2026 SEED MOCK CHEMISTRY PAPER 1

[00:00:04]Welcome, learners.

[00:00:08]My name is Mr. Pocholo La Once Again.

[00:00:13]I hope your exams are going on well. So far, so good.

[00:00:19]We just [clears throat] finished week number one of your exams.

[00:00:22]>> [snorts] >> And we look forward to a beautiful week next week.

[00:00:29]And the which is next week when you are going to write your chemistry paper one.

[00:00:41]For that reason, I decided to revise with you this paper. This is paper one, chemistry paper one from Southeast Education Division, SEED.

[00:01:01]This is >> [snorts] >> because we promised you that we are going to revise with you. We're going to be with you up to the end of the exams. So, revising and making more video lessons.

[00:01:19]Once again, if you are writing MYSCE next year, 2027, or you if you are in form three this time, consider enrolling for lessons on the Black Slides Online Secondary School.

[00:01:46]We make sure we cover all topics which are exam-focused in order to boost your confidence.

[00:01:58]This is the place to have best preparations aimed at producing the best results.

[00:02:06]Welcome, learners.

[00:02:08]So, once again, without wasting your precious time, um I'll see that subscribing if you have not done so already.

[00:02:19]And also hit the notification bell so that every time we upload a new video material, you become the first to know.

[00:02:30]Welcome, learners.

[00:02:34]So, [snorts] this is chemistry paper one from CD from South East Education Division.

[00:02:41][clears throat] >> [snorts] >> My name once again is Mr. Pochilola.

[00:02:54]Welcome to section A, which carries 70 marks.

[00:03:02]I'm sure we are supposed to have an instruction here.

[00:03:06]The standard money paper is supposed to have an instruction here that says, "Answer all questions in the spaces provided here." I think there's an omission there.

[00:03:20]Question number one, explain why bond breaking is endothermic and bond making is exothermic.

[00:03:29]In other words, we should explain why bond breaking requires an energy input [clears throat] while bond making a releases energy to the surroundings.

[00:03:42]>> [snorts] >> Explain why.

[00:03:48]So, bond breaking it requires energy.

[00:03:51]So, energy is required to overcome the attractive forces holding atoms together.

[00:03:59]So, energy is absorbed. Hence, it is endothermic.

[00:04:05]So, in in order to break bonds in a chemical reaction, energy must be absorbed.

[00:04:14]>> [snorts] >> Cuz there's need to break the or to overcome the attractive forces that hold atoms together. So, energy therefore must be absorbed.

[00:04:26]Therefore, bond breaking is endothermic, meaning that it requires an energy input.

[00:04:39]Bond making, on the other hand, the atoms move to a more stable lower energy state.

[00:04:48]For the atoms to move to a more stable lower energy state, >> [snorts] >> energy must be released. Therefore, bond breaking is exothermic.

[00:05:01]State any two importance of endothermic reactions in everyday life. Remember that endothermic reactions are the reactions which absorb energy from the surroundings.

[00:05:17]So, [snorts] these kinds of reactions I put on in everyday life.

[00:05:24]The common ones are photosynthesis.

[00:05:31]During photosynthesis, plants absorb energy from sunlight to convert carbon dioxide and water into glucose and the oxygen.

[00:05:43]>> [snorts] >> This is done in order to sustain life on the earth.

[00:05:47]So, the most important part of the answer is that plants absorb energy.

[00:05:53]That's endothermic.

[00:05:55]from sunlight in order to convert carbon dioxide and water into glucose.

[00:06:02]>> [snorts] >> And saying this, this is one mark.

[00:06:08]Another importance is seen in a humans in the process called the sweating or evaporation of sweat from your skin.

[00:06:22]So, the the body absorbs the heat energy from the skin to turn sweat into vapor which cools [snorts] the body down. So, that too is an endothermic process.

[00:06:35]Naturally, heat is needed for the skin produce water through sweat. Remember that water excess water must be removed from the body.

[00:06:47]So, that's done by a process called the sweating.

[00:06:51]So, during sweating, the body absorbs heat energy from the skin to turn sweat into vapor which cools the body down.

[00:07:01]That's also an important an importance of endothermic reaction.

[00:07:10]Another application importance of endothermic reaction is seen in instant cold packs.

[00:07:20]>> [snorts] >> In instant cold packs, chemicals inside react and absorb heat from the surroundings.

[00:07:27]>> [snorts] >> This is done in order to reduce swelling and pain in the injuries.

[00:07:33]>> [snorts] >> This too is application of an endothermic reaction or process.

[00:07:45]>> [snorts] >> We see an an importance of of endothermic reaction or process in melting of ice.

[00:07:56]For ice to melt, it must absorb heat from its surroundings, from the environment to turn into liquid water, which is useful cooling tricks drinks or preserving food.

[00:08:12]So, heat energy must be absorbed. All these are importance of endothermic processes in the everyday life. Take note that some of these are not necessarily chemical reactions.

[00:08:26]They're just processes.

[00:08:28]>> [snorts] >> Define a polymer.

[00:08:33]We define a polymer as a large molecule which is made of many small repeating units called monomers. I've seen this question from another paper, I think.

[00:08:45]>> [snorts] >> I can't just remember which paper it was, and we also made a video based on that.

[00:08:53]So, these monomers are linked together by converted bonds in a long chain.

[00:08:58]Once again, a polymer is a large molecule made up of many small repeating units which are called monomers.

[00:09:07]And these monomers are linked together by converted bonds in long chains.

[00:09:14]Defining um monomer.

[00:09:19]Just one mark. Remember that definitions in sciences attract only one mark.

[00:09:27]So, do your best never to miss a definition in the an exam.

[00:09:36]>> [snorts] >> Figure one shows polymerization process of one of the monomers.

[00:09:45]So, we have these monomers.

[00:09:50]When you have a chlorine attached to an alkene, that's called vinyl chloride. These These are vinyl chloride molecules. Vinyl chloride chlorides, also known as chloroethane, or monochloroethane.

[00:10:13]Ethene, sorry.

[00:10:16]Okay? But, [clears throat] suffice to call them vinyl chlorides. So, name the type of polymerization shown by the reaction above.

[00:10:25]>> [snorts] >> Since these molecules are identical, then this reaction of polymerization must be addition polymerization.

[00:10:37]And like condensation polymerization, the molecules being added up are different.

[00:10:44]Okay, that's how you'd know.

[00:10:46]But, also uh during condensation polymerization, a smaller molecule is produced apart from the main polymer.

[00:10:57]Name the polymer formed. So, name this polymer.

[00:11:01]Since it's just an alkane chain, a long chain with chlorine atoms bonded to the parent chain, then this must be polyvinyl chloride. Polyvinyl chloride, in short known as PVC.

[00:11:18]So, [snorts] this uh polymer formed is called polyvinyl chloride, in short PVC. So, this is PVC. You have many vinyl chlorides chlorides uh joined together but you can also call this polychloroethene.

[00:11:42]Polychloroethene.

[00:11:45]Either of these two, don't write both.

[00:11:48]Just write one which comes to your mind at that point.

[00:11:52]This is one mark.

[00:11:54]Describe how the polymer mentioned here above is formed. Two marks.

[00:12:01]All right, so that means we must produce two points also.

[00:12:05]Remember the number of marks indicated give you a clue as to how many points you need to give.

[00:12:14]So describe how the polymer mentioned above is formed. How PVC is formed.

[00:12:23]All right, so the double bond here, this one here, the double bond, the carbon-carbon double bond in the chloroethene has to be broken. So each of these bonds is broken and the monomers link together to form long chains with no small molecules [snorts] releasing being released.

[00:12:49]All right, so by saying the carbon-carbon double bond in chloroethene is break broken, that's one mark. And then [snorts] monomers link together to form long chains, that's another mark.

[00:13:04]Two marks all together. So this part here which says no small molecules are formed, that's that may not be important. It doesn't add anything to the number of marks.

[00:13:16]Yeah, but the marks which will come from saying the carbon-carbon double bond in chloroethene breaks, that's one mark.

[00:13:25]Monomers link together to form longer chains.

[00:13:28]>> [snorts] >> Another mark.

[00:13:30]Or you can say long chains of PVC.

[00:13:35]>> [snorts] >> Welcome, learners.

[00:13:40]>> [snorts] >> Figure two shows allotropes of carbon labeled P and Q.

[00:13:45]So, these are allotropes of P and Q.

[00:13:48]I've also seen this question quite a lot. And I think uh some of the videos we have made so far had a question like this uh of this one in one way or another.

[00:14:00]Remember that allotropes are different forms of the same element. So, carbon, for example, can exist in two forms. It can exist as diamond, which is this structure here, and graphite, which is this structure here.

[00:14:19]Identify the allotrope that conducts electricity.

[00:14:22]So, the allotrope that conducts electricity is this one here, which is Q, and is graphite. So, using the letters, you write Q there.

[00:14:38]Don't write the name.

[00:14:40]Okay, I've just put the name in brackets for you to be able to tell or to record that this is uh graphite, but the answer is Q.

[00:14:50]One mark. Which allotrope is used to make surgery instruments? [snorts] Okay, by the way, we are saying Q because uh graphite exists in layers of hexagonal shapes.

[00:15:06]Okay? And also, each carbon in a uh >> [snorts] >> in the graphite is bonded to three others, leaving one electron free for conduction.

[00:15:19]For conduction. So, graphite conducts electricity.

[00:15:23]So, graphite is Q.

[00:15:25]Which allotrope is used to make surgery instruments?

[00:15:29]Remember that um diamond is such a very strong material and it also non-corrosive material.

[00:15:40]Therefore, it is the best choice for making surgery instruments. Surgery instruments are those instruments that are used in a cutting out medical operations or surgical operations in the hospital.

[00:15:56]>> [clears throat] >> So, which allotrope is used to make surgery instruments?

[00:16:03]Most definitely P. Don't write the name.

[00:16:07]Like if you want, you can just put it brackets.

[00:16:10]So, P is diamond.

[00:16:12]But one of the uses of diamond is to make surgical instruments. Explain why the allotrope mentioned in 3A2, this one, is used to make surgical instruments. Why?

[00:16:27]All right. So, uh recall that uh diamond is extremely hard material.

[00:16:37]And it has a very high melting point and is chemically unreactive or inert.

[00:16:42]So, it does not corrode or react with body fluids.

[00:16:47]For that reason, it is the best choice of material for making surgery instruments or surgical instruments.

[00:16:57]Welcome learners. Just any one of them their response is given here will give you one mark. Don't write all of them.

[00:17:06]After all, the question is just one mark.

[00:17:15]Explain why ionic compounds conduct electricity in a molten state and not in a solid state.

[00:17:24]All right, ionic compounds do not conduct a uh when they're in solid state, but in a molten state when melted or when they they dissolved in water.

[00:17:41]And then they can be used to conduct electricity as electrolytes.

[00:17:46]>> [snorts] >> So, two marks. The two marks here are coming from you saying that um in the solid state ions are held tightly in a fixed lattice and cannot move freely. So, the ions in solid state are not so free to move.

[00:18:09]Therefore, solid state cannot conduct electricity because the electrons the ions are not so free to move. They're tightly held together in a fixed structure, which is called lattice. [snorts] So, when molten, the ionic lattice breaks down.

[00:18:30]Ions become free to move around and carry electric charge. Hence, ionic compounds conduct in molten state.

[00:18:42]Welcome, learners.

[00:18:48]State anyone use of Perspex Perspex in the type of glass. There are different types of glass.

[00:18:57]There's Perspex. There's also another type of glass, which is called borosilicate.

[00:19:04]>> [snorts] >> This particular one is the one that you see in the lab. The beakers, the test tubes, the measuring cylinders, especially those that are meant to be heat to be heated or to be used in heating experiments. They're called Perspex glass.

[00:19:24]So, Perspex glass is used to make lightweight shatter-resistant windows.

[00:19:34]It's used in headlights, display signs, spectacle lenses because I can put this uses many other materials you can remember.

[00:19:47]Just any one of these.

[00:19:53]>> [snorts] >> Define oxidation in terms of electron transfer.

[00:19:59]Oxidation is defined as the loss of electrons by a substance.

[00:20:06]So, when a substance loses electrons in a chemical reaction, that's called oxidation.

[00:20:13]When a substance gains the lost electrons, that's called the reduction.

[00:20:19]Now, there are different ways of defining oxidation and reduction.

[00:20:24]So, oxidation can be defined in three ways.

[00:20:27]You can define oxidation as the loss of electrons by a substance. You can also define oxidation as an increase in the oxidation number, also known as oxidation state.

[00:20:40]You can also define oxidation as the gain of oxygen.

[00:20:46]Welcome, learners.

[00:20:53]The electrode potential of copper and iron as shown is given below.

[00:21:01]The very first one here is iron ions gaining electrons to form iron atoms.

[00:21:08]This is reduction.

[00:21:12]And the potential associated with such a change is minus >> [snorts] >> point minus 0.44 volts.

[00:21:24]Cobalt ions gaining two electrons reduced to cobalt atoms is accompanied by an electrode potential of positive 0.34 volts.

[00:21:38]So, using this data we are asked to find the EMF, the electromotive force, which is basically voltage, of this cell obtained when cobalt iron are combined.

[00:21:51]So, when you combine these two, how much [snorts] voltage will the cell generate?

[00:21:58]All right.

[00:22:06]So, [snorts] from the first one we have uh uh iron being reduced to make or to form iron atoms.

[00:22:29]And this is more negative.

[00:22:33]So, if we reverse this, reverse this equation half equation, the better way to say that is a called half equations.

[00:22:41]So, if we reverse this, it means that the electrode potential associated with it is also going to reverse or change sign.

[00:22:53]So, if this was initially negative, when we reverse this one here, we shall have a positive volts.

[00:23:04]>> [snorts] >> So, if the forward process is a reduction process, then it means that the backward reverse uh process is going to be an oxidation. Remember, oxidation is an increase in the oxidation number.

[00:23:24]So, this uh zero here, when go back, is going to increase to 2+. Oxidation.

[00:23:31]So, upon reversal, this must be positive.

[00:23:45]All right. So, if we keep the second equation half equation [clears throat] unchanged, >> [laughter] >> you see that here we shall have copper ions gaining two electrons to form copper metal.

[00:23:59]And this then the electrode potential associated with it is 0.34 volts.

[00:24:06]>> [snorts] >> So, these are the two values that we are going to use in order to calculate the EMF of the cell of these two equations combined together.

[00:24:19]So, the EM of F of a cell is equal to the electron at cathode.

[00:24:25]Remember, uh at cathode is where uh reduction happens, minus the electrode uh the potential at the anode where oxidation takes place.

[00:24:42]All right. So, in this case, we shall have um [snorts] 0.34 volts, this one here for reduction, and the minus uh 0.00.44 for the oxidation process.

[00:25:12]So, simplifying this and adding, you get 0.78 volts.

[00:25:18]Welcome, learners. State any one use of propanone.

[00:25:22]Remember, propanone is an alkanone.

[00:25:26]We make use of this a lot in everyday life comes to uh uh practices.

[00:25:36]State any one use of propanone, which is an alkanone.

[00:25:42]This can be used to remove nail polish, especially the ladies.

[00:25:50]It can also be used as a solvent for paints.

[00:25:55]>> [snorts] >> It can be used in varnishes and glues.

[00:25:59]It can be used to dissolve grease and oil. Of course, it goes back to solvent.

[00:26:07]It can also be used in the manufacture of plastics and synthetic fibers.

[00:26:12]Define the term alloy. An alloy is defined as a homogeneous mixture.

[00:26:20]Not just a mixture, but you have to say homogeneous mixture. A homogeneous mixture is one that shows a consistent mixture.

[00:26:30]It does not show separate layers of the mixtures, the components of the mixture.

[00:26:36]So, it's a homogeneous mixture of a metal with other metals or nonmetals to improve its properties.

[00:26:46]Welcome, learners. So, that's the definition of an alloy.

[00:26:51]Which alloy is used in the manufacture of airplane parts?

[00:26:58]Airplane parts are made by alloys which are called duralumin.

[00:27:05]Duralumin and aluminum alloy.

[00:27:11]Okay.

[00:27:13]Duralumin and the aluminum alloy is made up of aluminum uh mixed with copper, magnesium, and manganese >> [snorts] >> to form this type of alloys used for airplane parts. Take note of that.

[00:27:36]Figure three is a diagram showing chromatograph process of ink spots X and Y.

[00:27:46]Let's just go look at the diagram together.

[00:27:50]So, we have ink spots X and Y placed here on the paper, chromatograph paper.

[00:27:59]Dipped in a solvent.

[00:28:03]So, the solvent can be an alcohol or something else.

[00:28:10]X separates into M and P.

[00:28:15]Well, Y separates into N and R.

[00:28:20]The distance traveled by N is [snorts] 4 cm.

[00:28:28]Well, the distance traveled by the solvent is 10 cm.

[00:28:37]Welcome, learners.

[00:28:40]So, these are called chromatograms.

[00:28:44]Just master the diagram very well.

[00:28:47]Let's look at the questions based on each.

[00:28:50]What two components are absorbed most strongly by the stationary phase? The stationary phase in this case is the paper, the chromatograph paper.

[00:29:03]I'll bring it right here.

[00:29:10]>> [snorts] [sighs] >> Which two components are absorbed most strongly?

[00:29:15]All right. So, in this case, we are saying the stationary phase is the paper itself.

[00:29:24]Because the paper is not involved in the movement.

[00:29:27]The mobile phase is the alcohol and the pigments.

[00:29:33]>> [snorts] >> So, which two components are absorbed most strongly?

[00:29:38]So, for you to tell that which components are absorbed most strongly, just look at the components that have moved the shortest distance.

[00:29:48]So, in this case, it's M and N. They have only moved 1.6 cm.

[00:29:54]And like P and R, P has moved much further than M and N.

[00:29:59]And R has moved the furthest of all.

[00:30:05]All right. So, the answer is M and N.

[00:30:11]The shorter the distance they travel, the more they are absorbed by the paper.

[00:30:18]All right. So, the explanation here is we are saying it's M and N because the components that travel the shortest distance are held most strongly by the stationary phase.

[00:30:31]Which component is most soluble in the solvent?

[00:30:35]The component that is most soluble is the one that has traveled a long distance, the The distance. So, in this case, R has traveled the longest distance of all from the starting point.

[00:30:48]That means it is more soluble.

[00:30:52]Most soluble. So, in this case, it's R. You're saying it's R.

[00:30:59]Because it has traveled the longest distance compared to the others, to M, N, and P.

[00:31:08]So, R traveled the farthest distance with the solvent showing high solubility in the mobile phase.

[00:31:19]Calculate the relative flow value known as RF value for component M.

[00:31:26]For component M, in order for us to calculate relative flow value, remember that is a ratio of the distance traveled by the component.

[00:31:39]In this case, M has traveled 4 cm divided by the distance traveled by the solvent.

[00:31:48]So, in this case, first of all, looking at the marks means you have to lay out your work in the three steps, three marks.

[00:31:58]All right. So, in this case, you first of all, write down the formula.

[00:32:03]Relative flow value, RF, is given or calculated by distance traveled by this component.

[00:32:13]In this case, the component is M has traveled a distance of 4 cm divided by distance moved by solvent.

[00:32:22]So, the distance moved by the solvent is 10 cm.

[00:32:28]The distance moved by the component is 4 cm.

[00:32:32]So, all you have to do now is to substitute and calculate.

[00:32:37][clears throat] So, substituting the values, distance traveled by component is 24.

[00:32:44]Distance traveled by solvent is 10.

[00:32:48]So, 4 / 10 this should give us 0.4.

[00:32:54]Take note that the relative flow value of a substance does not have units >> [snorts] >> because the centimeters cancel out by the end of the day.

[00:33:05]Welcome learners.

[00:33:07]In case you have questions to ask us relating to our school, don't hesitate to send us a message to the number given here.

[00:33:16]Welcome learners.

[00:33:21]Methane which is an alkane react completely with oxygen to produce carbon dioxide and water.

[00:33:31]Write a balanced chemical equation for the reaction, two marks.

[00:33:36]>> [snorts] >> Usually, marks come from two steps.

[00:33:39]Writing the equation correctly with the correct formula, you have one mark. And balancing it also another mark. That gives us two marks all together.

[00:33:50]Welcome learners.

[00:33:53]So, in this case, make sure you write the substances correctly. Methane is CH4.

[00:33:59]Just transfer this one down here. And oxygen is O2.

[00:34:05]This is combustion.

[00:34:09]All right.

[00:34:11]So, combustion produces carbon dioxide and water.

[00:34:16]And then you balance accordingly. For carbon, we assess we have one carbon there, we have one carbon here.

[00:34:23]But for oxygen oxygen, we have uh put it two over here. Two times two is two plus this two oxygen plus uh four oxygen atoms but this side we need to multiply two [clears throat] by oxygen to give us four again. Just practice how to balance equations.

[00:34:46]Uh balancing equations shouldn't be a problem. It's something that you you have learned from form one.

[00:34:53]If 3.2 g of methane reacted with oxygen, how much carbon dioxide was produced?

[00:35:01]So here we are given the reacting quantities.

[00:35:05]3.2 g of methane reacted with oxygen.

[00:35:10]How much carbon dioxide was produced?

[00:35:14]All right. So our basis shall be only molar masses given.

[00:35:24]So if you work out Let's work out the molar mass for methane.

[00:35:32]Since there's one carbon atom and four hydrogen atoms, so one carbon is 12 and one hydrogen is one and is one, so four multiplied by one is four plus 12 + 16. That's the molar mass of methane.

[00:35:54]So using [clears throat] the molar masses and the given mass, we can find the number of moles of methane that are involved there in the reaction.

[00:36:05]So number of moles is equal to the given mass divided by the molar mass. In this case, we have 3.2 g divided by 16, so 0.2 moles of methane reacted with oxygen there.

[00:36:24]Remember we are asked to say how much carbon dioxide is produced.

[00:36:30]So, the mole ratio of methane to carbon dioxide is 1:1.

[00:36:37]Where where we getting the one from?

[00:36:39]This one here, one mole.

[00:36:41]And there's also one mole there. So, the ratio of methane to carbon dioxide is 1:1.

[00:36:48]That means the number of moles of carbon dioxide that will be produced is also going to be equal to 0.2 moles, just like uh the number of moles in all of those methane, 0.2 moles.

[00:37:06]Welcome, learners.

[00:37:09]All right, so molar mass of carbon dioxide is 44.

[00:37:16]Where from?

[00:37:18]Carbon is 12.

[00:37:20]Two oxygen atoms is 2 * 16 which is 32.

[00:37:22]32 + 12 is 44 g per mole.

[00:37:30]All right, so the mass, how much carbon dioxide? Therefore, the mass is going to be equal to the number of moles which you worked out here multiplied by the molar mass which is going to be 8.8 g.

[00:37:44]>> [snorts] >> So, 8.8, not 8, 8 but 8.8 g of carbon dioxide will be produced under those conditions.

[00:37:55]Welcome, learners.

[00:37:59]Calculate the volume of oxygen required to burn 3.2 g of methane completely at the room temperature and pressure.

[00:38:07]>> [snorts] >> Given that the molar volume at this room temperature and pressure is going to be 22 cubic decimeters.

[00:38:21]One mole of any gas occupies this volume.

[00:38:26]24 cubic decimeters.

[00:38:32]So, we are asked to calculate the volume of oxygen that will be needed to burn the three Remember, remember this is a continuation of the same question.

[00:38:40]Uh to burn 3.2 g of methane.

[00:38:45]How much oxygen is needed?

[00:38:49]All right.

[00:38:51]So, once again, we need to find the molar mass of methane, which we already said is 16 g per mole.

[00:39:00]All right.

[00:39:02]And the the number of moles of methane is going to be molar mass Sorry, the given mass divided by molar mass, 3.2 divided by 16, which is 0.2 moles.

[00:39:17]Right. So, the mole ratio, once again, uh if you This time, we are looking at the methane and oxygen. I'll take you back to the previous slide.

[00:39:30]Here.

[00:39:33]The mole ratio of methane and oxygen is 1:2.

[00:39:39]That means for us to find the number of moles involved in this reaction for oxygen, we shall take the number of moles that we found for methane multiplied by two.

[00:39:52]All right. I hope we are together there.

[00:39:56]So, the mole ratio is 1:2.

[00:40:00]That's the ratio of methane to oxygen from the balanced equation.

[00:40:06]All right. So, the mole of oxygen therefore will be two multiplied by the moles of methane, which is 0.04.

[00:40:16]So, use the fact 1 mole of a gas buys a volume of 24 cubic meters. What about uh 0.4 moles?

[00:40:28]>> [snorts] >> All right, so you simply multiply the number of moles by the molar volume.

[00:40:37]So, in this case, multiplying 0.4 by 24 to give us a volume of 9.6 cubic meters. Define water pollution.

[00:40:49]We shall define water pollution as the introduction of harmful or toxic substances into water bodies. Water bodies here can mean lakes, dams, uh ponds, um seas, oceans, rivers, making the water unfit for use.

[00:41:11]So, the introduction of harmful or toxic substances into water bodies makes the water unfit for drinking, unfit for cooking, unfit for washing.

[00:41:23]State any two pollutants of water.

[00:41:27]Water is considered to be polluted if it contains any of the following things.

[00:41:34]Industrial chemicals and heavy [snorts] metals. Heavy Heavy metals are metals like lead, mercury, and the others.

[00:41:44]If it contains raw sewage and human waste such as excreta, I mean feces and urine, as well as blood, such water is considered to be polluted.

[00:41:56]Agricultural fertilizers and pesticides, oil and petroleum products, plastic wastes and solid rubbish, detergents and household chemicals. Any two of these is considered to be or considered to be pollutants.

[00:42:15]Mention any two useful benefits of hard water.

[00:42:19]Remember that hard water is one that has very high concentration of dissolved salts of magnesium and calcium.

[00:42:29]Like magnesium sulfate or calcium sulfate, such water is said to be hard water.

[00:42:37]So, it has high concentration of ions of magnesium and the calcium.

[00:42:50]All right, so such water is very very important in a number of ways. The question is requiring us to say two.

[00:42:59]So, provides calcium and magnesium ions needed for strong bones and teeth. If you drink such water, you will benefit from each calcium and magnesium ions into your body, which will make your teeth and bones strong.

[00:43:16]It also reduces risk of heart diseases.

[00:43:22]The ions such as calcium and magnesium are helpful to give the muscles in the heart the ability to contract and relax well.

[00:43:34]State any one advantage of reduced reduction at source.

[00:43:40]As a way of minimizing waste, reduction at source means that reducing the amount of waste generated at a place, particular place. So, reduction of waste has an advantage.

[00:44:04]Okay, and your examiners are looking for that.

[00:44:09]So, reduction [snorts] at the source reduces the cost of treating and disposing waste.

[00:44:18]While it is also at the same time preventing pollution before it occurs.

[00:44:23]That's [snorts and clears throat] one mark. Reduces the cost of treating and disposing waste.

[00:44:32]Welcome, learners.

[00:44:35]In case you have questions, don't hesitate to send us a text message on WhatsApp regarding black slides. We are going to come back with a response very immediately.

[00:44:50]Number eight.

[00:44:53]What is meant by empirical formula of a compound?

[00:44:58]We shall define an empirical formula as the simplest whole number ratio of atoms of each element present in a compound.

[00:45:14]In other words, we are looking at the simplest formula of a compound based on the the simplest whole number ratio [snorts] of atoms of each element.

[00:45:30]Work out the empirical formula of a compound that has the following percentages by composition.

[00:45:39]>> [snorts] >> Percentage composition by masses.

[00:45:43]All right. So, the compound has 40% carbon, 6.67 hydrogen, and the 53.33 oxygen.

[00:45:55]And we are Each element has given is given its relative atomic mass.

[00:46:02]So, you would take these to be equal to grams, these numbers.

[00:46:08]So, you have 40 g, 6.67 g for hydrogen, 40 g for carbon, and 53.33 g for oxygen.

[00:46:20]All right, so you lay out your work >> [snorts] >> in table form. This simplifies things.

[00:46:38]Okay, so the atoms present are carbon, hydrogen, and oxygen.

[00:46:43]And they each the percentage of each given here, carbon is 40 makes 40%, hydrogen makes 6.67% and oxygen makes 53.33%.

[00:47:01]The relative atomic mass of carbon is 40.

[00:47:06]>> [snorts] >> So, you can use these [clears throat] two numbers, the mass.

[00:47:11]Remember that I said that these values are basically masses. If you add them up, they must give you 100 g.

[00:47:19]So, this these are grams per hundred.

[00:47:22]All right, so to find the number of moles, you simply take the percentage.

[00:47:29]>> [snorts] >> Here's the mass divided by 12, divided by RAM.

[00:47:37]So, 40 / 12 is 3.33 moles of carbon are present in the compound.

[00:47:47]The number of moles in the hydrogen will be the given mass, which is the score of the percent, divided by one is 6.67.

[00:47:59]While the number of moles in oxygen is going to be 53.33 / 16 is equal to 3.33.

[00:48:11]So, these are moles.

[00:48:14]All right. So, in order to find the ratios in which the atoms are combined, you divide each of these numbers by the smallest number.

[00:48:26]>> [snorts] >> So, if you divide If you look at these numbers here, the smallest is 3.33. So, divide 3.33 by 3.33, you get a one.

[00:48:37]Divide 6.67 by 3.33, you get a two.

[00:48:43]This goes into that two two times.

[00:48:46]If it doesn't come out as number whole number, make sure you round it off to get a whole number.

[00:48:54]Remember that the definition of empirical formula is the simplest whole number.

[00:49:00]3.33 / 3.33, that's going to be one. So, the ratio in which carbon, hydrogen, and oxygen combine to form a compound is going to be 3.

[00:49:13]It's going to be 1 2 2 1 1 2 1 1 2 2 2 1.

[00:49:19]All right. Therefore, the empirical formula is going to be C H2 O. So, there's one carbon, two hydrogen, and one oxygen.

[00:49:32]Work out the elements.

[00:49:35]Let me ask you to revise on this, and also be able to write down the molecular formula of a compound when you are given enough information about its empirical formula.

[00:49:50]So, this table plus the formula is six marks.

[00:49:57]Welcome learners.

[00:50:01]State any two uses of phosphoric acid.

[00:50:06]Phosphoric acid finds its application in many places.

[00:50:12]It can be used in the manufacture of phosphate fertilizers like ammonium phosphate.

[00:50:19]Fertilizers, phosphoric acid is one of the raw materials.

[00:50:25]But, it can also be added to food and drinks as an acidulant.

[00:50:34]An acidulant is a substance that is deliberately added to food in order to enhance its flavor.

[00:50:43]Think of drinks like a Coke, Fanta, and other soft drinks. The acidic taste that you get from them is due to some level of phosphoric acid added to it.

[00:50:59]So, two of these, two marks.

[00:51:02]Below are condensed formula of some organic compounds.

[00:51:14]Okay.

[00:51:19]>> [snorts] >> All right. So, we are asked to name organic compounds A and B.

[00:51:28]A is simply a compound called three methyl pentan- pentan-2-one.

[00:51:39]In other words, A is is an alkanone.

[00:51:47]How do you turn an alkanone? An alkanone is identified by the carbonyl group.

[00:51:55]This part here is a carbonyl group, CO.

[00:51:59]So, both sides is within a structure, an alkane structure or alkyl structure. So, when you have CO in the middle of two carbohydrate, I mean, carbo hydrocarbon chains.

[00:52:20]When you have a carbonyl group, CO, between two >> [clears throat] >> hydrocarbon chains or alkyl groups, then that's an alkanone.

[00:52:33]If it was an aldehyde, the CO could be at the edge.

[00:52:39]Like here.

[00:52:41]This is an aldehyde.

[00:52:45]On the other hand, all right. So, you need to identify also the position of the double bond.

[00:52:52]The double bond is on the the the position of the Beg your pardon. The position of the carbonyl group is on carbon number three.

[00:53:03]1 2 3.

[00:53:09]All right. So, that's where we have this CO. So, it be CH3 C is 3-methylpentan-2-one.

[00:53:21]All right.

[00:53:23]But, you need to be careful here.

[00:53:28]>> [snorts] >> A little bit careful.

[00:53:33]This is 1 2.

[00:53:37]Yeah, this this CO is basically attached to carbon number two. Counting from the my left-hand side. This is one, two.

[00:53:47]So, it's there, this one, where this one is attached.

[00:53:54]That's the meaning of these two here.

[00:53:57]All right. On the other hand, this one, when you have something put in brackets, it means it's a branch.

[00:54:04]So, this branch occurs on carbon number three.

[00:54:18]Yeah, it will occur on carbon number three. Try to draw structure. This structure you then need to come out to come out clear to tell that where each one of them is located.

[00:54:30]In case you're not sure, please kindly send me a text message. I'll be glad to clarify what I've not clarified in the first place.

[00:54:41]So, this is methyl three. If you check, one, two, the main chain is one, two, three. Yeah.

[00:54:50]So, that means the branch is on a carbon number three.

[00:54:58]Which [snorts] makes sense.

[00:54:59]And the the what?

[00:55:08]The functional group, the carbonyl group right here. So, on carbon number two, one, two.

[00:55:15]This one is attached to this carbon.

[00:55:18]Just to draw it out and confirm what I'm saying.

[00:55:22]B is a compound called um three methyl pentanoic acid.

[00:55:32]So, you have one, two, three, four, five. The main is has five carbon atoms. The branch is on carbon number three, right here.

[00:55:44]That's why it's called three methylpentanoic acid.

[00:55:49]The chain the main chain The main chain is has 1 2 3 4 5.

[00:55:58]It has to be pentanoic acid. Similarly here, if you take the carbon the methyl group you have 1 [clears throat] 2 3 4 5.

[00:56:10]So, the methyl group is on carbon number three.

[00:56:16]And the C double bond, which is the carbonyl group >> [clears throat] >> lies on carbon number one.

[00:56:25]Draw the structural formula of compound which has a conformer to organic compound C. So, organic compound C This looks like I said earlier on, this is a an alkanal.

[00:56:39]It would please you not to know that this is butanal.

[00:56:43]How do we know that this is butanal?

[00:56:45]When the CHO is at the end like this, and there's no hydrogen chain this side. It's just on one side. That's an alkanal in the first place. And like here this CHO is between two alkyl groups.

[00:57:02]This tells you that it's a an alkanone.

[00:57:06]This is alkanal.

[00:57:09]All right. So, we are asked to draw the structures different structures of these. Remember conformers have basically the same formula and also the same structure.

[00:57:21]They only differ [clears throat] on the uh orientation of the carbon atoms at some points due to rotation.

[00:57:31]All right. So, this, which is butanal can be given by either this formula where the carbonyl group is at the end like here.

[00:57:43]But you can also have it like can twist this part can come this end to produce this.

[00:57:54]Basically, this is the same molecule.

[00:57:57]The only difference is on the orientation of the orientation of this last part here.

[00:58:09]>> [clears throat] >> It's like this has just been flipped this end to produce this. So, these are conformational isomers.

[00:58:20]Welcome, learners.

[00:58:24]Which organic compound can it react with compound B to produce ester?

[00:58:30]Remember that for an ester to be formed, an alkanol Excuse me. An alkanol and an alkanoic acid must react.

[00:58:48]>> [snorts and sighs] >> So, let's look at B.

[00:58:52]B is an acid. We are safe there.

[00:58:55]But which other one can it react with?

[00:59:00]>> [snorts] >> Can it react with B to produce ester?

[00:59:04]Remember an ester is also called an alkanoate.

[00:59:08]So, already B is alkanoic acid. How do we know that? It has COOH.

[00:59:15]So, we need a compound that has OH in it.

[00:59:19]Definitely, it's D.

[00:59:22]It has an OH here. That means this is an alcohol.

[00:59:26]So, B, which is an acid, will react with a D, which is an alcohol, to form an [clears throat] ester.

[00:59:36]So, which organic compound can react with compound D to produce an ester or alkanoate?

[00:59:44]So, in this case, to be compound D, which your compound which organic compound can give positive results with Tollens' test?

[00:59:53]Remember that Tollens' test has to do with formation of silver mirror in a test [clears throat] tube.

[01:00:01]When you add Tollens' reagent to sample in a test tube, the contents of the test tube are going to change to mirror.

[01:00:13]>> [snorts] >> mirror >> [snorts] >> mirror like color silver silver mirror image. So, I'll rather say silver mirror.

[01:00:28]Okay.

[01:00:31]So, it forms mirror silver mirror in the test tube.

[01:00:38]To show that it's an alkanal. So, the compound that gives a positive test with silver with this Tollens' reagent is an alkanal.

[01:00:50]So, we need to check which one which one it is. If you go back, an alkanal here is C.

[01:00:56]The one that has CHO at the end.

[01:00:59]That's how you tell that this is an alkanal.

[01:01:02]So, the response to that is compound C.

[01:01:10]It's only an alkanal that will give us a positive result of forming silver mirror with Tollens' reagent.

[01:01:18]Write the products of the following chemical equation.

[01:01:22]We have half ethanol. This is ethanol.

[01:01:26]Reacts with sodium.

[01:01:28]When ethanol reacts with sodium, by the way, it's a sodium reacts with sodium in the same way as sodium and water.

[01:01:38]One of the products is hydrogen.

[01:01:42]>> [snorts] >> But the other one it will simply be a compound in which the hydrogen is replaced with sodium.

[01:01:50]So in this case, the product will be sodium ethanoate.

[01:01:59]This is sodium ethanoate.

[01:02:05]Sodium ethanoate is a displacement reaction. Sodium is going to displace hydrogen.

[01:02:15]So it will form the sodium [snorts] ethanoate.

[01:02:18]You see a salt.

[01:02:23]Sodium ethanoate to also form hydrogen.

[01:02:30]Check that this equation is fully balanced.

[01:02:38]So the three marks are coming in for writing this correctly and this correctly and then balancing it.

[01:02:44]Check that the equation is balanced. You have 2 * 2 for carbon, you have four.

[01:02:49]Here 2 * 2 for four, we are safe on the carbon.

[01:02:53]Oxygen, we have 2 * 1, that's two oxygen atoms. Here 2 * 1 is two oxygen atoms.

[01:03:01]Hydrogen, here we have six * [clears throat] 2, that's 12.

[01:03:07]Here we have 2 * 5, that's 10 + 2, 12.

[01:03:11]We are good.

[01:03:13]>> [snorts] >> All carbon atoms are [snorts] counted This is the last question in section A.

[01:03:23]Figure four shows the graph of volume of gas collected against time for the reaction of manganese magnesium with hydrochloric acid at a certain concentration.

[01:03:36]So, figure four graph for the reaction of magnesium and hydrochloric acid against time.

[01:03:44]volume versus time What is the maximum volume of the gas collected from the reaction?

[01:03:58]At the start, the volume was zero.

[01:04:01]But as the reaction is progressing until it is stops, this flat part here represents the progress of the reaction when the reaction has stopped, has come to an end.

[01:04:16]Probably because one of the reactants is finished.

[01:04:21]So, all you have to do is >> [snorts] >> draw a dotted line, go to the Y axis, the volume axis. Right here.

[01:04:36]This will give you the maximum volume.

[01:04:40]So, the maximum volume, if you check, this is [clears throat] uh um roughly 25 or it can be 24.something.

[01:04:54]Expect that G students can either say 24 or 25.

[01:05:00]Over here, after all this grading, it's not a system of the grade.

[01:05:06]But you're going to try to estimate, take the difference between uh any two known values. This is the 10 / 10 / 4 is 2.2.

[01:05:28]So each division is 2.2.

[01:05:31]So you simply consider this is uh point So each division is 2 5 / 4. Okay, that's 5 / 4 is basically 1.

[01:05:51]Uh something like 0.2.

[01:05:54]So 1.2 take this times three that's going to be 3.6 something like it's close to 24. You can say 24 or 25.

[01:06:07]It's okay. After how many minutes did the reaction stop?

[01:06:13]So the reaction stopped somewhere somewhere here.

[01:06:22]The point when this graph is becoming flat that's where the reaction stopped here.

[01:06:31]This portion here represents a situation where the reaction has come to an end.

[01:06:35]So this happened somewhere roughly at 10 minutes.

[01:06:42]>> [snorts] >> Over here. Don't say here because there's still reaction going on though very slowly.

[01:06:50]So it's 10 >> [clears throat] >> minutes.

[01:06:53]But I would also give mark if anyone says 9.5 or something like that roughly but not eight.

[01:07:02]Eight is too far.

[01:07:04]It's this is where the graph start began to get flat since then.

[01:07:10]Yeah.

[01:07:12]So, the reaction stopped after 10 minutes.

[01:07:16]>> [snorts] >> Calculate the rate of reaction between 1 and 3 minutes.

[01:07:25]>> [snorts] >> Between 1 and 3 minutes.

[01:07:35]So, 1 minute is basically somewhere here.

[01:07:43]1 minute, 3 minutes is somewhere [snorts] there.

[01:07:51]So, what was the rate?

[01:07:54]All right. So, first to get to find the rate. Remember, rate is volume at 3 minutes which is volume at 3 minutes, volume here is roughly somewhere there.

[01:08:11]Volume at 3 minutes minus volume at 1 minute divided by time difference. So, the time difference is the time between 1 minute and 3. That should be uh 3 minus 1 >> [clears throat] >> which gives us two.

[01:08:36]So, volume at 1 minute, at 1 minute uh at 1 minute is depends, it's somewhere 12 cubic centimeters.

[01:08:51]>> [snorts] >> Well, the what these values are basically estimates because this graph is not to scale.

[01:09:00]I mean, well, there is scale but this great squeeze and issues.

[01:09:13]All right, [clears throat] just to estimate this could be two.

[01:09:18]And volume at 3 3 minutes 3 minutes.

[01:09:27]Okay, when it takes 3 minutes to minute here, 1 minute.

[01:09:33]Whatever the case.

[01:09:35]So, [snorts] 3 minutes will be somewhere there.

[01:09:45]The graph is not good enough.

[01:09:48]All right, so in this case um the time difference is just 2 - 1 2 - 1.

[01:09:55]The time here is subtract the time there. That's 3 time - 1, which is 2 minutes.

[01:10:01]And then therefore H is going to be 30 - 12, which is H 12 / 2. That's 4 cubic centimeters per minute.

[01:10:14]Explain how burning of fuels causes global warming.

[01:10:21]Remember, global warming is an average change in temperature of the earth.

[01:10:27]It's caused by a number of factors. One of them is burning of fuels.

[01:10:33]All right, so burning fuels basically releases This is two marks. Releases carbon dioxide gas and other greenhouse gases.

[01:10:42]That's by saying that you have one mark.

[01:10:46]So, these gases trap heat in the atmosphere leading to global warming.

[01:10:53]By saying these two points you have two marks, uh Welcome, learners.

[01:11:02]Describe how deforestation, which is the cutting down of trees without replacing them, increases the concentration of carbon dioxide in the atmosphere.

[01:11:14]Five marks means five points must be raised there.

[01:11:20]So, [snorts] trees and plants absorb carbon dioxide, that's what we know, from the air during photosynthesis.

[01:11:27]That's one mark.

[01:11:30]So, when forests are cut down, fewer trees are left to cover the texture carbon dioxide in, so less of this gas is removed [snorts] from the atmosphere.

[01:11:44]If the cut trees are burned, they will release carbon dioxide back into the atmosphere.

[01:11:56]First of all, like in the first place, we removed we cut trees, the amount of carbon dioxide decreases.

[01:12:12]The amount of carbon dioxide being removed by the plants decreases.

[01:12:17]And if the trees that we have cut are burned, they carbon dioxide is going to be added back to the air.

[01:12:29]All right, so even if trees are left to rot, decomposers, bacteria and others, will break down their wood and they release the stored carbon as carbon dioxide through respiration.

[01:12:44]These are very strong points. Make sure you understand what's going on here. So, the double effect in this case, that of respiration and burning, reduced absorption leads to reduced absorption and the extra release of carbon dioxide, the concentration increases in the atmosphere.

[01:13:11]Five marks, five points. Describe how nitrogen is isolated from air in that most in the laboratory.

[01:13:21]So, the first thing that needs to be done is to remove carbon dioxide from it. Remember that air contains 79% >> [snorts] >> So, 78% nitrogen, 20 or 21% oxygen, and 0.034 carbon dioxide.

[01:13:40]So, in order to get pure nitrogen, you must remove carbon dioxide and other gases.

[01:13:47]So, carbon dioxide is removed first by passing it through sodium hydroxide solution.

[01:13:54]So, in the process, there's also sodium hydroxide, which is >> [snorts] >> reacted reacted with carbon dioxide in order to add a salt water to remove it from the air.

[01:14:09]So, moisture is also eliminated when the air is then passed uh through concentrated sulfuric acid.

[01:14:17]Remember, sulfuric acid is a drying agent.

[01:14:21]So, it will remove moisture from the air.

[01:14:27]And oxygen is already acted away >> [clears throat] >> when the dry mixture is passed through overheated copper turnings.

[01:14:38]Remember that oxygen is an oxidizing agent on this one.

[01:14:43]All right. So, it will react with copper.

[01:14:46]Sorry. It is to pass through copper to form copper [clears throat] oxide. In so doing, oxygen is removed.

[01:14:57]So, the solid copper oxide is formed as oxygen is removed from the mixture.

[01:15:07]At the [snorts] end, pure nitrogen is collected as the only major unreactive gas left.

[01:15:13]That's how you separate nitrogen from the mixture in the lab.

[01:15:25]With the aid of a well-labeled diagram, describe how copper can be refined using electrolysis process.

[01:15:33]When a question requires you to use a diagram, [snorts] make sure the diagram is fully labeled.

[01:15:41]By drawing the diagram before you label it, you have one mark.

[01:15:48]And then, by labeling at least three parts, you have another mark.

[01:15:54]So, the diagram will give you two marks all together.

[01:15:57]Whatever you say down here, it must really refer directly to the diagram.

[01:16:06]So, fill an electrolytic cell with copper sulfate solution.

[01:16:11]If you what you want is copper metal, then the solution must be copper sulfate.

[01:16:17]>> [clears throat] >> We say that the ions, the atoms in the metal must be the same as those in the solution. So, here we must have copper ions in the solution. If the metal desired is going to be copper metal.

[01:16:43]So, take note. So, we have anode that part which is connected to positive end of the battery. And then the cathode the electrode that's connected at the negative end.

[01:16:54]Remember also that it is at the anode where oxidation happens.

[01:16:58]And it is at the cathode where reduction [clears throat] happens. Take note.

[01:17:06]>> [snorts] >> But just this is not a convention symbol for a battery.

[01:17:12]Beg your pardon there.

[01:17:16]So, fill an electrolytic cell with copper sulfate solution.

[01:17:22]Once that's done, connect the impure copper to the positive terminal of the DC power supply to act as the anode such as this one which needs to be purified.

[01:17:33]It's an impure [clears throat] copper metal.

[01:17:38]With some impurities.

[01:17:42]All right. So, connect a thin sheet of pure copper. Okay, let's start with um a pure sheet of um copper metal. We like the copper atoms to be deposited here.

[01:18:06]So, connect a thin sheet of pure copper to the negative terminal.

[01:18:11]Give us a sheet this Make sure you draw the correct version of the battery there.

[01:18:17]To act as the cathode.

[01:18:21]Switch on the power supply to allow electric current to pass through the electrolytes. So, close the switch here.

[01:18:30]So, the current will be flowing through the circuit.

[01:18:36]Then copper atoms from impure anode will dissolve into the solution as copper ions. This is oxidation.

[01:18:47]So, copper atoms are becoming copper ions into the solution.

[01:18:55]So, the copper ions move through the electrolyte.

[01:18:59]And then upon getting here, they'll gain electrons and copper metal, which is pure, is deposited there.

[01:19:10]So, insoluble impurities fall to the bottom of the cell as anode mud mud, while soluble impurities remain in the solution.

[01:19:28]So, you continue [snorts] doing this until the cathode gains a thick layer of pure refined copper and the anode becomes smaller and smaller.

[01:19:39]So, the pure melted copper is going to get into the solution in form of copper ions, which will gain electrons upon reaching there to form uh more copper, pure copper metal.

[01:19:54]Welcome, learners. So, check [snorts] three marks there for the setup. And then for the description, you should also have about seven marks. So, this is four.

[01:20:06]Then Yeah. So, here we assuming that the diagram without labels, this one, and the at least three labels, must be two.

[01:20:20]Welcome, learners.

[01:20:22]>> [snorts] >> My name is Mr. Bojjirola.

[01:20:26]>> [clears throat] >> Then reach me on the number shared here.

[01:20:31]>> [snorts] >> Welcome to the last question.

[01:20:37]In a laboratory, labels fell off from the bottles containing an alkane, alkene, alkanol, and alkanoic acid.

[01:20:47]Describe an experiment that could be done to identify the contents of the bottle.

[01:20:54]So, it's not known which bottle is which one.

[01:21:00]So, you're going to describe an experiment to tell which bottle contains what.

[01:21:05]Welcome, learners.

[01:21:09]So, first of all, label the four unknown liquids as A, B, C, and D.

[01:21:15]And place the samples in separate test tubes.

[01:21:20]So, together, you have four samples that are labeled A, B, C, and D.

[01:21:34]>> [snorts] [snorts] >> So, add a few drops of bromine water.

[01:21:48]Remember, bromine water is brownish.

[01:21:52]To samples A and B and shake gently.

[01:21:55]>> [clears throat] [snorts] >> All right.

[01:22:00]So, the sample that decolorizes bromine water, the one that removes color bromine color to colorless, must be an alkene. Remember that alkenes contain a double bond, and when added to bromine water, the brown bromine water will change color from being brown to colorless.

[01:22:25]The sample that does decolorize must be an alkane.

[01:22:31]Alkanes due to their saturated saturation Remember that alkanes are saturated >> [snorts] >> are saturated hydrocarbons.

[01:22:47]And when you add the bromine water to them to alkanes the color of the bromine does not go.

[01:23:01]It persists.

[01:23:03]>> [snorts] >> Then you prepare a pink sodium hydroxide solution by adding [snorts] a few drops of phenolphthalein indicator to sodium hydroxide solution.

[01:23:18]Remember that this gives us a pink solution.

[01:23:24]Then add equal volumes of samples C and D separately to the pink solution of sodium hydroxide solution.

[01:23:36]Observe You observe the color change of the mixture in each of the test tubes.

[01:23:46]So the sample that turns the pink solution colorless is the alkanoic acid because it neutralizes the sodium hydroxide in it.

[01:24:00]>> [snorts] >> The sample that does not change the pink color is the alkanol in this case.

[01:24:06]Assuming that both of both of them are identifiable using this approach.

[01:24:13]As all the four bottles now are identified as alkane, alkene, alkanol, and alkanoic acid.

[01:24:22]So make sure you write 10 points in order to get all the 10 marks.

[01:24:29]Welcome, learners. So, this has been your teacher, Mr. Boji Olara, having taken you through the [clears throat] entire paper.

[01:24:41]Let me urge you not to just memorize these things, because we don't guarantee that these are the questions that you're going to face.

[01:24:50]But, we want just to remind you the standard answers as required.

[01:24:55]Welcome, learners.

[01:24:57]Expecting to see you soon.

[01:25:00]Bye-bye.