Full transcript of “2024 STPM CHEMISTRY SEM 2 ANSWERING TECHNIQUES”. Click a timestamp to open the video at that moment. Need another video? Use the YouTube transcript generator.
0:09 hello everyone good afternoon so uh how are you I hope that you have already taken your lunch okay and is ready to um today okay okay so uh I'm sharing this right now you all can uh see the SL or not you all can see the slide or not can you all see the slide uh I mean the the screen in here uh I mean the presentation slide
0:55 not as the not as the this one I I'm not sure whether I'm sharing the correct things or [Music] not screen screen screen okay okay okay okay okay now this is correct okay okay so uh hello everybody so hope that you all are doing well in here so it's another day where we are going to uh talk about answering technique for semester two okay so I hope that you're ready for this okay so uh without further Ado
1:42 Shall We Begin okay so uh as usual let's go to a little bit of self introductions in here okay so let's move on okay so uh as you already know I'm Mr Chu so I hope that um you have already uh know me already we have been together since semester one now we are in semester two already so little bit Origins to those who don't know me uh uh I'm Mr Chu yeah currently 43 years old now teaching ink t utar
2:27 johu uh okay I've been teaching 21 years already how time flies Okay and hopefully for more years to come right okay so let's move on to the next one um okay so uh teaching experience I've been teaching 20 years already uh I'm ex National trainer presenter uh authors of this absolute guide and then last time you have this uh LP books okay which you can find uh I I not recently see that in shoy also they have this soft copy I don't know but it's not me who's selling if Mr CH sells one it's definitely the original copy uh but how to put in words I don't
3:14 know whether it's a good thing for them to do that or not uh so if you purchase from them I don't know because they sell in a very cheap price might as well you ask from me I give you for free so that you don't have to buy from them okay I can share my M okay I I believe some of your teachers also they have so ask for your teachers okay don't buy online support original okay so um uh then let's go to the content shall we okay so today we are going to talk about the scheme of assessment first uh even though we have already go through a few times already but still I would like to highlight a few things that um I feel that you all need to know as a stpm students okay where uh in here you have
4:01 to know that uh oh okay Point okay so uh you have already SE in semester one okay you have already sit in semester one now you are in semester two already um so uh it is still the same 15 objective two structure two essay and then throughout our talks I will try my best to get you on how to uh this one how to answer in the best way because uh most of you I figure out that even my old school student they are still very weak in time management now um because stpm you must be very fast you see you
4:47 cannot afford to really waste a lot of time here and there so that is why uh it is important for you to learn the time management for uh how to answer the stpm so uh I will share my experience with all of you if I can and hopefully you'll be you'll be benefits from it okay oh okay okay uh I'm running out battery already I also don't have battery oh God is this happening okay never mind okay so uh okay never mind okay so let's continue okay okay so we got go to the next one okay so this are this is what you can do
5:32 so make sure that you did in uh read in details all all the possible uh and create your own short notes based on your understanding and check with your teachers or reference book or online sources um I I already share with you isn't it the short notes of the uh semester two and if you buy my module I also have short notes isn't it so um my short notes is definitely not 100% but it will cover at least 60 70% um of the content in here okay so I hopes that um you'll be able to create your own short notes and write your own ideas based on it okay okay so you must also do stpm paration uh so far in stpm semester 2 we have a total of 22 set from 2013 until 2023 so total is 22 set
6:22 in my school I me forc all my student to do all the 22 sets so and that is what we are discussing right now uh and um hopefully you all will be able to uh you know understand accordingly all right thank you very much e okay and then do a study group with your friends so a study group actually is the best way to study so sometimes it's very hard to study alone um and then when you share the knowledge with each other you'll be able to actually catch better isn't it okay so I hope that um you'll be able to uh understand uh how to study in group okay so even if you go to university later you still need need it all right
7:09 okay so uh shall we go on to the main component of our talks today so I hope that you have already solved the questions that I prepare for you pre-prepare for you uh before the uh our talks today uh so yeah uh without further Ado let's begin so the answer you can see from the the YouTube link down there and then I also forwarded to your school teachers I believe I think I forwarded to your school teacher and uh most of you have already uh this one so uh I hope that um you have already have the answer with you do you all have the answer already okay okay so
7:56 uh hopefully you have the answer with you already yeah okay so without further Ado let's begin our discussion of the past year uh of the questions so uh we shall begin with the section A so you have to answer all the question in this section so uh in here if you look at the if you look at the this one is 15 Mark 15 Mark 30 marks okay so if you have 15 Mark in part a 15 Mark in Part B uh then 30 Mark in part C so you should be divided equally into uh 15 Mark 15 marks in here okay I I don't know how to write in the second screen so that is why it's a little bit like handicap when I do that
8:41 okay okay never mind okay so uh yeah okay so you should be able to uh divide the time equally section a 20 minute section B 20 minute section B can even shorter actually and then section c2525 because it's essay you need more time to write okay so you must be very good in your time management and I hope that you really train for that okay all right so without further Ado let's together begin our discussion for um the questions shall we okay so let's start from question number one the following uh some chemical equation carry out under standard condition which of the following equation are exothermic
9:27 okay so let's together understand one by one what does each of these eny change represent so number one mg solid plus half O2 give mg o this is a what it can be T it can be formation and it also can be combustion of the oxygen so it is an exothermic reaction okay okay and then the second one is O minus plus e minus give O2 minus this is second electron affinity so second electron affinity is an endothermic process so it is uh endothermic and then uh number three you have mg2+ to mg Aquas so this is analy change of hydration so eny change of hydration is
10:12 exothermic and then last but not least is mg give mg2 plus gas plus O2 minus this is a reverse of LTI energy so reverse of ltis energy will make it to become Endo so in here which one is EXO number one and number three are exothermic therefore the uh answer is C okay okay so uh can can can understand any questions so in here students you have to be familiarized that is why I said that you need to uh memorize all the 10y change we have 10 eny change what are the 10y change that we are talking about number one you have formation combustion uh neutralization uh atomization bond energy
10:59 ionization energy electron affinity L energy hydration and solution so you have 10 eny change that you are you must be able to familiarize with so that um you'll be able to answer accordingly okay so hopefully uh you are when you are asked to questions of how to uh answer it so you'll be able to know this process is endo or EXO okay so that's why I said short notes is important for you to identify this okay all right so that is for question number one immediately we goes to question number two shall we so nitrogen monoxide is formed through Rapid Heating and Cooling of the gases in the atmosphere within a lightning bolt so you have half N2 plus half O2 give n o so Delta H is positive 91 kilj per Mo Center eny of atomization
11:47 of N2 and O2 are positive 473 and positive 249 Kil per Mo respectively what can we deduced from the informations in here okay so before we go into the detail let's start let let's have a look at how to calculate the bond energy or the atomization of nitrogen monoxide shall we okay so from the equation we have half and2 this is the atomization so you have half N2 give n gas so this is the standard of atomization for nitrogen and then you have half O2 give o this is a standy change of atomization for oxygen and then you have the formation as given to you earlier half N2 plus half O2 give n o so because now you want to find bond energy isn't it so the bond energy in here or atomization actually these are
12:33 two same things so it is n o give n gas plus o gas so Delta h of atomization is positive 6 631 kilj per mole so in here D is automatically wrong okay so it's here so for here um as you can see change of the uh reaction is the endothermic so that means equilibrium does not favor to the right uh you must always remember exothermic react for formation exothermic is always more stable than endothermic so endothermic means that this reaction is not stable hence nitrogen is not readily to oxidize so a is incorrect so as for atomization of nitrogen it is more noo than oxygen hence we say that no o has a stronger
13:19 calent Bond uh therefore B is also wrong and then uh this one we say that half N2 plus half O2 give n o it is also actually known as the bond energy here so for your information how does the uh nitrogen looks like so you can uh see that the structure of the nitrogen monoxide actually looks like this this is a radical okay so you have a triple bond in here actually triple bond o okay you have oxygen okay uh you have a lone pair electron and the oxygen in here so this is how the nitrogen monoxide actually looks in terms of L structure since n o has triple bond so that is why it is more stronger you require more energy to
14:06 break them so bond energy is basically Breaking the coent Bond in here okay so when you break the coent bond in here you form uh nitrogen and also oxygen isn't it okay so that is why at the same time we say that this is atomization it can also be the uh bond energy also at the same time so the best answer in here is C okay okay so K any question so far okay so no problem that is for number uh two okay okay so immediately we goes to number three shall
14:52 we okay so let's go to number three okay number three is a bond table cycle of a white solid is formed so the bowling cycle is show below so you have barium half oxygen give Pao so uh you must know by yourself that uh this is the formation this is the eny change of the formation okay uh this is the formation and then this is the atomization okay and then this is the ionization energy this is the ionization energy I'm so sorry here because my mouse uh the battery depleted already so I didn't have extra battery to replace my mouth so I have to draw using this
15:39 one I actually have this uh B but um too bad I don't know how to use in this presentation mode so that's why I'm not able to draw effectively okay if you Zuma or whatsoever maybe yes I can but in this second screen I'm not good at it okay okay so uh sorry before that I have this one okay so the uh sing wait for a while we go to that question only then we answer to your questions okay okay so uh and then from the thermochemical data what is the value of x so in here you should use the eny change of formation is equals to the summation of everything so you must know the formation is enty change of the
16:27 formation okay uh over over everything is equals to the summation of everything sorry because my I don't have a mouse to write so T energy plus everything in there okay okay so uh this is how we are going to uh this one okay so that is for the uh this one so if you press your calculator what is your value of x so your value of x is positive 751 okay so that means your answer will be C in here
17:14 okay okay so that is how you are going to answer for question number uh three in here all right so any questions that you want to ask for this okay so I hope that is clear enough because uh it's quite straightforward these questions just that you will take sometimes for you to calculate so you must be very good in your calculations okay okay so let's continue number four the following is a half cell and their standard reduction potential so the following reactant undergo Redux reaction spontaneously uh under standard condition except so uh you have to know what is a pair that you need to be used
17:59 in order to deduce the spontaneity now whenever you check inside equation here you must know that one of your species is at the left another species is at the right okay you have to know that one of the species is at the left so for example in B you have H2O2 and also mn2 plus let's say inside your C okay so when you know that your H2O2 and mn2+ so your mn2 one species must be at the right hand side of the equation and then another species must be at the left hand side so for example in C you use equation three and equation four to compare and you reverse the equation three to find out the spontaneity okay
18:45 so let's have a look one by one how you're going to solve so for a there are no suitable equations in here because if you notice carefully both uh M4 and also cl2 are at the right hand side left hand side sorry you see that M4 and cl2 are both at the left hand side so there's no reduxx reactions in here so when there is no Redux reactions so that is why we do not have a suitable equation for a okay as for B you use equation three and reverse reverse equation one okay so you you use equation three and reverse equation one so I'm going to highlight with different color so that you can see very clearly so for B I'm going to use blue color color so it is the M4 minus and also
19:32 H2O2 so if if in here you know that M4 is doing reduction so your H2O2 must be doing oxidation so the oxidation of H2O2 is the equation in here okay so that is why you choose the equation one and reverse equation two so it's positive 1.52 plus 0.68 you get positive 0.84 therefore reaction is spontaneous okay as for C as I already highlighted you use uh MN uh equation 4 and reverse equation three so you get positive 1.77 and minus a plus negative 1.52 you still get a positive 0.25 so that is why the reaction is spontaneous and finally for equation number four in here so you should have uh equation
20:21 number uh right see this one okay so you have the CL minus which is the reverse of the equation two okay with H2O2 so H2O2 must use this one okay so you have the equation uh reverse equation two and also equation four so uh inverse equation 2 equation 4 you have 1.36 plus positive 1.77 you get a positive 0.41 so that is why the reaction is spontaneous so since the reaction is spontaneous then then the the reaction is possible so the best answer in here for number four is a okay
21:08 okay can understand so you must be very good in choosing your pair of the reactants right okay so that is for number four okay are you ready for number five okay so let's goes to number five okay an electrochemical cell is made by connecting a standard hydrogen electrod to a stand uh to a half consist a half cell to a half cell consisting of a magnesium electrode and saturated solution of mg2 so when the switch is on volt meter reading show 2497 volt which of the following statement is not true okay so uh in here electron flow for magnesium to the Platinum which is
21:54 actually correct because magnesium is the anod and the uh St hydrogen electrod is the cathod okay and then number two the solubility product of the mg2 is 5.06 -12 so this one you don't zero back your calculator if you zero back your calculator then you'll not get this answer okay so first you calculate your e cell equals to e cell plus 0.059 /2 log it's actually minus not plus sorry for that it's actually minus uh I think I made a mistake in here it's minus not Plus minus not plus sorry for that okay - 0.059 /2 * x * 1.0 over 1 squ which is the stand hydrogen electron Therefore
22:40 your concentration of mg2+ which is your x equals to 1.08 * 1044 then you uh as the solubility product ng2 solid give mg2 plus plus 2 O minus therefore X give X and 2 X so therefore your concentration or your KSP of the mg2 is 1.08 104 * 2.16 104 s so your KSP is 5.06 10 -12 so this one if you straight away press your calculator now 1.08 time4 * 2 * 2.16 4 squ you get around 5.43 if I'm not mistaken which is not accurate so this one if you want to
23:26 calculate accurately don't zero back your calculator because if you zero back your calculator then you'll never get this answer okay so in here number B is also correct okay see a field draw of sodium hydroxide is added to the solution so what happen when sodium hydroxide is added so when you added sodium hydroxide to the mg2 so the equilibrium will shift to the right so the equilibrium shift to the right uh this will make the concentration of the O minus increase hence cause the equilibrium shift to the right so equilibrium shift the right so the voltmeter reading is going to increase so that is why C is also true and then D concentration of the O mg in here as you can see in here concentration of O minus is 2.16 *
24:11 10^4 so D is wrong okay so the best answer in here is donkey D okay okay can can understand okay so that is for uh how you're going to answer for number five okay it's a little challenging because you require some calculation and then some understanding also at the same time so I would I wouldn't advise students to focus on this question too much skip to other question first okay okay so if no problem then we go teacher how to get the concentration of the
25:00 how to get the concentration of mg and O that already calculated for you you use the nuns equation in here do you see that okay I already use this equation to show you already okay so you see you use the N equation use the nun equation to calculate okay so you substitute accordingly to this equation minus 0.059 / 2 log x * 1.0 / by 1.0 squ so you are able to get this okay slowly calculate your your eel is your eel is 2.47 your eel is 2.47 your e not cell is
25:46 2.38 in here I'm so sorry I didn't substitute uh accordingly uh this one is uh two 479 497 okay and then this one is uh 2.38 sorry yeah I I may be a little bit handicapped in the drawing in here because my mouse don't have B already I didn't change my batter for the
26:31 mouse I know that is it sorry for that okay uh in this case is two in this case the N equation in here is two okay n equation is two okay uh try to usually in structure essay three marks goes to substitute uh equation substitution answer three marks in here usually it's like that so I would really encourage you all to write the N equations okay okay so that is for number five okay so let's move on to number six let's not focus too much on the questions number six two electroly are connected in the series shown in the diagram so if the mass of the increase
27:16 in cathode in a is 15.6 what is the mass of the gaseous product form at the anot in the electrolytic cell okay okay so uh no one is for silver one is for silver but if you compare to standard hydrogen electrod usually it's two electron it won't be one trust me okay okay so uh how to calculate this one first you need to calculate the mole of copper so 15.6 ided by 63.5 you get 2.47 uh 2 four 5 S so since copper 2 plus plus 2 minus keep copper so more of electron is equals to 0.494 Mo and then uh you use your anot 2
28:04 h 2 H2 give 4 H+ plus O2 plus 4 eus now I'm so sorry this one sometimes the objective does not give you you know uh by by hook or by cro you have to at least memorize the half equation at anod for and cathod especially for the electrolysis of water try your best to uh remember okay or not okay um because sometimes uh objective question can be like this you know they they don't give you all the equation they expected you to know by yourself so uh if you don't know then you won't be able to solve this question okay yeah there are few years also like that you can look back at the past equation and you see that so for an the half equation
28:49 you use is the uh 2 H2O plus give 4 H+ plus O2 plus 4 e minus whereas at k so the cod in here uh the half equation is the 2 H2O I'm going to be slow a little bit in here because uh I I don't know how to use the uh B on the writing screen uh plus two minus this is the problem with the uh using the b in here because they cannot write on the second screen very well okay uh so I have to use my mouse pad and write slowly okay sorry for that for all the
29:36 inconvenience cost okay uh if anyone know about how to use this stuff can you can you teach me so that next in our next semester I'll be abl to show you better because Mr CH is very clumsy in Tech old man is like that okay so that is the half equation okay okay so some of you demanded uh teacher if uh when na is added o react with hydrogen ion and the hydrogen ion concentration uh in this case it usually goes to um I scroll back to the question okay in this number C1 usually
30:23 this question is referring to uh mg2 definitely not definitely not H+ it's definitely going for mg2 uh of course um of course it's my it's my fault also because I didn't specifically say that if any is added to the Magnesium half cell my fa here okay yes yes you are correct J okay uh it will not react with a h yes you're correct okay teacher how to get the value of 0.491 4 you mean for number six is it this one stric 2 to one time two time two *
31:09 two time two okay time two times two okay okay so uh with that your mass of the oxygen is 01229 * 2 * 16 you get 1.97 which is your answer here okay okay so with that that is how you're going to answer for uh question number six I hope that is clear enough the explanation okay okay so uh for the Cod this is the half equation try your best to remember okay this is a half equation at COD uh sometime you have to memorize
31:56 this is a half equation for not so you have to try your best to remember some okay uh okay uh teacher can I ask about how and not is to determine which equation you use in here um I'm I I I now for the selectivity of the Electro in here I will recommend you to go to my YouTube uh you search inside my YouTube channel uh I teaches uh you all how to select the uh the selectivity inside Aqua Solutions I explain that in details so you might want to go to my YouTube and learn more I'm so sorry inside here because of the time constraint I'm not able to express everything so sorry for that okay you can go to there okay
32:43 number seven diaphragm cell is used to electrolyze brain to obtain chlorine gas and useful side product which is true about the diaphragm cell okay okay so this is the typical diaphragm cell as you can see here okay so uh in the anod compartment you have the uh liquid must be in the anod must be higher than the cathode to prevent back pressure because if you have back pressure then the uh what the NaCl will be diluted so when NCL is diluted then uh your selectivity will be influenced okay so that is why uh it must be higher then number two uh in anod you use titanium whereas cathode you use steel okay so that is why your B
33:31 standless steel is used as an not is wrong and not you use titanium okay and then see the useful side product so the product of electrolysis are cl2 and the anod while H2 and NaOH is produced at the cathode about how the H2 and NaOH is produced at the cathode you can always refer to my diagram in here okay okay and then finally for the is used to the diaphrag to allow the C to pass through so in here the best answer is okay so that is how we are going to soloft for question number seven okay can any questions that you want so application T sometime can be
34:17 also quite important try your best to remember okay okay so uh can grai ED as uh electrod or not in this case no uh in this case specific specifically is for uh you must memorize anod is for titanium cathod is for steel okay okay okay number eight the following graph is obtained when a physical properties is plotted against proton number of the period tree element what is this property so this property is showing what class this property is showing first ionization energy okay in the meanwhile how should the electrical conductivity graph looks like for period Tre how should the electrical
35:04 conductivity graph looks like for period tree it shows increase then decrease like that so this is electrical conductivity okay what about melting point so this is the melting point you look at the red line not the Blue Line the blue line is for period two okay blue line is for period two red line is for period three yeah okay and then last but not least uh Atomic uh going across radius uh period so atomic radius decrease okay so these are the graph of how they are supposed to looks like for the physical trend of the period tree for this uh few things so try your best to remember again okay so that is how we are going to explain for uh number
35:52 eight 10 okay so let's go to number nine the temperature of decomposition temperature of magnesium and calcium are 4 540 and 900 respectively which statement best explain the differences of the temperature uh difference in temperature uh section c later I will explain in section c s please be patient I'll go to section c and we discuss together okay a magnesium carbonate has a stronger LTI energy than calcium carbonate is it true so these are actually all the points in here so uh magnesium L energy of magnesium carbonate yes it is in this stronger than calcium carbonate it is true this statement is true however it is irrelevant to why is the decomposition
36:38 temperature of the magnesium carbonate is lower than that of the calcium carbonate okay so irrelevant okay and then uh number B mg ion has a higher polarizing power which is correct because you have a smaller charge okay High charge small ionic radius means that you have high charge density High density will lead to high polarizing power so therefore B is correct C mg atom has a smaller atomic radius our statement is correct but we are not using atomic radius to compare we are using ionic radius so C is irrelevant to explain in this and then D mg2 plus has a smaller charge density which is wrong mg2 plus has a higher charge density so therefore the best answer in here is bo b okay very
37:24 straightforward okay so uh if you were to explain about the thermal decomposition temperature these are the basic five point that you have to use to explain about uh how are going to explain about the termal stability in here okay okay so with that that is for number nine so teacher for number six or is it d uh I think I made a mistake in here so it's uh D number six is D sorry for that it's 3.93 1. 1229 time 32 yeah yeah yeah 3.93 yes yes yes sorry answer is D donkey D yes yes yes sorry for that I highlighted wrongly the answer is donkey D okay but
38:10 I I show you the calculation already so you should be understand okay thank you very much for that Fatam thank you very much okay so between CO3 2 minus NO3 which is the most stable and which is the least stable it depends on the C it depends on the cion see on how you are comparing with if you are comparing using magnesium okay uh if magnesium the thermal decomposition temperature of the magnesium hydroxide is the highest okay whereas uh this one you can use check GPT to check if you want to know okay you can use the cheat GPT to check the temperature so in this case it seems that magnesium hydroxide should be the highest oh magnesium carbonate magnesium
38:57 carbonate should be the highest followed by magnesium hydroxide then magnesium nitrate that is a way of how you are going to handle this so uh in stpm there's only one year that ask the anion actually anion is not really inside our C okay so don't bother too much about it right okay then number 10 berum and aluminium show diagonal relationship in the periodic table so uh which of the following statement is true in regard to the diagonal relationship one ber oxide aluminum oxide are coent no they are both ionic compound yes I know they sometime they compare Anon but only one one or two times only out of 22 par question so I'm not sure why are they
39:42 doing that because it is not part of the C Revere okay so uh yeah I'm not sure why okay you just focus on more important part this small small part try to ignore it okay okay uh if it's really come out in stpm then no choice because um it is outside cerse to compare the an it is outside the cers in the in the cers you have to compare a group two cation but not the in between the anion no okay okay then back to the question two verium chloride and aluminium chloride form Dio gous space which is correct so they exist as B2 cl4 and L2 co6 in even in the gaseous phase three radium Chlor hydroxide and hydroxide are inic which
40:29 are correct so the equation do I write no I didn't write okay and then number D solt of the B h242 plus and L h263 plus undergo hydrolysis to form acidic solution not basic solution so in here two is correct three is correct so the answer in here is point B okay okay so that is for number uh 10 okay I'm so sorry to say this to you XY Z that sometime they compare the an yes they do compare the an yes they do compare I never said that they didn't but uh like I said it is out of the cus context so I'm so sorry for not able to discuss in here
41:15 okay so uh there's only 2019 or 20 that year one year one time only they ask instructure and then 2017 ask a one mark question on that okay so don't be bothered too much by it um it does ask but not very frequently okay okay number 11 which of the following true is true regarding to the inner pair effect by L so inner pair effect is a result of large energy gaps between the 6s and 6p that causes the two electron in the 6s to be nonreactive hence uh in so as a result it caused the plus two to be more stable in plus4 okay so uh in here uh a the first four ionization of energy are of Le are lower than
42:02 te actually ionization first ionization of first ionization energy of lead is slightly higher than Tina why due to in effective screening of the 4f orbital and then number B inner te effect is caused by the ineffective screening so uh which is of five and 5 P which is wrong the ineffective screening is more to 4f orbital okay and then C the plus two oxidation state is less stable than the plus4 oxidation state uh for L which is uh correct incorrect because plus two is more stable not less stable and the answer D is correct let four oxide and decompos to Le three oxide when heated strongly okay it will form okay it will form to to oide P when you heit it
42:49 strongly okay so with that that is how you're are going to answer question number 11 uh teacher is inner pair effect only show for Pb or group specifically for Pb yes specifically for Pb okay number 12 Boral Cate glass is one of the most common glass used to the uh of typical following statement is true regarding the Bor c glass so it main component is uh S2 and B which is wrong uh is actually b23 and silicon dioxide okay and which is used in the laboratory as Regent bottle uh with low coefficient Thal expansion so that means if you want to hit the chemical inside a
43:34 let's say glasswar okay so uh you will require this Boral silicate glass as the glass wear to prevent the cracking because Boral silicate glass has a relatively low coefficient of the turbo expansion okay okay so uh in here then uh the answer in here is dony is resistant to terminal attack yeah thermal shock that means it can absorb heat very well so that is why it is suitable to be used in a lot of the laboratory apparatus okay okay so any questions that you want to ask for question number 12 okay so hopefully no problem so if no problem let's goes to question number 13
44:20 shall we okay chlorine react differently with aqual sodium hydroxide under different condition which is the f is true when chlorine is bble through hot concentrated sodium hydroxide so when uh so chlorine disproportionate in hot NaOH according to the equation 3 cl2 plus 6 n give 5 n CL plus 5 N3 plus 3 H2O so the changes of oxidation state is from 0 to minus one and + five okay so a chlorine under goes reduction which is wrong because chlorine under goes this proportion one of the product can decompos in the sunlight to form uh oxygen gas uh there is no such product inside the mixture here and then C oxidation state of chlorine after reaction is plus minus one and plus five
45:07 which is actually correct okay and then one of the F Primal one of the product is mainly used as a bleaching agent so uh it is also inappropriate okay okay so uh with that that is for question number Don sorry 30 C 30 C okay so with that that is how you are going to uh this one thank you Mr go for helping me to explain okay then number 14 a fi drop of potassium tan is added to a test containing an EO solution of Iron Tree
45:53 sulfate which the following statement is true regarding to the reac so in here you have to remember that the equation goes like this f263 plus plus minus give f252 plus scn2 plus plus H2 last is a displacement reaction so uh in here the color changes from yellow to blood red nowadays they start to accept that F3 plus is Browner but still uh the general statement of the iron is still yellow the general statement of iron 3 plus is still yellow according to the stpm because you have two years stating that already so uh but in the real situation now if you look at fe2 plus is actually Brown so um but still they they insisted they want to say yellow okay so
46:40 nevertheless never mind so the BR rat solution is caused by the F2 so in here oxidation number still remain three both of them are still plus three oxidation state whereas both of these complexes has the coordination number is equals to six because they have six ative Bond so uh the best answer in here is C okay okay so with that that is how you're going to solve question number 14 okay okay so any questions in here okay so try your best to remember this equation because this equation is extremely common in the stpm uh this equation for the blood bre
47:28 this one okay okay so if no problem we goes to the last question 15 many transition metal are used as Catalyst for various chemical reaction which of the following is the reason for you to use platinum as the Catalyst for the hydrogenation of alken so Platinum has empty orbital uh for reactant to form temporary Bond bya chem aborption duee to platinum they have active sight and weaken the HH Bonner causing the C2 H4 to form C2 H5 first before forming C2 H6 okay okay so the best answer in here is Don a a because they have de orbital to form a temporary chemical B okay okay so with that that is all for the 15 objective okay let's have a look at some
48:14 of the question uh teacher can I ask Will splitting of the orbital still occur for 3D zero and 3d10 actually yes uh splitting of the D orbital is the result of the valance Ron in the uh 3D uh is repelled by the uh Lian so it is the electron electron repulsion that cause the D orbital to split okay it's just that uh when uh because D orbital is empty or D orbital is f complete fi there are no DD electronic transition occur okay okay
49:00 so that is why it is still splitting okay it is still splitting okay and then is it acceptable to use displacement we use substitution we use substitution or lean exchange also can legan exchange also can okay okay how are you going to determine the coordination number the coordination number depends on the complexes uh if there is a six native Bond then coordination number is six okay uh I've already explained this in one of my YouTube so you can go to the YouTube and have a look okay okay so uh thank you Mr go for helping us in here okay okay thank you very much okay anything else that you want to ask for this objective question before we move
49:45 into the structure okay so uh hope hopefully no problem for the uh objective because they are quite straightforwards and um hopefully uh you all can understand that I've already explained them in quite details so uh yeah okay okay so let's go to the structure so uh in the structure my advice to student is you are as advice to spend 15 to 20 minute of answering so the answer should be short and clear and then do not over focus on one question ah nowadays uh stpm is your section B also
50:31 have to write on the test P okay uh your section B also have to write on the test pad so you have to make sure that you transfer all your answer into a test pad and don't forget to com to bind together I when I go for ination sometime I see student forget to bind answer sheet together so please be careful double check before you submit your paper okay this one some some of the student they make careless mistake so please be extra careful all right okay so uh my advice in here is to spend 15 to 20 minutes for these questions H okay okay let's have a look at the structure question number 16 D sulfur D floride S2 F2 is a liquid often used as an insecticide to uh uh
51:18 fumigan in the wooden structure to control termites a write the thermochemical equation for the eny change of formation and atomizing for d sufur d fluide so the standard change for formation is 2 s solid plus F2 give sf2 liquid the question already said so state of matter is a m uh teacher can I use S8 or not now you can also use S8 so if you use S8 uh your answer will be one over4 S8 okay so both of them are accepted as per stpm 20 2016 I don't know if there is the latest uh this one for it but so far both s and S8 are accepted in stpm
52:06 okay and the standard Endy change of the atomization is S2 F2 liquid give to S gas plus 2 F gas so it will atomize to become gous atom is for a compound okay so that is why uh please be able to distinguish in between atomization for El and atomization for compound atomization of element is a standard one what do I mean by standard if chlorine you have to write half cl2 gives CL so that is the standard atomization for element but for compound when you want to atomize compound um you atomize the whole molecule to become the gaseous atom separately like this okay okay B by
52:52 using H law calculate the standy change of the formation of d sufur d fluide if the CH atomization of sulfur Florine and d sufur d fluide are positive 128 positive 109 and positive 6 78 Kil per more respectively so you must be very good in writing thermochemical equation because the question already said using H laww so you have to State using H law okay so these are the respective thermochemical equation that you can have so uh then what now what do you want to find you want to find the uh formation of the sf2 so you reverse the uh you first equation times two second equation also times two third equation
53:38 you reverse okay so at the end you get two s solid plus F2 gas give S2 F2 liquid so eny change of the reaction is -204 K per mole okay so pretty straightforward n okay any questions if you give me one minute excuse me set for a while I get my drinks
54:30 sorry yeah I go and get my drinks okay okay let's answer some of the question so uh teacher atomization uh question write short form of f atom atom enough already like like how I show to you on the screen is enough already actually to be honest we don't really very strict on that okay so if I can can I ask in cycle energy level diagram we need to put the value of positive or not uh usually we we we put usually we put okay okay uh hey I think I answer to really atom is enough already atom is enough already
55:16 okay okay so uh okay so the answer for this is uh -24 okay write the positive write the positive 12 okay just write the value enough already because the value itself will represent the eny change the value itself will represent the eny change okay okay so now uh can okay so we goes for C based on the information of B construct energy level diagram now I'm so sorry L uh maybe different teacher they have different way of teaching so so uh in here I use uh my own way of the energy level diagram so um if you uh can can
56:05 accept then you'll just accept if you cannot accept then the maybe you have your own methods to try that uh maybe other school teachers they teaches you I'm not sure about that uh but the energy level diagram that I use in my discussion is always one kind only which is I use this uh this one so all the informations that you need to relate in here is atomization of the sulfur 2 s solid give 2 s gas you also need to relate F2 give to F gas so positive 218 so you also have to relate the atomization S2 F2 liquid give 2 F gas and 2 F gas and then 2 s solid plus F2 give S2 F2 liquid so the formation is -204 so in here um teacher how do you
56:52 know which uh you construct the energy level so uh in this case uh you relate uh 256 218 and 204 you sum up this three okay this one two three so the total in here we don't don't look at the positive negative don't look at the positive negative so the total of this equation one equation two and equation four in here will still equals to like I said don't just look at the value don't look at the positive negative you get six 78 in here okay you also get 678 so in another words the S2 F2 is at one side and then uh the rest is on the another side okay
57:40 so in here this is how we are going to construct that is why Bond H Cycles mention the following things eny change what does Bond H cycle mention b h cycle mentioned that eny change of the reaction is the same no matter the reaction take one step okay or series of Step so that is what series of Step In Here means in the H law in here okay so the enty change should be the same so that is that is why uh I uses this diagram to this uh teach you and then in the y axis in here we use a enalia here you write eny in kilo and then just write the value is enough already uh
58:25 always remember remember if it is endo reactant low product High okay endothermic process reactant low product High okay so the arrow goes up in here uh so this is the energy level diagram that I usually use to teach my student maybe like I said different teacher they have different methods so I will uh it's not up to me to say whether it's accepted or not but this is the one that um I I I use for my own school okay okay so uh with that that is all for the question for number 16 zero for the element
59:12 okay okay so uh actually it's uh not zero I I don't know but for from the angle sh label zero at the AIS is actually no need no need to um zero because enthalpy if zero means that uh you don't have differences okay so uh usually we don't encourage put zero okay okay teacher how do we know we need to multiply the value it has more than one more sometime the question already multiply the eny change but sometime they don't this one uh I'm so sorry shikin that
1:00:00 uh okay in stpm usually they uh usually they will give in the what uh usually they will give in the like this question you know usually they just give you statement like this the equation you have to come out by yourself the equation you have to come out by yourself and when it is given to you like this uh it you should be following the standard the standard way so uh for example uh atomization so for atomization is uh uh s s uh s solid give S gas uh half F2 give F gas so this is all standard uh equations these are all standard equations in here so uh you
1:00:46 multiply by two like in this case uh you you you you multiply this one by two isn't it uh uh you multi ly this one by two then this one you also multiply by two uh it is in order for you to get and then this one uh you reverse it okay you reverse this one so it is in order for you you do all these changes uh it is in order for you to get to your answer you know so that is why uh hlaw allows you to do all that in order to get to your final answer what you really desire so your your desire equation is 2s solid okay 2s solid give
1:01:33 F2 okay give uh S2 F2 uh liquid in here okay so you have to this one so teacher this is Bond H cycle no no no bond H cycle is another already this is also a form of energy level diagram this is a form of energy level diagram actually Bond H cycle is also a energy level diagram okay B cycle is also an energy level diam okay okay so uh anything else but like I said um I like I said I I don't know any other teachers uh you you you you have to follow your the way that your teacher teaches you but if you were to follow my
1:02:20 lesson so far I only teach one Ty of the uh energy level diagram which which is this one so um yeah I'm so sorry so maybe different teacher they have different way so uh nevertheless I I I I have no right to say that my answer is the only way no you just follow to whatever your teacher teaches you okay so um this is the way uh how to differentiate it uh in one H cycle is usually used to relate L energy Hydra uh formation I ization energy electron affinity and also formation okay but like I said actually Bond H cycle is also a part of energy level diagram also
1:03:06 okay okay so with that that is for question number 16 okay any else question okay so uh hopefully it's clear enough uh Mr ch is there a mistake in the head cycle uh so far I don't think so no mistake no mistake okay so far there is no mistake if there is any mistake please let me know I may be wrong okay I may be wrong uh 256 is yeah yeah yeah yeah yeah I think I I put wrongly somewhere or already uh Salah okay you help me to identify it
1:03:56 six seven I think I this one already uh yeah yeah yeah you are correct you're correct you are correct you are correct my thank you very much for helping me uh I think IAL put this two already this one should be another way around uh this one should be 2s this one should be 2s sorry yeah reactant High product low you're right my dear thank you so much for helping me uh I give you the wrong uh diagram already thank you very much uh you are correct My Dear Mr CH also sometime very clumsy okay uh yeah correct not correct can you help me to correct this one please uh ladies and
1:04:43 gentlemen and then this is two sorry for that okay okay so this is the correct one sorry I put in okay yes yes yes thank you very much for helping yeah okay okay because uh 204 uh 204 is from uh 2 s two 204 is from 2 s solid give uh s 2 F2 so reactant High product low okay and
1:05:31 then this one is the atomization okay equation three in here equation three in here is the uh I use different in so that you can distinguish it okay equation three in here is F2 S2 F2 give 2 s and also 2 F so it's a S2 F2 uh endothermic so endothermic is reactant uh low product High okay cor right thank you very much child CH child okay thank you very much for helping me Mr also human also sometime
1:06:18 make mistake okay okay so that is how you're are going to sketch the uh cycle uh teacher can I can write as uh can I write as the one4 S8 instead of 2 s yes you can yes you can I think I already explain in the previous slide in here uh you can also write one over four S8 yes you can okay okay okay what else okay so thank you very much and shall we move on to question number 17 okay so uh so sir energy level di can be draw in another pattern right if the equations are correct
1:07:05 hey can can I I I to be honest I don't know I I cannot say for sure but what I can say for sure is uh when when I don't know I don't know different teacher they have different approaches but for me this is the safest energy level diagram because in energy level diagram if it is uh this this diagram clearly shows you about how does the process actually take place in here if if this process is EXO EXO means that reactant High product low okay so um and then if Endo is reactant low product High okay so um I I don't know I'm so sorry if there is any other methods already but so far this is what
1:07:52 I use to teach my students H so if they any mistake that I did so um you just follow your teachers teaching okay but this is my way okay okay uh can if H can use one cycle can can can yes you can okay okay so uh with that that is how you're are going to answer for number 16 there are many questions here okay very good okay okay number 17 Colin al2 SI I25 o4 is a hydrated alumino silicate mineral which is able to absor C in Aqua Solution on the surface according to the following
1:08:39 equation so M plus AAS plus al2 s205 O3 o give this is reversible give al2 s25 O3 oom so you can see from the equation in here this is the somehow the replacement of the hydrogen ion by the metal so it is the cite that is Ables to absorb the C this is how the cite to absorb the C in here okay so a state the oxidation number of the Silicon in the colite so it is 2 Al plus 2 SI plus 4 plus 5 plus 4 o equals to zero so aluminium is plus three si is what you want to find so 5 * -2 Plus 4 * -1 = Z
1:09:27 so SI is +4 so we actually only one oxidation state is plus4 so uh if you show this answer then it's good enough already just write plus four then enough already okay okay B State why basic Clay is not suitable for Mo most plant why is it basic clay not suitable because in alkaline condition concentration of the cion will be low why do we say if concentration of the C is low because uh the oh minus will react with the H+ okay uh to uh the if you use alkalina H+ will react with the oh minus so as a result concentration of the H+ will decrease if
1:10:15 concentration of the H+ decrease equilibrium shift to the right so that is why we say that concentration of the cation is slow because equ equilibrium shift to the right okay since the H+ decrease already okay so actually this is the modification from stpm 2003 which is uh 21 years ago questions yeah I just take from there I I I don't know because there are not many questions that ask in regards to the Zite so um I can only make use of all these very old questions to ask you all and then if you uh if you have seen my schools uh trial examinations we also have fuel cell fuel cell has never been asked in stpm so far
1:11:02 so I would also hope that you'll take notes on the FI cell a little bit um especially for the chapter two okay so uh there there are a lot of things that haven't asked in stpm especially when it comes to the near end of the stpm system cver system so you have to write Okay so so uh okay we we finish C first okay or not Zoe we finish C first then I'll go to your questions okay okay suggest use of cite with reference of the above equation either you say a c converter resine or ion converter okay so this is you can convert the C uh teacher can I set cion absorber yes
1:11:50 you can you can also State C absorber can ion exchange rine ion exchange rine is also C exchange rine C convert rine also can either one okay now this is the one that you want to ask so uh we we go for the explanation first by stating the repeating unit of the in the silicate sheets describe the common characteristic presence in here so the repeating unit of silicate sheet is s252 minus and then each silicate Share three of it oxygen atom with other cicat in here and then State a physical properties either you say electrical neutral or slippery okay so some of you ask uh teacher although here don't need
1:12:36 to draw may I ask how to draw the structure of the CATE sheets okay uh this is asked in stpm 2021 if I'm not mistaken that requires the students to draw the uh this one isn't it so uh in here you can uh draw like this you know uh I my my drawing may not be very beautiful but this is how we are going to draw a repeating unit of the Silicon so you draw s IO uh okay this is minus okay so you draw s IO okay uh this is 4 SII
1:13:32 okay is also minus uh this is four okay so uh you'll see on how uh one structure of the cicon C share with three other silicon atom so this is how you're going to draw the structure of the silicate sheets you can draw like that so from here you can see SI 2052 minus okay okay uh teacher do we need to write the state of the equation if possible try to write State noways they stress the state very
1:14:17 much already yeah it's repeating unit the same as basic unit basic unit is Si i44 minus different is different B is s i44 minus okay what else okay so that is how we are going to draw the uh Cate sheets repeating unit like how I how you wanted to know so uh this is a recommended uh structure okay okay okay so with that that is for question number 17 uh do we have to draw the Sate structure we need to label it a label as in
1:15:04 what because you're answering the question so no need to label R but uh you have to try your best to learn to draw that uh because it's negative so you want to show how SI i2i 2 minus one you want to show this formula or what okay so try s252 minus this how you share draw okay okay can okay so no problem then we goes to the essay question shall we okay so last but not least section c 30 marks so uh try your best to answer only two questions so you are advised to spend 20 to 25 minutes uh in this session uh do your best to present in an essay full
1:15:50 with content related to the questions Point form table is allowed but not advisable unless you really confident nowadays when they come to essay they really wanted you to write like essay long so I don't really recommend this one okay uh teacher can you show how to determine the repeating unit of the diagram given uh this one uh I think in my I think in my YouTube I explained that isn't it can you look back at the YouTube I think I show that before teacher if you write the correct equation but the wrong physical state we be right uh it is wrong if your physical state is wrong then it's wrong but nowadays stpm is very good in the sense
1:16:37 of a lot of equations that give you what state of matter already unless it's a very basic one okay okay so never mind we as we go through then we go through each of the questions again okay let's go to the first essay question like in this case electromotive forces for the four cell is positive 0.15 so you have PT H2 gas H+ AAS sn4 plus sn2 plus okay just now in the earlier on I see one question uh I go back to you because you are the one who asked this uh ah this one okay so sir may I ask does the cell notation or cell diagram need to include the more of decimeter Cube actually no need like this one is very standard this one is
1:17:23 very standard so no need to inut the this one question usually they will give you unless if it is a concentration cell if it is a concentration cell yes you need to write the concentration yes concentration cell then you need to write the concentration okay but usually in the cell diagram normal cell diagram you do not require to re write the concentration as long as the state of matter everything is written then should be okay already unless they really question sometime users concentration to ask yes but when you comes to write the standard one they usually do not require student to write the concentration okay okay uh never mind one draw a label diagram for the chemical cell indicate the electron flow from the diagram so
1:18:09 this is the recommended okay so uh you have to draw two half cell each of them so try your best to label everything so this is 1.0 moal H2 1 atmosphere 25° C okay and then soft Bridge Platinum if possible this is you put anot okay you label this is your anod this is your cathod okay full full sentence and then you also have to show the flow of electrons as in here so try your best to show the flow of electron must label e minus r to show that this electron flow so uh yeah this how you should be drawing the diagram complete label ah
1:18:55 complete label diagram here okay okay so uh that is for the uh cell diagram in here okay so let's go for the second one okay so calculate the eot for the sn4 plus sn2 plus equal solution so you use inot cell equals to not Cod minus not anot so 0.15 is uh try your yourself to try to uh label substitute even if the positive sign try to write the positive positive 0.15 equals to not Cod minus 0 so you get not Cod equals to positive 0.15
1:19:42 okay okay so that is for number A1 and also A2 okay okay so uh can okay so that is for a and shall we go for B already okay number three if the acid in the cell above is replaced by sulfuric acid electromotive for of the cell become positive 0.158 uh calculate the concentration of the sufc acid and pH of the solution is the hydrogen cell half cell this is actually from 2021 past year I also didn't know that but never mind let's continue so
1:20:32 uh okay uh okay sorry yeah I go back to the previous question okay Mr Chu is there any part of the drawing that must require us to draw a ruler and should be dashes by draw in a soft Bridge indicating K solution inside a w so detail I don't know I so far we don't really go to that detail I I remember that in a very old cver SL they insisted that the students should write saturated KCl yes it was very very old sers nowadays the you just label s bridge is far more than enough already okay as for the part of using lurer or not ruler is foras for tidiness so I don't think they will really stress too
1:21:17 much on must use r or not but uh I may be wrong I may be wrong but so far we don't we don't really say that must use ruler to draw very nicely no is if you draw ruler draw nicely is for the presence of the examiner to mark your answer so it's for the presence of the examiner to Mark the paper okay all right so let's go back to the question so uh try to write your overall equation so your overall equation is the uh sn4 plus if possible Right the state of matter but in this case the the the uh overall equation is uh it's not but I really recommend you all to write the uh what's uh overall equation for
1:22:06 this one sorry for the very slow handwriting okay so this is the overall equation plus the hydrogen gas Mr CH very bad in the TCH I do have the things with me but when it comes to a live streamer I don't know how to use the bot already if it is in uh other platform like I use Zuma or Zoom very convenient okay but when it comes to this one Mr CH also become handicapped already so just bu Mr CH okay Sabal okay
1:22:58 okay uhq AQ okay plus 2 H+ okay okay so this is the complete equation so from the overall equation uh you'll be able to uh come out with the nun equation where the next equation is e cell equals to EO cell minus 0.059 over n log concentration of now uh in this case you also have to express for the partial pressure of the hydrogen you must Express like this please okay you must Express like this okay okay so uh then you substitute positive 0.15 AAL to positive 0.15 - 0.052 log 1.0 time
1:23:45 concentration of H+ squ 1.0 * 1.0 so then you'll get your answer for the H+ is 0.732 Mo but that's the cube then your pH is negative 0.732 you get positive 0.135 for three significant figure because the question use positive 0.158 which is also three significant figure so your answer should also be in three and then your concentration of the h204 is the half of the concentration of the H+ because one h204 give 2 H+ okay so this is your concentration of the h204 in here uh unit don't forget to uh add to add add in your unit right okay so um that is for question
1:24:34 number uh three yeah okay okay so with that that is for 18a so we go for 18b okay so 18b sounds like this in laboratory chlorine can be prepared by reacting manganes for oxide and hydrochloric acid with the reference of the standard electrod potential value below predict the spontaneity of the reaction between manganese 4 oxide and hydrochloric acid under standard condition so your under standard condition is 25 de CSUS one atmosphere so in here I give you a we will give you a series of the half equation so you must be able to choose by yourself which one is the appropriate so your reactant in now is manganese for oxide reacting
1:25:20 with hydrochloric acid so in here you want to prepare chlorine gasma okay you want to prepare chlorine gasma so it's actually between CL minus to become cl2 so HCL will produce CL minus so your M manganes for oxide is M2 not M4 minus M4 minus is manganes uh manganate 7even manganate 7even ions manganes 7even ion so manganese for oxide is M2 so you are choose in between equation two and equation three and then you reverse this equation okay you reverse this equation okay okay so
1:26:06 uh okay so question so teacher is it advisable to give more DV yes yes yes in the halfway of calculation give more uh SF for accuracy purpose yes I will definitely recommend that yes it is only the final answer then we run off to the suitable significant figure yes you are right okay right okay so shall we go back to the question so uh now you know that uh you should use this half equation so M O2 plus 4 h++ 3 minus give mn2 plus plus 2 H2O 2 CL minus give cl2 plus 3 minus so your overall equation you must write the overall M2 plus 4 H+ plus 2 CL minus give cl2 plus mn2+ plus 2 H2O so uh this is the uh overall equation then your
1:26:53 inot cell is equal equal to positive 1.23 + 1.36 you get Negative 1.0.13 volt so since in a cell is negative reaction is nonspontaneous so supposedly uh under standard condition your reaction is nonspontaneous so uh uh so how to make this reaction happen so that is why if you still remember in your four five last time you use concentrated HCl so if you increase the concentration of the CL minus this will somehow make equilibrium shift to the right and cause your uh chlorine to be produced in here okay okay so um usually in this case uh when
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