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Paper 5 is the easy one. You don't even have to study that much.Paper 5 is the biggest trouble which will not let me end with nice grades. Help me. I usually end to get 80 in paper 4 and 15 in paper 5![]()
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Paper 5 is the easy one. You don't even have to study that much.Paper 5 is the biggest trouble which will not let me end with nice grades. Help me. I usually end to get 80 in paper 4 and 15 in paper 5![]()
What is the minimum and maximum score u got in it?Paper 5 is the easy one. You don't even have to study that much.I guess you just need practice and basic logic.
I had only two tests in p5 ... both of which were my mocks.What is the minimum and maximum score u got in it?
Can u link that papers here.I had only two tests in p5 ... both of which were my mocks.
Scored 24 in one and 19 in the other.
But it's true that sometimes we mess up easily in p5 ... that problem is there.
Can u link that papers here.
True.
I am not sure I have them.Can u link that papers here.
True.
You should attempt a few paper 5s and compare your answers with the mark schemes. Most of the questions are logical and only require basic chemistry concepts.I haven't even started P5
Can someone guide me on what I should do
what are the specific p5 topics that I must cover?
pls help
Guide me for physics & bio p5 too
qwertypoiu
Hats off man! ^_^You should attempt a few paper 5s and compare your answers with the mark schemes. Most of the questions are logical and only require basic chemistry concepts.
In question 1 you'll be asked a few questions then asked to devise an experiment. You just need to be careful t mention the independent, dependent, and control variables. You must make clear how you'll change the independent variable. You must specify all the apparatus to be used. You must explain how the control variables will be controlled. They'll guide you on these points. Then they'll ask you a few more questions like precautions. You just have to think logically.
In question 2, you'll be given data. You just have to draw graphs and analyse. Even simpler.
For physics it's mainly same. In question 2 there is uncertainties involved and you'll need to learn to draw "line of worst Fit". Again a few mark schemes and examiner reports will make it very clear what they want.
Bio is also same.
You can write without brackets as well in this case, no problem.Hi guys can I ask if
CH3CHClCHClCH3 is the same as CH3CH(Cl)CH(Cl)CH3, cause the mark scheme specifies CH3CH(OH)CH(OH)CH3 for the -OH functional group
You can write without brackets as well in this case, no problem.![]()
To be at safer side include brackets. In my As level I did without brackets. ^_^Would CIE accept though?
The silver ions will form the following precipitates:View attachment 60111
So precipitate with aquous silver nitrate means that could be any of the halogens but what does solubility with aquous ammonia tell about?
thank u so much I get that...here is anotherThe silver ions will form the following precipitates:
AgCl - white ppt
AgBr - creamy ppt
AgI - yellow ppt
These ppt will dissolve as follows:
If dilute ammonia is added to AgCl, it dissolves.
If concentrated ammonia is added to AgBr, it dissolves.
Even conc. NH3 is not able to dissolve AgI.
Therefore, in your question, since they said dilute ammonia cannot dissolve the ppt, we know that X cannot be a chloro-compound.
No!thank u so much I get that...here is another
Question is :
Compound A is a liquid which does not react with 2,4-DNPH reagent or with aqueous bromine. Suggest two structural formula of A.
(We found the molecular formula in previous part which was C2H4O2) Here is what the markscheme says
View attachment 60113
It's (d)
I know ethanoic acid is formed but isn't HCO2CH3 just another way of writing CH3CO2H?
wow, you just cleared my concept. thank you! Okay, just one more pleaseNo!
HCO2CH3 is in fact an ester!
The name is methylmethanoate.
It can be formed by esterification of methanoic acid (HCOOH) and methanol (CH2OH).
If we wanted to write ethanoic acid with the acid group on the left, we'd write:
HO2CCH3.
Because otherwise an R1-CO2-R2 is interpreted as an ester.
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