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    (Original post by djmans)
    what about sodium carbonate
    alcohols don't react with sodium carbonate, they are slightly acidic and this won't be enough to liberate CO2
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    (Original post by shiney101)
    Can someone help me with q9? It's about percentage error, and they've said in the answer that the titre will be 24.4 so you can use that.
    I've forgotten what the formulae for percentage error is ://
    http://filestore.aqa.org.uk/subjects...T-P10-TEST.PDF

    Thanks
    Well percentage error is (uncertainity or error)/total titre*100
    and since this gives percentage error in each reading so when doing this for burrette we multiply uncertainity by 2 as we take two readings (initial and final)
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    (Original post by Louisss)
    why must the solvent be hot though?


    Posted from TSR Mobile
    You should definately watch a few videos on this if you haven't done it in labortary yourself. One i'd suggest is https://youtu.be/qJLvB6NFnoA
    And you'll see that to dissolve the solute we heat the solvent x)
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    can some one help?
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    (Original post by djmans)
    can some one help?
    In a (i) Look for the peak with the highest value for m/z, and that value is the relative formula mass of the compound.
    a(ii) we add O and H molar mass (16+1) and then deduct it from 88 =71 so the CH3 group at one end has Mr 15. Deduct it from 71=56,now divide 56 by 14 (as -CH2 has Mr 14) and you get 4. So the x=5 (4+1 in methyl group at the terminal end) and y=11 as (4*2+3 in the methyl group)
    b. In three as it doesn't oxidise so its a tertiary alcohol with formula CH3-CH2-CH2-CH2-CH2-OH.
    Hope you understood. Better see the markscheme!
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    Good evening i am having some serious bother with this question-describe the significance of enthalpy during iron production in the blast furnace whilst explaining how heat exchange can be put to good use. Any help would be great, thank you
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    I need help with a double indicator titration.

    I've got methyl orange and phenolpthalein, but I can't figure out how you work out the number of moles of each substance.

    I understand that the first end point = half of na2co3 and naoh
    Second end-point = 2nd half of na2co3, but where does CO2 come into it?


    Eqn: 2NaOH + CO2 --> Na2CO3 + H2O
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    (Original post by Confused_Soul_)
    I need help with a double indicator titration.

    I've got methyl orange and phenolpthalein, but I can't figure out how you work out the number of moles of each substance.

    I understand that the first end point = half of na2co3 and naoh
    Second end-point = 2nd half of na2co3, but where does CO2 come into it?


    Eqn: 2NaOH + CO2 --> Na2CO3 + H2O
    HCl + Na2CO3 --> NaHCO3 + H2O
    HCl + NaHCO3 --> NaCl + CO2 + H2O
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    (Original post by Confused_Soul_)
    I need help with a double indicator titration.

    I've got methyl orange and phenolpthalein, but I can't figure out how you work out the number of moles of each substance.

    I understand that the first end point = half of na2co3 and naoh
    Second end-point = 2nd half of na2co3, but where does CO2 come into it?


    Eqn: 2NaOH + CO2 --> Na2CO3 + H2O
    To work out the concentration of the analyte you need to work out the concentration of what you're titrating it against, so use the concentration of your standard solutions to do it.

    The moles would be with this equation: No. moles = (concentration x volume (cm3))/1000 or N = C x V (dm3).
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    (Original post by MiracleLeaf)
    To work out the concentration of the analyte you need to work out the concentration of what you're titrating it against, so use the concentration of your standard solutions to do it.

    The moles would be with this equation: No. moles = (concentration x volume (cm3))/1000 or N = C x V (dm3).
    It's being titrated with HCL and i've got the volumes for both end-points.
    i know that the second end-point minus the first-end point gives you the volume needed to react with half of the na2co3. Times it by two to give the volume needed for all the Na2CO3. Right?

    What i dont understand is the HCl titre --> how do i calculate the number of moles of HCL and would i use the following equation to calculate the moles of CO2?

    Na2CO3 + 2HCl --> 2NaCl + H2O + CO2?

    thank you
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    Hi !!!!! I do ocr chemistry and was wondering if anyone has 2015 f321 f322 f324 f325 papers
    Feel free to pm me 😜😜
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    (Original post by djmans)
    can some one help?
    Hi there.

    Question is one of simple maths really. The formula mass of the alcohol is 88. So subtract the mass of the OH functional group from the formula mass of the alcohol. This leaves you with 71. This must be the formula mass of the alkyl group represented by carbon and hydrogen in their relative quantities. Now it is simply a question of finding how many carbon atoms will make up as much of the 71 as possible. As carbon has a relative atomic mass of 12, this means there are 5(12×5=60). This means the remaining 11 must be all hydrogens. Hence the alkyl group must have the formula C5H11

    Hope this helps and good luck
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    With E cell if it's negative does that mean the half cell reaction is oxidising?
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    (Original post by Ladymusiclover)
    With E cell if it's negative does that mean the half cell reaction is oxidising?
    The one that's more negative is the one being oxidised, yes.
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    Can someone please help
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    (Original post by PlayerBB)
    Can someone please help
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    Posted from TSR Mobile
    Hey!:hugs:
    Ive given it a shot:

    1. First I calculated both the moles of HCl and NaOH..
    moles of NaOH: 2.54 X 10-3
    moles of HCl: 5 X 10-3

    because we titrated the acid with the alkali we can take the number of mole of HCl from NaOH. so (5X10-3)- (2.54X10-3)= to get us 2.46 X 10-3 moles used.

    Because it is a 2:1 ratio we divide the number of moles by 2 to get us 1.23 X10-3 in 25cm3.
    We can then multiply the number of moles by 10 (to get it in 250cm3): 0.0123

    Now we can calculate the mass: Mass= moles x Mr:
    0.0123 x 132.1= 1.62483g

    % by mass: 1.62483/3 multipied by 100: 54.12%
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    (Original post by haj101)
    Hey!:hugs:
    Ive given it a shot:

    1. First I calculated both the moles of HCl and NaOH..
    moles of NaOH: 2.54 X 10-3
    moles of HCl: 5 X 10-3

    because we titrated the acid with the alkali we can take the number of mole of HCl from NaOH. so (5X10-3)- (2.54X10-3)= to get us 2.46 X 10-3 moles used.

    Because it is a 2:1 ratio we divide the number of moles by 2 to get us 1.23 X10-3 in 25cm3.
    We can then multiply the number of moles by 10 (to get it in 250cm3): 0.0123

    Now we can calculate the mass: Mass= moles x Mr:
    0.0123 x 132.1= 1.62483g

    % by mass: 1.62483/3 multipied by 100: 54.12%
    Heey

    I have solved it exactly as you said and I keep getting 54.16% which is really annoying and then on my calculator, I have put your numbers and it keeps giving me 54.16 so how you're getting 54.12% ?

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    (Original post by PlayerBB)
    Heey

    I have solved it exactly as you said and I keep getting 54.16% which is really annoying and then on my calculator, I have put your numbers and it keeps giving me 54.16 so how you're getting 54.12% ?

    Posted from TSR Mobile
    Loool fail😂🙈🙈 yes you're right it is 54.16% I'll try to figure it out and get back to you, sorry! X
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    Anyone got tips on how to do and answer nmr questions? (Ocr)

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    (Original post by haj101)
    Loool fail😂🙈🙈 yes you're right it is 54.16% I'll try to figure it out and get back to you, sorry! X
    Lool thank you so much!! I've been stuck on it for ages!
 
 
 
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