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BedokFunland JC's A Level H2 Chemistry Qns (Part 2)

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    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
    UltimaOnline's Avatar
    14,146 posts since May '05
    • Memo asked :

      To a 1dm3 saturated solution of magnesium carbonate, 50.0 cm3 of 1.00moldm-3 aqueous magnesium chloride was added. Find the mass of the precipitate formed. Given : Ksp of Magnesium carbonate is 1.00 x 10^-5 mol2dm-6

      Yes, Tetrahedron has posted the correct solution (pun intended)!

      Procedure :

      From the Ksp of MgCO3, find molarity of Mg2+ ions in the 1dm3 saturated solution (before MgCl2(aq) is added), hence find moles of Mg2+ and CO3 2- ions present then.

      Find the moles of Mg2+ ions present in the 50.0 cm3 of 1.00moldm-3 aqueous magnesium chloride.

      Based on the new volume of the solution after the MgCl2(aq) is added, find the new number of moles and hence new formal (ie. initial) molarities of both the Mg2+ and CO3 2- ions present (in the new solution), before precipitation occurs.

      Using algebra, let the moles of MgCO3 precipitated out be x. Hence the new equilibrium molarities of both Mg2+ and CO3 2- ions can be found (both terms will contain x, since the new initial molarities of both ions are known, the change will be minus x for both due to a 1 : 1 stiochiometry)

      Using the Ksp formula and value of MgCO3, and plugging in the new equilibrium molarities of both the Mg2+ and CO3 2- ions, solve for x.

      From moles of MgCO3, multiply by molar mass of MgCO3 to find the mass of the precipitate.

      Edited by UltimaOnline 12 Oct `17, 2:38AM
    • Q1)
      When ammonia is converted into nitric acid on a commercial scale, the following reactions can occur.
      In which reaction does the greatest change in oxidation number of the nitrogen occur?

      A 4NH3 + 5O2 → 4NO + 6H2O
      B 3NO2 + H2O → 2HNO3 + NO
      C 2NO + O2 → 2NO2
      D 4NH3 + 6NO → 5N2 + 6H2O

      Answer is A.


      Q2)
      Why is ethanoic acid a stronger acid in liquid ammonia than in aqueous solution?

      A Ammonia is a stronger base than water.
      B Ammonium ethanoate is completely ionised in aqueous solution.
      C Ammonium ethanoate is strongly acidic in aqueous solution.
      D Liquid ammonia is a more polar solvent than water.

      Answer is A.

       

      Comments on Q1. From -3 to +2, that's an increase of FIVE huge OS units.

      Comments on Q2. The strength of an acid is measured by how much % of it dissociates protons. In water, only a small % of the carboxylic acid molecules present will release / dissociate its protons. In ammonia, almost 100% of the carboxylic acid molecules present will release / dissociate its protons. Simply because ammonia is a stronger Bronsted-Lowry base than water. So the calculated acidic strength, ie. Ka or pKa, of any species is actually relative to its solvent.

      Edited by UltimaOnline 12 Oct `17, 2:41AM
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      Edited by UltimaOnline 12 Oct `17, 2:42AM
    • KickMe asked :

      Regarding the more accurate way of calculating pH at equivalence point, for example, the first equivalence point of HOOCCH2COOH,

      I have 2 options:

      1) -OOCCH2COOH + H2O --> -OOCCH2COO- + H3O+
      2) -OOCCH2COOH + H2O --> HOOCCH2COOH + OH-

      How then do I know which one is the correct one?


      Here (and in point of fact, for most cases), this (ie. considering only either the Ka or the Kb alone, instead of considering both simultaneously vis-a-vis the amphiprotic formula) would be the less accurate way of determining pH.

      As long as the Ka (proton dissociation) and Kb (base hydrolysis) of the amphiprotic species are comparably large (and in a majority of cases they will indeed be), then the amphiprotic formula will be more accurate : pH = 1/2 (pKa1 + pKa2)

      But in rare cases, such as in the conjugate acid of hydrazine, ie. N2H5+, where the Ka value of this (theoretically) amphiprotic species is much larger than the Kb value, or vice-versa, then you should (obviously) use whichever value is larger.

      Consider the diprotic acid H2A.

      If Ka of HA- (ie. Ka2 of H2A) is >>> than Kb of HA- (ie. Kb2 of A2-), then regard HA- as only acidic and use the Ka of HA- (ie. Ka2 of H2A) to determine pH.

      If Kb of HA- (ie. Kb2 of A2-) is >>> than Ka of HA- (ie. Ka2 of H2A), then regard HA- as only basic and use the Kb of HA- (ie. Kb2 of A2-) to determine pH.

      But as long as the Ka and Kb of that species (eg. HA-) are comparably large (ie. less than 1 million times magnitude difference from each other), then this species is indeed (for all practical considerations) truly an amphiprotic species, in which case the amphiprotic formula pH = 1/2 (pKa1 + pKa2) would obtain a more accurate pH answer, because it takes into consideration the fact that this amphiprotic species would actually undergo hydrolysis both ways : as a Bronsted-Lowry acid and also as a Bronsted-Lowry base, simultaneously.

       

      KickMe remains terribly confused at this point, and no, at this point he still doesn't see it. At least not yet.

      Now I see! We consider the value of the Ka and Kb to determine if it is effectively an acid or a base. And in the case of Ka/Kb<10^7, and Kb/Ka<10^7, we use pH=1/2(pKa1+pKa2)

      However, what if it is the 2nd equivalence point of a diprotic acid that I am talking about? I will only have the value of one pKa to use. Then I will not be able to successfully determine the pH at the 2nd equivalence point using 1/2(pK1 + pK2)

      For example, given HOOCCH2COOH,

      pKa1=2.65
      pKa2=5.70

      For the first equivalence point, since Ka is more than one million times larger than Kb, we treat it as an acid, hence use -OOCCH2COOH --> -OOCCH2COO- + H+

      For the second equivalence point, Ka/Kb=1000 hence use 1/2(pK1 + pK2) But theres not pK2 in this case.

      And when I attempt to use the approach of "treat it as acid or base", since Ka is larger than Kb, I treat it as an acid, which in this case is wrong as I am suppose to treat it as a base.

      Once the 1st equivalence point is reached, you have HOOCCH2COO- (which is the only species present), which is clearly an amphiprotic species, hence pH = 1/2 (pKa1 + pKa2).

      (The only time you only need to compare Ka of a species, with Kb of that same species (ie. not Ka of an acid with Kb of that acid's conjugate base), is if the amphiprotic formula does not appear to be mathematically viable, ie. if the species' acidic capacity far outweighs that of its basic capacity, or vice-versa.)

      Once the 2nd equivalence point is reached, you have the dinegative -OOCCH2COO- (which is the only species present), which is clearly only basic (and not acidic at all, since there are no protons available to dissociate, obviously) and therefore you have no choice but to use the Kb (and not Ka, which applies only to acidic species that have protons available to dissociate) of -OOCCH2COO-, which could be labeled as Kb1, to calculate the [OH-], hence pOH, hence pH.

      When we define the various Ka and Kb values such that Ka1 > Ka2, and Kb1 > Kb2, then :
      Ka1 x Kb2 = Kw and Ka2 x Kb1 = Kw
        

      For now, the KickMe saga finally comes to a conclusion :

      I understood the second equivalence point. The only doubt I have now is why HOOCCH2COO- is "clearly an amphiprotic species". By calculating the Ka and Kb, I found the ratio of Ka/Kb=50 million, shouldn't then I treat it as effectively an acid?

      pKa1 = 2.65 hence Ka1 = 2.2387 x 10^-3 hence Kb2 = 4.4669 x 10^-12
      pKa2=5.70 hence Ka2 = 1.9953 x 10^-6 hence Kb1 = 5.0118 x 10^-9

      For the amphiprotic species, Ka2 and Kb2 are relevant.
      Since Ka2 is only 446,669 times larger compared to Kb2, hence this is indeed an amphiprotic species, and the amphiprotic formula for pH applies with validity.

      Edited by UltimaOnline 10 Nov `12, 8:59PM
    • I thought that was the end of it, but apparently not yet...

      KickMe persisted doggedly :

      Why do you get Kb2 from pKa1 and Kb1 from pKa2?

       

      Because as long as we consistently define Ka1 > Ka2 > Ka3 etc, and Kb1 > Kb2 > Kb3 etc, then :

      Let H2A represent a weak molecular diprotic acid (eg. the conjugate acid of A2-)
      Ka1 is the 1st proton dissociation, H2A --> H+ + HA-
      Ka2 is the 2nd proton dissociation, HA- --> H+ + A2-

      Hence,

      Let A2- represent a diprotic base (eg. the dinegative conjugate base of H2A).
      Kb1 is the 1st base hydrolysis, A2- + H2O --> OH- + HA-
      Kb2 is the 2st base hydrolysis, HA- + H2O --> OH- + H2A

      By observing the relevant species involved, notice that :

      Ka1 x Kb2 = Kw, since both Ka1 and Kb2 deal with the same conjugate acid-base pair of H2A / HA-.
       
      Ka2 x Kb1 = Kw, since both Ka2 and Kb1 deal with the same conjugate acid-base pair of HA- / A2-.

      The amphiprotic species here is obviously HA-.

      If HA- were to function as a Bronsted-Lowry acid,
      then it is Ka2 that is relevant, since HA- --> H+ + A2-

      If HA- were to function as a Bronsted-Lowry base,
      then it is Kb2 that is relevant, since HA- + H2O --> OH- + H2A

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

      The above equations and formulae regarding the various proton dissociations or base hydrolyses, can always be used, regardless of the symbols used, eg. say if the acid was the conjugate acid (eg. H2B2+) of a weak molecular diprotic base (eg. B), and the amphiprotic species would then be HB+.

    • Drbiology asked :

      Do you think that alkyl and hydride shift is likely to be tested in H2 syllabus? There are far-reaching implications if they chose to do so. For instance we must consider the Friedel-craft acylation instead of alkylation due to instability of carbocation intermediate formed.

      Anyway, am I allowed to use reagents I learned in H3 for synthesis questions in paper 3? For instance PCC, DIBAL-H, etc. My school chem teacher told me not to do so, but am I allowed to?

       

      Regarding the application of reagents or knowledge beyond that which is usually taught within the H2 syllabus, in answering H2 A Level exams, I still advise the same exam-smart strategies :

      Whenever you see the possibility of equally appropriate alternative answers, go ahead and write them all. Of course, if the most appropriate answer (ie. what you suspect Cambridge is most likely looking for) can be found within the H2 syllabus (and this is most usually the case), then go ahead and write that H2 answer in first.

      Thereafter, if you feel it appropriate (eg. if you feel a better answer would be one that you've learnt outside the H2 syllabus), then go ahead and add that in as well.

      Despite what some JC teachers might warn you about, writing in alternative answers in the A Level exams is an exam smart thing to do, as long as you carefully qualify your answers, and as long as none of these answer contradict each other in any way (in which case the marker would think you're trying to cheat and of course you'll get zero marks for the question).

      ( Examples of this would be to write all 4 of the following non-contradictory answers, when asked to describe the type of reaction between an alcohol and a carboxylic acid or acyl halide, to generate an ester :
      esterification, condensation, nucleophilic acyl substitution, addition-elimination )

      Since the H2 Chem papers are increasingly testing beyond the basic H2 syllabus, this issue is all the more relevant and pertinent. Bottonlime : When faced with such questions, go ahead and write in both the H2 answer as well as the better H3 answer, with an appropriate statement so the Cambridger marker understands you, eg. "we could use this H2 method OR this H3 method to synthesize this product, but the H3 method has the following advantages...". Just be careful that your answers don't contradict each other in any way.

      Usually, if the desired answer by Cambridge, involves concepts beyond that which is typically taught in H2 syllabus, they will give the necessary info on the concept as part of the question, ie. a data-based question. This will likely be the case for concepts such as carbocation rearrangements via alkyl and hydride shifts.

      If so, all the more power to those of you who made the effort to study beyond the basic H2 syllabus. But it's still fair to those who didn't (ie. the typical within-syllabus H2 student doesn't lose out too severely to the H3 or Olympiad or BedokFunland JC student, because all the necessary data will likely be provided in the question, or the answer can be extrapolated from the basic syllabus with a bit of thinking, etc).

      Yes, carbocation rearrangement can be asked by Cambridge in a H2 exam, in the form of "Carbocation rearrangements can occur as follows... Clemmenson reduction involves the following... The acylium ion is resonance stabilized as follows... Suggest why Friedal-Crafts alkylation to generate propylbenzene using propyl chloride has a low yield, and suggest how this difficulty can be overcome" or something similar.

      My BedokFunland JC Challenge Qn titled "H2Chem_Clemmenson_Reduction" deals with this problem and gives the solution.

       

      Dr Biology replied :

      Very insightful, thank you Ultima! But I guess there is still the 'time' factor which could really prevent most able students from writing the most appropriate answers with extensive line of reasoning. This would, more often than not, limit the answers to the perceived accurate answer.

       
      No problem at all.

      And yes you're right, time constraints will always be a factor of being exam-smart, as well. In this regard, my advice to students is to always skip any and all questions that you suspect will be time consuming (eg. difficult equilibria calculation qns, or Deductive Elucidation qns, etc), and/or questions on topics that (for whatever reason) you're ill-prepared to handle (eg. particularly so for the Planning Qn of Paper 2 ; for instance in last week's paper, unless you happened to have prepared very well for this Kinetics clock reaction Planning topic, it would have been exam-smart of you to have skipped it to focus on the other questions more productively first).

      Edited by UltimaOnline 12 Nov `12, 12:18AM
    • Memo asked :

      Speaking of Deductive Elucidation qns which will be appearing in P3. Is it wise to skip that question which consists of 8/9/10m organic elucidation qn? and choose the other 4qn to do?

       

      Depends. Have a quick read and spend a minute or so on the Deductive Elucidation qn. If you feel confident you've a rough idea on the skeleton of the structure, and can flesh it out pretty quick, then go ahead and go right into the question. But if there are too many details that confuse you, or if after a couple of minutes into the question you're stuck, then skip it and proceed to the next question.

      Killer Deductive Elucidation qns do not just appear in Prelim papers. In the last few years, Cambridge has been setting increasingly sadistic questions in this regard as well, particularly on conditions that involve isomeric branching and chiral carbons.

      And as always, keep an open mind for cyclo-compounds and dienes that can be oxidatively cleaved to generate ethandial that's further oxidized to ethandioic acid that's further oxidized to carbonic(IV) acid, that exists in equilibrium with and hence can decompose into, carbon dioxide and water.
       
      Also beware that "hot, acidified, KMnO4(aq)" does more than just oxidation : it can also protonate any basic groups (eg. amines), and can also hydroyse groups such as esters, amides and nitriles. Similarly for "hot, alkaline...".

      Also note that "a gas that turned moist red litmus paper blue" may not necessarily be ammonia, but can be any amine (primary, secondary or tertiary), or diamine, or triamine, etc, with a low boiling point.


      Edited by UltimaOnline 12 Oct `17, 2:43AM
    • KickMe asked :

      Just realised that there isnt any instructions on the no. of s.f. to give in the A levels. How do we know how many s.f. we should give?

       

      For H2 Chem purposes :

      Rule #1
      Follow the question's data, particularly if the question instructs "Give your answer to an appropriate number of significant figures or decimal places". Whether the question's data gives to 1 significant figure or 10 decimal places, follow suit.

      Rule #2
      Rule #1 takes priority, but if Rule #1 is not relevant or specified, then all intermediate workings should be given to 5 significant figures, final answer should also be given to 5 significant figures first, then rounded off and presented to 3 significant figures.

      Some JC teachers instruct their students to work with 3 significant figures all the way, which is risky and not advisable.

      Edited by UltimaOnline 13 Nov `12, 12:10AM
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    • Question :

      3-methylpentane can undergo reaction with chlorine to form monosubstituted compounds that are optically active. How many possible isomers, including stereoisomers, can be formed in the reaction?

       

      Answer :

      8 mono-halogenated isomers may be generated, which are :

      R-1-chloro-3-methylpentane
      S-1-chloro-3-methylpentane
      2R,3R-2-chloro-3-methylpentane
      2R,3S-2-chloro-3-methylpentane
      2S,3R-2-chloro-3-methylpentane
      2S,3S-2-chloro-3-methylpentane
      3-chloro-3-methylpentane
      3-chloromethylpentane

    • MCQ 2007 "A" levels
      Qn24.
      A solid compound Z dissolved readily in water to give a weakly alkaline solution. On evaporation of the water, Z was recovered unchanged.

      What could Z be?

      A:CH3NH3+Cl-
      B:CCl3-CO2-Na+
      C:C6H5O-Na+
      D:H2NCH2CO2H

       

      Solution :

      This is a bit of an unfair question by Cambridge, because of the subjective "weakly" adjective. Both B and C are arguably correct, but when faced with two possible answers, you gotta choose the better answer.

      If option B was simply ethanoic acid, the question would be more problematic. But here, due to the strong electron-withdrawing by induction effect of the 3 Cl atoms, the strength of trichloroethanoic acid is increased by a tremendous magnitude.

      Compare the pKa values :
      http://en.wikipedia.org/wiki/Phenol vs http://en.wikipedia.org/wiki/Trichloroacetic_acid

      When the pKa value of an acid is very low, the acid is very strong, which means that the conjugate base does not undergo hydrolysis to any significant extent.

      If you've understood everything I said so far, the best answer to this MCQ should now be obvious to you.

    •  A 100cm^3 solution contains 0.2 mol/dm^3 MgCl2 and 0.1 mol/dm^3 CuCl2. A solution of sodium hydroxide is added to the mixture. Mg(OH)2 starts precipitating when 40 cm^3 of sodium hydroxide has been added.

      Ksp of Mg(OH)2 is 6.3 x 10^-10, Ksp of Cu(OH)2 is 2.2 x 10^-20

      What is the concentration, in mold/dm^3, of Cu^2+ in the solution when Mg(OH)2 just precipitates?

      Solution :

      This question has several red herrings.

      From Ksp of Mg(OH)2, and new molarity of Mg2+, find molarity of OH- at that point. Plug this molarity into the Ksp of Cu(OH)2, find molarity of Cu2+.

      Answer is 4.99 x 10^-12 mol/dm^3

    • 1. Which of the following statements contains one mole of the stated particle?
      A Molecules in 19.0 g of fluorine gas.
      B Electrons in 24.0 dm3 of hydrogen gas at room temperature and pressure.
      C Neutrons in 1.00 g of helium gas.
      D Protons in 2.02 g of neon gas.



      2. An experiment is conducted to investigate the kinetics of reaction between bromopropane and 0.1 mol dm-3 sodium hydroxide.

      The rate equation is as follows:

      Rate = k [bromopropane] [OH-]

      The half-life of bromopropane in one of the experiments is t minutes.

      What is the new half-life (in minutes) of bromopropane when the concentration of bromopropane is doubled and concentration of sodium hydroxide is reduced to 0.01 mol dm-3?

      A 0.05t
      B 0.1t
      C 5t
      D 10t



      3.Consider the following equilibrium system:

      H2 (g) + I2 (g) 2HI (g) H = +53 kJ mol-1

      Which of the following change is incorrect?

      A Numerical value of Kp is not equal to Kc at 25 C.

      B Increasing the mass of H2 will not cause the equilibrium constant to increase.

      C Increasing temperature increases the rate constant and equilibrium constant.

      D Rate of forward reaction is equal to rate of backward reaction when equilibrium is reached.



      4. The solubility product of iron(II) carbonate is 2.1  10–11 while that of silver carbonate is 8.1  10–12 at 25°C.

      Which of the following statements is true?

      A Addition of silver nitrate increases the solubility of silver carbonate.

      B Addition of sulfuric acid to a solution containing iron(II) carbonate increases the solubility product of iron(II) carbonate.

      C Iron(II) carbonate precipitates first when sodium carbonate is added to a solution containing equal concentrations of iron(II) and silver ions.

      D The solubility of iron(II) carbonate is higher than the solubility of silver carbonate.



      5. Liquid E has an H of vapourisation of 10.0 kJ mol1 and a boiling point of
      266 K.

      What is the S of condensation of vapour E?

      A –26.6 J mol1 K1

      B –37.6 J mol1 K1

      C +26.6 J mol1 K1

      D +37.6 J mol1 K1



      6. 2-methylpropane can react with bromine in the presence of sunlight to give two monosubstituted halogenoalkanes, 1-bromo-2-methylpropane and 2-bromo-2-methylpropane.
      Given the relative rates of abstracting H atoms are:
      Type of H atom primary secondary tertiary
      Relative rate of abstraction 1 4 6

      What is the expected ratio of 1-bromo-2-methylpropane to 2-bromo-2-methylpropane formed?

      A 9 : 1 C 6 : 1


      B 3 : 2 D 1 : 1



      7. In which sequences are the molecules quoted in order of decreasing boiling points?

      1 CH3(CH2)3CH3, (CH3)2CHCH2CH3, CH3C(CH3)2CH3

      2 AlBr3, AlCl3, AlF3

      3 SO2, SiO2, CO2



      8. A cell consisting of a V2+ (aq), V3+ (aq) | Pt (s) half-cell and a Au3+ (aq) | Au (s) half-cell is shown below using conventional notation.
      Pt (s) | V2+ (aq), V3+ (aq) || Au3+ (aq) | Au (s) Ecell = +1.76 V
      Which of the following statements is true?
      1 The mass of the Au electrode increase.
      2 The negative electrode is the Pt electrode.

      3 The standard electrode potential for Au3+ (aq) | Au (s) is +2.02 V.



      9. Oxytetracycline is a class of broad-spectrum antibiotics used to treat a variety of infections.

      Which of the following statements about oxytetracycline is correct?
      1 One mole of oxytetracycline reacts with three moles of thionyl chloride.
      2 One mole of oxytetracycline reacts with two moles of hot sodium hydroxide to liberate one mole of ammonia gas.

      3 One mole of oxytetracycline reacts with six moles of ethanoyl chloride.



      10. The mass percentage of magnesium in a mixture of magnesium chloride and
      magnesium nitrate was found to be 21.25%. What mass of magnesium chloride
      is present in 100 g of the mixture?

      A 47 g
      B 51 g
      C 53 g
      D 56 g



      11. Methane was burned in an incorrectly adjusted burner. The methane was converted into a
      mixture of carbon dioxide and carbon monoxide in the ratio of 98:2, together with water
      vapour.
      What will be the volume of oxygen consumed when y dm3
      of methane is burned?

      A 1.99y dm3
      C 0.995y dm3
      B 1.995y dm3
      D 0.99y dm3



      12. To identify an oxide of nitrogen, 0.10 mol of the oxide was mixed with 10 dm3
      of hydrogen gas and passed over a heated catalyst. At the end of the reaction, 0.4 dm3
      of hydrogen gas remained. The ammonia produced required 125 cm3 of 1.6 mol dm-3
      HCl for neutralisation. All gasoues volumes were measured at room temperature and pressure.

      What is the formula of the oxide of nitrogen?

      A NO C NO2
      B N2O D N2O4



      13. Which atom has the highest ratio of unpaired electrons to paired electrons in its ground
      state?

      A boron C nitrogen

      B carbon D oxygen



      14. Use of the Data Booklet is relevant to this question.

      Based on bond energies listed in the Data Booklet, what are the possible products of the
      following reaction?

      •CH3 + CH3CH2Cl ?

      A CH4 + CH3CHCl C CH3CH2CH3 + Cl•
      B CH3CH2CH2• + HCl D CH3CH2CH2Cl + H•




      15. Which of the following quantities is equal to the Avogadro constant?

      1 The number of oxygen atoms in 49.9 g of allactite, Mn7(AsO4)2(OH)8, of molar mas
      798 g mol^-1

      2 The number of aqueous chloride ions in a solution containing 0.5 mol of the comple
      [Cr(H2O)5Cl]Cl2

      3 The number of ions in 168 g of Reinecke’s salt, NH4[Cr(NH3)2(SCN)4], of molar mas
      336 g mol^-1





      16. Which of the following is hydrogen bonded in the liquid state?

      1 CH3NH2

      2 CH3CHO

      3 CH2F2



      17. Which of the following mixture produce ND3 gas upon heating?
      [D = an isotope of hydrogen]

      1 CaO (s) and ND4Cl (s)

      2 CH3CN and NaOD in D2O


      3 CH3CONH2 and NaOD in D2O

       

      (Partial) Solutions :

      Q1. From sample mass of neon gas, find moles of neon gas. Then find moles of protons (1mol of Ne contains 10mol of protons).

      Q2. Changing the molarity of a reactant does not alter its half-life. But changing the molarity of other reactants, will.

      Q3. Kp and Kc have the same mathematical value, under standard conditions. Alternatively, this MCQ can be solved by elimination.

      Q4. From Ksp, find molar solubility. Fe(OH)2 will have a smaller molar solubility.
      Note : usually for such questions, you can find the actual value of [OH-] at the point when the less soluble compound precipitates out. Plugging this molarity into the Ksp for the other (more soluble) compound, you can obtain the required molarity of the cation, which will be found to be mathematically larger than the existing molarity at this point.

      Q5. At boiling point (or melting point, for that matter), the system is at equilibrium and delta G = 0.

      Q6. Combine the two factors mathematically (by multiplying them together)
      - number of H atoms subtitutable
      - stability of alkyl radical intermediate

      Q7. For the first case, branching decreases the surface area available for van der Waals attractions.
      For the second case, the greater the number of electrons present, and the greater the molecular size, hence the more polarizable the electron charge clouds, hence the greater the magnitude of dipoles and partial charges, hence the stronger the electrostatic attractions, therefore the stronger the van der Waals.

      Q8. From the cell notation, Au is the cathode, Pt is the anode. Cell potential = Reduction potential @ Cathode + Oxidation potential @ Anode.

      Q9. The amide N atom and the enolic OH groups do not react with SOCl2.
      1 mole of OH- is required for (base-promoted) hydrolysis of the terminal / primary amide, generating NH3. Another 1 mole of OH- is required to substitute away the -NR2 group, generating NR2H (where R = CH3).

      Q10. Use algebra (only 1 algebraic variable required, since it's out of 100%) to solve.

      Q11. Write two separate equations, one for complete combustion, one for incomplete combustion. Add up total O2 required.

      Q12. From data, you can find the ratio of N to O, and also the actual moles of the oxide. Hence solve for x and y (in NxOy).

      Q13. For N, each p orbital is only singly filled and hence unpaired.

      Q14. Weaker bonds are more readily broken than stronger bonds, and stronger bonds are more readily formed than weaker bonds. This is why you cannot iodinate an alkane via free radical substitution.

      Q15. Avogadro constant = 1 mole.

      Q16. CH3NH2 can both accept and donate H bonds, while CH3CHO and CH2F2 can only accept H bonds.

      Q17. It's all about the mechanism.

      1 - CaO (s) and ND4Cl (s). D+ proton transferred from ND4+ to O2-, generating ND3 and OD-.

      2 CH3CN and NaOD in D2O. OD- attacks C, pi bond shifts to N, N grabs D+ proton; repeat until you obtain a geminal triol (which dehydrates into a carboxylic acid) and ND2-, which grabs a D+ proton from the carboxylic acid to generate ND3.

      3 CH3CONH2 and NaOD in D2O. OD- attacks C, pi bond shifts up, reforms carbonyl group, forming RCOOD while NH2 is eliminated as NH2-, which grabs a D+ proton from RCOOD to generate RCOO- and NH2D.

    • KickMe asked :

      Why is it that C2H6 has a greater entropy standard molar entropy that C2H4? Why does the stronger id-id interactions in C2H6 not result in a smaller entropy than C2H4?

      Answer :

      Regarding the van der Waals attractions, interestingly ethane does indeed have a slightly higher boiling point, but ethene has a slightly higher melting point (because melting point has to do with symmetry / stackability as well).

      As long as the state (eg. gaseous versus gaseous) is the same, for entropy we do not look at the strength of the intermolecular attractions, but rather we consider the total number of permutations and combinations possible.

      The more atoms are present, the more ways to permutate and combine, in terms of atom rotations, bond rotations, bond stretchings, bond bendings, etc. Hence entropy is greater for the molecule with greater number of atoms.

    • Originally posted by hoay:

      NaCl is not hydrolyzed by water instaed water just hydrates it. AlCl3 is hydrolyzed by water producing acidic solution. CH3Cl in hydrolzsed by NaOH (aq) and by water to form alcohol... Hydrolysis is the reaction with water or NaOH(aq)?? Please calrify and define hydrolysis.


      Technically, hydrolysis refers to the chemical reaction with water (don't worry about the "lysis" part, as either water or the other reactant will inevitably be broken up or "lysed" when the reaction occurs).

      However, because water is a weaker nucleophile, therefore to save time and money, we use a stronger nucleophile OH- in nucleophilic substitution reactions involving water, as the richer the electron-rich species, the stronger its nucleophilic strength, the lower the Ea required, the faster the rate of reaction.

      Furthermore, the final intended product is the same. For instance :

      R-X reacts with water to generate R-OH2+, which loses a proton to X-, hence generating R-OH and H-X as the final products. Higher Ea, slower reaction.

      R-X reacts with Na+OH- to generate R-OH, with Na+ and X- as counter ions, generating R-OH and Na+X- as the final products. Lower Ea, faster reaction.

      On a related note, notice that physical & inorganic chemists are lazier compared to organic chemists in some ways (eg. physical/inorganic chemists' quick dative bond arrowhead versus organic chemists' full curved-arrow mechanism), but are stricter in other ways (eg. physical/inorganic chemists say "hydration refers to physical interaction with water, while hydrolysis refers to chemical reaction with water"; but organic chemists say, "since dehydration means removing water from alcohol to alkene, then let's just call adding water to alkene as hydration; no need to specify hydrolysis rather than hydration"). 

    • EmpireCity asked :

      AlCl3 reacts with AlH4 and (CH3)3N to give (CH3)3NAlH3.
      Which statement about (CH3)3NAlH3 is correct?
      A It contains hydrogen bonding
      B It is dimeric
      C The Al atom is electron deficient
      D The bonds around Al atom are tetrahedrally arranged.

      Why is B untrue when the product has dative bonding?

       
      Mechanism :
       
      Hydride ion (H-) transfers from AlH4- to AlCl3, generating AlH3 (a Lewis acid or 'electrophile'), which is then available for trimethylamine (CH3)3N (a Lewis base or 'nucleophile') to attack, forming the Lewis acid-base product of (CH3)3NAlH3 with AlCl3H- as a byproduct.

      This (arguably) isn't considered 'dimerization' because an H- ion is lost from the two 'monomers' when forming the 'dimer' product. (It's a bit of an ambiguous area because you can also argue that polymerization can be either addition polymerization or condensation polymerization). So it's still a rather lousy option to include in this weak question.

    • _

      Edited by UltimaOnline 12 Oct `17, 2:46AM
    • The rate of metabolic removal of paracetamol from the human body is a first order reaction with k = 0.26 / hour. How much time will have elapsed, between taking a paracetamol pill, until only 25% remains in the body?

      Solution :

      [Final] / [Initial] = ( 1 / 2 ) ^ n            where n = number of half lives
      1 / 4 = ( 1 / 2 ) ^ n
      n = 2

      Since half-life = ln2 / k = 2.666 hours

      Hence time elapsed = (number of half lives) x (duration of half life) = 2 x 2.666 = 5.332 hours

    • The mass percentage of magnesium in a mixture of magnesium chloride and
      magnesium nitrate was found to be 21.25%. What mass of magnesium chloride
      is present in 100 g of the mixture?

      Solution :

      % by mass of Mg in MgCl2 = 2.5498 x 10^-1
      % by mass of Mg in Mg(NO3)2 = 1.6386 x 10^-1

      Since % by mass of Mg in mixture = 21.25%, hence :
      21.25 / 100 = X (2.5498x10^-1) + (1-X) (1.6386x10^-1)
      X = 53.38 g

      Edited by UltimaOnline 26 Nov `12, 6:54PM
    • Rate = k [Br2] [NO]^2

      With 3 mol/dm3 of NO, time taken for the molarity of Br2 to decrease to 1/8 its original molarity, is 15 hours. If 6 mol/dm3 of NO is used, what's the time taken for the molarity of Br2 to decrease to 1/8 its original molarity?

      Solution :

      Since 15 hours elapsed to reduce molarity from 1k to 1/8k,

      [Final] / [Initial]   =   (1/2) ^ n
      1/8 = (1/2) ^ n
      n = 3

      Since 3 (old) half-lives = 15 hours, hence 1 (old) half-life = 5 hours.

      Since molarity of the second-order reactant is doubled, the rate of reaction increases by (change in molarity) ^ (order of reactant) = 2 ^ 2 = 4 times.

      Hence (new) half-life of the other reactant is the (old) half-life decreased by 4 times, ie. 5 / 4 = 1.25 hours.

      Since molarity of Br2 is to be reduced from 1k to 1/8k,

      [Final] / [Initial]   =   (1/2) ^ n
      1/8 = (1/2) ^ n
      n = 3

      Since (new) half-live = 1.25 hours, hence time elapsed to decrease molarity of Br2 from 1k to 1/8k = 3 x 1.25 = 3.75 hours.

      Edited by UltimaOnline 27 Nov `12, 12:03PM
    • 2012 H2 Chemistry Paper 1 (Answers)

      Q1. C
      Q2. C
      Q3. D
      Q4. B
      Q5. C
      Q6. A
      Q7. D
      Q8. C
      Q9. A
      Q10. D
      Q11. D
      Q12. C
      Q13. B
      Q14. A
      Q15. A
      Q16. D
      Q17. D
      Q18. C
      Q19. C
      Q20. C
      Q21. D
      Q22. B
      Q23. B
      Q24. D
      Q25. C
      Q26. A
      Q27. C
      Q28. B
      Q29. B
      Q30. A
      Q31. B
      Q32. D
      Q33. A
      Q34. B
      Q35. B
      Q36. B
      Q37. D
      Q38. A
      Q39. B
      Q40. B

      Edited by UltimaOnline 28 Nov `12, 2:59PM
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