Stankovic123's chem q's

[vox nihili]

Came across a question which asked me to draw the structural formula of 1-chlorohex-3-ene???
How is that possible, shouldn't the double bond take precedence over functional groups when naming hydrocarbons??

That's my understanding of it. Though, sometimes people don't fret as much as IUPAC would like them to do. Alternatively, we could both be wrong and no doubt someone will tell us soon enough!

[zvezda]

Hey,
In the data book, it says that it gives the specific hear capacity of water as 4.18 J g^-1 K^-1
Whereas, in heinemann, instead of kelvins it gives degrees.
Whats with that?
Especially considering this q from vcaa where the solutions didnt really consider this, assuming that the specific heat capacity was in terms of degrees:
"A foam cup calorimeter containing 100ml of water is calibrated by passing an electric current through a small heater placed in the solution. Assuming that all measurements are accurate, which one of the following is the most likely calibration factor (in J C^-1) for the calorimeter and contents?
The answer is 480.

[Aurelian]

Hey,
In the data book, it says that it gives the specific hear capacity of water as 4.18 J g^-1 K^-1
Whereas, in heinemann, instead of kelvins it gives degrees.
Whats with that?
Especially considering this q from vcaa where the solutions didnt really consider this, assuming that the specific heat capacity was in terms of degrees:
"A foam cup calorimeter containing 100ml of water is calibrated by passing an electric current through a small heater placed in the solution. Assuming that all measurements are accurate, which one of the following is the most likely calibration factor (in J C^-1) for the calorimeter and contents?
The answer is 480.

It doesn't matter whether heat capacities/calorimeter constants are given in kelvin or degrees celsius because ultimately they concern changes in temperature, and changes in temperature will have the same values whichever of the two units are used.

E.g. consider:
T_i = 20^oC = 293K
T_f = 25^oC = 298K
∆T = 5^oC = 5K

Hope that helps!

[zvezda]

It doesn't matter whether heat capacities/calorimeter constants are given in kelvin or degrees celsius because ultimately they concern changes in temperature, and changes in temperature will have the same values whichever of the two units are used.

E.g. consider:
T_i = 20^oC = 293K
T_f = 25^oC = 298K
∆T = 5^oC = 5K

Hope that helps!

Ahhh yes of course. Thanks for that.

Also, im having a bit of trouble with vcaa 2009 q10 from exam 2. Im not sure why theyve multiplied 4.18 by 100

[scribble]

the 100 is because you have 50ml of HCl and 50ml of KOH which adds to become 100ml.
you assume that 1ml=1g and since E=m*c*delta(T), E=100*4.18*3.5
to get deltaH (in Jmol-1), divide by 0.025

[zvezda]

the 100 is because you have 50ml of HCl and 50ml of KOH which adds to become 100ml.
you assume that 1ml=1g and since E=m*c*delta(T), E=100*4.18*3.5
to get deltaH (in Jmol-1), divide by 0.025

Surely that cant be assumed? 1mL=1.00g for water doesnt it? Not anything else?

[vox nihili]

Surely that cant be assumed? 1mL=1.00g for water doesnt it? Not anything else?

It's an approximation and is close enough.

[scribble]

its an assumption you're allowed to make.
in the aqueous solution of HCl and KOH, theres going to be waywaywaywaywaay more water than there will be HCl and KOH, so effectively the HCl and KOH won't really make much of a different to the heat capacity of the solution, so just work with 4.18JK-1mol-1

[lzxnl]

For comparison, [H2O] = n(H2O) / v(H2O)
In one litre of water, the mass is very close to a kilogram and the molar mass is 18 g/mol, so we're talking 55.6 moles of water every litre.

How often do you come across a 55.6 M solution of ANYTHING?

[zvezda]

I know this is a long time afterwards lol. But, what if there was something like a 5M solution? We still use 4.18?


Also, not sure if you guys are familiar with thushans guide, but theres a q on there of which im not 100% sure.
To summarise it (as its got several parts):
-we deal with the combustion of methane
-we initially found the molar enthalpy of the reaction having been ignited in a calorimeter and the temperature of the calorimeter changed by 8.5 K. However, we used the temperature change and assumed that water (200g) absorbed it all.
-we then ignited benzoic acid and figured out the calibration fact having known the molar enthalpy of the benzoic combustion.
-in the end, the molar enthalpy calculated using the calibration factor was higher than the one calculated using the specific heat capacity of water.

My question is more concerned with the experimental set up. In the heinemann book, its 2 images of calorimeters involve thermometers in water. How do you actually measure the temperature of the calorimeter itself?

Thanks

[zvezda]

Hey,
When calculating the energy absorbed from dissolution, and we have say 10g of a substance we are dissolving in 100g of water. When calculating the energy absorbed, when plugging in the mass of water, do we use (for this example) 100g or 110g?
Thanks

[lzxnl]

Hey,
When calculating the energy absorbed from dissolution, and we have say 10g of a substance we are dissolving in 100g of water. When calculating the energy absorbed, when plugging in the mass of water, do we use (for this example) 100g or 110g?
Thanks

Water mass is still only 100g. You may have to factor in the heat capacity of the dissolved substance, but if the question doesn't give you the info to do so, you don't have to worry.

[zvezda]

Water mass is still only 100g. You may have to factor in the heat capacity of the dissolved substance, but if the question doesn't give you the info to do so, you don't have to worry.

I thought so. I think in the AN chem guide they include the added mass of solute.

[zvezda]

Was just thinking about titrations (as you do), and was windering about the immediate colour change in the aliquot before the colour returns back to normal. Why does this happen?
Shouldnt the change in colour of the aliquot be proportional to the amount of solution added from the burette?
Thanks

[Limista]

Was just thinking about titrations (as you do), and was windering about the immediate colour change in the aliquot before the colour returns back to normal. Why does this happen?
Shouldnt the change in colour of the aliquot be proportional to the amount of solution added from the burette?
Thanks

If by 'normal' you mean it goes clear again, it's because you are really close to the equivalence point, but you haven't passed it yet to get to the endpoint where the colour change is permanent. 

I don't think change in colour of the aliquot has any correlation with amount of solution from burette.