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dont have word :C just got this new mac. if u copy pasta it ill give it a go.
Reply 2
Attachment is not working for me either.
Oo, sorry. hold on:smile:
Malsi101
Attachment is not working for me either.


how about now..
Reply 5
LittleMissTwinkle
how about now..



nah

it's not working; maybe it's just my comp.

it's making me 'find' it but then microsoft something or other comes up.

gah
originalname
dont have word :C just got this new mac. if u copy pasta it ill give it a go.



Method This reaction is very fast and so we can measure the maximum temperature rise directly. You are going to take the average temperature of the acid and alkali at the start as the start temperature.

1) Place a clean, dry polystyrene cup into an outer plastic cup to increase insulation and stability.

2) Pipette 25 cm3 of the 1.0 mol dm-3 hydrochloric acid into the polystyrene cup.

3) Place a thermometer into the cup through a cardboard lid within a tripod.

4) Pipette 25 cm3 of the 1.0 mol dm-3 sodium hydroxide solution into a second clean, dry polystyrene cup.

5) Give both solutions a few minutes to allow their temperatures to settle and then record them. Make sure the thermometer bulb is clean and dry before transferring from one solution to the other.

6) Pour the alkali into the acid and place the lid on the cup.

7) Stir well and record the maximum temperature reached.


Results 8) Record all necessary temperatures in a suitable table here.

ACID ALKALI
START/ DEGREES 18.8 19.0
MAX / DEGRESS 24.5










Analysis 9) Calculate the heat evolved in this reaction. Assume that the solution has a density of 1.0 g/cm3 and that the specific heat capacity is 4.18 J g-1 K-1.

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10) Calculate the moles of acid and alkali that reacted.

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11) Calculate the molar enthalpy change for the reaction.

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12) Calculate the percentage apparatus error in the final result
25 cm3 pipette  0.1 cm3 thermometer  0.1 C

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Evaluation 13) The quoted value for the enthalpy change for this reaction is –57.1 kJ mol-1. Find the difference between your value and the actual value as a percentage of the actual value.

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14) Comment on the accuracy of your results

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15) If your value is outside the apparatus error, suggest specific reasons why it is too exothermic or not exothermic enough. If your value is within apparatus error, give reasons why somebody else’s value may not be exothermic enough. Give ways to improve the experiment to reduce these problems.

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Malsi101
nah

it's not working; maybe it's just my comp.

it's making me 'find' it but then microsoft something or other comes up.

gah


Malsi, i've copied and pasted it above=)
If this part of your coursework?
Reply 9
ACID ALKALI
START/ DEGREES 18.8 19.0
MAX / DEGRESS 24.5


Sorry-I can't decipher this although some might argue there's nothing to decipher:redface:
Witches_Rave
If this part of your coursework?


Nope, it's practise questions..
Reply 11
for 11 I don't think you have to convert it to kJ again as you did that with the first question so the answer is -46.8 kjmol^-1
Malsi101
ACID ALKALI
START/ DEGREES 18.8 19.0
MAX / DEGRESS 24.5


Sorry-I can't decipher this although some might argue there's nothing to decipher:redface:


at the start the temperature of the acid was 18.8 degrees
at the start the temperature of the alkali was 19.0

step 7) record the max. temp. reached = 24.5
anyone.....?
Reply 14
LittleMissTwinkle
anyone.....?

If you look above I did post a post.

Please, though, do your own homework. It will really benefit you more.
Think about what could have gone wrong and I'm sure you'll bag the marks available.
But don't just put no effort into the rest of them; try them yourself.

Also remember your units. It's exothermic thus your answers require the negative sign.
Malsi101
If you look above I did post a post.

Please, though, do your own homework. It will really benefit you more.
Think about what could have gone wrong and I'm sure you'll bag the marks available.
But don't just put no effort into the rest of them; try them yourself.

Also remember your units. It's exothermic thus your answers require the negative sign.


Yeah true. i've worked most of them out now, but not all. if i do get stuck, i will ask for help on that question and not be lazy in posting all the homework.xx
Reply 16
9) 1.2 kJ
10) 0.025 mol HCl / 0.025 mol NaOH
11) Using (9), -48 kJ mol-1
12) Kinda long answer...

Not taking into account the uncertainties of the assumed specific heat capacity and the density of water:

T(inicial and average) = 18.9 °C ± 0.1 °C
T(final) = 25.5 °C ± 0.1 °C
ΔT = 5.6 °C ± 0.2°C (Percentage of uncertainty = 4%)

V of NaOH solution = 25.0 cm3 ± 0.1 cm3
V of HCl solution = 25.0 cm3 ± 0.1 cm3 (Percentage uncertainty = 0.4%
V total = 50.0 cm3 ± 0.2 cm3 (Percentage uncertainty = 0.4%)


Mass of water = (50.0)(1.0) = 5.00 x 10^1 g ± 0.2g (Percentage of uncertainty = 0.4%)


Concentration = 1.0 mol dm-3 ± 0.1 mol dm-3 (Percentage uncertainty = 10%)
(This one is assumed due to significant figures)

Moles of HCl/NaOH = (0.0250)(1.0) = 0.025 mol HCl/NaOH (Percentage uncertainty = 10%(10% + 0.4%))

Q = mcΔT = (50.0)(4.18)(5.6) = 1.20 kJ ± 0.05 kJ (Percentage of uncertainty = 4% + 0.4% = 4.4% = 4%)

ΔH = -(1.20)/(0.025) = -48 kJ mol-1 ± 7 kJ mol-1 (Percentage of uncertainty = 4% + 10% = 14%)
Reply 17
Qwaps
ΔH


Off topic, but how did you get the delta? :tongue:
Kinkerz
Off topic, but how did you get the delta? :tongue:


I get them from here, copy and paste :cool:
Reply 19
EierVonSatan
I get them from here, copy and paste :cool:


Δ :tongue:

Cheers :grin: