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Originally posted by theophilus:
LOLOL
But essentially they are 2 different papers altogether ah and im not even gonna mention the help availed. So only have 3 conclusion
1. They just hantam everything into one bell curve and ignore the difference
2. paper 5 candidates and paper 4 candidates individual bell curve
3. Or somehow they bell curve only the theory papers together and separate paper 5 and 4 for inidividual bell curves and add them back separately
Yeah, but it doesn't really matter, does it? Regardless of which case is true (a mere technicality), the only thing that's in your power, is you do the best you can. The rest is out of your hands.Though as an ex-MOE teacher, I can tell you there isn't a separate bell-curve for private candidates. The entire Singapore cohort is graded together. Cambridge sends MOE-SEAB the raw score, and MOE-SEAB assigns the grades based on the bell-curve of the Singapore cohort.
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haixxx i guess you're right, just that i hear from afew people here and there la. wanted some assurance. Thanks though !! haha good luck for As which I may or may not be sure if you are taking LOL
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Originally posted by theophilus:
haixxx i guess you're right, just that i hear from afew people here and there la. wanted some assurance. Thanks though !! haha good luck for As which I may or may not be sure if you are taking LOL
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must a molecule(or part of it, the atoms that are of interest to resonance) be planar to have resonance?
e.g. benzene, amide , histidine, pyridine , etc
is it becos the overlap of u.P orbital must overlap in the same plane if not the e- cannot be delocalised?
or if u.P orbital is not on the same plane ,cannot overlap at all?
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Originally posted by Flying grenade:
must a molecule(or part of it, the atoms that are of interest to resonance) be planar to have resonance?
e.g. benzene, amide , histidine, pyridine , etc
is it becos the overlap of u.P orbital must overlap in the same plane if not the e- cannot be delocalised?
or if u.P orbital is not on the same plane ,cannot overlap at all?
Obviously. Finally you get this. After all this time. Feels good to finally understand, eh? -
Hi, sorry to disturb, I would like to check on 1 question :
MJC 2014 Prelim P2 Q6 iii
I calculated the rate to be 4.00, and I'm pretty sure the rate is 4.00. The answer key states 8...did I go wrong somewhere??
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Originally posted by supercat:
Hi, sorry to disturb, I would like to check on 1 question :
MJC 2014 Prelim P2 Q6 iii
I calculated the rate to be 4.00, and I'm pretty sure the rate is 4.00. The answer key states 8...did I go wrong somewhere??
I regret to inform u that the MJC teacher owns ur ass, SuperCat. Correct answer is indeed 8x, not 4x. Try to figure out for urself why. -
Hi, for CJC prelim 2014 P3 qn 2bii, the answer key gave the set up as
http://m.imgur.com/IZvuFAD
Shouldn't it be the other way around? In standard practice, anode is on the left, cathode on the right, right??
And for the equation at the anode and cathode, may I ask if arrow should be full arrow (-->)?
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Originally posted by UltimaOnline:
I regret to inform u that the MJC teacher owns ur ass, SuperCat. Correct answer is indeed 8x, not 4x. Try to figure out for urself why.Oh ouch Haha ok let me try again!!
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Hi, for CJC prelim 2014 P3 qn 3 d,
Why can't E and F be these?
http://m.imgur.com/IDks6TX
Regarding presentation of answers, could I write "ZnCO3 < CaCO3 < PbCO3" for the A level exam? Or do I have to replace < with "less than"?
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all molecules are subjected to the thermodynamic propensity for all molecules/reactants to gain a greater stability through chemical reaction to form more stable products, by releasing energy
what are the factors that determines stability of molecules?
and how eager/spontaneous a reaction will take place?
based on the energy of reactants and products only? cant think of anything else
the degree of how a molecules being polar only affect IMF ?
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wouldnt the degree of how polar a molecule is affects its reactivity too?
like alkanes alkenes benzene less reactive than other polar functional groups
(ok i do recognise that the above is too great a generalisation, there are other factors that explain why some of the above mentioned functional groups are more reactive than others, due to difference in electron density(like delta minus delta plus) , electronegativity , ? or something ? idk)
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Originally posted by UltimaOnline:
I regret to inform u that the MJC teacher owns ur ass, SuperCat. Correct answer is indeed 8x, not 4x. Try to figure out for urself why.ohhh i see !!! they loan reference(euphemism for copy) from *the source*..
wow, no wonder U.O. wasn't surprised when he saw the same qn from the source. lmao
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Originally posted by Flying grenade:
all molecules are subjected to the thermodynamic propensity for all molecules/reactants to gain a greater stability through chemical reaction to form more stable products, by releasing energy
what are the factors that determines stability of molecules?
and how eager/spontaneous a reaction will take place?
based on the energy of reactants and products only? cant think of anything else
the degree of how a molecules being polar only affect IMF ?
U bloody coconaden, u spamming again. Want to ask all these kind of questions, u pay me $$$ and ask me during tuition. Or else go ask ur school teacher. I will henceforth only answer queries on exam questions (like what *everyone else* on the forum asks me, only u Mr Coconaden keep spamming ur endless coconaden qns!), and ignore ur endless conceptual coconaden qns. -
Originally posted by supercat:
Hi, for CJC prelim 2014 P3 qn 2bii, the answer key gave the set up as
http://m.imgur.com/IZvuFAD
Shouldn't it be the other way around? In standard practice, anode is on the left, cathode on the right, right??
And for the equation at the anode and cathode, may I ask if arrow should be full arrow (-->)?
For cell diagram (aka cell notation), anode has to be on the left. For diagram of the cell, it doesn't matter. "Cell Diagram" is not the same as "Diagram of the Cell".You should use single arrow, not double-half arrow.
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Originally posted by supercat:
Hi, for CJC prelim 2014 P3 qn 3 d,
Why can't E and F be these?
http://m.imgur.com/IDks6TX
Regarding presentation of answers, could I write "ZnCO3 < CaCO3 < PbCO3" for the A level exam? Or do I have to replace < with "less than"?
Your F is wrong (no chiral C atom), hence your E is also wrong.It's fine to use > or <, as long as it is made clear the characteristic that you're basing your ranking on, eg. specify clearly in words : "from most basic to least basic" or "from least basic to most basic", etc. In fact, after writing this, you can use comma to separate the species, instead of the < or > sign, which may accidentally confuse yourself and screw up your answer accidentally.
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Ok, I understand, thank you!
For A level 2006 P1 Q8, what is the difference between polished vs platinised electrode??
Q13, Don't Cambridge mean "addition of limited water"? Both AlCl6 and MgCl2 can undergo hydrolysis to give Cl-, which can react with H2O to give HCl fumes correct?
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Originally posted by supercat:
Ok, I understand, thank you!
For A level 2006 P1 Q8, what is the difference between polished vs platinised electrode??
Q13, Don't Cambridge mean "addition of limited water"? Both AlCl6 and MgCl2 can undergo hydrolysis to give Cl-, which can react with H2O to give HCl fumes correct?
Q8 : Polished = low surface area, not suitable for electrolysis.
Platinized = coated with (melted-on) platinum powder to give a high surface area, suitable for electrolysis.Q13 : Yes, limited water. Since excess water can't give you an answer, so be exam-smart and think "limited water". Always choose the best answer, even if all 4 options are not totally correct. With limited water, only AlCl3 gives Al(OH)3 and HCl(g), MgCl2 is too weakly acidic to generate HCl(g) (which only occurs for strongly acidic species like the other 3 options).
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Hi UltimaOnline,
I've 1 question to ask here:
N14/P2/Q6b(ii)
In this case, there are 2 chiral centres, which means there are 4 stereoisomers. At the same time, however, it is also possible to for the compound to display geometric isomerism (E/Z) owing to the 3-D position of the Br & methyl substituents across the ring. Why aren't there 8 stereoisomers in all then?
My thinking is that there can't be 8 stereoisomers because there can only be 4 permutations with which the substituents can be arranged around C1 and C2. However it seems odd that the number of stereoisomers is only 4 given that there are already 2 chiral centres - (and the geometric isomers seemingly don't contribute/coalesce with the optical isomers).
Thank you! :) -
Hi, I have a question from JJC prelims
http://m.imgur.com/nwWQpfl
Why is sentence 3 true? Higher solubility means SrF2 is more likely to dissolve? Since Ksp of SrCO3 > SrF2, shouldn't statement 3 be false??
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Originally posted by gohby:
Hi UltimaOnline,
I've 1 question to ask here:
N14/P2/Q6b(ii)
In this case, there are 2 chiral centres, which means there are 4 stereoisomers. At the same time, however, it is also possible to for the compound to display geometric isomerism (E/Z) owing to the 3-D position of the Br & methyl substituents across the ring. Why aren't there 8 stereoisomers in all then?
My thinking is that there can't be 8 stereoisomers because there can only be 4 permutations with which the substituents can be arranged around C1 and C2. However it seems odd that the number of stereoisomers is only 4 given that there are already 2 chiral centres - (and the geometric isomers seemingly don't contribute/coalesce with the optical isomers).
Thank you! :)
Hi Gohby,This stereochemistry involved in this question goes beyond the H2 syllabus, so notice that Cambridge doesn't specifically ask the candidate how many stereoisomers exist in total, or ask the candidate to draw all of them out.
The reason for this, is that for cyclic molecules like this, some of the geometric isomers overlap with the optical isomers (eg. a trans isomer may also be an R isomer, with the exact same stereochemistry), and you cannot double count them.
As an analogy, if you have an apple, a (red) pen, and a (round) pineapple, and you're asked to count how many objects you have, while it is true that apple is red, and the apple is round, that doesn't mean you have 4 objects (ie. you cannot double count apple as "red apple" and also as "round apple"), because you only have 3 objects in total (ie. there is overlap between the "red" and the "round" properties).
In conclusion, due to overlap between optical and geometrical isomerism for this molecule (which H2 students needn't worry too much about, as it's beyond the syllabus), there are only 4 stereoisomers in total, not 8.
No prob, Gohby :)
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Originally posted by supercat:
Hi, I have a question from JJC prelims
http://m.imgur.com/nwWQpfl
Why is sentence 3 true? Higher solubility means SrF2 is more likely to dissolve? Since Ksp of SrCO3 > SrF2, shouldn't statement 3 be false??
Tis a common trick question.You have to use molar solubility to compare solubilities. You can only directly use solubility products as a shortcut to compare solubilities, provided the stoichiometry of the ions in the ionic compound are the same.
Since SrCO3 (ie. 1 cation : 1 anion) has a different stoichiometry of ions, compared to SrF2 (ie. 1 cation : 2 anions), you cannot simply use solubility products as a shortcut to compare solubilities, but must instead separately calculate the molar solubilities of each ionic compound, to compare solubilities.
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Yo got a chem question from TYS
TYS 2012 P2 Q3biii
When conc HCl is added to a solution of PbCl2, how come it dissolves ? I checked the answer from different sources. One said the Pb2+ is oxidized to Pb4+, the other says the a (Pb(Cl)4) complex is formed which reduces the conc of Cl and Pb2+ ions. Not sure which is right !
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Okay wait I figured it out, but how can Pb form complexes ? isit just something that is taken to be known ?
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Originally posted by theophilus:
Okay wait I figured it out, but how can Pb form complexes ? isit just something that is taken to be known ?
All metal cations with high cationic charge densities can form coordination complexes, certainly not limited to transition metals.If you're an A grade student who has practiced hundreds of TYS and Prelim papers to expose yourself to a large variety of questions, you would already be familiar with such coordination complexes across the periodic table, even those not mentioned within the basic syllabus requirements.