BedokFunland JC's A Level H2 Chemistry Qns (Part 1)
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BedokFunland JC's Social Responsibility to Support Students on Financial Assistance :
If you cannot afford Private Tuition, you can post your Chemistry questions here on SgForums, and I will help you out with all your A Level & O Level Chemistry questions free-of-charge.
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This thread is the older (archived) thread of my collection of Chemistry questions, and is no longer being updated (ie. no new questions will be added to this thread). Please view my other thread for regular updates (ie. for new questions to be added regularly).
http://sgforums.com/forums/2297/topics/392932
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Question Ambiguity in the 'O' or 'A' Level Exam
If you encounter ambiguous questions in the exams, write both answers (yes it'll take up more time to do so, but at least your marks are secure). Write a short note to explain to the examiner/marker how the question could be interpreted in both ways, then proceed to address the question in both ways.
Example, updated 20 Apr 09 :
davidche posted : When 20 cm3 of a gaseous hydrocarbon were sparked with 150cm3 of oxygen and the residual gases cooled to rtp, a contraction of 60cm3 occured. A further contraction of 80cm3 took place when the residual gases were subjected to aq NaOH. All the vols were measured at 20 degrees. Determine the formula of the hydrocarbon.
Chemfreak022 posted : Vol of H2O produced = initial contraction = 60 cm3
Vol of CO2 produced = 80 cm3 after NaOH
Therefore vol of CO2 : H2O = 4:3
2CxHy + O2 ---> 8CO2 + 6H2O
Therefore hydrocarbon is C4H6
SBS261P posted : CxHy + (x + y/4) O2 ---> x CO2 + y/2 H2O.
Amount of CO2 absorbed by aq. NaOH = 80cm^3.
Ratio of CxHy : CO2 = 20 : 80 = 1:4
=> x = 4.
There is a contraction of 60cm^3 => final volume is 110cm^3, of which 80cm^3 is CO2. The other 30cm^3 is excess oxygen (since no water left)
=> 120cm^3 of oxygen used.
Ratio of CxHy : O2 = 20 : 120 = 1:6
i.e. x + y/4 = 6
since x = 4
y/4 = 6 - 4 = 2
y = 8
Hence your hydrocarbon is C4H8.
For the record, Chemfreak022 and SBS261P are both correct, depending on how you choose to interpret this ambiguous question (contraction of product gases or contraction of reactant gases?).
In the actual 'A' or 'O' level exam, should such critical abiguity be found in a question, the intelligent candidate will write out both alternative workings and answers, but with qualification and explanation. "Dear Examiner/Marker Sir/Mdm, I regret to inform that the question has failed to unambiguously state whether the contraction of gases referred to the product gases or the reactant gases. As such, I am left with no choice but to provide alternative workings and answers below. On the left below, I assume contraction of product gases. On the right below, I assume contraction of reactant gases. Thank you very much Sir/Mdm, for your kind understanding in this critical matter."
Other examples of exam question ambiguity, and how the intelligent candidate should respond :
Eg. "Dear Examiner Sir, the question did not specify clearly the exact reaction conditions used. If the temperature is below X deg C, then the products are such-and-such. If the temperature is above X deg C, then the products are such-and-such."
Eg. "Dear Examiner Sir, the question did not specify clearly whether the KMnO4(aq) used on the alkene was "alkaline, cold, dilute" or "acidified, hot, concentrated".
If "alkaline, cold, dilute" KMnO4 was used, then the resulting product is...
If "acidified, hot, concentrated" KMnO4 was used, then the resulting product is...
Eg. "Dear Examiner Sir, the question did not specify clearly whether the body temperature had decreased due to ambient environmental temperature, or due to physiological homeostatic response. If the former, then the following explanation applies... If the latter, then the following explanation applies..."
Eg. "Dear Examiner Sir, the question did not specify clearly which categories of isomers are to be considered.
If structural isomers only, then the answers are...
If stereoisomers only, then the answers are...
If structural and stereoisomers, then the answers are...
You'll get your marks in this way (because after showing the examiner that the question's ambiguity is at fault, you've more importantly demonstrated to the examiner you know all your facts/concepts that the question is asking for, whichever interpretation turned out to be the one the question setter or mark scheme had in mind), while candidates who only write one interpretation have a 50% chance of getting zero marks.
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Cambridge Mark Schemes for Cambridge International Examinations (CIE) and Oxford, Cambridge and Royal Society of Arts (OCR) Examination Papers :
www.cie.org.uk
www.ocr.org.uk-------------------------------------------
Balancing Redox Half-Equations ('A' levels)
Here is how you balance half-equations :
Hydrogen peroxide is reduced to water (chemical name : hydrogen hydroxide aka dihydrogen monoxide aka hydrogen(I) oxide).
Hydrogen peroxide is oxidized to molecular oxygen gas.
(Acidified) Manganate(VII) ions are reduced to Manganese(II) cations.
Nitrogen dioxide is oxidized to Nitrate(V) anions.
[reduction]
H2O2 --> H2O
To balance oxygen,
H2O2 --> 2H2O
To balance hydrogen,
2H+ + H2O2 --> 2H2O
To balance charges,
2H+ + 2e- + H2O2 --> 2H2O
[oxidation]
H2O2 --> O2
To balance hydrogen,
H2O2 --> O2 + 2H+
To balance charges,
H2O2 --> O2 + 2H+ + 2e-
[reduction]
MnO4 - --> Mn2+
To balance oxygen,
MnO4 - --> Mn2+ + 4H2O
To balance hydrogen,
8H+ + MnO4 - --> Mn2+ + 4H2O
To balance charges,
8H+ + MnO4 - + 5e- --> Mn2+ + 4H2O
[oxidation]
NO2 --> NO3 -
To balance oxygen,
H2O + NO2 --> NO3 -
To balance hydrogen,
H2O + NO2 --> NO3 - + 2H+
To balance charges,
H2O + NO2 --> NO3 - + 2H+ + e-
To obtain overall balanced redox equations, we ensure no. of electrons lost = no. of electrons gained (by multiplying one or both half-equations to obtain the lowest common multiple for coefficient of electrons), then add up the half equations (left hand side + left hand side, right hand side + right hand side), and finally cancel away any common species on both sides, invariably including all electrons.
For the reaction of using acidified KMnO4(aq) on H2O2, we have
[reduction]
8H+ + MnO4 - + 5e- --> Mn2+ + 4H2O
16H+ + 2MnO4 - + 10e- --> 2Mn2+ + 8H2O
[oxidation]
H2O2 --> O2 + 2H+ + 2e-
5H2O2 --> 5O2 + 10H+ + 10e-
[Balanced Redox]
(16H+ + 2MnO4 - + 10e-) + (5H2O2) --> (2Mn2+ + 8H2O) + (5O2 + 10H+ + 10e-)
6H+ + 2MnO4 - + 5H2O2 --> 2Mn2+ + 8H2O + 5O2
For the reaction of bubbling NO2 into H2O2, we have
[reduction]
2H+ + 2e- + H2O2 --> 2H2O
[oxidation]
H2O + NO2 --> NO3 - + 2H+ + e-
2H2O + 2NO2 --> 2NO3 - + 4H+ + 2e-
[Balanced Redox]
(2H2O + 2NO2) + (2H+ + 2e- + H2O2) --> (2NO3 - + 4H+ + 2e-) + (2H2O)
2NO2 + H2O2 --> 2NO3 - + 2H+
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Did you know?
Two common errors in Chemistry are :
"The 3 isotopes of hydrogen are 1) hydrogen 2) deuterium 3) tritium", and
"Isotopes are atoms with the same no. of protons but different no. of neutrons."
Actually, the 3 isotopes of hydrogen are in fact protium, deuterium and tritium; atoms of which should all be written as 'H' (but with different mass numbers indicated), but the common (though technically erroneous) practice of writing 'D' and 'T' for atoms/ions of deuterium and tritium isotopes, is acceptable by IUPAC.
In addition, the proper definition of isotopes (which means "types" of any given element) is "Isotopes of elements have atoms and ions with the same number of protons, but different number of neutrons", or alternatively, "Different isotopes of the same element, have atoms/ions that have the same number of protons but different number of neutrons."
But the (technically erroneous) common definition of isotopes which presume they are "atoms" is commonplace and acceptable at 'O' and 'A' levels.To reiterate :
Isotopes are NOT atoms. They are "types" of an element (and elements are defined by their unique number of protons).
The correct idea of isotopes is :
"Different isotopes of the same element, have atoms/ions that have the same number of protons but different number of neutrons."
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Balancing Chemical Equations (both 'O' and 'A' levels)
I find the following equation to be good practice for my students (both 'O' levels and 'A' levels). Note that this is a simple, non-ionic, non-redox equation.
At 'O' levels, an equation you should be familiar with, in regard to how nitrogen dioxide contributes to acid rain, is
4NO2(g) + O2(g) + 2H2O(l) --> 4HNO3(aq)
Similarly, for sulfur dioxide contributing to acid rain :
2SO2(g) + O2(g) + 2H2O(l) --> 2H2SO4(aq)
But for the purpose of the exercise of balancing chemical equations, let's say you're given the following equation to balance :
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(mercilessly silenced, and replaced with a Jan 2010 UltimaOnline's post on a Stomp forum)
WhySoSerious28 asked me :
>>> I noticed that you are a former school lecturer-thats cool. Mind sharing why you left the teaching service? <<<Ho ho ho! The million dollar question. If I had a dollar for every time I was asked this question...
But seriously, while I could go on and on about the frustrations of being within the system and how I resolutely reject to be part of a flawed education mechanism that I feel does not augur well for future generations and the well being of humanity; but essentially and ultimately these are only a part of my list of dozens of reasons why I resigned - in nature some more personal, some more professional, some more ideological, some more practical, and so on. And they (these multiple reasons) would *ALL* be true, relevant and contributive (only to differing extents or in differing ways) to my ultimate decision to resign from the MOE teaching service.
I certainly do not regret joining MOE when I did (and still would have done so given the chance to do it all over again), just as I recognize the crucial timeliness, opportuneness and congruousness of resigning and leaving the system exactly when I did. (Perhaps exactly why I left *when* I did, involves a more specific set of reasons, more personal in nature, than the dozens of reasons why leave at all; or to be precise why I knew it was only a matter of time that I inevitably leave).
Am I bitter about the education system or anti-MOE? Not at all, and quite the contrary. I do sincerely and heartily encourage students interested in a teaching career or joining the MOE teaching profession, to give it their best shot to do so. In fact, I'm hoping to see more young people with a sincere passion in teaching and in helping others learn (prerequisites for being a good teacher, whether as a private tutor, or as an MOE teacher), choose teaching as a career over say, the financial sector.(We shall refraining from hijacking and digressing this thread into an intellectual (and forseeably) emotional discussion over which group of people contribute significantly more to the betterment, progress and advancement of humanity - the finance / economics / bankers / business admins / stock-market / wall-street, versus scientists / researchers / engineers / doctors; and how motivated by material and financial greed, we see an increasing trend (notably in Singapore, but also worldwide including UK and USA) of technological-to-finance brain drain whereby the top scholars and brightest sparks of the next generation are in increasing numbers choosing the banking & finance sector, over the science & technology sector, reflected by way of an increasing disparity in popularity of University degrees, pursued careers and long-term focus. Like I said, we'll not go there.)
So to all of you young folks out there who have a genuine passion for particular academic subjects (eg. Chemistry) *and* a genuine passion for teaching and helping others learn, do give due thought and consideration to a teaching career as an MOE teacher.
If, like myself, you find your professional life as an education officer inside MOE to be incompatible (in principle or in practice or in any irreconciliable fashion) to your own personal objectives or values in life, then fret not (for your rice bowl or bread & butter), because (on the mandatory assumption that you're a genuinely effective teacher/tutor, of course) giving private tuition to students in today's merciless education system (and unforgiving paper chase societal dogma) in Singapore, will always remain a beneficial and ethical service, and one in which you can (reasonably) reliably find feasible employment, with a bonus of having the personal fulfillment of being self-directed (as your own boss).
All The Best!
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'O' Level Qn.
How to determine limiting reactant :
1) Compare Experiment Mole Ratio (of 1st reactant to 2nd reactant) versus Equation Mole Ratio (of 1st reactant to 2nd reactant). Eg. hydrogen:oxygen (experiement) versus hydrogen:oxygen (equation).
2) Having calculated numerical values for the two ratios (eg. 4.5 vs 4.3), write the inequality sign between them (eg. 4.5 > 4.3).
3) Highlight (eg. circle) the name of the numerator reactant (eg. "hydrogen"), for both experiment and equation mole ratios.
4) Interpret your findings, eg. "The hydrogen we have in the experiment is greater than the hydrogen we need in the equation, hence hydrogen is the excess reactant and therefore oxygen is the limiting reactant".
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UltimaOnline's Solution :
NO2 + H2O ---> N2O + HNO3
Let coefficient of H2O be x.
NO2 + xH2O ---> N2O + HNO3
Then coefficient of HNO3 becomes 2x.
NO2 + xH2O ---> N2O + 2xHNO3
Let coefficient of N2O be y.
NO2 + xH2O ---> yN2O + 2xHNO3
Then coefficient of NO2 becomes 2x+2y.
(2x+2y)NO2 + xH2O ---> yN2O + 2xHNO3
Looking at oxygen, we have :
2(2x+2y) + x = y + 6x
Simplifying, we get :
x = 3y
Notice that x is the larger value here, and y the smaller value.
Hence, let y be the smallest possible integer, ie. 1.
Consequently, y = 1, x = 3.
Substituting these values, we obtain the balaced equation :
(8)NO2 + (3)H2O ---> (1)N2O + (6)HNO3
This is the systematic way of balancing equations that I teach to my students. I always remind my students, "Algebra is your friend who is here to make your life easier, so use it whenever you can in Chemistry calculations!"
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'O' & 'A' Level Qns.
1)
A sample of 0.43g of an organic compound containing only carbon, hydrogen and oxygen, was burnt in excess oxygen. The combustion produced 1.10g of carbon dioxide and 0.45g of water.
a) Calculate the empirical formula of the compound.
b) Given that this compound has a relative molecular mass of 250g, deduce its molecular formula.
Ans : C15 H30 O3
2) 20cm3 of a gaseous hydrocarbon was mixed with 100cm3 of oxygen so that the hydrocarbon was completely burnt. The volume of gas remaining at the end of the combustion was 70cm3. After passing through soda lime, the volume was reduced to 10cm3. All gases were measured at r.t.p. Determine the formula of the hydrocarbon.
Ans : C3 H6
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%
Solution :
4) When Fe and Fe3+ are mixed together, a reaction occurs in which Fe2+ is produced. What is the ratio of Fe to Fe3+ required to produce equal moles of Fe2+ and Fe3+ when the reaction is complete?
Ans : Ratio is 1:5
Solution :
Write half equations, then write overall balanced Redox equation.
Write ICF table (Initial, Change Final).
Fe + 2Fe3+ --> Fe2+
Initial 1 : x : 0Change -1 : -2 : +3
Final 0 : x - 2 : 3
Since "equal moles of Fe2+ and Fe3+", hence x - 2 = 3 ; x = 5
Therefore, required ratio is 1 : x which is 1 : 5.
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'O' & 'A' Level Qn.
Here is how you balance half-equations :
Hydrogen peroxide is reduced to water (chemical name : hydrogen hydroxide aka dihydrogen monoxide aka hydrogen(I) oxide).
Hydrogen peroxide is oxidized to molecular oxygen gas.
(Acidified) Manganate(VII) ions are reduced to Manganese(II) cations.
Nitrogen dioxide is oxidized to Nitrate(V) anions.
[reduction]
H2O2 --> H2O
To balance oxygen,
H2O2 --> 2H2O
To balance hydrogen,
2H+ + H2O2 --> 2H2O
To balance charges,
2H+ + 2e- + H2O2 --> 2H2O
[oxidation]
H2O2 --> O2
To balance hydrogen,
H2O2 --> O2 + 2H+
To balance charges,
H2O2 --> O2 + 2H+ + 2e-
[reduction]
MnO4 - --> Mn2+
To balance oxygen,
MnO4 - --> Mn2+ + 4H2O
To balance hydrogen,
8H+ + MnO4 - --> Mn2+ + 4H2O
To balance charges,
8H+ + MnO4 - + 5e- --> Mn2+ + 4H2O
[oxidation]
NO2 --> NO3 -
To balance oxygen,
H2O + NO2 --> NO3 -
To balance hydrogen,
H2O + NO2 --> NO3 - + 2H+
To balance charges,
H2O + NO2 --> NO3 - + 2H+ + e-
To obtain overall balanced redox equations, we ensure no. of electrons lost = no. of electrons gained (by multiplying one or both half-equations to obtain the lowest common multiple for coefficient of electrons), then add up the half equations (left hand side + left hand side, right hand side + right hand side), and finally cancel away any common species on both sides, invariably including all electrons.
For the reaction of using acidified KMnO4(aq) on H2O2, we have
[reduction]
8H+ + MnO4 - + 5e- --> Mn2+ + 4H2O
16H+ + 2MnO4 - + 10e- --> 2Mn2+ + 8H2O
[oxidation]
H2O2 --> O2 + 2H+ + 2e-
5H2O2 --> 5O2 + 10H+ + 10e-
[Balanced Redox]
(16H+ + 2MnO4 - + 10e-) + (5H2O2) --> (2Mn2+ + 8H2O) + (5O2 + 10H+ + 10e-)
6H+ + 2MnO4 - + 5H2O2 --> 2Mn2+ + 8H2O + 5O2
For the reaction of bubbling NO2 into H2O2, we have
[reduction]
2H+ + 2e- + H2O2 --> 2H2O
[oxidation]
H2O + NO2 --> NO3 - + 2H+ + e-
2H2O + 2NO2 --> 2NO3 - + 4H+ + 2e-
[Balanced Redox]
(2H2O + 2NO2) + (2H+ + 2e- + H2O2) --> (2NO3 - + 4H+ + 2e-) + (2H2O)
2NO2 + H2O2 --> 2NO3 - + 2H+
The Question :
300 cm3 of a mixture of dinitrogen monoxide and nitrogen dioxide, at r.t.p. conditions is bubbled through 75 cm3 of 0.10 mol/dm3 acidified hydrogen peroxide solution. The nitrogen dioxide is oxidized to nitrate(V) ions, while the inert dinitrogen monoxide does not react. The remaining hydrogen peroxide in 50.0 cm3 of the resulting solution is then titrated with a 0.050 mol/dm3 acidified potassium manganate(VII) solution, of which 22.0 cm3 was required. Calculate the % by volume of dinitrogen monoxide in the gas mixture.
Final Answer :
46%
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(Drinking Carl's Jr's milkshake at Playground @ Big Splash... mmmm.... nicer than McDonald's milkshake!)
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'A' Level Qn (Combining Acid-Base Equilibria with Solubility Equilibria)
Calculate the molarity of NH3 (aq) needed to initiate the precipitation of Fe(OH)2 (s) from a 0.003 mol/dm3 solution of FeCl2 (aq). Given Ksp Fe(OH)2 = 1.6 X 10-14 and Kb NH3 = 1.8 x 10-5.
Ans : [NH3] > 2.6 X 10-6 mol/dm3
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^_^
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The following formula is not usually taught in most JCs, but it's good-to-know should Cambridge one day ask on it :
pH of a solution containing only an amphiprotic species = 1/2 (pKa1 + pKa2)
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'A' Levels Qn.
Based on the Data Booklet, identify 4 reducing agents that, under standard conditions, will reduce Cr3+ to Cr2+ (without further reduction to Cr). Include all relevant reduction potentials in your answer.
Ans :
1 - Fe to Fe2+
2 - Ni and NH3 to hexaaminenickel(II) ion.
3 - Co and NH3 to hexaaminecobalt(II) ion.
4 - Fe(OH)2 to Fe(OH)3 -
'A' Level Qn.
Draw the (Kekule or Lewis or dot-&-cross) structures of
a) triiodide ion, I3 -
b) HF2 -
Ans :
a) Nucleophilic I- ion attacks instantaneous dipole (delta +ve) of one of the I atoms in the I-I molecule; the attacked iodine atom still retains 3 lone pairs in addition to the newly formed dative bond from the iodide ion. The negative formal charge hence shifts to the central I atom.
electron geometry - trigonal bipyramidal (ie. hybridization sp3d)
molecular geometry - linear (I-I-I, with 3 lone pairs & 2 bond pairs about central I atom)
Note : in the 'A' level H2 exam, the VSEPR geometry asked is always the molecular geometry; but you should (ie. it is easiest and smartest to) first mentally figure out (don't memorize blindly, which is boring and unreliable) the electron geometry first (based on number of electron pairs, don't care lone pair or bond pair), then considering how many are lone pairs and how many are bond pairs, figure out the molecular geometry.
b) [ F- ~ H-F ]-
where F- ion is hydrogen bonded (~ represents hydrogen bond) to the H (which is delta +ve) which is covalently bonded to F.
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'A' Level Qn :
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Why does a saturated aqueous solution of 0.713moldm-3 phenol not cause evolution of carbon dioxide when added to 1.0moldm-3 sodium carbonate solution?
Carry out short calculations to aid your explanations.
Ka: H20 +CO2 <---> H+ HCO3- 4.5x10^-7
Ka: HCO3- < -----> H+ CO32- 2.0x10^-4
pKa of phenol = 10.0-------------------------------------------
Soln :
I spotted an error in the question, which makes it impossible to do the question correctly. Common sense dictates that the Ka value for the monohydrogen carbonate ion should be smaller (ie. weaker acid) than the Ka value for dihydrogen carbonate aka carbonic acid (obviously a stronger acid).
At 25 deg C, the correct Ka values should be 4.2 x 10^-7 for carbonic acid, and 4.8 x 10^-11 for the monohydrogen carbonate acid. The correct Kb values should hence be 2.4 x 10^-8 for the monohydrogen carbonate base, and 2.1 x 10^-4 for the carbonate ion.
At 25 deg C, the Ka value for phenol should be 1.3 x 10^-10, and the Kb value for phenoxide ion is 7.7 x 10^-5.
Notice that the strongest acid is carbonic acid, then phenol, then monohydrogen carbonate ion.
Notice that the strongest base is carbonate ion, then phenoxide ion, then monohydrogen carbonate ion.
Based on the dilution effect,
the new molarity of carbonate ions would be 0.5 mol/dm3.
the new molarity of phenol would be 0.3565 mol/dm3.
Imagine hypothetically that, disregarding Ka value (for the sake of explaining an important concept here), ALL of the acidic protons from phenol (the only significant source of acidic protons) were dissociated into solution; we would have 0.3565 mol/dm3 of protons that would be snatched up by two competing bases - 0.5 mol of carbonate ions (from strong electrolyte aqueous sodium carbonate) and 0.3565 mol of phenoxide ions.
Since carbonate ions are stronger bases than phenoxide ions, all of the acidic protons would be snatched by the carbonate ions to form monohydrogen carbonate ions (which being a weaker acid than phenol, would not transfer protons over to the phenoxide ions to form phenol). In fact, there would still be 0.5 - 0.3565 = 0.1435 mol of carbonate ions present, that are stronger bases than the 0.3565 mol of phenoxide ions present.
That being the case (the fact that there is still a significant percentage of carbonate ions (the strongest base of the 3 species) that are still unprotonated), you would certainly not expect any of the monohydrogen carbonate ions to have the opportunity to be protonated to form dihydrogen carbonate aka carbonic acid, which is our only hope of obtaining / releasing from solution carbon dioxide guess.
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'O' & 'A' Level Qn.
Describe and explain how the QA test for nitrate(V) anion works.
Solution :
Adding aluminium (or zinc) reduces nitrate(V) anion to ammonium cation. (Prove this to yourself by checking the oxidation state of nitrogen in these ions.)
NO3- + 10H+ + 8e- --> NH4+ + 3H2O (standard reduction potential = +0.87V)
The hydroxide ion (from aqueous sodium hydroxide added) then combines with the ammonium ion in an equilibirum reaction as follows :
NH4+(aq) + OH-(aq) <---> NH3(aq) + H2O(l)
NH3(aq) <---> NH3(g)
The ammonia gas liberated turns moist red litmus paper blue. The process is the direct opposite of the above - ammonia undergoes hydrolysis ("moist" litmust) to produce OH- ions, which are responsible for turning red litmus paper blue.
Heating provides activation energy to overcome the hydrogen bonding to allow aqueous ammonia to vapourize into gaseous ammonia.
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'O' & 'A' Level Qn.
Qn : Explain why it is necessary to acidify aqueous silver(I) nitrate(V), AgNO3(aq), before testing a solution for chloride ions.
Ans : There are 2 reasons.
Firstly, to avoid false positive ppt formed, from any carbonate ion that might be present. All carbonates are insoluble (ie. they form ppts) except Na+, K+, NH4+. Adding acid will protonate any carbonate ions present, removing them from solution in the form of CO2(g). Hence if a ppt is still formed, it has to be silver chloride.
2H+(aq) + CO3 2-(aq) --> H2CO3(aq) --> CO2(g) + H2O(l)
Secondly, to avoid false positive ppt formed, from any hydroxide ion that might be present. Hydroxide ions will combine with silver ions, to form silver(I) oxide, a brown ppt. (To balance the equation that forms this ppt, add water molecules on RHS). Adding acid will remove any hydroxide ions present, forming water. Hence if a ppt is still formed, it has to be silver chloride.
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'O' & 'A' Level Qn.
Ans :
x, y, z, n = 5, 4, 2, 4
Coefficients of balanced eqn = 1, 5, 4, 5Solution :
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Hi UltimaOnline
Want to help me in ExamWorld?
I'm doing a collection of Questions for students as well...
It's sort of what we are already doing here.... Just that I'm categorizing it properly
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Originally posted by eagle:
Hi UltimaOnline
Want to help me in ExamWorld?
I'm doing a collection of Questions for students as well...
It's sort of what we are already doing here.... Just that I'm categorizing it properly
Hi Eagle,I'll PM you.
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'A' Level Qn :
In an episode of "The Simpsons", Bart mixes nitric acid with ethanol, with the result of a gas being produced. Indeed, this would occur; a total of 5 different possible products, involving 2 separate reaction pathways, is to be expected.
Describe and write equations for these 2 reaction pathways, and draw the Kekule structures of the 5 products. (Bonus Qn : Draw reaction mechanisms.)
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.
.
.
.
.
.
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.
.
Ans :
Redox reaction pathway products :
NO2(g) + CH3CHO(aq) + CH3COOH(aq)
Nitrogen dioxide, ethanal, ethanoic acid.
Proton Transfer, Nucleophilic Substitution, and Condensation reaction pathway products :
H2O(l) + CH3CH2ONO2 (aq)
Water, ethyl nitrate.
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'O' & 'A' Level Qn.
(For 'O' levels, info should be given that the oxidation product of the ethanedioate ion C2O4 2- is CO2. 'A' level students familiar with Organic Chemistry must be able to figure this out by themselves.)
A solution contains a mixture of anhydrous ethanedioc acid, and sodium ethanedioate. 25 cm3 of this solution required 20.0 cm3 of 0.10 mol/dm3 sodium hydroxide for neutralization using phenolphthalein as indicator, and 28.0 cm3 of 0.02 mol/dm3 of potassium manganate(VII) solution for complete oxidation at 60°C in the presence of excess sulfuric acid. Calculate the molarity of the acid and the salt.Ans :
0.04 mol/dm3
0.016 mol/dm3 -
'A' Level Qn.
At sea level, standard atmospheric pressure (ie. 1.01325 x 10^5 Pa) causes mercury in a dish to rise 760 mm up a glass column. A mixture of two alkanes (with molar masses 16.0 g and 30.0 g respectively) is stored in a container at 294 mmHg. The gases undergo complete combustion to produce CO2 that has a pressure of 356mmHg when measured at the same temperature and volume as the original mixture. Calculate the percentage composition of the mixture.
Ans :
78.9% and 21.1% -
'A' level Qn.
Qn : What is the (electron orbital) hybridization of the nitrogen atom in the ammonia molecule?
Ans / Hint :
For (electron orbital) hybridization (eg. sp, sp2, sp3, etc), look at the electron geometry, rather than the molecular geometry. If you don't know the difference between the 2, here's an example : the electron and molecular geometries of the tetrachloromonoiodide ion ICl4-, is octahedral and square planar respectively. The negative formal charge lies on the central iodine atom.
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OMG are you sure that's an A level question?!?! It wasnt taught in school...
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Originally posted by Bellofemme:
OMG are you sure that's an A level question?!?! It wasnt taught in school...
Yep, it was a 2008 Prelim Qn from one of the top JCs.If you're worried about any possible gaps your teacher/JC may have left out, well it's not too late to have some last minute crash-course tuition, even if your 'A' Levels are a month away.
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Q1) What is the correct order for the enthaply of combustion from the least exothermic to the most exothermic?
But-1-ene, trans-but-2-ene, cis-but-2-ene
Suggestion :
The larger the magnitude difference between the stability of reactants and products, the more exothermic (or endothermic) the reaction.
Ask yourself which is more stable, trans or cis isomer? Cis isomers have bulky side chains which suffer from steric or van der Waals repulsion. You should be aware that healthy cis-fatty acids are altered to form toxic, cancer-causing trans-fatty acids when foods are cooked, worst of all deep fried (google Aajonus Vonderplanitz for more info on diet and health). And the position of the double bond in alkenes, relates directly to stability of the alkene : electron donating alkyl groups stabilize the double bond in alkenes. So does but-1-ene have more alkyl groups next to the C=C, or does the cis/trans isomers of but-2-ene?
Q2) Ethanal reacts with CN- from HCN in the presence of a weak base.
In a similar reaction, -CH2COCH3 ions are generated when CH3COCH3 reacts with a strong base. Which one of the following compounds is the product when ethanal reacts with -CH2COCH3?
A. CH3CH(OH)CH2COCH3 B (CH3)2C(OH)CH2CHO
C (CH3)2C(CHO)CH2OH D (CH3)2C(OH)COCH3
Suggestion :
The carbanion nucleophile generated attacks the delta-positive carbonyl carbon in an nucleophilic addition reaction. Draw the Kekule structures and mechanism to figure out which option is the answer.
Q3) Which of the following processes are both the enthalpy change H and S are positive.
A C2H4(g) + H2(g) -->C2H6(g)
B H2O(s) ---> H2O(g)
C H2O2(l) -->H2O(l) + 1/2O2(g)
D NH4NO3 (s) + aq --> NH4+(aq) + NO3-(aq)
Suggestion :
Entropy - Compare LHS vs RHS. Gases have highest degree of disorderliness, followed by aqueous, lastly solid. When both sides have the same state, compare no. of moles.
Enthalpy - For some reactions, use the Data Booklet bond enthalpies/energies data to determine endo or exo. For others, like (B), changing of state, is pretty obvious whether endo or exo. For others, like (D), you should know that by Hess Law, Solution enthaply = (endothermic) Lattice Dissociation enthalpy + (exothermic) Hydration enthalpy (which is the ion-dipole interactions between polar water molecule and the cations & anions).
[Updated 20 Sept 08 : Regarding option 4 (dissolving or solution of ammonium nitrate into water).
This is a well known endothermic reaction (ie. your hand feels cold when holding the test tube containing the reaction mixture).
What about entropy change? This is trickier than at first glance.
Entropy would both increase and decrease (there will be of course either a net increase or decrease, overall) for the following reasons :
1) Entropy would be expected to increase because solid state to aqueous state involves an obvious increase in disorderliness.
2) Entropy would be expected to decrease because of ion-dipole interactions between water molecules and cations & anions (charge density and hence ion-dipole interactions is always higher & stronger for cations than anions, do you know why?). Disorderly water molecules now arrange themselves in an orderly fashion around the ions, particularly those of high charge density (eg. Al3+, Fe3+). For the NH4+ ion, each H atom is capable of forming extra strong (since the positive formal charge on N is strongly electron withdrawing by induction) hydrogen bonds with the solvent water molecules, decreasing entropy.
3) Entropy would also be expected to decrease slightly because of a lowered temperature (due to the reaction being endothermic), and hence less kinetic energy results in less disorderliness of all species present, ions and molecules.
Overall, entropy entropy change would have to take into consideration both the simultaneous increase and decrease in entropy (as discussed above).
But because we know all nitrate(V) compounds and all ammonium compounds are soluble (which means of course, ammonium nitrate is certainly soluble!), meaning that the solution (dissolving) is a feasible & spontaneous process, so by Gibbs Free Energy formula (delta G = delta H - T delta S), we surmise that the entropy change must be positive (ie. favourable) and large enough to overcome the unfavourable (endothermic) enthalpy change.
Hence, both options B and D are correct.
Q4) The reaction between X and Y is given as 2X + 2Y --> 2XY. The mechanism of the reaction is as shown below:
2X <---> X2
X2 + Y --> XY + X(radical) slow
X(radical) + Y --> XY
Which of the following statement is true?
1. Doubling the concentration of X doubles the rate of reaction
2. Doubling the concentration of X and Y increases the rate of reaction by 8 times
3. The overall order of the reaction is 3.
Suggestion :
- The slow step is the rate determining step in which the stoichiometry of the elementary equation will reveal the order of the reaction with regards to the reactants. (note that you can't use stoichiometry for overall equation to determine order of reaction, you can only use stoichiometry on the rate determining elementary equation to determine order of reaction).
- The formation of the free radical is usually the slow step, and the subsequent reaction of the free radical is the extremely fast step, because free radicals are highly unstable and thus highly reactive.
- If the rate determining step involves intermediates (eg. X2), we have to substitute another expression for X2 into the rate equation, because we do not want intermediates as part of the rate equation. From the 1st step (the association of two X atoms to form an X2 molecule), we obtain the expression k1[X]^2 = k2[X2] ==> [X]^2 = [X2](k2)/(k1). Substituting this expression into the rate equation of Rate = k3 [X2] [Y], we obtain the final rate equation of Rate = k [X]^2 [Y]. In other words, the reaction is 2nd order in X and 1st order in Y.
Q5) Which of the following are species of electrophilies?
1. Br2, NO2+
2. AlCl3, HNO3
3. Na+, (CH3)C+
Suggestions :
Electrophiles are formal positive or partial positive charged species that invite attack by nucleophiles (species that have at least one available lone pair for donation). As for diatomic molecules like Br2, there can be instantaneous or induced dipoles that cause a shift in electron density between the atoms, resulting in a delta-positive or delta-negative, the delta-positive functioning as an electrophile to be attacked by nucleophiles such as the pi-bond of an alkene.
Note that electrophiles must have energetically accessible orbitals for attack by nucleophiles. For instance, NH4+ is not an electrophile because being in Period 2, nitrogen does not have empty d orbitals to expand its octet, thus it cannot be attacked by nucleophiles, even if it has a positive formal charge on the nitrogen.
That being said, there can be instances where resonance will result in energetically accessible orbitals being freed up in an electrophile to enable nucleophilic attack. For instance, draw the Kekule structure of NO2+. Initially, the main resonance contributor (ie. the main Kekule structure) is O=N+=O, two double bonded oxygens with a central positive formal charged nitrogen. Although N is in Period 2 and cannot expand its octet, when the nucleophile (eg. pi-electrons of benzene ring) attacks the N, one of the pi-bonds between N and O will shift over to become a lone pair on O, resulting in a negative formal charge on that O. The final NO2 attached to the benzene ring (for instance, if this was an electrophilic aromatic substitution reaction), has 2 formal charges which cancel out : a +ve formal charge on N (because it only has 4 valence electrons from 4 bond pairs, but it is in group V; but note that it certainly has a stable octet), and a -ve formal charge on one of the Os.
Futher comments on this Qn :
1. Br2, NO2+
2. AlCl3, HNO3
3. Na+, (CH3)C+
Br2, NO2+, (CH3)C+ are hence functional electrophiles as explained above. AlCl3 functions as an Lewis acid by accepting an electron pair from a nucleophile (since the Al does not yet have a stable octet in AlCl3, and is willing to accept an electron pair from a Cl2 molecule, forming -AlCl4 and a Cl+ electrophile). The Kekule structure of HNO3 has a negative formal charge on the O, and a positive formal charge on the N. However, HNO3 or nitric acid, functions as an acid (proton donor) or a base (proton acceptor, in the case of reacting with sulfuric acid in the nitrating mixture), an also as a strong oxidizing agent (eg. with Bart Simpson adding ethanol to nitric acid), but not so much as either a nucleophile or an electrophile (note that it has both a positive and a negative formal charge side by side on the N and O, which repels incoming electrophiles and nucleophiles respectively. Moreover, if HNO3 were to behave as an electrophile, and a nucleophile successfully attacks the positive formal charged N, to avoid violating the octet rule, one of the pi-bonds with oxygen will have to become a lone pair on the oxygen, resulting in a species with too many negative formal charges to be stable; bearing in mind that being a strong acid, the acidic proton would also dissociate resulting in yet another negative formal charge for yet another O in the species, which is too unstable).
As for Na+, it has neither the inclination nor tendency to accept electron pairs - Na desperately wants to be oxidized to Na+ (enegretically stable octet), which will resist efforts to be reduced back to Na.
A carbocation has 3 bond pairs and 0 lone pairs, it certainly functions well as an electrophile. And the 3 methyl groups are electron donating by induction, stabilizing somewhat the carbocation allowing it to exist long enough as an electrophile species.
Q6) Manganese (IV) oxide acts as a catalyst in the decomposition of H2O2 to O2. What alteration to the original experimental conditions would increase the volume of O2 produced over time?1. adding more H2O2
2. lowering the temperature
3. adding more H2O
Suggestion :
Adding more H2O2 reactant would obviously result in more O2 product over (a long period of) time.
The decomposition of hydrogen peroxide is exothermic (determine this using bond enthalpies in Data Booklet), so heat is regarded as a product; hence lowering the temperature (ie. reducing the availability of a product) will result in position of equilibirum to be shifted to the RHS, product favoured reaction, so more O2 will be produced at equilibrium, assuming this is an equilibrium reaction.
But is the decomposition of hydrogen peroxide an equilibrium reaction? One of the products, O2, is gaseous, which means that it will leave the reaction mixture, and hence pull over the position of equilibrium completely to the RHS. Hence decomposition of H2O2 is certainly *not* an equilibrium reaction. Hence, the decomposition of hydrogen peroxide would be complete (after a long period of time) anyway, and hence lowering temperature does not make a difference (over a long period of time).
Adding more water does two things - increases molarity of product water, and dilutes molarity of reactant hydrogen peroxide. In regard to kinetics, you'd expect lowerig molarity of reactant would result in a slower rate of O2 produced. And again, assuming it is an equilibirum reaction, you'd expect increasing molarity of RHS product would shift position of equilibriuim to the LHS, resulting in less O2 produced. But because the gaseous product O2 pulls the position of equilibrium completely to the RHS, hence over a long period of time, adding water doesn't affect the final volume of O2 produced (apart from having an additionally small percentage of oxygen gas dissolving into a greater quantity of water).