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

last SG · 2010-07-26T23:40:35+00:00

With Singapore having the highest incidence of diabetes in the world, every Singapore should take Berberine. Berberine helps to prevent diabetes if you don'...

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http://papers.gceguide.com/A%20Levels/Chemistry%20(9701)/9701_w13_qp_43.pdf

http://papers.gceguide.com/A%20Levels/Chemistry%20(9701)/9701_w13_ms_43.pdf

http://papers.gceguide.com/A%20Levels/Chemistry%20(9701)/9701_w13_er.pdf

In Q6 part f(i) , i dont understand 2 things.

First of all, i dont understand why G and H are not similar to F...if we rotate both the left and right hand side of the molecule, shouldnt we get F.

Secondly, how does rotating right hand side of J by 60o clockwise make ot identical to F as suggested by the examiner report. And why is Chirality of Left hand carbon of both G and H opposite to that of F.

After all this, how to solve f(iii) ie drawing the fourth optical isomers.

Is this related to R,S configurations or is this something else...as to where can i read more about this?

If you could explain this i detail, i would be grateful..thanks.

This is the toughest question in the paper for 2 reasons : because it's slightly beyond the A level syllabus, and also because it's extremely difficult to visualize (solely using imagination) the rotation of 3D structures.

Going beyond the syllabus, ie. at Uni level, the correct way to approach such questions is to assign priorities to the 4 different groups based on the Cahn–Ingold–Prelog priority rules, and concordantly label each of the 2 chiral C atoms present as either R (Rectus) or S (Sinister). The 4 optical stereoisomers would then be (2R,3R), (2R,3S), (2S,3R) and (2S,3S), where (2R,3R) & (2S,3S), and (2R,3S) & (2S,3R), are the 2 sets of enantiomers, while (2R,3R) & (2S,3R), (2R,3R) & (2R,3S), (2S,3S) & (2S,3R), and (2S,3S) & (2R,3S), are the 4 sets of diastereomers.

Alternatively, or simultaneously, if you can obtain (or make for yourself) a 3D 'ball-&-stick' model of the molecule, and rotate each end, it'll help you to better visualize and solve this question. Of course, you won't be allowed to bring in such tools or toys in the A level exams, so you'll have to settle with using your hands : orientate your fingers (using both hands) in a tetrahedral geometry, with 1 finger towards you (the wedge), 1 finger away from you (the hash or dash), and 2 fingers (1 from each hand) neither towards or away (the normal bonds), then slowly rotate your hands. While simultaneously mentally keeping track of which finger representing which group of atoms, *and* the priority of each group (if you're assigning R vs S configuration to help you answer this question). Not an easy task, unfortunately.

You could say this question is unfair to be asked at A levels. But it's a bell-curve anyway, so since it's unfair for everyone, it's arguably still fair. Such questions are intended for only the top few % of the A level cohort to be able to answer correctly, otherwise too many people would score A grade. Regardless of where you're ranked in the cohort, exam-smart students would always skip such time-consuming questions first, complete the rest of the paper, then go back to such 'skipped' questions last. It'll be penny-wise-pound-foolish to waste 15 min struggling on this 3 marks question, then find out at the end of the exam that you only have 3 min left with 15 marks of much easier questions left unanswered.

Lets assume that i know how to assign R and S priorities

To assign these, the lowest priority group has to be oriented away from us..in this case, the lowest priority group is the Hydrogen atom but the problem is that on the Left Hand Carbon atom for each of the structures, the H-atom is shown to lie on the plane rather than pointing out or into the plane...so assuming that we know how to assign R and S, how can we get past this issue?

By visualizing the structure vividly in 3D space using your imagination, than slowly rotating (not just the single sigma bond, but move the entire structure) it in your mind, until the lowest priority substituent, is pointing away from yourself. And try using your fingers (orientated in a tetrahedral geometry, eg. left hand 3 fingers in a trigonal pyramidal tripod stand, right hand's forefinger representing lowest priority facing up, before rotating it till it's away from you) to help.

Also see these videos :
https://www.youtube.com/watch?v=auphHXYK8tE
https://www.youtube.com/watch?v=kFD6hzLseVs

UltimaOnline SG

As a chemist, it's annoying to read stuff like, "Singapore recorded its highest temperature in a decade this month, as the mercury soared to 36.7 deg C". - http://www.straitstimes.com/singapore/wednesday-hottest-day-in-a-decade

Helloooo, due to its toxicity, the use of mercury in thermometers has been phased out. Ergo, the journalistic use of "mercury" to mean "temperature" is outdated. Hmmph.

Did you know? Research studies have shown that dentists generally suffer from lower IQ compared to medical doctors, as a result of cumulative brain damage from long-term exposure to mercury in their dental work. Needless to say, if you've mercury teeth fillings, you've been suffering from cumulative brain damage all these decades. How do you think the phrase "Mad Hatter" came about? Be sure your children don't suffer the same tragic fate (especially tragic because it's totally preventable, if only more people knew).

http://www.toxicteeth.org/mercuryfillings.aspx

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What are the causes of amphoterism?

Tried to research online, but to no avail ! The closest relevant info i can find is 'Amphoterism depends on the oxidation states of the oxide' from wiki! But still, this piece of info doesnt help much!

Prior to researching i only know Al2O3 is amphoteric,

but after researching now i know Al(OH)3 , Zn (OH)2, Be(OH)2 are also amphoteric species.

I encountered a qn asking if SrO is amphoteric.

We only learnt and memorised that AL2O3 is amphoteric. So how can we determine which species is amphoteric is a qn asked?

A level shortcut : covalent = acidic, ionic = basic, hence partial covalent partial ionic = amphoteric.

BedokFunland JC (ie University) level : to understand identify and understand why & how a species is amphoteric, and why the A level shortcut works (it's not an explanation, but merely an indirect indicator), you first need to understand why & how an acidic species is acidic, and why & how a basic species is basic, and you *must* draw out the curved-arrow electron-flow reaction mechanisms. It's too much detail to go into on the forums, and my BedokFunland JC students can ask me to teach them these mechanisms, while the rest of you can go ask your school teacher or private tutor if interested.

So do you think strontium oxide qualifies to be amphoteric? Btw, amphiprotic is a subset of amphoteric, but not vice-versa. Don't confuse these terms. Go ask your school teacher or private tutor if interested.

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Hmm i try,

O is more electronegative than N. O withdraws some e- density from N. Magnitude of delta -ve on O decreases. The O-H bond in hydroyxlamine is less polar compared to O-H bond in water. Hence strength of H bond between hydroxylamine molecules is weaker than strength of Hbond between water molecules.

Hence water has a higher B.P. than hydroxylamine even though hydroxylamine can from more Hbonds per molecule than water.

Indeed, the O-H bond in hydroxylamine is less polar, compared to in H2O or in alcohols, due to the electronegative N atom. So far so good. But here's where your school cocked up : the boiling point of hydroxylamine should actually be higher, not lower, than H2O, as number of H bonds formed is more important than strength of H bonds. Simply look at their melting points, and you'll see I'm right. However, in practice, hydroxylamine never gets to reach its theoretical boiling point (ie. above 100 deg C), because it readily decomposes (at 58 deg C) into water, ammonia, and either dinitrogen monoxide or nitrogen gas (depending on mechanism pathway and reaction conditions).

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'Maximum buffer capacity' section

Is it that for an acidic buffer or alkaline buffer, theoretically, the buffer can have the pH of the solution remains(almost) unchanged, when acid or base of concentration of up to 10 times the concentration of the weak acid / weak base, added to the buffer solution?

Yes, mathematically, a solution can be considered a valid buffer as long as the ratio of molarities of conjugate acid-base pairs lies within 1:10 and 10:1, ie. pH = pKa (+1 or - 1)

Woah damn cool add 5moldm-3 of acid/base to a buffer solution comprising of 0.5moldm-3 weakacid/base , ph still remains unchanged

A buffer solution's pH changes *slightly* when a *small* amount of strong acid or base is added to it. Note that highlighted words. And Cambridge will require you to write equations to explain exactly how a buffer works. The exact equations will of course, depend on exactly what is the conjugate acid-base pair of the buffer solution.

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Hi ultima, hope you could lend help for this qn,

Qn 1 in here : https://www.dropbox.com/s/3o7vr03sfgpyhm6/20160418_193429.jpg?dl=0

Qn and attempted answer is in the pic. Suggested soln is C, but i cant figure out why. I dun have solution for mcq, only final answer alphabet

Thanks

2015 ACJC P1 Q9

Work out the Kc value for original conditions.

Kc = (183x183)/90 = 372.1

Apply the same Kc value for new conditions.

Concordantly, equilibrium partial pressure of hydrogen gas = Square root of (150x372.1) = 236.25 kPa.

How would u know the values for CO (g) ??

By avogadro's law? But n= v/vm , vm same for all gaseous products, since same temp and pressure, all gaseous products is 1 mol from this eqn, wouldn't volume be different?

So next time i just use the same partial pressure value of a gaseous product, of same mol, for the partial pressure value of a unknown gaseous molecule of same mol ??

I've already given you detailed solution liao. If you still can't figure out this little bit by yourself (even if it didn't occur to you before reading my solution, after reading my solution you should now be able to figure out why), you'll have to go ask your school teacher or private tutor. Try to refrain from repeatedly asking too much on a single question, especially after I've already given you the answer on the forum.

UltimaOnline SG

Hi ultima, hope you could help me here,

For qn 2

https://www.dropbox.com/s/ses61cnec6rtqpo/20160418_201120.jpg?dl=0

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Cs toh advanced study guide page 238 "group VII elements are typical non metals ( each element has 7 valence electrons) "

I know by observation that non metals have large number of valence e-, but is large number or valence e- a *characteristic* of non metals?

From wiki " ...chemically, they tend to have high ionization energy and electronegativity values, and gain or share electrons when they react with other elements or compounds. "

By definition, metals are electropositive, non-metals are electronegative. To be electronegative, you need more protons in your nucleus vis-Ă -vis the other elements in your same period. Condordantly, as a non-metal you need to be on the right side of the periodic table. Ergo, you'll have a large number of valence electrons. So it's not a definition of non-metals, it's just a characteristic. Transition metals also have a large number of valence electrons, because both the (n-1)d and ns electrons (eg. 3d and 4s), are both considered valence electrons.

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thinning of ozone layer *reduces* global warming, not increases it!

Ideally ozone layer should be thin ,without holes, is it?

But if the ozone layer is thin, it will have holes. You can't have your cake and eat it too. You can't lepak for A levels and still score distinction A grade (unless you're me, Mr BedokFunland JC).

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Hi I would like to ask something about Senior Chem Olympiad (for JC1) and whether I should take it

So recently, i somehow got selected by my school for SChO training, altho im like not very good, just a decent chem student and the ppl who got selected are all quite beast in their studies, like half of them are china scholars.

So i went for my first session and apparently for the olympiad, the whole syllabus for A level chem needs to be covered as well as some other university chem modules, which is pretty scary given that the trainer only has now until end of Nov to finish it. And my school also not very good at chem olympiads, for last years olympiad 3 silver, 3 bronze, 1 merit

So right now im not very sure if i should continue as i need to dedicate 2h per week for training and perhaps even more to complete the homework/readings that the external trainer gives. And im not very sure if it helps with my A levels Chem or is it too far fetched. Im afraid it would affect my time for studies and other stuff as in JC i find myself always havng v little time.

So any advice on whether to go for it would be appreciated!

Don't bother going for formal Olympiad Chem training or H3 Chem lessons in school, it just takes up valuable time you yourself said you can not afford. Olympiad and H3 are only for students who are already scoring all As in all their H2 subjects, are confident they have the time to do so, and wish to apply for Medicine and/or scholarships.

If you like, read up and self-learn on Olympiad or H3 stuff yourself, for giving yourself the extra edge in H2 Chem. Or if you (ie. anyone reading this) join my BedokFunland JC tuition in the future, I'll teach you only the Olympiad and H3 concepts and strategies that will help you to get A grade in your H2 Chem.

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Atoms(e.g. Cl) of Halogen molecules(e.g. Cl2) already have complete octet, why are they still reactive?

Update : (Noticed bonds between halogens are quite weak, as seen from bond strength from data booklet), but still, principal question above^

Because the Cl atom in covalent molecules (ie. bond-Cl), and the Cl- ion in ionic compounds, also have a stable octet, plus an even more stable OS of -1 (instead of 0 in Cl2).

Why O.S. of -1 more stable?

Doesn't mean the more -ve o.s. is, more stable the atom is correct?

Ya know, it's annoying when you (ie. Singapore JC students) say things like that, and you'll lose marks in the A level exams for such ambiguous phrasing. They way you phrase it, makes you appear simple minded (ie. oversimplifying matters due to lack of depth of understanding) or dogmatic (ie. blindly memorizing school notes without deeper understanding), because it makes you seem to think "the more -ve o.s. is, more stable the atom is" holds true for all elements.

Obviously (at least it's obvious to me and should be obvious to anyone who truly understands Chemistry, but apparently this doesn't necessarily include the majority of Singapore JC students or even some teachers and tutors, does it?), metals being electropositive will find positive OSes more stable, and non-metals being electronegative will find negative OSes more stable.

As for the exact unique OSes preferred by each individual element, you can go figure out for yourself, and/or research online, and/or ask your school teacher and/or private tutor. At A levels, if you really understand your A level Chemistry, then you'll begin to get a glimpse of understanding as to the myriad of chemistry factors (some opposing each other) contributing to why different elements (eg. different transition metals) have different unique preferred OSes.

UltimaOnline SG

When do we write equation/reaction thermodynamically/entropically favourable and thermodynamic/entropically feasible and spontaneous ?

When delta H is -ve,=, or delta S is +ve, then the -ve delta H, or +ve delta S, is called thermodynamically favorable (but reaction may or may not be thermodynamically feasible). When delta G is -ve, then the reaction is thermodynamically feasible.

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So for maximum buffer capacity (half neutralisation point) always can skip icf and ice table?

First, if you're not sure it's maximum buffer capacity (eg. because the question guai-lan and give different volumes and different molarities of both acid and base), do the ICF table to confirm it's maximum buffer capacity. With Cambridge questions getting tougher each year, it may be near, but not exactly maximum buffer capacity. So when in doubt, just do ICF table (which is arguably more important than ICE table, especially since Cambridge has stated they will accept the approximation of equilibrium molarity back to initial molarity).

If it's really exactly maximum buffer capacity, obviously neither ICF nor ICE table is required, and not even the Henderson-Hasselbalch equation is required (since log 1 = 0). For maximum buffer capacity, pH = pKa1 or pKa2 or pKa3 (depending on which maximum buffer capacity it is during the titration of a polyprotic or multiprotic acid or base).

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T-T my answer wrong means my working epic fail means my understanding epic fail T_T

Pls help

https://www.dropbox.com/s/3codbenfirr3luk/20160425_190417.jpg?dl=0

Q8) A student dissolved 10.5 g of sodium fluoride in 250 g of water. The following data is obtained:

Lattice energy of NaF â918 kJ mol-1
Enthalpy change of hydration of F- â457 kJ mol-1
Enthalpy change of hydration of Na+ â390 kJ mol-1
Given that the specific heat capacity of water is 4.2 J g-1 K-1, what would be the final temperature of the solution, if the initial temperature of water is 30 oC?

BedokFunland JC guidance :

Calculate mol of NaF.
Hence calculate total enthalpy change for the following process :
NaF(s) --> Na+(g) + F-(g) --> Na+(aq) + F-(aq)
Plug Q, m, c into formula Q = m x c x delta T
Solve for delta T.
Hence obtain the final temperature, and answer for this MCQ.

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Is it right that Exothermic deltaHsoln (delHsoln <0 means soluble in water? ) means water gets more hot(gain energy hence temp rise) , as ions release energy(which outweighs LEdissociation) on hydration?

Solution Enthalpy = (endothermic) Lattice Dissociation Enthalpy + (exothermic) Hydration Enthalpy

Solution Entropy = (positive) Lattice Dissociation Entropy + (negative) Hydration Entropy

Gibbs Free Energy of Solution = Solution Enthalpy - Temperature x Solution Entropy

A species is soluble in water (ie. solution process is thermodynamically feasible, ie. when energetically and entropically considered at a given temperature) only if Gibbs Free Energy change is negative.

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Page 334 advanced cstoh guide

What does 'esterification can be used as a test for alcohol - the ester formed floats on water and has a sweet smell' means?

Isnt this a test of presence of ester instead?as we deduce ester present from sweet smell?

You can test for the presence of alcohols by adding an acyl halide. If a sweet smell is observed, meaning that an ester is generated, which in turn means the original analyte must have contained an alcohol.

UltimaOnline SG

In case you're wondering why potassium iodide is often prescribed in cases of radiation exposure, and how it works (only in a very specific and limited way).

http://hps.org/publicinformation/ate/faqs/ki.html

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Hi Ultima,

https://www.dropbox.com/s/2i79w0ow9hq7bsp/1111.PNG?dl=0

for part ii) , why is this thought process wrong : i want find [H+] . i use ka formula. tried, but didnt get the correct answer.

(p.s. , i've internalised the 2 methods/approach that was taught to me , but want to know why above concept wrong. )

is it because, inside the solution, its not just the weak acid inside, there's the cj base too? Help my concept please T-T

thanks ultima

Being a buffer soln, you *can* indeed use Ka, or even Kb. It's just that the Henderson-Hasselbalch equation saves you time as you can work in moles instead of molarities. If you didn't get the correct answer, it means your molarities (ie. moles / new total volume of soln) of both HA and A- at that point (double-equivalence) are wrong. Obviously you need to plug in the molarities of both conjugate acid and conjugate base into the Ka (or Kb) equation, and not all questions will be as kind as this question to let you have maximum buffer capacity, so you better start practicing tougher questions in which the moles or molarities of both conjugate acid and conjugate base are different. Then try using all 3 formulae : Ka, Kb, and Henderson-Hasselbalch equation, to prove *to yourself* that all 3 formulae will give you the same final answer of the correct pH.

UltimaOnline SG

Is it okay/fine/chemically correct to write

E.g.

For Mg(OH)2 to be precipitated, Qsp (Mg(OH)2) >Ksp (Mg(OH)2)

Looking at 2015 jc prelims solutions, all wrote I.P. or just Solubility = ____

None wrote Qsp

My school uses I.P. , but sometimes i forgot, i wrote Qsp a number of times already ,but now trying to get the habit back of writing I.P. But is it wrong to write 'Qsp' ?

Btw, SAJC uses ICF table, lol, cool. that makes BJC students, and them ,that know about it

Both are acceptable by Cambridge, use whichever you prefer, IP or Qsp, to be compared with Ksp. Personally, I'll recommend Qsp (since it's closer in form to Ksp). For solubility (which is not the same as Ksp. ie. solubility product), always specify "molar solubility" (mol/dm3) or "mass solubility" (g/dm3).

UltimaOnline SG

DHS/2015/P2/Q3 part iv )

https://www.dropbox.com/s/x2gzb3vp1ekxl3c/20160502_012359-1.jpg?dl=0

I understand the solution.

But i dun understand the qn, Why to determine solubility of CaC2O4 at high pH, use NaOH? What are the alternative method?

I was thinking, CaC2O4 ăăCa2+ + C2O42- , let solubility of CaC2O4 be x, hence solubility is just root(Ksp)

Thanks ultima

Your molar solubility based strictly on Ksp ignores the effect of pH, which is the usual case within the H2 syllabus. This DHS question goes beyond the H2 syllabus (which Cambridge often does) to include the effect on pH, which is why your answer is inadequate and thus wrong. If you do not add NaOH, how are you going to make the pH high (ie. alkaline)? That's why you add NaOH, which increases solubility by precipitating out the cation (in contrast, adding acid increases solubility by protonating the anion). There's no alternative method ; Wilbur's equation only covers acidic to neutral pH.

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