NEED HELP FOR SL CHEM

[wahyan32]

Am I finding the change of heat correctly? I am not so sure...please help!
And i think i am totally off with the units...= = do we have to change celsius into kelvin and is the final answer j or kj
Thx a lot ><


Mass of the Salt = The Salt’s mass with the weighting boat – Mass of the weighting boat
Mass of the Salt = (5.86) g ±0.01 – (1.01) g ±0.01
Mass of the Salt = (4.85) g ±0.02

Mass of the water = Volume of water x Density of water
Mass of the water = (98.0) mL ±0.01 x (1.00) g/mL
Mass of the water = (98.0) g ±0.01

∆ Temperature = Initial temperature of the solution – Final temperature of the solution
∆ Temperature = (23.0) ℃ ±0.01 – (19.6) ℃ ±0.01
∆ Temperature = (3.40) ℃ ±0.02

Assuming that the reaction is 100% efficient:
∆ Heat = Mass of water x Heat Capacity of Water x ∆ Temperature
∆ Heat = (98.0) g ±0.01 x (4.18) J∙g^(-1)∙mol^(-1) x (3.40) ℃ ±0.02
∆ Heat = ((98.0) g ±1.02 x 〖10〗^(-4)) x ((4.18) J∙g^(-1)∙mol^(-1)) x ((3.40) ℃ ±2.94 x 〖10〗^(-3))
∆ Heat = (13.9 x〖10〗^2) J ±5.98 x 〖10〗^(-3)



Number of moles of Salt used = Mass of the Salt x (One mole)/(Mass in one moles of the Salt (NH_4 Cl))

Number of moles of Salt used = (4.85) g ±0.02 x (1 mol)/((53.49) g)

Number of moles of Salt used = (0.09067) mol ±0.412%



Molar heat = (Energy released)/(number of moles)

Molar heat = ((13.9 x〖10〗^2)J ±0.598%)/((0.09067) mol ±0.412%)

Molar heat = ((13.9 x〖10〗^2)J ±0.598%)/((0.09067) mol ±0.412%)

Molar heat = (15.3 x 〖10〗^3)J ±1.01%

[Tarz]

This seems right to me :/
Just one technical correction:


∆ Heat = (98.0) g ±0.01 x (4.18) J∙g^(-1)∙mol^(-1) x (3.40) ℃ ±0.02

should be


∆ Heat = (98.0) g ±0.01 x (4.18) J∙g^(-1)∙K^(-1) x (3.40) K ±0.02

Just use kelvin for everything... It shouldn't matter if you're talking about the change of temperature, but it's good to keep it consistent. (since the specific heat capacity is usually defined in J/gK)

[moneyfaery]

[quote name='Tarz' post='39238' date='Mar 6 2009, 08:15 PM']Just use kelvin for everything... It shouldn't matter if you're talking about the change of temperature, but it's good to keep it consistent. (since the specific heat capacity is usually defined in J/gK)[/quote]
It actually makes no difference as a change in 1 degree K is equal to a change in 1 degree C. So it's really up to personal preference.

[quote name='wahyan32' post='39173' date='Mar 6 2009, 05:52 AM']∆ Heat = (13.9 x〖10〗^2) J ±5.98 x 〖10〗^(-3)[/quote]
H = mct = 98(4.18)(3.4) = 1392 J g[sup]-1[/sup] = 1.39 kJ kg[sup]-1[/sup]
It's 1.39 x 10[sup]3[/sup] J g[sup]-1[/sup] [b]or[/b] 1392 J g[sup]-1[/sup] [b]or[/b] 1.39 kJ kg[sup]-1[/sup]. What you have is incorrect scientific notation.

I'm sorry to say this but your presentation is pretty ugly. Can you think of a way to remove the clutter? Also, split up the calculations and error propagation... putting the two together makes it messy and hard to read.

The final answer should be in kJ kg[sup]-1[/sup] mol[sup]-1[/sup].

[wahyan32]

[quote name='Irene' post='39244' date='Mar 7 2009, 01:52 AM']It actually makes no difference as a change in 1 degree K is equal to a change in 1 degree C. So it's really up to personal preference.


H = mct = 98(4.18)(3.4) = 1392 J g[sup]-1[/sup] = 1.39 kJ kg[sup]-1[/sup]
It's 1.39 x 10[sup]3[/sup] J g[sup]-1[/sup] [b]or[/b] 1392 J g[sup]-1[/sup] [b]or[/b] 1.39 kJ kg[sup]-1[/sup]. What you have is incorrect scientific notation.

I'm sorry to say this but your presentation is pretty ugly. Can you think of a way to remove the clutter? Also, split up the calculations and error propagation... putting the two together makes it messy and hard to read.

The final answer should be in kJ kg[sup]-1[/sup] mol[sup]-1[/sup].[/quote]

i noe....do u have nay suggestion to present those uncertainties more clearly?

[moneyfaery]

[quote name='wahyan32' post='39335' date='Mar 7 2009, 09:31 AM']i noe....do u have nay suggestion to present those uncertainties more clearly?[/quote]
Split it up into two columns.