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Need help with measuring enthalpy [neutralisation] Watch

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    So I know how to solve it, but in part (i), for mass aren't we suppose to include both volumes of the acid and alkali.

    So it's mass x temperature change x4.18
    We know the volume of HCl which is 50 but we don't know the volume of Ca(OH)2 , so I thought we'll have to calculate the volume of the alkali, then add 50 to get the mass then multiply 4.18 x 5.5

    But in the markscheme, they only included the volume of HCl so 50 x4.18 x 5.5 ..
    Whys that?!


    I know how to solve it,
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    Im assuming its the mass of the reacting solution hence 50. It only mentions solid mass for calcium hydroxide. That is my guess anyway
    Tbf that has always confused me :/
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    (Original post by Adorable98)
    So I know how to solve it, but in part (i), for mass aren't we suppose to include both volumes of the acid and alkali.

    So it's mass x temperature change x4.18
    We know the volume of HCl which is 50 but we don't know the volume of Ca(OH)2 , so I thought we'll have to calculate the volume of the alkali, then add 50 to get the mass then multiply 4.18 x 5.5

    But in the markscheme, they only included the volume of HCl so 50 x4.18 x 5.5 ..
    Whys that?!


    I know how to solve it,
    You are told that it's a SOLID
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    (Original post by charco)
    You are told that it's a SOLID
    Oh, So when it's a solid I only include one volume, but if the alkali was not a solid, then I should have calculated its volume right!
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    (Original post by Pentaquark)
    Im assuming its the mass of the reacting solution hence 50. It only mentions solid mass for calcium hydroxide. That is my guess anyway
    Tbf that has always confused me :/
    The mass of calcium hydroxide is not mentioned but as Charco pointed out, I only calculate the volume of the acid.
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    It doesn't look like an OCR A Q, but if it were, the Q would state that the specific heat capacity of an aqueous solution is 4.18 J g-1 K-1. And, since the Ca(OH)2 becomes part of the solution, you should include the mass of the solid. You would therefore have to work out the mass of the Ca(OH)2, which considering you know the formula (and hence Mr) and the amount, wouldn't be too much effort.

    If the solid did not dissolve and was in excess, OCR A would expect you to not include the mass of the solid in Q=mcDT. Even if some of the solid reacted and became part of the solution. There are plenty more issues with how OCR A does/did it, but my dinner in nearly ready.
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    (Original post by Pigster)
    It doesn't look like an OCR A Q, but if it were, the Q would state that the specific heat capacity of an aqueous solution is 4.18 J g-1 K-1. And, since the Ca(OH)2 becomes part of the solution, you should include the mass of the solid. You would therefore have to work out the mass of the Ca(OH)2, which considering you know the formula (and hence Mr) and the amount, wouldn't be too much effort.

    If the solid did not dissolve and was in excess, OCR A would expect you to not include the mass of the solid in Q=mcDT. Even if some of the solid reacted and became part of the solution. There are plenty more issues with how OCR A does/did it, but my dinner in nearly ready.
    I'm doing Edexcel so would that be any different? So when it's in excess I wouldn't need to mention its mass or volume, whether it's a solid or not, right?
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    I can't say for certain as I've never taught Edexcel, but it looks like the Q is trying to steer you away from including the mass of the solid (by just giving you the amount).

    Best thing to do is go through all the Edexcel past papers you can access - there's bound to be a similar Q, which you can check the MS to see how it should have been answered.
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    (Original post by Adorable98)
    I'm doing Edexcel so would that be any different? So when it's in excess I wouldn't need to mention its mass or volume, whether it's a solid or not, right?
    You cannot include a solid as you do not know it's specific heat capacity. You have to accept that it is an experimental inaccuracy.
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    (Original post by charco)
    You cannot include a solid as you do not know it's specific heat capacity. You have to accept that it is an experimental inaccuracy.
    Oh, I see!! Thaaank you!!
 
 
 
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