About Moderator UltimaOnline's BedokFunland JC Tuition Services :

(UltimaOnline is an ex-MOE Teacher, now professional full-time Tuition Teacher)

http://infinity.usanethosting.com/Tuition

 

Chemistry and Biology 

'A' and 'O' Levels - JC 1, JC 2, Sec 3, Sec 4 (all same rates/fees)

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$300 (per person) for 8 group sessions per month (twice a week)

$200 (per person) for 4 group sessions per month (once a week)

$600 (per person) for 8 solo sessions per month (twice a week)

$400 (per person) for 4 solo sessions per month (once a week)

 

All fees must always be paid in advance on 1st session, for every month.

All sessions are 2 hours each.

All sessions held at my place (Lagoon View, located exactly between Victoria JC and Temasek JC).

Students are advised to find a couple of friends to form their own group, and to go for the best deal of $300 (per person) for 8 group sessions per month (twice a week), which works out to be only $18.75 per hour.

 

My contact number is 8189-4871.

 

My Dad (also an ex-MOE Teacher now Tuition Teacher) offers Mathematics tuition for 'A' levels and 'O' levels, at similar rates to mine and also at our place of residence (between Marine Parade and Bedok). For more details, please visit http://infinity.usanethosting.com/Tuition

 

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Recommended books available from Popular Bookstore.

  

Chemistry 'A' Level Study Guide

  

Chemistry Practice Questions (Ten Year Series)

 

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Memoirs as an ex-MOE Teacher

I recall times (very often, actually) when the principal gave hour-plus-long speeches (no exaggeration) during daily morning assembly period, and when he was finally done, he always received tremendous applause (sometimes a standing ovation) from all the students (because they were tremendously relieved the speech was over) and indeed (applause) from a number of (us) teachers as well. Incidentally, the principal always believed the tremendous applause was for his (self-believed) brilliant oratorical skills, and always enjoyed the applause. - UltimaOnline

 

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For 'O' Level School Leavers...

Can't decide between Poly and JC? Read this thread here.
Ultimately, trust that whatever path you've chose and/or you're lead to take, there's always a unique pro (pros vs cons) and unique advantage to that path in life.

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Singapore Examinations and Assessment Board (SEAB)

http://www.seab.gov.sg

- Download 'O' & 'A' Level Syllabuses

- Register Online as private candidate

- View your Examination Results Online

- Contact SEAB at their website for further inquiries

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'A' Level Grades Requirements for Local Universities :

National University of Singapore (NUS)
http://admissions.nus.edu.sg/sprogramme-igp.html

Nanyang Technological University (NTU)
http://www3.ntu.edu.sg/oad2/pdfs/COP.pdf

Singapore Management University (SMU)
http://www.smu.edu.sg/admissions/downloads/pdf/Samp%20Notif%20(SMU)%20COP%20FINAL%20with%20FAQs.pdf

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For 'A' level students interested in Science...

 

For all students who are keenly interested and/or are seriously considering pursuing a lifelong career in Science (eg. medicine, engineering, research & development, etc), and/or are weighing your options of Singapore-vs-Overseas academic paths, the following educational and informative articles are well worth a read.

 

A*STAR - Agency for Science Technology and Research
http://www.a-star.edu.sg/

About doing Science in Singapore
http://takchek.blogspot.com/2006/07/about-doing-science-in-singapore.html

Biomedical Engineering and US Medical School
http://takchek.blogspot.com/2007/01/biomedical-engineering-and-us-medgrad.html

'A' levels and intelligence as predictors of medical careers
http://www.bmj.com/cgi/content/full/327/7407/139

Intellectual aptitude tests and 'A' levels for medical school
http://www.pubmedcentral.nih.gov/articlerender.fcgi?artid=1200591

"Stressed-out varsity applicant? Check"
http://takchek.wordpress.com/2008/03/15/stressed-out-varsity-applicant-check/

The Incomplete Guide to Financial Aid for Singaporeans
http://igfas.wiki.zoho.com/

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Christopher Hendryx's "Oxygen" (an animated video)
http://vimeo.com/4433312

 

Professor Fink explains Cellular Respiration (H2 Biology) on YouTube :
http://www.youtube.com/watch?v=WxQeKBHAdn8
http://www.youtube.com/watch?v=Gfo7df5tiKU
http://www.youtube.com/watch?v=6AhdTZ03Mvg
http://www.youtube.com/watch?v=Zc_rhzn3tPA
http://www.youtube.com/watch?v=dCvfrc4yJ8w
http://www.youtube.com/watch?v=7lkA5yaqY7c
http://www.youtube.com/watch?v=D68uKTG6H0o

Alternatively :
http://www.youtube.com/watch?v=GYG8PgGeedU

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Medical Science is essentially the application of the Chemical, Biological, and technological sciences in an effort to serve and improve the quality of human life. Note, however, that medicine is an evolving science, and many doctors (being humans) would disagree with each other on many aspects of medicine.

 

A book strongly recommended for anyone interested in medicine, health and disease, is "We Want To Live" by Aajonus Vonderplanitz. His all-natural dietary guidelines (ie. no drugs required) has been successfully applied by many people to support healing with all types of diseases, including Cancer, Multiple Sclerosis, Lou Gehrig's disease, Alzheimer's disease, Crohn's disease, Lupus, AIDS, muscular atrophy/dystrophy, nerve degenerative diseases, etc.

 

Episode Reviews by Polite Dissent (a real life MD)
http://www.politedissent.com/house_pd.html

"Dr House" DVDs available from Amazon.com
http://www.amazon.com/House-Seasons-1-4-Collection/dp/B001AV3BY0/

 

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UltimaOnline on Question Ambiguity in the 'O' or 'A' Level Exam

 

As an ex-MOE teacher with experience setting and marking examinations, I advise you : 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.

 

Prime (Best!) 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 following mechanism pathway (which I've drawn) on the left applies. If the temperature is above X deg C, then the following mechanism pathway (which I've drawn) on the right applies."

 

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...

(Note : at 'A' levels, it is not fair for the question to use combinations such as "acidified, cold, concentrated", "alkaline, hot, dilute", etc. Because the adjectives are not quantified (how "hot"? how "dilute"? etc), and the different mechanism pathways as a result of different temperatures/concentrations/acidified-vs-alkaline are only explored at University level. However, here's a sneak peak : temperature is actually the most important factor. Cold = diol, hot = oxidative cleavage) 

 

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...

(Note : Structural isomers are futher subdivided into functional group isomers vs positional isomers vs chain isomers. Stereoisomers are futher subdivided into enantiomers (optical isomers (d vs l; r vs s; (+) vs (-))) vs diastereomers (conformational isomerism vs geometric isomers (cis vs trans; E vs Z))).

 

You'll get your marks in this way (because after showing the examiner that the question's ambiguity is at fault, you've most 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 (assuming the question really is 50-50% ambiguous, not that uncommon, even in the actual 'O' and 'A' levels).

 

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Regarding "Steel cans vs Aluminium cans", have a read :

http://answers.yahoo.com/question/index?qid=20071027185514AA18Knx

"Paul Ding" makes a good point - in the real world, it's not just science which determines the way things are run. This is especially true for medical science.

 

Everyone should be aware about "Dr Royal Raymond Rife and the Cure for Cancer" :

http://www.rense.com/health/rife.htm

  

  

'A' Level Challenge Qn.

 

This question was first asked by a NYJC dragon boater student of mine, Calvin Seah Yong Yeh, who scored a distinction 'A' grade in his 2008 'A' Level Exam for H2 Chemistry under my tuition guidance. Due credit goes to him for coming up with this question, because it is indeed an excellent question (no 'A' level student has so far been able to solve Calvin's Riddle on his/her own, without my help).

 

I'll reveal the answer to Calvin's Riddle only to students who come for my tuition.  Hehe. Just kidding. If you think you know the answer, do NOT post it here, but Private Message me. If you're right, I'll confirm it and say, "Congrats". If you're wrong, I'll tell you, "Keep on persevering, the Esplanade wasn't built in a day."

  

Calvin's Riddle

(aka Calvin's Cunningly Cryptic Chemistry Conundrum Challenge)

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According to the oxidation states of Mn in KMnO4 and Mn2+, the d orbitals of the former should be empty and the latter should be partially filled. So why is the former "purple" and the latter "colourless"?

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Solution :

To find out, either join my tuition ('A' & 'O' Level Chemistry & Biology), or figure it out yourself and Private Message me to check your answer. Have fun!  

 

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The Periodic Table

 

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Standard Redox Potentials

 

Note that you can obtain the oxidation potential by changing the -ve/+ve sign. For instance,

 

The standard reduction potential for Ag+ to Ag is +0.8V.

The standard oxidation potential for Ag to Ag+ is -0.8V.

 

The standard reduction potential for Zn2+ to Zn is -0.76V.

The standard oxidation potential for Zn to Zn2+ is +0.76V.

 

 

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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 very 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 (including by UCLES) 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."

But the (technically erroneous) common definition of isotopes which presume they are "atoms" is commonplace and acceptable at 'O' and 'A' levels (and in fact the majority of schools, teachers, textbooks and even UCLES commonly commit this error).

 

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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, one equation you should be familiar with, in regard to how nitrogen dioxide contributes to acid rain, is

 

2NO2(g) + H2O(l) --> HNO2(aq) + HNO3(aq)

 

(For the mechanism of the hydrolysis reaction represented by the equation above, see page 5 of this thread.)

 

But for the purpose of the exercise of balancing chemical equations, let's say you're given an alternative equation to balance :

 

NO2(g) + H2O(l) --> N2O(g) + HNO3(aq)

 

Try it.

(mercilessly silenced)

'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".

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!"

 

(Challenging) 'O' Levels Qns.

(Standard) 'A' Levels 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  :  0

Change  -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.

 

 

 

(Standard) 'A' Levels Qn.

(Challenging) 'O' Levels 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%

(Drinking Carl's Jr's milkshake at Playground @ Big Splash... mmmm.... nicer than McDonald's milkshake!)

'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 K_sp Fe(OH)₂ = 1.6 X 10^-14 and K_b NH₃ = 1.8 x 10^-5.                  

                                                                                                                                    

Ans : [NH₃] > 2.6 X 10^-6 mol/dm³

 

'A' Level Qns (Hess Law and Energetics)

 

 

 

 

Calculate the enthalpy change for     N₂O₃ (g) + N₂O₅ (s) ® N₂O₄ (g) 

 

            NO(g) + NO₂(g) ® N₂O₃(g)                            ΔH_r = -39.8 kJ/mol

NO(g) + NO₂(g) + O₂(g) ® N₂O₅(g)              ΔH_r = -112.5 kJ/mol

2NO₂(g) ® N₂O₄(g)                                        ΔH_r = -57.2 kJ/mol

2NO(g) + O₂(g) ® 2NO₂(g)                            ΔH_r = -114.2 kJ/mol

N₂O₅(g) ® N₂O₅(s)                                        ΔH_r = -54.1 kJ/mol

 

 

Solution :

http://infinity.usanethosting.com/Tuition/NO2_HessLaw_Solution.gif

 

 

 

 

Solution :

http://infinity.usanethosting.com/Tuition/ALevelH2_EnergeticsQ33b_soln.jpg 

 

 

 

 

 

 

 

Solution :

http://infinity.usanethosting.com/Tuition/ALevelH2_EnergeticsQ33c_soln.jpg

 

 

'A' Level Qn.

 

Qn : 25 cm3 of the diprotic acid phosphorous acid, H₃PO₃, of initial molarity 0.1mol/dm3 was titrated against NaOH, of molarity 0.1mol/dm3.

Given the following equations and corresponding Ka values:-

H₃PO₃      -->     H₂PO₃^-    +    H⁺               Ka₁ = 1x 10^-2
H₂PO₃^-     -->     HPO₃^2-     +    H⁺               Ka₂ = 2.5 x10^-7

Sketch the shape of the pH curve during this titration, labelling significant pH values and relevant volumes of NaOH on the axes of the curve.

 

 

Solution :

 

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)

'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.

 

 

'A' Level Qn :

 

-------------------------------------------

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. 

Qn Level : 'O' & 'A' Levels both.

 

 

Qn : Describe and explain how the QA test for nitrate anion works.

 

 

Ans :

 

Adding aluminium (or zinc) reduces nitrate 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.

 

Heating increases the rate of reaction, so that the test results are quickly obtained.

 

 

Qn Level : 'A' lvl and 'O' lvl both.

 

 

Qn : Explain why it is necessary to acidify silver nitrate 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.

 

(Standard) 'A' Levels Qn.

(Challenging) 'O' Levels Qn.

 

 

 

 

 

 

Ans :
x, y, z, n = 5, 4, 2, 4
Coefficients of balanced eqn = 1, 5, 4, 5

 

 

Solution :

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

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.

'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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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.

'A' & 'O' Level Qn.

(For 'O' levels, info should be given that the oxidation product of the ethanedioate ion C2O4 2- is CO2. 'A' level students 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.

OMG are you sure that's an A level question?!?! It wasnt taught in school...

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.

I will give comments and suggestions, rather than work out the answer for you. As an 'A' & 'O' Levels Chemistry tuition teacher now (ex-MOE teacher), I guide my students to the correct thought processes to help them work out the answers for themselves. It's more helpful this way, generally.

 

 

 

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? 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+).

3) Entropy would be expected to decrease 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 change would be near zero (points 1 would cancel out the effects of points 2 & 3); and without further data given, impossible to state with any certainty if net entropy change is positive or negative.

 

For the purpose of this particular question, by elimination, because we know option B is the obvious answer, we can deduce that (assuming this question is correct, valid and accurate) net entropy change for dissolving of ammonium nitrate is negative.

 

 

 

 

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.

 

(Read the following 2 paragraphs very carefully, it might be a little difficult for some students to understand).

 

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).