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Edexcel - Chemistry Unit 2 - 4 June 2013 watch

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    (Original post by Jayqwe)
    Is there a particular structure for damp red phosphorous like P4 or is it just 2P +3I2=2PI3 for the mixture of a 3(halogenoalkane) with PI3=alkane(I)+H3PO3?
    Red phosphorus is just monatomic P
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    can anyone help on june 2011 question 16
    The enthalpy change of neutralization of an acid by an alkali is measured by adding 10 cm3 of HCl to 10cm3 of NaOH. 10 cm3 pipettes are used with an accuracy of +/- 0.04cm3.

    The overall percentage error in measuring the total volume of the reaction mixture is
    A 0.04%
    B 0.08%
    C 0.4%
    D 4.0%

    answer in C but why?
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    (Original post by edex123)
    can anyone help on june 2011 question 16
    The enthalpy change of neutralization of an acid by an alkali is measured by adding 10 cm3 of HCl to 10cm3 of NaOH. 10 cm3 pipettes are used with an accuracy of +/- 0.04cm3.

    The overall percentage error in measuring the total volume of the reaction mixture is
    A 0.04%
    B 0.08%
    C 0.4%
    D 4.0%

    answer in C but why?
    I have an issue with this one too.... apparently your supposed to do this:

    0.04 x 2 / 20 x 100 = 0.4 %

    I see why you do this, but in physics I'm used to calculating percentage error in each measurement first & then adding those errors together.... don't seem to get the same answer though :confused:

    ( 0.04/ 10 + 0.04/10 ) x 100 = 0.8%
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    is it possible that someone could possibly draw out all the diagrams we need to know? pretty please
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    Hey guys,

    I'm stuck on this question from the May 2011 paper .

    10) There would be a major peak in the mass spectrum for butan-1-ol, CH3CH2CH2CH2OH, but not for butan-2-ol, CH3CH2CH(OH)CH3, at m/e value :

    A) 15

    B) 17

    C) 29

    D) 43

    The answer is D but I don't understand why. All replies are much appreciated.
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    (Original post by yousefshah77)
    Hey guys,

    I'm stuck on this question from the May 2011 paper .

    10) There would be a major peak in the mass spectrum for butan-1-ol, CH3CH2CH2CH2OH, but not for butan-2-ol, CH3CH2CH(OH)CH3, at m/e value :

    A) 15

    B) 17

    C) 29

    D) 43

    The answer is D but I don't understand why. All replies are much appreciated.
    Butan-1-ol can have CH3CH2CH2 fragment whereas butan-2-ol can't...

    the atomic mass of this fragment is 15+14+14= 43

    so the answer is D
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    (Original post by SophieL1996)
    drying agent for ethanol and how should it be used?
    Use any of the common drying agents, like anhydrous magnesium sulphate, calcium oxide, or calcium chloride. Add this anhydrous drying agent to ethanol, let it stand for a while then just filter it out, the water will be gone from ethanol
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    Anyone knows why tertiary alcohols are not oxidised?
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    (Original post by posthumus)
    I have an issue with this one too.... apparently your supposed to do this:

    0.04 x 2 / 20 x 100 = 0.4 %

    I see why you do this, but in physics I'm used to calculating percentage error in each measurement first & then adding those errors together.... don't seem to get the same answer though :confused:

    ( 0.04/ 10 + 0.04/10 ) x 100 = 0.8%
    i got 0.8% too !
    But i guess it makes sense to times the error by 2 since we are calculating total error
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    (Original post by Kurraiyo)
    Anyone knows why tertiary alcohols are not oxidised?
    For an alcohol to be oxidised there has to be at least one hydrogen directly bonded to the carbon atom which is also attached to the hydroxide. You need it as water gets produced and uses one hydrogen from te hydroxyl and one from the carbon.


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    (Original post by geor)
    No C-H bonds to break, there's a good explanation on chemguide
    What about "Steric Hinderance", I thought it was something to do with that too ? :confused:
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    (Original post by posthumus)
    What about "Steric Hinderance", I thought it was something to do with that too ? :confused:
    High steric hindrance means that the carbon attached to the group is surrounded by bulky alkyl groups, the better explanation in this case is the lack of C-H bonds but steric hindrance is good for comparing SN1 vs SN2
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    guys what test is the Nh4cl? i've never come across this one!
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    also for halogenoalkanes, why do you occasionally have to acidify before they react? does it mean adding an alcohol
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    (Original post by charlieejobson)
    guys what test is the Nh4cl? i've never come across this one!
    To test for the presence of a hydrogen halide. dip a glass rod in the solution (HCl) and hold in ammonia and dense white smoke of NH4Cl will form.
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    (Original post by charlieejobson)
    also for halogenoalkanes, why do you occasionally have to acidify before they react? does it mean adding an alcohol
    I'm not sure but i would guess that adding an acid protonates the haloalkane leading to it being more soluble in water. However i've never heard of it and it doesn't make much sense considering the reagents for haloalkane reactions are all alkali
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    Looking forward to this exam. It seems we all know our stuff inside out.

    105/120 and Chemistry is a sealed deal for me.
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    (Original post by James A)
    Looking forward to this exam. It seems we all know our stuff inside out.

    105/120 and Chemistry is a sealed deal for me.
    wanting to apply to cambridge is so soul destroying... i need(ish) 120/120 :'(
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    (Original post by geor)
    Same looool. We need about 78/80 for a safe 120. Tricky stuff
    I know the chemistry i just don't like the ''How Science Works'' rubbish. Thats where i always lose marks
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    (Original post by Goods)
    High steric hindrance means that the carbon attached to the group is surrounded by bulky alkyl groups, the better explanation in this case is the lack of C-H bonds but steric hindrance is good for comparing SN1 vs SN2
    (Original post by geor)
    Not in this case I think! Because even if the carbon was attached to smaller groups but there were still no C-H bonds, the oxidation still would not occur.
    Thanks guys I actually never knew a C-H bond was required to break from the same carbon as that attached to the halogen. I thought maybe the Hydrogen came from the acidic conditions...

    .. I guess I'll just have to remember this but I still hope it does not come up
    "There is no C-H bond attached to the carbon-halogen bond, therefore oxidation is not possible in tertiary alcohols"
    Does that sound like an answer that could get me the 2 marks ?

    (Original post by James A)
    Looking forward to this exam. It seems we all know our stuff inside out.

    105/120 and Chemistry is a sealed deal for me.
    Wow If I got that, then I would need to get only a high D in Unit 5 (and that's if I don't improve my 68 UMS from Unit 4!)....to achieve my B
 
 
 
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