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Originally posted by UltimaOnline:
Figure it out yourself. Think step by step what happens in the series of reactions.alkaline hydrolysis of that two amide product all colorless
idk color of phenylamine relative to what
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moreover wiki writes phenylamine melts at -4°c
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Originally posted by Flying grenade:
alkaline hydrolysis of that two amide product all colorless
idk color of phenylamine relative to what
Give u extra face for this becoz of the titration qn ; so help u out 1 last time. In future, if u still make this kind of mindless (ie. not mindful of the situation) error again, I won't spoonfeed u the answer liao.Phenylamine in the absence of solvent is a colorless or pale yellow liquid.
In aqueous solution (bear in mind you just carried out hydrolysis, be mindful of the situation dah dey!), phenylamine (ie. beyond the very small amount that is soluble or miscible in water), will appear as a emulsion or suspension or 'precipitate' (even if it's technically not solid at rtp).
If the volume of phenylamine is comparable with the volume of water (ie. water is no longer a solvent, because solvent by definition must be in large excess), then eventually the 2 immiscible species will separate into 2 immiscible layers, based on differing density.
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Originally posted by UltimaOnline:
Give u extra face for this becoz of the titration qn ; so help u out 1 last time. In future, if u still make this kind of mindless (ie. not mindful of the situation) error again, I won't spoonfeed u the answer liao.Phenylamine in the absence of solvent is a colorless or pale yellow liquid.
In aqueous solution (bear in mind you just carried out hydrolysis, be mindful of the situation dah dey!), phenylamine (ie. beyond the very small amount that is soluble or miscible in water), will appear as a emulsion or suspension or 'precipitate' (even if it's technically not solid at rtp).
If the volume of phenylamine is comparable with the volume of water (ie. water is no longer a solvent, because solvent by definition must be in large excess), then eventually the 2 immiscible species will separate into 2 immiscible layers, based on differing density.
OMG OMG OK i understand liao thanks God
wa really need be mindful
yeah but if i were them i wouldn't write white ppt, i would write phenylamine forms a distinct layer(based on differing density )
OMG thanks god mindfulness master time to drink more cooling coconut juice
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coconadens eating and drinking coconut juice mixed with nata de coco complement with cococrunch
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Hi, i have a couple of questions from tjc paper 3 2016. for question 1b) the answer said ethanedioic acid is a stronger acid, since stabilisation of the monoanion is formed by intramolecular hydrogen bonding with the unionised cooh group. Since we are talking about monoanion, shouldnt the answer be intramolecular hydrogen bonding between coo- and cooh group instead? For question 1c, for compound B, i drew cooh-ch2-ch2-co-cooh instead. Is it correct? And for 1di, i said sodium burns vigorously instead of readily. Will they penalise me? thank you very much!
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Originally posted by ArJoe:
Hi, i have a couple of questions from tjc paper 3 2016.
for question 1b) the answer said ethanedioic acid is a stronger acid, since stabilisation of the monoanion is formed by intramolecular hydrogen bonding with the unionised cooh group. Since we are talking about monoanion, shouldnt the answer be intramolecular hydrogen bonding between coo- and cooh group instead?For question 1c, for compound B, i drew cooh-ch2-ch2-co-cooh instead. Is it correct?
And for 1di, i said sodium burns vigorously instead of readily. Will they penalise me?thank you very much!
Yes, you're right. For A level exams, best to draw out the structure including the H bond, even if the question doesn't specifically ask you to do so.No, you're wrong. The question specifies that B is symmetrical, yours isn't.
Yes, that's acceptable.
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Hi, I have questions from 2014 AJC Prelim P1.
For Qn 29, may I ask if this Sn2 reaction mechanism is right? The structure looks a little weird to me...
http://imgur.com/ERMU03G
For Qn 40, option 2, will the hydrogen bond be disrupted? I don't think it will, but the answer key had B as the answer.
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Originally posted by supercat:
Hi, I have questions from 2014 AJC Prelim P1.
For Qn 29, may I ask if this Sn2 reaction mechanism is right? The structure looks a little weird to me...
http://imgur.com/ERMU03G
For Qn 40, option 2, will the hydrogen bond be disrupted? I don't think it will, but the answer key had B as the answer.
This AJC MCQ is flawed, because the real correct mechanism is : partial electrophilic addition, followed by nucleophilic addition, followed by deprotonation. So all 4 options are actually wrong.But in the A level exams, you still gotta choose the best answer. So in which case, yes your mechanism is the one the AJC teacher thought of, so it's acceptable.
Yes you're right, statement 2 is false.
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Hi, regarding CJC 2014 Prelim P1 Q6,
How to derive the shapes of the graphs? I used pV = nRT.
Since I has higher Mr than J, I is not as ideal as J. Here are my thoughts :
Option A : Answer key gave vertical lines as the answer...why are the lines vertical? pV = nRT (constant) When P is plotted against PV, shouldn't everything be constant too?
Option B : I think the shapes of the graphs are right, since P is inversely proportional to (1/V)
Option C : Using pV = (nR)T, shapes of graph are right too.
Option D : I don't really know how to manipulate the graph into P vs. (1/T) from pV = nRT...any tips?
It's more of the manipulation part that I can't understand...how to deduce the shapes of the graphs?
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Originally posted by supercat:
Hi, regarding CJC 2014 Prelim P1 Q6,
How to derive the shapes of the graphs? I used pV = nRT.
Since I has higher Mr than J, I is not as ideal as J. Here are my thoughts :
Option A : Answer key gave vertical lines as the answer...why are the lines vertical? pV = nRT (constant) When P is plotted against PV, shouldn't everything be constant too?
Option B : I think the shapes of the graphs are right, since P is inversely proportional to (1/V)
Option C : Using pV = (nR)T, shapes of graph are right too.
Option D : I don't really know how to manipulate the graph into P vs. (1/T) from pV = nRT...any tips?
It's more of the manipulation part that I can't understand...how to deduce the shapes of the graphs?
Option A : Lines are vertical because for ideal gases, PV remains constant. P can vary (the larger the P, the lower the V), but PV remains constant.Option B : If the x-axis was V instead of 1/V, then the graphs would be correct.
Option C : Yes, they're right, which is why C is the correct answer.
Option D : The graph would be correct if x-axis was T instead of 1/T. As temperature increases, so should pressure.
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Hi, for ajc 2016 paper 3 qn 2c, can i use acidified kmno4(aq) and heat, hco2ch2ch3 will give efferverscence of co2, due to hcooh further oxidising to form co2, whereas ch3cooch3 will not?
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Originally posted by ArJoe:
Hi, for ajc 2016 paper 3 qn 2c,
can i use acidified kmno4(aq) and heat, hco2ch2ch3 will give efferverscence of co2, due to hcooh further oxidising to form co2, whereas ch3cooch3 will not?
Your method will NOT work to "to distinguish between 2-hydroxybutanedioic acid (malic acid) and 2-hydroxypropanoic acid (lactic acid)", as neither species will generate CO2 when heated with KMnO4 (even if it did, you still have to elaborate on how you confirm it's CO2). -
I explained that hco2ch2ch3,in the presence of acidic hydrolysis, produces hcooh which will undergo further oxidation with kmno4 to produce efferverscence of co2? Is this correct?
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Originally posted by ArJoe:
I explained that hco2ch2ch3,in the presence of acidic hydrolysis, produces hcooh which will undergo further oxidation with kmno4 to produce efferverscence of co2? Is this correct?
Oops, my bad. Mistook ACJC for AJC.. In which case, your answer may (or may not) just score 1 mark out of 2 marks. Because "effervescence" is not necessarily observable (that only reliably happens with acid carbonate and some acid metal reactions).
And depending on the amount present (usually insufficient), using other means to test for the presence of the (small amount of) CO2 generated, eg. extinguishes a lighted flame, turns moist blue litmus paper red, etc. are all highly unreliable in this acidic environment.
Therefore, to score full marks, only the iodoform test is acceptable for this question.
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Hi, regarding DHS prelim 2014 P1 Q7,
Why is D wrong? The H-CR3 BE is lowest at +390kJ/mol, so it should be broken and the Cl will be added there, right?
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Originally posted by supercat:
Hi, regarding DHS prelim 2014 P1 Q7,
Why is D wrong? The H-CR3 BE is lowest at +390kJ/mol, so it should be broken and the Cl will be added there, right?
1stly, you cannot just look at which is the weakest bond, but the thermodynamic favorability in terms of exothermicity (since entropy change is approximately the same for all options) for the overall enthalpy change of reaction (ie. endothermicity of bond dissociation + exothermicity of bond formation).2ndly, for option D, you're breaking the strongest H–CH2R bond, not the weakest H–CR3 bond.
So you've actually made 2 separate errors here in a single MCQ.
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Oh no ok let me re-draw everything out.
For Qn 12 (same paper), could you please take a look at my working? Where did I go wrong?
http://imgur.com/XnR22rs
I put B as my answer, because it's slightly more than 10cm3.
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Originally posted by supercat:
Oh no ok let me re-draw everything out.
For Qn 12 (same paper), could you please take a look at my working? Where did I go wrong?
http://imgur.com/XnR22rs
I put B as my answer, because it's slightly more than 10cm3.
Maleic acid, being a weak diprotic acid, can be represented by H2A. Note that the x-axis specifies volume of weak acid added, not strong base added.At 20cm3 of H2A added, you have 1st equivalence point, hence you only have HA-, which being amphiprotic, *can* function as a buffer.
(Fyi, 2nd equivalence point to exactly fully deprotonate all of H2A, is at 10cm3 of H2A added against 20cm3 of NaOH.)
However, at 40cm3 of H2A added (ie. 20cm3 excess H2A), you have equal moles hence equal molarities of HA- and H2A, hence you have *maximum* buffer capacity. Concordantly, D is the answer.
Options A and B are obviously wrong, because you have a significant excess of the strong base NaOH, which means all of the 1st acidic protons have been removed, and most of the 2nd acidic proton as well. In fact, for option A, all of H2A is completely deprotonated to A2-, which means it is only basic, hence not a buffer at all). While for option B (which at 12cm3 of H2A against 20cm3 of OH-, leaving behind only a little bit of HA- left) is only effective at buffering against incoming strong acids, but almost totally useless at buffering against incoming strong bases. Thus options A and B are obviously wrong.
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AJC 2013 P3
qn 1aiv)
NAD+ is reduced to NADH, while ethanol is oxidised to ethanoate ions
shouldnt deltaG be -ve150kjmol-1 instead of +150kjmol-1 ?
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I got a dumb chem question to ask but its like a terrible itch I need to scratch.
how come eggs dont just dissolve in water if they got so many components that can form favourable interactions with water??
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Originally posted by theophilus:
I got a dumb chem question to ask but its like a terrible itch I need to scratch.
how come eggs dont just dissolve in water if they got so many components that can form favourable interactions with water??
Because those hydrophilic components are chemically bonded to components that are hydrophobic. -
Originally posted by Flying grenade:
AJC 2013 P3
qn 1aiv)
NAD+ is reduced to NADH, while ethanol is oxidised to ethanoate ions
shouldnt deltaG be -ve150kjmol-1 instead of +150kjmol-1 ?
+ not - -
Thanks !!
Also for practical what do they provide ah? like do they pass give you the QA results data booklet all ?