J2026 ZIMSEC O LEVEL CHEMISTRY SECTION B

[00:00:00]All right, it's Niyaki. It's a Niyaki osteo zone. And today we are simply going to revise the June 2026 O level chemistry from the Zimse sec exam board as you can see on the screen. Right. So if you're doing Zimse, pay attention to these instructions. And today we're simply going to focus with section B. So in our previous tutorial, we did section A. If you haven't checked on our playlist, you need to do so. And always remember always remember to subscribe so that you'll be notified whenever whenever we post. Right. So we are simply going to dissect a section B step by step and we are simply going to discuss all the questions for the sake of revisions. Right? In section B we are required to answer a total of four questions but we are simply going to revise all the questions in section B.

[00:00:41]Right? So you should pay attention you should pay attention is we are simply going to discuss this paper step by step. Right? So we're starting with number six. So number six the first part says draw the dots in cross diagram to show the electronic structure of ammonia. So we're required to draw the during cross for ammonia. One it is the formula of ammonia. Right? And then this one it is what you call the ammonium ion. Right? So this one is ammonia. This one is ammonium the ion. Right? The cation. Right? So I'm simply going to draw that one of ammonia and also to explain that one of ammonium ion. Right?

[00:01:13]Because it is one of the frequently examined question where you're simply going to be examining to give the den cross diagram of ammonium ion. Also to explain the types of bonding present in ammonium ion. Are we together? So here we're simply going to have ammonia. In ammonia we're simply going to have nitrogen as the central atom bonded to three hydrogen atoms. Right? So having three hydrogen atoms, right? And then here is our nitrogen. Hydrogen hydrogen hydrogen. Then you have the dot and cross dot and cross dot and cross. And then we're simply going to be left with a one lone pair of electron. So you should highlight the key cross electron from nitrogen and then dot electron from hydrogen. Right? And we are done. And then in terms of ammonium ion we are having four hydrogen atoms and then here we're having three. This one is a neutral molecule. This one is a charged ion the cution. Right? So we're simply going to have nitrogen as the central atom again. And then here we're having this plus sign indicating that we're having one proton. Right? So here we're having one hydrogen atom which is coming to bonding without an electron. Right?

[00:02:14]So the equation we're having ammonia reacting with the proton to give the ammonium ion. So this one is the equation which is occurring. So we're having this hydrogen atom right. So the hydrogen atom contains one electron. It lost the electron. So it is coming to bonding without any electron. So it is coming only as a proton without any electron right. So here we're having this um proton having a dative bond to this lone pair of electron. So we having what you call the dative the dative bonding. So the bonding is a type of bonding whereby one molecule donates the electron and the other one only receive the electron. Right? So one is a donor and then the other is a receiver. So it is differs from coalent bonding where we're having the sharing of electrons.

[00:02:51]Here in dative bonding there's no sharing one is the receiver and then one is the don of electron. So here having this lone pair of electron and bonded electron right electrons which are not involved in bonding. So this lone pair is simply going to be donated to this proton. Right? So you have what you call the dative bonding or coordinate bonding are together. So here we're simply going to have uh so let me draw here. So we're simply going to need another proton right. So here we have another proton H and then we indicated that we're having both the electrons from from nitrogen right. So here this one is the one for ammonium ammonium ion right we having this dative bonding right so here you're having this dative bonding and then you're having this coalent bonding right and then it have this positive ch let me clear here right so let me clear them so that it becomes more clear right so here it has this positive sign so this positive sign meanly going to have also ionic ionic bonding right so here this one is positively charged so we're simply going to have this one attracted to another anion negative charge ion.

[00:03:52]First you have the ionic bonding. So in terms of ammonium chloride we're simply going to have ionic bonding simply going to have this coalent bonding and then this negative bonding. So you must know the bonding in ammonium ion inside inside out. Are we together? And then let us go back to the question. Let us go back to the to the question. Right?

[00:04:10]So here we're told that we're having a right and then you're having natural natural gas. Right? and then you're having B and then you're having A and then they react in the reaction chamber and then we're going to have ammonia. So from A simply going to nitrogen from A right so nitrogen is obtained by fractional distillation of liquefied air fractional distillation distillation of liquefied liquefied air. So a is simply going to be compressed in cooled at [music] minus 200°C. Right? So here you should know fractal distillation inside out. You should know the steps. We're simply going to have the purification is the first one. Purification. And then we're simply going to have compression whereby the purified air is simply going to be compressed to minus 200°C. [music] Right? Atus 2 200°C a is simply going to be turned into liquid. And then we're simply going to have fraction of dissolation. So in distillation the main idea of distillation we are separating mixtures based on their difference boiling point. So we're simply going to have the main constituents we're simply going to have nitrogen we're simply going to have oxygen and then we still going to have argon. So nitrogen boils at - 196°C and then oxygen at minus at minus 183°C and then argon atus 186°C.

[00:05:17]So these are the three main gases you must know by heart. You must know the temperatures at which these gases are simply going to boil. Simply going to obtain nitrogen because it is the one with the lowest boiling point and etc. Right? So you must know also the uses of these gases. So nitrogen is used in [music] the production of ammonia. So like here nitrogen is simply going to use it in the production of of ammonia and then it can also be used in food preservation. So food preser preservation and then also in the manufacturing of fertilizers. Right? And then in terms of oxygen can be used in terms of hospital right. So in medicinal supports. So we [music] can use oxygen in hospitals and also in welding and steel production. So it's welding and steel steel production. And then in terms of agony we're simply going to use argon light bulbs in welding. So argon is used in light light bulbs and in in welding again right so you must know the uses you must know the temperatures which these gases are simply going to to boil. Right? So you must know the fractional dissolation of liquefied liquefied A by R A is simply going to be purified first comprised to minus 200°C turned into liquid and then simply going to car out the distillation process whereby we're simply going to separate mixtures based on their different boiling point. The boiling points are these ones right and then we must also know the uses of these of these gases.

[00:06:27]Are we together? So this is how we simply going to obtain nitrogen from air. Right? And then we now go to the production of of hydrogen. And then to obtain hydrogen from the natural gas, we're simply going to use what you call the steam reforming reaction. Right? So we have what you call the steam reforming reaction of natural of natural gas. So we simply going to give this natural gas methane mixed with steam.

[00:06:48]Right? It's simply going to react with steam to give carbon monoxide and nitrogen. So we are simply going to have the reaction occurring at a temperature of about 700°C to 120°C under the nickel catalyst. Right? So this one is the equation for the reaction. So it's methane plus team. Right? So we must indicate that it is steam to give carbon monoxide plus hydrogen gas. Then to balance [music] we need to have a three here and then the equation is now is now balanced. Right? So you should know the conditions of the steam reforming the steam reforming reaction we're simply going to have the catalyst and then the temperature from 700 up to 1,200°. So together so here from this reaction we need also to proceed to because carbon monoxide is very toxic right it reacts with hemoglobin to form caroxymoglobin which reduce the oxygen carrying capacity of blood. Right? So you should know in terms of biology. So we're simply going to have what you call the water gas shift reaction. So it is the water water gas shift reaction whereby we're simply going to react this carbon monoxide plus water to give carbon dioxide and also we are going to have hydrogen gas. Right? So this reaction reduces the concentration of carbon monoxide which is a harmful gas. Right?

[00:07:53]So it's simply going to produce this carbon dioxide and then to balance this reaction this reaction is already is already balanced. Right? So this produced hydrogen is then further purified or for or you see. So in the uses of hydrogen we're simply going to have the manufacturer of ammonia is indicated in this in this reaction in the hydrogen fuel cells hydrogenation of vegetable oils and also petroleum refining. Right? So you should know these processes the steam reforming and also the fractional dissolation of liquefied air. Right? So here we're having nitrogen and then here we're having hydrogen. Right? So let me clear this so that you can now concentrate with with the equation. Right? So here we are having at the reaction chamber where we having nitrogen reacting with hydrogen and then we're simply going to have this reversible reaction and then we having this [music] ammonia and then the reaction is exothermic reaction indicated by this end. So you should know this one from your understanding of industrial processes. Right? So this reversible sign indicates that the reaction is proceeding in either in either direction. Right? So you're simply going to have a two here and then to balance again with a three here. Then these are all in the gaseous the gaseous phase. Right? And we are we are done with the equation. You should know this equation by by heart, right? And then in terms of the conditions, we're simply going to have a temperature of 150°C 200 atm of pressure. And then powdered is powdered ion is the catalyst. So why is the ion state to be powdered, right? So powdered ion simply going to have iron being powdered to increase to increase it surface area. Right? So we did this in reaction kinetics where we said powdered powdered particles got a larger surface area compared to these solid clustered particles. Are you together?

[00:09:25]So the year we're simply going to have increment in the what you call the conduct points. So these are the contact points right? So here in this solid particle we're simply going to have this peripheral surfaces that's where we only having the contact points right. But here all the ion particles are simply going to be exposed in this powdered ion. So we're still going to have the increment in the contact in the contact [music] points. So you must know this one in terms of reaction kinetics. Why we are going to use powdered ion instead of this solid ion. Are we together? And then we now go to the questions [music] right. So we now go to the to the questions. So the question says identify substance A. So A we say is nitrogen, right? And then B we say is is hydrogen, right? And then state two conditions. So these are the conditions, right? And then number B says a chemistry student prepared 500 cm of 1 mo decium dromate.

[00:10:14]So this one is ammonium dromate. So you should know this one is our ammonium ion. Then this [music] one is the dromate and the dromate ion 2 minus. So you should know that we're simply going to have the formula. This one is the formula of ammonium droate. And then the next part says calculate the mass of ammonium droate. So we're given the volume and then the concentration. So we can easily calculate the the number of moles using this equation. N is equal to CV. [music] Right? So here we need to convert this one into cubic dec. So 1 m equ= to 1,00 c. This simply mean to say we're going to have 0.5 dec. Right? So it's n is equal to c v and then our n is unknown 1 is equal to c here v 0.5 then we're going to have 0.5 moles right so these ones are the number of moles right so we are required to give the mass right so we have the number of moles then we want the mass so we going again to this equation n is equal to mass over m so we want [music] to have this this mass so we're simply going to have to rearrange this equation to make m the subject of the formula so it's n by m is equal to mass right so we have this one then the M we're simply going to add the atomic masses of these atoms in the ammonium dromate formula right so we simply going to add these atoms in the in the formula right so a is atomic mass defined as the mass of an atom measured relative to one mass of carbon 12 is together so here for nitrogen we having nitrogen is 14 that one of hydrogen 1 chromium 52 and then oxygen we're having oxygen is 16 right so how many oxygen atoms do we have we are having seven so it's by seven and then how many chromium is two having by two and then how many nitrogen having two by two and then hydrogen by 8 [music] having eight right so we're simply going to add this uh this area so it's 16 by 7 or 16 by 7 then we're going to have 112 and then 52 by 2 we're going to have 1 4 then here we're having 8 and then here we're having 28 so [music] it's this one plus 112 + 28 + 8 and Then we're going to have 2 52 right so this one is the the MR right so we want to have this formula so it's n* m equal to mass we have the number of moles n which is equal to 0.5 so this one by 0.5 going to have the the mass so the mass [music] is equal to 126 here is 126 g right and we are and we are done and then we now moving on to [clears throat] the next part which is number number C so number C [music] says fig 6.2 Two shows the label on a container of chemical waste. Right? So we're having the warning the hazardous chemicals. So this one contains ion 3 chloride. So this one ion 3 chloride and then concentrated hydrochloric hydrochloric acid. Right? So explain why the chemical waste is said to be. So we going to have this one hydrochloric acid. So hydrochloric acid is highly corrosive to living tissues and surfaces and also it contains this ion 3 chloride which is a heavy metal and then it is toxic nonbiodegradable right so simply going to have this heavy metal so heavy metal which is toxic and non biodegradable right biiogradable and then simply going to have this hydrogen chloride which is highly highly corrosive so it's [music] highly corrosive to living to living tissues and services All right. So this is why this one is said to be zadas. And then we go to the next part. So the next part says describe the most suitable methods [music] of disposing the chemical the chemical waste. Right? So we're simply going to use the neutralization and the precipitation method. So in terms of the neutralization, we're simply going to neutralize this hydrochloric acid by reacting with the base. So we can [music] react with sodium sodium hydroxide, right? So we're simply going to react this one with a base before disposal, right? Before disposal, right?

[00:14:03]So we're simply going to ne utilize this one before disposal. And then before disposing this iron 3 chloride, we are simply going to use the precipitation the precipitation method. We simply going to precipitate the iron 3 out. So to precipitate this out, we're simply going to react with sodium sodium hydroxide again. Right? So sodium hydroxide will give a rusty precipitate or a reddish brown [music] precipitate.

[00:14:23]So we can have it as a rusty PPT precipitate or a red red brown precipitate with sodium hydroxide or we can also use ammonia ammonia solution.

[00:14:32]So in both cases we're simply going to have the same observation the rust precipitate which is insoluble in excess right and then if we are to use ion 3 right so in this question we're having ion 3 right so if we having ion 2 we are simply going to obtain a green precipitate so ion 2 produces a green precipitate with sodium hydroxide in ammonia solution so we simply going to precipitate these heavy metals using these suitable precipitating reagents are we together right so let us proceed to the next [music] part let us proceed to the the next part, right? So the next part says suggest the environmental effects of improper disposal of of chemicals. Right? [music] So we're simply going to have acidification of our soils, destroying vegetation and also destroying ecosystems. [music] Right? So we can have acidification.

[00:15:16]So it's acidification of of soils and also destruction of the eco of the ecosystems and then also destruction of the ve vegetation. Are we together? And then we can also have contamination of water bodies. Contamination contamination of water water bodies. And then we can also have these heavy metals leeching into the water bodies thereby increasing toxicity and also killing the aquatic environment. And then we can also have pollution of ground water by these metals pollution of of groundwater by this by [music] these metals right and we are and we are done. And then we have go to the next part which is number seven. So number seven says a mass of 0.27 27 g of aluminium we reacted with excess ion 2 sulfate right. So the chemical equation for the reaction is shown. So here we're having this equation right and then name the type of reaction right. So this one we're having aluminium reacting with this ion to sulfate and then we're having aluminium sulfate and then this ion right. So here this is what you call the displacement the displacement reaction. So displacement reaction is whereby a more reactive element this aluminium displaces a lesser reactive element this ion from its sulfate. So we having this one as a displacement reaction which is an example of a redux reaction. So a redux reaction is defined as a reaction whereby we're going to have reduction and oxidation at the same time occurring simultaneously. Right? [music] So here we're having this aluminium being oxidized. Here the oxygen state is zero.

[00:16:41]Then here is plus three. And then this ion being reduced here is plus2. [music] Here is zero. So ion has been reduced.

[00:16:46]Aluminium has been oxidized. So that's why it's said to be in redox [music] reaction. So the correct answer for this one it was displacement reaction which is an example of redux [music] reaction.

[00:16:55]You can then further elaborate are you together and then calculate the number of moles of aluminium that reacted. So given the mass of aluminium that reacted. So it's n [music] is equal to mass over m. So you want the number of moles we having the mass 0.27 over the ar the of aluminium from the periodic table we having 27. Right? So here is [music] 27. So we're simply going to the calculator. Uh, so it's 0.

[00:17:18]So it's uh 0.27 / 27 0.27 / 27 and then we're going to have 0.01.

[00:17:29]So these are the number of moles of aluminium that reacted. And then the next part says calculate the mass of aluminium sulfate that was formed. So here we are having two moles producing one mole. So the number of moles that reacted we're having 0.01. So two moles simply going to produce one mole of aluminium aluminium sulfate. Right? two [music] we produce one and then how many number of moles of aluminium sulfate are simply going to be produced when 01 moles. So we're going to have less. So this is 0.01 [music] will give us less.

[00:17:57]Right? So here we're having 0.01 / 2 * 1 and then we're simply going by 2 and then we're simply going to have 0.5 [music] moles of aluminium sulfate which were produced. So we want the mass of aluminium aluminium sulfate. So we simply going to have this equation again, right? We're simply going to use N is equal to mass over MR. So we want this this [music] mass, right? So we're simply going to have N* M is equal to to the mass. So we have the the number of moles being equal to 0.005 and then the MR we're simply going to add the within this [music] this formula. Right? So here we're simply going to have aluminium having the of aluminium having 27. How many aluminium atoms having two and then for sulfur we're having 32. For sulfur main sulfur atoms we're having three. Right? this [music] three here and then so the formula of aluminium sulfate is 2 S O4 3 right so this one is the clear formula [music] right and then here we're having three and then in terms of oxygen we're having 12 so that one of oxygen is 16 by by 12 right so here we're simply going to have 16 by 12 and then we're going to have 1 92 and then we're going to have 32 by 3 and then we're going to have 96 and then 27 by 2 we're going to have 54 so here is 54 24 + 96 [music] + 1 92 and then we're going to have uh 3 42. Right? So here we're going to multiply this one by 3 42 to give us the mass. Right? So this one by 0.5 and we're going to have the mass being equal to 1.71 g. So this one is the mass which was simply going to be to be produced. Are we together? And then we now move on to the next part. [music] So let me clear here so that we can easily go to the to the next part. Right? So here we're simply going to have number number B. So number B says the flowchart shows some of the stages involved in the processing of of diamonds. Right? So we are simply going to have the processing of of diamonds. So we're having the raw diamond. We're having sorting, cutting, polishing and then we're having our processed diamond here. So the first part says describe what happens during the sorting the sorting process. Right?

[00:20:08]So in terms of the sorting we are simply going to have the raw diamonds being accepted and being categorized based on their quality, size, color and clarity.

[00:20:16]Right? So we're simply going to have the row the raw diamonds are being inserted right and then further categorized categorized based on what? Based on their quality, based on their size, based on their color and then also based on their clarity. Right? So this one is the sorting the sorting method the sorting process. Right? And then we go on to the cutting the cutting process.

[00:20:40]So here on cutting we're simply going to have the diamond being splitted along the strategic structural line to give a shape of the geometrical forms. Right?

[00:20:48]So, we're simply going to have the diamond being splitted right along the strategic the strategic structural structural lines to uh to shape to shape it into the desired geometric forms, right? Desired geometric forms, right? So, these ones uh this one is the cutting the cutting process. And then we go on to the polishing. We're simply going to have the faces of the diamond.

[00:21:19]Uh the diamond being smoothened being smoothened out and then refined using the polishing wheel. using the polishing the polishing wheel to bring out the brilliance to bring out the maximum brilliance and shine from the diamond and shine from from the diamond. Right? So this one sorting the cutting and then the polishing stage and we are and we are done and then we now go on to the next part which says suggest any two benefits of processed diamond. Right? So the benefits of processed diamond, we're simply going to have a significant increase in market and commercial trade value. So we're simply going to have the increase in in value, right? And then also we simply going to enhance the industrial usability of the diamond in terms of durability pressures and also in cutting two tips, right? So we're simply going to have [music] enhance enhanced industrial visibility, [music] right? So these are the benefits of processed diamonds, right? So here we increase the value the market value in the commercial trade value and then here we also increase it high durability and then we go on to the next part where we given bronze and brass is two alloys of copper. So an alloy is defined as a as a chemical substance formed when a metal is mixed with either a metal or a nonmetal to give a certain compound with the desired properties. So we're starting with a metal. It always contains a metal mixing it with either a metal or a [music] nonmetal. So it's either metal plus either a metal or a nonmetal to give a product in desired properties. Are we together? So here we are going to have bronze and in brass.

[00:23:06]So bronze is brocoin and then brass is bra right. So broco bro for bro for bronze co for cobba tin for tin. So bronze contains copper and tin and then brose brass contains copper and zinc. So [clears throat] you should know this one by heart, right? So state the composition of bronze and brass. We've fully highlighted this one. And then we go on to the next part which says [music] state in two advantages of using alloys.

[00:23:32]So the alloys are harder, right? So the alloys are hard. So let us say here we're having this one is the latice structure.

[00:23:39]The latice structure of cobraite is this one. Right? We're having regular arrangement in the latice structure of copperite. So if we are to insert to introduce either a metal or nonmetal to give the alloy we're simply going to introduce of different sizes right so here we're introducing these ones are larger right so here we have distorted the regularity of this lat structure right so this one is no longer regular [music] right so this the layers are no longer able to slide past over each other so this gives the alloy its hardness the alloy is simply going to be hard and and strong so these are the main advantages of of alloys and also are corrosion resistance corrosion resistant Are we together? So these are the advantages of of ions of alloys. Are we together? And then I we go on to the next part which is number number eight. So number eight says a fig 8.1 shows the flowchart of how ion is produced in the blast furnace. Right?

[00:24:28]How ion is going to be produced in the blast [music] in the blast furnace.

[00:24:31]Right? So we going to have x and then ion and then coke. Right? And then here having wasted gases slion and then here we're having water in.

[00:24:40]Right? for having slag meaning say this one is obviously going to be a calcium carbonate which acts as a flask right so this one it acts as a flask it removes impurities in [music] the plastics right so here we're simply going to have calcium carbonate thermally decomposing right to give calcium oxide plus carbon dioxide so this one it is what you call the quick the quick l so the quick lime is simply going to react with impurities and the main impur we are simply going to have is silicon dioxide right so it's simply going to have calcium oxide but silicon dioxide side [music] to give calcium silicate which is called sl together and then um we are done on X and then here we're having wasted gases.

[00:25:19]So in terms of the wasted gases we can have carbon dioxide we can have carbon monoxide and then we can also have nitrogen nitrogen nitrogen gas [music] right nitrogen gas. So these are the wasted gases and then here obviously we're going to have water and then molten ion is simply going to be collected [music] right. So here coke is the reducing agent in the blast you need to know on that one reducing agent and then you should also know the definition of a reducing agent. [music] So a reducing agent is the one which becomes oxidized after the end of the reaction.

[00:25:48]So it takes oxygen from other molecules.

[00:25:50]It gives other atoms other molecules electrons. It donates hydrogen there by itself is simply going to be oxidized at the end of the reaction. It donates hydrogen. It donates electrons. It takes away oxygen from other other molecules.

[00:26:04]That's a reducing a reducing agent together. You should know the definitions [music] of these reducing agents and then the definition of this oxidizing agent. Right? So let's go back to the question. Right? So the question says identify the substance X. [music] So we said X is calcium carbonate. And then the use we say this one is a flask.

[00:26:21]It decomposes to give calcium oxide which reacts with silicon dioxide to give a slake which is calcium [music] silicate right there by removing the impurities. Then one gas present the wasted gases. We have already highlighted highlighted that one. We have carbon dioxide and then carbon monoxide and then we can also have nitrogen gas right. And then number four says part four says ion 3 oxide is reduced to ion according to this equation. So we're having this equation.

[00:26:48]So uh this equation uh is not balanced properly right we supposed to have a two here this one is a typo we are supposed to have a two [music] here not a three right to balance this equation right so number of iron we're having two here we're having two and then in terms of oxygen we're having 3 + 3 [music] to give us six and then having six and then three and then the equation is not balanced so we need to have a two here not a three right [music] and then the question says calculate the mass of ion produced from 3.2 and 2 g of ion. Right?

[00:27:16]So from the from the stochometrical ratios, one mo of ion 3 oxide gives two moles of iron, right? So one will give us two moles of of ion. So we're given the mass of the iron 3 oxide that reacted, right? So here we're keeping saying ion 3 oxide. Why are we not saying ion? Right? So in the previous question we had aluminium. We had aluminium sulfate. Right? We didn't say aluminium aluminium 3 sulfate. But in terms of ion we are saying that [music] ion 3 sulfate. We're highlighting to the oxygen oxygen states. Why are we highlighting to the oxygen state?

[00:27:46]Because it is a transition and transition elements. So transition elements they've got a they've got variable oxygen states. So we have ion 3, we have ion 2. So whenever we are dealing with transition states, always try to highlight which oxygen state are you referring to. So when you're dealing with copper, it's copper 1, copper 2.

[00:28:02]You're dealing with amanganesees all the vanadium all the transition element always remember to highlight the oxidation state. But when you are dealing with the S block and the P block elements, you only need to you don't need to highlight the oxygen state all together because they doesn't have variable oxidation state. Right? So here we are given the mass of the ion 3. You want to have the number of moles that are reacted. [music] So here we having N being equal to mass over MR. Right? So here we have the mass being equal to 3.2. And then the MR of this ion 3 we simply going to have ion the ion is 56.

[00:28:34]How many? We have two. And that one of 16. How many we have? we have three. So we simply going to add so here it's 56 by 2 and then we're going to have 1 2 and then this one is 48 [music] going to add 48 and then we're having 160 right and we're still going to calculate the number of moles.

[00:28:51]So it's 3.2 divided divided by 160 and then we're going to have 0 0.02 moles of iron 3 oxide. So in stometry you should know these are two chief equations in stoometry. This one N is equal to mass over MR and then N is equal to CV. You must know these two [music] equations by heart in terms of stoometry and the mole concept. Right? So here we have the number of moles of iron 3 oxide that reacted. Right? So here from this stoometrical relationship we know that one mo is simply going to give us two moles. Right? So here one mole of iron 3 is simply going to give us two moles of iron. So what about 0.02 moles? We're simply going to have to have less.

[00:29:30]Right? So we're simply going to have 0.02 over 1* 2 to give us 0.0 04 moles, right? So these ones are the number of moles of ion which we're simply going to be produced. And then the question requires us to calculate the mass. We required to have the mass. And then here we're having the number of moles which were produced. So in order for us to calculate the mass, we are going back to this equation again. We want to have the mass and then we given the number of moles and the mass right? So in terms of ion is an atom. So we having a r atomic atomic mass. Right? So [music] here is we are simply going to make m the subject of the formula. So it's ni a r is equal to to mass. Right? And then here we are having the number of moles being equal to 0.04* which is equal to 56 is equal to the mass we simply going to have 2.224 g. So this one is the mass of iron which we are simply going to to obtain are together and then we now move on to the next the next part. So let me clear here so that you can easily move to the next part. Right. So the next part says table 8.1 shows the physical properties of some of the organic organic compounds.

[00:30:33]Right? So having the organic compound the boiling point and then solubility in water. So we're having XN octane and then ethanol and then the caroxilic acid. Right? So these two are soluble in water. These ones are not soluble in water. Right? So these ones are non-polar. They are nonpolar. Right?

[00:30:48]Then these ones they are polar. Right?

[00:30:49]Polar. So in all level chemistry you don't want [music] to discuss on the formation of the hydrogen bonding and etc. You want to discuss simple stuff.

[00:30:56]Right? so that you can easily extract uh the information from this video. Right?

[00:31:00]So here we're having the boiling point.

[00:31:01]So this one it tends to have a lower boiling point compared to this one.

[00:31:04]Right? So here as you go down the organic chain in terms of the alken as you go down the as we move from one carbon atom going down we simply going to have this increment in the strength of the forces holding the molecules.

[00:31:16]Thereby the forces are simply going to be strong and then we require a lot of energy to separate the molecules.

[00:31:20]Thereby we simply going to have high melting and high boiling point. Right?

[00:31:23]So the strength of the forces is what governs the boiling and the melting point. So if we're having stronger forces, we're simply going to have high melting and high boiling point. That's what you need to know whenever describing the difference in boiling point. The strength of the of the forces in the molecules is what determines the at the boiling point in the melting in the melting point. So this simply to say here having we forces compared to this one because they in the same family because of this suffix, right? So this one contains eight eight carbon atoms.

[00:31:52]this one6 [music] right and then we go to ethanol and caroxilic acid right so these two um they've got what you call the hydrogen bonding right you have this O group so for you guys you don't need to know on the hydrogen bonding just know that here we're having stronger forces and then here we're having weaker forces this one is soluble so in terms of solubility um you can know this phrase this one [music] is good for you it's like dissolves dissolves like right so like dissolves like so if we to like forces between the molecules like forces. So like forces between the molecules we're simply going to have the molecule dissolving right. So here in ethano we're having like forces in ethanol with the forces in water. They are like forces. So like dissolves like right? So water is a polar solvent. So here we say this one is a polar one.

[00:32:40]Right? So like dissolves like so going to have this one dissolving into water.

[00:32:44]Right? So you should know this one in terms of all level chemistry. That's what you only need to know. You don't need to know the interaction of hydrogen bonding and etc. You only need to know like dissolves like. So if we're having similar intermolecular forces between the substance in the solvent, we're still going to have the substance being soluble like dissolves like are together. So we go now to the questions, right? So let us go to the questions.

[00:33:07][music] Suggest a reason the most suitable procedure of separating exen octane and ethanol. Right? So these three are missible. Right? So these three are missible liquids. Right? So we're simply going to use fractional fractional dissolation to separate these three based on their difference boiling boiling [music] points. Right? We're simply going to use these three based on their difference boiling point. So given the boiling points so this one evaporate first and then it is collected as the distate followed by this one and lastly we're going to have this one as the residue. Right? And then we're simply going to the next one which says caroxilic acid and so the imissible liquids right. Yeahible emissible liquids. Right? So we're simply going to have two separate >> [music] >> layers or the one of Xin and the other one of the caroxilic acid. So we're simply going to use a separating funnel to separate this one. [music] So we are going to have a separating a separating funnel to separate these two imissible imissible liquids are together and then we go to the next one. So the next question says the cells shown in table two table 8.2 we connected to the bulb.

[00:34:11]Right? So we having the cell and then the electro being used. So A having magnesium and copper and then B we're having iron and copper and then C [music] we're having zinc and copper.

[00:34:20]Right? So here the first part says identify with the reason the cell that will give the brightest [music] light.

[00:34:25]So in order for us to have the brightest light we need to have a larger different in the voltage. So here we need to have the activity series. Let's say we're having a here and then we're having Z here. So if the distance between these two atoms is larger we're simply going to have the brightest voltage. We're simply going to have the larger the voltage which is simply going to be produced. Soon we have the production of the brightest light. [music] So the distance between these two in the reality series is what carvin the outcome voltage. So here we're having this one here is the largest distance followed by by this one and then lastly we have this one. So here is number one number two [music] and then this one number three all together and then we now move on to the next part. So the next part says suggest a suitable electrolyte that can be [music] used for simply going to make use of dilute sulfuric acid. So it's dilute not concentrated. You need to take note on that one. Dilute sulfuric acid not concentrated sulfuric acid. So why not concentrated sulfic acid? Concentrated sulfuric acid is not an acid. It acts as an oxidizing oxidizing agent. You need to understand that one. Confiric acid acts as an oxidizing agent not acid.

[00:35:23]Right? Together. So you should also remember why copper doesn't react with dilute acids. So copper [music] it is below hydrogen in the reactivity series.

[00:35:31]We're having hydrogen more reactive than copper. So between sulfuric acid this one plus copper we simply going to have copper being a lesser reactive metal so it cannot displace hydrogen from this compound to give the salt but if we to use concentrated sulfic acid it acts as an oizing agent we're simply going to have the reaction okay so I want someone in the comment section to give us the products of the reaction whereby coba is simply going to react with concentrated sulfuric acid not dilute acid so cobba doesn't react with dilute acids are all together and then we now move on to the next part which is number number nine.

[00:36:04]So number nine says the most common waste disposal method in schools are incineration, land fuse and recycling and then I state one advantage and one disadvantage of each of these waste disposal disposal method. So we are going to give the advantage and the disadvantage of each one. Right? So we are starting with incineration. So in terms of the advantage we're simply going to give it. Let me add another page so that we can create a table.

[00:36:27]Right? So let me add another page.

[00:36:30]Right? So let me add another page here.

[00:36:32]So we're simply going to have incineration. So we're starting with incineration and then we go on to the advantages and then we go on to the disadvantages. Are you together? So in terms of the advantages of incineration but incineration is highly effective at reducing total waste to volume and it also destroys harmful pathogens. Right?

[00:36:52]So it's h it is highly effective at reducing at reducing total total waste volumes right and also destroys harmful harmful pathogens. [music] This one is the advantage and then which comes with the disadvantages it releases toxic air pollutants and also greenhouse gases into into the atmosphere. So it's going to have the production of toxic gases into the atmosphere. So toxic gases of sulfur the nitrogen so greenhouse the greenhouse gases we can have carbon dioxide and then we can have methane.

[00:37:28]These are the greenhouse [music] gases right and then we go on to the next one.

[00:37:31]So the next one we having the first one we having incineration. So the next one is land fuel and then recycling right.

[00:37:37]So in terms of len land filling in terms of land filling the advantages it is simple to implement. So it is simple to implement. So here in terms of disadvantages it pollutes the underground uh the ground water because of the formation of leech right. So it's simply going to have lieet which contaminate the ground the ground ground water and then also it requires the use of valuable land. It requires the use of valuable valuable land. Are we together?

[00:38:06]And then we go to the last one which is re recycling. Right? So lastly we're simply going to have recycling. Right?

[00:38:12]So the advantages of recycling it conserves the the raw materials. Right?

[00:38:17]It [music] conserves the the raw materials and also minimize the overall manufacturing energy usage. Right? And then we go to the disadvantages. It can be it can be economically costly to sort to clean and to reprocess the materials safely. Right? It can be economically economically costly to sort to clean and also to re reprocess materials safely. Right? So these ones are the disadvantages and the advantages of these three of these three right so you must know these ones by by art right and then let us go to the next one to the next question. So the next question says a symbol a symbol of water was purified at a school laboratory name the methods that can be used to purify water symbol.

[00:39:05]So we can use filtration we can use distillation again right so these are the methods which we can use to purify to purify water and then explain how each of the methods in B improves the pury so purification so filtration we're simply going to remove the insoluble particles suspended solids by trapping them over the porous filter medium right so we're simply going to have this one is the final and then we have the porous filter medium and then we're simply going to pour and then the residue are simply going to be suspended here and then we collect the water here. So we're simply going to use it to separate insoluble insoluble materials. Right?

[00:39:38][music] And then in terms of dissolation, we're simply going to remove dissolved mineral source. We can use remove microorganisms. You can also remove nonvolatile impurities by using their different boiling points. Right?

[00:39:50]So it's simply going to use different boiling points in in distillation to remove mineral mineral source microorganisms and also the nonvolatile the nonvolatile impurities. Are you are you together?

[00:40:05]And then we go on to the next uh the next question which says the exhaust films contains many pollutants. Name two pollutants that are there present. We're simply going to have carbon monoxide and then the oxides of of nitrogen. Right.

[00:40:17]and then state any two effects of the named pollutants to the environment.

[00:40:20][music] Right? So carbon monoxide affects the living organisms by combining with respiration to form what you call the caroxy caroxymoglobin. Right? So it forms the caroxy hemoglobin which reduces the carrying capacity of oxygen of the blood. Right? So caroxymoglobin it is a very stable compound. It bonds with the hemoglobin. Thereby we are not having thereby in the blood we are no longer having any free hemoglobin to to carry the oxygen to the to the cells right then we go on to the effects of the nitrogen oxides. So nitrogen oxides we are still going to have this one forming acidic rain which corros the infrastructure and then in the water bodies we're still going to have a death of aquatic ecosystem. Are we together?

[00:41:01]And then we go to the next the next part says suggest any one way of minimizing air pollution by exhaust gas. So we're simply going to make use of what you call the catalytic catalytic converters.

[00:41:11]Right? So catalytic converters we are simply going to oxidize to oxidize this produce carbon dioxide carbon monoxide into carbon carbon dioxide and then these unbent hydrocarbons to give us carbon dioxide plus water and then these o oxides of nitrogen to be reduced to give nitrogen gas. So these are the reactions in the catalytic converter and then in terms of the catalyst you're simply going to have platinum have palatium and then you're simply going to have rodium right and rodium right so these are the the catalysts present in the calic converters you must know this one by you must know the equations you must know the catalyst presence platinum platium and rodium right and then platinum rodium is also used in the word process where we simply going to the production of nitric acid you must know this one from your understanding of the industrial processes right and then now we Now go to the next uh the next question. So let me clear here so that you can have the next question more more clear. Right? So number 10 says draw the structural formula of propin. So prop is a prefix for three carbon atoms and then E N E. It indicates that we're having carbon to carbon double double bond the alken functional group. Right? So here we're having three. So the general formula is C N H2 N. So here we're having N being equal to three. So C3 H6.

[00:42:27]Right? So here we're simply going to have this one carbon to carbon double bond and then here [music] so you need to know that carbon is the maximum forms the maximum number of bonds four is the maximum number of of bond right and we are and we are done. So here is 1 2 3 4 1 2 3 4 1 2 3 4 and we are and we are done [music] and then name the type of reaction under propene when it reacts with hydro hydrogen. So this one is what we call addition addition reaction. So we're going to have hydrogen being added across this double bond. So in addition reaction we're simply going to add [music] A plus B to form one single product which is equal to to C. We're having one single product. One single product.

[00:43:05]Yeah. So we're having this one propane is the is the single product. Right. So we simply going to convert this alken into an alkan. Here we going to have propane. So this one is our propane.

[00:43:15]Right? Having hydrogen hydrogen hydrogen hydrogen hydrogen hydrogen hydrogen.

[00:43:20]Right? And we are done. And then the next part says describe how bromine water can be used [music] to distinguish between propane and propane. So bromine water simply going to have bromination of propane and propin of propane and propin. Right? So the brmination we're simply going to car out this distinguishing reaction in the darkness.

[00:43:40]You need to highlight that you're having this reaction in the darkness. Why?

[00:43:43]Because in the presence of uim both the alken and the alken are simply going to decolorize broine water. But in the presence in the darkness in the absence of light we are only having the alken declarizing this broine word. So in the presence of UV in terms of this um the propane we simply going to have what you call the free radical substitution reaction. Right? So this one we simply going to discuss this one later when we dealing with A4 chemistry. But for you guys you only need to highlight that we are using darkness as the condition. It is the only way to distinguish [music] between these two because in the presence of UV they both decolorize bromine water. But in the darkness only the alken they [music] decorize broine broine water. Are we together? Together together together. And then we now go on to the next part which is number number B. Right. So number B says the empirical formula of the organic compound R [music] is this one. So empirical formula is the simplest one number ratio of atoms present in a compound. So we're given the organ the empirical formula of R which is this one. And then they use the molecular formula. So molecular formula shows the actual number of atoms present. So this one it is a ratio the one number ratio of atoms. This one shows the actual actual number of atoms present. Right? So in other words, empirical formula the relationship between empirical formula and molecular formula. Molecular formula is the multiple of empirical formula right is equal to xip by the empirical empirical formula [music] right. So so we can then further deduce that the m is equal to the molecular mass is equal to x by the empirical empirical mass. So here we can simply deduce the empirical mass. So we have this one the empirical formula c h2 right. So the empirical mass having 12 having two and then 16 right. So we can have this one = 30. So the empirical mass is equal to 30. So we [music] said m= x by the empirical mass. So m= 60 is equal to [music] x by the empirical mass which is equal to 30. So you want to have the value of of x. So here x= 30 / 30. So x= 2 together. So here we have the value of x. So we're still going back to this equation. So we say molecular formula is equal to x by the empirical formula. So the empirical formula we are given is this one C H2O * X which is equal to 2. So here we still going to have 2 by this one C2 two by this one H4 two by this one O2. So this one is the molecular molecular formula together part right. So the next part so let me clear here. So the next part says um draw the structural formula. So the structural formula we're simply going to have. So here C2 right? So going to have ethaninoic acid. So it's ethaninoic acid. So this one is our ethaninoic ethaninoic acid as you can see right. So having 1 2 3 4 hydrogen atoms 1 2 o atom two carbon atoms. So this one is the formula and then state the homologous series to this one belongs to this one belongs to the caroxilic caroxilic group right caroxilic acids right. So we are simply going to have this caroxilic group right. So ma series is defined as the family of organic compound with the same functional group. So we having this one as the functional group same general formula same similar chemical properties and similar [music] physical properties.

[00:46:49]So this one is the definition of the orgas family. And then the definition of the functional group this is an atom or group of atoms or arrangement of bonds.

[00:46:59]Arrangement of bonds that determines the chemical nature of an organic of an organic compound together. [music] And we are and we are done. So we are now moving on to the next part. We are now moving on to the next the next part which is number number C. So fig 10.1 shows one of the uses of some of the organic compounds. So we are given this one. So we given this a fet acid reacting with this sodium hydroxide in the presence of sodium chloride in it.

[00:47:23]And then we're having the solid Y. Then we're having glycerol. Right? So um this one glycerol is not inconsistent with the equation. Right? So glycerol so this equation they wanted to for us to have sonification whereby we're still going to have the alkaline hydrolysis of triglycellides which contain the esta linkage right so we're still going to have the presence of this estester linkage so we can deduce so that we can have our glycerol which is an alcohol together so here we're still going to have this one so this one was the compound which we were supposed to have glycerol right so here we're having this linkage then we go again to have another esester the linkage and then the [music] other group and then we also go again to have the estester linkage and the other group so that we can easily have we can easily have our glycer right so here we can easily join this [music] then we're having H and then H here then here we're having H then this one we're having H and then we're having H so this one is the formula of the triglycerides so that we can hydraulize this [music] estage and then we are simply going to have um we are simply We're going to have the alkalco which is glycerol and also the caroxilic caroxilic group which can be then further reacted with this alkaline medium to give the sodium salt which is sodium steroid are [music] together which is sodium the sodium steroid right which is salt. So, so this one is the subonification reaction all together.

[00:48:51]So, here if you have glycerol, it is not consistent with the equation because here we're not having any linkage to neutralize. We're only having this caroxilic group. So, we can't neutralize you can't idolize the caroxilic group for us to have this alcohol. Right? So, so here this diagram we're going to have a typo in the diagram. Right? So, here the first part says name the solid name the solid Y. So, here the solid Y it is soap. Right? Or we [music] can say the sodium salt because here having sodium hydroxide and sodium chloride, right? So we can say the sodium the sodium salt.

[00:49:21]Are we together? And then name the process. So this one is subonification.

[00:49:25]They wanted us to have subonification like we have alluded to before. So the next part says describe a simple chemical test for the solid Y. So why we said it is the soap, right? So we're simply going to dissolve the soap in water and then we're going to shake till we have a stable form and then we [music] add dilute acid to form a white precipitate of the of the of the aid acid. Right? So we're simply going to add [music] we're going to add water and then we we shake till we have the stable till we form the stable form right and then we're simply going to add dilute dilute acid so acid is represented by this H+ the protons right so here this is the acid so when we to add the acid we're simply going to form a white precipitate of the the fatty acid right so of the fatty of the fatty acid right so we're going to add water till we form the stable form and then we add the acid acidic medium till we form the Y precipitate and then we go on to the next uh the next part. Let me clear here that we can [music] have the next part more clear. I just need two methods of adding value to to Y. So two methods of adding value to to the soap. Why is the soap right? Two methods of adding value to the soap. We're simply going to um add perfumes. So we're simply going to add perfumes to the soap. We're simply going to color the soap. So coloring and then we're simply going to add antiseptics. Anti septics and moisturize. Right? Moisturize. Right.

[00:50:42]And then we're simply going to add attractive packaging. Right? So we are going also to have attractive pack packaging. Are we together? So these are the ways in which we're simply going to add value to the sol. Right? So you should take note of this the error in this question. We are not having any estester linkage here to for us to have this glycerol. Right? So for us to have the glycerol we need to have the estester linkage so that we can hydrayze the esester in the presence of this sodium hydroxide. So we are having alkaline hydrolysis to form the sodium salt which is the soap and then the glycerol which is the the alcohol.

[00:51:11]Right? So you should take note on this uh the area right and we are done. So always remember always remember to subscribe so that you'll be notified whenever whenever we post right. All right. So this one is our Nyaki online tutotoring as you can see on the screen.

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