BedokFunland JC's A Level H2 Chemistry Qns (Part 2)

last SG · 2010-07-26T23:40:35+00:00

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UltimaOnline SG

What is the distinction between equilibrium shifts (to a side ) and lies (to a side) ??

Is it that for *shift* ,

Eqm alr shifted?

While *lies* to a side is the eqm is more inclined towards that side?

When to use which term ?? ):

If you didn't change conditions, and Qc < Kc, then you say position of equilibrium LIES to the RHS. If you didn't change conditions, and Qc > Kc, then you say position of equilibrium LIES to the LHS.

If at first Qc = Kc, ie. system is at equilibrium, but then you changed experimental conditions, then you say position of equilibrium SHIFTED to either RHS or LHS.

If you increased or deceased the moles or molarities or partial pressures of reactants or products, then it's Qc which changes, not Kc.

If you or increased or decreased temperature, then it's Kc which changes, not Qc.

If Qc now < Kc, then you say position of equilibrium has SHIFTED to the RHS.
If Qc now > Kc, then you say position of equilibrium has SHIFTED to the LHS.

All these your own JC never teach you meh? That proves that you (yo JC students reading this) should come for my BedokFunland JC tuition. *hmmph*

UltimaOnline SG

On dissolving silver halide precipitates when adding NH3(aq)

Eqn 1 - Ag+(aq) + Cl-(aq) --> AgCl(s)
Eqn 2 - Ag+(aq) + 2 NH3(aq) ---> [Ag(NH3)2]+(aq)

When you add excess NH3(aq), position of equilibrium for Eqn2 shifts to RHS, hence position of equilibrium for Eqn1 shifts to LHS, hence AgCl(s) halide precipitate dissolves.

With Br-, you need concentrated NH3(aq) because the Ksp for AgBr is much smaller, ie. less soluble. With I-, the Ksp for AgI is so damn small, that no amount of concentrated NH3(aq) or even NH3(l) can dissolve the AgI(s) halide precipitate.

As for why the solubility of AgX decreases down the group, read Jim Clarke's related article on solubility of Group II hydroxides vs carbonates / sulfates :
http://www.chemguide.co.uk/inorganic/group2/problems.html

UltimaOnline SG

2014 P3 Q4

I still dont get it. Why only coo- group of aspartic acid can interact with the amino acid di) residue given by the qn

Val and tyr aren't polypeptide chains, their discrete amino acids

Aren't the groups on val and tyr considered R groups?

Aspartic acid's R group is a COO- group. Valine's R group is an alkyl group. Tyrosine's R group is a phenol group. The question isn't asking for interactions between discrete amino acids. It's asking for interactions between R groups of these amino acid residues within a polypeptide chain. Don't lose your marks carelessly like this, it'll cost you at least 1 grade (with steep bell-curves).

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Update

Is my understanding correct, and can help improve my explanation?

Sian, think question interpretation might also be a problem

Is it because :in this qn, the amino acids are part of a polypeptide chain, e.g. val-tyr-asp- etc etc

Then, both val and tyr's nh2 and cooh is used up in amide bond linkage

Only aspartic acid has the extra cooh group, hence can interact with the qn?

But then wouldnt it be Hbond interactions?

I thought of another :

Is it because ph 7, so means its in water? ( thats why the acid and basic group will deprotonate and protonate??)

Thats why will have ionic interactions instead of h bonding?

But if its in water, then hydrolysis would occur, then wouldnt the nh2 and cooh group on val and tyr able to interact? ?

Help!

1st of all, hydrolysis only occurs partially, it's an acid-base *equilibrium* remember?

2ndly, technically (Singapore JCs don't teach this well so it's not surprising JC students don't really understand this), while COOH and NH2 groups are only capable of (both donating and accepting) H bonds, COO- and NH3+ groups are actually capable of participating in both H bonds (COO- can only accept H bonds, while NH3+ can only donate H bonds) as well as ionic bonds, depending on who the other party is.

And yes, Cambridge expects you to automatically first deprotonate COOH groups at pH7, and protonate NH2 groups at pH7.

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If ph7 in water, will it cause the polypeptide chain to undergo hydrolysis and cause it to break into small portions of amino acid residues?

Oi!!! Wah biangz eh, if at pH7 and room temperature, proteins spontaneously "break into small portions of amino acid residues", then you won't be able to remain in 1 piece typing this post with your phone liao (you won't even be able to hold your phone), you'll disintegrate instantly into a sludge. Sludge man!

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Can tyrosine's phenol group interact with nh3+?

If cannot, is it because phenol doesn't ionise in water?

Hydrogen bonding is possible, between the partial positive H atom of NH3+ (the H bond donor), and the partial negative O atom of phenol (the H bond acceptor).

Cambridge already stated in the question, regarding extent of deprotonation of phenol at pH 7. Read the question carefully lah!

UltimaOnline SG

You need to either heat under reflux or use biological enzymes, plus at strongly acidic or strongly basic pH lah. That's how your strongly acidic stomach digests the meat you eat, with proteases enzymes.

Acidic COOH groups and basic NH2 groups do ionize via hydrolysis to generate COO- groups and NH3+ groups.

Different from hydrolysis of the protein's peptide bond within the amide group in the polypeptide chain of a protein (the word "peptide" refers to *either* the peptide bond within the amide group in a polypeptide chain of a protein, *or* the amino acid residues in a polypeptide chain of a protein). Because the amide group within a polypeptide chain bonds two amino acid residues or 'peptides' together, the amide group in proteins is called the peptide group, and the C-N bond within the amide group is called the peptide bond, linking two 'peptides' or amino acid residues together.

Hydrolysis means "chemical reaction with water" (where chemical reactions involve breaking or forming of sigma bonds, while resonance involves breaking or forming of pi bonds only), while hydration means "physical interaction with water" (physical, inorganic & organic chemistry) or "addition of water" (in some organic chemistry contexts, eg. hydration of alkenes to alcohols).

UltimaOnline SG

Pcl5 reacts with acid to form acyl chloride

Reacts with alcohol to from halogenoalkane

^both via nucleophilic substitution

Correct.

PCl5 (and for that matter, SOCl2 as well) doesn't react with phenol to generate aryl halides though, for 2 reasons.

1stly, the unlike aliphatic alcohols, the lone pair on the phenolic sp2 O atom is delocalized by resonance to form a pi bond with the sp2 C atom of the benzene ring in some resonance contributors), exactly as is the case with aryl and vinyl halides, and thus the resonance hybrid's phenolic O atom is electron-deficient (has a positive formal charge on some contributors, and hence a partial positive charge in the resonance hybrid), and thus is non-nucleophilic and cannot attack the electrophilic P atom of PCl5 (or the electrophilic S atom of SOCl2).

2ndly (even for the small % of phenols which happen to possess sufficient activation energy to attack PCl5 or SOCl2), the electron-rich Cl- nucleophile (eliminated from PCl5 or SOCl2 after reacting with ROH) is electrostatically repelled by the pi electrons of the electron-rich benzene ring nucleophile, and thus PCl5 and SOCl2 cannot complete the mechanism to convert phenol to aryl halides (which therefore does not happen).

If descriptions of the elementary steps are provided in the question, Cambridge could ask you (eg. in this year's A level exam) to draw the (partial or full) mechanism for PX5 or PX3 or SOX2 reacting with alcohols and carboxylic acids to generate alkyl halides and acyl halides respectively. So for those of you interested (and ready ; if any of these confuses you, then you're not ready and you should skip this entirely), here are the mechanisms

(The use of SOCl2 is often favoured over PCl5, because the SOCl2 reaction has an important advantage over the PCl5 reaction, can you guess what it is? Cambridge could ask this question sooner or later.

Ans : The SOCl2 reaction generates 2 gaseous products, while PC5 reaction only generates 1 gaseous product. Hence, the Gibbs free energy is expected to be more negative (ie. thermodynamically feasible) for the SOCl2 reaction compared to the PCl5 reaction, due to thermodynamically favourable positive entropy change.

Furthermore, since SO2 is a gaseous product, which readily leaves the solvated reaction mixture, hence the position of equilibrium is pulled over (or if you prefer, shifts more) to the RHS, as predicted by Le Chatelier's principle.)

Mechanism for SOCl2 and PBr3 reacting with alcohols to generate alkyl chloride and alkyl bromide :
https://www.khanacademy.org/science/organic-chemistry/alcohols-ethers-epoxides-sulfides/reactions-alcohols-tutorial/v/preparation-of-alkyl-halides-from-alcohols

Mechanism for SOCl2 reacting with carboxylic acids to generate acyl chloride :
https://www.khanacademy.org/science/organic-chemistry/carboxylic-acids-derivatives/reactions-carboxylic-jay/v/preparation-of-acyl-acid-chlorides

Mechanism for PCl5 reacting with carboxylic acids to generate acyl chloride :
https://upload.wikimedia.org/wikipedia/commons/7/71/Phosphorus_pentachloride_mechanism.png

Note that the phosphoryl chloride generated can be further hydrolyzed, this is covered in the H2 syllabus under the inorganic chemistry topic : periodicity. In fact, the mechanism for both water H2O (inorganic chemistry) and alcohol ROH (organic chemistry) reacting with PCl5, is exactly the same (up to a point), because did you notice that water (H-O-H) is a 'type' of alcohol (R-O-H) in which the R group is a H atom, and thus it should make natural and logical sense to you (did your school teach this to you?) that the reaction mechanism (up to a point) is exactly the same : HOH with PCl5 gives POCl3 and 2HCl, while ROH with PCl5 gives POCl3 and RCl + HCl.

You can try drawing the mechanism for further hydrolysis of POCl3 to give H3PO4 + 3HCl (notice that the OS of all elements remain unchanged during hydrolysis reactions, so based on the starting OS, you can deduce the final OS which will help you deduce the structure or formula of the final hydrolysis product). The mechanism involves 3 water molecule nucleophiles (one by one) attacking the electrophilic P atom of POCl3, thereby eliminating the 3 Cl- ion leaving groups one by one, followed by proton transfers to generate the final products of H3PO4 and 3 HCl. Cambridge could ask you to draw out this fairly simple mechanism (of which the balanced equation you have to memorize under the Periodicity topic anyway, so might as well you draw out the mechanism to make sense of it for yourself instead of blindly memorizing the equation like most JC students do... which is just sad. Learning should always be fun and meaningful.)

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Btw, the following is something extra, a bonus for those of you (eg. Flying Grenade) who may be interested in studying Chemistry at the Uni level. Did you know that other than SN1 (racemization) and SN2 (inversion of configuration), there's a 3rd mechanism called SNi (retention of configuration) whose secret existence, bewildered chemists figured out, when trying to figure out why some reactions like SOCl2 with alcohols mysteriously resulted in alkyl chlorides with retention of configuration instead of the expected racemization or inversion of configuration (ie. meaning it was neither SN1 nor SN2). If hearing this excites you, then yup, you should certainly consider studying Chemistry at Uni level. Enjoy! :)

SOCl2 and the SNi Mechanism
http://www.masterorganicchemistry.com/2014/02/10/socl2-and-the-sni-mechanism/

UltimaOnline SG

There is Shielding effect provided by electron filled orbitals (x/y/z orbital) experienced by the valence e- , albeit a smaller/weaker effect than a filled principal e- quantum shell is it?

Idk tat why i ask

Help!!!

"Is shieding effect provided from orbitals insignificant, only shielding effect provided by principal quantum electron shell significant?"

1st error : shielding effect is from electrons, not orbitals or shells.
2nd error : electron shells consist of orbitals, so you can't say shielding from shell more powerful than shielding from orbitals.
3rd error : there is no such thing as a "principal quantum shell". (Did your school teach you that? *sheads head*). There is only such a thing as "principal quantum number" and "electron shell".

UltimaOnline SG

Cant find both cie and moe-cambridge mark schemes online thought they're restricted/confidential

U have? U are godly

Yes I am, but on a separate note, you can download the UK CIE papers from XtremePapers.com. The Singapore A level Mark Schemes are highly confidential and illegal to access.

UltimaOnline SG

When describing trends of ionization energy and electronegativity across a period, Just keep it clear and simple and write, "Valence shell electrons are shielded from the positively charged nucleus most significantly from *inner shell* electrons; hence shielding effect remains approximately constant across a period." to get 1 mark when explaining the trend of ionization energy and electronegativity across a period.

The 2nd mark is stating "The number of protons in the nucleus increases from left to right across a Period, hence valence shell electrons are more strongly electrostatically attracted to the positively charged nucleus, and atomic radius decreases." and conclude with "Therefore, ionization energy becomes increasingly endothermic from left to right across a Period because of the increasing magnitude of energy required to remove an electron across a Period."

When asked to explain electronegativity trend across a period, the 2nd marking point is slightly modified to "The number of protons in the nucleus increases from left to right across a Period, hence valence shell *bonding* electrons are more strongly electrostatically attracted to the positively charged nucleus, and atomic radius decreases." and conclude with "Therefore, electronegativity increases from left to right across a Period because of the increasing effective nuclear charge.".

When describing and explaining ionization energy or electronegativity trends down a group, the common error taught by Singapore JCs, and for which Cambridge penalizes thousands of Singapore JCs students every year, is to say "effective nuclear charge decreases down a group". The moment you write that, you get zero marks for the question.

Instead, write "Ionization energy decreases down a Group despite increasing nuclear charge, for 2 reasons : due to the increasing atomic radius as a result of increasing number of electron shells, (not "principal quantum shells"!!!) hence as predicted by Coulomb's law, less energy is required to remove the increasingly further-from-nucleus valence shell electrons going down the Group; in addition, the valence shell electrons are also increasingly shielded from the positively charged nucleus by an increasing number of inner shell electrons going down the Group, thereby further reducing the magnitude of energy required to remove the valence shell electron to carry out ionization."

Of course, you don't have to phrase it exactly the way I did. Unlike Singapore JC teachers who anally insist you phrase your answer exactly the way they phrased it in their own lecture notes or else they'll penalize you in your Promo/MidYear/Prelim exams (and therefore you can expect JC teachers to also heavily penalize other JC teachers, both within a JC and across to other JCs, so few JC teachers will be able to get an A grade for Prelim papers when marked by other teachers from other JCs), Cambridge is *reasonable* (not lenient, but reasonable, there's a critical difference). So as long as you have the required Mark Scheme points, phrased in your own way (with some degree of creative freedom), you'll get the marks.

UltimaOnline SG

Whats the diff bet side product and by product?

Go to my website by googling "BedokFunland JC", then look under my list of BedokFunland JC's A Level H2 Chemistry Questions. It's one of my questions, answers are provided therein.

UltimaOnline SG

25 cm3 of NaHCO3 and NaOH titrated against 0.5M HCl. 21.5cm3 and 15cm3 of HCl required to reach 1st and 2nd equivalence points respectively. Find the molarities of NaHCO3 and NaOH.

Solution :

Molarity of NaOH = moles of HCl which neutralized NaOH / 25cm3 volume = (0.5(21.5/1000)) / (25/1000) = 4.3x10^-1 M

Molarity of NaHCO3 = moles of HCl which neutralized NaHCO3 / 25cm3 volume = (0.5(15/1000)) / (25/1000) = 3x10^-1 M

UltimaOnline SG

Whats the difference between L (levo) and D (dextro) , And sinister and dexter isomers???? And also R and S isomers??

To answer your question, read :
https://en.wikipedia.org/wiki/Chirality_(chemistry)#Naming_conventions

For H2 Chemistry, if asked to name the enantiomers, Cambridge will accept either (+)/(-) or d/l labelling based on optical activity, or the R/S labelling based on the Cahn–Ingold–Prelog priority rules (which are also used for E/Z labelling system for alkenes, which is more versatile than the limited cis/trans system, and is also accepted by Cambridge for H2 Chemistry) though the R/S and E/Z labels are not compulsory for H2 Chemistry as it's not within syllabus, and only required for H3 Chemistry. The D/L labelling is used in biochemistry (H2/H3 Biology).

UltimaOnline SG

Hi UltimaOnline,

For oxytocin, there is no C terminus, because the C terminus (belonging to glycine) has reacted (in a nucleophilic acyl substitution, addition-elimination, condensation reaction) with NH3 to become the amide group you see at the bottom-left corner of oxytocin. You *must* write it the correct way around to get the 1 mark.

https://en.wikipedia.org/wiki/Oxytocin#Structure

UltimaOnline SG

2010 p3 qn 1d
Sir ultima can help check my drawings

For cpd A, the cl is substituted at that position is it because it is most stable there because attached to 3 mildly electron donating C group ah? Markovnikov’s rule ah?

For D, im asking this as an extension
When the Br is being eliminated, are there 2 other by-product as drawn?

https://www.dropbox.com/s/lo7wynbasjbmo7k/20151…

Eh don't anyhow misuse his name. Markovnikov (Chinese for "horse cough you also cough") would turn in his grave if he heard you say that. Markovnikov's rule is only for hydrohalogenation via electrophilic addition across a double bond (if you know why it usually works to predict the major product, you'll also know when it'll fail, for challenging tricky qns). This is halogenation via free radical substitution, nothing to do with Markovnikov (he'll say 关我屁事!). Because free radical substitution cannot be controlled, you'll get a mixture of many monohalogenated products. To determine which monohalogenated intermediate was used for this qn, observe the structures of the final product upon dehydrohalogenation (ie. this qn isn't asking you to state which is the major product, but rather identify the intermediate that could give you the final products drawn). But assuming you're jus curious if that's the major product of halogenation via free radical substitution, it *is* the major product, based solely upon the factor of stability of alkyl radical intermediate (in actuality, you need to mathematically combine both factors at a specified temperature to determine the actual major product). Then yes, your reasoning (though you wrongly labeled your explanation using Markovnikov's name) is correct : tertiary alkyl radical is more stable than secondary alkyl radical is more stable than primary alkyl radical. This factor is more pronounced for Br, less for Cl, and even less so for F (I is irrelevant here, because iodination via free radical mechanism cannot be carried out; you know why? asking you to explain why, was a Singapore TYS A level exam qn).

No, the dienes you drew cannot be generated. Why? Asking you to explain why, was 2014's Singapore TYS A level exam qn. Hints (any or all of these terms may be used in your explanation) : orbital hybridization, electron geometry, ring strain, angle strain, thermodynamic instability, thermodynamic infeasibility.

Ohh!!

I only thought of that, due to the factor of ring strain.

Becos sp hybridised should have a angle of 180°, however in that case its bent at an angle.

Hence the diene cannot be formed

So yes, only 1,3diene can be formed right?

Can help improve and add in the rest of the factors into the answer pls

Thank u ultima!

You got the idea.

Cumulated dienes can indeed be generated, just not within a small ring like this molecule, because of the resultant ring strain due to angle strain (due to the significant deviation from the ideal bond angles about the sp hybridized C atom as predicted by VSEPR theory, which is to maximize thermodynamic stabilities by minimizing electron pair electrostatic repulsions) making it thermodynamically unstable and thus its formation thermodynamically unfeasible.

Conjugated dienes (such as the final product in this Cambridge qn) are thermodynamically favored due to the resonance stabilization energy (for the same reason why only benzene, but not the non-resonance mythical version, cyclohexa-1,3,5-triene, exists), which results in the resonance hybrid enjoying overall stronger C-C bonds (ie. 3 C-C bonds all with partial double bond characters, is overall still stronger and still has a more exothermic formation enthalpy compared to 2 full double bonds and 1 strictly single bond), a more exothermic formation enthalpy, and thus more thermodynamically stable and formation more thermodynamically favored.

Which 2 factors?? Alkyl stability and temperature uh?

1st factor : No. of H atoms substitutable

2nd factor : Stability of alkyl radical intermediate, aka, reactivity of primary vs secondary vs tertiary H atoms.

Your own JC school teacher never teach you meh?

Go google up on how these factors influence product distribution during halogenation via free radical substitution, and the reactivity-selectivity principle (in the context of halogenation via free radical substitution, which requires an understanding of the Hammond postulate).

UltimaOnline SG

Another godly answer.., thank you so much..!!!

Iodine is such a big molecule so it's unreactive?

The equation c2h5• +I• -> c2h5I +HI , is endothermic is it?

Oi, just saying "Iodine is a big molecule so it's unreactive" is a sure way of making the Cambridge examiner shake his/her head, "Singapore students only know how to smoke", don't xia suay Singapore hor.

Yes, correct, you must link to the fact that the bonds formed are weaker than the bonds broken (quote Data Booklet values whenever relevant), and hence the reaction is thermodynamically unfeasible.

Isnt no. Of H atoms available for substitution same as tertiary, secondary or pri alkyl radical intermediate?

Is temperature a factor for FRS?

Not the same. In fact, the two factors are opposing. Take propane for example. Based on each factor, what is the expected major product? 1 factor predicts 1-halopropane as the major product, while the other factor predicts 2-halopropane as the major product.

If the Cambridge A level exam question requires you to calculate out the overall product distribution mathematically (which has appeared in past CIE A levels and Pre-U papers), the question has to provide you with the relative reactivities of a pri vs sec vs tert H atom (corresponding to the relative stabilities of a pri vs sec vs tert alkyl radical) in numerical values, for you to apply mathematically (ie. multiply both factors together). Such values differ across halogens, and across temperatures.

If your school JC teacher didn't teach you about these 2 factors (see the benefits of a Chemistry education @ BedokFunland JC ?), you can google up yourself, and/or read the following webpages :

http://orgchem.chem.uconn.edu/2443s2011/2443-041811.pdf

http://www.masterorganicchemistry.com/2013/10/31/selectivity-in-free-radical-reactions-bromine-vs-chlorine/

http://www.mhhe.com/physsci/chemistry/carey/student/olc/ch04radical.html

UltimaOnline SG

pH lower than pka, compound protonates

pH greater than pka, compound deprotonates

Which chapter is this??

Acid base equilibria?

Does this have any link to equilibria Le chatelier's principle shift left shift right?

I just memorise ^ only!

Hope i can draw links to reinforce

Yes of course it's acid-base equilibria, and your school JC teacher *did* teach you this, didn't he/she? It's basic (no pun intended) acid-base equilibria. Of course any equilibrium can be linked to Le Chatelier's principle, and also to Organic Chemistry (eg. acid-base equilibria of amino acids). Everything can be linked, once you understand Chemistry deeply enough.

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2010 p3 qn 2b

Need to draw all the protein structures to answer this qn?

Diagrams of structures are optional (as indicated in the question phrasing). Cambridge will award you the marks if the required points are present in your answer, whether in words, in balanced equations, in your diagrams, or all 3. Unless you're supremely confident you can include all the required points are present in words alone, or equations alone, or diagrams alone, it's usually a good idea (assuming your time management is good enough) in such questions to use at least 2 out of the 3 forms in your answers (even if the question doesn't compulsorily require it), to ensure all the required points are present in your answer to secure full marks. So to address your question directly, no need, as long as the required marking points are present (go see a TYS publisher's answer for this).

UltimaOnline SG

I have a couple of O Level Science (Chemistry) MCQs on hand - would greatly appreciate if anyone could shed some light on these questions - cheers! :)

Answer: A

Remarks: Why is the answer A? I thought all of the statements are correct.

Remarks: The answer is B. However, how do I know that the [Fe3+]>[Fe2+], given that the question did not say that the precipitates were left to stand, thereby allowing for oxidation of Fe2+ to take place?

Answer: B

My postulation: The change in water level would mean that the gaseous pressure has increased. The answer can’t be D because ammonium chloride would be formed (so presumably the gaseous pressure exerted on the LHS would decrease since an ionic salt is formed). However, what will happen when CO or N is pumped into the set up? Am I right in saying that the answer is B because methane does not react with ammonia?

Q1. Diffusion also occurs in the solid and aqueous states.

Q2. Fe(OH)3 has a higher molar mass than Fe(OH)2.

Q3. Only CH4 has a lower molar mass than NH3. At constant temperature, the rate at which gas particles diffuse is inversely proportional to their molar mass.

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Hi UltimaOnline,

Thank you for your help :)

For 1 - how does diffusion take place in solids when the atoms are arranged in fixed positions? As for 3 I agree with what you mentioned, but how does it relate to the question?

Hi Gohby, no prob, as always :)

Q3. CH4 (being lighter) diffuses into the porous container faster than NH3 (being heavier) can diffuse out, resulting in an increase in gaseous pressure within the porous container.

Q1. This isn't taught at either O levels or A levels, but Truth is Truth, Chemistry is Chemistry, and diffusion *does* occur in solids (the thermodynamic motivation for diffusion, is of course, thermodynamically favorable positive entropy change, which is also the thermodynamic motivation for Le Chatelier's principle).

https://en.wikipedia.org/wiki/Mass_diffusivity#Solids
https://en.wikipedia.org/wiki/Atomic_diffusion
https://en.wikipedia.org/wiki/Effective_diffusion_coefficient
https://en.wikipedia.org/wiki/Lattice_diffusion_coefficient
https://en.wikipedia.org/wiki/Kirkendall_effect

UltimaOnline SG

Hi UltimaOnline,

I would like to seek your opinion/concurrence for some of the questions where my answers diverged from the answers in the answer scheme :) :

Remarks: The answer is B but I got A. Given the number of moles of ammonium nitrate used is 0.1, the number of moles of ammonia produced should be 0.1 too, so I think the volume produced should be 2.4dm³. Btw, what is the role of the aluminium powder in the reaction?

Remarks: The answer is A. What should I be looking out for?

Remarks: The answer is A but I think the answer is B. As greater number of moles of lead (II) nitrate was present in Experiment II , there should be more lead (II) iodide formed in experiment 2, thus the total mass of lead (II) iodide is smaller in I.

Remarks:The answer is B but I think the answer is D. I would think this is an electrophilic addition reaction of an alkene rather than an FRS reaction.

Hi Gohby,

Your own answer for the last question about electrophilic addition of hydrogen halide across an alkene double bond is correct. Don't put too much trust in Singapore secondary schools' teachers' answers, or Singapore assessment practice books' answers, or Singapore TYS publishers' answers.

Q1. Al is oxidized to Al3+, and NO3- (OS of N = +5) is reduced to NH4+ (OS of N = -3). Thus a total of 0.1 + 0.1 = 0.2 mol of NH4+ is deprotonated by the OH- to generate 0.2 mol of NH3, which works out to be 4.8 dm3.

Q2. You need to use insoluble bases (ie. hydroxides and carbonates), as soluble alkalis are corrosive and will damage your mouth, throat and esophagus before it can reach your stomach to reduce acidity. (Medical note : aluminium is toxic and causes brain damage, thus shouldn't be used in drugs or additives, yet its use continues to be widespread, contributing to Alzheimer's disease and other related dementia medical conditions)

Q3. Either A or B could be correct, depending on the molarity of KI. If the molarity of KI is sufficiently low, it's *possible* for the mass of ppt generated in experiment 2 to be *less* than in experiment 1 (despite the moles of Pb2+ in experiment 2 being more than in experiment 1), because the total volume of solution in experiment 2 is greater. The secondary school teacher who set this question was being a smart-ass and must have been proudly confident of having set the trickiest O level Chemistry prelim paper question among all the secondary schools for that year (there's always an informal competition to set the toughest prelim papers, just look at the top JCs' prelim papers, most JC teachers themselves wouldn't be able to get an A grade if they sat for the top JCs' prelim papers within the allocated time). Unfortunately, he/she didn't realize that since the molarity of KI wasn't specified, if the molarity was sufficiently high, then even with a larger volume of solution, 99.9% of the moles PbI2(s) could be precipitated out in both experiments, then *your* answer would be correct. So this question has insufficient data to conclusively rule out either A or B.

UltimaOnline SG

2010 p3 qn 4 gii

Photo is here

https://www.dropbox.com/s/morxcgb8519dx60/20151105_145329.jpg?dl=0

How can secondary alcohol oxidise to Carboxylic acid??

Thought secondary alcohol can only oxidise to ketone at maximum?

There are only 3 ways for a ketone to be further oxidized :

If it's part of a benzene ring side chain (with a benzylic H atom).

If you use a chemical oxidizing agent stronger than KMnO4, eg. concentrated HNO3.

If you burn it.

UltimaOnline SG

TYS 2011 Paper 2 question 4e

how do you determine which half equation to use for the V3+ "in a solution containing V3+" like what hints tell u that you are supposed to consider the VO2+ half equations and that there is further reaction?

First, you should be aware (if you didn't know, now you know liao) that the OS of V in the common V containing ions are 0, +2, +3, +4, +5 and +5, in V, V 2+, V 3+, VO 2+, VO2 + and VO3 - respectively.

(Special Note : The equilibrium between the dioxovanadium(V) ion and vanadate(V) ion isn't a redox reaction, but an acid-base equilibrium, similar to the equilibrium between chromate(VI) and dichromate(VI) ; you can figure out exactly which species with this OS is present under acidic vs alkaline conditions, by looking at whether H+ is present on the LHS or RHS).

Observe PbCl4 : obviously the strongest redox effect of PbCl4, is its oxidizing power (since Pb4+ is unstable and wants to be reduced to Pb2+, which is more stable due to lower anionic charge density and/or the inert pair effect).

Hence, Let Pb4+ be reduced to Pb2+, and let V be oxidized from V3+ to VO 2+ (ie. OS change from +3 to +4). Prove using Cell potential = Reduction potential @ Cathode + Oxidation potential @ Anode, that this redox reaction is thermodynamically feasible.

Next, notice that VO +2 (OS of +4) *might* be able to be further oxidized to VO2 + (OS of +5) under acidic conditions. Again prove using Cell potential = Reduction potential @ Cathode + Oxidation potential @ Anode, that this redox reaction is thermodynamically feasible.

Therefore, the observable color change is from green V3+ to blue VO 2+ to yellow VO2 +, using Pb4+ reduced to Pb2+ as the oxidizing agent in both (thermodynamically feasible) redox reactions. Warning : between blue VO 2+ to yellow VO2 +, a green colour can be observed. This is not due to an intermediate OS between +4 and +5 (which is impossible), but it's due to a mixture of blue VO 2+ and yellow VO2 +, which we humans perceive as green (ie. blue + yellow = green), a common exam trick question.

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TYS 2011 Qn 4c(ii) can the explanation be that CCl4 is non polar hence unable to form intermolecular bonds with water molecules that are strong enough to overcome the hydrogen bonding In water hence CCl4 is inert?

Totally wrong. Zero marks. This is basic H2 syllabus stuff that you need to memorize (with understanding of course), how can you not know this when Paper 2 is just a few hours away?!? Be careful not to lose easy marks like this, whether in P2 or P3!

Chan Kim Seng + Jeanne Tan : Physical / Inorganic book page 376.

George Chong : Inorganic book page 59.

Jim Clark : simpler A level explanation : http://www.chemguide.co.uk/inorganic/group4/chlorides.html

Rod Beavon : more advanced A level / University level explanation : http://rod.beavon.org.uk/hydrolys.htm

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