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Edexcel - Chemistry Unit 2 - 4 June 2013 Watch

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    can anyone explain what instantaneous-induced dipole is ? and an example im getting so confused
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    (Original post by Gunner121)
    Can anyone please tell what do we have to know about carbon nanotubes ?
    Uses like caging drugs
    Properties like high strength and conductivity


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    (Original post by Tuya)
    Uses like caging drugs
    Properties like high strength and conductivity


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    is carbon nanotubes and buckminister fullerene the same thing
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    (Original post by Priya08)
    can anyone explain what instantaneous-induced dipole is ? and an example im getting so confused
    It's a tad late to learn about those now, coz half the paper might be based on those. They are also called London forces.
    The theory is that in a molecule, electrons are likely to be on one side of the molecule. So that side is slightly negatively charged and the other end is slightly positively charged. This is an instantaneous dipole. Because of these, the electrons in other molecules of the substance are repelled or attracted ie an induced dipole. Overall, these are attractive forces which is why non polar molecules don't just scatter away.


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    (Original post by Priya08)
    is carbon nanotubes and buckminister fullerene the same thing
    Kind of. The nanotubes and buck minister belong to a class of carbon molecules called fullerenes. The buck minister fullerene is specifically a C60 molecule in the shape of a sphere. Nano tubes are also fullerenes except they're tube shaped.


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    (Original post by Priya08)
    can anyone explain what instantaneous-induced dipole is ? and an example im getting so confused
    Same as London forces/Van der Waals...called many different things. They are found between every molecule, the weakest intermolecular force and are caused by the uneven distribution of electrons which create this instantaneous dipole in a molecule.

    So like in a Cl---Cl bond one of the chlorines will be slightly positive the other will be slightly negative, if alligned to an adjacent Cl---Cl then an attraction will occur between the two dipoles.
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    Little question : 1-iodobutane can me made by reacting butan-1-ol with phosphorus(III) idodie, PI3, formed by reacting moist red phosphorus with iodine.

    Complete the equation for the formation of 1-iodobutane.

    PI3 + ...C4h9OH -) ?

    I looked at the answer in the mark scheme but don't understand
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    (Original post by marseille_h)
    little question : 1-iodobutane can me made by reacting butan-1-ol with phosphorus(iii) idodie, pi3, formed by reacting moist red phosphorus with iodine.

    Complete the equation for the formation of 1-iodobutane.

    Pi3 + ...c4h9oh -) ?

    I looked at the answer in the mark scheme but don't understand
    pi3 + 3 c4h9oh ----> 3ch2h5i + h3po3

    EDIT: Why on earth does it end up going lower-case
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    (Original post by marseille_h)
    Little question : 1-iodobutane can me made by reacting butan-1-ol with phosphorus(III) idodie, PI3, formed by reacting moist red phosphorus with iodine.

    Complete the equation for the formation of 1-iodobutane.

    PI3 + ...C4h9OH -) ?

    I looked at the answer in the mark scheme but don't understand
    Just remember this...Name:  ImageUploadedByStudent Room1370276126.215431.jpg
Views: 192
Size:  189.8 KB


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    Hi, is there a way to know if a reaction is nucleophilic or electrophilic, or do we just have to remember it for each reaction and does anyone have an easy way of remembering it?? Thank you!
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    (Original post by Tuya)
    Just remember this...Name:  ImageUploadedByStudent Room1370276126.215431.jpg
Views: 192
Size:  189.8 KB


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    Thanks , i went through the revision guide for 2days learning every single thing but it dosen't come up in my mind when doing past papers.
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    Anyone know how to calculate this?? The ans is D but not sure how...?

    The enthalpy change of neutralization of an acid by an alkali is measured by adding
    10.0 cm3 of hydrochloric acid to 10.0 cm3 of sodium hydroxide. 10.0 cm3 pipettes with
    an accuracy of ±0.04 cm3 are used to measure out both solutions.
    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%
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    Why's Chemistry so hard? fml.
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    (Original post by maryam1996)
    Hi, is there a way to know if a reaction is nucleophilic or electrophilic, or do we just have to remember it for each reaction and does anyone have an easy way of remembering it?? Thank you!
    In general it's not so easy to tell whether an electrophile was involved first or a nucleophile, you have to be able to see the mechanism in your head when you look at the reactants/products or reactants/catalysts. But for AS level I think the only time it is electrophilic is when it is addition onto a C=C double bond. And if it is addition then it will be electrophilic, if it is substitution of any atom other than H it will be nucleophilic (common nucleophiles are -OH, X-, -C triple bond N, -NH2, etc.), if it is substitution of H it will be free radical substitution.
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    (Original post by marseille_h)
    Little question : 1-iodobutane can me made by reacting butan-1-ol with phosphorus(III) idodie, PI3, formed by reacting moist red phosphorus with iodine.

    Complete the equation for the formation of 1-iodobutane.

    PI3 + ...C4h9OH -) ?

    I looked at the answer in the mark scheme but don't understand

    PI3 + 3 C4h9OH - 3C4H9I + H3PO3
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    http://www.edexcel.com/migrationdocu...e_20100607.pdf q19 iv why do you times it by 10 ?? thanks you
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    (Original post by Khurc003)
    Anyone know how to calculate this?? The ans is D but not sure how...?

    The enthalpy change of neutralization of an acid by an alkali is measured by adding
    10.0 cm3 of hydrochloric acid to 10.0 cm3 of sodium hydroxide. 10.0 cm3 pipettes with
    an accuracy of ±0.04 cm3 are used to measure out both solutions.
    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%
    I seem to remember from doing this past paper that the answer is actually C? Though I can't explain why. I got 0.8% and then decided that 0.4% is closest so wrote that one down. (0.04/10.0)*2=0.8%.
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    (Original post by Big-Daddy)
    I seem to remember from doing this past paper that the answer is actually C? Though I can't explain why. I got 0.8% and then decided that 0.4% is closest so wrote that one down. (0.04/10.0)*2=0.8%.
    I think you do; (0.04 x 2)/20, then times by 100 to get it in % which is 0.4%
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    I swear fullerenes and nanotubes are Unit 1?
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    http://www.edexcel.com/migrationdocu...e_20100607.pdf 19 f i ??
 
 
 
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