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    Normally high 50s/low60s range dependent on difficulty


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    (Original post by Meganward123)
    can someone please explain the concentration 6 marker?
    I worked out that there were 0.0025 moles, 0.0025/(20/1000) equals 0.125 mol/dm3
    I used the equation- Volume (acid) x Concentration (acid)= Volume (alkali) x Concentration (alkali)
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    1920 cm^ 3 for the volume question using 24000cm3

    0.08 moles

    0.08 was the conc of sodium hydroxide in the 6 marker at 20 cm was the end point
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    (Original post by Swagswag1963)
    1920 cm^ 3 for the volume question using 24000cm3

    0.08 moles

    0.08 was the conc of sodium hydroxide in the 6 marker at 20 cm was the end point
    Yes I agree.

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    (Original post by Rjhepkenhdjd)
    do you know what the grade boundaries for it are normally like in general? ty !!
    Going off last year, it was 60 for an a*, 50 for an A, decreasing each time by 10 marks.
    C1,2,3 was 61 for an A*
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    (Original post by Meganward123)
    can someone please explain the concentration 6 marker?
    I worked out that there were 0.0025 moles, 0.0025/(20/1000) equals 0.125 mol/dm3
    concentration of acid x volume acid = concentration of alkali x volume alkali

    This is the equation u use for titrations.

    So u fill it in

    0.1*40=?* 25

    ?= (0.1*40)/25
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    (Original post by esmithxx)
    I used the equation- Volume (acid) x Concentration (acid)= Volume (alkali) x Concentration (alkali)
    Thanks Emily!
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    (Original post by Meganward123)
    can someone please explain the concentration 6 marker?
    I worked out that there were 0.0025 moles, 0.0025/(20/1000) equals 0.125 mol/dm3
    I did it like so:
    Moles = concentration*volume
    Moles HCl = 0.1 * 20/1000 (convert to dm3) = 0.002

    HCl:NaOH
    1:1
    0.002:0.002

    Concentration = moles/volume
    Concentration NaOH = 0.002/(25/1000)
    = 0.002/0.025 = 0.08


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    what was the answer to how much pesticide was used for country E
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    (Original post by JickDee)
    what was the answer to how much pesticide was used for country E
    1,260,000,000 ???

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    (Original post by mxskaan)
    1,260,000,000 ???

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    Didn't you have to multiply that by 0.18
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    (Original post by mxskaan)
    1,260,000,000 ???

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    I got something like that too
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    (Original post by happiness12)
    I got something like that too
    Yhh tht was the answer
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    (Original post by ellamoon_22)
    Yhh tht was the answer
    I got 0.16 for the titration six marker
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    (Original post by ellamoon_22)
    Yhh tht was the answer
    Didn't you have to multiply that by 0.18
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    (Original post by haithamnhaque)
    I got 0.16 for the titration six marker
    It was 0.08
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    (Original post by haithamnhaque)
    I got 0.16 for the titration six marker
    Sme bit everyone sed end point was 20 idk
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    What was the answer to the question where it asked what mass would be produced with 20.6 amps instead of 10.3 for 120 minutes ???
    i got 60 g
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    (Original post by ellamoon_22)
    Yhh tht was the answer
    I just read the titration page and you are supposed to use the point of neutralization. That means it is not 0.16
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    (Original post by happiness12)
    What was the answer to the question where it asked what mass would be produced with 20.6 amps instead of 10.3 for 120 minutes ???
    i got 60 g
    Yeah I think I got that too.
 
 
 
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