CHEM UNIT 2!

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Can someone post the diagrams and apparatus mostly used in unit 2.....or just name it plzz ???:unsure:
 
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Not really good tbh! :/ I havent got enough time to do all the pastpapers.. i think ill do the latest exams only.... what about you?

same here..... only 1 day left.... and i still have some great ambiguities!!!!!:)
from which centre r u gonna appear and where?:sneaky:
 
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Can some one plz give an explaination to these questions
The attached file Q6 , Q11d , Q13 ( how to know where is the top layer and which is the bottom using the density )
 

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Can some one plz give an explaination to these questions
The attached file Q6 , Q11d , Q13 ( how to know where is the top layer and which is the bottom using the density )

6) The answer is B
CH4 is definitely not because it only contains london forces, the other 3 contain hydrogen bonds.
The answer is between HF or H2O and the answer is H2O because each H2O molecule forms 2 hydrogen bonds per molecule but HF forms only 1 per molecule
b)A
The molecular formula of the structure is C4H9I, C4H9Cl is not the answer because it has fewer electrons, hence weaker Van Der waals. So, this is between 1-iodobutane or 2-methyl-2-iodopropane. 1-iodobutane has a larger surface area than the more spherical structure of 2-methyl-2-iodopropane, hence the answer is 1-iodobutane

11d) Initiators are usually used for polymerization reactions, so i think it's B but i am not sure

13) The density of hydrocarbon layer is 0.660 g/cm^3 so it is less dense
An organic and an aqueous layer forms in the seperating funnel.
The density of the aqueous layer is close to the density of water which is about 1g/cm^3 so it is more dense than the density of the hydrocarbon layer, so it is below the hydrocarbon layer.
 
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6) The answer is B
CH4 is definitely not because it only contains london forces, the other 3 contain hydrogen bonds.
The answer is between HF or H2O and the answer is H2O because each H2O molecule forms 2 hydrogen bonds per molecule but HF forms only 1 per molecule
b)A
The molecular formula of the structure is C4H9I, C4H9Cl is not the answer because it has fewer electrons, hence weaker Van Der waals. So, this is between 1-iodobutane or 2-methyl-2-iodopropane. 1-iodobutane has a larger surface area than the more spherical structure of 2-methyl-2-iodopropane, hence the answer is 1-iodobutane

11d) Initiators are usually used for polymerization reactions, so i think it's B but i am not sure

13) The density of hydrocarbon layer is 0.660 g/cm^3 so it is less dense
An organic and an aqueous layer forms in the seperating funnel.
The density of the aqueous layer is close to the density of water which is about 1g/cm^3 so it is more dense than the density of the hydrocarbon layer, so it is below the hydrocarbon layer.

Thank u sooooooooooo much :D
 
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guys wat do u ppl say will iodometric titration cum 2moro or acid base......
wat r ur predictions and how difficult da ppr is gonna be???//
 
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can someone plzzz help me with some questions:

Q10 Q11 Q15 and Q16

http://www.edexcel.com/migrationdocuments/QP GCE Curriculum 2000/June 2011 - QP/6CH02_01_que_20110527.pdf

One more thing, plzzzz provide me with notes on "Mass and infrared spectrometry" i don't understand it at all :( :'(


Q10. D-- because butan-1-ol can be fragmented at CH3CH2CH2 to give an m/e value at 43, which is not possible in butan-2-ol

Q11. C-- not too sure why, but I guess it's because there are 3 main elements, hence there will be 3 molecular ion peaks, since both isotopes of Bromine combine to form one major molecular ion peak for Bromine

Q15. B-- because ethylamine cannot form when the reactant is a methane group. The methane can be repeated when it replaces one H atom from NH3, but it will increase by methyl groups only. 1-methyl, dimethyl, trimethyl etc.

Q16. C-- since the TOTAL percentage error has to be calculated (for measuring out NaOH and acid), multiply the accuracy of the pipette by 2 (i.e. 0.04*2= 0.08). Then divide by the total amount that is measured out (i.e. 10 cm3+ 10 cm3) and multiply by 100, to give +-0.4%

Hope you could understand :p
 
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