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2004 - Written Paper - Q8(a) - HSN forum

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2004 - Written Paper - Q8(a)


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#1 dfx

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Posted 30 May 2005 - 05:40 PM

How do you work this out?

I've got to the bit where you do mcdeltaT using 0.04 Kg and 6.5 centigrade and 4.18 JK/KgK, but then what? Any help appreciate.d. cheers.

#2 Soph.E.

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Posted 30 May 2005 - 05:45 PM

I have no idea either so someone PLEASE reply quickly!!! The answer booklet doesn't tell you anything either!

#3 dfx

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Posted 30 May 2005 - 05:50 PM

Just managed to figure it out.

After doing mcdeltaT you get 1.0868 KJ released.

Calculate the no. of moles of acid using n = cv, and you get 0.02 (20/1000 x 1)

so 1.0868 KJ released by 0.02 mol
x released by 1 mol

And you cross multiply as usual and end up with -54 KJ/mol (-ve because its exothermic - resulting NaCl is higher temp. than the reactant solutions).

#4 Gavers

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Posted 30 May 2005 - 05:57 PM

I know this!!! biggrin.gif

Heres what I got, from the start

Eh = C m delta.gif t
= 4.18 (specific heat capacity of water) * 0.04 (mass of the two liquids) * 6.5 (Temp change in the liquids)

= 1.0868 kJ


0.02 moles give us 1.0868 kJ
1 mole gives us x kJ

Cross multiply to find the answer, so it is...


0.02x = 1 * 1.0868

x = 1.0868
0.02 (<-- That's supposed to be underneath the other thing)

x = 54.34 kJ/mol, and thats yer answer smile.gif

Hope ya followed it okay, I actually made a daft mistake right at the start and got 4.5 as the temperature hange, d'oh!! wacko.gif

Gavers =)

#5 dfx

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Posted 30 May 2005 - 05:58 PM

Cheers m8.. thanks biggrin.gif

#6 Gavers

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Posted 30 May 2005 - 06:00 PM

No probs, but ya beat me to it tongue.gif

I forgot about working out the concentration, I didn't even put that calculation down rolleyes.gif

#7 Soph.E.

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Posted 30 May 2005 - 06:20 PM

Aaahh so THAT'S how you do it... rolleyes.gif silly mistake really - always, always I do not look at the question properly! Thank you!





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