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Edexcel Official Chemistry Paper1:Core Inorganic and Physical Chemistry - 27th of May watch

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    (Original post by clairebear101)
    Really!! I thought I messed it up? Who did??

    Sorry for not being clear - on the exam paper, I'm pretty sure? I know they wrote it as "ClF3" but I also think, somewhere in the paragraph, they did write "Chlorine triflouride". I remember thinking about it by name, and I'm not great at chemistry, so I'd only have done that if I'd read the name somewhere
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    (Original post by clairebear101)
    Really!! I thought I messed it up? Who did??
    Also, like someone else said, it was the inorganic paper - so it's unlikely to be a molecule with carbon in
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    (Original post by loveire&song)
    Sorry for not being clear - on the exam paper, I'm pretty sure? I know they wrote it as "ClF3" but I also think, somewhere in the paragraph, they did write "Chlorine triflouride". I remember thinking about it by name, and I'm not great at chemistry, so I'd only have done that if I'd read the name somewhere
    Okay i hope so, i assummed too as this compound is commmon and we have talked about it in class before
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    (Original post by sc67)
    Has anyone made or started making an unofficial mark scheme for this paper ?
    Remembering the questions would be the hard bit, we could try and make a list?
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    (Original post by richpanda)
    Remembering the questions would be the hard bit, we could try and make a list?
    Question 1 as I remember but I could be wrong!
    Spoiler:
    Show
    1) a) Chlorine is a gas and Iodine is a solid at room temperature. [1]
    b) Astatine would be a solid due to it having stronger London forces than chlorine (or you could have referred to the trend down the group) [1]
    c) Reaction of chlorine with cold aqueous sodium hydroxide.
    Cl_{2} + 2NaOH \rightarrow NaCl + NaClO + H_{2}O [1]
    This is a disproportionation reaction. [1]
    d) In the reaction of hydrogen bromide gas with ammonia white smoke would be observed. [1]
    e) - Electronegativity is the tendency of an atom to attract electron towards itself. [1]
    - In chlorine trifluoride, fluorine is more electronegative than chlorine [1]
    - Therefore the electrons are more attacted to fluorine resulting in a polar bond (refer to dipole) [1]
    f) Was a tickbox here?
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    (Original post by Cryptokyo)
    Question 1 as I remember but I could be wrong!
    Spoiler:
    Show
    1) a) Chlorine is a gas and Iodine is a solid at room temperature. [1]
    b) Astatine would be a solid due to it having stronger London forces than chlorine (or you could have referred to the trend down the group) [1]
    c) Reaction of chlorine with cold aqueous sodium hydroxide.
    Cl_{2} + 2NaOH \rightarrow NaCl + NaClO + H_{2}O [1]
    This is a disproportionation reaction. [1]
    d) In the reaction of hydrogen bromide gas with ammonia white smoke would be observed. [1]
    e) - Electronegativity is the tendency of an atom to attract electron towards itself. [1]
    - In chlorine trifluoride, fluorine is more electronegative than chlorine [1]
    - Therefore the electrons are more attacted to fluorine resulting in a polar bond (refer to dipole) [1]
    f) Was a tickbox here?
    yep I agree with all of this, the next one was the dodgy experiment one with the maths and the volume of hydrogen?
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    For the magnesium reaction one.
    Spoiler:
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    Equation Mg + 2HCl \rightarrow MgCl_{2} + H_{2}
    a) 0.04g of magnesium of RAM 24.3gmol-1 reacted. So for moles of magnesium:
    \frac{0.04}{24.3} = 1.646\times 10^{-3} moles
    b) Therefore for moles of hydrogen 1.646\times 10^{-3} moles
    c) Gas measured at 24 centigrade (297 kelvin), 101,000 Pa.
    PV=nRT
    V=\frac{nRT}{P}
    V=\frac{1.646\times 10^{-3}\times 8.31\times 297}{101000} = 4.022\times 10^{-5} m^{3}
    =40.22 cm^{3}
    d) Can't remember 6 marker.
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    (Original post by Cryptokyo)
    For the magnesium reaction one.
    Spoiler:
    Show
    Equation Mg + 2HCl \rightarrow MgCl_{2} + H_{2}
    a) 0.04g of magnesium of RAM 24.3gmol-1 reacted. So for moles of magnesium:
    \frac{0.04}{24.3} = 1.646\times 10^{-3} moles
    b) Therefore for moles of hydrogen 1.646\times 10^{-3} moles
    c) Gas measured at 24 centigrade (297 kelvin), 101,000 Pa.
    PV=nRT
    V=\frac{nRT}{P}
    V=\frac{1.646\times 10^{-3}\times 8.31\times 297}{101000} = 4.022\times 10^{-5} m^{3}
    =40.22 cm^{3}
    d) Can't remember 6 marker.
    I can't remember getting that for the moles of H2, but I got 40.22 for the volume and so must be correct! I rounded my answer for the moles but kept the un rounded answer for my volume calculation, lucky I did that! Otherwise I would have got 40.3! I think the next question was bout the relative mass?
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    (Original post by Cryptokyo)
    Question 1 as I remember but I could be wrong!
    Spoiler:
    Show
    1) a) Chlorine is a gas and Iodine is a solid at room temperature. [1]
    b) Astatine would be a solid due to it having stronger London forces than chlorine (or you could have referred to the trend down the group) [1]
    c) Reaction of chlorine with cold aqueous sodium hydroxide.
    Cl_{2} + 2NaOH \rightarrow NaCl + NaClO + H_{2}O [1]
    This is a disproportionation reaction. [1]
    d) In the reaction of hydrogen bromide gas with ammonia white smoke would be observed. [1]
    e) - Electronegativity is the tendency of an atom to attract electron towards itself. [1]
    - In chlorine trifluoride, fluorine is more electronegative than chlorine [1]
    - Therefore the electrons are more attacted to fluorine resulting in a polar bond (refer to dipole) [1]
    f) Was a tickbox here?
    Yes I would agree. We can probably miss out some of the easily forgettable 1 markers, as most people probably get them correct.
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    (Original post by richpanda)
    It was still stupid of them though, but I also did it as chlorine trifluoride. Wasn't the question asking about the bond polarities? What did you put? I said that the bonds would be polar due to electronegativity difference, and that Cl would be delta positive as it is less electronegative. I also quoted the Pauling values from the data book.
    That's all I did too. It said bonds, not the overall molecule
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    (Original post by clairebear101)
    I can't remember getting that for the moles of H2, but I got 40.22 for the volume and so must be correct! I rounded my answer for the moles but kept the un rounded answer for my volume calculation, lucky I did that! Otherwise I would have got 40.3! I think the next question was bout the relative mass?
    For the relative atomic mass question. There was 5.000g of lithium of which 0.460g was Lithium-6 and the rest is Lithium-7.
    Molar mass of Li-6: 6.015
    Molar mass of Li-7: 7.016
    Spoiler:
    Show
    Let A be moles of Li-6:
    A=\frac{0.460}{6.015}=0.076475..  . [1]

    Let B be moles of Li-6:
    B=\frac{4.540}{7.016}=0.64709... [1]

    Therefore RAM of Lithium is
    RAM=\frac{5}{A+B}=6.9012gmol-1 [1]
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    Does anyone know the acid used in the titration? I will post a solution to the titration once I know the acids and the concentrations.
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    (Original post by Cryptokyo)
    For the relative atomic mass question. There was 5.000g of lithium of which 0.460g was Lithium-6 and the rest is Lithium-7.
    Molar mass of Li-6: 6.015
    Molar mass of Li-7: 7.016
    Spoiler:
    Show
    Let A be moles of Li-6:
    A=\frac{0.460}{6.015}=0.076475..  .

    Let B be moles of Li-6:
    B=\frac{4.540}{7.016}=0.64709...

    Therefore RAM of Lithium is
    RAM=\frac{5}{A+B}=6.9012gmol-1
    well i didnt do that methid but i think i got that answer or something like that, defo was 6.9 something!
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    (Original post by Cryptokyo)
    Does anyone know the acid used in the titration? I will post a solution to the titration once I know the acids and the concentrations.
    It was nitric acid. Mean titre was 19.95 cm3. NaOH was in burette of conc 0.08moldm-3? 25cm3 aliquots from larger 250 cm3, make up from 10cm3 pure. Thats all I remember
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    (Original post by clairebear101)
    well i didnt do that methid but i think i got that answer or something like that, defo was 6.9 something!
    Perhaps this way? But I am not sure if this is correct as it gives a different answer.
    \frac{0.460\times 6.015 + 4.540\times 7.016}{5}=6.923
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    (Original post by SolomonP)
    It was nitric acid. Mean titre was 19.95 cm3. NaOH was in burette of conc 0.08moldm-3? 25cm3 aliquots from larger 250 cm3, make up from 10cm3 pure. Thats all I remember
    Cheers, sounds about right! You are correct just checked.
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    (Original post by Cryptokyo)
    Perhaps this way? But I am not sure if this is correct as it gives a different answer.
    \frac{0.460\times 6.015 + 4.540\times 7.016}{5}=6.923
    it was 6.9 something. don't fuss over a few decimal places.
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    For titration:

    Average titre of 19.95cm^3 of 0.08 moldm^3 NaOH, so 0.08 x 0.01995 = 0.001596mol of acid in 25cm^3.

    Therefore 0.01956 mol of acid in 250cm^3. This is the diluted bit of acid. They added 10cm^3 of pure acid, so concentration is 0.01956/0.01 = 1.596 moldm^3.

    Multiply this by the molar mass of nitric acid (63) to get 100.548 gdm^3.

    Therefore the acid is suitable.
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    My solution to the titration with help of SolomonP. The question was that 10.00cm^3 of nitric acid was made into 250cm^3 using distilled water. It was then titrated using 0.08moldm^-3 NaOH in 25cm^3 aliquots.
    Spoiler:
    Show
    Equation:
    HNO_{3} + NaOH \rightarrow NaNO_{3} + H_{2}O

    a) Mean titre: 19.95cm^{3}

    b) Moles of NaOH:
    0.08 \times 19.95 \times 10^{-3} = 1.596 \times 10^{-3}

    Moles of HNO3 in 25cm^3 aliquot:
    1.596 \times 10^{-3}

    Moles of HNO3 in 250cm^3:
    1.596 \times 10^{-3} \times 10 = 1.596 \times 10^{-2}

    Molar mass of HNO3 is 63gmol^-1. So mass of HNO3 is
    1.596 \times 10^{-2} \times 63 = 1.00548

    Mass per dm-3 of HNO3 in sample:
    \frac{1.00548}{10 \times 10{-3}}=100.548gdm^{-3}

    Therefore HNO3 in sample is suitable for aging wood as it needed to be 100gdm-3 and the value above is very close to it.



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    (Original post by Cryptokyo)
    Perhaps this way? But I am not sure if this is correct as it gives a different answer.
    \frac{0.460\times 6.015 + 4.540\times 7.016}{5}=6.923
    yep i did it this way so did everyone else i know
 
 
 
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