hi everyone this is a video walk through of AQA AS chemistry June 2024 paper 1 if you do find this video useful please do give it a like And subscribe if you have any questions you can add them to the comments Below in the first question a student does a series of reactions with aqu Solutions of some pottassium halides of equal concentration each solution contains a different halide ion chloride bromide or iodide the student adds three drops of bromine water to three drops of aquid solutions of the potassium halide the student also adds three drops of water of the Broman water to three drops of water table one shows the students observations so over here the purpose of having bromine being added to water is to be a control so what that means it is to show the scientists who's doing the experiment what bromine water looks like upon no reaction so just by itself so that tells us that with Solutions P and R bromine did not successfully react with said halide therefore there is no visible change and it stays orange however Q does show visible change where where it is a brown solution now and that brown solution would indicate the presence of iodine right so we can't tell whether or not p is PID iodide and the same thing works for R because both of them stay orange in the first part it is to identify the Hal ion present in Q and to write the ionic equation now we said that that is iodine but the Hal ion is the iodide ion the ionic equation is when iodine react sorry iodide reacts with bromine to form bromide ions and iodine and then you have to add TW in front to balance that out in 1.2 to explain in terms of oxidizing ability so in terms of strength of oxidization why the observations from these reactions do not allow the student to identify the ion present in P and R so we know that we can't identify it specifically because it both stay orange now the reason for that could be because bromine is not strong enough to oxidize chlorine and it is also well not strong to Ox not strong enough to oxidize it itself that's how you get the two marks over here you could have also said that bromine has the same oxidizing ability as itself therefore you would see no visible change with that either in 1.3 the student does a second experiment to determine the Hal ion present in each P and R the student adds a few drops of AC silver nitrate solution to 2 cm cubed of potassium halight solution table two table two shows the students observations so we see a precipitate being formed with each one now you should know from P knowledge that upon addition of silver nitrate you get three precipitates depending on the halide that is silver chloride produces a white precipitate and silver bromide produces a cream precipitate and silver iodide would have produced a yellow one but that's beyond what the question's asking right so white and cream if you look at them they're pretty much the same shade of color really it's very difficult to distinguish them visually so a follow-up test would be adding a different solution that solution being ammonia so if you add dilute ammonia to the sample and it dissolves it indicates that that's silver chloride if you add concentrated ammonia that would dissolve the silver bromide so in this test over here we'd want to start off with the dilute ammonia because if you add concentrated both would dissolve and that would not be a clear indication so we start off by adding so sorry that would be add dilute specific you have to be clear with this it is dilute ammonia solution and upon adding the dilute ammonia to this it would dissolve the silver chloride precipitate and you would see no visible change with um silver bromide so no visible change with silver bromide precipitate all right that's it for the three marks question two is about elements in group two explain the explain why the third ionization energy for burum is much higher than the second ionization energy for buril so buril being in group two have two electrons in the outer shell and the third electron would be on a shell that is closer to the nucleus so because the third electron is closer to the nucleus it require it experiences a higher nuclear attraction which makes it more difficult now when you're answering this you can also incorporate like subshells in your answer so what I mean by that is the first two electrons that is being removed from the highest um shell which is the 2s shell right so the one that we're removing the third electron from is from the 1 s shell and 1 s just happens to be closer to the nucleus than the 2s is so to get three marks over here is to say that the electron is removed from the 1 s shell rather than 2 s and that is also lower in energy than 2 s or you could also say that it is closer to the nucleus or there's less shielding that would have also given the mark of here and lastly that would mean that they stronger attra ction between the nucleus and outer shell electron so the key word being outer over here in part two um magnesium reacts very slowly with cold water but rapidly with steam compare these reactions in terms of the products formed you should directly identify one similarity and one difference between these reactions so the similarity is both of them actually before we start with that let's write down the equations of cold water and wi steam so magnesium when it reacts with cold water H2O cold it produces magnesium oxide which is a solid and hydrogen gas whereas when magnesium reacts with steam instead that makes magnesium hydroxide and hydrogen gas so going back to the question then the similarity would be the formation of hydrogen gas and the difference would be what the other product is that is formed so the difference is that you make magnesium hydroxide with steam versus the magnesium oxide solid that is formed with um cold water all right that brings us to 2.3 which is the reaction of calcium with water is a reduxx reaction explain in terms of oxidation States why this reaction involves both oxidation and reduction so calcium reacting with water would produce calcium hydroxide ca2 and hydrogen gas right and then two in front of here to balance that right so calcium which is an element by itself has zero oxidation however in the compound which is calcium hydroxide it is plus two and we know that because calcium is a Group Two element right can only form a plus two charged ion and um hydrogen well in the form when it's in water it is+ one and when it's by itself it is zero so in terms of why this is a reduxx reaction it's because the calcium has lost electrons therefore is oxidized and the hydrogen has gained electrons there is set to be reduced so this is why it's redo because calcium is oxidized and the hydrogen has been reduced right and that brings us to question three which is about structure and bonding the first part is to Define electr negativity now this is the ability of an atom to attract a pair of electrons in a coent bond pair of electrons in a coent bond when when it comes to definitions I highly recommend you just sort of making a stack of flash cards and flipping through them ever so often that's the best way to um revise definitions right that is what the definition is um the next one is to explain why the C CL bond is polar now polar meaning um one element in that Bond would have a slight negative charge but whereas the other one be slight negative Char sorry positive charge so in CCL the chlorine you just have to know in just s of have to memorize this really is it's Delta negative has a slight negative charge because it is more electronegative whereas carbon is Delta positive because it's less electronegative so the the theory behind this is on the periodic table the most electronegative element is Florine and everything surrounding that is also pretty high negativity electro negativity so chlorine right next to it so that's pretty high oxygen and nitrogen are sort of second in command of there of also having very high Electro negativities and just so further away it gets from Florine the electro negativities just sort of take a hit and keep decreasing the the elements with the lowest electr negativity sit towards the bottom of groups 1 and two specifically the group one right so furthest away therefore the least electr negative now that's besides the point the point is to explain why that bond is polar so like I said earlier the reason why it is polar is due to chlorine being more electr negative thus the electro sorry the electron density is unsymmetrical y that would be the answer so let me just write that down as well the next one is question 3.3 which is although the C the carbon chlorine bond is polar C4 is a non-polar molecule so C4 would have to be a tetrahedral shape and the reason why that's not not polar is because as being in a tetrahedral shape there's some degree of what do you call that um symmetry to it sorry there's some degree of symmetry to it and that symmetry allows for the dipoles to cancel out leaving it with an overall negative charge sorry overall um no charge so it's a nonp molecule apologies so um to put that in words that is that ccl4 is um tetrahedral or you could say it is tetrahedral sorry so it's tetrahedral or you could say it is symmetrical and because it is symmetrical the dipoles cancel out so question 3.4 is there are vendor wals forces between non-polar molecules explain what causes vendor walls forces so these forces are usually caused by like the electron clouds the electron densities that sit around the um nucleus when they move so for example let's say we had um two chlorine molecules next to each other now these chlorine molecules have sort of like an electron cloud or you could just say the shells around it now at any given moment what could happen is this chlorine molecule will approach the other one and in doing so it would cause the electron cloud of this one to shift this way so that the electron density shifts towards the um chlorine atom towards the left hand side making the left hand side Delta negative and the right hand side Delta positive now because this is Delta positive this electron cloud will sort be more attracted to it again resulting in this one being Delta negative and that one being Delta positive so the the idea over here again is that there is movement of electrons and this forms a temporary dipole forms temporary dipole these induced dipoles again so these induce again they further induce dipoles in in another molecule and these dipoles across the different molecules in different molecules they attract each other and that is it for the three marks so M electrons form temporary dipoles and these dipoles sort of interact with one another or rather more technically attract one another right barium reacts with oxygen to form barium oxide which has a high melting point and an ionic lce structure similar to that of sodium chloride draw a 3D diagrams to show how particles are arranged in barium oxide lce you should draw eight particles so um it would have to look something like this where there's a plus two charge or rather let's do it as barium with a plus two charge that ion is going to sit next to an oxygen ion which again sits next to a barium ion that is next to an oxygen ion right something like this now they want eight molecules and remember this is like a lce structure so it's more like cuboidal it's very difficult for me to draw this because I'm not very good at Art but I'm just going to draw on the side again show you what I mean so this sort of thing so I hope you can see how this is more cuboidal and at each end you've got a barium and then and oxygen so when you are drawing this please do put put it down like I did earlier but like burum and it's sorry barium and it's charge and then oxygen and it's charge except for me doing it that way makes it more difficult for me to actually draw it so I'm just going to show it to you this way but if you are better at Art of destroying it please do so correctly yeah so that is the 3D diagram question four is that a student is provided with separate unlabel samples of four different solutions for analysis the four Solutions are known to be ammonium nitrate pottassium sulfate sodium carbonate and magnesium nitrate for the student not know which sample is which outline a series of test Tu reactions that the students should use to identify each of these Solutions include expected observations and ionic equation sorry ionic equations for each reaction so sodium nitrate and the sodium nitrate and the magnesium nitrate both of these require um sodium hydroxide solutions to be added one under like normal conditions the other one with slightly more warmed sodium sulfate sorry pottassium sulfate on the other hand that requires barium chloride because with berium chloride that would produce a white precipitate which barium sulfate and then the sodium carbonate well that would have to you'd have to react that with any acid so strong acid like hydrochloric acid or even uh nitric acid for one and that would F of s uh which produces a gas if you collect that gas and bubble it through lime water the line water should go cloudy which is a positive test for carbon dioxide gas right so we can start this off by saying to add sodium hydroxide just normal sodium hydroxide and this should form a white precipitate with magnesium yes so with a magnesium nitrate so forms a white precipitate with magnesium which is magnesium nitrate right and the ionic equation for this is the Magnesium reacts with the hydroxide ions two of them to form magnesium hydroxide and you should know that magnesium hydroxide is pretty un um it's insoluble because Going Down group two solubility increases magnesium sits towards the top of group two yes it sits towards the top therefore it is insoluble which is why forms a precipitate right so solid state symbol for that one now what you do then is you can also add this sodium hydroxide again to another test tube or you could sort of do it sequentially besides the point now again if you add um sodium hydroxide and you warm it and warm what that should do is it produces a gas upon reaction with the ammonium nitrate the gas that forms is going to be ammonia gas and you can test that by holding red litmus paper around like the neck of the test tube and if you hold it over there it should turn blue so uh warm that forms ammonia gas uh which can be tested with red lmus paper held at the mouth of the tube of the test tube and that red lus paper it turns blue right the ionic equation for this reaction is where nh4 plus reacts with hydroxide to produce NH3 and so that's the gas that is formed and water just H2O uh yes this is an acid base reaction the acid being nh4 the base being o minus and again you produce a base which is the ammonia so the thing of here is when you're doing this experiment you must not add the red limus paper to the solution itself because that would automatically turn blue considering you're adding well sodium hydroxide so you have to hold it at the mouth or the opening of the test tube and then the gas is what turns the r lius paper blue then for the potassium sulfate part yep that would be with barium chloride so You' add acidified barium chloride now if you're wondering why it's acidified it's usually to get rid of impurities that would be present in the sample and prevent it from interfering with what you actually testing but you don't technically need a c just can put that in Brackets so if you add barium chloride to this um it would form a white precipitate so add barium chloride and that forms white precipitate with the potassium sulfate yeah potassium sulfate but I'm putting potassium in Brackets because what it's producing the precipitate WID is actually the sulfate part okay so produce a white precipitate with um the sulfate so the ionic equation for that would be B a2+ plus4 2 minus that gives us ba A2 plus sorry that is B so4 salt which is precipitate finally that leaves us with the carbonate so for to that you could add um HCL or what's the other acid I believe that was not yeah nitric acid or any other acid for that matter so you add either Hydrochloric or nitric acid and that should F of s and you could bubble gas through lime water and then it turns cloudy indicating uh that it is yeah carbon dioxide carbon dioxide gas right that should be sufficient oh no never mind missing the ionic equation for that so that would be CO3 2us plus H+ gives you CO2 and H2O yeah we need two in over there y that should be baled now so that should be worth six marks question five is sf6 and sf3 plus have different shapes and different Bond angles deduce the shapes of sf6 and sf3 plus State the bond angles and S of 6 and S of 3 plus justify the bond angles by referring to electron pairs now s of6 that is octahedral and the reason why it's octahedral is because it has um maximum bonding electrons sorry bonding pairs so six bonds because sulfur is in group six so we can hold um can make six bonds with the Florine and with sf3 well it's making three bonds with the Florine leaving behind one lone pair so it's got like one loone pair I believe yes and then three bonding pairs so remember lone pairs repel more than bonding pairs do and the degree in which they repel the bonding pairs is by 2.5 de so for the explanation now the first thing would be that sf6 that is octahedral and octahedral have Bond angles of 90° and all and the reason why it's 90° is because all bonding pairs repel equally pairs repel equally whereas for SF 3+ now remember we said three bonding Pairs and one lone pair so that would mean that it is something like this so it's sulfur lone pair and then three Florine surrounding it type of thing and in this case the bond angle remember so we count four things around it and anything with four things around it is tetrahedral so when I say four things I mean three bonding Pairs and one Lo pair so tetrahedral and we know that tetrahedral is usually 109.5 now the lone PA over there would decrease that by another 2.5 so you could say about 107 would be the bond angle so a shape that looks like that we call um trigonal pyramidal or just pyramidal for that matter pyramidal right like we said earlier 107° and the reason being is because uh lone pairs repel more than bonded Pairs and that brings us to question six which is this question is about atomic structure and Mass spectrometry give the full electron configuration of bromine so bromine sits over here so reading that of the periodic table together is 1 s to 2 2 s to 2 2 p to 6 3 S to 2 3 p to 6 4 S to 2 3D to 10 and then 4 P to 5 so it ends in 4 P to 5 okay so just as a quick reminder that the 4S subshell fills first then the three d subshell f second now on the mark scheme you could have had it either way so you could have either put the 4S first or the 3D first it did not matter which way you put it around but just to tell you 4S does f first anyway the sample of bromine is analyzed by mass spectrometer the sample is ionized using electron impact ionization give the equation including state symbols for the process that occurs and the ionization of bromine so electron impact is when you bombard the sample with electrons that in turn kicks off one electron from the sample itself so yeah that would be bromine so bromine uh will be bombarded with electrons giving the bromine molecule a positive charge just like that so you could have left it like this you could have put the two over here to indicate uh both the electrons the one that's bombarding with and then the one that loses but if you have the two you don't have the two doesn't matter because the mark scheme accepts either one so question three is bromine exists as two isotopes 70 79 and 81 which exist in equal abundances figure one shows the incomplete Mass Spectrum of the s of bromine complete the Spectrum by adding labels to each axis and adding further Peaks you would expect to see so bromine is diatomic so in one bromine molecule you could have either both being 90 sorry 79s or you could have both bromines being 81s or you could have again um each bromine being different so 79 and 81 now it tells us that oh right both of these exist in equal abundances and they've already drawn out the peak they've already drawn out one Peak which is 158 so the 158 is a bromine molecule which has two 79s in it so basically this one this is represented by this line so the bromine which has two um 79 Isotopes if you want to call it that now we know that the 79 and 81 exist in equal abundances therefore the one with two 81s should be of a similar height so that would have to be um this one over here right so this let me just do that in a different color so the 162 is indicating a molecule that has 2 81s so that's 25% each so total of 50 that leaves us with 50% left over the only other thing it can be is a one that exists as a mixture of the two isotopes so this one over here at 50% indicating the two is one all right um part four is to State how the detector enables a relative abundances of each to be determined and the way that that is based on the current right so the abundance is proportional to the size of the current so the higher the current that means that that isotope of that fragment would be the most abundant 7.4 is that some Runners take tablets to help muscle recovery after long ex after long races these tablets contain magnesium oxide a student wants to find the percentage by mass of magnesium oxide in tablets magnesium oxide reacts with hydrochloric acid to form magnesium chloride the equation shown in an experiment student adds excess hydrochloric acid to some tablets the suan does a titration using sodium hydroxide to find how much acid is left that equation has also been provided the student follows this method so step one place a Bea on a balance and record the mass they added six tabls to the Bea then record the mass again add it to 25 cm cubed of 2 m per decim cubed of hydrochloric acid to the beaker and then s until all of the magnesium oxide has reacted the mixture is made up to 250 with distilled water in a volumetric flask 25 of this diluted mixture is put into a conical flask three drops of a suitable indicator is added and then 0.09 M perm Cub of sodium hydroxide is added from a buet until the indicator changes color steps 5 to7 is repeated until concordant results are obtained now concordant results are results that sort of vary not1 yeah so not .1 from each other um the results are shown below so the mass of six tablets is a total of 2.14 g mean tighter they've already calculated that for us that is 2038 CM Cub the reading form the balance has an uncertainty of plus orus 0.5 G calculate the percentage uncertainty using the balance in this experiment so the way you calculate percentage uncertainty there is an equation for that so that would be um the number of readings times the uncertainty divided by the final reading so sorry final reading so in this case um using the balance we've taken two readings right so we place a vehle on the balance and then recorded the mass and then added the tablet recorded the mass again so that's two readings therefore that would be 2 * 0.005 divided by the final reading which we know is 2.14 G unless they told us of another value no it would just be the 2.14 so if we do this we should get and you should get 0.46 7% as the answer now let's say for instance that in the question they just um added the sort of be of the container to the balance they teared it so they got it to zero and then added the tablet now the percentage error for to calculate the percentage error over there it would just be one times the uncertainty because you've only taken one reading and not two I hope that makes sense if you stick to reusing the equation you should be okay no matter what question they throw at you 4 7.2 it is to calculate the amount and moles of hydrochloric acid that was added to the tablet in step three give your answer to an appropriate precision so in step three over here we've been told that it's 25 CMB of hydrochloric acid and 2 MMB is the concentration so to work out the moles that is simply concentration time volume should be 2 * 25 1,000 because you want to convert that to decim cube first and in doing so that should be 0.05 moles right so that simple one Mark over there 0.05 is the answer now they said appropriate number sorry appropriate Precision so in order to do that we need to go back to the question and see how many decimal places or significant figures they had used so in the entirety of the question highest number of significant figures or decimal places rather is this one over here which is 0.0 9 0 0 so that's what we're going to leave our answer with so just adding two more zeros after that so it sort of is consistent with the question 7.3 is to use the answer from 7.2 and the information given to calculate the percentage by mass of magnesium magnesium oxide in the tablets all right so looking back at the method again it seems like the they had the six tablets react with the hydrochloric acid until all of the magnesium oxide had reacted then they made up that mixture to 250 with distilled water and then reacted part of that 250 which was 25 of it with sodium hydroxide so what this should indicate is that the hydrochloric acid would have to be in excess so all of the magnesium oxide has reacted with it but there's some moles of hydrochloric acid Left Behind which is then what you dilute and react with the sodium hydroxide so what that would mean is if we work out the moles of sodium hydroxide and we subtract it or actually the M ratio thing and then subtracted by the moles we worked out in 7.2 which was that of the hydrochloric acid it should tell us how much of it reacted with the magnesium oxide itself right so if that's a bit confusing let's start off with it first so the first thing to do would be to find the moles of sodium hydroxide and to do that that is just the concentration time volume so in the question the the concentration was9 and its volume was 2038 and that would give 0.18 3 4 2 moles now you don't have to write everything down when you're doing the question itself you could just put like um like dot dot dot to indicate that the number continues however on your calculator make sure you save the entire number because then that provides rounding errors going forward right so now that we have the moles of that the M ratio between the sodium hydroxide and the hydrochloric acid over here we can see it's a 1:1 ratio so that would mean that whatever moles of sodium hydroxide we have would be equal to the moles of the hydrochloric acid the moles of hydrochloric acid is also 0.183 Etc and again that is because of 1:1 ratio so now that we have that we could subtract this mole value from the mole value we worked out earlier and that should be an indication of how much of it had initially reacted with the magnesium oxide all right so oh so just to clarify again the 0.183 thingy that is in reference to how much hydrochloric acid was left behind after reacting it with the magnesium oxide so to find out the moles of hydrochloric acid that reacted with magnesium oxide just have to subtract the two so that would be the initial Mo value of it subtracted by whatever was left behind and reacted with the sodium hydroxide and that would be 0.031 658 now that we know the moles of hydrochloric acid that reacted with the oxide we could use the mol ratio again or the equation to find out the moles of magnesium oxide and we can see it's a 2 to one ratio so the moles of the magnesium oxide uh would be half of this so moles of mg is equal to 0.031 165 8 / 2 and that is 0.015 829 moles now this mole value represents um six tablets because that's what they had added initially so now that we know what the moles of six tablets are we have to work out the percentage by mass of magnesium oxide in all of the tablets together so what we could do now is to convert this into GRS so the mass of magnesium oxide is the moles multipli by it's Mr so for magnesium oxide that's 16 + 24.1 is you could say so 40.3 roughly would be the Mr and that gives us a mass of 0.637 G so the percentage by mass then would be 06379 ID the total mass um yeah 2 2.14 G * 100 and that results in 29.8% as the final answer in question eight um this is about silver nitrate Define the standard enthropy of formation so standard enthropy formation is the enthalpy change you could say energy change when one Mo of a substance is formed from its elements in their standard conditions or rather States would be more appropriate yeah so it's the formation of one Mo of substance from its element in their standard States or yeah standard States not standard conditions so for question 8.2 is that silon nitrate 5 is formed in silver nitrate 3 under go thermal decomposition here's the equation for it standard enthalpy of silver nitrate 3 is 123 the standard eny of formation of nitrogen monoxide is postive 90.4 we have to determine the standard enthalpy for the formation of silver nitrate 5 which happens to be one of the which happens to be the reactant of here now to do this you could conduct a h cycle however I find it a lot easier to use an equation that being Delta HF is equal to the products minus the reactants so we have Delta HF already and that Delta HF happens to be um postive 65.2 we have the one for the products I believe yes we do so we don't consider silver because it's an element by itself so in formation we leave that out we do have ag3 that's 123 and we also have the formation of nitrogen monoxide which is positive 90.4 but we are finding what the reactants are so that is the question so this is two lots of ag2 so if we sort of substitute everything into this that is positive 56.2 is equal to the product which are 123 + 90.4 minus 2x then to rearrange this that is um would be equal to 88.8 and then X by itself is that divid 2 which is- 44.4 K per Mo for question 8.3 to suggest why the enthropy change for the thermal decomposition of solid silver nitrate 3 is difficult to determine experimentally so in thermal decomposition we're actually heating up the reactant itself so what we are doing is we're applying heat to the silver nitrate 5 and then we watch that decompose into the products now in order to work out the enthalpy change of something that would be to work out the energy it rather gives off or takes in depending on the type of reaction but in this case it would be difficult to do because you're applying heat to one of the reactants now I hope you can see that if you're applying heat to it be very difficult for you to measure the energy change that is taking place right so the answer here is because the silver nitrate has to be heated therefore um cannot measure the temperature change silver nitrate VI solution reacts with solid sodium chloride the equation is shown below the student does an experiment to determine the enthropy change for this reaction the student follows this method so they measured out 50 cm cubed of .1 AC silver nitrate 5 using a clean dry measuring cylinder they pour the silver nitrate 5 solution into a glass speaker weighed out 2 G of solid sodium chloride which is in excess used a weighing bolt and tip the solid into the silver nitrate 5 solution reweigh the weighing bolt to determine the mass of the sodium chloride at it then add a lit to the beaker that has two small holes for stirring rod and for therometer stir the mixture with a plastic stirring Rod while recording the temperature with the thermometer record the maximum temperature reached identify three aspects of this method that could cause inaccurate results so looking back at the method over here we're measuring temperature change um and we're using a glass speaker now that would cause so the inaccuracy that would cause is that it would allow for like heat loss to the surroundings we want something that's more insulated such as a polystyrene cup so that would be the first one and for the second inaccuracy well according to this we're just recording the maximum temperature that is reached it makes no indication of the initial temperature and the reason why we need it is to work out the temperature change right so the second in accuracy is that we need to um calculate not calculate record the initial temperature right and then for the third in accuracy that would have to be related to okay so they've said they've stirred it so that cannot be one uh perhaps maybe okay so they've mentioned that the sodium chloride we've got two gs of that and it's in excess if you look back at the sort of concentration of silver nitrate that's pretty low so if you have a low concentration of the silver nitrate and you're reacting with an excess of sodium chloride the temperature change you would get wouldn't be as high and but we want a high temperature change CU always remember that the higher the values the higher the thing in general the lower the percentage error that's just a general rule of them so same thing when you're doing titration if there is a larger volume of one solution that you're using that results in a smaller percentage error okay so I believe that could be the third inaccuracy which is um to use a larger concentration or higher concentration of the silver nitrate right so the inaccuracy in this case is that they've losed a low concentration for the next part of the question it is oh never mind I was expecting calculation anyway um question n is to state in terms of electrons what is meant by the term oxidizing agent so oxidizing agent in terms of electrons is an electron acceptor right give the half equation to show the oxidation of copper to 2 plus ions so copper is getting oxidized 2 copper 2 plus so that means it has lost two electrons give the half equation to show the reduction of nitrate ions in acidic solution to nitrate so nitrate ions is NO3 minus and that is in acidic solution so adding hydrogens and that becomes N2 right now to balance this um we have to add in water to balance the hydrogen's on this side right and once we have done that it is just a matter of balancing charges and hydrogen because it looks like nitrogen are equal on either side oxygen we've got three over here on the product side as well it's just the hydrogen that are left over so two over here to account for the two hydrogen on the product side and we need one more electron to get this down to a neutral charge so we've got two positive charges one negative charge from the nitrate and the second negative charge from the electron so use your answer to part two and part three to determine the overall equation for the reduction of NO3 minus ions by copper so I'm just going to write both down again now if you want to combine both equations you need to cancel out electrons between both of the equations and to do that you'd have to multiply everything over here by two in order to get that two electrons and this be canceled so that becomes four two over here cancel out the electrons and just rewrite everything else so that becomes 23 minus plus 4 hydrogens produces 2 N before that plus the copper now that produces copper 2+ plus 2 N2 and two um water molecules that would be the overall reaction section B the multiple choice question section um question 10 is what is percentage atom economy for the formation of sodium nitrate and the reaction between sodium carbonate and nitric acid so for this to work out percentage percentage atom economy it is a relative formula mass of the desired product over the sort of sum of relative formula mass of all the reactants right so to do that then um the desired product is sodium nitrate so this one over here and two lots of it that's 2 * um that which is 23 for sodium nitrogen is 14 and Oxygen [Music] 16 yes Oxygen 16 nren 14 right so together all of that would be 170 divid by the total mass of all of the re um formula masses of all of the reactants which is 232 multiply this by 100 and that should give approximately 7327 is yeah so over here option D would be the answer question 11 is which involves the formation of a dative calent bond so in the first one this is only coent bonds because um phosphorus over here is in group five so can form a maximum of five bonds this was coent this one over here is ionic because it's a positive and a negative charge coming together magnesium chloride is also ionic because it's between a metal and a non-metal um this last one would have to be our answer as dative coent so in this case the third Bond pair comes from the lone pair on hydrogen which is why the overall molecule has a negative charge question 12 is the table shows results from the titration what the correct mean tighter so mean to calculate the mean tighter you're going to use values out concordant results and concordant results are values that differ by 0.1 um of each other so in that case we are looking at well not the trial tighter that's out of the question it would have to be these two 25.9 and 26 and the middle the middle way point between the two is 25.95 and that is the answer 13 is which species is never formed during a reaction of chlorine with water now chlorine reacting with water could be like under two circumstances the first one is like normally and then with sunlight so normally chlorine reacts with water to form chloric acid which is hclo and hydrochloric acid and this is an example of disproportionation because chlorine goes from zero oxidation it's an element by itself to positive one in chloric acid and negative 1 in hydrochloric acid and the second one is in the presence of sunlight so chlorine reacting with water over here would produce four hcls and 1 O2 so going back to the question that is what is never formed so we know that chloride ions are formed because um hydrochloric acid is a strong acid which dissociates completely to form the chloride ions we know that chlorate ions are also formed that is in reference to this bit over here and then hydrogen um no it doesn't look like hydrogen is being formed so that would be our answer cuz that is never formed we know oxygen is formed um in the reaction with sunlight so answer is C question 14 is which statement is correct so calcium oxide is us to remove sulfur dioxide from flu gases I don't think I've ever heard that before uh so just going to put a pin in it over there first checking out the other options calcium has a larger atomic radius than barium so looking at the periodic table calcium sits above barium therefore it has a smaller atomic radius in the question it says larger therefore that is not our answer magnesium has a lower electro negativity than barium so again magnesium is above barium therefore it has a higher electr negativity remember furthest away from Florine means a lower Electro negativities so these guys are lower so magnesium would be higher than barium therefore not the answer because that has to be higher and lastly magnesium is used to oxidize titanium chloride in the extraction of titanium that is true so is magnesium reacting with titanium for chloride and that makes magnesium chloride and titanium so in titanium chloride titanium's oxidation state is plus 4 and Titanium by itself is zero so we're not oxidizing titanium instead we are reducing it therefore this option is not correct which should mean that the first one is correct by process of elimination now if you are familiar with what they're referring to in option A if you just put that in the comments below um referring to a textbook or aate you have use would be extremely helpful question 15 is which element has the lowest melting point now to do this we could refer back to the periodic table so um out of all of our options it is more so likely going to be the group one element so we can sort of scratch out the group two elements of here because the reason why it would be the it would not be the group two elements is the group two elements they both make positive two charges and a positive two charge attracts the delocalized electrons more strongly than a one plus charge and if there's a stronger attraction this requires more energy to break resulting in a higher melting point so it can't be the group twos so between the group ones we've got sodium and uh potassium so looking back over here we can see that sodium has fewer electrons than the potassium fewer electrons meaning fewer metallic bonds if you want to call it that uh between the positive ions and the delocalized electrons which is why the answer would be sodium because as fewer electrons therefore fewer um uh metallic bonds thus lower melting point which correctly shows the general Trends in properties across period 3 so across period 3 or just across a period in general what is happening is we in the same shell so in Period 3 that would be the third shell so all of the electrons are being added to the third shell but the proton number or the atomic number is increasing and because of that the nuclear charge is a lot stronger pulling those outer shell electrons in more closely so across the period the atomic radius decreases right and the first ionization energy would increase because again the um atomic number increases so nuclear charge increases so it have to be option A for this one question 17 is that ammonia is oxidized as shown which whole number of values for WX Y and Z balance the equation so you could really start anywhere in this equation I'd like to start with elements that only appear once such as the nitrogens because it only appears once in the reactant and the product side and the hydrogen also appear once on the reactants and the product side oxygen is not not a good place to start because it appears twice in the products if that makes sense so in this case I would say let's start with a h because there happen to be more of them which would be a good idea to start with that so we've got three two over here and three over here smallest multiple between the two of that would be six so to get that six y would have to be three and W over here would have to be two so now that we've done that we can start balancing the nitrogens so we have two nitrogens on the reactant side so Z over here would also have to be two then for the oxygen we've got a total of five oxy on this side and to get this to five it would have to multiply it by 2.5 now we want whole numbers according to the question so if we multiply everything by two that would be 4 6 five and four again so four five 6 volt so option C is the answer question 18 is what is the empirical formula of an oxide of chlorine that contains 42.5% by mass of chlorine right to find empirical formula it is the percentage which would you sort of take to be the mass because it says percentage by mass so we've got chlorine we've got oxygen so for chlorine that is 42.5 dividing by its Mr 35.5 gives us a moles of 1.1 1971 and we can do the same for oxygen so 42.5 is chlorine the rest of it would be 57.5 dividing that by 16 which is the Mr of oxygen that is 3.59 375 then we divide both of these by the smallest mole number which in this case it would be that of chlorine so that's a ratio of one we divide this one two that gives us three so C3 and that is option b 19 is which of these solid sodium halides does not reduce um concentrated sulfuric acid so the theory behind this over here is that as you go down group seven the sodium halides sorry so down the group would mean that the sodium halides they increase in reducing power that's just something you have to memorize and have to learn there a bunch of equations that are in the Oxford press textbook I can link that textbook in the description later so you can check that out um yeah so if you refer to the textbook over there I'll put up a picture of what chapter this is from um there are a bunch of equations that you do need to know you need to know what is formed in the equations you need to know the other equations that are linked to it Etc so just going off of this principle though down the group means stronger reducing um agency so the answer in this case would have to be chlorine because that sits towards the top of the group therefore not as strong of a reducing agent question 20 is samples of four different substances are analyzed using time of flight mass spectrometry in each case the samples are ionized to form ions with a single positive charge the ions are accelerated to the same kinetic energy which samples give ions with the shortest time of flight so shortest time of flight would mean that that sort of molecule or sample of fragment has the lowest Mr so the lighter or lowest Mass if you want to use that so the lighter it is the faster it moves therefore it takes a shorter time to make it through the time of flight tube so looking at our options we can automatically cancel those that say electron spray ionization and the reason why we can't it can't be any of these answers is because during electron spray ionization you dissolve the sample in this um volatile solution of solvent and you um squeeze the sample out through a very fine needle and as you're doing that actually the sample tends to gain a proton and if we are gaining a proton that would therefore affect its mass right which would slow it down in the time of flight tubing so which is why B and C cannot be our answer so the answer would be between A and D so the Mr of carbon dioxide is 44 we know that the Mr of um Scandium of here is 45 the lighter one would have to be option D for question 21 that is which isotope has two more protons and three more neutrons than an atom of a cadmium 112 So cadmium 112 um we would have to increase that by five because two protons and three neutrons it should be 117 the only options that have that are between C and D so looking back the periodic table for C atomic number it is over here so cadmium is 48 so two more protons would have to be in it is 50 so answer would be C question 22 is which equation shows the process that occurs during the second ionization of magnesium the second ionization means we're making a 2 plus ion so it cannot be options a or D between B and C it is option C that is the correct answer because in order to do the second ionization the ion molecule or whatever has to already had its first ionization and that is what option C is displaying 23 is which of these practical steps will improve the accuracy of the titration so um using a 10 cm cuet instead of 25 well no because remember the larger the volumes the greater so sorry the larger the volume is the smaller the inaccuracy it's not option a um B is rinsing the size of the conical flask with water and um rinting the buet with water before filling it's actually rinting the conical flask with water that helps improve the accuracy of that and as for the um indicator drop stuff that makes no change to it so the answer is option b and then the final question is which atom has the greatest number of unpaired electrons we've got phosphorus vadium ion and copper so it's phosphorus vadium ion and copper so if we do the filling of the orbitals over here we can compare them so starting with phosphorus phosphorus would end in 2p to sorry 3 p to three so the P orbitals you you have three p orbitals they first fill singly and then doubly so for 3p that means it has three unpaired electrons the nium sits over here that ends in 3 D to 3 and we know that the D subshell has five orbitals right and 3D to 3 would mean something like this so it only has three unpaired electrons ion ends in 3d2 six so this one would have four unpaired electrons and lastly copper is over here which is 3d2 3D to 9 so that has only one unpaired electron so our answer then would be iron and that is it for the paper I hope you find 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