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Paper 5 Tips !! :)

FRENZYAMU

FRENZYAMU

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xhizors said:
Measuring enthalpy change of combustion!!
Take an empty spindle weigh it and then pour any alcohol and re weigh it
setup a dry wrick and cap the spindle to avoid evaporation ofuel
take 150 cm3 metal can pour 50 cm3 water and place it over spindle on a table alike
measure the initial temp of water
light the spindle and keep measuring temperature until their is a rise of 10*C
recap spindle allow it to cool remove cap and wrick and reweigh it
substract mass after heating and before heating u'll get the mass which is burnt in excess oxygen
use q=mc4T and u'll get enthalpy change of combustion
Click to expand...
we use mass of water in this case or mass of alcohol burnt ? btw no need to answer this after today since i'll be done with the paper tommorow
 
knowitall10

knowitall10

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But in the end, i wrote that the the experiment that shows the greatest change in volume would be correct (2.2) because 24dm3 is the vol of gas at rtp.
is that right?
 
X

xhizors

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FRENZYAMU said:
Question : might sound stupid but does mixing 30g of NaOH and 50g of H20 make up the mass of solution to 80g because doesnt molecules like dissolve in it nd do not really increase vol of water ? i knw stupid, but its really bugging me
Click to expand...
1g = 1cm3 of water so 80g of water = 80cm3
now just learned it and dont get confuse by the bugs
 
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g
knowitall10 said:
But in the end, i wrote that the the experiment that shows the greatest change in volume would be correct (2.2) because 24dm3 is the vol of gas at rtp.
is that right?
Click to expand...
greatest vol changed will be by 2.1
also u also used up examiner portion :p
 
AdeleintheDeep

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Ashique said:
@kazi umayer alim

In this experiment we get a precipitate. The prediction that we're putting into test is that the moles of of the precipitated copper hydroxide increases with the increasing conc. of CuSO4.
  • So keeping our hypothesis in our mind, we should know that the conc. of CuSO4 is the independent variable, which means it is in our control. Hence, you need to prepare a range of concentrations of CuSO4. You're also aware of the fact that the solutions saturates at 1.39 mol/dm3. So the range of concentrations that you prepare must not exceed 1.39 mol/dm^3. So your range is going to be 0-1.39 mol/dm^3. You need to prepare at least five diluted solutions.
  • Say you're going to prepare 100 cm^3 of each of the following conc. range: 0.01, 0.05, 1.00 1.10, 1.20 and 1.39. You need to tell the examiners exactly how you're going to prepare them. Since you're given SOLID hydrated CuSO4, you're going to have to dissolve a certain mass in water. So say you want to make the 0.01 mol/dm^3 solution. You need to know how many moles are are to dissolved in a 100 cm^3 solution. So Moles= CV= 0.01*100*10^-3= 0.01 mol. So you need to dissolve 0.01 mol of hydrated CuSo4 in 100 cm^3 of water. To find out the mass= Moles*Mr= 0.01*249.6= 2.50 g. So you dissolve 2.50 g of the salt into 100 cm^3 of water to make a 0.01 mol/dm^3 solution. The rest of the concentrations are to be prepared in the same way- but you don't need to show the working. One calculation for conc. should be enough.
  • After you've added the NaOH into the solution a precipitation of Cu(OH)2 will occur. You need to filter this (Or centrifuge it,and then decant).You could leave it out to dry, or add water and propanone.
Thats about it! Hope it helped!
Click to expand...
Ashique , what about part 3 for this question? Where did u calculate the molar concentration?
Im confused because part 1 and 3 seem related to me. We'd have to find the concentration of CuSO4 in both parts. :/
 
FRENZYAMU

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Ashique said:
@kazi umayer alim

In this experiment we get a precipitate. The prediction that we're putting into test is that the moles of of the precipitated copper hydroxide increases with the increasing conc. of CuSO4.
  • So keeping our hypothesis in our mind, we should know that the conc. of CuSO4 is the independent variable, which means it is in our control. Hence, you need to prepare a range of concentrations of CuSO4. You're also aware of the fact that the solutions saturates at 1.39 mol/dm3. So the range of concentrations that you prepare must not exceed 1.39 mol/dm^3. So your range is going to be 0-1.39 mol/dm^3. You need to prepare at least five diluted solutions.
  • Say you're going to prepare 100 cm^3 of each of the following conc. range: 0.01, 0.05, 1.00 1.10, 1.20 and 1.39. You need to tell the examiners exactly how you're going to prepare them. Since you're given SOLID hydrated CuSO4, you're going to have to dissolve a certain mass in water. So say you want to make the 0.01 mol/dm^3 solution- you need to know how many moles are are to dissolved in a 100 cm^3 solution. So Moles= CV= 0.01*100*10^-3= 0.01 mol. So you need to dissolve 0.01 mol of hydrated CuSo4 in 100 cm^3 of water. To find out the mass= Moles*Mr= 0.01*249.6= 2.50 g. So you dissolve 2.50 g of the salt into 100 cm^3 of water to make a 0.01 mol/dm^3 solution. The rest of the concentrations are to be prepared in the same way- but you don't need to show the working. One calculation for conc. should be enough.
  • After you've added the NaOH into the solution a precipitation of Cu(OH)2 will occur. You need to filter this (Or centrifuge it,and then decant).You could leave it out to dry, or add water and propanone.
Thats about it! Hope it helped!
Click to expand...
they say in some marking schemes that dissolve solid in lil' water first and then make up the solution to for eg, 250 cm^3 ,,,,,we dont do that here ? mixing solid in lil' water first and then making up the sol. to the required volume ?
 
Gémeaux

Gémeaux

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knowitall10 said:
I'm sorry xhizors the internet was playin arnd gettin me pissed :/ Finally, here, my solubility table.. and yeah, i like ur idea of discussing papers:)
Jazak Allahu Khair brother:)
Click to expand...
For column F you use, C-B. Doesn't that give the mass of solution, and not the mass of water in the solution? :unsure:
 
knowitall10

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knowitall10

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Gémeaux said:
For column F you use, C-B. Doesn't that give the mass of solution, and not the mass of water in the solution? :unsure:
Click to expand...
The mass of solution right here is the mass of water:)
1cm3 = 1g try using this principle to understand what i did. because obviously, there isnt any other way to get values less than 100 now is there?
 
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