Chemistry and Physics AS paper 12 MCQS

[hope4thebest]

1 cannot be as molecules have to join. The molecules can join in any fashion so it can be any number less than 1.

thanks

 

[hope4thebest]

HELP required

 

[IGCSEstudent2012]

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_s03_qp_1.pdf
QUES 2 pls explain...i so bad in organic chem:/

 

[IGCSEstudent2012]

View attachment 28696

HELP required

v=f*lambda
speed of light divided by wavelength of uv waves
(3*10^8)/ (1*10^-7) = 3*10^15
is B the ans??

 

[1597.2217]

Hi,

I've been following this thread since over 15 hours (yes, that's right I didn't sleep in over 30 hours) and it had cleared many of my problems, not all. I now wish to help others wherever I can. Good luck with your exams everyone.

 

[hope4thebest]

v=f*lambda
speed of light divided by wavelength of uv waves
(3*10^8)/ (1*10^-7) = 3*10^15
is B the ans??

yes B thanks

 

[ShaanSiddiq090909]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w09_qp_11.pdf
Question no 4

 

[mehranshaikh]

thia question drove me crazy and the answer was so simple.
Thank you mehranshaikh

You're welcome. Examiner report clarified that for me.

 

[1597.2217]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s03_qp_1.pdf
QUES 2 pls explain...i so bad in organic chem:/

After about 40 minutes of scribbling, I've concluded to the answer.

First residue: CH3(CH2)4CH=CHCH2CH=CH(CH2)7CO2
Second residue: CH3(CH2)7CH=CH(CH2)7CO2
Original branch: CH3(CH2)3CH=CHCH=CHCH=CH(CH2)7CO2

First, I omitted out the unchanged part of molecules resulting in:

First residue: (CH2)4...CHCH2CH...
Second residue: (CH2)7...

For first residue: [Looking at the breaking of double bonds]
One CH2 molecule broke up and moved to change (CH2)3 to (CH2)4. Hence this part is justified.


=CHCH=CHCH= (Break the middle double bond by adding 1 hydrogen atom on each carbon). [Total 2 Hydrogen atoms.]
This results in: =CHCH2CH2CH=
Take one CH2 away since it was already added to (CH2)3
Result: =CHCH2CH=

For second residue: [The end part remains same, so we look at breaking the double bonds again.]
-CH=CHCH=CHCH= (Break the first two double bonds by adding 2 Hydrogen atoms on each carbon.) [Total 4 Hydrogen atoms].
-CH2CH2CH2CH2CH= (Take three CH2 molecules to original (CH2)3 making it (CH2)7 again justifying this part.)

Therefore the first residue required 1 mole of Hydrogen and second required 2 moles. However, the second residue is formed with 2 branches hence it multiply the moles by 2, giving us 1 Mole + 2 Moles + 2 Moles = 5.

5 Moles of Hydrogen is required, which is (B).

 

[IGCSEstudent2012]

After about 40 minutes of scribbling, I've concluded to the answer.

First residue: CH3(CH2)4CH=CHCH2CH=CH(CH2)7CO2
Second residue: CH3(CH2)7CH=CH(CH2)7CO2
Original branch: CH3(CH2)3CH=CHCH=CHCH=CH(CH2)7CO2

First, I omitted out the unchanged part of molecules resulting in:

First residue: (CH2)4...CHCH2CH...
Second residue: (CH2)7...

For first residue: [Looking at the breaking of double bonds]
One CH2 molecule broke up and moved to change (CH2)3 to (CH2)4. Hence this part is justified.


=CHCH=CHCH= (Break the middle double bond by adding 1 hydrogen atom on each carbon). [Total 2 Hydrogen atoms.]
This results in: =CHCH2CH2CH=
Take one CH2 away since it was already added to (CH2)3
Result: =CHCH2CH=

For second residue: [The end part remains same, so we look at breaking the double bonds again.]
-CH=CHCH=CHCH= (Break the first two double bonds by adding 2 Hydrogen atoms on each carbon.) [Total 4 Hydrogen atoms].
-CH2CH2CH2CH2CH= (Take three CH2 molecules to original (CH2)3 making it (CH2)7 again justifying this part.)

Therefore the first residue required 1 mole of Hydrogen and second required 2 moles. However, the second residue is formed with 2 branches hence it multiply the moles by 2, giving us 1 Mole + 2 Moles + 2 Moles = 5.

5 Moles of Hydrogen is required, which is (B).

lol ...40 mins is a big heck of time spent on 1 mcq!! i think chemistry shd be for nerds only:/
anyways THANK YOU for ur time

 

[IGCSEstudent2012]

P1 O/N 2003 chemistry ...ques 15 please?
thnx in advance!

 

[1597.2217]

Helping others is my pleasure.

P1 O/N 2003 chemistry ...ques 15 please?
thnx in advance!

First, n=V/24
To get V=24 as asked in the question, number of moles should be 1.

This means upon heating we should get 1 mole of gas. Write and balance the equation for thermal decomposition of each compound, and you will find that thermal decomposition of MgCO3 would produce 1 mole of CO2. Therefore answer is B.

 

[IGCSEstudent2012]

thnx..pls the same paper qus 22 and 23

 

[1597.2217]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w03_qp_1.pdf

Q22-

Mole ratio is given for the products. You can form an equation:

X ---> 5 C2H4 + C3H8 + C4H8
You have total of 17 C and 36 H. Option (B) is C17H36 which is your answer.

Q23-

You need to form 1,2-Dibromo-3-chloropropane. Let's look at the options given to us:
A) CH3CH2CH2Cl + 2Br2→ DBCP + 2HBr [This is not possible because Cl is more reactive than Br and it will not react.]
B) CH3CHBrCH2Br + Cl2→ DBCP + HCl [This is not possible either because Cl will replace both Br atoms and give you CH3CHClCH2Cl.]
C) CH2=CHCH2Cl + Br2→ DBCP [This is quite possible where the Br will attack the double bonds and form 1,2-Dibromo-3-chloropropane.]
D) ClCH2CH=CH2+ PBr5→ DBCP + PBr3 [PBr5 is acidic and will not react with the other reactant.]

So your answer is (C)

 

[1357913579]

thanks b

since the C:H ratio is high, we can predict that benzene is present.
Since it is not readily oxidised by mild oxidising agents, we say that it is a tertiary alcohol. draw the structures and the only compound which produced by the dehydration of X is D.

 

[1357913579]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s08_qp_1.pdf
FOR THIS PAPER NUMBER-25 HOW IS THE ANSWER (c) its written in the book that for secondary halogen hylides both sn1 and sn2 happens but question ass for ONLY BY AN SN1 MECHANICSM please help
thanks

 

[1597.2217]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_1.pdf
FOR THIS PAPER NUMBER-25 HOW IS THE ANSWER (c) its written in the book that for secondary halogen hylides both sn1 and sn2 happens but question ass for ONLY BY AN SN1 MECHANICSM please help
thanks

The answer is D, not C.
D is a tertiary haloganoalkane. C is a secondary haloganoalkane.
D can ONLY follow Sn1 mechanism, whereas C can follow both Sn1 and Sn2 mechanism.

 

[1357913579]

The answer is D, not C.
D is a tertiary haloganoalkane. C is a secondary haloganoalkane.
D can ONLY follow Sn1 mechanism, whereas C can follow both Sn1 and Sn2 mechanism.

thanks bro soory just realized my book misprinted the question answer

 

[1357913579]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w10_qp_12.pdf
number-40 please

 

[kingos3111]

oct nov 2004 q10 please abt eqbm pressure how to solve this