BedokFunland JC's A Level H2 Chemistry Qns (Part 1)

last SG · 2008-09-19T13:00:14+00:00

'O' & 'A' Level Qns. Qn 1. When 1.01g sample of potassium nitrate(V) is heated above its melting point, oxygen gas is evolved. The mass of the sample dec...

UltimaOnline SG

(moved to 1st post on 1st page)

tinuviel07 SG

hi ultima, any recommendations where you get your resources and references from?

UltimaOnline SG

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)!

BadzMaro SG

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").

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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.

ohyeah193720 SG

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 )

Garrick_3658 SG

kns why all of you last minute one. like me :P

http://sgforums.com/forums/2297/topics/333447

ohyeah193720 SG

yup i got the answer(i got around 73.7% in 3 sig. fig.) thx a lot

UltimaOnline SG

'O' Level Qn.

Question :

A sample of nitrogen gas contains the same number of atoms as found in 4.0g of methane gas. What is the mass of the sample of nitrogen gas?

Answer :

(2)(x/28) = (4/16)(5)

x = 17.5g

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'O' level and 'A' level Qn.

Draw the Dot-&-Cross diagram of Urea (Chemistry 'O' Level Qn), given that urea is CO(NH2)2.

Solution :

Wikipedia Image of Urea displayed structural formula

So your diagram should have :

between O and C, double bond, so "xoxo"

between C and each N, single bond, so "xo"

between N and each H, single bond, so "xo"

make sure your O has two lone pairs, ie. "xx" and "xx" (not bonded with C)

make sure your N has one lone pair, ie. "xx" (not bonded with H or C)

check that total no. of valence electrons (both bond pairs and lone pairs) = 12 pairs = 24 valence electrons

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'O' and 'A' level Qn.

Methane was burnt in an incorrectly adjusted burner. The methane was converted into a mixture of carbon dioxide and carbon monoxide in the ratio of 99:1, together with water vapour. What will be the volume of oxygen consumed when 'y' dm3 of methane is burnt?

Solution :

Write and balance the equations :

Carbon dioxide
CH4 + 2O2 --> CO2 + 2H2O

Carbon monoxide
CH4 + 3/2 O2 --> CO + 2H2O

Based on stoichiometry and info given in qn, we find

Carbon dioxide
CH4 + 2O2 --> CO2 + 2H2O
0.99y 2(0.99y) 0.99y (don't care)

Carbon monoxide
CH4 + 3/2 O2 --> CO + 2H2O
0.01y 3/2(0.01y) 0.01y (don't care)

Total vol. of oxygen required =
vol of O2 needed for 1st equation + vol of O2 needed for 2nd equation
= 2(0.99y) + 3/2(0.01y)

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'O' level & 'A' level Qn.

For instance, given that average molar mass of Cl is 35.5, and that there are 2 isotopes, Cl-35 and Cl-37, find the relative frequency of each isotope.

Solution :

(x/100)(35) + ((100-x)/100)(37) = 35.5

x = 75

Hence 75% of all chlorine atoms have a molar mass of 35g, and 25% of all chlorine atoms have a molar mass of 37g.

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Soln to Q8.

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.]

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'A' level Qn :

How is it that transition metals can function as catalysts?

Solution :

Transition metals and their compounds function as catalysts either because of their capacity for variable oxidation states (hence allowing them to readily transfer electrons; this provides an alternative pathway with lower activation energy (as compared to the uncatalyzed pathway), hence increasing the rate of reaction) which is usually the modality of homogenous catalysts; or because of their capacity to adsorb/adsorp other substances (specifically, the reactants) on to their surface, and activate them in the process (transition metals have partially filled d-orbitals and hence can have different number of bonds, and can bond to a wide variety of ions and molecules; and so the catalyst can form temporary bonds with the reactant molecules, which weakens the remaning bonds within the reactant molecules (eg. from double to single bonds), eventually causing them to dissociate into highly reactive atoms, and/or by holding the reactant molecules in closer proximity with each other on the catalyst surface as compared to in free gaseous state, all of which lowers activation energy as well as increasing surface area concentration of reactants available for reaction), which is usually the modality of heterogenous catalysts.

A case of the former (homogenous catalyst; partially filled d-orbitals with capacity for variable oxidation states) would be the redox reaction (catalyzed by Fe²⁺/Fe³⁺ ions) between peroxydisulphate(VI) ions S₂O₈^2-, and iodide ions I-; to generate sulfate(VI) ions SO₄^2- and iodine molecules I₂.

Overall Redox : 2I- + S2O8 2- --> I2 + 2SO4 2-

Step 1 : S2O8 2- + 2e- <---> 2SO4 2-

Reduction potential = +2.01V

Step 2 : Fe3+ + e- <---> Fe2+

Reduction potential = +0.77V

Step 3 : I2 + 2e- <---> 2I-

Reduction potential = +0.54V

The uncatalyzed reaction (steps 1 and 3) is energetically feasible (according to redox potential values); however, the reaction is slow because the negatively charged S2O8 2- and I- anions repel each other.

In the presence of Fe2+/Fe3+ ions, a alternative pathway (that is also energetically feasible, but in addition) with much lower activation energy (since the reactants are oppositely charged ions), and with a much faster rate of reaction, can now occur :

S2O8 2- + 2Fe2+ --> 2SO4 2- + 2Fe3+

2Fe3+ + 2I- --> I2 + 2Fe2+

Two examples of the latter (heterogenous catalyst) would be nickel acting as a catalyst for the hydrogenation reaction between ethene and hydrogen in the presence of a nickel catalyst; as well as iron acting a catalyst for the synthesis of ammonia in the Haber process.

For heterogenous solid metal catalysts in the hydrogenation reaction of alkenes, the mechanism is as follows :

A similar mechanism occurs for the Haber process :

First the H2 and N2 molecules form temporary bonds with the surface of the iron catalyst (ie. adsorbtion/adsorption); this interaction with the catalyst weakens the covalent bonds within the molecules and eventually causing the H2 and N2 molecules to dissociate into atomic hydrogen and nitrogen; which are highly reactive, and rapidly combine with each other to form NH3 molecules, which then leave the surface of the iron catalyst.

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'A' level Qn.

hydrogen iodide, hydrogen bromide, hydrogen chloride, hydrogen fluoride.

Arrange (and explain your arrangement) the above 4 hydrogen halides according to

a) their boiling points, from lowest to highest

b) their acidity when dissolved in water, from strongest to weakest

Solution :

The lowest to the highest boiling points :

Hydrogen chloride (-85.1°C) , hydrogen bromide (–66.38°C), hydrogen iodide (–35.36 °C), hydrogen fluoride (19.54 °C).

Hydrogen fluoride has the highest boiling point due to strong hydrogen bonding between molecules (fluorine is the most electronegative element in the periodic table, hence is capable of the strongest hydrogen bonding).

From hydrogen chloride down the Group to hydrogen iodide, the boiling points increase due to the increasing numbers of electrons in the molecules that in turn leads to an increase in the frequency and strength of (induced dipole - induced dipole) van der Waals forces.

The strongest to weakest acids :

HI > HBr > HCl > HF

This is because the bond energies for the hydrogen halides further down the group are lower, so they dissociate more easily (see Data booklet values below). Furthermore, the conjugate base halide ions increase in stability down the Group, due to the negative charge being spread out over a larger ionic radius (directly related to number of electron shells).

Bond energies

H-F 562 kJ/mol

H-Cl 431 kJ/mol

H-Br 366 kJ/mol

H-I 299 kJ/mol

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Solution to Qn 9.

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.

Solution :

1) Hydroxide ion nucleophile attacks delta +ve carbonyl carbon.

2) Electron density shifts up; pi-bond pair becomes a lone pair on oxygen.

3) When electron density shifts back down (lone pair becomes pi-bond pair, reforming the carbonyl group), the trichloromethyl group leaves as an anion.

4) The trichloromethyl anion is capable of being a leaving group (ie. it is stable enough to exist on its own if only briefly), only because of the 3 highly electronegative and hence strongly electron-withdrawing Cl groups.

5) However, carbanions are still more unstable compared to alkoxide or alkonoate ions, hence the next step is

6) proton transfer (ie. acid-base reaction) from methanoic acid to the carbanion (ie. lone pair on carbanion becomes bond pair with proton (H⁺); bond pair between proton (H⁺) and oxygen becomes lone pair on oxygen), forming

7) trichloromethane and sodium methanoate (negative charge of HCOO^- counterbalanced by Na⁺; note that these ions are aqueous meaning dissolved in water, having a hydration shell, ion-dipole interactions with polar water molecules).

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'A' levels.

Qn.

Describe the bond enthalpies of the halogen molecules down the periods. Explain any anomaly present in the trend.

Ans.

(Quote Data Booklet values here.)

Down the group, due to increasing number of electron shells, the bond lengths increase and hence bond strengths and bond enthalpies decrease.

An anomaly presents itself in that the bond enthalpy of F-F is significantly lower than the bond enthalpies of Cl-Cl and even Br-Br.

The reason for this anomaly, is that due to the very short bond length of F-F, the two F atoms are brought exceedingly close together, that there is significant electrostatic repulsion between the lone pairs of both F atoms; consequently decreasing the effective bond enthalpy of F-F.

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'A' Level 2008 Qn.

Qn. In another separate experiment, a new alkane F, C6H14, was produced. When reacted with bromine under ultraviolet light, F produced only two isomeric monobromo compounds G and H, with the formula C6H13Br. Compound H was chiral. Suggest the structures of F, G and H, explaining your reasoning.

Solution :

The only C6H14 alkane that when monobrominated would yield only 2 isomers one of which is a chiral compound, is 2,3-dimethylbutane.

Notice that 2,3-dimethylbutane has a mirror plane of symmetry. Substituting away H on the 4 terminal C atoms (1st, 4th, 2 methyl and 3 methyl carbons) are equivalent. That leaves the 2nd and 3rd C atoms, which are also equivalent thanks to this mirror plane.

To obtain the chiral compound, substitute away H on (any of the 4) terminal carbons. The 2nd (or 3rd) C atom would then be the chiral carbon.

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H2 'A' level 2008 paper :

List and explain the following alkyl halides (halogenoalkanes) according to their boiling points :

Alkyl bromide, alkyl iodide, alkyl chloride.

Solution :

There are two opposing factors, which you have to describe and explain in the exam to score full marks.

Electronegativity and hence strength of permanent dipole-dipole interactions would point to (highest boiling point) alkyl chloride, alkyl bromide, alkyl iodide (lowest boiling point).

However, no. of electrons and hence strength of induced dipole-dipole van der Waals interactions would point to (highest boiling point) alkyl iodide, alkyl bromide, alkyl chloride (lowest boiling point).

As it turns out, based on experimental evidence, van der Waals interactions trumps/owns/pwnz/outweighs permanent dipole-dipole interactions (specifically in THIS context of alkyl iodide versus alkyl bromide versus alkyl chloride; because we're talking about a lot of electrons here).

Hence, the correct answer is (highest boiling point) alkyl iodide, alkyl bromide, alkyl chloride (lowest boiling point).

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