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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)
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    • 2017 BedokFunland JC H2 Chemistry Challenge Qn

      Draw the full curved-arrow electron-flow multi-step reaction mechanism for the hydrolysis followed by dehydration of (i) aluminium chloride and (ii) silicon tetrachloride, including an appropriate representation of the transition from simple molecular reactant, to ionic product (for i) and giant covalent lattice product (for ii).

      Also draw the full curved-arrow electron-flow multi-step reaction mechanism for the complete hydrolysis of magnesium nitride to generate a solid and a gas. The correct stoichiometry must be accurately reflected in the reaction mechanism.

      Answer :

      BedokFunland JC students can ask me during tuition. Other students can go ask your school teacher or private tutor.

      Edited by UltimaOnline 11 Mar `17, 7:17PM
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    • Originally posted by hoay:

      A gardener fertilizer is said to have a phosphorous content of 30.0% soluble  in water.

      what is the % by mass of the phosphorous in the fertilizer?

      A     6.55 %          B   13.1 %          C   26.2 %            D   30.0 %


      Ans ... the balanced reaction is  P2O5   +   2H2O............... 2H3PO4

      Now 30.0 % of P2O5 (Mr = 142) is 42.6. so mass of phosphorous will be 

      142g..........62g in H3PO4

      42.6..........x g in H3PO4

      x = 18.6 g of Phosphorous

      since only 30.0% is soluble so out of total mass of H3PO4 (196) only 58.8 is available

      18.6/58.8 x 100 = 13.8 %

      Is the working correct?










      The original Cambridge question is different from the one you paraphrased.

      For the original question, P2O5 takes up 30% (by mass) of the fertilizer.

      Calculate % (by mass) of P in P2O5, ie. (2 x Mr of P) / (Mr of P2O5).

      Final answer = (% P in P2O5) x 30%

      The "solubility" is just a red-herring, irrelevant to the qn, so just ignore it.

      Originally posted by hoay:

      I got 13.1 % which is the correct answer. 

      This  was from Nov 2003 / CIE / P1. I copied the question as i have it here. Anyway thank you for your prompt answer.



      Your original post left out "P2O5" in your question, which changes the question entirely. In future, for all CIE questions, please link directly to the question online, whenever available.


      Working is (as I explained in my previous post) (30 / 100) x ( ( (2 x 31) / ( (2 x 31) + (5 x 16) ) ) = 0.131 or 13.1%

      Edited by UltimaOnline 17 Mar `17, 7:38PM
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    • Originally posted by Carychidestar:

      Hi ultima,could you help me to solve this question?

      I got negative values....

      "0.7g of a hydrocarbon K was burnt completely in excess dry oxygen.2.2g of co2 is formed.100cm3 of K required 600cm of o2 gas,measured at same temp and pressure for complete combustion.Determine the molecular formula of K."

      Hi Carychidestar,

      (x + y/4) = 6

      [ 0.7 / (12x + y) ] x = [ 2.2 / 44 ]

      Solve the simultaneous equations.

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    • Originally posted by glitter58:


      for Oxidation state questions, when do i use the O level formula and when do i draw out the structure and figure out the OS? 

      since sometimes both methods give different answers

      Depends on the question and the molecule / ion involved, ie. simple vs complicated structure. If in doubt, draw out the structure to work out the OS of the individual atom in the individual resonance contributor (using my BedokFunland JC / University level formula or method for working out OS using structure), then give the more relevant answer (eg. for Organic Chem, in the oxidation of alcohols to carbonyl compounds to carboxylic acid, obviously the O level method, ie. an average OS per element, won't suffice), or if in doubt, give both answers (with explanation of each working).

    • Originally posted by keefay:

      Hi UltimaOnline, is S in SO3 an expanded octet? Thank you! (: 

      Yo keefay,

      There are 7 major resonance contributors for SO3.

      For A level H2 purposes, you can take it that the required resonance contributor (ie. the structure that you need to draw if asked in the A level exams) is the one without formal charges, ie. S has 6 bond pairs, 0 lone pairs. Hence in terms of a stable octet, the S atom (in this resonance contributor) has 12 valence electrons, ie. expanded octet.

      But as a BedokFunland JC / Olympiad / H3 Chem student, I expect you pple to be cognizant of the other resonance contributors, in 3 of which the central dipositively formal charged S atom has exactly a stable octet with 1 doubly bonded no formal charged O atom and 2 singly bonded uninegatively formal charged O atoms, as well as another 3 resonance contributors in which the central unipositively formal charged S atom has an expanded octet with 2 doubly bonded no formal charged O atoms and 1 singly bonded uninegatively formal charged O atom.

      In reality, the resonance contributor without formal charges, though usually taught in Singapore JCs, is actually the most minor of the resonance contributors, in spite of it having no formal charges (can you figure out why? My BedokFunland JC students can ask me during tuition, all other students can go ask their school teacher or private tutor).

      Concordantly, all resonance contributors considered, you should be able to elucidate the resonance hybrid. But for A level H2 purposes, unless otherwise specified by the question, it will suffice to draw the simplest resonance contributor without formal charges.

      Bonus BedokFunland JC challenge question : identify the orbital hybridization involved in all 4 atoms in the SO3 molecule, and hence describe the exact nature (ie. what type of overlap involving which specific orbitals) of each of the sigma and pi bonds (for the resonance contributors that have pi bonds).

      No prob, keefay! :)

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