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

2297

9469421

10769560

381008

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

124947

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

2297

9468602

10768112

380953

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

124947

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

2297

9467796

10766793

320107

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

124947

1 Both are acceptable at 'O' and 'A' level exams. However, for the sake of technical accuracy, it makes better chemistry sense to use water as the species oxidized (and/or reduced) in aqueous solutions, as it's molarity far exceeds that of OH- ions (or H+ ions). Unless the solution is acidic (then H+ rather than water, is reduced at the cathode) or alkaline (then OH- rather than water, is oxidized at the anode).

Both are acceptable at ‘O’ and ‘A’ level exams. However, for the sake of technical accuracy, it makes better chemistry sense to use water as the species oxidized (and/or reduced) in aqueous solutions, as it’s molarity far exceeds that of OH- ions (or H+ ions). Unless the solution is acidic (then H+ rather than water, is reduced at the cathode) or alkaline (then OH- rather than water, is oxidized at the anode).

2009-11-02T19:21:03+08:00

2297

9462021

10757632

380623

2009-11-02T19:21:07+08:00

124947

1 No. Sulfate(VI) ions are not reduced. Protons from the acid are reduced to hydrogen gas.

No. Sulfate(VI) ions are not reduced. Protons from the acid are reduced to hydrogen gas.

2009-11-02T19:14:30+08:00

2297

9462011

10757618

380622

2009-11-02T19:14:33+08:00

124947

1 A layer of insoluble BaSO4(s) coats the Ba metal strips, preventing complete reaction between the acid and the metal.

A layer of insoluble BaSO4(s) coats the Ba metal strips, preventing complete reaction between the acid and the metal.

2009-11-02T18:39:27+08:00

2297

9461947

10757520

380622

2009-11-02T18:39:30+08:00

124947

1 To be precise, SiO2(s) doesn't react with aqueous alkalis, but reacts only with fused/liquid/molten alkalis at high temperatures.

To be precise, SiO2(s) doesn’t react with aqueous alkalis, but reacts only with fused/liquid/molten alkalis at high temperatures.

2009-11-01T15:35:03+08:00

2297

9459332

10753457

380437

2009-11-01T15:35:07+08:00

124947

1 Correct. There are actually many factors involved in determining number of ligands, geometry of the complex ions, and stability of the complx ions, many of which are beyond the requirements or understanding of the 'A' level syllabus. The chemistry of complex ions, can be... complex. You have availability of energetically accessible low-lying vacant orbitals of the transition metal ions to consider. You have thermodynamic and redox stability to consider, including enthalpy and entropy. You have size (sterics) of ligands to consider. You have ('invisible') water ligands to consider. You have co-ordination bond lengths and electron geometries to consider. You have crystal field splitting theory to consider. You have ligand exchange/substitution/displacement/replacement to consider. You have Lewis-base strength and crystal field splitting strength of the ligands to consider. Etc etc. For instance, on the matter of 'invisible' water ligands, the deep blue tetraaminecopper(II) ions are actually tetraaminediaquocopper(II) ions, hence making their (complex)ionic geometry square planar rather than tetrahedral. In contrast, other complex ions with ammonia ligands are linear (eg. diammine silver(I) ions) or octahedral (eg. hexaaminenickel(II) ions) with no water ligands. In some cases, there is ambiguity and even disagreement among chemists on the number of ligands or their geometry, for some complex ions. The Cambridge Mark Scheme should be fair (but this does differ from year to year, depending on the question and mark scheme authors), in that for instance, if only 4 ligands are indicated for a relatively obscure complex ion, both "tetrahedral" and "square planar" may be acceptable. For 'A' levels, it's more practical and convenient to simply be familiar (ie. memorize) a list of common transition metal complex ions and their geometries, rather than go into greater depth on the subject than is required by the 'A' level syllabus. For this topic on Complex Ions : 1) Jim Clark's website is excellent and thoroughly suffices for 'A' levels. http://www.chemguide.co.uk/inorganic/complexmenu.html#top 2) For users of CS Toh's 'A' Level Study Guide, his 'Advanced' Study Guide (a thicker version covering multiple Chemistry syllabuses including H2 Singapore 'A' levels, International Baccalaurate Chemistry, and other non-local 'A' level syllabuses) is actually more recommended, specifically or only for this topic on Complex Ions. (For all other topics, the original 'A' Level Study Guide will suffice).

Correct. There are actually many factors involved in determining number of ligands, geometry of the complex ions, and stability of the complx ions, many of which are beyond the requirements or understanding of the ‘A’ level syllabus. The chemistry of complex ions, can be… complex. You have availability of energetically accessible low-lying vacant orbitals of the transition metal ions to consider. You have thermodynamic and redox stability to consider, including enthalpy and entropy. You have size (sterics) of ligands to consider. You have (‘invisible&#8217<img title="Wink" src= "/images/emoticons/classic/icon_wink.gif" alt="Wink" /> water ligands to consider. You have co-ordination bond lengths and electron geometries to consider. You have crystal field splitting theory to consider. You have ligand exchange/substitution/displacement/replacement to consider. You have Lewis-base strength and crystal field splitting strength of the ligands to consider. Etc etc. For instance, on the matter of ‘invisible’ water ligands, the deep blue tetraaminecopper(II) ions are actually tetraaminediaquocopper(II) ions, hence making their (complex)ionic geometry square planar rather than tetrahedral. In contrast, other complex ions with ammonia ligands are linear (eg. diammine silver(I) ions) or octahedral (eg. hexaaminenickel(II) ions) with no water ligands. In some cases, there is ambiguity and even disagreement among chemists on the number of ligands or their geometry, for some complex ions. The Cambridge Mark Scheme should be fair (but this does differ from year to year, depending on the question and mark scheme authors), in that for instance, if only 4 ligands are indicated for a relatively obscure complex ion, both “tetrahedral” and “square planar” may be acceptable. For ‘A’ levels, it’s more practical and convenient to simply be familiar (ie. memorize) a list of common transition metal complex ions and their geometries, rather than go into greater depth on the subject than is required by the ‘A’ level syllabus. For this topic on Complex Ions : 1) Jim Clark’s website is excellent and thoroughly suffices for ‘A’ levels. <a href= "http://www.chemguide.co.uk/inorganic/complexmenu.html#top" rel= "nofollow">http://www.chemguide.co.uk/inorganic/complexmen…</a> 2) For users of CS Toh’s ‘A’ Level Study Guide, his ‘Advanced’ Study Guide (a thicker version covering multiple Chemistry syllabuses including H2 Singapore ‘A’ levels, International Baccalaurate Chemistry, and other non-local ‘A’ level syllabuses) is actually more recommended, specifically or only for this topic on Complex Ions. (For all other topics, the original ‘A’ Level Study Guide will suffice).

2009-11-01T10:24:02+08:00

2297

9458857

10752682

379808

2009-11-01T10:24:05+08:00

124947

1 http://thestar.com.my/columnists/story.asp?col=insightdownsouth&file=/2009/10/31/columnists/insightdownsouth/5013452 Saturday October 31, 2009 Recession hits Singapore unevenly Insight Down South By SEAH CHIANG NEE While the rich few rejoice, many Singaporeans are making do quietly. YOUNG Singaporeans, who were raised in an era of affluence, have been indulging in a spending binge that appears out of line with economic realities. The splurge, which followed signs of a mild recovery in recent weeks, resulted in a strong price run-up in stocks, properties and cars, taking many economists by surprise. It was so strong that people were ironically fearful of an asset bubble building up during a weak economy. This apparent over-indulgence appears to ignore repeated warnings from government leaders and economists that more job cuts are in store and the recession could return. One of its sovereign wealth funds, Temasek Holdings, said that as far as it was concerned, Singapore is still in crisis. Despite these, expensive restaurants are once again packed with weekend diners and private clubs, once quiet, are again buzzing. Most of the big spenders are young professionals, married and singles, and the wealthy. Their buying has caused prices of resale public flats and private condominiums to soar. And despite the downturn, car usage in Singapore – one of the costliest in the world – has risen at the expense of public transport. Analysts have, however, pointed out that the consumer splashing is unlikely to last and is only one aspect of life in a recession. It is confined largely to the upper-middle class and irrationally exuberant professionals, who appear unfazed by the severity of it all. Growing up in a golden era with years of news screaming about more good times ahead, many Singaporeans seem oblivious to their country’s vulnerability to world turmoil. The bigger story is of a struggling middle class (some two-thirds of the population) that is too badly affected to be able to buy luxury items. When I mentioned it to an old friend, a businessman and former human resource manager, he said the wider picture is different. For the majority of workers, jobs have been lost and take-home pay diminished. Almost all Singaporeans, rich and poor, have lost out in the recession, the worst being the poorer class. “I admire these people very much. Mostly old and little skilled, they struggle on silently. No time to talk about their plight, just carry on working,” he said. During the past two years, almost every Singaporean had to dip into his own savings to sustain himself, like the government did with its reserves in an effort to protect jobs. The picture is different for the rich, whose number has been growing substantially through immigration. The crisis has decimated fortunes, but the bulk of high-asset owners have enough financial muscle (again like the country itself) to ride out the storm or even prosper from it. It is largely the spending habit of this group that fuelled the recent indulgence. Years later, if writers looked back at the current severe downturn to ask what lasting impact the global crisis had on this society, one answer would be the erosion of the middle class. The trend was first detected in Japan, and to a lesser extent in Hong Kong and Taiwan, as these middle-class societies prospered. The theory, known as the M-shaped society, was enunciated by Japanese strategist Kenichi Ohmae. He observed that in Japan’s “M-shape” class distribution, very few middle-class people may climb up the ladder into the upper class, while the others gradually sank to the lower classes. These people suffered a deterioration in living standards, faced the threat of unemployment, or their average salary was dropping, he said. Gradually, they could only live the way the lower classes lived: taking the bus instead of driving their own car, cutting their budget for meals instead of dining at better restaurants, and spending less on consumer goods. Kenichi said all this might take place while the economy enjoyed remarkable growth and overall wages rose. However, the wealth increase may concentrate in the pockets of the very few rich people in society. The masses cannot benefit from the growth, and their living standard goes into decline. For many middle-class Singaporeans, these sound uncomfortably like home. The government, which relies on middle-class voters to keep itself in power, has vowed to make the closure of the economic gap a national priority. It is a doubly tough job given the economic crisis which is widening – rather than narrowing – the differences. Minister Mentor Lee Kuan Yew seems to find this gap an inevitable feature here. Singapore has the second-highest income gap with a Gini score of 42.5 among developed economies after Hong Kong, according to the UN Development Programme report, (The Gini Coefficient index measures the income gap between the poor and the rich in any country with zero denoting absolute equality.) Lee was speaking at a forum with undergraduates when he rejected a minimum wage for workers to narrow this gap, saying it was more important to keep jobs. “Never mind your Gini coefficient,” he said. “If you don’t have a job you get zero against those with jobs.” In other words it is better to have a job with lower pay than no job. Such remarks would obviously be more acceptable, albeit grudgingly, to the previous generation of poorer citizens than the current one. It was such logic that helped turn Singapore into the richest country in South-east Asia, with a per capita GDP income rising steadily in four decades to S$53,192 (RM130,048) in 2008. But in today’s high-cost city with Singaporeans finding it harder and harder to cope with the crisis, his words have neither helped nor dispelled many concerns. http://thestar.com.my/columnists/story.asp?col=insightdownsouth&file=/2009/10/31/columnists/insightdownsouth/5013452

<a href= "http://thestar.com.my/columnists/story.asp?col=insightdownsouth&#38;file=" rel= "nofollow">http://thestar.com.my/columnists/story.asp?col=...</a>/2009/10/31/columnists/insightdownsouth/5013452 Saturday October 31, 2009 Recession hits Singapore unevenly Insight Down South By SEAH CHIANG NEE While the rich few rejoice, many Singaporeans are making do quietly. YOUNG Singaporeans, who were raised in an era of affluence, have been indulging in a spending binge that appears out of line with economic realities. The splurge, which followed signs of a mild recovery in recent weeks, resulted in a strong price run-up in stocks, properties and cars, taking many economists by surprise. It was so strong that people were ironically fearful of an asset bubble building up during a weak economy. This apparent over-indulgence appears to ignore repeated warnings from government leaders and economists that more job cuts are in store and the recession could return. One of its sovereign wealth funds, Temasek Holdings, said that as far as it was concerned, Singapore is still in crisis. Despite these, expensive restaurants are once again packed with weekend diners and private clubs, once quiet, are again buzzing. Most of the big spenders are young professionals, married and singles, and the wealthy. Their buying has caused prices of resale public flats and private condominiums to soar. And despite the downturn, car usage in Singapore – one of the costliest in the world – has risen at the expense of public transport. Analysts have, however, pointed out that the consumer splashing is unlikely to last and is only one aspect of life in a recession. It is confined largely to the upper-middle class and irrationally exuberant professionals, who appear unfazed by the severity of it all. Growing up in a golden era with years of news screaming about more good times ahead, many Singaporeans seem oblivious to their country’s vulnerability to world turmoil. The bigger story is of a struggling middle class (some two-thirds of the population) that is too badly affected to be able to buy luxury items. When I mentioned it to an old friend, a businessman and former human resource manager, he said the wider picture is different. For the majority of workers, jobs have been lost and take-home pay diminished. Almost all Singaporeans, rich and poor, have lost out in the recession, the worst being the poorer class. “I admire these people very much. Mostly old and little skilled, they struggle on silently. No time to talk about their plight, just carry on working,” he said. During the past two years, almost every Singaporean had to dip into his own savings to sustain himself, like the government did with its reserves in an effort to protect jobs. The picture is different for the rich, whose number has been growing substantially through immigration. The crisis has decimated fortunes, but the bulk of high-asset owners have enough financial muscle (again like the country itself) to ride out the storm or even prosper from it. It is largely the spending habit of this group that fuelled the recent indulgence. Years later, if writers looked back at the current severe downturn to ask what lasting impact the global crisis had on this society, one answer would be the erosion of the middle class. The trend was first detected in Japan, and to a lesser extent in Hong Kong and Taiwan, as these middle-class societies prospered. The theory, known as the M-shaped society, was enunciated by Japanese strategist Kenichi Ohmae. He observed that in Japan’s “M-shape” class distribution, very few middle-class people may climb up the ladder into the upper class, while the others gradually sank to the lower classes. These people suffered a deterioration in living standards, faced the threat of unemployment, or their average salary was dropping, he said. Gradually, they could only live the way the lower classes lived: taking the bus instead of driving their own car, cutting their budget for meals instead of dining at better restaurants, and spending less on consumer goods. Kenichi said all this might take place while the economy enjoyed remarkable growth and overall wages rose. However, the wealth increase may concentrate in the pockets of the very few rich people in society. The masses cannot benefit from the growth, and their living standard goes into decline. For many middle-class Singaporeans, these sound uncomfortably like home. The government, which relies on middle-class voters to keep itself in power, has vowed to make the closure of the economic gap a national priority. It is a doubly tough job given the economic crisis which is widening – rather than narrowing – the differences. Minister Mentor Lee Kuan Yew seems to find this gap an inevitable feature here. Singapore has the second-highest income gap with a Gini score of 42.5 among developed economies after Hong Kong, according to the UN Development Programme report, (The Gini Coefficient index measures the income gap between the poor and the rich in any country with zero denoting absolute equality.) Lee was speaking at a forum with undergraduates when he rejected a minimum wage for workers to narrow this gap, saying it was more important to keep jobs. “Never mind your Gini coefficient,” he said. “If you don’t have a job you get zero against those with jobs.” In other words it is better to have a job with lower pay than no job. Such remarks would obviously be more acceptable, albeit grudgingly, to the previous generation of poorer citizens than the current one. It was such logic that helped turn Singapore into the richest country in South-east Asia, with a per capita GDP income rising steadily in four decades to S$53,192 (RM130,04<img title="Cool" src= "/images/emoticons/classic/icon_cool.gif" alt="Cool" /> in 2008. But in today’s high-cost city with Singaporeans finding it harder and harder to cope with the crisis, his words have neither helped nor dispelled many concerns. <a href= "http://thestar.com.my/columnists/story.asp?col=insightdownsouth&#38;file=" rel= "nofollow">http://thestar.com.my/columnists/story.asp?col=...</a>/2009/10/31/columnists/insightdownsouth/5013452

2009-10-31T13:24:55+08:00

2297

9456786

10749575

259804

2009-10-31T13:25:04+08:00

124947

1 If you want to take Science in JC or Poly, you should take Pure Science, to be precise - Pure Chemistry and Pure Physics. Biology isn't anywhere as important at any level - Secondary, JC and University.

If you want to take Science in JC or Poly, you should take Pure Science, to be precise – Pure Chemistry and Pure Physics. Biology isn’t anywhere as important at any level – Secondary, JC and University.

2009-10-30T00:46:41+08:00

2297

9454009

10745304

380263

2009-10-30T00:46:44+08:00

124947

1 China, Shanghai. Girl gangster beats up girl victim. The fact that the girl victim (and her boyfriend) didn't dare to resist the violent beating at all, tells a lot about the terror she/they must have been suffering : http://www.youtube.com/watch?v=dCyDAxBdxIw

China, Shanghai. Girl gangster beats up girl victim. The fact that the girl victim (and her boyfriend) didn’t dare to resist the violent beating at all, tells a lot about the terror she/they must have been suffering : <a href="http://www.youtube.com/watch?v=dCyDAxBdxIw" rel= "nofollow">http://www.youtube.com/watch?v=dCyDAxBdxIw</a>

2009-10-28T22:15:38+08:00

2297

9451569

10741522

259804

2009-10-28T22:16:27+08:00

124947

2 Yes, the nature of chromophores (the structure responsible for colour) arise primarily due to metal complexes (crystal field splitting of d orbitals) or conjugated pi systems (for organic compounds with alternating single and multiple bonds). The latter is an example of how phenolphthalein and many pH indicators work, and does not involve metal complexes.

Yes, the nature of chromophores (the structure responsible for colour) arise primarily due to metal complexes (crystal field splitting of d orbitals) or conjugated pi systems (for organic compounds with alternating single and multiple bonds). The latter is an example of how phenolphthalein and many pH indicators work, and does not involve metal complexes.

2009-10-25T09:55:18+08:00

2297

9445145

10730915

379808

2009-10-25T09:55:21+08:00

124947

1 Yes, it works too if bromine is used, but the iodine is preferred as triiodomethane product is less soluble (and is hence a more easily detected precipitate) than tribromomethane. Heating will increase the rate of reaction and is routinely carried out for most electrophilic aromatic substitutions; but because room temperature will suffice for halogenation in the presence of a Lewis acid catalyst, you're required to state "room temperature" in the exams to demonstrate that you're aware room temperature is indeed sufficient (unlike other reactions in which heat is mandatorily required). And as far as 'A' levels (and equivalent, eg. International Baccalaurate) are concerned, heating won't result in di or tri-substitution of the benzene ring. Notice that each halogen added on further deactivates the benzene ring (as halogens withdraw electrons inductively more significantly than donate electrons by resonance). If di or tri-substitution is desired, you will need to first place an activator (a substitutent that donates electrons by resonance) onto the benzene ring. Molecular bromine, like all molecular halogens, are only slightly soluble in water. Indeed, so-called "aqueous bromine", similarly to "aqueous iodine", are not actually molecular, but exist in complexed forms, most notably X3- (eg. Br3- and I3-). But because the complexed and the molecular forms exist in equilibrium, it's fine (for 'A' levels and equivalent) to write as "Br2(aq)" or "I2(aq)". Think of the complexed form, eg. I3-(aq), as a "temporary delivery agent" for the I2 molecule in aqueous conditions.

Yes, it works too if bromine is used, but the iodine is preferred as triiodomethane product is less soluble (and is hence a more easily detected precipitate) than tribromomethane. Heating will increase the rate of reaction and is routinely carried out for most electrophilic aromatic substitutions; but because room temperature will suffice for halogenation in the presence of a Lewis acid catalyst, you’re required to state “room temperature” in the exams to demonstrate that you’re aware room temperature is indeed sufficient (unlike other reactions in which heat is mandatorily required). And as far as ‘A’ levels (and equivalent, eg. International Baccalaurate) are concerned, heating won’t result in di or tri-substitution of the benzene ring. Notice that each halogen added on further deactivates the benzene ring (as halogens withdraw electrons inductively more significantly than donate electrons by resonance). If di or tri-substitution is desired, you will need to first place an activator (a substitutent that donates electrons by resonance) onto the benzene ring. Molecular bromine, like all molecular halogens, are only slightly soluble in water. Indeed, so-called “aqueous bromine”, similarly to “aqueous iodine”, are not actually molecular, but exist in complexed forms, most notably X3- (eg. Br3- and I3-). But because the complexed and the molecular forms exist in equilibrium, it’s fine (for ‘A’ levels and equivalent) to write as “Br2(aq)” or “I2(aq)”. Think of the complexed form, eg. I3-(aq), as a “temporary delivery agent” for the I2 molecule in aqueous conditions.

2009-10-25T03:14:07+08:00

2297

9444992

10730652

379808

2009-10-25T03:14:11+08:00

124947

1 Disubstituted carbon takes priority over monosubstituted carbons, as far as the name with the lower numbering is preferred. So 1,1 & 3 is preferred over 1 & 3,3. IUPAC naming allows for a certain measure of flexibility (eg. phenylamine = aniline, methylbenzene = tolune, etc), because senseless dogma for the senseless purpose of senseless dogma aside, see the bigger picture - the IUPAC rules of nomenclature were designed to help, not hinder, humanity; as long as non-ambiguity is achieved, that's the entire point for the rules.

Disubstituted carbon takes priority over monosubstituted carbons, as far as the name with the lower numbering is preferred. So 1,1 & 3 is preferred over 1 & 3,3. IUPAC naming allows for a certain measure of flexibility (eg. phenylamine = aniline, methylbenzene = tolune, etc), because senseless dogma for the senseless purpose of senseless dogma aside, see the bigger picture – the IUPAC rules of nomenclature were designed to help, not hinder, humanity; as long as non-ambiguity is achieved, that’s the entire point for the rules.

2009-10-25T01:09:57+08:00

2297

9444802

10730370

379761

2009-10-25T01:10:00+08:00

124947

1 Since all sessions are held at my place (Bedok / Marine Parade, TJC / VJC area), so it's not feasible for Lord dejavu. Perhaps dkcx can tutor him?

Since all sessions are held at my place (Bedok / Marine Parade, TJC / VJC area), so it’s not feasible for Lord dejavu. Perhaps dkcx can tutor him?

2009-10-24T21:03:09+08:00

2297

9444001

10729264

379775

2009-10-24T21:03:12+08:00

124947

1 With regards to Le Chatelier's principle, the more positive the redox potential (could be oxidation or reduction potential, depending on whether an oxidation or reduction half-equation is written next to the redox potential value), the more the position of equilibrium lies to the right. The more negative the redox potential, the more the position of equilibrium lies to the left. Hence, adding LHS species will result in the reduction/oxidation potential becoming more positive / less negative (as position of equilibrium shifts to the right); and adding RHS species will result in the reduction/oxidation potential becoming less positive / more negative (as position of equilibrium shifts to the left).

With regards to Le Chatelier’s principle, the more positive the redox potential (could be oxidation or reduction potential, depending on whether an oxidation or reduction half-equation is written next to the redox potential value), the more the position of equilibrium lies to the right. The more negative the redox potential, the more the position of equilibrium lies to the left. Hence, adding LHS species will result in the reduction/oxidation potential becoming more positive / less negative (as position of equilibrium shifts to the right); and adding RHS species will result in the reduction/oxidation potential becoming less positive / more negative (as position of equilibrium shifts to the left).

2009-10-23T00:03:58+08:00

2297

9441184

10724614

379474

2009-10-23T00:04:01+08:00

124947

1 Yes, 'O' level students should be able to give a reasonable answer to this question, which to be fair, already comes with hints. When oxygen is blown over the molten iron containing carbon impurities, these carbon impurities are combusted (and combustion is obviously exothermic) away in the form of carbon dioxide. This is a reasonable deduction 'O' level students should be able to make. It's not 'essential knowledge' in the sense that it's found in all 'O' level textbooks or 'O' level notes, and 'O' level students must memorize about it; because much (or ideally, all) of chemistry, including this question, is all about (chemistry-type) common sense and reasonable deductions, not blind memorization.

Yes, ‘O’ level students should be able to give a reasonable answer to this question, which to be fair, already comes with hints. When oxygen is blown over the molten iron containing carbon impurities, these carbon impurities are combusted (and combustion is obviously exothermic) away in the form of carbon dioxide. This is a reasonable deduction ‘O’ level students should be able to make. It’s not ‘essential knowledge’ in the sense that it’s found in all ‘O’ level textbooks or ‘O’ level notes, and ‘O’ level students must memorize about it; because much (or ideally, all) of chemistry, including this question, is all about (chemistry-type) common sense and reasonable deductions, not blind memorization.

2009-10-21T23:08:34+08:00

2297

9439377

10721604

379146

2009-10-21T23:08:37+08:00

124947

1 ~ Isn't it always preferred to just use reduction potential for both? Even at uni level we do not use the oxidation potential for any calculations. ~ Not true, actually. It's only a matter of convention (ie. standardization for the sake of convenience, not because it's more 'correct') that reduction potentials are more often used. Conceptually, oxidation and reduction potentials, are every bit as valid, correct, important and scientific, as oxidation and potential itself. In terms of calculations, regardless of whether oxidation or reduction potentials are used, there is no difference in the calculated result. There are actually professional chemists and university professors & lecturers, who even favour the use of oxidation potentials over reduction potentials. Personally for myself, I find it more meaningful to use both. The vast majority of students who use only reduction potentials (for which most JC students and even JC teachers often wrongly call "eeee-nought value"; that's not a "nought" symbol, it's a "standard" symbol! Different meaning!) don't truly realize or appreciate the meaning (ie. they just blindly memorize it!) the formula for calculating cell potentials. I encourage students to appreciate deeper meaning in their chemistry, their work, their life, their relationships. Most people don't, at all. Pity. ~~~ so, it means that the answer given is wrong?? since the cell potential becomes less positive if strictly following the question? and about the relation between the potential values and concentration, am i supposed to memorise that by hard?? and isn't the first equation supposed to be oxidation since carbon is oxidised?? ~~~ Don't bother with the question; it's far more important for you now, to expose yourself to, understand and appreciate the correct chemistry behind the carbon monoxide detector. Use Wikipedia to support your self-learning, Wikipedia - One of the Best Things that Happened to Mankind. As for redox potentials and concentration, just apply Le Chatelier's principle for a qualitative understanding (which is all that's required for 'A' levels). As for your last question, don't bother with it. Since the question is erroneous, it'll just confuse you further if you bother with it. But the idea is, by comparing redox potentials (if you know the reduction potentials, you also know oxidation potentials, and vice-versa), you can decide which is the reduction half-equation at the cathode, and which is the oxidation half-equation at the anode.

~ Isn’t it always preferred to just use reduction potential for both? Even at uni level we do not use the oxidation potential for any calculations. Not true, actually. It’s only a matter of convention (ie. standardization for the sake of convenience, not because it’s more ‘correct&#8217<img title="Wink" src= "/images/emoticons/classic/icon_wink.gif" alt="Wink" /> that reduction potentials are more often used. Conceptually, oxidation and reduction potentials, are every bit as valid, correct, important and scientific, as oxidation and potential itself. In terms of calculations, regardless of whether oxidation or reduction potentials are used, there is no difference in the calculated result. There are actually professional chemists and university professors & lecturers, who even favour the use of oxidation potentials over reduction potentials. Personally for myself, I find it more meaningful to use both. The vast majority of students who use only reduction potentials (for which most JC students and even JC teachers often wrongly call “eeee-nought value”; that’s not a “nought” symbol, it’s a “standard” symbol! Different meaning!) don’t truly realize or appreciate the meaning (ie. they just blindly memorize it!) the formula for calculating cell potentials. I encourage students to appreciate deeper meaning in their chemistry, their work, their life, their relationships. Most people don’t, at all. Pity. ~ so, it means that the answer given is wrong?? since the cell potential becomes less positive if strictly following the question? and about the relation between the potential values and concentration, am i supposed to memorise that by hard?? and isn’t the first equation supposed to be oxidation since carbon is oxidised?? ~~ Don’t bother with the question; it’s far more important for you now, to expose yourself to, understand and appreciate the correct chemistry behind the carbon monoxide detector. Use Wikipedia to support your self-learning, Wikipedia – One of the Best Things that Happened to Mankind. As for redox potentials and concentration, just apply Le Chatelier’s principle for a qualitative understanding (which is all that’s required for ‘A’ levels). As for your last question, don’t bother with it. Since the question is erroneous, it’ll just confuse you further if you bother with it. But the idea is, by comparing redox potentials (if you know the reduction potentials, you also know oxidation potentials, and vice-versa), you can decide which is the reduction half-equation at the cathode, and which is the oxidation half-equation at the anode.

2009-10-21T23:02:05+08:00

2297

9439363

10721587

379474

2009-10-21T23:02:09+08:00

124947

1 2)A domestic carbon monoxide detector is based on an electrochemical cell consisting of the electrodes: CO2(g) + 2H+(aq) + 2e- --> CO(g) + H2O(l) E= -0.10V H2O(l) --> 1/2O2(g) + 2H+(aq) +2e- E= +1.23V ~~~ Which statements about the cell are true? 1. In the cell, carbon is oxidised 2. Under standard conditions the cell potential is +1.33V 3. As the concentration of carbon monoxide in the detector increases, the cell potential becomes more positive. (Ans: all three) for one i know carbon is oxidised because it changes from CO to CO2. and 2. is right cause E(red) - E(oxi) = +1.23 - (-0.10) = +1.33V but i can't find anything concerning 3. in my notes. o.o ~~~ If based strictly upon the equations and information given by this question, then : When molarity of CO increases, position of equilibrium shifts left, ie. oxidation potential becomes more positive ( > +0.1V ), or the reduction potential becomes more negative ( < -0.1V ). Because the 2nd equation given is written as an oxidation half-equation, which has a positive value of +1.23V, it can be deduced that the 2nd equation (oxidation) occurs at the anode. Which means that the 1st equation (reduction) occurs at the cathode. Cell potential = oxidation potential at anode + reduction potential at cathode. Since the reduction potential becomes more negative, the cell potential becomes less positive. This is what Wikipedia has to say about "Electrochemical Carbon Monoxide Detector" : This is a type of fuel cell that instead of being designed to produce power, is designed to produce a current that is precisely related to the amount of the target gas (in this case carbon monoxide) in the atmosphere. Measurement of the current gives a measure of the concentration of carbon monoxide in the atmosphere. Essentially the electrochemical cell consists of a container, 2 electrodes, connection wires and an electrolyte - typically sulphuric acid. Carbon monoxide is oxidised at one electrode to carbon dioxide whilst oxygen is consumed at the other electrode. For carbon monoxide detection, the electrochemical cell has advantages over other technologies in that it has a highly accurate and linear output to carbon monoxide concentration, requires minimal power as it is operated at room temperature, and has a long lifetime (typically commercial available cells now have lifetimes of 5 years or greater). Until recently, the cost of these cells and concerns about their long term reliability had limited uptake of this technology in the marketplace, although these concerns are now largely overcome. Therefore, the correct half equations would be : Anode : CO(g) + H2O(l) --> CO2(g) + 2H+(aq) + 2e- E = +0.10V (assuming original question data given by Charmy is valid) Cathode : O2(g) + 4H+(aq) + 4e– --> 2H2O(l) E = +1.23V Therefore, when molarity of CO(g) increases, oxidation potential at anode increases (by Le Chatelier's principle). Since cell potential = oxidation potential at anode + reduction potential at cathode, hence overall cell potential increases. ~~~ on a side note though, why wouldn't i need to flip the equations since its going from CO to CO2? and if i really flip, 2. would be wrong. [E(red) - E(oxi) = -1.23 - (0.10) = -1.33V] and for some questions, the E(red) - E(oxi) equation doesn't apply. like you have to add up the E values instead. i'm confused. ~~~ To avoid confusion, choose either one formula and stick with it : EITHER : Cell potential = oxidation potential at anode + reduction potential at cathode. OR : Cell potential = reduction potential at cathode - reduction potential at anode.

2)A domestic carbon monoxide detector is based on an electrochemical cell consisting of the electrodes: <acronym title="g">CO2</acronym> + 2H+(aq) + 2e- <del>-> CO(g) + <acronym title="l">H2O</acronym> E= -0.10V <acronym title="l">H2O</acronym> -</del>> 1/2O2(g) + 2H+(aq) +2e- E= +1.23V ~~ Which statements about the cell are true? 1. In the cell, carbon is oxidised 2. Under standard conditions the cell potential is +1.33V 3. As the concentration of carbon monoxide in the detector increases, the cell potential becomes more positive. (Ans: all three) for one i know carbon is oxidised because it changes from CO to CO2. and 2. is right cause E(red) – E(oxi) = +1.23 – (-0.10) = +1.33V but i can’t find anything concerning 3. in my notes. o.o ~ If based strictly upon the equations and information given by this question, then : When molarity of CO increases, position of equilibrium shifts left, ie. oxidation potential becomes more positive ( > +0.1V ), or the reduction potential becomes more negative ( < -0.1V ). Because the 2nd equation given is written as an oxidation half-equation, which has a positive value of +1.23V, it can be deduced that the 2nd equation (oxidation) occurs at the anode. Which means that the 1st equation (reduction) occurs at the cathode. Cell potential = oxidation potential at anode + reduction potential at cathode. Since the reduction potential becomes more negative, the cell potential becomes less positive. This is what Wikipedia has to say about “Electrochemical Carbon Monoxide Detector” : This is a type of fuel cell that instead of being designed to produce power, is designed to produce a current that is precisely related to the amount of the target gas (in this case carbon monoxide) in the atmosphere. Measurement of the current gives a measure of the concentration of carbon monoxide in the atmosphere. Essentially the electrochemical cell consists of a container, 2 electrodes, connection wires and an electrolyte – typically sulphuric acid. Carbon monoxide is oxidised at one electrode to carbon dioxide whilst oxygen is consumed at the other electrode. For carbon monoxide detection, the electrochemical cell has advantages over other technologies in that it has a highly accurate and linear output to carbon monoxide concentration, requires minimal power as it is operated at room temperature, and has a long lifetime (typically commercial available cells now have lifetimes of 5 years or greater). Until recently, the cost of these cells and concerns about their long term reliability had limited uptake of this technology in the marketplace, although these concerns are now largely overcome. Therefore, the correct half equations would be : Anode : CO(g) + <acronym title="l">H2O</acronym> <del>-> <acronym title= "g">CO2</acronym> + 2H+(aq) + 2e</del> E = <ins>0.10V (assuming original question data given by Charmy is valid) Cathode : O2(g) + 4H</ins>(aq) + 4e–—> 2H2O(l) E = +1.23V Therefore, when molarity of CO(g) increases, oxidation potential at anode increases (by Le Chatelier’s principle). Since cell potential = oxidation potential at anode + reduction potential at cathode, hence overall cell potential increases. ~ on a side note though, why wouldn’t i need to flip the equations since its going from CO to CO2? and if i really flip, 2. would be wrong. [E(red) – E(oxi) = -1.23 – (0.10) = -1.33V] and for some questions, the E(red) – E(oxi) equation doesn’t apply. like you have to add up the E values instead. i’m confused. ~~ To avoid confusion, choose either one formula and stick with it : EITHER : Cell potential = oxidation potential at anode + reduction potential at cathode. OR : Cell potential = reduction potential at cathode – reduction potential at anode.

2009-10-21T20:34:14+08:00

2297

9439058

10721113

379474

2009-10-21T20:34:14+08:00

124947

1 Q5. The word "mole" refers to 6.02 x 10^23. So 1 mole of ethanol contains 6.02 x 10^23 of ethanol molecules, and hence 9 x 6.02 x 10^23 atoms. Q4. BaSO4(s) is insoluble, and adding HCl won't dissolve it. Unlike adding excess NH3(aq) to Cu(OH)2(s), which would dissolve the Cu(OH)2 ppt since positition of equilibrium shifts as Cu2+ is used up to form the soluble complex ion tetraaminediaquocopper(II) ion. Q3. Oxides are stable, but peroxides and superoxides are easily polarized and distorted by cations of higher charge densities, and hence peroxides and oxides are only commonly present with cations of lower charge densities. Q2. An ether functional group simply means R-O-R, and the R groups may or may not be saturated (ie. it's a different matter). An unsaturated compound refers to double or triple bonds present between two adjacent C atoms. In the list of ethers given, none of them are unsaturated. Q1. You're correct. No matter what the yield at varying temperatures would be, repeated fractional distillation would need to be carried out. Wikipedia industrial alcoholic fermentation for more info.

Q5. The word “mole” refers to 6.02×1023. So 1 mole of ethanol contains 6.02×1023 of ethanol molecules, and hence 9×6.02×10^23 atoms. Q4. BaSO4(s) is insoluble, and adding HCl won’t dissolve it. Unlike adding excess <acronym title="aq">NH3</acronym> to Cu(OH)2(s), which would dissolve the Cu(OH)2 ppt since positition of equilibrium shifts as Cu2+ is used up to form the soluble complex ion tetraaminediaquocopper(II) ion. Q3. Oxides are stable, but peroxides and superoxides are easily polarized and distorted by cations of higher charge densities, and hence peroxides and oxides are only commonly present with cations of lower charge densities. Q2. An ether functional group simply means R-O-R, and the R groups may or may not be saturated (ie. it’s a different matter). An unsaturated compound refers to double or triple bonds present between two adjacent C atoms. In the list of ethers given, none of them are unsaturated. Q1. You’re correct. No matter what the yield at varying temperatures would be, repeated fractional distillation would need to be carried out. Wikipedia industrial alcoholic fermentation for more info.

2009-10-21T16:55:40+08:00

2297

9438750

10720599

379467

2009-10-21T16:55:43+08:00

124947

1 Ammonia has the capacity to act as a Bronsted-Lowry base, due to the availability of a lone pair on N to accept a proton, for instance, from water solvent, thereby generating the hydroxide ion; hence where the discriminating student decides the question requires the involvement of the hydroxide ion (where "aqueous ammonia" is stated"), the student should first write the equation for the hydrolysis of the ammonia : NH3 (aq) + H2O (l) (insert equilibrium double half arrow here) NH4+ (aq) + OH- (aq) or simply NH4OH (aq) Subsequently, the remaining required equation(s), eg for the formation of a blue ppt when aqueous ammonia is added to an aqueous solution containing Cu2+ ions : Cu2+ (aq) + OH- (aq) (insert equilibrium double half arrow here) Cu(OH)2 (s) and in excess aqueous ammonia : NH3 (aq) + [Cu(H2O)6]2+ (aq) (insert equilibrium double half arrow here) [Cu(NH3)4(H2O)2]2+ (aq) Furthermore, there are instances, in which the student may optionally write the equation in which ammonia is directly the Bronsted-Lowry base, OR the equation in which ammonia first undergoes hydrolysis to generate a separate Bronsted-Lowry base - the hydroxide ion, which accepts the proton from the Bronsted-Lowry acid; as an example, both of the following are acceptable at 'O' levels and 'A' levels : Either : <nr> NH3 (aq) + HCl (aq) (insert equation arrow here) NH4Cl (aq) Or : NH4OH (aq) + HCl (aq) (insert equation arrow here) NH4Cl (aq) + H2O (l) Technically though, the former is more accurate. The reason being, because the hydroxide ion OH- is a stronger base than ammonia NH3 (or put in another way, NH4+ is a stronger acid than H2O), the position of equilibrium lies strongly to the left : NH3 (aq) + H2O (l) (insert equilibrium double half arrow here) NH4+ (aq) + OH- (aq)

Ammonia has the capacity to act as a Bronsted-Lowry base, due to the availability of a lone pair on N to accept a proton, for instance, from water solvent, thereby generating the hydroxide ion; hence where the discriminating student decides the question requires the involvement of the hydroxide ion (where "aqueous ammonia" is stated"), the student should first write the equation for the hydrolysis of the ammonia : NH3 (aq) + H2O (l) (insert equilibrium double half arrow here) NH4+ (aq) + OH- (aq) or simply NH4OH (aq) Subsequently, the remaining required equation(s), eg for the formation of a blue ppt when aqueous ammonia is added to an aqueous solution containing Cu2+ ions : Cu2+ (aq) + OH- (aq) (insert equilibrium double half arrow here) Cu(OH)2 (s) and in excess aqueous ammonia : NH3 (aq) + [Cu(H2O)6]2+ (aq) (insert equilibrium double half arrow here) [Cu(NH3)4(H2O)2]2+ (aq) Furthermore, there are instances, in which the student may optionally write the equation in which ammonia is directly the Bronsted-Lowry base, OR the equation in which ammonia first undergoes hydrolysis to generate a separate Bronsted-Lowry base - the hydroxide ion, which accepts the proton from the Bronsted-Lowry acid; as an example, both of the following are acceptable at 'O' levels and 'A' levels : Either : NH3 (aq) + HCl (aq) (insert equation arrow here) NH4Cl (aq) Or : NH4OH (aq) + HCl (aq) (insert equation arrow here) NH4Cl (aq) + H2O (l) Technically though, the former is more accurate. The reason being, because the hydroxide ion OH- is a stronger base than ammonia NH3 (or put in another way, NH4+ is a stronger acid than H2O), the position of equilibrium lies strongly to the left : NH3 (aq) + H2O (l) (insert equilibrium double half arrow here) NH4+ (aq) + OH- (aq)

2009-10-21T00:21:01+08:00

2297

9437841

10719128

379146

2009-10-21T00:28:34+08:00

124947

5 ~ but how to use the information of reduction of 30cm3 in the residual gas?? ~ CO2 is an acidic gas, it is hydrolysed and neutralized by alkalis.

~ but how to use the information of reduction of 30cm3 in the residual gas?? ~ CO2 is an acidic gas, it is hydrolysed and neutralized by alkalis.

2009-10-15T00:05:49+08:00

2297

9426588

10700169

378838

2009-10-15T00:05:53+08:00

124947

1 (edited - dkcx has already replied TS liao). Spectroscopy's not my area (moreover that it's not relevant to my work now), so I have nothing to add. As an aside, in addition to any school notes for any students (eg. IB, 'A' Level H2, Poly Diploma, etc), always take advantage of the world's most wonderful encyclopedia (don't think just because it's free means it's lousy, it's actually more up to date than any other encyclopedia available to the general public) Wikipedia : http://en.wikipedia.org/wiki/Spectroscopy http://en.wikipedia.org/wiki/Atomic_absorption_spectroscopy http://en.wikipedia.org/wiki/Ultraviolet-visible_spectroscopy

(edited – dkcx has already replied TS liao). Spectroscopy’s not my area (moreover that it’s not relevant to my work now), so I have nothing to add. As an aside, in addition to any school notes for any students (eg. IB, ‘A’ Level H2, Poly Diploma, etc), always take advantage of the world’s most wonderful encyclopedia (don’t think just because it’s free means it’s lousy, it’s actually more up to date than any other encyclopedia available to the general public) Wikipedia : <a href="http://en.wikipedia.org/wiki/Spectroscopy" rel= "nofollow">http://en.wikipedia.org/wiki/Spectroscopy</a> <a href= "http://en.wikipedia.org/wiki/Atomic_absorption_spectroscopy" rel= "nofollow">http://en.wikipedia.org/wiki/Atomic_absorption_…</a> <a href= "http://en.wikipedia.org/wiki/Ultraviolet-visible_spectroscopy" rel="nofollow">http://en.wikipedia.org/wiki/Ultraviolet-visibl…</a>

2009-10-15T00:02:58+08:00

2297

9426571

10700176

378835

2009-10-15T00:06:43+08:00

124947

4 Answers to Phosphorous acid Qn Relevant pH values (at various volumes of NaOH added) : 1.5 (0 cm3) 2.0 (12.5cm3) 3.95 (25cm3) 6.6 (37.5cm3) 9.5 (50cm3) approaches 13.0 (at large excess of NaOH) Answers to Ammonia Qn : Mass of ammonium chloride added = 23.41g Change in pH (upon adding acid) from 8.9 to 8.868 is minimal due to buffer action. Significant pH values of graph : 11.27 (0cm3), 9.24 (25cm3), 5.12(50cm3), approaches 0.699 (large excess of HCl).

Answers to Phosphorous acid Qn Relevant pH values (at various volumes of NaOH added) : 1.5 (0 cm3) 2.0 (12.5cm3) 3.95 (25cm3) 6.6 (37.5cm3) 9.5 (50cm3) approaches 13.0 (at large excess of NaOH) Answers to Ammonia Qn : Mass of ammonium chloride added = 23.41g Change in pH (upon adding acid) from 8.9 to 8.868 is minimal due to buffer action. Significant pH values of graph : 11.27 (0cm3), 9.24 (25cm3), 5.12(50cm3), approaches 0.699 (large excess of HCl).

2009-10-14T20:23:51+08:00

2297

9426092

10699433

378777

2009-10-14T20:23:55+08:00

124947

1 Hi Darkness_hacker99, Perhaps modify (ie. update with your new mobile no) the graphic file and reupload to same URL? Less troublesome for you than changing all urls to a new uploaded graphic.

Hi Darkness_hacker99, Perhaps modify (ie. update with your new mobile no) the graphic file and reupload to same URL? Less troublesome for you than changing all urls to a new uploaded graphic.

2009-10-14T18:49:28+08:00

2297

9425979

10699219

378714

2009-10-14T18:49:32+08:00

124947