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Discussion in 'International A And AS Level' started by XPFMember, Sep 29, 2011.
Whenever it says file size too big just take a screenshot of the picture you want then post it.
Got it! thank you.
Can you please help me with my question though?
Sorry I don't know the answer.
Hi, can anyone help me solving this question? This is taken from 2017 MJ 12. The answer should be D.
Pls help me with paper 34 May June 2016 chemistry. How to find the dilution factor?
Qs 1 c part
Where can I get notes that follow the syllabus and explains each learning outcome without adding extra unnecessary information. If there is for chemistry, is there also for biology and physics?
Try Znotes, they'd probably have it all. As for chemistry chemguide is pretty good!
Heating an alcohol with H2SO4 would dehydrate it and result in alkene formation. If you heat A,B or C , you have 2 options for a double bond in all of them ( double will come between the OH carbon and the carbon next to it, That can be on either side whicj would result in mixture of both compounds ). However, in D there is only one carbon beside OH carbon so there's just one possibility.
Do you have others? Also can you get chemguide as a pdf file?
which ions are present in a solution of ethanol in an excess of concentrated sulphuric acid?
Not sure ummm but 2&3?
1) is definitely out as it requires the OH group to act as an acid and release a H+ . which is not possible in the presence of H2SO4
Can anyone explain to me how to arrive to the answer .... https://drive.google.com/file/d/1BkV12DgTZwm7fl42yz1lTP0m7SrHlKRq/view?usp=drivesdk
Can anyone tell me why in a(ii) they multipled experiment 1's initial rate with 27 to get K?
where that 27 came from?
The two equations are:
CH4 + 2O2 --> CO2 + 2H2O
CH4 + 3/2O2 --> CO + 2H2O
multiply equation 1 by 99 and add to equation 2
99CH4 + 198O2 --> 99CO2 + 198H2O
CH4 + 3/2O2 --> CO + 2H2O
100CH4 + 199(1/2)O2 --> 99CO2 + CO + 200H2O
So, burning 100 moles needs 199(1/2) moles of oxygen
Therefore burning y moles needs 2y - y/200
or expressed as a decimal fraction:
2y - (0.01/2)y
i.e. answer A
Reaction is first order wrt O2 and 2nd order wrt NO. So concentration of NO is increased 3 times as well as [O2]. You get (3^2) x 3 considering the orders which'd give you 27 times the rate in reaction 1.
Edit : An alternative way would be to calculate the rate constant using any experiment of your choice and then using it along with concentrations in the 4th experiment to get your desired rate.
bro where did this 3 came from?
O2 is first order and it's concentration is tripled ^1 will be 3
Please help me with this one too.. ?
What is the procedure of building these sort of equations and then telling which is the slow one?
Separate names with a comma.