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    Let me know what your answers were below, so we can make an unofficial markscheme.

    1)

    Define a standard solution (1 mark): A solution, where an exact mass is dissolved - so that you know the precise concentration and volume (something along those lines will do).

    Define an alkali (1 mark): A base that is soluble in water (e.g. metal hydroxide or ammonia), that releases OH- ions into solution.

    Calculate the mean tire (4 marks): You had to simply record the readings. Probably 2 marks for all correct readings. Then you had to calculate the mean titre, from the concordant tire (so using only 2 of the titres). You also had to be very careful about decimal places.

    Percentage error (1 mark): 0.34% (something like that).

    How should the solution have been prepared (1 mark): They should have used a beaker to dissolve the solute first, and then they should have used a volumetric cylinder (rather than a "beaker" ). Then invert, etc.

    Calculate the molar mass of the acid (4 marks): The answer, was 118 gmol^-1.

    Suggest the structure of this acid (1 mark): It was butanoic acid I think - 4 carbon atoms, with TWO carboxylic acid groups at either end.

    2)

    Write the ionic equation for bromine displacing iodine (1 mark): Br2(g) + 2I-(aq) >> 2Br-(aq) + I2(g)

    Why is bromine more reactive than iodine (3 marks): Iodine has a greater atomic radius, and more electron shielding. So therefore Iodine has less nuclear attraction to it's outer electrons, than Bromine does. Halogens gain electrons during reactions - but iodine gains electrons less easily than bromine. Hence, it is less reactive.

    Draw a 'dot and cross diagram' for Na2S (1 mark): Draw 2 Na- ions, and the S2- ion.

    Write the electron configuration, for Se (1 mark): 1s22s22p63s23p63d104s24p4

    State the toxic gas formed, and write the equation (2 marks): Toxic gas: H2Se. Equation: Na2Se + 2HCl >> H2Se + 2NaCl

    3)
    Outline how the alcohol is heated, and how the acid impurities are removed (6 marks):
    You had to draw out a reflux setup. So pear shaped flask, vertical condenser, correct water in/out points, thermometer (at top), screw tap adapter. Possibly also a mark for drawing the separating funnel. The organic product then had to be shaken with sodium carbonate, but allow the gas to escape - and then put through a separating filter. Top up with water. Run each separate layer through the tap, into separate containers. Then add anhydrous CaCl2, to remove any excess water. Then decant the final solution.


    Describe how to product the maximum yield/equilibrium question (5 marks):
    A low temperature is required, as the forward reaction is exothermic - so the equilibrium position shifts to the right.
    A low pressure is required, as the right side has fewer moles of gas particles. The equilibrium position will shift, in order to minimise the change. A higher temperature may actually be used, to increase the rate of reaction (whilst compromising a slightly higher yield). A higher pressure may also be used, to increase the rate of reaction - and potentially reduce operating costs and possible safety concerns.

    4)

    Draw the enthalpy progress graph (3 marks): You had to draw the products below the reactants. Draw Ea, enthalpy change, products and reactants.

    Calculate the energy released, from x tonnes of NH3 (4 marks): calculated to be 6.79x10^7 KJ (to 3 s.f.)

    State the equation, for the enthalpy change of vaporization of bromine (1 mark):

    Is the enthalpy change of vaporization endothermic or exothermic? (1 mark):Endothermic, because you are breaking the (inter molecular) forces/bonds, which requires energy.

    5)

    Calculate the rate of reaction, after 30 minutes (2 marks): You had to draw a tangent, and then calculate the gradient. I got 2.7*10^-3 (any acceptable range).

    Calculate the mass formed, from a reaction with percentage yield of 61% (2 marks): Calculated to be 12.6 g.

    7)

    Analyse the data. Draw out the compound F's structure (6 marks):
    The empirical formula needed to be calculated. Then determine the molecular formula (from the M+ peak). Both turned out to be C4H6O.
    The molecular ion at an m/z of 41, was COHCH+ (I think). Possibly also CH3CH2C+ (as it worked both ways).
    Quote the IR range, for both the alkene functional group AND the C=O carbonyl group.
    The compound can show trans stereoisomerism, impying it cotains the C=C functional group.
    The final structure was but-2-ene-1-al (google crotonyl acid). BUT the name was not required, you just had to draw the structure.

    Vote here how you found it, so the grade boundaries can be estimated:

    https://www.strawpoll.me/15765060/r
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    (Original post by RedGiant)
    Let me know what your answers were below, so we can make an unofficial markscheme.
    12.6 for percentage yield
    and what did you get 4 naming the compound
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    (Original post by IGCSEKID2017)
    12.6 for percentage yield
    and what did you get 4 naming the compound
    I got 12.6g too.
    The last question? I got butan-2-ene-al I think (although I think the 'an' bit wasn't correct).
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    I put a solution of a known conc. for standard and for alkali i put it reacts with an acid to form salt and water is that wrong?

    12.6 for yield
    Dy/dx got like -0.2/60
    5 marker you had to reduce pressue and temperature but you couldnt do it too much as the rate of reaction would be too slow
    6 marker reflux diagram and purifying was tapping off the 2 layers and using drying agent and so on
    Other 6 marker was c4h6o but i couldnt get the structure down.
    Got h2Se for the toxic gas B.
    Put a few more in the other thread if u wanna check them out
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    (Original post by RedGiant)
    I got 12.6g too.
    The last question? I got butan-2-ene-al I think (although I think the 'an' bit wasn't correct).
    yh i got that but i drew it instead of naming it will i lose marks?
    also you know that 1 marker for that bromo compound
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    was the gradient 3*10-3
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    (Original post by IGCSEKID2017)
    yh i got that but i drew it instead of naming it will i lose marks?
    also you know that 1 marker for that bromo compound
    It never asked for you to name the last compound, just predict the structure using the information given.
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    (Original post by IGCSEKID2017)
    was the gradient 3*10-3
    You are supposed to draw a tangent at 30mins then take 2 points, I got something like 2.4x10^-3
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    (Original post by CRM1337)
    It never asked for you to name the last compound, just predict the structure using the information given.
    oh ok and the one on addition and it was like 1 mark and you had 2 name it
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    (Original post by IGCSEKID2017)
    was the gradient 3*10-3
    I got like 3.08 x 10-3 depends on our tangents
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    The question about whether the enthalpy change is exothermic or endothermic was three marks. You had to say it was endothermic because heat energy is needed to change the liquid to a gas.
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    (Original post by PureDTJ)
    I got like 3.08 x 10-3 depends on our tangents
    yh i think they'll have a range of values
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    (Original post by lionike123)
    The question about whether the enthalpy change is exothermic or endothermic was three marks. You had to say it was endothermic because heat energy is needed to change the liquid to a gas.
    Yeah i got that but had no f'ing clue what the equation was haha
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    6 marker.
    Firstly you had to draw a reflux diagram (main things condenser, water pipes, bunsen burner). You had to place the solution into a seperating funnel add more water to see which one was the aqueous layer. Get a conical flask and release the aqeuous layer. Get anhydrous calcium chloride to absorb the water.

    5 marker
    Low temperature because the forward reaction was exothermic so the system will try to oppose the change and therefore the position of equilibrium will shift to the product side (Nitrogen Monoxide). Low pressure because there was more moles of gas on the product side then the reactant side. You had to make sure that there was a compromise because the rate at reaching equilibrium would be too slow and is not good for the economy.

    3 marker finding the x. You had to change tonnes to g and then divide by 17. Then you had to use molar ratio.
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    For the titration question I got the mass of A to be 118 to the closest whole number, then the formula for it: CH3CH(COOH)2
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    Guys remember to add re-distillation at the end of the 6 marker.
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    How should the solution have been prepared (1 mark): They should have used a beaker to dissolve the solute first, and then they should have used a volumetric cylinder (rather than a "beaker" ). Then invert, etc.

    Thats just extra. All you had to say he should have used a volumetric flask.
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    For the value of energy released I did:

    5.1 tonnes = 5,100,000 g

    5,100,000/17 to get moles of NH3 = 300000 mol

    300000/4 because the equation had 4NH3 per 905KJ of energy

    (300000/4)x905= 6.79x10^7 KJ to 3 sig.fig.
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    How many marks was that tonnes and percentage yield question. I know i definitely got what wrong.
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    (Original post by lionike123)
    How should the solution have been prepared (1 mark): They should have used a beaker to dissolve the solute first, and then they should have used a volumetric cylinder (rather than a "beaker" ). Then invert, etc.

    Thats just extra. All you had to say he should have used a volumetric flask.
    Extra is good

    (Original post by lyanu)
    Guys remember to add re-distillation at the end of the 6 marker.
    I don't think so, as the only product formed would be the haloalkane after one hour. Pretty sure the idea was to remove acid impurities. But using a seperating funnel works just as well for removing any potential excess alcohol.
 
 
 
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