2004 - Written Paper - Q8(a)
Posted 30 May 2005 - 05:40 PM
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.
Posted 30 May 2005 - 05:45 PM
Posted 30 May 2005 - 05:50 PM
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).
Posted 30 May 2005 - 05:57 PM
Heres what I got, from the start
Eh = C m 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
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!!
Posted 30 May 2005 - 06:00 PM
I forgot about working out the concentration, I didn't even put that calculation down
Posted 30 May 2005 - 06:20 PM
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