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Paper 5 Tips !! :)

M

MaxStudentALevel

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Silent Hunter said:
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w10_qp_53.pdf

How to do Q1 (d) ? xhizors MaxStudentALevel

or anyone?

JazakAllah :)
Click to expand...

You're lucky I just reviewed that! :)
-for the (I) part we just use a total fixed volume of H202 + H20 (I used 100cm3)
-And vary the volumes of each to vary concentration
- %of sol in overall = %of original concentration !!
-Just to get you started:
-ex: 40cm3 H202 + 60cm3 H20
-What's the %of H202 in final? 40/100 = 40%
-so what's the conc ?? initial = 2 so 40% of that
-(40/100) * 2 = 0.8 moldm-3 :)

As for the next part,
It's basically 3 mark planning! Just follow each step for each mark!
 
X

xhizors

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Wanderer said:
Won't the reaction start immediately after mixing the reactants...Aren't we supposed to heat the two solutions separately in a water bath until they come to equilibrium with the temp of water bath and then we mix them and start timing??
Click to expand...
exactly yes u warm thiosulfate to that extent and then tip the acid(also warmed acid)
my bad i didn't consider immediate start of reaction u guys rocks
JazakAllah guys
 
Silent Hunter

Silent Hunter

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MaxStudentALevel said:
You're lucky I just reviewed that! :)
-for the (I) part we just use a total fixed volume of H202 + H20 (I used 100cm3)
-And vary the volumes of each to vary concentration
- %of sol in overall = %of original concentration !!
-Just to get you started:
-ex: 40cm3 H202 + 60cm3 H20
-What's the %of H202 in final? 40/100 = 40%
-so what's the conc ?? initial = 2 so 40% of that
-(40/100) * 2 = 0.8 moldm-3 :)

As for the next part,
It's basically 3 mark planning! Just follow each step for each mark!
Click to expand...

should we always use 100 cm^3 ...... seems fine to me :p ?

thanks alot :) u giving exam this session? :)
 
X

xhizors

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MaxStudentALevel said:
You're lucky I just reviewed that! :)
-for the (I) part we just use a total fixed volume of H202 + H20 (I used 100cm3)
-And vary the volumes of each to vary concentration
- %of sol in overall = %of original concentration !!
-Just to get you started:
-ex: 40cm3 H202 + 60cm3 H20
-What's the %of H202 in final? 40/100 = 40%
-so what's the conc ?? initial = 2 so 40% of that
-(40/100) * 2 = 0.8 moldm-3 :)

As for the next part,
It's basically 3 mark planning! Just follow each step for each mark!
Click to expand...
a new approach thanks (y)
else u can use unitry method as posted before
 
X

xhizors

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Silent Hunter said:
should we always use 100 cm^3 ...... seems fine to me :p ?

thanks alot :) u giving exam this session? :)
Click to expand...
nopes its not necessary
you can use 50cm3 H2O2with 0cm3 of water u'll get the unchanged [H2O2]
then start decreasing volume of h2o2 and start inc vol of water by unitry u get dec concentration
sol A) u take 50cm3 of 2 mol/dm3 --> conc= 2mol each dm3
sol B) u add 10cm3 of Distilled water and 40cm3 of HCl in any container
50cm3-->2mol/dm3
40cm3--> x
by unitry method x = 1.6mol/dm3
 
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MaxStudentALevel

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xhizors said:
nopes its not necessary
you can use 50cm3 H2O2with 0cm3 of water u'll get the unchanged [H2O2]
then start decreasing volume of h2o2 and start inc vol of water by unitry u get dec concentration
sol A) u take 50cm3 of 2 mol/dm3 --> conc= 2mol each dm3
sol B) u add 10cm3 of Distilled water and 40cm3 of HCl in any container
50cm3-->2mol/dm3
40cm3--> x
by unitry method x = 1.6mol/dm3
Click to expand...


Yep! @SilentHunter
What he said!
 
Wanderer

Wanderer

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Can you guys also suggest a sensible diagram for NOV10/53 q1
What sort of internal separation device will be used here to separate H2O2 and the catalyst??
 
Ramsha007

Ramsha007

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xhizors

JazakAllah-o-Khair You have done a GREAT job MashAllah
Don't we have to place the plastic cup in beaker in enthalpy change questions?
 
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Hassi123

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xhizors said:
To find the solubility of any solid dissolved in any solution
Take 250 cm3 conical flask pour 50 cm3 of water
weigh a sample of solid and add it into water and stirr it until no more solid dissolves
then leave it for sometime to cool down to give crystals
filter the solution transfer the residue and discard the saturated solution
wash the residue with distilled water thoroughly and discard water then add propanone to absorb any water droplet left
warm a little or place the container under oven or sun (not bunsen burner else it will decompse solid) to evaporate any propanone left
now weigh the residue
repeat until constant mass is acheived
substract mass if residue with the initial mass so u'll get mass of solid dissolved to make a saturated solution,
now to find solubility use ( mass of solid use x 100)/mass of water
Click to expand...
Wait i might be wrong but for solubility dont you discard the residue (Not the saturated solution). You filter of the saturated solution and (Discard the residue). Then weigh the saturated solution. You then heat the saturated solution and you'll have a residue left, then you pour propanone on that (and not what you had initially). You can then warm this lightly via a water bath and carry this on till a constant mass attained. Then you just have to subtract this mass from the initial saturated solution mass and use the formula.
Atleast this is what i followed all this time.
 
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