the peroxide increases, the rate of reaction increases as well. So we would need to state that. So we just need to toggle between.

So we're flipping pages now. So we're going between pages. So we need to write that down as the concentration of. peroxide increases the rate of reaction increases right and then the next question they want us to determine the concentration of the hydrogen peroxide given that the rate of reaction is 0.045 per second. All right.

So we would need to go back and we'd find 0.045 per second on the y-axis. So we need to find 0.045 seconds per second rather on the y-axis. so let's use a different color to represent this and once we find it on the y-axis we're going to draw a line from it going across and wherever that line cuts our best fit line our curve we're going to draw a line or a straight line from that coming down to cut the x-axis wherever it cuts wherever it cuts the x-axis right did it come all the way we need to come all the way right and from what we see here this looks like it is zero point zero point two sorry zero point two five six so that we'll write it boldly so that is zero point two five six mole per dm cube that would be the concentration okay so we make a note of that and then so in part f we are to calculate the mass of the hydrogen peroxide present in the concentration identified in E2.

which is 0.25 moles per day cubed. So we're pretty much converting here from moles to, from moles to mass. Molar concentration, it's, it's the same as saying, so mass concentration here is the same as saying moles being converted to grams.

but the number of moles is in a set volume so we say moles over the m cube or moles per dmq now in order for us to move from moles to the mass we'd have to multiply by the molar mass which is in grams per mole and if we look at this we see that mole will cancel out mole and will be left with grams per dm cube so this will give us this will be 0.2 0.256 moles per dm cubed times 34.01 grams per mole and this looks like a lot of working out for one we're not just working we're not just looking for the answer we're looking at how we arrive at the answer so we're teaching you doing everything all in one so that's 0.256 times 34.01 and that will give us everything so far has been to three decimal places so this is and then we have three decimal places here so we'll just be consistent so this is 8.707 grams per bmq and they want us to know write a balanced chemical equation including state symbols show the decomposition of hydrogen peroxide all right so hydrogen peroxide you know pure hydrogen peroxide is a liquid but they were varying the concentration here so it means that it would be a a solution of a liquid in a liquid peroxide is the solute and the water is the solvent so we would write h2o2 very important it's not the same as water that second two makes a big difference don't take my word for it or you should take my word for it don't want you to try drinking this now all right so this will give us h2o which is a liquid plus we'll get o2 gas so to balance this would need to put a two oxygens are off we have three on the right side two on the on the left side now if you put a two here and a two here that would fix pretty much everything for everyone so that's the balance equation for three more marks and 25 marks full marks let's check our time did we spend more than 30 minutes on this question all right let's move on to the second question okay number two part one of a or a at atmospheric pressure water exists in three states of matter while carbon dioxide exists in only two states one list the three states of matter in which water can exist and this is for for one mark and as usual we're thankful for small mercies so those three states that water can exist in at atmospheric pressure would be solid liquid and gas. Then they want us to describe the energy of the particles in each of the three states listed in part one of a above. So let's go with the solid state. So in the solid state the energy that we're describing here is the kinetic energy that the particles would have as a result of their motion. So in a solid the particles are arranged in a fixed way held by very strong forces of attraction so they have nowhere to move to they can only vibrate.

So in a solid there is very little there is little. very little to no kinetic energy. Alright? In a liquid we could say there is a moderate amount and then in a gas the forces of attraction are very it's almost non-existent so these particles can move in any direction all about the place with a large amount of space between them so we say that there's large or enormous amount of kinetic energy present part three named the process which of course when carbon dioxide changes from one state to another. We have to appreciate the two states that it's in first.

It can be in the solid state, it can be in the gaseous state, so that would be sublimation. Part four, describe how the arrangement of the carbon dioxide particles changes as carbon dioxide undergoes the process named in the a3 above. its sublimation we'd have to decide which state we're moving from whether we're going from the gas to the solid or the solid to the gas all right so we're gonna start with the gas here so let's just describe what happens first when we're in the gaseous state well let's describe what happens first when we're in the solid state and then as we move from the solid to the gaseous state heat would have to be applied so let's go Alright, so we could mention here that it actually skips or they skip the liquid phase.

So in the solid state the particles in dry ice are very close to each other in a regular way. as heat is added or as heat is applied the particles begin to vibrate and quickly move into the state where they have large amounts of space between them that's into the into the gaseous state so that is the gaseous state and here they want us to use appropriate diagrams to illustrate the lattice structure of sodium chloride crystals and the giant molecular structure of diamond. I believe there's a neater way to do it than I'm doing it right now but as always this is not technical drawing we just want to show we just want to be able to draw something that looks like a square.

then we get our parallel lines so we can get it to look like a cube then we can take it from there all right so we're going to look at the lines at the back which are going down well that one that will join that it's not a perfect cube but we can actually work with this so all right so in this one i'm gonna put chloride at each vertex or what we call corner so chlorides are the vertices so i'm using the big one to represent chloride so we have alternating chloride and sodium so this would be let's do the key so this is chloride ion and then this little one will be the sodium ion na plus ion so sodium would be here Sodium would be here as well. Sodium here. Sodium here. And in the middle there would be a chloride and would draw a line to show that there there's you know electrostatic attraction between them. There's the bond right there, the ionic bond.

All right so sodium would be here. Here as well. Here. Chloride would be here. And we could have put sodium at each vertex.

What's important to note is that each chloride ion is surrounded by six sodium ion and each Sodium ion is surrounded by 6 chloride ions, whichever way we sample. All right, so let's continue. So here we have sodium, sodium, sodium, chloride.

Sodium would be here. Sodium would be here. Chloride would be there at the back.

we can line that up much better all right so line would be coming down through there one would be going across there there's a chloride at the one at the back line coming down line going across it probably should be blowing your mind now there is a chloride at the front i didn't focus on the front one so then right there and then there's one more so them would be in the middle a small one and then from the top going through at the bottom so that should give us our full mark then a much simpler one for diamond we're going to use carbon and we're going to show that one carbon is bonded to four more so this is our carbon in the center so each c represents a carbon atom and the line here represents a strong or we don't have to say strong we could say a covalent bond so that would give us three marks each and that would make it six so we're down to the last two in number two to make it up to 15 we have 13 so far part c in solution two metals a and b form a plus and b2 plus ions respectively so it means the a forms A plus and B forms B2 plus. Metal A displaces Fe from a solution containing Fe2 plus ions but metal B does not. Write a balanced ionic equation to show the reaction between metal A and Fe2 plus.

All right, kind of expecting something more like a reactivity series. based on this but this is where they stop it so we'll just take it and be thankful for small mercies again so all right so a is going to displace fe2 plus a is in is being treated as something in group one it has a valency of one so it will be it will be a and a here would be a solid so Fe2 plus would be an aqueous medium and from this we would end up with A plus plus Fe solid. But each time for every atom or every mole of A that reacts, we get one mole of A plus or we get one mole of electrons actually being liberated.

for each mole of the Fe2 plus we need two moles of electrons so it means then even though we're not seeing the electrons we would actually we'd actually have this is what we have happening A is actually reacting to give A plus plus an electron all right and then on so that is um oxidation right and then on the other side we have Fe2 plus actually gaining two electrons to be liberated as Fe the solid. So if we have two electrons on the left hand side we need two electrons on the right hand side. So to get that we would need two electrons but we can't go in and multiply the electrons by two without multiplying the A plus by two and anything we do to one side we have to do it to the other side.

So we'll end up with that. Then we remove the electrons and then we write back what we have. So we need 2A Right here. And we would need 2A plus.

And that would do it. And this would give us the additional two marks to make it up to 15. How many have you gotten so far? All right. Just like that, we've come to the end of this question. Continue with the others and see how many marks you've missed so far.

Couple later. Okay, so we're at number three. And this is an organic chemistry question. Figure three shows the fully displayed structures of three compounds, A, B, and C, which are from different homologous series. All right, so let's examine compound A.

like a saturated hydrocarbon all we have is a carbon carbon single bond and carbon hydrogen single bond all right so in compound b we have a functional group and that's the c double bond o o h and then in compound c we have a functional group that's the o h functional group so okay so for a saturated hydrocarbon only made of carbon and hydrogen so it's like uh hydrocarbon is in the true truest sense of the word so this is coming from the alkane series so that belongs to the alkanes compound b cooh that is our carboxylic group so this is coming from the carboxylic acid group or the carboxylic it's just called carboxylic acids or the alkanoic acids now they're saying compound c burns with a blue flame in oxygen yes it does that's some ethanol and they want us to write a balanced equation for this reaction so then let's just get right ahead and do that so ethanol here c2h5oh that's the liquid burns in oxygen whenever we write oxygen as an element like this we write it as o2 anything that ends in gen r in is diatomic so this will give us just like when any other organic compound burns this will give us carbon dioxide and water water here in the form of a vapor okay so straight up let's count we have two carbons on the left we have one on the right so to fix that and we're always balancing the order so we go with the c's first so we put that two there that fixes the two carbons for the right one the left so that's good when i look at the hydrogen cho so h comes next we have five here on the left plus one that's six over that side and over the right side we have you 2 so to make that equal to 6 we need to put a 3 right here right in front of the water and that makes 3 2 6 so the carbons the hydrogens are good so it's now a matter of the oxygen so this would be two twos four from the carbon dioxide and then three from the water that's seven a little oddish so we have one oxygen over here on the left from the ethanol and we have two here if we put a three here that will give us three to six plus the one over here in the ethanol that would make it seven so this would be our balance equation and then for part c state which of the two compounds a or c is more soluble in water give a reason for your response all right a or c well for one you straight up off the bat we're gonna go with compound c compound c is more soluble in water for one water is water is polar and compound c compound c so water is polar and we know that light dissolves light now compound c has an oh group you So the presence of the OH group means that this compound is polar, unlike compound A, where we have just carbon and hydrogen. So this is a non-polar compound. So compound A is polar. So we just need to state, OK, compound C has the OH group, so is polar. and dissolves in Water, which is also polar.

We could also state that, not to over-answer, but we could state that compound A is non-polar. And we know that light dissolves light. Okay. Part D. State whether compound B or compound C would react more vigorously with sodium metal and give a reason for your choice.

No. Both of them would react with... sodium here now it comes down to a matter of which of them would let go of that hydrogen more readily that hydrogen at the end well for one compound b is a weak acid compound c you know it can be as an acid but it's even weaker than um compound b so the one that would be more reactive in this case the one that would react more vigorously would be compound a B and so we should give we should give a reason for our answer now the hydrogen in compound B is more readily lost or is more readily replaced by the metal than that hydrogen in compound C so the hydrogen in compound B is more readily lost than the hydrogen in compound C.

And, you know, this is due to the fact that we have the presence of a carbon carbon, sorry, not a carbon carbon, carbon oxygen double bond, which adds to the polarity of the CO. Well, it adds to the polarity of the OH group in compound B. All right.

But that would be, that's probably even over answering. So we're just going to state exactly why that the We'll state that B is more polar with the C double bond O. So the H in the OH is more readily given up than the H.

in the OH. So the hydrogen in the OH of COOH in compound B is more readily lost than the hydrogen in the OH of compound C. And we'll just leave it right there.

So, so far for this question, that's 2, 4, 6, 8 marks. So let's go for the other 7 marks. Okay, in E, write a balanced equation for the reaction of compound C with sodium metal.

Okay, so compound C is ethanol. So we're going to go again that C2H5OH. And this is a liquid which is going to react with sodium, which is a solid. In this case, that H at the end is going to be acting as a replaceable hydrogen with a very...

metal like sodium so we'd end up with C 2 H 5 O and E and this is actually can put a LC this is an alcoholic you call it alcoholic alcoholic medium it's not aqueous no water is here we're talking about liquid ethanol and then this would liberate hydrogen but we cannot write h by itself it's diatomic so it would have to be h2 gas and then of course if we're going to do that we have to just finish off and ensure that we balance so for us to have h2 there we would need two right there and then that would make everything here two on the left you the ethanol so we just need to match that off with a two just ensure that this is visible two and then the sodium would need to be two and then that is it now part f describe one test that could be used to identify the gas that is produced in the reaction of compound c with sodium metal of course we'll have to tell them what we expect not just the test but what we expect to observe so we could to just hold a lighted splint at the mouth of the test tube and so all right so the gas present will put out the lighted splint with a pop it's hydrogen use the right terms so the pop that we hear is a mini explosion sometimes it's not really poppy it's squeaky sometimes And it just depends on the volume of the gas that is present. And that little pop or that squeak is just a mini explosion because hydrogen is explosive. Yes, that's right.

Part G. Compound B and compound C react together in the presence of a catalyst to form compound E. State the name of the catalyst. All right, just a reminder of what compound E is.

Compound E. propanoic acid that's compound b and compound c we said earlier is ethanol so they're asking for the name of the catalyst the name of the catalyst well would use um sulfuric acid sulfuric acid is our catalyst and they want the fully displayed structure of compound d now whenever compound B, this is propanoic acid. Whenever we're drawing or we're showing the product, the ester formed when an organic acid reacts with an alcohol, we draw the acid part first. Okay, so probably what we should do, perhaps what we could do, we could redraw this in blue. We're actually seeing where everything is coming from.

Retract the red, retract the blue. So we're going to pinch off. Let's use black to show or green to show what's happening.

We're going to join these together. We're going to pinch off the OH from the acid and the H from the alcohol, which will give us water. We're not so concerned about the water.

We want the fully displayed structure of compound D. All right. So we're going to go with red first. So we have a CH3 from the propanoic acid.

And we're just going to put. on the H's as we go along. Then we have a CH2.

Then we have C with the double bond O. And then from that now we're going to go on to the blue compound. So we're going to have now C. We have C double bond O from the red.

So we're going to have the blue. Then we're going to have a CH2. Then we're going to have a pH 3. They didn't ask us to name this.

So this is our compound. And whenever I draw, whenever I draw an ester, it's just second nature for me to highlight the carb, highlight the, for me to highlight the ester linkage. So that is our compound. They did not ask us to name it. It's ethyl propanoate nonetheless.

All right. just like that we've come to the end of the section how many months number four sulfur and magnesium are two elements in the same period of the periodic table the different properties of the oxides of these elements are presented in table two all right so we have the oxide of sulfur it's a gas that's the state the melting point is negative 72 degrees celsius and for the oxide of magnesium. It's a solid and the melting point is 2,852.

Whoa, that's a lot. That requires a lot of energy. Now, with reference to bonding, explain the difference in melting point between the oxides of sulfur and magnesium.

So we're going to be talking about the bonding that's there. Don't want to get too much into the structure. It's hard to talk about the bonding without talking about the structure, but we want to be very disciplined so we can get our full marks.

So let's just go. So for the oxides of sulfur, these are covalent compounds and they have covalent bonds. You could say the intramolecular bond is covalent.

Each molecule though is joined to another by weak intermolecular forces of attraction. As a result of this, because we have weak intermolecular forces of attraction, very little energy, very little energy. is required to break these molecules from these forces of attraction, holding them together in the solid state.

So they have very low melting point. Low melting point of negative 72 degrees Celsius. Now, for the magnesium oxide, this is an ionic compound.

They have ionic bonds present, and these arise from very strong electrostatic forces of attraction between the cations and the anions. The forces require lots of energy to break, well, to break them apart. To break them apart when they're in the solid state. So, magnesium has a melting point of 2852, or you could say a high melting point. That should give us our six marks.

Part 2 of 4A. Explain whether the oxides of sulfur and the magnesium will conduct electricity, and if so, Under what conditions? All right. So let's just tell them. So the oxides of sulfur.

So the oxides of sulfur will not conduct electricity under any condition. Now, these are covalent compounds. So that's that.

straight up no ions are present no ions are present there nothing nothing to carry a current if we were to apply a voltage no magnesium the oxide of magnesium so mgo mgo all right so we need to say here that the magnesium oxide it will conduct electricity when molten as that time the ions are set free and can move and carry a current for voltage. is applied now we need not say if it's molten because magnesium oxide is not soluble in water all right now we need not to go into the details here you probably need to look back over your solubility of um your oxides so magnesium oxide is insoluble in water so we need not to need not talk about um when it's dissolved in aqueous medium this is enough right here for the four marks Okay, so part B for the win. Figure 4 is a partial diagram of the apparatus a group of students intend or propose to use to investigate whether ethanol, aqueous ammonia, and aqueous lead to nitrate would conduct electricity.

So part 1, complete the diagram in figure 4 in order to make it a circuit that is suitable for achieving the aim of this experiment. So we have the battery. and they're showing us the positive terminal. They're showing us electrode A connected to the positive terminal, B connected to the negative terminal.

So what we didn't know is some sources will say it's a cell, or some will just say it's a container. So we just would need that to be present, and we'd need to have our electrolytes, so our electrodes must be immersed. into the electrolyte or not the electrolyte whatever the liquid we're testing to see if it's an electrolyte if it's an electrolyte then it will conduct so we would put our ethanol in the container then at another time or aqueous ammonia and at another time or aqueous lead to nitrate so we've done that so part two now that should give us the two mark so we could state that we're putting the liquid here liquid to be tested and what we would need as well would need since this is all we have we could put in a a lamp that would glow of course if we um get if we get um current passing that's if the liquid is if the liquid conducts electricity so they want us to know classify the three liquids given as conductors or non-conductors know our knowledge of you know we we call it snow weak acid weak bases should chip in but they said conductors are non-conductors so even if it's a weak base or a weak acid we don't have any weak acid here once we'll get conduction occurring we'd have to classify the substance as a conductor we can't say weak conductor it's just a conductor or a non-conductor so we'll start with the most well obvious one conductors aqueous lead nitrate would be a conductor so we would put that here The other thing that will conduct aqueous ammonia. And then our non-conductor will be ethanol. That would give us the other three marks to top it up to 15. How are you faring so far?

Leave a comment. Did you get this right? What did you put for this?

What do you have for your answer here? Okay. all right so this is number five part well 5a anaerobic fermentation occurs when yeast is used in the production of wine yeast very important all right part one define the term anaerobic fermentation all right so fermentation we know once um when something ferments it's forming an alcohol and if they say anaerobic we'll have to tell them what anaerobic means aerobic air and means without so we're looking at the formation of alcohol in the absence of oxygen so if we put that we'll get our full marks so we could say it's the conversion of glucose to alcohol in the absence of oxygen all right and they said here you That's it.

when yeast is used. So when they ask us now why high temperatures are not suitable for anaerobic fermentation in part two, we have to bear in mind that in yeast would have an enzyme and high temperatures would actually destroy the enzyme that is in yeast. That would actually cause them to become, you could say, denatured.

But if you say high temperatures will kill the enzymes in yeast or will destroy the enzyme in yeast, then that will give you the marks. So the high temperatures will destroy the yeast, making the enzyme inactive. And they want us to write a balanced equation for the anaerobic fermentation of glucose. That's part three.

So that's C6H12O6. aqueous medium. This we're not using any air, it's anaerobic so we need to just write this form in ethanol C2H5. Perhaps ethanol is a theme for this paper because this is the third time we're encountering ethanol or we have to do something with ethanol.

So this will be ethanol here in aqueous medium. I guess we'd have to distill it, we'd have to distill it to actually get the ethanol. from the water it's not like our equation earlier where we saw sodium a very reactive method um a very active metal reacting with um ethanol so we'd get this would get co2 and to balance we have six carbons on the left we would need a change of color here we would need two here and that would make four and then you know you can look at it to put two right here that would fix it because we need six twelve hydrogen so two five ten plus two times one there two so that would give us our twelve and then for the oxygen we'd have two times this one here which is two and then two two's four so everything is checked so that will give us our full mark so there we go we have five already let's go for the other ten part b of five Soaps are formed from the alkaline hydrolysis or saponification of natural oils and fats, esters.

Compound E, shown below, is an ester which is hydrolyzed by acrosodium hydroxide. Draw the fully displayed structures of the hydrolysis products. So what you could do if you're at this point and I don't want to say clueless, I'm recommending that you check the link in the description below that outlines well it that outlines reactions of esters all right so if you if you had covered that before the exam then you should be you should be good all right so this is an ester so whenever we're going to um hydrolyze we're going to have hydrolysis of an ester we always cut right here right so cut right there so we can draw our pair of scissors right there that's where we cut And when we cut there, we're splitting water into OH and H, and we're putting the OH back onto the acid portion, which is the part normally drawn first. That's the part with the C double bond, O. And then we're putting the H back onto our alcohol.

So in this case, we're going to get the salt of the acid here. Okay, so if we're putting on the OH, the OH would have to now react. with the sodium hydroxide even though i'm getting too much into the mechanism and that would give us the salt of the acid so we would have ch3 then there's a c double bond o and then we would have o n a even if the oh goes back on it will not have to react with the sodium hydroxide to give us the salt and we would end up with the corresponding alcohol and the alcohol here would have the oh group again It's ethanol. So it's the fourth time we are seeing ethanol on this paper overall.

All right. So that is that. That should give us four marks. So let's keep racking them up.

Okay. Part one of C. Name the byproduct of the saponification of fats and oils.

Now, earlier they told us that the saponification of fats and oils give us a soap. So. We cannot say about soap here.

We'd have to give them the other thing that we'd get. all right so we'd actually get um would actually get a special alcohol which is called glycerol and as i said before yet be sure to check out the link in the description below to the video that actually breaks down reactions of esters and now they're asking us so i guess since we're talking about soap they're looking at applications of soap and sopless detergents all right state one difference part two one difference between the effect of using soaps and sopless detergents on hard water okay and it's just one mark so we're just gonna hit it right on the head so so when we use soap in um hard water soaps do not lather well In hard water, we get the formation of a scum. So, it's a deformed scum.

Alright, and then sopless detergents on the other hand. So, we would not get any scum with our sopless detergents. Part D. Figure 5 shows the structure.

of an amino acid and propene right amino amino we have the NH group there and then acid portion we have the COOH and you can check out another link in the description below that will look at amino acids and how they're joined to actually give us polyamides you can look at that addition versus condensation perhaps this question is leading us there but you can check out that video for more details on how the processes are work state the type of polymerization that amina acid in figure the amina acid in figure five would undergo all right so amina acids would undergo condensation polymerization state the general name for the type of polymer formed from the amino acid in figure five so it would be a polyamide all right be sure to check out that link to be on the safe side for the last two marks state two chemical tests that can be used to distinguish between propene and its polymer all right two marks so the polymer of propene would be polypropene you so test one test both with acidified amn o4 propene decolorizes it changes it from purple to colorless all right polymer does not all right the next test we could use we could test both with Both of them with Br2 liquid in CCl4. So bromine liquid in tetrachloromethane. Again, propene would decolorize this solution.

Propene would change. Propene changes this solution from reddish brown to brown. colorless and we could have used aqueous bromine that's br2 aqueous bromine in aqueous medium that would be yellow and then the propene would would decolorize that as well and the polymer would have no change on it so of the three tests um the propene would decolorize all of the substances and then the you polymer would have no effect on the testing reagents all right and that would give us our full marks so we're at number six part a farmer brown reaped a crop of vegetables and reported that the yield was far below what was expected an analysis of the soil revealed that there was a deficiency of magnesium oh no outline the importance of magnesium on plant health. now we have to read our question carefully now plant health in general why is magnesium importance of magnesium and plant health so we'd say all right so magnesium forms part of chlorophyll and chlorophyll is needed to me is needed to make um food so that's very important for the plant and then the magnesium it aids in protein formation all right and that's that's two they're more like it's used in energy transfer to get a little more technical but those will give us the marks in part two now we're supposed to explain our magnesium or magnesium deficiency can result in the low yield of vegetables all right so we said earlier that magnesium forms a part of chlorophyll let's make the link now If there's a deficiency in magnesium, then there will be a low amount of chlorophyll.

If there is a small amount of chlorophyll, then there's going to be a reduced ability, you know, in the plant making food. And then if there's little food being manufactured, then there just won't be much to store. And so there will be low yield.

All right, let's put that in words. All right, so if a lot of food is manufactured, then there will be a lot of food to be stored. The yield will be high. So magnesium forms a part of chlorophyll. Plants use chlorophyll to trap sunlight and make food.

If there's a deficiency in magnesium, then it means that we'll have very little chlorophyll. If there's little chlorophyll, then we'll have little food being manufactured. If little food is being manufactured. then there would be little food available to be stored, and so we'll have a low yield. That should give us our marks.

Part 3 of A of 6A. State two other metal ions which are important to plant growth and the result of each deficiency. So straight up, just like in us, yes, we're similar to plants, calcium ions. the deficiency stunted growth and then we could give potassium ion and then if we have a deficiency we'll end up with weak spindly leaves first finely leaves also we can get undersized seeds or poorly developed seeds all right so that's um that's two i think we could have put iron this would give us many yellow yellow leaves but we just want um we just want two so we'll work with those two and that would give us four months okay part b um this one speaks about um the pollution plastic pollution and its effect on the environment um be sure to check out um a video that i did on this in celebration of world ocean day june 8th please check the link in the description for that one we're just going to get straight to the point here but if you're really interested in looking after the environment just having an idea having you know more insight into how pervasive this problem of plastic pollution of the marine waters or of our oceans is then you can just check that out let's just get straight to the point the caribbean sea and many oceans around the world have been found to be polluted by solid waste.

mainly in the form of plastics. Part 1 states two examples of plastic waste that are commonly found in the sea and oceans. So we could plastic bags, plastic bottles, all right, that will do it. We could do fishing gears. probably not as common but common still discuss two harmful effects of plastics on marine life and the possibilities here the possibilities are endless we're just going to give um two so all right so harmful effects um harmful is you know it doesn't have to necessarily mean death but it actually happens that that can be one of the effects So sometimes these animals, they can become trapped or become entangled, you know, in these plastic materials that are left.

And they can actually be cut. They can get, you know, bad wounds, lacerations. So they can cause deep lacerations.

So you can end up with some organisms being wounded. We're going to just give like two, for example. You can get wounding of the whales and turtles.

Not only those, but entanglement can cause de-placerations. And then sometimes when these animals eat the plastic, for example, or turtles, they will mistake a plastic bag for a jellyfish. And they'll eat that, and then their guts become filled with plastic. It gives them the sense of them being full.

And so they end up... dying of hunger later on. And I mean, some of them know when they become trapped, it reduces their ability to actually swim.

And so they now become, you know, easy pickings for predators. They can't escape danger. or they can't swim and find food and eventually they die so we'll just leave it there remember to check out the link in the description below or you can check out the card um above just to actually look at that presentation in honor or in celebration of world ocean day remember use what's in your hand take that bottle recycle it reduce reuse recycle and just like that we've come to the end of this session i hope this was informative i hope you actually were able to work the questions with me and you actually messed up the marks just like um i did here i hope you actually did very well right remember to like share and consider subscribing if you find value in what i do here on couple later

Transcript for:CXC CSET Chemistry Exam Insights

Transcript for:
CXC CSET Chemistry Exam Insights