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

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    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
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    15,015 posts since May '05
    • 'O' & 'A' Level Qn.


      "If carbon emissions are not cut by 25 percent to 40 percent by the year 2020, higher ocean temperatures could kill off vast marine ecosystems and half the fish in them, according to the WWF." - Straits Times 14 May 09


      Source :

      "Climate change threatens 100M in SE Asia"



      Explain why increasing ocean temperatures (eg. from 20 deg C to 25 deg C; nowhere near denaturation temperatures for enzymes yet) will kill off corals and fishes.



      Solution :


      Solubility of dissolved oxygen decreases significantly with temperature. Gaseous oxygen has more energy than aqueous (dissolved) oxygen. The forward reaction for O2(aq) to O2(g) is endothermic. Hence increasing temperature pulls the position of equilibrium even further over to the right (in the first place, oxygen is rather insoluble in water because it is non-polar, and it doesn't ionise in water either, hence remains largely insoluble. Elitist/snobbish water would rather hydrogen bond with itself (ie. other water molecules), than bother with weaker permanent dipole - induced dipole interactions with oxygen. "It's beneath our social class." "Not a profitable/worthy thermodynamic investment.").


      Increasing temperature provides increased kinetic energy to the aqueous oxygen molecules, to overcome the weak permanent dipole - induced dipole interactions between the non-polar oxygen molecules and the polar water molecules. The increased kinetic energy of the oxygen molecules, results in dissolved oxygen molecules escaping from the warmed ocean waters as oxygen gas.


      This is the reason why boiling water removes dissolved oxygen gas, and why fishes put in boiled water will die of oxygen deprivation. When you boil water, the bubbles of gas you see, are not merely dissolved oxygen, but all dissolved gases, including (especially near boiling point), primarily water vapour itself, H2O(g).



      (Edit : Added the following. Acknowledgements : Raymond Chang's Chemistry 9th Edition.)


      There is one other interesting point to note in this discussion, a biological one. Fish, being cold blooded, will have their metabolism (ie. biochemical reactions in the body, invariably involving enzymes) increase accordingly with the temperature of the ambient water. The speedup of metabolism caused by global warming of the oceans, increases the fishes' need for oxygen, at the same time that the supply of dissolved oxygen decreases because of its lower solubility in heated water.


      The decreased solubility and (hence) availability of dissolved oxygen for aquatic flora and fauna as a result of heated water and their consequent death, dwindling population and species extinction (eg. from global warming, from dumped waste water used for cooling in electric/nuclear power stations, etc), is known as thermal pollution.

      Edited by UltimaOnline 15 May `09, 3:14AM
    • 'A' & 'O' levels qn.


      Stainless steels contain varying amounts of chromium in addition to the iron and carbon in seel. Explain the chemistry of how the addition of chromium makes stainless steel less vulnerable to corrosion, as compared to regular steel.


      Solution :

      "High oxidation-resistance in air at ambient temperature are normally achieved with additions of a minimum of 13% (by weight) chromium, and up to 26% is used for harsh environments.[8] The chromium forms a passivation layer of chromium(III) oxide (Cr2O3) when exposed to oxygen. The layer is too thin to be visible, and the metal remains lustrous. It is impervious to water and air, protecting the metal beneath. Also, this layer quickly reforms when the surface is scratched. This phenomenon is called passivation and is seen in other metals, such as aluminium and titanium." - Wikipedia entry http://en.wikipedia.org/wiki/Stainless_steel

      Edited by UltimaOnline 18 May `09, 10:57AM
    • 'A' Level Qn.


      In kinetics, a reaction's rate constant is called a constant, because it doesn't change as long as temperature doesn't change. Yet, when you add a catalyst, at that same temperature, the rate of reaction increases, implying that rate constant changes. Explain this seeming paradox.


      Solution :

      The rate constant for the uncatalyzed reaction indeed does not change, as long as temperature remains unchanged. However, in the presence of a catalyst, the reaction goes by a different pathway altogether, and is hence regarded as a completely different reaction, with a completely different (and larger) rate constant at any given temperature. Notice also that the rate law or rate equation for the catalyzed reaction includes the concentration of the catalyst in it, in addition to having a different rate constant. 

    • 'O' & 'A' Level Qn.



      Originally posted by anpanman:


      May I know if the reaction between Na3N and HCl produces NH3 and NaCl? It looks like it but in my question, they said that the reaction produces solution S. And S, when boiled with NaOH, gives a gas G which turns damp litmus from red to blue.


      Name the salt present in S that reacts with aq NaOH to form gas G. I think it is the NO3- ions since this test is for presence of NO3- ions. But the products don't seem to have any compound that has the ion. Thanks.


      Solution :


      1st step is (tetra)protonation of the aqueous trinegative nitride anion N3- (4 lone pairs, no bond pairs), easily achieved since all sodium salts are completely soluble / strong electrolytes :

      N3- + 4H+ --> NH4+


      The +ve charges from 3 Na+ cations and 1 NH4+ cation, are counterbalanced by the 4 Cl- ions.


      Na3N + 4HCl --> NH4Cl + 3NaCl


      A mixture of NaCl(aq) and NH4Cl(aq) is obtained (ie. Na+, NH4+ and Cl- ions present).


      Proton transfer (ie. Bronsted-Lowry acid-base reaction) occurs with aqueous hydroxide ions.


      Ionic :

      NH4+ + OH- --> NH3 + H2O


      Chemical :

      NH4Cl + NaOH --> NH3 + H2O + NaCl


      Gaseous ammonia undergoes hydrolysis on the moist red litmus paper to produce OH- ions which turns red litmus blue :


      NH3 + H2O --> NH4+ + OH-


      Notice that this is exactly the reverse of the earlier proton transfer reaction (see above). 

      Edited by UltimaOnline 26 May `09, 7:42PM
    • 'O' and 'A' Levels Qn.


      Why is it unsafe to consume food from dented food cans?


      Solution :

      (Thanks to Darkness_hacker99 for the hyperlink)


      Source :



      One of the things that I have enjoyed on this trip is doing the cooking. I really like to cook, both for the fun of making things that taste good and for the science involved. We went into Whangarei to get groceries, and while we were shopping, I noticed some dented cans. That gave me the idea for this week's experiment. Why is it a bad idea to buy dented cans? Because they don't look as nice? Because some of the food inside might be dented too? To find out, you will need:

      a penny or other copper coin
      some aluminum foil
      some catsup
      a plate or saucer

      Tear a strip of foil about an inch wide and 3 inches long. Place it on a plate. Put the penny on one end of the strip. Squirt a blob of catsup on top of the penny and then fold the other end of the foil over, so that it squishes into the catsup. Crease the fold enough so that it will stay in place. This should give you sort of a foil and penny sandwich. Looking from the side, you should have foil on the bottom, with a penny on top of that, then catsup and then the other end of the foil. Set the plate in a warm place where it will not be disturbed.

      The next day, rinse the catsup off the foil and examine it carefully. In the area where it was touching the catsup, it is full of holes! Oh no! If catsup does that to aluminum foil, think what it is doing to my stomach! I can never eat French fries again!

      Yes, the catsup does contain acid. ACID!!!! Don't get too excited. The acid is vinegar. Read the label and you will see. How did vinegar do that? Well, the acid is not responsible, or at least, not directly. The culprit is electricity.

      Electricity? I did not tell you to plug it into the wall. DON'T DO THAT!!! It would be very dangerous, and besides, you don't need to. This experiment makes its own electricity. The copper penny and the aluminum foil, connected by a conducting solution (catsup) form an electric cell, which will produce about 1/2 of a volt of electricity. You can't light a light with it, as the amperage is very small, but it does produce an electric current. If you cut through the foil strip in the center and connect a probe from a volt meter to each piece of foil, you should be able to measure about 1/2 of a volt.

      In the process of producing the electric current, the penny is stealing electrons (tiny, negatively charged pieces of atoms) from the aluminum. This changes the aluminum and lets it dissolve in the acid in the catsup.

      OK, so now we know how to dissolve holes in aluminum, but what does that have to do with dented cans. Cans are made of steel. The problem with steel is that it rusts when it comes in contact with water and oxygen. Since most of the foods that we put into cans are wet, they would quickly rust the can. To solve this, the steel cans were plated. That means that the inside was covered with a thin coating of the metal tin. Tin is not strong enough to make a good can, but it does not rust. The tin coating on the inside keeps the can from rusting and the steel give the can its strength.

      If a can is dented, the tin layer can be cracked, letting the liquid come in contact with the steel. Now you have two metals (steel and tin) in contact with a conducting solution (the liquid in the can). The can of food begins to produce electricity. It also starts to oxidize (rust) the metal in the can, producing a hole and letting the food inside spoil. That is why we don't buy dented cans.


      From Robert Krampf's Science Education Company
      PO Box 60982
      Jacksonville, FL 32236-0982
      [email protected]

    • 'A' Level Qn.


      Convert 1,2-dimethylcyclohex-1-ene to 1,6-hexandioc acid.



      Solution :

      1) Oxidation with hot, concentrated, acidified KMnO4.

      2) Reduce with LiAlH4 (in dry ether), followed by protonation/hydrolysis.

      3) Heat with I2(aq) and NaOH(aq) (triiodomethane test).

      4) Acidify / protonate.


      Edited by UltimaOnline 13 Jul `09, 9:51PM
    • 'O' & 'A' Levels Qn.



      10cm3 of a hydrocarbon, 85.7% carbon by mass, undergoes complete combustion to produce an acidic oxide, which requires 16.65cm3 of 0.2mol/dm3 of NaOH for complete neutralization. What is the density of the hydrocarbon at r.t.p?



      Ans : 2.33g/dm3

    • 'O' & 'A' Levels Qn.


      State and explain the effect of adding a salt (eg. sodium chloride) on the freezing and boiling points of water.


      Solution :

      Adding salt depresses (decreases) the freezing point, and elevates (increases) the boiling point. The presence of cations and anions presents electrostatic (ion-dipole) interactions with the polar water molecules, hence

      (i) disrupting the orderly arrangement (including fixed hydrogen bond lengths and angles) of the hydrogen bonded tetrahedral structure of ice, thus requiring a lower temperature to freeze;

      (ii) increasing the cohesion of water molecules (as a result of attractive ion-dipole interactions between ions and water molecules), thus requiring a higher temperature to boil (since boiling refers to overcoming the intermolecular attractive forces (hydrogen bonds) between water molecules for them to be far apart in the gaseous state).



    • 'A' Level Qn.


      Liquid butane is commonly found in canisters used for portable gas stoves during camping trips. The standard enthalpy change of combustion of butane is -2887 kJ/mol. A camper found a left-over canister of butane and estimates that there was 60 g of liquid butane left.

      a) Calculate the mass of water, at r.t.p, that he can boil before he uses up the butane.


      b) Using a Hess Law energy cycle diagram, and given that the standard enthalpies of formation of CO(g) and CO2(g) are -111kJ/mol and -394kJ/mol respectively, calculate the standard enthalpy change of reaction for the incomplete combustion of butane to produce water and carbon monoxide only.




      Answers :

      a) 9.49 kg

      b) -1745 kJ/mol 

      Edited by UltimaOnline 09 Jul `09, 2:08PM
    • 'A' Level Qn.


      A compound Y has the formula C6H10. One mole of Y reacts with 2 moles of aqueous bromine. On reaction with hot, concentrated, acidified KMnO4(aq), compound Y yields an acidic gas and a compound Z, which when reacted with LiAlH4 in dry ether (and followed by protonation/hydrolysis), yields the only primary alcohol that gives a positive triiodomethane (iodoform) result. Deduce and elucidate the structure of Y.


      Answer :



    • 'O' & 'A' Level Qn.



      Within a molecule of hydrogen sulfate(VI), also known as sulfuric(VI) acid (but in its protonated, unionized, undissociated form), state :

      (i) the number of electrons involved in bonding

      (ii) the number of electrons not involved in bonding



      Answer :

      Formula is H2SO4 (oxidation state of sulfur is +6).

      Diagram (either Kekule or dot-&-cross) should be drawn.

      Bonding electrons = 16

      Non-bonding electrons = 34


    • 'A' Level Qn.


      For compounds with the molecular formula C3H3Br3,

      (i) Do two structural isomers exist, one of which has optical isomerism, and the other has geometric isomerism? If yes, draw them.

      (ii) Does a particular structural isomer exist which has both optical isomerism and geometric isomerism? If yes, draw it.



      Answers (partial) :

      (i) Yes.

      (ii) No.



      (If you'd like me to check your answers, scan and upload your diagrams and Private Message me the url)

      Edited by UltimaOnline 16 Jul `09, 7:04PM
    • 'O' & 'A' Level Qn.


      'O' level version :
      Write a balanced equation to represent the reaction between magnesium nitride and water.


      'A' level version :
      Draw the electron flow mechanism to represent the hydrolysis of magnesium nitride.



      Solution :
      The N3- ion is a strong Bronsted-Lowry base which abstracts 3 protons from 3 water molecules, generating ammonia gas and magnesium hydroxide precipitate.

    • 'O' & 'A' Level Qn.


      When aqueous sulfuric(VI) acid is electrolysed using graphite anode, a mixture of water, carbon monoxide and carbon dioxide is obtained at the anode. 30cm3 of gas is obtained at the cathode, while 17cm3 of gas is obtained at the anode. When the anode gas was bubbled through excess soda lime (soda = sodium hydroxide; lime = calcium hydroxide), its volume was reduced to 9cm3.


      Determine the composition of the anodic gas.



      Solution :

      5cm3 molecular oxygen, 4cm3 carbon monoxide, 8cm3 carbon dioxide.

    • 'O' & 'A' Levels Qn.


      A primary alcohol X, when heated at 170 deg C with excess concentrated sulfuric(VI) acid, produced an organic compound with molar mass 42g. Identify X, and hence write the relevant half-equations and overall balanced equation to represent the reaction between X and acidified potassium dichromate(VI) solution.



      Solution (partial) :

      Coefficients of the overall balanced equation are 2, 3, 16, 4, 11, 3



    • 'A' Level Qn.


      Draw the Kekule or dot-&-cross structure of nitrous acid (latin name) aka nitric(III) acid (stock name), and hence state the electron and molecular geometry about the N atom in this acid.



      Solution :

      Electron geometry is trigonal planar, molecular geometry is bent/v-shape/non-linear. In other words, the central N atom has 1 lone pair and 2 electron charge clouds (1 double bond and 1 single bond).


      A common mistake made by students/candidates is to state the molecular geometry as trigonal planar, because their erroneous Kekule (or dot-&-cross) structure either has an octet-violated N atom, or has at least one formal (hence ionic) charge that is not cancelled out; note however that nitrous acid is an electrically balanced compound. 


    • 'A' Level Qn.


      State the total number of, and draw all, possible isomers with the name dichlorophenol.



      Answer (highlight with mouse) :


    • 'A' Level Chemical Equilibria Qns



      1) For the Haber Process, when 1 mole of nitrogen gas and 3 moles of hydrogen gas were mixed and passed over a catalyst at 500 deg C, an equilibrium pressure of 200atm was registered. The equilibrium composition of ammonia was 25.3% by volume. Calculate Kp at this temperature.


      Ans : acceptable range* between (4.7 to 4.9) x 10^-5



      2) For the following reaction : 2A(g) + B(s) <--> C(g) + 3D(g)

      When 1.2 moles of A was reacted with 1.0 mole of B in a 2.0dm3 vessel to form C and D, it was found that 30% of A had been converted. If the total pressure at equilibrium was 200kPa,  calculate Kp at this temperature.


      Ans : 660 kPa2 (3 sig fig)



      3) For the Haber Process, when nitrogen and hydrogen are mixed in a 1:3 ratio at 673K to produce ammonia, the Kp for this reaction is 2.18 x 10^-6.

      a) Calculate the total pressure required to achieve a 45% yield of ammonia at equilibrium at this temperature.

      b) Calculate the total pressure at equilibrium where ammonia is 90% dissociated into its elements at 400 deg C.


      Ans : 1.2 x 10^3 atm; 122 atm.



      4) When gaseous iodine and gaseous hydrogen are mixed in a 2:3 mole ratio at 400 deg C, there is a 42% conversion of hydrogen gas to gaseous hydrogen iodide at equilibrium.

      a) Calculate Kp for this reaction if the total pressure of the equilibrium mixture was found to be 5 x 10^7 Pa.

      b) A 1dm3 vessel contains 0.3 mol of I2, 0.4 mol of H2 and 0.1 mol of HI at initial conditions. If the temperature is kept constant and Kc for this temperature is 0.191, what is the number of moles of H2 that must be introduced into the vessel at initial conditions in order to increase the equilibrium concentration of HI to 0.15 mol?


      Ans : 4.93; 0.0534 mol



      5) At high temperatures, molecular bromine gas dissociates into gaseous bromie atoms. A 0.57g sample of Br2 is placed in a 1dm3 container and heated to 1550 deg C, after which the total pressure at equilibrium is found to be 0.75 atm. Calculate Kp for the equilibrium at 1550 deg C.


      Ans : acceptable range* between (0.585 to 0.592)



      6) For the reaction between gaseous water and carbon to produce hydrogen gas and carbon monoxide gas, use Le Chatelier's Principle and/or a Qc vs Kc comparison, to predict and explain the effect of
      a) adding helium gas to the equilibrium mixture at constant pressure

      b) adding helium gas to the equilibrium mixture at constant volume


      Ans : (highlight with mouse)

      shift right; no change.




      *Note : reasons for a possible range of values obtainable involve different significant figures of conversions used by different students (eg. 1 atm = 1.01325 bar; 1 bar = 10^5 Pa; gas constant R may be 8.31 (data booklet 3 sig fig) or 8.314 (4 sig fig) if units for pressure and volume are Pa and m3, or 0.0821 (3 sig fig) or 0.08206 (4 sig fig) if units for pressure and volume are atm and dm3.

    • 'A' Level Qn.


      The Kp for the following reaction at 25 deg C is 4.


      2HI(g) --> H2(g) + I2(g)


      When a sample of HI is decomposed in a sealed container at 25 deg C, and the total pressure of the equilibrium mixture is 1.0 atm, what is the equilibrium partial pressure (in atm) of H2 in the tube?



      Solution :

      Let y be the initial partial pressure of HI. Let x be the equilibrium partial pressure of H2. Based on info given, write two equations in x and y. Solve for simultaneous equations. You will end up with a quadratic equation in x2. Solve for x. Final answer is 0.4 atm.



    • 'A' Level Qn.


      A sparingly soluble calcium salt ionizes according to the equation :


      Ca3X2(s) <---> 3Ca2+(aq) + 2X3-(aq)


      If the solubility product Ksp of Ca3X2 is S, what is the molarity of Ca2+(aq) at equilibrium?


      Answer :

      3((S/108)^1/5) which may be simplified to (9S/4)^1/5

    • 'A' Level Qn.


      An ester Q, was refluxed with aqueous sodium hydroxide and the resulting mixture then distilled. The distillate was subjected to hot, acidified KMnO4 which decolourized. The residue in the distillation flask, upon acidification, gave a white ppt.


      Which of these could be Q?


      A) CH3CO2C6H5

      B) C6H5CO2CH3

      C) CH3CO2CH2C6H5

      D) CH3CH2CO2CH2C6H5



      Answer (highlight with mouse) :



    • _

      Edited by UltimaOnline 12 Oct `17, 12:12AM
    • _

      Edited by UltimaOnline 12 Oct `17, 12:11AM
    • ‘A’ & ‘O’ Level Qn.

      125cm3 of a solution of phosphoric(V) acid were mixed with 75cm3 of 0.25mol/dm3 of barium hydroxide. A 50cm3 aliquot of the resultant mixture required 7.5g of sodium carbonate for exact neutralization, in which a gas was produced.

      (a) Calculate the volume of the gas produced, at (i) r.t.p, and at (ii) s.t.p.

      (b) Describe and explain (with equations) how the identity of this gas may be ascertained.

      (c) By calculating the molarity of phosphoric(V) acid used, calculate the sample mass of solid phosphoric(V) acid that needs to be dissolved in 125cm3 of pure water to prepare this molarity.

      (d) Draw the structural formulae for (i) phosphoric(V) acid, (ii) barium hydroxide, and (iii) sodium carbonate.

      Answers :
      (c) The sample mass of solid phosphoric(V) acid that needs to be dissolved is 19.7g (to 3 significant figures).
      Note : getting© correct means you already have parts (a) and (b) correct. Check Wikipedia for (d).

      Edited by UltimaOnline 03 Sep `09, 11:51AM
    • ‘A’ Level Qn.

      ------- With reference to the data booklet predict and describe the observations when FeCl3 (aq) is added to the following solutions. In each case, calculate the electromotive force or standard cell potential and write balanced redox equations. i) Mn2(SO4)3 ii) CuI How am I supposed to know what will happen? Help needed thanks!

                              Standard Cell Potential = Standard Reduction Potential at Cathode + Standard Oxidation Potential at Anode.
                              If the cell potential is positive, then the redox reaction is feasible, and the current will flow (for galvanic/voltaic cells).
                              Note that for a standard solution of FeCl3 (for this question asked),
                              Standard Reduction potential for Fe3+ to Fe2+ is + 0.77V
                              Standard Oxidation potential for 2Cl- to Cl2 is - 1.36V
                              For experiment (i)
                              Standard Reduction potential for Mn3+ to Mn2+ is + 1.49V
                              Standard Reduction potential for SO4 2- to SO2 is + 0.17V
                              Standard Oxidation potential for 2SO4 2- to S2O8 2- is - 2.01V
                              Conclusion :
                              Under standard conditions, Redox reaction between Mn3+ and Cl- (to produce Mn2+ and Cl2) is feasible.
                              For experiment (ii)
                              Standard Reduction potential for Cu+ to Cu is + 0.52V
                              Standard Oxidation potential for Cu+ to Cu2+ is - 0.15V
                              Standard Oxidation potential for 2I- to I2 is - 0.54V
                              Conclusion :
                              Under standard conditions, Redox reaction between Fe3+ and both Cu+ and I- (to produce Fe2+, Cu2+ and I2) is feasible.
                              (Note : there are of course other species to consider as possible products of additional possible redox reactions, but unless the exam question specifies the redox potentials for such, otherwise based on the Data Booklet given in the exams, the above are the only directly relevant considerations).
      Edited by UltimaOnline 12 Oct `17, 12:11AM
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