Chemistry and Physics AS paper 12 MCQS

[Manobilly]

Chemistry oct 2003 q 1,2,3.

 

[Mairaxo]

Chemistry oct 2003 q 1,2,3.

Q1- 3*10^-21 * 6.02*10^23 = 1806
Q2- P in P2O5 = 62/62+80=0.43*30=13.1%
Q3- C6H12 + 9O2 --> 6CO2 + 6H2O
mass of water= 18*6=108
mass of CO2= 44*6=264
108/264=0.41

 

[papersxtreme]

physics june 2012/12 mcq 13
Help.

 

[Alice123]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s03_qp_1.pdf
no 26, 27, 35, 38
your help will be highly appreciated....
Thanks

 

[hope4thebest]

u

A and B have H-bonds as well. In D Vander Waals forces are becoming stronger as its solidifiacation. In C there are only Vander Waals forces and its melting so the bonds need to be overcome.

Thanks I understand it now

 

[hope4thebest]

syed1995 please help! Thanks

 

[mariamh]

Q14- A because max height means it covered half the distance so the energy also gets halved.
Q16- C. first find input power. output/input *100=efficiency. so 400000000/80=5MW
then find I using P=IV, 5MW/25kV=200A
Q21- find the area under the graph. Make it a triangle 1st. A=0.5*100*(2*10^-3)=0.1J but the area is not exactly a triangle. It is slightly more than a triangle so we assume it to be 0.11J

OK thank you so much

 

[mariamh]

Physics October November 2010 Paper 12
9, 14, 21, 27

 

[mariamh]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s03_qp_1.pdf
no 26, 27, 35, 38
your help will be highly appreciated....
Thanks

35. These are the properties of suphur always remember it is used to kill bacteria as it is a reducing agent. And the second property is also to be memorized 3 is wrong as it has a smell and large quantities is not good.
38 1 is correct as the nuclophile oh- replaes br-. 2 is correct as u know that the reaction of halogenoalkanes with hot aqueous NaOh and Koh results in formation of an alcohol. Three is wrong as the Nacl should be concentrated but not aqueous!

 

[Mairaxo]

Physics October November 2010 Paper 12
9, 14, 21, 27

Q9- momentum=mass*velocity= 2mv
if its elastic, momentum after collision will also be 2mv so the total collision would be 4mv. here its inelastic so it is not 4mv. Its not 2mv as well because this would mean it did not rebound. so it is 3mv
Q14- find the component of the force. it will be sin30=200/force. force=400N then multiply 1.5 to get the work = 600J add frictional force and you get 750
Q21- E=FL/A*extension. E is the same and F is same so Lp/Ap*extensionp=Lq/Aq*extensionq
2L/4A*extensionp=L/A*extensionq
extensionp/extensionq= 2LA/L4A = 0.5

 

[snowbrood]

View attachment 28330
syed1995 please help! Thanks

well PV=nRT
we know number of moles wont change before and after the gases are mixed. T is also constant R is always constant so
P1V1 for helium P2V2 for Neon P3V3 for mixed gases.
P1V1+P2V2=P3V3
(5*12)+(6*10)=P3*15
solving this we get P3 as 8

 

[sweetjinnah]

i need help in these qts plz....mcq 21,37 nd 39....
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w06_qp_1.pdf

 

[sweetjinnah]

well PV=nRT
we know number of moles wont change before and after the gases are mixed. T is also constant R is always constant so
P1V1 for helium P2V2 for Neon P3V3 for mixed gases.
P1V1+P2V2=P3V3
(5*12)+(6*10)=P3*15
solving this we get P3 as 8

can i ask u one thing how is V3 15....??

 

[mariamh]

Q9- momentum=mass*velocity= 2mv
if its elastic, momentum after collision will also be 2mv so the total collision would be 4mv. here its inelastic so it is not 4mv. Its not 2mv as well because this would mean it did not rebound. so it is 3mv
Q14- find the component of the force. it will be sin30=200/force. force=400N then multiply 1.5 to get the work = 600J add frictional force and you get 750
Q21- E=FL/A*extension. E is the same and F is same so Lp/Ap*extensionp=Lq/Aq*extensionq
2L/4A*extensionp=L/A*extensionq
extensionp/extensionq= 2LA/L4A = 0.5

Thanks alot. Can u please also explain question 27 of this same paper

 

[mariamh]

October november 2010/ paper 11
Question 9 and 37
Please help !

 

[mariamh]

May June 2011 Paper 11 Physics
question 13

 

[sweetjinnah]

October november 2010/ paper 11
Question 9 and 37
Please help !

37...: the voltmeter is connected across both resistors R and Q hence it shows a reading of 4 V even if the slider position is changed....

 

[hope4thebest]

well PV=nRT
we know number of moles wont change before and after the gases are mixed. T is also constant R is always constant so
P1V1 for helium P2V2 for Neon P3V3 for mixed gases.
P1V1+P2V2=P3V3
(5*12)+(6*10)=P3*15
solving this we get P3 as 8

Thanks for helping!!

 

[hope4thebest]

I thought the question would be easy but couldnt solve this please help!

 

[ahmed abdulla]

snowbrood