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    (Original post by britishtf2)
    Sounds about right. I can't speak for the rounding.
    what did you round to?
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    No i meant the bromine isotopes one?
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    (Original post by SANTR)
    It won't be below 80% unfortunately.
    I don't think the paper was too bad but where have you based this prediction off?
    In my opinion the paper was harder than previous years
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    wouldnt the precipitate formed be magnesium sulphate ?
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    for the gas mc question, was the total amount of gas 30cm3? You start with 30 of A (can't remember), 20 of B and 0 of product. The mole ratios were 1:1:1, so 20 of the product forms but as 10 of A is in excess it remains so 20+10=30??
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    (Original post by DaVinciGirl)
    Me too, I got the same, but I have strong feeling we have may lost a mark by not writing 0.70g as most of the values were like that.
    my chemistry teacher once gave us a test in which all the values given were to 3sf, except for one which was 100cm^3. the whole class gave their answer to 3sf because they assumed that 100cm^3 was also 3sf, but he told us that it's not clear that it is so we should have answered to 1sf. (SO IF WE LOSE THE MARK FOR THIS I'M GONNA BE VERY MAD AT HIM LOL)
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    (Original post by britishtf2)
    Oh you won't lose all the marks for that. First error lose 1 or 2 (if they are being mean) probs. Second error lose 1, I would think.
    god, that would be a relief! I would get nearly half my marks back thank you, you're a lifesaver
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    (Original post by RIDDHI7)
    No i meant the bromine isotopes one?
    Ahh. Because it is [Br2]+ you have 4 possible outcomes from 2 isotopes:
    81-81
    81-79
    79-81
    79-79.

    Therefore, on avg it will be 25% each for 158 and 162, with 50% 160.
    I think 1 mark for 3 lines on correct numbers, 1 mark for correct peaks.
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    (Original post by rosemondtan)
    god, that would be a relief! I would get nearly half my marks back thank you, you're a lifesaver
    No problem - any time.
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    I think I've literally lost like 30 marks 😭😭😭😥 but at least it doesn't count for me and won't influence my predicted grade I suppose....
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    (Original post by haes)
    my chemistry teacher once gave us a test in which all the values given were to 3sf, except for one which was 100cm^3. the whole class gave their answer to 3sf because they assumed that 100cm^3 was also 3sf, but he told us that it's not clear that it is so we should have answered to 1sf. (SO IF WE LOSE THE MARK FOR THIS I'M GONNA BE VERY MAD AT HIM LOL)
    I also study Physics and Biology. In both I have learnt that to round to the appropriate significant figure you find the value given in the least significant figure and use that.
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    (Original post by haes)
    what did you round to?
    I cannot remember from memory. If I were to hazard a guess, I think I either put 0.69 or 0.695. I have nooooo clue.
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    (Original post by Kantth)
    I can understand how you worked it out. I used another method by working out the number of moles using moles=mass/mr of baso4 then i wrote the equation bacl2+h2so4=baso4+2hcl and everything is a 1:1 ratio so the same moles of bacl2 were used. I worked out the mass of bacl2 from this using the mass=mr x moles. Then i took this away from the mass of both bacl2 and mgcl2 to work out how much mgcl2 there was. My answer was also 35.5% 👍🏿👍🏿
    Yup same, agreed
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    (Original post by rumanah1)
    wouldnt the precipitate formed be magnesium sulphate ?
    Sulphate trend is: down the group, less soluble.

    "Barium sulfate is the inorganic compound with the chemical formula BaSO₄. It is a white crystalline solid that is odorless and insoluble in water"
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    (Original post by Kantth)
    I can understand how you worked it out. I used another method by working out the number of moles using moles=mass/mr of baso4 then i wrote the equation bacl2+h2so4=baso4+2hcl and everything is a 1:1 ratio so the same moles of bacl2 were used. I worked out the mass of bacl2 from this using the mass=mr x moles. Then i took this away from the mass of both bacl2 and mgcl2 to work out how much mgcl2 there was. My answer was also 35.5% 👍🏿👍🏿
    I got 35.5% too. Got a bit nervous when some people said 29.5%, but i think that method was correct.

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    Looking at some of the answers on page one, some of the content you do at AS level is similar to highers we do in Scotland. I sat my higher chemistry exam last week.


    Posted from TSR Mobile
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    (Original post by Kantth)
    I can understand how you worked it out. I used another method by working out the number of moles using moles=mass/mr of baso4 then i wrote the equation bacl2+h2so4=baso4+2hcl and everything is a 1:1 ratio so the same moles of bacl2 were used. I worked out the mass of bacl2 from this using the mass=mr x moles. Then i took this away from the mass of both bacl2 and mgcl2 to work out how much mgcl2 there was. My answer was also 35.5% 👍🏿👍🏿
    First person I have seen to get the same answer as me! Didn't ask anyone after the exam though, to be fair.
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    (Original post by vignesh1234567)
    Guys for the gas mc question, was the total amount of gas 30cm3? You start with 30 of A (can't remember), 20 of B and 0 of product. The mole ratios were 1:1:1, so 20 of the product forms but as 10 of A is in excess it remains so 20+10=30??
    did the question ask for the volume of product? i thought it asked for the volume at the end of the reaction, and because it wasn't an equilibrium reaction i just added them together and said it was 50 :/
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    I am going out for a while: post any questions you have and I will get back to you on them. Ciao!

    Edit: Make sure to reply to this post with the questions so I see it.
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    (Original post by haes)
    did the question ask for the volume of product? i thought it asked for the volume at the end of the reaction, and because it wasn't an equilibrium reaction i just added them together and said it was 50 :/
    It asked for the volume at the end of the reaction. I thought as 20 of A reacts with 20 of B to form 20cm3 of product, at the end of the reaction there is 20cm3 of product + the excess 10cm3 of A.
 
 
 
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