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    I genuinely can't wait for my essay subjects to be done so I can revise Chemistry. It's literally a relief to not be learning Shakespeare or Historical facts/figures by heart.
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    (Original post by Gulzar)
    Wow that's really encouraging of him. My teacher literally said I had no chance of getting an A

    What kind of AS questions are u getting wrong?
    Well I just watched the video Bluedate put up about Spectator Ions etc etc. which was great!!

    So
    1. Ionic Equations (might be cleared up have to re-do papers to check)
    I think I just didn't get that if something isn't forming a New State Compound then it is removed. I was just confused as they seemed to be jumping steps, but now I think I get it.

    2. Molar Equations, (AS, might be cleared up to, think I'm better)


    3. Water of Crystallization or something confusing like that

    You can do it bro, we've got a few days to get to that Peak Performance level - Lets do this!!
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    (Original post by Member427)
    can someone help me on qs 6 (e) in the jan 2013 paper please, i dont even understand the question.
    Cr^3+ (aq) +3e -----> Cr (s)

    X (s) -----> X^2+ (aq) + 2e

    In the question it said that the Cr metal gained in mass, so in the first equation I made it the product, while in the second equation the metal X is going from a solid to aqueous solution as it is losing mass. We know X is 2+ because in XSO4, sulfate is 2- (the sulfate cancels out in the equations btw)

    Anyway, you can now make an overall equation by X 2 the first equation, and X 3 the second equation to get the electons same number, then combine them:

    2Cr^3+ + 3X -----> 2Cr + 3X^2+

    So we know the ratio of X:Cr is 3:2

    We can then use the information we have to first find the no. moles of Cr
    So mass/ mr is 1.456/52 = 0.028 mols
    Use this to find X mols, so 0.028 X 3/2 (1.5) = 0.042mols

    Now you have the number of moles and the mass from X from the question, you can now find the Mr
    So m/n = 1.021/0.042 = 24.3
    Look at the periodic table and you'll see that 24.3 is the Mr for Mg
    X = Mg

    I hope that makes sense
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    (Original post by bluedate)
    Cr^3+ (aq) +3e -----> Cr (s)

    X (s) -----> X^2+ (aq) + 2e

    In the question it said that the Cr metal gained in mass, so in the first equation I made it the product, while in the second equation the metal X is going from a solid to aqueous solution as it is losing mass. We know X is 2+ because in XSO4, sulfate is 2- (the sulfate cancels out in the equations btw)

    Anyway, you can now make an overall equation by X 2 the first equation, and X 3 the second equation to get the electons same number, then combine them:

    2Cr^3+ + 3X -----> 2Cr + 3X^2+

    So we know the ratio of X:Cr is 3:2

    We can then use the information we have to first find the no. moles of Cr
    So mass/ mr is 1.456/52 = 0.028 mols
    Use this to find X mols, so 0.028 X 3/2 (1.5) = 0.042mols

    Now you have the number of moles and the mass from X from the question, you can now find the Mr


    So m/n = 1.021/0.042 = 24.3
    Look at the periodic table and you'll see that 24.3 is the Mr for Mg
    X = Mg


    I hope that makes sense
    In the question it uses the word loss, and you. Think it involves some sort of subtraction?
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    (Original post by otrivine)
    In the question it uses the word loss, and you. Think it involves some sort of subtraction?
    Loss of mass from X, which is gained by Cr, you're given the masses in the questions, so no subtraction to be calculated
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    (Original post by bluedate)
    Cr^3+ (aq) +3e -----> Cr (s)

    X (s) -----> X^2+ (aq) + 2e

    In the question it said that the Cr metal gained in mass, so in the first equation I made it the product, while in the second equation the metal X is going from a solid to aqueous solution as it is losing mass. We know X is 2+ because in XSO4, sulfate is 2- (the sulfate cancels out in the equations btw)

    Anyway, you can now make an overall equation by X 2 the first equation, and X 3 the second equation to get the electons same number, then combine them:

    2Cr^3+ + 3X -----> 2Cr + 3X^2+

    So we know the ratio of X:Cr is 3:2

    We can then use the information we have to first find the no. moles of Cr
    So mass/ mr is 1.456/52 = 0.028 mols
    Use this to find X mols, so 0.028 X 3/2 (1.5) = 0.042mols

    Now you have the number of moles and the mass from X from the question, you can now find the Mr
    So m/n = 1.021/0.042 = 24.3
    Look at the periodic table and you'll see that 24.3 is the Mr for Mg
    X = Mg

    I hope that makes sense
    thank you so much that makes sense now
    and I have another question how do you know what products are formed in qs 7 b ii in the same paper?
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    Hi
    why is there two equations to solve from for buffer solutions and what about the equilibrium of the buffer
    thank you
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    (Original post by Better)
    Well I just watched the video Bluedate put up about Spectator Ions etc etc. which was great!!

    So
    1. Ionic Equations (might be cleared up have to re-do papers to check)
    I think I just didn't get that if something isn't forming a New State Compound then it is removed. I was just confused as they seemed to be jumping steps, but now I think I get it.

    2. Molar Equations, (AS, might be cleared up to, think I'm better)


    3. Water of Crystallization or something confusing like that

    You can do it bro, we've got a few days to get to that Peak Performance level - Lets do this!!
    Totally right man
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    Hi

    just wondering if people are learning all the colours of transition metals in different oxidation states? the table on page 205?
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    (Original post by samcooke343)
    Jan 2013 with Mark Scheme if needs it

    Attachment 223567
    Attachment 223568
    Nice one mate
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    (Original post by rosie66)
    Hi

    just wondering if people are learning all the colours of transition metals in different oxidation states? the table on page 205?
    no
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    (Original post by Member427)
    thank you so much that makes sense now
    and I have another question how do you know what products are formed in qs 7 b ii in the same paper?

    Sure, well in the question it says that Mg is oxidised, so it's going to go from Mg to Mg^2+

    Reaction with Ca(HSO3)2 - you know that SO3 is ^2- whilst both Mg and Ca are ^2+

    Mg (s) + Ca(HSO3)2 (aq) ------> MgSO3 (aq) + CaSO3 (aq) + H2 (g)

    Ionic equation is:

    Mg + Ca^2+ + 2H+ + 2SO3^2---> Mg^2+ + SO3^2- + Ca^2++ SO3^2- + H2

    Cancel out the spectator ions, so overall it should be:

    Mg + 2H+ ----------------> Mg2+ + H2


    I hope that's okay, if there's anything else just ask
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    Can someone help me with this initial rates question? =\
    Question 3

    The mark scheme says to divide 0.010 by 250 but I don't understand where the 250 came from?

    Thanks in advance ^-^
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    (Original post by Sarangtaec)
    Can someone help me with this initial rates question? =\
    Question 3

    The mark scheme says to divide 0.010 by 250 but I don't understand where the 250 came from?

    Thanks in advance ^-^
    I think it is because if you draw a tangent from t=0 you should find that the x-intercept is at around 250.
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    (Original post by Lehal)
    Hi there, the way I worked this out was as follows:

    Step 1: Simple, combining the 2 equations given as the copper is reacting with the HNO3.. so, u just balance and combine them,
    Cu(s) + 4HNO3(l) --> Cu2+(aq) + 2NO3-(aq) + 2NO2(g) + 2H20(l)

    Step 2: It says that the sodium carbonate, is neutralised by any acid..
    You should know that.. when acid + carbonate --> salt + CO2 + H20
    So, I did.. NaCO3 + 2HN03 (as any acid, this was given so I used this- it makes no difference.. btw, u see I am writing out the equation first just to make it a bit easier to understand.. from the equation, u can then deduce the ionic equation), so NaCO3 + HN03--> Na(NO3)2 + C02 + H20

    hence, ionic equation.. Na+ CO32- +2H+ + 2NO3- --> Na+ + 2NO3- + CO2 + H20

    cancelling, gives.. CO32- + 2H+ --> CO2 + H20

    u follow same format for equation 4, this is just how I worked it out.. try equation 4 and see if u get
    Hi
    I'm still a little confused, in Na(NO3)2 , why are there two NO3-, when Na is 1+

    and in the mark scheme there is another equation...?
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    Can anyone help me with Q3cii) on June 2012 paper? :confused:

    http://www.ocr.org.uk/Images/131289-...d-elements.pdf
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    (Original post by mayurforlife)
    Hi
    why is there two equations to solve from for buffer solutions and what about the equilibrium of the buffer
    thank you
    Not sure exactly what you mean, so I'll just write what I think

    pH of Buffer is H+ = Weak acid/ salt of weak acid
    then -log(H+)

    Shifting of equilibrium:
    Example:
    CH3CH2COOH + NaOH -----> CH3CH2COO-Na+ + H2O

    Equilibrium: CH3CH2COOH <------> CH3CH2COO- + H+

    Buffer formed when excess acid left over reacts with its salt

    When excess acid is added to equilibrium, the equilibrium shifts to the left as the salt reacts with H+

    When excess alkali is added, the equilibrium shifts to the right as the acid reacts with the added alkali

    Hope that helps, anyone correct me if I'm wrong
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    (Original post by bluedate)
    Sure, well in the question it says that Mg is oxidised, so it's going to go from Mg to Mg^2+

    Reaction with Ca(HSO3)2 - you know that SO3 is ^2- whilst both Mg and Ca are ^2+

    Mg (s) + Ca(HSO3)2 (aq) ------> MgSO3 (aq) + CaSO3 (aq) + H2 (g)

    Ionic equation is:

    Mg + Ca^2+ + 2H+ + 2SO3^2---> Mg^2+ + SO3^2- + Ca^2++ SO3^2- + H2

    Cancel out the spectator ions, so overall it should be:

    Mg + 2H+ ----------------> Mg2+ + H2


    I hope that's okay, if there's anything else just ask
    ahh that makes so much sense now thank you so much, so if a question says for example potassium is oxidised we assume it turns into k^1+ ? And also will it be the same for chlorine as well if it says cl is reduced it goes to cl^1- ?
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    4Ci) jan 13
    "how does S2082- and I- catalysed by Fe2+?
    - I got the first one as Fe2+ has higher E theta than S208

    but in the mark scheme, the second one is 2Fe3+ + 2I- --> 2Fe2+ + I2

    Surely ^^ is being catalysed by fe3+ and not fe2+?
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    (Original post by bluedate)
    Can anyone help me with Q3cii) on June 2012 paper? :confused:

    http://www.ocr.org.uk/Images/131289-...d-elements.pdf
    Hi

    Okay what you do is this:-

    1) You can use the formula for buffers, I use [H+]=Ka x [HA]/[A-]
    2) We have Ka, thats fine , we need to get the concentrations of HA and A-
    3) Before we do that, we need to get the moles of the butaneoic acid and the NaOH
    4) The moles of butanoic acid is 0.0125 and the moles of NaOH is 2.5x10-3
    5) now using your knowledge of buffers, we know that some acid remains in solution and hence, the acid is in excess and so we substract the moles of butanoic acid and NaOH to get the moles of butanoic acid that has reacted with NaOH.
    6) so the new mole of butanoic acid is 0.01 and the moles of NaOH is the same 2.5 x 10-3, but we have to get the concentration so we divide by the volume which would be 100cm3(0.1dm3) , this is because there is 50cm3 of butanoic acid and 50cm3 of NaOH , so we add the volumes.
    7) Now that you got the concentration just sub the values into the Ph buffer formula and you get the value of [H+] and then do the -log[H+] which should give 4.22 correct to 2 decimal place


    please tell me if you are unsure
 
 
 
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