Source : http://frankcomment.blogspot.com/2009/11/how-singapore-appreciates-prc-talent.html All the talk about PRC scholars remind me of my classmate in NUS. He was born and bred in China. He had arrived earlier with his parents and attended JC and did well. Unlike the PRC scholars, he spoke decent English and he had no problem mingling with Singaporeans. In fact, he prefers to hangout with us than those fresh from China, even though he is very brilliant and very driven, and that made him more like the PRC scholars than the Singaporean slackers (like me). In fact, he was so brilliant he was the "go to" guy for PRC scholars who needed help with school work. And, even though he is strictly speaking a first generation PR, he served National Service. An exemplary foreign talent, don't you think? A model new citizen? And for all the hard work he put in, what did Singapore offer him? Well, Singapore offered to screw him. Since he was not recruited via "rigorous interviews" held in China, he was not entitled to apply for the PRC scholarship, even though he proved himself worthy by aceing the "A" levels. This is not the case, for example, for ASEAN scholarships. Irregardless of the route you took to enter NUS, as long as you are a non-Singaporean citizen of ASEAN, you can apply for the ASEAN scholarship. And either because his Singapore citizenship did not arrive fast enough, or because some scholarship boards referred him to the PRC scholarships, he did not manage to get any scholarship at all. I had trouble knowing the details because he was always eye-bulging mad when he talked about it. This is the stark reality of Singapore. Be a non-committal tourist like Zhang Yuan Yuan and enjoy the best of both worlds, or embrace Singapore and get screwed, like my friend. He was incredibly bitter about his situation throughout his four years in NUS, and immediately upon graduation, found a great job in US and never returned since. So, what is the lesson learnt here? Don't be stupid and serve NS, since Singapore will not appreciate it? The biggest benefit of the Singapore citizenship, ironically, is that it allows you to find a job and migrate to developed countries easily? Honestly, I cannot help but feel that our whole foreign talent policies are being run by people who are not thinking very far. Posted by Jimmy Mun at 8:30 PM Source : http://frankcomment.blogspot.com/2009/11/how-singapore-appreciates-prc-talent.html

Source : <a href= "http://frankcomment.blogspot.com/2009/11/how-singapore-appreciates-prc-talent.html" rel= "nofollow">http://frankcomment.blogspot.com/2009/11/how-si…</a> All the talk about PRC scholars remind me of my classmate in NUS. He was born and bred in China. He had arrived earlier with his parents and attended JC and did well. Unlike the PRC scholars, he spoke decent English and he had no problem mingling with Singaporeans. In fact, he prefers to hangout with us than those fresh from China, even though he is very brilliant and very driven, and that made him more like the PRC scholars than the Singaporean slackers (like me). In fact, he was so brilliant he was the “go to” guy for PRC scholars who needed help with school work. And, even though he is strictly speaking a first generation PR, he served National Service. An exemplary foreign talent, don’t you think? A model new citizen? And for all the hard work he put in, what did Singapore offer him? Well, Singapore offered to screw him. Since he was not recruited via “rigorous interviews” held in China, he was not entitled to apply for the PRC scholarship, even though he proved himself worthy by aceing the “A” levels. This is not the case, for example, for ASEAN scholarships. Irregardless of the route you took to enter NUS, as long as you are a non-Singaporean citizen of ASEAN, you can apply for the ASEAN scholarship. And either because his Singapore citizenship did not arrive fast enough, or because some scholarship boards referred him to the PRC scholarships, he did not manage to get any scholarship at all. I had trouble knowing the details because he was always eye-bulging mad when he talked about it. This is the stark reality of Singapore. Be a non-committal tourist like Zhang Yuan Yuan and enjoy the best of both worlds, or embrace Singapore and get screwed, like my friend. He was incredibly bitter about his situation throughout his four years in NUS, and immediately upon graduation, found a great job in US and never returned since. So, what is the lesson learnt here? Don’t be stupid and serve NS, since Singapore will not appreciate it? The biggest benefit of the Singapore citizenship, ironically, is that it allows you to find a job and migrate to developed countries easily? Honestly, I cannot help but feel that our whole foreign talent policies are being run by people who are not thinking very far. Posted by Jimmy Mun at 8:30 PM Source : <a href= "http://frankcomment.blogspot.com/2009/11/how-singapore-appreciates-prc-talent.html" rel= "nofollow">http://frankcomment.blogspot.com/2009/11/how-si…</a>

2009-11-20T11:12:13+08:00

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1 Yes, that's what my students said. Odd though, that this year Cambridge put electrochem only in P1.

Yes, that’s what my students said. Odd though, that this year Cambridge put electrochem only in P1.

2009-11-17T16:18:55+08:00

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2009-11-17T16:18:58+08:00

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1 For the next half an hour, revise your notes (CS Toh's 'A' Level Study Guide or your own JC lecture/tutorial notes) on (i) Inorganic Chemistry (Periodicity, Grp II, Grp VII, Transition Metals). (ii) ElectroChemistry. If you're stuck at any question, skip it and return to it later to save precious time. Sleep for approximately 7 hours. (ie. sleep at about 11pm, wake at 6am). Enjoy a delicious breakfast before you sit for your paper (sufficient glucose for your brain).

For the next half an hour, revise your notes (CS Toh’s ‘A’ Level Study Guide or your own JC lecture/tutorial notes) on (i) Inorganic Chemistry (Periodicity, Grp II, Grp VII, Transition Metals). (ii) ElectroChemistry. If you’re stuck at any question, skip it and return to it later to save precious time. Sleep for approximately 7 hours. (ie. sleep at about 11pm, wake at 6am). Enjoy a delicious breakfast before you sit for your paper (sufficient glucose for your brain).

2009-11-16T22:19:14+08:00

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1 Yes, that’s correct on both counts (N atom has 7 electrons in terms of an octet, and 4 electrons in terms of formal charge). If the question tasks you to draw NO2, it might also require you to draw N2O4 in the same qn. The unpaired electron in one NO2 molecule combines with another unpaired electron in another NO2 molecule to form N2O4.

Yes, that’s correct on both counts (N atom has 7 electrons in terms of an octet, and 4 electrons in terms of formal charge). If the question tasks you to draw NO2, it might also require you to draw N2O4 in the same qn. The unpaired electron in one NO2 molecule combines with another unpaired electron in another NO2 molecule to form N2O4.

2009-11-16T19:13:50+08:00

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2009-11-16T19:21:28+08:00

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2 A couple years ago, my student was just about the only one in her cohort who could correctly draw the structure of NO2 in a major exam. My point is (not about boasting, but about) most JC teachers don't focus (not blaming them though, they have it tough enough) on teaching the relevant, fundamental and crucial aspects of Chemistry that empower students to understand, appreciate and enjoy the Science as an Art. JC students most often fail to allow themselves to do the above (not blaming them though, they have it tough enough), and instead blindly memorize their way through JC. Specifically (for this question), I refer to the concept of formal charges and resonance (that is ill-taught in JCs), without an understand of these, drawing structures become senseless and hopeless. To answer your question : Central atom is N (with a +ve charge), single (and dative) bonded to an O atom (with a -ve charge), and double bonded to a (neutral) O atom. The N atom thus has 3 bond pairs and an unpaired electron, the (-ve charged) O atom has 3 lone pairs and 1 bond pair, the (neutral) O atom has 2 lone pairs and 2 bond pairs. The electron geometry is trigonal planar, the molecular geometry is v-shape or bent or non-linear. The resonance hybrid sees each of both N to O bonds as having bond lengths & bond strengths equivalent to 1.5 bonds. The charge on each O atom is an averaged -1/2 charge. For 'A' levels exams, it will suffice to draw a single resonance contributor.

A couple years ago, my student was just about the only one in her cohort who could correctly draw the structure of NO2 in a major exam. My point is (not about boasting, but about) most JC teachers don’t focus (not blaming them though, they have it tough enough) on teaching the relevant, fundamental and crucial aspects of Chemistry that empower students to understand, appreciate and enjoy the Science as an Art. JC students most often fail to allow themselves to do the above (not blaming them though, they have it tough enough), and instead blindly memorize their way through JC. Specifically (for this question), I refer to the concept of formal charges and resonance (that is ill-taught in JCs), without an understand of these, drawing structures become senseless and hopeless. To answer your question : Central atom is N (with a +ve charge), single (and dative) bonded to an O atom (with a -ve charge), and double bonded to a (neutral) O atom. The N atom thus has 3 bond pairs and an unpaired electron, the (-ve charged) O atom has 3 lone pairs and 1 bond pair, the (neutral) O atom has 2 lone pairs and 2 bond pairs. The electron geometry is trigonal planar, the molecular geometry is v-shape or bent or non-linear. The resonance hybrid sees each of both N to O bonds as having bond lengths & bond strengths equivalent to 1.5 bonds. The charge on each O atom is an averaged -1/2 charge. For ‘A’ levels exams, it will suffice to draw a single resonance contributor.

2009-11-16T17:50:56+08:00

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2009-11-16T17:50:59+08:00

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1 That's right. Hydrogen chloride gas would dissociate to ionize during hydrolysis forming hydrochloric acid.

That’s right. Hydrogen chloride gas would dissociate to ionize during hydrolysis forming hydrochloric acid.

2009-11-15T23:45:08+08:00

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1 Depending on the exact molarity, very dilute NaCl(aq) will see mainly water or hydroxide ions oxidized at the anode, rather than chloride ions.

Depending on the exact molarity, very dilute NaCl(aq) will see mainly water or hydroxide ions oxidized at the anode, rather than chloride ions.

2009-11-15T17:31:59+08:00

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2009-11-15T17:32:49+08:00

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2 2009 'A' Level Exam Qn. Methylbenzene requires concentrated sulfuric(VI) and nitric(V) acids and 30 degrees celsius to undergo nitration whereas phenol only requires dilute nitric acid. Give a reason for the difference in conditions and reagents. Solution : Phenolic hydroxy group is electron donating by resonance and is hence a strong activator, enriching the pi electron density of the benzene ring, making it a much stronger nucleophile (than ubsubstituted benzene). Hence only dilute nitric(V) acid is required for electrophilic aromatic substitution and nitration to occur. Methyl group is electron donating by induction (and hyperconjugation) and is hence a weak activator, only slightly enriching the pi electron density of the benzene ring, making it only a slightly stronger nucleophile (than ubsubstituted benzene). Hence a nitrating mixture of concentrated nitric(V) and sulfuric(VI) acids are required to generate the nitronium cation (NO2+) for electrophilic aromatic substitution and nitration to occur.

2009 ‘A’ Level Exam Qn. Methylbenzene requires concentrated sulfuric(VI) and nitric(V) acids and 30 degrees celsius to undergo nitration whereas phenol only requires dilute nitric acid. Give a reason for the difference in conditions and reagents. Solution : Phenolic hydroxy group is electron donating by resonance and is hence a strong activator, enriching the pi electron density of the benzene ring, making it a much stronger nucleophile (than ubsubstituted benzene). Hence only dilute nitric(V) acid is required for electrophilic aromatic substitution and nitration to occur. Methyl group is electron donating by induction (and hyperconjugation) and is hence a weak activator, only slightly enriching the pi electron density of the benzene ring, making it only a slightly stronger nucleophile (than ubsubstituted benzene). Hence a nitrating mixture of concentrated nitric(V) and sulfuric(VI) acids are required to generate the nitronium cation (NO2+) for electrophilic aromatic substitution and nitration to occur.

2009-11-15T13:27:26+08:00

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2009-11-15T13:27:30+08:00

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1 2009 'A' Level Exam Qn. Methylbenzene requires concentrated sulfuric(VI) and nitric(V) acids and 30 degrees celsius to undergo nitration whereas phenol only requires dilute nitric acid. Give a reason for the difference in conditions and reagents. Solution : Phenolic hydroxy group is electron donating by resonance and is hence a strong activator, enriching the pi electron density of the benzene ring, making it a much stronger nucleophile (than ubsubstituted benzene). Hence only dilute nitric(V) acid is required for electrophilic aromatic substitution and nitration to occur. Methyl group is electron donating by induction (and hyperconjugation) and is hence a weak activator, only slightly enriching the pi electron density of the benzene ring, making it only a slightly stronger nucleophile (than ubsubstituted benzene). Hence a nitrating mixture of concentrated nitric(V) and sulfuric(VI) acids are required to generate the nitronium cation (NO2+) for electrophilic aromatic substitution and nitration to occur.

2009-11-15T13:26:37+08:00

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2009-11-15T13:26:42+08:00

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1 2009 'A' Level Exam Qn. When methylbenzene is reacted with Cl2 and AlCl3, a monochloro compound K is formed. Treatment of K with more Cl2 in the presence of light produces compound L. When L is heated with NaCN in ethanol, compound M, C8H6ClN is formed. M can be converted into an acidic compound N by heating under reflux with dilute H2SO4. Heating L with NaOH(aq) produces compound P, C7H7ClO. When a mixture of N and P is heated with a small amount of concentrated H2SO4, compound Q, C15H12Cl2O2, is produced. Elucidate K to Q. State the type of each reaction described above. Solution : For questions such as these, where it is not clearly specified what they mean by "type of reaction", students are advised to give both the type of reaction in terms of the mechanism, and in terms of the end product. You will never be penalized for giving both, unless one of them is wrong and/or contradicts the other. To produce K, the reaction is : electrophilic aromatic substitution resulting in chlorination of benzene. To produce L, the reaction is : free radical substitution resulting in chlorination of alkyl side chain. To produce M, the reaction is : SN2 nucleophilic substitution (cyanide ion as nucleophile) resulting in the formation of a cyano compound. To produce N, the reaction is : repeated nucleophilic substitions during hydrolysis (water as nucleophile) to generate a carboxylic acid. To produce P, the reaction is : SN2 nucleophilic substitution (hydroxide ion as nucleophile) resulting in the formation of a primary alcohol. To produce Q, the reaction is : addition-elimination (not nucleophilic substitution) resulting in the removal of water (ie. condensation reaction) and the formation of an ester (ie. esterification reaction).

2009 ‘A’ Level Exam Qn. When methylbenzene is reacted with Cl2 and AlCl3, a monochloro compound K is formed. Treatment of K with more Cl2 in the presence of light produces compound L. When L is heated with NaCN in ethanol, compound M, C8H6ClN is formed. M can be converted into an acidic compound N by heating under reflux with dilute H2SO4. Heating L with NaOH(aq) produces compound P, C7H7ClO. When a mixture of N and P is heated with a small amount of concentrated H2SO4, compound Q, C15H12Cl2O2, is produced. Elucidate K to Q. State the type of each reaction described above. Solution : For questions such as these, where it is not clearly specified what they mean by “type of reaction”, students are advised to give both the type of reaction in terms of the mechanism, and in terms of the end product. You will never be penalized for giving both, unless one of them is wrong and/or contradicts the other. To produce K, the reaction is : electrophilic aromatic substitution resulting in chlorination of benzene. To produce L, the reaction is : free radical substitution resulting in chlorination of alkyl side chain. To produce M, the reaction is : SN2 nucleophilic substitution (cyanide ion as nucleophile) resulting in the formation of a cyano compound. To produce N, the reaction is : repeated nucleophilic substitions during hydrolysis (water as nucleophile) to generate a carboxylic acid. To produce P, the reaction is : SN2 nucleophilic substitution (hydroxide ion as nucleophile) resulting in the formation of a primary alcohol. To produce Q, the reaction is : addition-elimination (not nucleophilic substitution) resulting in the removal of water (ie. condensation reaction) and the formation of an ester (ie. esterification reaction).

2009-11-15T12:46:50+08:00

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2 In the exams, if the temperature is not stated, write both with qualifications. That is, don't just write the equations, but write the equations with the relevant temperatures clearly stated in each case.

In the exams, if the temperature is not stated, write both with qualifications. That is, don’t just write the equations, but write the equations with the relevant temperatures clearly stated in each case.

2009-11-15T12:31:08+08:00

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1 'A' & 'O' Level Qn. In the industrial electrolysis of brine (concentrated sodium chloride solution), a diaphragm cell is used. Write equations that occur at the cathode and anode. What are the ratio of useful products generated in this process? What is the purpose of the diaphram? Write equations for the undesirable reactions that would occur if the diaphram was absent. Solutions : Cl2, H2 and NaOH are produced in the molar ratio of 1:1:2. This industrial process is used to manufacture 3 different useful products of chlorine, hydrogen and sodium hydroxide. As to the purpose of the diaphragm : “In diaphragm cell electrolysis, an asbestos (or polymer-fiber) diaphragm separates a cathode and an anode, preventing the chlorine forming at the anode from re-mixing with the sodium hydroxide and the hydrogen formed at the cathode.” - http://en.wikipedia.org/wiki/Chlorine_production Hydrogen reacts with chlorine to form hydrogen chloride gas (which undergoes hydrolysis to form hydrochloric acid). Chlorine reacts with sodium hydroxide to form sodium chlorate(I) (at 15 deg C) and sodium chlorate(V) (at 70 deg C), and additionally reforms sodium chloride and water.

‘A’ & ‘O’ Level Qn. In the industrial electrolysis of brine (concentrated sodium chloride solution), a diaphragm cell is used. Write equations that occur at the cathode and anode. What are the ratio of useful products generated in this process? What is the purpose of the diaphram? Write equations for the undesirable reactions that would occur if the diaphram was absent. Solutions : Cl2, H2 and NaOH are produced in the molar ratio of 1:1:2. This industrial process is used to manufacture 3 different useful products of chlorine, hydrogen and sodium hydroxide. As to the purpose of the diaphragm : “In diaphragm cell electrolysis, an asbestos (or polymer-fiber) diaphragm separates a cathode and an anode, preventing the chlorine forming at the anode from re-mixing with the sodium hydroxide and the hydrogen formed at the cathode.” – <a href= "http://en.wikipedia.org/wiki/Chlorine_production" rel= "nofollow">http://en.wikipedia.org/wiki/Chlorine_production</a> Hydrogen reacts with chlorine to form hydrogen chloride gas (which undergoes hydrolysis to form hydrochloric acid). Chlorine reacts with sodium hydroxide to form sodium chlorate(I) (at 15 deg C) and sodium chlorate(V) (at 70 deg C), and additionally reforms sodium chloride and water.

2009-11-15T12:28:03+08:00

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1 The oxidation potential of Na is too positive (ie. the reduction potential of Na+ is too negative); hence Na+ will not be reduced to Na. Even in a concentrated solution, it is still a solution, and the solvent water molecules will be reduced at the cathode to generate hydrogen gas and hydroxide ions. Cl2, H2 and NaOH are produced in the molar ratio of 1:1:2. This industrial process is used to manufacture 3 different useful products of chlorine, hydrogen and sodium hydroxide. As to the purpose of the diaphragm : "In diaphragm cell electrolysis, an asbestos (or polymer-fiber) diaphragm separates a cathode and an anode, preventing the chlorine forming at the anode from re-mixing with the sodium hydroxide and the hydrogen formed at the cathode." http://en.wikipedia.org/wiki/Chlorine_production#Diaphragm_cell_electrolysis_.28Bipolar.29 You should also be able to write the equations for these undesirable reactions that would occur, should the diaphragm be absent, if the question asks for it.

The oxidation potential of Na is too positive (ie. the reduction potential of Na+ is too negative); hence Na+ will not be reduced to Na. Even in a concentrated solution, it is still a solution, and the solvent water molecules will be reduced at the cathode to generate hydrogen gas and hydroxide ions. Cl2, H2 and NaOH are produced in the molar ratio of 1:1:2. This industrial process is used to manufacture 3 different useful products of chlorine, hydrogen and sodium hydroxide. As to the purpose of the diaphragm : “In diaphragm cell electrolysis, an asbestos (or polymer-fiber) diaphragm separates a cathode and an anode, preventing the chlorine forming at the anode from re-mixing with the sodium hydroxide and the hydrogen formed at the cathode.” <a href= "http://en.wikipedia.org/wiki/Chlorine_production#Diaphragm_cell_electrolysis_" rel= "nofollow">http://en.wikipedia.org/wiki/Chlorine_productio…</a>.28Bipolar.29 You should also be able to write the equations for these undesirable reactions that would occur, should the diaphragm be absent, if the question asks for it.

2009-11-15T12:09:10+08:00

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1 That's coz 'O' levels are kept simple (but not necessarily accurately reflective of industrial practices in the real world). Since at 'O' levels, you're briefly exposed to the idea of alkane cracking (the chemistry of which is really considerably more complex than is implied at 'O' levels), which does generate small amounts of hydrogen, so at 'O' levels it's ok to say "cracking is used to generate hydrogen". Theoretically, you could crack all larger alkanes until you obtain methane, which is still the main industrial source for hydrogen. However, cracking is an inefficient source of hydrogen, and more feasible industrial methods of obtaining methane are used. See Wikipedia : http://en.wikipedia.org/wiki/Methane

That’s coz ‘O’ levels are kept simple (but not necessarily accurately reflective of industrial practices in the real world). Since at ‘O’ levels, you’re briefly exposed to the idea of alkane cracking (the chemistry of which is really considerably more complex than is implied at ‘O’ levels), which does generate small amounts of hydrogen, so at ‘O’ levels it’s ok to say “cracking is used to generate hydrogen”. Theoretically, you could crack all larger alkanes until you obtain methane, which is still the main industrial source for hydrogen. However, cracking is an inefficient source of hydrogen, and more feasible industrial methods of obtaining methane are used. See Wikipedia : <a href="http://en.wikipedia.org/wiki/Methane" rel= "nofollow">http://en.wikipedia.org/wiki/Methane</a>

2009-11-13T02:26:04+08:00

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1 You're right. Ignore the book's answer.

You’re right. Ignore the book’s answer.

2009-11-12T20:53:33+08:00

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1 Neither. Both are lousy methods. The standard industrial way to obtain hydrogen is from methane. See Wikipedia : http://en.wikipedia.org/wiki/Hydrogen_production However, for the 'O' and 'A' level syllabuses, which does not teach the abovestated industrial process, you could always offer electrolysis of water (eg. as aqueous sodium chloride) to generate the required hydrogen gas.

Neither. Both are lousy methods. The standard industrial way to obtain hydrogen is from methane. See Wikipedia : <a href="http://en.wikipedia.org/wiki/Hydrogen_production" rel= "nofollow">http://en.wikipedia.org/wiki/Hydrogen_production</a> However, for the ‘O’ and ‘A’ level syllabuses, which does not teach the abovestated industrial process, you could always offer electrolysis of water (eg. as aqueous sodium chloride) to generate the required hydrogen gas.

2009-11-12T17:30:39+08:00

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1 Q1. Reason why you're having difficulty with these, is because you need a minimum of 'A' levels or diploma to understand all the different types of isomers possible. Your teacher and the TYS qn are both correct, because they're talking about DIFFERENT types of isomers. For 'O' levels, any two compounds with the same molecular formula but different structural formula are called isomers. See Wikipedia for more info : http://en.wikipedia.org/wiki/Isomers Q2. Polyethene is a common plastic. Not all plastics consist of alkene monomrs. See Wikipedia for more info : http://en.wikipedia.org/wiki/Plastics Q3. You're right. Q4. The oxide ion of CaO will accept protons from water to form hydroxide ions, and also accept protons directly from ammonium cations to form ammonia gas. Q5. You're right. Q6. If you mean "only" when you say "completely", then of course it's not possible since all charges need to be balanced. Presence of counterbalancing ions (even if only spectator) are required. Q7. Molarity x Volume = No. of Moles. No. of moles x molar mass (of HYDRATED salt) = sample mass. Q8. Boron is a metalloid, and doesn't gain electrons to form anions. Q9. For a covalent compound changing from solid to liquid, or liquid to gas, you're overcoming ('breaking') intermolecular forces, not breaking covalent bonds (if you write 'bonds' the marker will assume you're referring to covalent bonds and you'll lose marks). And yes, it's still endothermic. In the other equation you gave, there is obviously both bond forming and bond breaking, so there's no way to tell for sure without further relevant enthalpy information. However, by considering the stability of the reactants vs products, you can make a reasonable deduction. CO2 and H2O are obviously very stable compounds, hence the reaction is likely exothermic.

Q1. Reason why you’re having difficulty with these, is because you need a minimum of ‘A’ levels or diploma to understand all the different types of isomers possible. Your teacher and the TYS qn are both correct, because they’re talking about DIFFERENT types of isomers. For ‘O’ levels, any two compounds with the same molecular formula but different structural formula are called isomers. See Wikipedia for more info : <a href="http://en.wikipedia.org/wiki/Isomers" rel= "nofollow">http://en.wikipedia.org/wiki/Isomers</a> Q2. Polyethene is a common plastic. Not all plastics consist of alkene monomrs. See Wikipedia for more info : <a href="http://en.wikipedia.org/wiki/Plastics" rel= "nofollow">http://en.wikipedia.org/wiki/Plastics</a> Q3. You’re right. Q4. The oxide ion of CaO will accept protons from water to form hydroxide ions, and also accept protons directly from ammonium cations to form ammonia gas. Q5. You’re right. Q6. If you mean “only” when you say “completely”, then of course it’s not possible since all charges need to be balanced. Presence of counterbalancing ions (even if only spectator) are required. Q7. Molarity x Volume = No. of Moles. No. of moles x molar mass (of HYDRATED salt) = sample mass. Q8. Boron is a metalloid, and doesn’t gain electrons to form anions. Q9. For a covalent compound changing from solid to liquid, or liquid to gas, you’re overcoming (‘breaking&#8217<img title="Wink" src= "/images/emoticons/classic/icon_wink.gif" alt="Wink" /> intermolecular forces, not breaking covalent bonds (if you write ‘bonds’ the marker will assume you’re referring to covalent bonds and you’ll lose marks). And yes, it’s still endothermic. In the other equation you gave, there is obviously both bond forming and bond breaking, so there’s no way to tell for sure without further relevant enthalpy information. However, by considering the stability of the reactants vs products, you can make a reasonable deduction. CO2 and H2O are obviously very stable compounds, hence the reaction is likely exothermic.

2009-11-12T13:39:41+08:00

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2009-11-12T13:39:45+08:00

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1 Q2 In the Hydrazine fuel cell, hydrazine is oxidized to molecular nitrogen at the anode (release electrons to the external wire), and molecular oxygen is reduced to water at the cathode. You'll need enough H+ present at the beginning, it's unrealistic and inefficient to expect the cathode to rely solely (eg. if you used deionized water) on the new H+ generated at the anode from the oxidation of hydrazine. Q5 Phenols are more acidic than alcohols, since the conjugate base (phenate or phenoxide ion) is stabilized by having the negative formal charge of the C6H5O- delocalized by resonance into the benzene ring. The conjugate bases of alcohols are alkoxide ions, which are particularly unstable (thus are strong bases and nucleophiles) because the alkyl group is electron donating by induction and hence intensifies and destabilizes the negative formal charge of the RO-. Q6 Yes, you need to memorize the disproportionation redox reactions of molecular chlorine with aqueous sodium hydroxide at 15 deg C (to generate sodium chloride, sodium chlorate(I), and water) and at 70 deg C (to generate sodium chloride, sodium chlorate(V), and water).

Q2 In the Hydrazine fuel cell, hydrazine is oxidized to molecular nitrogen at the anode (release electrons to the external wire), and molecular oxygen is reduced to water at the cathode. You’ll need enough H+ present at the beginning, it’s unrealistic and inefficient to expect the cathode to rely solely (eg. if you used deionized water) on the new H+ generated at the anode from the oxidation of hydrazine. Q5 Phenols are more acidic than alcohols, since the conjugate base (phenate or phenoxide ion) is stabilized by having the negative formal charge of the C6H5O- delocalized by resonance into the benzene ring. The conjugate bases of alcohols are alkoxide ions, which are particularly unstable (thus are strong bases and nucleophiles) because the alkyl group is electron donating by induction and hence intensifies and destabilizes the negative formal charge of the RO-. Q6 Yes, you need to memorize the disproportionation redox reactions of molecular chlorine with aqueous sodium hydroxide at 15 deg C (to generate sodium chloride, sodium chlorate(I), and water) and at 70 deg C (to generate sodium chloride, sodium chlorate(V), and water).

2009-11-09T11:18:09+08:00

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2009-11-09T11:18:14+08:00

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1 1) It's more worth it (more 'wu hua' instead of 'bo hua' in hokkien) for the universe to expand the 1st and 2nd endothermic ionization energies required to generate dipositive Grp II cations, because the resulting exothermic lattice energy compensates adequately and even profitably (ie. 'wu hua'). If the universe were to expand less energy ('cheapskate' efforts/investments yield 'cheapskate' results/profits) to ionize only to generate unipositive Grp II cations, the much smaller exothermic lattice energy that results either doesn't adequately compensate the endothermic ionization energy expanded in absolute values (ie. profits don't cover costs, therefore net loss), or doesn't compensate as much (as in the case of dipositive Grp II cations), (ie. why invest only $10 to make only $20 profit, when we can invest $20 and make $80 profit!). 2) Dinitration would occur, in the two positions para to the phenolic groups. Each phenol group donates strongly by resonance, so even if the nitro group withdraws electrons (by both induction and resonance) and hence deactivates the benzene ring (ie. making it a weaker nucleophile), but two phenol groups activates the benzene ring (ie. making the it a stronger nucleophile) sufficiently for dinitration to occur. Para positions (instead of ortho) because of steric hinderance. Not meta because phenol donates by resonance (further explanation lookup my Collection of Chem qns thread; not required at 'A' levels). 3) Amines (primary amines have less steric hinderance and thus are more effective nucleophiles) and phenols are both nucleophilic. Iodine is a good leaving group because iodide ions are very stable (due to electronegativity and large ionic radius to stabilize the negative formal/ionic charge). Hence SN2 nucleophilic substitution occurs (not SN1 because methyl carbocation is too unstable). 4) Sulfur dioxide reduces dichromate(VI) to chromium(III), and is itself oxidized to sulfate(VI). No other available ligands, so water functions as the ligands to generate the hexaaquochromium(III) ion. 5) You're using standard redox potentials. Obviously we're not gonna use standard molarities to carry out these redox reactions. So yes, bromide ions are oxidized to molecular bromine, and are able to reduce sulfate(VI) ions to sulfur dioxide (aka sulfur(IV) oxide). Notice the change in O.S. of sulfur from +6 to +4. In contrast, with iodide ions (larger ionic radius results in greater reducing power), the O.S. of sulfur is reduced from +6 all the way to -2 in hydrogen sulfide gas.

1) It’s more worth it (more ‘wu hua’ instead of ‘bo hua’ in hokkien) for the universe to expand the 1st and 2nd endothermic ionization energies required to generate dipositive Grp II cations, because the resulting exothermic lattice energy compensates adequately and even profitably (ie. ‘wu hua&#8217<img title= "Wink" src="/images/emoticons/classic/icon_wink.gif" alt="Wink" />. If the universe were to expand less energy (‘cheapskate’ efforts/investments yield ‘cheapskate’ results/profits) to ionize only to generate unipositive Grp II cations, the much smaller exothermic lattice energy that results either doesn’t adequately compensate the endothermic ionization energy expanded in absolute values (ie. profits don’t cover costs, therefore net loss), or doesn’t compensate as much (as in the case of dipositive Grp II cations), (ie. why invest only $10 to make only $20 profit, when we can invest $20 and make $80 profit!). 2) Dinitration would occur, in the two positions para to the phenolic groups. Each phenol group donates strongly by resonance, so even if the nitro group withdraws electrons (by both induction and resonance) and hence deactivates the benzene ring (ie. making it a weaker nucleophile), but two phenol groups activates the benzene ring (ie. making the it a stronger nucleophile) sufficiently for dinitration to occur. Para positions (instead of ortho) because of steric hinderance. Not meta because phenol donates by resonance (further explanation lookup my Collection of Chem qns thread; not required at ‘A’ levels). 3) Amines (primary amines have less steric hinderance and thus are more effective nucleophiles) and phenols are both nucleophilic. Iodine is a good leaving group because iodide ions are very stable (due to electronegativity and large ionic radius to stabilize the negative formal/ionic charge). Hence SN2 nucleophilic substitution occurs (not SN1 because methyl carbocation is too unstable). 4) Sulfur dioxide reduces dichromate(VI) to chromium(III), and is itself oxidized to sulfate(VI). No other available ligands, so water functions as the ligands to generate the hexaaquochromium(III) ion. 5) You’re using standard redox potentials. Obviously we’re not gonna use standard molarities to carry out these redox reactions. So yes, bromide ions are oxidized to molecular bromine, and are able to reduce sulfate(VI) ions to sulfur dioxide (aka sulfur(IV) oxide). Notice the change in O.S. of sulfur from +6 to +4. In contrast, with iodide ions (larger ionic radius results in greater reducing power), the O.S. of sulfur is reduced from +6 all the way to -2 in hydrogen sulfide gas.

2009-11-08T17:29:55+08:00

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2009-11-08T17:29:57+08:00

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1 Q2. Even if the two electrodes are compartmentalized, they need to be linked with a salt bridge, and an electrolyte capable of conducting electricity (in the form of mobile ions) still need to be present. If the two compartments were truly separate, charge would buildup and current would not flow. Q3. Correct. Use oxidation potentials together with reduction potentials (for any given redox reaction) to consider feasibility/spontaneity. Q4. The mechanism involves free radicals. It's "homogenous" because the catalyst is in the same state as the reactants. A possible catalyst for this reaction would be a halogen radical. Q5. Benzoic acid is most acidic, followed by para-methylphenol, followed by phenylmethanol. Explain using stabilities of conjugate bases after proton dissociation, and/or polarity/strength of conjugate base - proton bond before proton dissociation. Use electron donating/withdrawing by resonance/induction arguments. Q6. Effectively, disregarding spectator ions, the O.S. of chlorine changes from +5 and -1 on LHS, to 0 and +4 on RHS. So it is still considered disproportionation as chlorine is simultaneously reduced (from +5) and oxidized (from -1).

Q2. Even if the two electrodes are compartmentalized, they need to be linked with a salt bridge, and an electrolyte capable of conducting electricity (in the form of mobile ions) still need to be present. If the two compartments were truly separate, charge would buildup and current would not flow. Q3. Correct. Use oxidation potentials together with reduction potentials (for any given redox reaction) to consider feasibility/spontaneity. Q4. The mechanism involves free radicals. It’s “homogenous” because the catalyst is in the same state as the reactants. A possible catalyst for this reaction would be a halogen radical. Q5. Benzoic acid is most acidic, followed by para-methylphenol, followed by phenylmethanol. Explain using stabilities of conjugate bases after proton dissociation, and/or polarity/strength of conjugate base – proton bond before proton dissociation. Use electron donating/withdrawing by resonance/induction arguments. Q6. Effectively, disregarding spectator ions, the O.S. of chlorine changes from +5 and -1 on LHS, to 0 and +4 on RHS. So it is still considered disproportionation as chlorine is simultaneously reduced (from +5) and oxidized (from -1).

2009-11-08T16:54:44+08:00

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2009-11-08T16:54:47+08:00

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1 Wheref is correct. As for the other part of Qn : As predicted by Markonikov's Rule, the reaction pathway involving the more stable secondary carbocation intermediate resulting in the generation of 2-chloropropane, will be the major product. The reaction pathway involving the less stable primary carbocation intermediate resulting in the generation of 1-chloropropane, will be the minor product. As for the rest of the questions, I'll give my further comments later tonight (only if there is still any need for that, seeing how there are lots of helpful folks around).

Wheref is correct. As for the other part of Qn : As predicted by Markonikov’s Rule, the reaction pathway involving the more stable secondary carbocation intermediate resulting in the generation of 2-chloropropane, will be the major product. The reaction pathway involving the less stable primary carbocation intermediate resulting in the generation of 1-chloropropane, will be the minor product. As for the rest of the questions, I’ll give my further comments later tonight (only if there is still any need for that, seeing how there are lots of helpful folks around).

2009-11-08T10:53:37+08:00

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2009-11-08T10:53:40+08:00

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1 'A' Level Qn. 1. Solid NaCl and conc. H2SO4 can be used to generate HCl in situ for the reaction with alkenes to produce choloroalkanes. (a) Explain why 2-chloropropane is formed as the major product instead of 1-cholorpropane. (b) Will a high yield of 2-iodopropane be obtained by a similar method by replacing NaCl with NaI? explain your answer with appropriate equations. Solution : (a) As predicted by Markonikov's Rule, the reaction pathway involving the more stable secondary carbocation intermediate resulting in the generation of 2-chloropropane, will be the major product. The reaction pathway involving the less stable primary carbocation intermediate resulting in the generation of 1-chloropropane, will be the minor product. (b) A redox reaction occurs between the iodide ions and sulfuric(VI) acid : H2SO4 + NaI ---> NaHSO4 + HI 8HI + H2SO4 --> 4I2 + H2S + 4H2O Hence, a lower yield of 2-iodopropane will result. (Notice that the O.S. of sulfur decreases from +6 to -2, due to the stronger reducing power of iodide ions. If bromide ions were used instead (eg. H2SO4 + NaBr), the O.S. of sulfur decreases only from +6 (H2SO4) to +4 (SO2). The difference in reducing powers of the various halide ions, indicated by their different redox potentials in the Data Booklet (quote these values in the exam!), is explained by the difference of the distance between the positively charged nucleus and the valence shell, for chloride vs bromide vs iodide.)

‘A’ Level Qn. 1. Solid NaCl and conc. H2SO4 can be used to generate HCl in situ for the reaction with alkenes to produce choloroalkanes. (a) Explain why 2-chloropropane is formed as the major product instead of 1-cholorpropane. (b) Will a high yield of 2-iodopropane be obtained by a similar method by replacing NaCl with NaI? explain your answer with appropriate equations. Solution : (a) As predicted by Markonikov’s Rule, the reaction pathway involving the more stable secondary carbocation intermediate resulting in the generation of 2-chloropropane, will be the major product. The reaction pathway involving the less stable primary carbocation intermediate resulting in the generation of 1-chloropropane, will be the minor product. (b) A redox reaction occurs between the iodide ions and sulfuric(VI) acid : H2SO4 + NaI <del>—> NaHSO4 + HI 8HI + H2SO4 -</del>> 4I2 + H2S + 4H2O Hence, a lower yield of 2-iodopropane will result. (Notice that the O.S. of sulfur decreases from +6 to -2, due to the stronger reducing power of iodide ions. If bromide ions were used instead (eg. H2SO4 + NaBr), the O.S. of sulfur decreases only from +6 (H2SO4) to +4 (SO2). The difference in reducing powers of the various halide ions, indicated by their different redox potentials in the Data Booklet (quote these values in the exam!), is explained by the difference of the distance between the positively charged nucleus and the valence shell, for chloride vs bromide vs iodide.)

2009-11-08T10:50:17+08:00

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2009-11-08T10:50:20+08:00

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1 Take both Chemistry and Physics, since a background in both are required for the majority of University Science courses, including Engineering courses. If you take only one but not both, you will be limiting yourself very severely in the University courses you can apply for.

Take both Chemistry and Physics, since a background in both are required for the majority of University Science courses, including Engineering courses. If you take only one but not both, you will be limiting yourself very severely in the University courses you can apply for.

2009-11-06T16:29:57+08:00

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2009-11-06T16:30:00+08:00

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1 Note the time of Coolplay's post above. Notice that 6 hours 11 minutes earlier, SBS n SMRT had precognitively predicted Coolplay's post, and had posted the required hints to Coolplay's problem, within my "Collection of Chemistry Qns" thread (scroll to the bottom). http://www.sgforums.com/forums/2297/topics/320107?page=8#posts-9467798 If anyone wants to know what happens to the world in 2012, ask SBS n SMRT to share his prophetic vision of the future.

Note the time of Coolplay’s post above. Notice that 6 hours 11 minutes earlier, SBS n SMRT had precognitively predicted Coolplay’s post, and had posted the required hints to Coolplay’s problem, within my “Collection of Chemistry Qns” thread (scroll to the bottom). <a href= "http://www.sgforums.com/forums/2297/topics/320107?page=8#posts-9467798" rel= "nofollow">http://www.sgforums.com/forums/2297/topics/3201…</a> If anyone wants to know what happens to the world in 2012, ask SBS n SMRT to share his prophetic vision of the future.

2009-11-05T23:55:13+08:00

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2009-11-05T23:55:16+08:00

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1 'A' Level Qn. Explain the differences in the stereostructures of the complex ions formed for the cobalt(II) ion when complexed with water ligands, versus with chloride ligands. Solution : The hexahydroxocobalt(II) ion has an octahedral stereostructure, while the tetrachlorocobaltate(II) ion has a tetrahedral stereostructure. This is due to the steric considerations : the O atoms (of water ligands) are significantly smaller than the Cl- ion ligands; hence (six) water ligands can pack closer to the Co2+ ion (to give an octahedral stereostructure), while there is not enough space to pack six Cl- ion ligands to give a similar octahedral structure.

‘A’ Level Qn. Explain the differences in the stereostructures of the complex ions formed for the cobalt(II) ion when complexed with water ligands, versus with chloride ligands. Solution : The hexahydroxocobalt(II) ion has an octahedral stereostructure, while the tetrachlorocobaltate(II) ion has a tetrahedral stereostructure. This is due to the steric considerations : the O atoms (of water ligands) are significantly smaller than the Cl- ion ligands; hence (six) water ligands can pack closer to the Co2+ ion (to give an octahedral stereostructure), while there is not enough space to pack six Cl- ion ligands to give a similar octahedral structure.

2009-11-05T16:20:18+08:00

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2009-11-05T16:20:26+08:00

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