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2017 AQA A-Level Chemistry 7405/1 Unofficial Markscheme
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Yoo listen people, i'd post this on the thread but it would get lost in the replies, so here's my answers for today's paper.
I don't remember the units, but if u got same answers thumbs up this post so i know how i did!
-869
1.05
4.84 <----- (my answer's wrong, correct answer is 4.50)
7.27
12.95
8.14*10^-26
9.612*10^-7 seconds [TOF]
-238 KJ Mol-1
9.41*10-7 kPa^-2 [KP]
56.6%
[Fe(C2O4)3]^3-
Shape = Octahedral
Isomerism = Optical Isomerism
Bond Angle = 90 degrees
The Copper colorimeter one = In the complex, copper has a full 3d sub-shell, electrons can't absorb wavelength of visible light and get excited (get promoted from ground state), therefore no light transmitted.
I don't remember the units, but if u got same answers thumbs up this post so i know how i did!
-869
1.05
4.84 <----- (my answer's wrong, correct answer is 4.50)
7.27
12.95
8.14*10^-26
9.612*10^-7 seconds [TOF]
-238 KJ Mol-1
9.41*10-7 kPa^-2 [KP]
56.6%
[Fe(C2O4)3]^3-
Shape = Octahedral
Isomerism = Optical Isomerism
Bond Angle = 90 degrees
The Copper colorimeter one = In the complex, copper has a full 3d sub-shell, electrons can't absorb wavelength of visible light and get excited (get promoted from ground state), therefore no light transmitted.
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#7
(Original post by wannabe_lawyer)
****, I forgot to convert my entropy value to kJ
****, I forgot to convert my entropy value to kJ
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#8
Method for the 8 Marker
Titration 1
Titration 2
Calculation
Thus 56.6%
Titration 1
- Work out moles of Manganate ions
- Multiply by 5/2 to get number of moles of C2O42- ions
Titration 2
- Work out moles of OH-
- Divide by 2 to get number of H2C2O4 moles
Calculation
- Subtract number of H2C2O4 moles from number of moles of C2O42- ions
- Multiply by 10 (as these moles are only in 25cm3 but total solution was 250cm3)
- Multiply moles by Mr of the sodium ethanoate to get it's mass
- Divide this mass by the total mass (1.9g)
- Multiply by 100
Thus 56.6%
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#9
(Original post by DarkEnergy)
Method for the 8 Marker
Titration 1
Titration 2
Calculation
Thus 56.6%
Method for the 8 Marker
Titration 1
- Work out moles of Manganate ions
- Multiply by 5/2 to get number of moles of C2O42- ions
Titration 2
- Work out moles of OH-
- Divide by 2 to get number of H2C2O4 moles
Calculation
- Subtract number of H2C2O4 moles from number of moles of C2O42- ions
- Multiply by 10 (as these moles are only in 25cm3 but total solution was 250cm3)
- Multiply moles by Mr of the sodium ethanoate to get it's mass
- Divide this mass by the total mass (1.9g)
- Multiply by 100
Thus 56.6%
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#13
(Original post by EvilScientist)
What was up with Question 10.3?
What was up with Question 10.3?
Fe3++ e---> Fe3+
So instead of being logical and telling us to correct the second Fe3+ to Fe2+ they basically stopped us from getting an extra two easy marks
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#14
How did you get 4.5?
I converted everything to moles
Subtracted moles of added sodium hydroxide away from ha, added sodium hydroxide moles to a-
Worked out h+ then worked out ph
I got 4.78
I converted everything to moles
Subtracted moles of added sodium hydroxide away from ha, added sodium hydroxide moles to a-
Worked out h+ then worked out ph
I got 4.78
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#18
(Original post by wannabe_lawyer)
****, I forgot to convert my entropy value to kJ
****, I forgot to convert my entropy value to kJ
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#19
(Original post by Bwile12)
How did you get 4.5?
I converted everything to moles
Subtracted moles of added sodium hydroxide away from ha, added sodium hydroxide moles to a-
Worked out h+ then worked out ph
I got 4.78
How did you get 4.5?
I converted everything to moles
Subtracted moles of added sodium hydroxide away from ha, added sodium hydroxide moles to a-
Worked out h+ then worked out ph
I got 4.78
Just find the concentration of naoh.,,
Add that to the conc of ethanoate... this is the new conc of ethanoate
Then subtract the concentration of naoh from the conc of acid... this is the new conc of acid,,,
Then using ka from previous part
Rearrage to find conc of h+
And then -log it and get 4.5
Thats what i did
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