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AQA A-level Chemistry 7405 - Paper 1 (Inorganic & Physical Chem) - 04th June 2019

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Reply 80
chem1.JPGchem 2.JPG
Can someone help, I got the first couple steps, but then they divided by three, and I don't really understand
Does anyone have the links to the specimen set 2 papers and mark schemes (paper 1, 2 and 3)?
has anyone got exampro
Could you share the link for the question papers please
Original post by TommieP123
Does anyone have the links to the specimen set 2 papers and mark schemes (paper 1, 2 and 3)?
The complex has 3 C2O42- ions. You need to find just one so you divide by 3
Original post by smaiu
chem1.JPGchem 2.JPG
Can someone help, I got the first couple steps, but then they divided by three, and I don't really understand
Reply 87
Thank you, I understand it now.
Original post by thewaterbottle12
The complex has 3 C2O42- ions. You need to find just one so you divide by 3
chem18 q2.pnghow?
So they tell you that the total pressure was 150 kPa.
The mole fraction of NH3 is 0.80 so if you do 0.80 multiplied by 150 you get a partial pressure of 120kPa for nitrogen.
There are still 30 kPa left unaccounted for, and they tell you that the ratio is 1:3. So you divide 30 by 4 (because there are 4 parts in total, 1+3=4) and then you get an answer of 7.5 kPa. This the partial pressure of nitrogen. To get hydrogen you do 7.5 times 3 (because 1:3 ratio) and you get 22.5 kPa for hydrogen.
Hope it helps
Original post by xxxtentacion..
chem18 q2.pnghow?
tysm!
Original post by who_am_i08
So they tell you that the total pressure was 150 kPa.
The mole fraction of NH3 is 0.80 so if you do 0.80 multiplied by 150 you get a partial pressure of 120kPa for nitrogen.
There are still 30 kPa left unaccounted for, and they tell you that the ratio is 1:3. So you divide 30 by 4 (because there are 4 parts in total, 1+3=4) and then you get an answer of 7.5 kPa. This the partial pressure of nitrogen. To get hydrogen you do 7.5 times 3 (because 1:3 ratio) and you get 22.5 kPa for hydrogen.
Hope it helps
Reply 91
I got 408m3 too so why would it be B (408dm3)?
Original post by Chez 01
i found the question paper, it says (1.01 x 10^5) in the question not (1.01 x 105)

https://www.thestudentroom.co.uk/showthread.php?t=914063&page=2

the same question was asked in this thread.
this is the worked solution from this thread

n = mass / Mr = (1000 x 10^3) / (28 + 16 + 16) = 16666.67 mol

pV = nRT

V = nRT / p = (16666.67 x 8.31 x 298) / (1.01 x 10^5) = 408m3

hope this helps
No they divided by 2 then times by 5 to get moles of C2O7 then u divide it by 5 to get One mole of C2O7 because its 5 moles in the equation ( 5C2O7) so divide it by 1 to work out percentage purity of one mole of the compound instead of 5 moles of it, then do M= Mr x N to work out mass 2.16/2.29 x 100= 94.4% :smile:
Original post by smaiu
chem1.JPGchem 2.JPG
Can someone help, I got the first couple steps, but then they divided by three, and I don't really understand
(edited 4 years ago)
do we need to know about fragmentation of molecular ions in mass spectrometers? like the equation for fragmentation
Original post by pigeontree27
do we need to know about fragmentation of molecular ions in mass spectrometers? like the equation for fragmentation


just put your charge and radical dot. There's no "equation".

Its assumed knowledge, not exactly listed on the spec
A student added 627 mg of hydrated sodium carbonate (Na2CO3.xH2O) to 200 cm3 of
0.250 mol dm–3 hydrochloric acid in a beaker and stirred the mixture.
After the reaction was complete, the resulting solution was transferred to a volumetric
flask, made up to 250 cm3 with deionised water and mixed thoroughly.
Several 25.0 cm3 portions of the resulting solution were titrated with 0.150 mol dm–3
aqueous sodium hydroxide. The mean titre was 26.60 cm3 of aqueous sodium
hydroxide.
Calculate the value of x in Na2CO3.xH2O
Show your working.
Give your answer as an integer.


I get how to do the question, but why is the 25cm^3 scaled up to 250 and not 200? because there was originally 200cm^3 of HCl, not 250. and the reaction happened with 200.
The total volume was 250cm^3, you need to consider the final concentration after the addition of the water. That's why you use 250 and not 200.
Original post by shfuewygthf7uw
A student added 627 mg of hydrated sodium carbonate (Na2CO3.xH2O) to 200 cm3 of
0.250 mol dm–3 hydrochloric acid in a beaker and stirred the mixture.
After the reaction was complete, the resulting solution was transferred to a volumetric
flask, made up to 250 cm3 with deionised water and mixed thoroughly.
Several 25.0 cm3 portions of the resulting solution were titrated with 0.150 mol dm–3
aqueous sodium hydroxide. The mean titre was 26.60 cm3 of aqueous sodium
hydroxide.
Calculate the value of x in Na2CO3.xH2O
Show your working.
Give your answer as an integer.


I get how to do the question, but why is the 25cm^3 scaled up to 250 and not 200? because there was originally 200cm^3 of HCl, not 250. and the reaction happened with 200.
Screenshot 2019-05-28 at 13.32.45.png
Anyone understand part b? I did it but got 3cm3, the right answer is 4cm3.
do we need to remember how to draw the structure of EDTA for paper 2? Bc i really cba with that.
work out moles of MnO4- , which is 0.0005
multiply by 8 because of the ratio - giving you 0.004
but, sulfuric acid is diprotic so will release two protons - therefore divide by 2 - giving you 0.002
divide by 0.5 ( concentration)
x1000 to convert from dm3 to cm3- should give you 4cm3
Original post by thepewds_
Screenshot 2019-05-28 at 13.32.45.png
Anyone understand part b? I did it but got 3cm3, the right answer is 4cm3.

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