Chemistry (Science Paper 2) 2024 GCE | Past Exam Questions & Detailed Solutions | Full Revision

[00:00:00]Greetings science students.

[00:00:03]In this video, we are going to revise chemistry or science paper 2 of 2024 GCE.

[00:00:13]All right. So, we we have to answer all the questions section B.

[00:00:20]Okay. All the questions in section B was um just 45 marks. So let's let's begin with the first one.

[00:00:31]A box of chemicals in a laboratory has the following symbol attached to it.

[00:00:38]Okay, this is the symbol attached to a bottle of chemicals.

[00:00:44]Now describe the nature of chemicals in the box related to the symbol. Okay. So whenever you're going to you see this symbol just know that uh it means that those chemicals can easily catch fire or catch what catch fire easily.

[00:01:05]Okay. So you say um describe the chemicals in the box the chemicals.

[00:01:14]Okay.

[00:01:16]They say chemicals um easily catch what catch fire. Okay. Sometimes they can be written highly flammable.

[00:01:31]Highly what? Highly flammable. It's just one and the same thing easily catching what? Catching fire. Okay.

[00:01:41]Now write a laboratory rule which that is observed when handling chemicals in this box.

[00:01:51]Okay. So if it catches fire then automatically it needs to be so it must be kept okay in a cool and dry place. Okay. In a cool.

[00:02:17]All right. And dry uh place.

[00:02:25]Okay. Cool and dry place. So when you say cool and dry place, it means uh with low uh temperature. Okay.

[00:02:37]With low temperature.

[00:02:43]All right, let's proceed. [snorts] When a bottle of ammonia is opened, when a bottle of ammonia is opened, within a few minutes, it smells it it smell spreads through the laboratory throughout the laboratory. Name the process responsible for the evenly spreading of the smell of ammonia um solution. So the process here is called diffusion not osmosis or active transport but it is what diffusion.

[00:03:27]Okay, diffusion because diffusion is just the movement of particles from the region of high concentration to the region of lower what lower concentration. Okay, lower concentration um concentration gradient. Now describe the process responsible for the evenly uh for the even spreading of the smell of ammonia uh ammonia solution.

[00:03:59]Describe the process responsible. So we have to describe what diffusion. Okay.

[00:04:07]So diffusion.

[00:04:19]So diffusion of ammonia particles move from particles diffusion of ammonia. Okay. Uh during diffusion, let me just say during diffusion of ammonia, particles move from the region of high concentration.

[00:05:05]Okay.

[00:05:08]High high concentration concentration to the region of lower concentration.

[00:05:26]Lower concentration.

[00:05:29]Okay. The region of lower concentration.

[00:05:31]Let's just end here. It's just a mark.

[00:05:34]Let's proceed. relate the evidence which um which the process named D in question Aman numera one provided to the kinetic what is provide provides to the kinetic theory of matter. Now we need to know what the kinetic theory of matter states. So the kinetic theory of matter states that matter is made up of 10 particles that are in continuous random random motion. So here you can just say um particles particles of matter uh in any in continuous.

[00:06:31]Okay.

[00:06:33]Continuous.

[00:06:35]Uh-oh. Continuous random motion.

[00:06:44]Okay. Then continuous random random motion. So we're done with question.

[00:06:50]This is question what? Question one.

[00:06:53]Let's go to question what? Question two.

[00:06:58]Okay.

[00:07:02]So question two here we are we are given an apparatus they're saying the following diagram shows a piece of apparatus used in the laboratory.

[00:07:14]All right so this is the apparatus here with of the stopper and the tap here.

[00:07:19]The first question is name the piece of apparatus shown in the diagram. So this apparatus here is what is a separating funnel.

[00:07:33]Okay, separating funnel. A separating funnel is an apparatus or a separation technique or method that is used to separate imissible two imissible liquids.

[00:07:50]Liquids that do not mix completely like oil and water.

[00:07:55]Okay. So, question B1, identify identify one property of a mixture separated by the piece of operators named in one. Yes. Um, you can say immissible um imissible liquids. Okay.

[00:08:18]I emissible.

[00:08:22]Okay. I emissible liquids.

[00:08:29]Okay. Emissible liquids. Liquids that do not mix completely that just mix.

[00:08:34][snorts] Okay.

[00:08:36]Now, suggest the names of two liquids that can be separated by the piece of shown in the diagram. So, you can say water and oil.

[00:08:52]Mhm.

[00:08:54]Water and oil. Last question.

[00:08:58]On question two, suggest suggest two precautions that need to be taken to uh to separate the mixture in the in B2.

[00:09:11]Okay. Now, two precautions are that number one, you need this tab needs to be closed.

[00:09:21]Okay. Before filling the funnel. So, close the tap.

[00:09:29]Close the tap before filling the funnel.

[00:09:46]Filling the funnel with a mixture. Okay.

[00:09:49]>> [snorts] >> Then number two um suggest I mean the second one is that you need to to clean this must be clean. Okay.

[00:10:06]Clean the apparatus.

[00:10:14]Okay. Apparators.

[00:10:16]Clean the word the operators which is the separating what? Separating funnel.

[00:10:28]Okay. Which is the separating funnel.

[00:10:31]All right. So that's all. So we given two questions but it's a mark. Oh okay.

[00:10:38]Maybe maybe there was a mistake. Let's go to question three.

[00:10:48]Question three.

[00:10:50]So question three is saying the following diagram shows the electronic configuration of the ion of the element of an element M. So this is an an ion of an of element M. Then the first question now before we go to the question look at this when you look at this this is the ion and since it's two plus since it's two plus it means that it is giving out uh it is donating okay so there's a transfer of electrons two electrons so the only so how we can know that this is this atom is um this element is by counting the by counting the number of electrons. So you have 2 + 8 10 plus another 8 18 plus the two electrons 20.

[00:11:44]Then we go to the periodic table on the periodic table and check element number 20. We're going to find that it is calcium. Okay, it is calcium.

[00:11:55]All right. So based on that knowledge we can proceed and look at the questions.

[00:12:03]Draw the atomic structure and write the electronic configuration of the neutral atom. You see neutral what? Neutral atom.

[00:12:13]Okay. So they're asking us to come up with a neutral neutral atom. So the neutral atom it is the element before the before the transfer of electrons or before getting electrons. Okay. by accepting electrons or before involving any chemical bonding. Okay. So, [clears throat] so we're going to say so this is the space that we're given.

[00:12:45]We draw here. So, we're going to say this is the nucleus. Let's represent the nucleus with this one. Okay. The first shell will occupy how many electrons?

[00:12:55]Two electrons. Okay. The second shell will occupy eight electrons. 1 2 3 4 5 6 7 8. Okay. The third shell will occupy occupy 1 2 8 electrons as well. 6 7 8. Okay.

[00:13:21]Now because of these two electrons that we are lost we can we have another another shell. Okay like this one two [clears throat] okay so that's it.

[00:13:38]Now you should know that this one is a cation.

[00:13:43]Cation have fewer number of electrons. You should not take note of that few number I mean this one is an atom but this one is a cation that's what I wanted to say okay so cation has a few number of what of electrons than the element check if you count the number of electrons here will be the number of electrons will be 18 but here the number of electrons in the atom is 20 so that is a cation Then an ion since it's gaining you don't have an ion but I'm just explaining an ion is gaining electrons it will have more number of electrons than the original atom okay than the neutral atom because it is gaining electrons.

[00:14:37]So you should have knowledge of chemical bonding. Let's go let's proceed. So the electronic configuration the first shell occupies up to two. So it will be two.

[00:14:47]Second shell eight then another eight then two. So what do we know about this?

[00:14:54][snorts] The number of shells determine the period determines the period of an element on the periodic table. So since it is 1 2 3 4 four shells then it is in period 4.

[00:15:08]Then the number of electrons in the outermost shell determines the group in which the element is on the periodic table. So this element here.

[00:15:21]Okay, let's proceed. Give the name group and period of element M. The name here is causam.

[00:15:33]So you check on the predict table element number 20. Okay, it is kam group.

[00:15:41]It is in group two. Okay, in group two.

[00:15:45]So usually use Roman numerals for groups. Okay, period. It is in period what? 1 2 3 4. It is in period 4. Okay, it is this simple. Now compare the reactivity of a compare the reactivity of an atom M and an ion M. The reactivity you should know that atom M atom of M. So this is say an atom of M An atom of M is reactive.

[00:16:35]Okay.

[00:16:37]Is reactive.

[00:16:41]Wow.

[00:16:42]Or whereas whereas the ion of M is an reactive because because Because because it is stable.

[00:17:16]It is now stable. This is reacting because it is what? It is not stable.

[00:17:21]But this one is stable because that's why it is not what? It is not reacting.

[00:17:30]Okay, let's go to question four.

[00:17:36]So question four salts are substances that have many methods of preparation.

[00:17:45][snorts] So we have ion to chloride. Ion to chloride can be prepared by reacting substance X and dilute hydrochloric acid. Okay.

[00:17:59]Now which reactant should be in excess?

[00:18:04]Okay. Which reactant should be what? In excess. So the reactant that should be in excess is this substance here. This metal. I don't know if it's metal or what. This substance. So we can just say substance X should be should be in excess.

[00:18:30]Okay.

[00:18:32]Oh, we are even asked to give a reason.

[00:18:35]Okay. Why should the named reactant in a one above or in A1 be in excess? Okay.

[00:18:44]So, this one should be in excess in order.

[00:18:49]Okay. Or in order or to ensure to ensure that this acid hydrochloric acid to ensure the acid is used completely.

[00:19:15]Okay. During the during the reaction.

[00:19:22]during the reaction. Okay.

[00:19:25]During the reaction. Then other questions. Question.

[00:19:36]Question B1. Suggest a suitable reactant as substance X.

[00:19:41]Okay. Suggest a suitable reactant as what? substance X.

[00:19:47]So now look at this. You need to be very careful here. So since this one is a product, okay, this one is a salt that can be prepared by reacting substance X and hydrochloric acid. So you can hear here you can see here that there's hydrochloric acid. So whatever the substance X is, there has to be a metal.

[00:20:12]Okay? There has to be a metal. Let me show you uh here substance X must be one of the constituents of this metal this I mean this salt. So since the hydrochloric acid here there is chloride then ion should be substance X or substance X should be ion metal. Check hydrochloric acid reacting with ion this ion. Okay. So one of the chemical properties of acids is that they react they react with some reactive metals to produce a salt and hydrogen gas. So the chlorine and ion here will be ion to chloride.

[00:21:04]Yeah.

[00:21:05]Because here we told that ion has a valance of what? Two. So we can just say iron 2 chloride and hydrogen what hydrogen gas let's just I'm just explaining anyway [snorts] check this is ion to chloridehon to chloride then this is hydrogen what hydrogen gas [snorts] hydrochloric acid ion so the answer here is ion metal so you can just say here ion metal Very good. Let's go to the next substance.

[00:21:45]Write a balanced chemical equation for the reaction between substance X suggested in B1 or here with dute hydrochloric acid.

[00:21:56]Nice. We've already we've already written this answered this question here. Okay. Let's just copy this side and balance the equation. The equation.

[00:22:07]Okay. So we can say substance X since we said this ion so we can just say substance X reacting with hydrochloric acid to form ion to chloride. Since this ion has a valance of two chlorine is group seven has a valence of one electron valence of one. So the two will come here and hydrogen gas. Okay, this one is a solid.

[00:22:39]It's a metal. It's a metal solid. This is aquas. This is a salt. Aquas.

[00:22:45]Um, let's check before we say it's a summary.

[00:22:53]Yes. Then this one is a gas. Let's balance the equation. So, balancing the equation.

[00:23:02]ion one the number of ion atoms atoms of ion on the reactant side here is one on the product side is also one um now when you look at the number of hydrogen atoms here is one here the number of hydrogen atoms is two so what you do since when balancing use a pencil because the coefficients tend to change okay so let's put a two there um the number of chlorine atoms is two because of this two which is in front so also here two so this equation is what balanced thank you so much let's go to the next question which is question um question question five B5 okay now now B5 if you are still together up to this question B5 consider Consider subscribing and liking the video if you're not part of the Nostage Academy family.

[00:24:08]So this question is under what question is this? And um more concept.

[00:24:20]Okay. So 10.5 g of sodium hydroxide was reacted with was reacted with um dilute hydrochloric acid according to the equation. So this is the equation here. Wow. So the first question is balance the given equation. Well, this is simple.

[00:24:44]So we're going to say sodium hydroxide since it's in aquas here reacting with sulfuric acid H2SO4 aquas as well to form sodium sulfate okay aquas and water H2 or liquid then balancing this equation Okay, passing this equation is very simple. We're going to say we're going to say how many number of atoms of sodium do we have on the reactant side? This side it's one. This side two. So we're going to change this one to a two.

[00:25:35]Okay, like this we put a two. Then we go to the next. How many? So we go to the next element which is oxygen. How many how many atoms of oxygen do we have on the reactant side? We have this is 2 + 4. So we have six on the reactant side.

[00:25:56]How about the product side? We have 4 + 1 5. So we can put a two here where there is water here in front. So that it becomes six as well. Okay. Take note that the number of hydrogen atoms will be affected. What? No problem.

[00:26:15]Let's just proceed.

[00:26:18]Um the number of [snorts] the number of atoms of sulfur here is one here is also one. So what hydrogen has been affected only? Hydrogen has been what? Affected.

[00:26:37]So what we can do what we do here is check since here there are four atoms of whom is it number of hydrogen has been have been affected.

[00:26:52]No so it is balanced. No it has balanced.

[00:26:57]So hydrogen 2 + 2 4 here 2 + 2 4. So everything now is balanced. Let's go to the next what? Next question.

[00:27:09]Okay, next question. B one. Use the balanced chemical equation to calculate the mass of sodium to to calculate the mass of sodium sulfate formed. Wow.

[00:27:28]Now the balanced chemical equation. So let's get the two compounds that are involved.

[00:27:36]We are given 10.5 g of sodium hydroxide.

[00:27:39]So we're going to use 2 moles of sodium hydroxide.

[00:27:47]Okay. Two. What else? We asked to calculate the mass of sodium sulfate.

[00:27:52]Sodium. So it be sodium sulfate.

[00:27:57]Okay. Like this. Now you have you have you have to be very careful here.

[00:28:05]We're [snorts] going to say first we need you need we need to know the number of um the the relative molecular mass. Okay, the relative molecular mass of sodium hydroxide and sodium sulfate.

[00:28:28]So when you look at relative molecular mass of sodium hydroxide since we have a two here it's going to be hydrogen how many atoms of hydrogen because of these two you have two all right 2 * the mass number which is 1.

[00:28:45]Then we go to oxygen oxygen we have also two the mass number is 16.

[00:28:52]Sodium is also 2 * 23.

[00:28:57]>> So when you add this side, you're going to have a 2. When you multiply, I mean 2 * 16 32 4 2 * 2 * 23 is um 46.

[00:29:13]And then we add.

[00:29:16]When we add here, we're going to have this is 2 + 32 + 48.

[00:29:24]This is giving me 80.

[00:29:27]Okay, 80.

[00:29:30]So, here we're going to use 80.

[00:29:35]Okay, 80. How about for sodium sulfate?

[00:29:41]Okay, let's do as well for sodium. Let's calculate for sodium sulfate as well.

[00:29:47]So we have one here. So we just only consider the number of atoms that are here.

[00:29:55]So 4 * 16 then sulfur is 1 * is it 32?

[00:30:05]Let me check on the predict table so that you don't make mistake.

[00:30:11]Okay.

[00:30:13]The sulfur yes it is 32.

[00:30:17]Sodium it is this is two. So it will be 2 * 23.

[00:30:25]Okay. So when we multiply here and add we're going to have we're going to have 142 since because of space here we're going to have 142. Uh okay. So 16 * 4 is 64. This is 32. This is 46. When you add, you're going to have 142. So we're going to use 142 here. 142. Then listen and listen very carefully. Here you ask yourself, what mass of sodium hydroxide are we given here? You're given 10 10.5 g.

[00:31:11]Okay. Then what mass of sodium sulfate we don't know is the one we want what we are calculating. Then we cross multiply.

[00:31:22]When you cross multiply this times this it will be 80x.

[00:31:26]Okay. It will be what? 80 x = to 10.5 g [clears throat] multiplied by 142.

[00:31:37]Then we divide by 80. Even this side by what? By 80. The x, the eight and the 80 will cancel. Then [clears throat] you have x is equals to. So when you punch this on the calculator, we're going to have 18 6 4 correct to two decimal places. 6.4 g.

[00:32:04]Okay. This is grams of this is the mass of sodium sulfate.

[00:32:11]Okay, sodium sulfate. Okay, so that's it to this question. So I was explaining this I've taken time here explaining how we get the 80 and the 40 142 how to get the one this relative atomic mass molecular mass I mean or formula formula mass the question 2 B2 what type of reaction is involved in the equation this is a base and this is an acid so this is neutral neutralizing ation reaction.

[00:32:47]Okay. So it is neutral neutralization reaction.

[00:32:57]Neutralization reaction. Okay. That's what question B5.

[00:33:03]Let's go to B6.

[00:33:06]B6. This is under periodic table.

[00:33:10]Uh this is under periodic table.

[00:33:13]So the question reads an element an element A is in group one of the periodic table. Okay. Another element N is in group seven.

[00:33:26]Nice. Write one property of the compound formed of the compound formed when element A and N react.

[00:33:39]Now you need to know first the type of compound that is formed. the name of a compound. Oh yes of the type of compound I mean so elements A are metals and element N are nonmetals. So when nonmetals when metals and nonmetals are involved in a chemical bonding the bond the bonding is called ionic bonding or electrovalent what electrovalent bonding.

[00:34:05]Now the compound formed has a lot of properties. Okay. Properties. Number one, you can say it is made up of positively and negatively charged ions or yes, negatively and positively charged ions.

[00:34:23]You can say the compound formed in conducting electricity in m state or aquacy state. It has high boiling and melting point. Okay. high boiling and um so they're a lot so we can just write one. So let's just say let me just say the compound the compound has high boiling and melting points.

[00:35:17]Okay, there a lot. Okay, you need to know a number of properties of ionic compound. What type of bonding exist in the compound formed when the two elements react? Chemistry is very simple guys. As long as you understand the the topic or understand the the the lesson okay you go you understand everything go through different past papers and chemist will be nice so you see as I was explaining here I've even answered this question so the type of bonding is ionic bonding uh bonding ionic bonding you can say electro or electrovalent what bonding let's just write ionic bonding since you know this is group one elements and metals and group uh seven elements are nonmetals.

[00:36:13]How does the reactivity of elements elements in group one differ from elements of group that of um from that of elements in group one [clears throat] in group seven. You should know that um group one elements okay group one elements their reactivity increases as you go down the group. The trend in reactivity of group one elements increases as you go down the group. So group one and elements um I would have said okay it's okay group and elements their reactivity reactivity increases okay increases down the group increases is down the group.

[00:37:29]Okay.

[00:37:33]Where is he?

[00:37:36]Where is he?

[00:37:42]Group seven increases increases up the group up the group. So meaning the most reactive element in group one is the element which is down in B in in group one which is fancium. Then the element which is most reactive in group seven it is florine. Okay florine. Are we done with this question? Yes. Just before let's go to question. What question is this? Question seven.

[00:38:30]Question seven. Now question seven reads.

[00:38:36]Metallic bonding accounts for most of physical properties of metals. Wow. With reference to metallic bonding.

[00:38:45]Okay.

[00:38:47]With reference to metallic bonding.

[00:38:51]Explain why most metals have high noting what have high noting point points I mean. Okay.

[00:39:03]So you can say metals.

[00:39:15]Okay. They have high melting point.

[00:39:24]Melting points.

[00:39:26]Um okay. Have high melting. They have high melting points because um because of because of the strong the strong electro the strong electrostatic forces electrostatic Force of attraction.

[00:40:11]Attraction.

[00:40:15]Okay.

[00:40:17]Electrostatic forces of attraction.

[00:40:22]Between between The negatively what is this between the negatively and [music] positively ions.

[00:40:50]Okay.

[00:40:52]Charged ions.

[00:40:56][clears throat] All right.

[00:40:59]charged ions. So the number number two which is B draw the diagram to show metallic bonding of zinc. Oh this is okay. So metallic bonding of zinc. So you metallic bonding of zinc metal just the zinc metal you can say. So you can start like this zinc has a valance of two. So it' be 2 plus then electron okay which is negative. Then you also have you can write zinc plus like this. Then you you you continue the the process the decolorization of electrons. Okay. This one this is the zinc ion. This one is what I mean not deolarized deoized.

[00:41:53]Okay. Deoized electrons. Okay. So these are deoized electrons. So you can have um this deoized electron just negative then zinc metal. All right. Zinc ion.

[00:42:10]Okay. Or ion of zinc. Then like this. Um let me just end here. Zinc 2+ electron zinc. Zinc 2 plus. Okay. So you can just put the zinc ion in what right in circle like this. Okay. So we've just shown that um this this zinc this is a zinc ion.

[00:42:44]Okay. Then these are deoized electrons.

[00:42:47]Okay. Deoized electrons. So therefore this bond is not eas easily broken. is not easily broken. It requires a lot of heat for this bond to break. Let's proceed.

[00:43:03][clears throat] [snorts] Elena carried out the following experiments.

[00:43:09]In experiment one, 5 g of powder of zinc and in experiment two, 5 g of granules of zinc.

[00:43:21]Okay, that's the chops or pieces of zinc were were each reacted with equal volumes of deladoric acid both of the same concentration for 15 seconds? The reactions were not were not yet completed at 15 seconds. In [snorts] which in which experiment was a large volume of hydrogen produced? Give a reason for your answer. Ah, this is simple. So the powder exper in experiment one. Okay.

[00:43:56]Experiment what? Experiment one.

[00:44:00]So experiment experiment one. Okay. Which is what?

[00:44:09]Zinc powder or powdered zinc.

[00:44:13]Powdered zinc. [clears throat] Listen the reason it is very simple.

[00:44:22]Okay.

[00:44:25]Powdered [snorts] powdered zinc this um experiment one. Suppose that zinc has a large surface area surface area for chemical or they just say for the chemical reaction.

[00:45:01]Okay, for the chemical reaction this is very simple that's the reason what would be the effect of reducing the temperature of the acid in the reaction what what is the effect of reducing the temperature of the acid in the reaction.

[00:45:19]Um so you know the temperature is one of is one of um the factors of rate of reaction that increases the rate of reaction. So if the higher the temperature the higher the rate of reaction. So when the temperature is reduced automatically the reaction will be slower.

[00:45:41]Okay. The reaction will be slower or be reduced. Okay.

[00:45:57]Will be reduced.

[00:45:59]Which question? What question is this?

[00:46:01]Question seven. Let's go to question eight. We almost done just remaining with uh just in two questions. Just remaining with two questions. This and the last one which is uh organic chemistry. Yes, organic chemistry.

[00:46:22]Okay. So this is the question.

[00:46:26]The question reads, I uh the following diagram shows the laboratory preparation of a gas and its collection. [snorts] Mhm.

[00:46:39]Now listen and listen very carefully. Let's first study the diagram. There is dute hydrochloric acid here.

[00:46:49]direct hydrochloric hydrochloric acid and a granu of met or pieces of metal Y here.

[00:46:58]Now listen very carefully from here you can uh uh come up with something you can know you can predict the results we've answered a question um what question was that let me just check this question here where an acid reacts with a metal and acid reacts with a metal hydrogen gas is produced and a salt so from here you can even say I mean you can know that this gas is hydrogen gas.

[00:47:35]Okay. So what's the first question? Identify one error in the setup uh one area in the setup in the diagram shown. Okay. In the diagram shown.

[00:47:53]Now look at this. Here there has to be a tap.

[00:48:02]Here there has to be what? A tap on the funnel. Okay. For opening and closing or preventing the escape as gas is escaping here that the reaction is taking place here. So there has to be a tap here to close um um the substance. So this is just one error. So you guys you can just say the tap the tab is missing.

[00:48:37]The top is missing.

[00:48:40]Okay.

[00:48:42]On the funnel on the this funnel is the seaw funnel.

[00:48:52]Okay. [clears throat] Orto final.

[00:48:57]Okay.

[00:48:59]Which is above the reaction substance here. The reacting substances.

[00:49:04]So it's missing. [snorts] Let's just proceed.

[00:49:11]Okay. Proceed. Again, if you've noticed, there is also another reason here.

[00:49:19]This This funnel is not dipped into um into the reacting substances here.

[00:49:29]It it just ends here. Anyway, since we've already discovered one error, then we can proceed.

[00:49:40]Identify gas X. Gas X is hydrogen gas.

[00:49:45]Okay.

[00:49:49]Hydrogen gas. The form of hydrogen gas is H2. It exists as a diatomic molecule. All right.

[00:50:00]Predict what will be observed if metal Y is replaced with copper. H. If metal Y is replaced with copper, you should also know that copper does not react with the dilute acid.

[00:50:14]So no reaction.

[00:50:17]Okay.

[00:50:25]There'll be no reaction or we can just say there.

[00:50:43]So there would be there would be no bubbles formed.

[00:50:51]Okay. There would be no bubbles uh formed. I'm just write from here formed.

[00:51:01]Okay.

[00:51:03]Then this last question here which is two marks. With reference to air pollution, compare the use of ethanol and diesel fuse.

[00:51:14]Okay, [clears throat] this is simple. Now, ethano ethano burns to produce less carbon dioxide. So ethano ethano bs to produce less carbon. The two type of gases two type of gases that are produced here. There is carbon dioxide and the carbon mono monoxide.

[00:52:05]Okay.

[00:52:08]wireless or while when diesel Okay. When disobys, it produces it produces a lot of a lot of carbon dioxide and carbon mono or monoxide.

[00:52:52]Okay, that's it this question. Let's go to the last question.

[00:52:59]Mhm. The last question.

[00:53:09]So the last question B9 the question read the following table shows some of the fractions of hydrocarbons obtained from fractional distillation of crude oil.

[00:53:26]All right of crude oil. Nice.

[00:53:32]Um so fraction A B up to C up to D. Then these are the boiling points ranges. So this one has the lowest boiling point the range of the lowest range of boiling point.

[00:53:47]Um this one has the highest. So expect this one to have the highest chain of hydrocarbons or carbon chain. What's the first question? Which fraction contains the longest chain?

[00:54:02]Oh, give a result for the answer. Nice.

[00:54:05]So we can just say fraction D fraction D.

[00:54:14]Okay. The reason is very simple because it has it has um it has the highest highest range of boiling point.

[00:54:48]Highest range of boiling what? Boiling point because it has the highest range of boiling point 200 to 660.

[00:55:03]Okay. Next question is question um what's question this?

[00:55:11]Question B. Decane dec22 can be cracked to produce hexen C 6H124 and one other hydrocarbon T. Nice.

[00:55:31]Nice. This is okay. This is good. Now listen and listen very very carefully.

[00:55:40]cracking results in formation of an alkan and and an alken a saturated and then saturated hydrocarbons. So since this one is an hexen then what is remaining is an alken. Now how can you know that this is what is remaining?

[00:56:00]Just subtract. Look at this.

[00:56:05]Let me show you here.

[00:56:08]C10 H22.

[00:56:13]From here you come up with what? When you come up, you come up this one is cracked to form C uh uh C6 C6 H14.

[00:56:29]So you ask yourself how many how many carbon atoms are remaining you are remaining out of 10 you subtract six remaining with what four then how many hydrogen 20 - 14 uh 20 - 14 is what is it 20 or 22 is 22 okay 22 - 14 is 8 so [snorts] the formula for t Here is C C4 H8 C4 H8 that's how you calculate okay C4 H8 just subtract since it's cracking just subtract what you are given there if you want to come up with this just add these two you come up with what dec [clears throat] well what is the name of this substance this is butinine H this is butin see this is butin this is an alken now we asked to draw the display structure structure formula of T.

[00:57:44]Okay, no problem. We're going to say the first carbon to carbon will be a double bond since this is an alken. Then you have since we have of three uh three what carbon atoms so four carbon atoms.

[00:58:00]So it will be C double bond to C then C to C then single single bond is single bond. Okay. So here we're going to have this like this they display the structure hydrogen atoms. hydrogen atoms. Hydrogen atoms there like this like this. So since here this is a single bond we have hydrogen here. Don't have add hydrogen here because there's a double bond here. You need to break this double bond for you to add another to react for this to react with other elements. Okay. So this is what have I done?

[00:58:42]Um, what have I done? 1 2 3 4 5 6 7 8 9. No, no, I've made a mistake. So, remove this hydrogen.

[00:58:57]This is a common mistake that is usually made. Okay, since we have hydrogen here closing, so you don't have you don't even have to have hydrogen here. So, this is the display structure.

[00:59:11]Okay.

[00:59:15]All right. Let's proceed to question the last question of section B.

[00:59:24]The last question section B is saying, would you expect hydrocarbon T to undergo polymerization?

[00:59:31]Explain your answer.

[00:59:35]Yes.

[00:59:37]Yes.

[00:59:42]Um, absolutely.

[00:59:49]Okay. [clears throat] Absolutely.

[00:59:53]Absolutely.

[00:59:55]Um, element T is it hydro carbon T and what?

[01:00:10]can undergo polymerization because it contains its contains a double bond a double carbon to carbon coalent bond.

[01:00:38]Um double double bond. Okay, double bond. Look at this. This is question nine.

[01:00:47]Then section C. Question one.

[01:00:53]So section C you have three questions and you only you have to ask answer two questions. So these are the questions in section C. Now um this is the last question. Yeah. So, we we'll answer section C in a different video. I'm going to I'll make a different video for section C only. So, guys, if you are still together up to this point, consider subscribing to the channel and liking the video. Thank you so much. Okay.