BedokFunland JC's A Level H2 Chemistry Qns (Part 1)
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'A' Level Qn.
Elements in the same group react similarly, eg. Grp VI elements O and S, in their reaction with H.
Dihydrogen monoxide (prefix name) aka hydrogen(I) oxide (stock name) aka hydrogen hydroxide (alternative name) aka water (common name) aka "sui3" (chinese name) aka "air" (malay name) has a H-O-H bond angle of approximately 104.45 deg.
Would you expect the H-S-H angle in dihydrogen monosulphide (prefix name) aka hydrogen(I) sulphide (stock name) to be exactly the same as 104.45 deg, smaller than 104.45 deg, or greater than 104.45 deg?
Explain your answer.
Solution (by Roberto Gregorius)
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In regards to to above question,
UltimaOnline's addition to Roberto Gregorius' solution :
The purpose of hybridization, is to achieve ideal geometrical bond angles which minimize electron pair repulsion.
Hence, in H2O as well as in H2S, the hybridization would be sp3 (tetrahedral), since we have 4 electron pairs in total (2 bond pairs, 2 lone pairs ===> tetrahedral electron geometry, bent/v-shape/non-linear molecular geometry).
However, because the atomic radius is S is one electron shell larger than the atomic radius of O, hence in H2S there is significantly less electron repulsion (visualize a tennis ball with pencils sticking out, in contrast with a soccer ball with pencils sticking out) as compared to in H2O.
Consequently, there is lesser need and hence lesser extent of hybridization for the bonding electron orbitals of H2S as compared to in H2O.
Therefore, we would expect the bond angles in H2S to be between 104.5 deg (ie. hybridized bond angles) and 90 deg (ie. unhybridized bond angles). Experimental evidence indicates the bond angle to be approximately 92 deg.
Similarly, you can use the above arguments to compare the bond angles of NH3 versus PH3.
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^_^
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(04 October 2008) A Collection of 'A' level Qns by UltimaOnline.
Q1) Given equal volumes of equal molarities of an acid solution and a buffer solution, both of the same pH; and using only a pH data logger and another beaker of water, how could you identify which was the acid solution and which was the buffer solution?
Q2a) its boiling point (ie. 100°C), which of the following is true regarding pure (ie. distilled) water?
1 pH < 7
2 pH = 7
3 pH > 7
4 pOH < 7
5 pOH = 7
6 pOH > 7
7 pKw < 14
8 pKw = 14
9 pKw > 14
Q2b) At its freezing point (ie. 0°C), which of the following is true regarding pure (ie. distilled) water?
1 pH < 7
2 pH = 7
3 pH > 7
4 pOH < 7
5 pOH = 7
6 pOH > 7
7 pKw < 14
8 pKw = 14
9 pKw > 14
Q3) Explain whether the rate of nucleophilic addition reaction would be faster for an aldehyde or a ketone, using the same nucleophile.
Q4a) Explain why you get different results when reacting phenol with Br2 (in CCl4) versus Br2 (aq).
Q4b) Explain why ortho-para-ortho (or 2-4-6) tribromophenol is less soluble than phenol.
Q5) For the reaction H2O(g) + C(s) --> H2(g) + CO(g), enthalpy change = +131 kJ/mol
What happens to the position of equilibrium when pressure is increased?
What happens to the position of equilibrium when temperature is increased?
What happens to the position of equilibrium when helium gas is added at constant pressure?
What happens to the position of equilibrium when helium gas is added at constant volume?
Q6a) What is the difference between the way these 2 reducing agents work? Potassium Iodide versus Lithium Aluminium Hydride? Illustrate your answer with suitable examples.
Q6b) What is the difference between using LiAlH4 versus NaBH4 as a reducing agent?
Q7) Whenever the quesiton states "using hot, concentrated, acidified KMnO4", what functional groups must you watch out for?
Q8) Using the following information, draw the mechanism for the reduction of an amide to an amine, using LiAlH4 in dry ether.
[Description : after the hydride ion attacks the δ+ carbon and the pi bond electron density gets kicked up to become a lone pair on the oxygen; the oxygen then attacks AlH3 (again because Al in AlH3 lacks a full octet), and subsequently becomes a leaving group as the lone pair on the nitrogen forms a pi bond with the δ+ carbon; finally, another hydride ion attacks the δ+ carbon, and the pi bond electron density shifts back to become a lone pair on the nitrogen.]
Q9) When trichloroethanal is heated with aqueous sodium hydroxide, two products are formed in a single reaction pathway. One of the products is trichloromethane. Draw the mechanism for this pathway and hence identify the other product.
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Solution to Q1.
Q1) Given equal volumes of equal molarities of an acid solution and a buffer solution, both of the same pH; and using only a pH data logger and another beaker of water, how could you identify which was the acid solution and which was the buffer solution?
Solution :
Add equal amount of water to both, and use pH data logger to monitor. The acid solution will show a rise in pH (due to a fall in molarity of the acid and hence of the protons; since molarity is defined as no. of moles per dm3 volume of solution), while the buffer solution will remain approximately the same pH.
Let's say we're using a solution of a weak acid and it's conjugate base, eg. ethanoic acid and sodium ethanoate. Because Ka of acid > Kb of conjugate base, the pH of the buffer solution will depend on Ka.
So Ka = [H+][A-]/[HA]
Rearranging the equation above, we obtain the Henderson-Hasselbach equation :
pH = pKa + log [A-]/[HA]
When you dilulte a buffer solution using water, the ratio of the conjugate acid to its conjugate base remains unchanged.
From the equation, this proves that the pH of the buffer solution does not change upon dilution, since the molarity of both the acid and its conjugate base is affected equally by the dilution, and hence the ratio of the acid to its conjugate base remains the same.
Cut-To-The-Chase :
If this question, "Explain your expected observations of any changes in pH when a buffer solution is diluted with water" is asked in your 'A' level examination, answer this question simply by re-arranging and re-writing the Ka expression as the Henderson-Hasselbach equation, and state,
"From the Henderson-Hasselbach equation (shown above), we deduce that the pH of the buffer solution does not change upon dilution, since the molarity of both the acid and its conjugate base is affected equally by the dilution, and hence the ratio of the acid to its conjugate base remains the same."
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Solution to Q2.
Q2a) its boiling point (ie. 100°C), which of the following is true regarding pure (ie. distilled) water?
1 pH < 7
2 pH = 7
3 pH > 7
4 pOH < 7
5 pOH = 7
6 pOH > 7
7 pKw < 14
8 pKw = 14
9 pKw > 14
Q2b) At its freezing point (ie. 0°C), which of the following is true regarding pure (ie. distilled) water?
1 pH < 7
2 pH = 7
3 pH > 7
4 pOH < 7
5 pOH = 7
6 pOH > 7
7 pKw < 14
8 pKw = 14
9 pKw > 14
Solution :
Dissociation of H2O into H+ and OH- ions is an endothermic process.
Everyone knows that at 25 deg C, Ka x Kw = Kw = (1 x 10^-7) x (1 x 10^-7) = 1 x 10^-14.
Hence at 100 deg C, even more H+ and OH- ions will be generated, hence Ka and Kb and Kw increases, and pKa and pKb and pKw decreases. (FYI, pH of pure water at 100 deg C is 6.14)
Similarly hence, at 0 deg C, less H+ and OH- ions will be generated, hence Ka and Kb and Kw decreases, and pKa and pKb and pKw increases. (FYI, pH of pure water at 0 deg C is 7.47)
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Solution to Q3.
Q3) Explain whether the rate of nucleophilic addition reaction would be faster for an aldehyde or a ketone, using the same nucleophile.
Solution :
The reactivity, as well as rate of reaction, is slightly higher for aldehydes than for ketones, for 2 reasons.
#1 - Electronics. Assuming no substituents, in ketones the 2 (electron donating by induction) alkyl groups stabilize to a greater extent and reduce the magnitude of the partial positive charge of the carbonyl carbon, as compared to only 1 alkyl group in aldehydes.
# - Sterics. Two bulky alkyl groups in ketones present a greater steric hinderance to an incoming nucleophile, as compared to 1 alkyl group in aldehydes.
For the reasons described above, aldehydes are relatively more reactive towards, and experience a relatively faster rate of reaction for, nucleophilic attacks, as compared to ketones.
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Kelecthor posted :>>> Does anyone know how to write ionic equation? For example.... Sodium oxide + hydrochloride acid, wads the ionic equation? <<<My reply :Whenever you (the candidate) feels there is any ambiguity in exactly what the question is asking for, go ahead and write all possible alternative forms of the answer, as long as they are all correct and do not contradict each other.
For instance, if you're asked what is the type of reaction that occurs when H2 is reacted with an alkene in the presence of nickel catalyst, and you're not sure if "hydrogenation" or "addition reaction" is required, write both "Hydrogenation, which is a type of addition reaction." You'd get your marks for whichever answer the mark scheme is looking for (commonly, the mark scheme would accept either answer).
But of course, if you write contradictory answers, eg "Addition reaction, condensation reaction.", that's considered cheating (obviously), and you'll be marked wrong and get no marks, even though "addition reaction" may have been acceptable.
As another example, if the question is ambiguous, and it leads to two different alternative answers, give both answers, but with qualification and explanation.For instance, for this qn :
>>> Does anyone know how to write ionic equation? For example.... Sodium oxide + hydrochloride acid, wads the ionic equation? <<<
(That is probably intended to be hydrochloric acid, btw, which is aqueous ionized hydrogen chloride; other similar acids include hypochlorous acid, chlorous acid, chloric acid, perchloric acid, etc).
Smart Candidate's Solution :
(you'd get the full marks regardless of which alternative ionic equation solution the mark scheme was looking for. Now, the following is an extreme example for illustration purposes. In most cases, you usually don't need to write so much to express the answer fully to ensure you obtain full marks).Sodium oxide is soluble in aqueous solutions, forming aqueous sodium hydroxide, which consequently reacts with hydrochloric acid in an acid-base (proton transfer) reaction.The following chemical equations could be written to represent this reaction :
Na2O(s) + H2O(l) --> 2NaOH(aq)2NaOH(aq) + 2HCl(aq) --> 2NaCl(aq) + 2H2O(l)Summarily, the above chemical reactions could be written as :
Na2O(s) + 2HCl(aq) --> 2NaCl(aq) + H2O(l)Hence, the relevant ionic equations could be written as :
Na2O(s) + H2O(l) --> 2Na+(aq) + 2OH-(aq)2OH-(aq) + 2H+(aq) --> 2H2O(l)Summarily, the above ionic equations could be written as :
Na2O(s) + 2H+(aq) --> 2Na+(aq) + H2O(l)
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The point being, if the question appears to be ambiguous with regard to exactly what the expected answer is, then an exam smart candidate (that's who you want to be, of course) would intelligently qualify & explain his/her answers (particularly when the question is ambiguous with alternative answers possible), and give all possible alternative forms of the answer, as long as they are all correct in their own way. -
Solution to Q4.
Q4a) Explain why you get different results when reacting phenol with Br2 (in CCl4) versus Br2 (aq).
Q4b) Explain why ortho-para-ortho (or 2-4-6) tribromophenol is less soluble than phenol.
Solution :
a) With Br2 (in CCl4), and at room temperature, phenol is mono-brominated at either the ortho or para positions (since OH substituent is electron donating by resonance). With Br2 (aq) and at room temperature however, phenol is tri-brominated at the ortho and para positions (2,4,6 carbon atoms).
The reason for the greater extent of electrophilic aromatic substitution, has to do with the polar water molecules inducing a stronger dipole in the Br2 molecule, hence generating a stronger electrophile (as compared to the weaker instantaneous dipole - induced dipole in Br2 with a non-polar solvent such as CCl4).
b) Tribromophenol is less soluble than phenol for 2 reasons :
Reason #1 - The electron withdrawing by induction effect outweighs the electron donating by resonance effect of the Br substituents, resulting in a shift of electron density away from the OH substituent of the phenol; consequently weakening the strength of any hydrogen bonding of the phenol with solvent water molecules.
Reason #2 - The greater no. of electrons, as well as increased polarizability, of the 3 bromine substituents, results in a significantly increased induced dipole - induced dipole van der Waals non-polar hydrophobic interaction with other tribromophenol molecules.
For non-polar hydrophobic interactions, the non-polar groups are first 'pushed away' by the polar solvent water molecules, which would rather hydrogen bond with each other (macham elite-class snobs) than with the non-polar tribromophenol molecules. Subsequently, the now stronger (thanks to increased no. of electrons and increased polarizability, due to the bromine substituents) induced dipole - induced dipole van der Waals non-polar hydrophobic interactions between the tribromophenol molecules cause them to 'stick together', hence precipitate out of aqueous solution.
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Summarily, the above ionic equations could be written as :
Na2O(s) + 2H+(aq) --> 2Na+(aq) + H2O(l)Sodium oxide has a state symbol of (s)? Are you certain? All sodium salts are soluble, hence shouldn't it be (aq)?
I believe all neutralisation reactions have the same ionic equation:
H+ (aq) + OH-(aq) ---> H2O (l)
It applies to this reaction as well.
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Originally posted by Garrick_3658:
Sodium oxide has a state symbol of (s)? Are you certain? All sodium salts are soluble, hence shouldn't it be (aq)?
I believe all neutralisation reactions have the same ionic equation:
H+ (aq) + OH-(aq) ---> H2O (l)
It applies to this reaction as well.
Anhydrous sodium oxide is an ionic solid, which is soluble in water. Since the question gave you as sodium oxide, obviously it is a solid. Otherwise, the question would have specified "sodium hydroxide", since sodium oxide cannot be aqueous (if aqueous, it exists as sodium hydroxide).In other words, there is no such thing as Na2O(aq) (you'd be penalized in the exam if you wrote this). Sodium oxide can only exist as a solid, Na2O(s); which when dissolved in water (that's the meaning of (aq)), forms aqueous sodium hydroxide, NaOH(aq).
If you had read and understood my solution, you'd notice I gave all 3 ionic equations (all of which are valid and correct alternative answers) as part of the answer (to be submitted by the candidate in an exam).
Hence, the relevant ionic equations could be written as :
Na2O(s) + H2O(l) --> 2Na+(aq) + 2OH-(aq)
2OH-(aq) + 2H+(aq) --> 2H2O(l)
Summarily, the above ionic equations could be written as :
Na2O(s) + 2H+(aq) --> 2Na+(aq) + H2O(l)
Which is the point of my post - an exam-smart candidate would give all possible alternative answers, as long as all of these alternatives are all correct.
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Thanks for clarifying.
What about the other SPA salts, such as potassium oxide (does this even exist?) and ammonium oxide? Do I have to write their state symbols as (s) too?
If the question only stated sodium OXIDE with hydrochloric acid, can i use this equation?
2OH-(aq) + 2H+(aq) --> 2H2O(l)
(BTW I just realised the "2" in front of each reactant, signifying its mole ratio, is redundant, or is it?)
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Originally posted by Garrick_3658:
Thanks for clarifying.
What about the other SPA salts, such as potassium oxide (does this even exist?) and ammonium oxide? Do I have to write their state symbols as (s) too?
If the question only stated sodium OXIDE with hydrochloric acid, can i use this equation?
2OH-(aq) + 2H+(aq) --> 2H2O(l)
(BTW I just realised the "2" in front of each reactant, signifying its mole ratio, is redundant, or is it?)
>>> What about the other SPA salts, such as potassium oxide (does this even exist?) and ammonium oxide? Do I have to write their state symbols as (s) too? <<<
Yes, sodium oxide and potassium oxide oxides do exist of course, but only in the solid state. They dissolve in water to form hydroxides.
Ammonium oxide as well as ammonium hydroxide, however, do not exist in the solid state. In fact, ammonium oxide is not stable and does not exist in any state, or at least not in the empirical form as 'O' level students might conceive - as "(NH4)2O".
Ammonium hydroxide on the other hand, exists in equilibrium with aqueous ammonia (see the equation below), but does not exist as a species that can be isolated, simply because the OH- is a much stronger (Bronsted-Lowry) base as compared to NH3.
NH3(aq) + H2O(l) <<<----> NH4+(aq) + OH-(aq)
For this reason, you may come across people/teachers/textbooks/websites who argue that "ammonium hydroxide does not exist! the chem lab bottles should only be labelled as aqueous ammonia!"
>>> If the question only stated sodium OXIDE with hydrochloric acid, can i use this equation?
2OH-(aq) + 2H+(aq) --> 2H2O(l)
(BTW I just realised the "2" in front of each reactant, signifying its mole ratio, is redundant, or is it?) <<<
You can indeed submit the equation H+(aq) + OH-(aq) --> H2O(l) as the answer to the question, but you should first explain that Na2O(s) dissolves in water to form NaOH(aq).
In other words, the best answer is the one I gave earlier :
Hence, the relevant ionic equations could be written as :
Na2O(s) + H2O(l) --> 2Na+(aq) + 2OH-(aq)
2OH-(aq) + 2H+(aq) --> 2H2O(l)
Summarily, the above ionic equations could be written as :
Na2O(s) + 2H+(aq) --> 2Na+(aq) + H2O(l)
Finally, you're welcome if my posts helped you, and you're welcome to freely ask more questions on any Chemistry topic if you've any, but (for this and all future cases) I'd ask you to please start a new thread on the Homework Forum with an appropriate title, eg. "Does sodium oxide exist in the aqueous state?", rather than having a discussion within this thread which is meant to be, "'A' & 'O' Level Chemistry Qns (A Collection)".
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Solution to Q5.
Q5) For the reaction H2O(g) + C(s) --> H2(g) + CO(g), enthalpy change = +131 kJ/mol
What happens to the position of equilibrium when pressure is increased?
Answer : Shift to the LEFT.
What happens to the position of equilibrium when temperature is increased?
Answer : Shift to the RIGHT.
What happens to the position of equilibrium when helium gas is added at constant pressure?
Answer : Shift to the RIGHT.
What happens to the position of equilibrium when helium gas is added at constant volume?
Answer : NO CHANGE.
Comment : Pretty straightforward, except for the last two questions, the toughest question being "the effect of adding an inert gas at constant pressure".
The following is the explanation for "the effect of adding an inert gas at constant volume" resulting in no change to position of equilibrium, and "the effect of adding an inert gas at constant pressure" resulting in a shift of position of equilibrium to the right.
Q. What happens to position of equilibrium when an inert gas (eg. helium) is added at constant volume (eg. in a steel vessel)?
A. Adding helium to the equilibrium mixture at constant volume increases the total gas pressure and decreases the mole fractions of all 3 gases, but the partial pressure of each gas given by the product of its (reduced) mole fraction and the (increased) total pressure does not change. Hence, position of equilibrium does not change.
Q. What happens to position of equilibrium when an inert gas (eg. helium) is added at constant pressure, but not constant volume?The important effect is on the partial pressure (mole fraction x total pressure) of reactants versus products, and since the mole fractions of the gases are now reduced (on addition of helium), their partial pressures are also reduced (since total pressure remains the same).
Another way to think about it, is that with increased volume, the concentrations/molarities of the reactant and product gases, all decrease.
Since on the RHS of equation (or numerator of Qc or Kc expression) has 2 gaseous terms (each unimolar), while the LHS of equation (or denominator of Qc or Kc expression) has only 1 (unimolar) gaseous term, in effect, the decrease in RHS or numerator is greater than the decrease in LHS or denominator, which translates into Qc < Kc, and the position of equilibrium (Kc) now is shifted to the right of Qc, so the net reaction is the forward reaction to produce more moles of gas until Qc = Kc.
So to summarize, the effect on total pressure is not important, it is relative concentrations/molarities or partial pressures, of LHS vs RHS, which is determines positional shifts of equilibrium (ie. Qc vs Kc). -
Solution to Q6.
Q6a) What is the difference between the way these 2 reducing agents work? Potassium Iodide versus Lithium Aluminium Hydride? Illustrate your answer with suitable examples.
Q6b) What is the difference between using LiAlH4 versus NaBH4 as a reducing agent?
KI works as a reducing agent by donating electrons.
Eg. Redox reaction between Cl2 and KI.
2I- ---> I2 + 2e-
Cl2 + 2e- ---> 2Cl-
Cl2 + 2I- ---> l2 + 2Cl-
LiAlH4 works by donating hydride ions (H-).
Eg. Reduction of carboxylic acid to primary alcohol.
LiAlH4 ---> Li+ + AlH3 + H-
Description of mechanism ('A' level H2 students should be able to draw this)
A nucleophilic hydride ion (from LiAlH4) attacks the partially positively charged carbonyl carbon; causing the pi bond electron pair to shift up to become a lone pair on the carbonyl oxygen; the lone pair shifts back down to form a pi bond with the carbonyl carbon; to avoid violating carbon's octet (since it has no empty d-orbitals required to expand its octet) the OH becomes a OH- leaving group. Another hydride ion (from another LiAlH4) again the partially positively charged carbonyl carbon, causing the pi bond electron pair to shift up to become a lone pair on the carbonyl oxygen. This time, there is no suitable leaving group (H- is much more unstable than OH-), hence the electron density is stuck up there, leaving oxygen with a negative formal charge (3 lone pairs, 1 bond pair). We hence have an alkoxide (ie. alkyl oxide) ion. So far we've used LiAlH4 in dry ether. Now is the time to add water (as a source of protons). The alkoxide ion protonates itself (abstracting a proton from water, leaving water as a hydroxide ion), and the transformation to an alcohol is complete.
Comparing sodium borohydride and lithium aluminium hydride, we observe that aluminium (period 3) is larger and hence more polarizable than boron (period 2). Consequently, LiAlH4 is more willing to let go of its hydride ions, and is hence a more powerful (source of the hydride) nucleophile, and is therefore a more powerful reducing agent than NaBH4.
Sodium borohydride, NaBH4, will only reduce aldehydes and ketones.
Lithium aluminium hydride, LiAlH4, will reduce aldehydes, ketones, carboxylic acids, esters, amides, nitriles, etc.
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(moved to 1st post on 1st page)
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hi ultima, any recommendations where you get your resources and references from?
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Originally posted by tinuviel07:
hi ultima, any recommendations where you get your resources and references from?
Hi TinTin,Are you asking for general resource recommendations for your own tuition work (eg. for 'A' and 'O' level Chemistry)? Or are you asking specifically about the resources I'm using (eg. perhaps possibly because my materials/questions may seem somewhat unusual or "off the beaten track/path")?
If the former, then the market for 'A' level (H2/H1) and 'O' level sciences are already flooded and spoilt for choice; just head down to Bras Basar complex (Popular bookstore for sanctioned/authorized publications; other bookstores for (unsanctioned/unauthorized) JCs and Sec Schs' Prelim papers).
Choose publications that you can work well with (browse through their notes, questions and answers and see how compatible this book is with yourself, how much you agree with their answers, how useful you feel the books are, etc).
If the latter, then it's probably because I've always felt (for the purpose of the greater good for mankind, for the Universe) the importance of going beyond convention, dogma, orthodoxy, conforminity; if any significant breakthrough towards higher ground and higher gain is hoped to be made.
Specifically in the education context, being keenly aware (being an ex-MOE teacher) of the limitations of the education system, specifically that in their forced entry into the rat race of paper chase qualifications, students in Singapore don't get to enjoy what they study (particularly tragic for those students who actually would like to be able to deepen their understanding and appreciation of their chosen subjects). And part of this is because (and while I do not fault this or claim to offer any feasible alternatives, because such is necessary based on practical considerations; I certainly would want to do my part to work beyond such limitations, for the sake of students) the MOE/SEAB 'A' level and 'O' level curriculum does not allow (primarily due to time constraints, as well as a so-called 'results oriented' culture; the irony of it) students to truly understand, truly appreciate, truly enjoy their subjects.
Back to the question (if taken to be the latter interpretation), again while I don't mean to deliberately evade the answer, I think it not necessary to specifically name names, publications, online resources, etc. Suffice to mention that I obtain a variety of materials I deem to be useful and helpful for my (tuition) students, both from excellent sources on the internet, as well as publications both local and overseas. The local sources, as I've mentioned, may be obtained from Bras Basar (Popular and other bookstores). The overseas sources, include (multiple copies of) University level (Chemistry and Biology) textbooks (by multiple authors) that were obtained from Amazon.com, Kinokuniya, Clementi Book Store, NUS bookshops, NTU bookshops, etc. All in all, I use metal filing cabinets (the ones used by Mindef and MOE) to house my own private academic library of materials, textbooks and resources, which my (tuition) students may freely use and (not-so-freely) borrow. Approximate cost of resources (over the years, in total) : Sg$10,000.
>>> hi ultima, any recommendations where you get your resources and references from? <<<
In closing, I'm sure my reply was not what you were looking for, but I can only say every teacher (including tuition teacher) is unique and will work best with a unique set of materials/resources. And selecting and assembling your own private academic library is really lots of fun.
And as a colleague (being fellow tution teachers), here's looking forward to a great 2009 year ahead (for both students and tutors)!
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No.. NOT CHEMISTRY !! NOOO!!!!
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'O' level & 'A' level Qn.
Qn : What is the difference between "peptide" and "amide"? Is there a difference between "polypeptide" and "polyamide"?
Ans : "polypeptide" equals "polyamide", but "peptide" not equals "amide".
The peptide bond refers specifically to the C-N bond within the amide group (or amide linkage). So a macromolecule (eg. protein, nylon, etc) with lots of amide groups (ie. a "polyamide") hence also has lots of peptide bonds (ie. a "polypeptide"). -
Solution to Q7.
Q7) Whenever the quesiton states "using hot, concentrated, acidified KMnO4", what functional groups must you watch out for?
1) Acid-base (proton transfer) reaction.
Eg. a N atom with 3 bond pairs has a lone pair which can accept a proton (to form NH4+).
2) Acid hydrolysis.
Eg. Groups such as nitriles, amides, esters, etc.
3) Oxidation.
While K2Cr2O7 can only oxidize primary and secondary alcohols, and aldehydes; note that KMnO4 can also oxidize side-chains of benzene rings to benzoic acid; oxidative cleavage of alkene double bonds, etc.
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hello.
about qn :
3) A mixture of MgSO4.7H2O and CuSO4.5H2O is heated until a mixture of the anhydrous salts, is obtained. If 5.0g of the hydrated mixture when heated gives 3.0g of the anhydrous salts, calculate the % by mass of CuSO4.5H2O in the initial hydrated mixture.
Ans : 73.9%
how did you get this answer?i cannot get this answer.(instead i got the % of MgSO4.7H2O greater than that of CuSO4.5H2O )
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kns why all of you last minute one. like me :P
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yup i got the answer(i got around 73.7% in 3 sig. fig.) thx a lot