Chemistry and Physics AS paper 12 MCQS

[mariamh]

for Q3 to form ion2+ u need energy that is the sum of first and second ionisation
when u refer to databooklet u will realize aluminium needs 1820+577=2397
and cobalt also needs 757+1640=2397 see energies are same
for Q9 i will solve it i am tired at the moment

Thanks

 

[ahmed abdulla]

for Q3 to form ion2+ u need energy that is the sum of first and second ionisation
when u refer to databooklet u will realize aluminium needs 1820+577=2397
and cobalt also needs 757+1640=2397 see energies are same
for Q9 i will solve it i am tired at the moment

when you get recovered help with this too plz ​

 

[Aries_95]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w09_qp_12.pdf

q4,8,9,12,13,18,20,21,25

this paper was a tough one please help me with these?

Q.4:
Use the equation 2as=v^2-u^2
Where a=accelaeration= 0.2 ms^-2
S=distance=1km=1000,
U=initial velocity= 0 ms^-1
V=?
Putting the vaues in the equation, the answer is 20 ms^-1 ie A.
Q.18:
Pressure= 15200 Pa
Density=?
Height= 80 cm=80x10^-3 m
g=10ms^-2
Use the eq: P= densityxgxh
the answer is 1900kg/m^3 ie B
Q.20:
Use the formula F=kx
F=100N, x=2.0mm= 2x10^-3m hence calculate k (K= 5x10^4)
Next, F=1/2 kx^2
Put the values of k and x in the equation, the answer is 0.11 J
q.25:
d=10^3/300= 3.33m
3.33 sin 90= (n/2)x (450x10^-9)
We have divided n/2 since we have to find he no. of maxima).

The answer is 14.8 ie 15 (D).

 

[lubna1232]

Q.4:
Use the equation 2as=v^2-u^2
Where a=accelaeration= 0.2 ms^-2
S=distance=1km=1000,
U=initial velocity= 0 ms^-1
V=?
Putting the vaues in the equation, the answer is 20 ms^-1 ie A.
Q.18:
Pressure= 15200 Pa
Density=?
Height= 80 cm=80x10^-3 m
g=10ms^-2
Use the eq: P= densityxgxh
the answer is 1900kg/m^3 ie B
Q.20:
Use the formula F=kx
F=100N, x=2.0mm= 2x10^-3m hence calculate k (K= 5x10^4)
Next, F=1/2 kx^2
Put the values of k and x in the equation, the answer is 0.11 J
q.25:
d=10^3/300= 3.33m
3.33 sin 90= (n/2)x (450x10^-9)
We have divided n/2 since we have to find he no. of maxima).

The answer is 14.8 ie 15 (D).

Thank yooouuu! I really appreciate your help sorry to bother you but can you explain q34 of the same paper aswell?

 

[Aries_95]

Thank yooouuu! I really appreciate your help sorry to bother you but can you explain q34 of the same paper aswell?

It's ok I would love to help you
See for the potential divider, we have the general formula:
v=(r1/R1+R2)x E (in this case, we take it as Vo)
Now,
V=(R1/R1+R2)xVo
therefore:
(VoR1)/(R1+R2)
Ie B

 

[lubna1232]

It's ok I would love to help you
See for the potential divider, we have the general formula:
v=(r1/R1+R2)x E (in this case, we take it as Vo)
Now,
V=(R1/R1+R2)xVo
therefore:
(VoR1)/(R1+R2)
Ie B

you are an amazing teacher thanks again btw you skipped q 8,9,12,13 and 21 :/

 

[Aries_95]

you are an amazing teacher thanks again btw you skipped q 8,9,12,13 and 21 :/

Yeah because I am trying to solve them myself first
And thankyou for the compliment by the way

 

[lubna1232]

Yeah because I am trying to solve them myself first
And thankyou for the compliment by the way

haha good luck! post the answers when you understand them

 

[Aries_95]

haha good luck! post the answers when you understand them

Sure

 

[hassankhan]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf Someone plz help me with these questions Q4,Q6,Q10,Q15,Q22,Q23,Q24,Q25

 

[ahmed abdulla]

Q.4:
Use the equation 2as=v^2-u^2
Where a=accelaeration= 0.2 ms^-2
S=distance=1km=1000,
U=initial velocity= 0 ms^-1
V=?
Putting the vaues in the equation, the answer is 20 ms^-1 ie A.
Q.18:
Pressure= 15200 Pa
Density=?
Height= 80 cm=80x10^-3 m
g=10ms^-2
Use the eq: P= densityxgxh
the answer is 1900kg/m^3 ie B
Q.20:
Use the formula F=kx
F=100N, x=2.0mm= 2x10^-3m hence calculate k (K= 5x10^4)
Next, F=1/2 kx^2
Put the values of k and x in the equation, the answer is 0.11 J
q.25:
d=10^3/300= 3.33m
3.33 sin 90= (n/2)x (450x10^-9)
We have divided n/2 since we have to find he no. of maxima).

The answer is 14.8 ie 15 (D).

what if they asked no. of minima ? + for the strain energy question u get 0.10 not 0.11 ??

my doubts if possible and u can answer

no . 29 >>>http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w07_qp_1.pdf​

no . 17 also > http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w06_qp_1.pdf
thanks​

 

[snowbrood]

when you get recovered help with this too plz ​

no problem
For Q1
Sn1 is two step reaction
sn2 is one step reaction
even in 3rd option u have cl attached to primary carbon so it is Sn2 reaction
since option 2 is incorrect so we are left with 1 being correct
intermediate means there is a two step reaction
for Q2
well for option b
u do know that aldehyde is oxidized and a silver mirror is formed right
silver mirror is Ag atoms
silver has +1 in ion
the change of +1 to 0 is reduction right?
so there is redox

 

[snowbrood]

what if they asked no. of minima ? + for the strain energy question u get 0.10 not 0.11 ??

my doubts if possible and u can answer

no . 29 >>>http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w07_qp_1.pdf​

no . 17 also > http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w06_qp_1.pdf​

thanks​

well for q29
since power is same so
12^2/Rx=6^2/Ry
12^2/6^2=Rx/Ry
4=Rx/Ry

for Q17
well put in formula mgh
80*10*(2.5*6)(as there are 6 stairs)
u get 12000 D is correct

 

[Aries_95]

what if they asked no. of minima ? + for the strain energy question u get 0.10 not 0.11 ??

my doubts if possible and u can answer

no . 29 >>>http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w07_qp_1.pdf​

no . 17 also > http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w06_qp_1.pdf​

thanks​

Q.29
Since power of both resistors is equal, P1=P2
V^2/Rx=V^2/Ry
12^2/Rx=6^2/Ry
12^2/6^2=Rx/Ry
4/1=Rx/Ry
hence the answer is D

 

[Aries_95]

what if they asked no. of minima ? + for the strain energy question u get 0.10 not 0.11 ??

my doubts if possible and u can answer

no . 29 >>>http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w07_qp_1.pdf​

no . 17 also > http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w06_qp_1.pdf​

thanks​

Q.17:
See as far as i understand the question, it states that out of 100% of electrical energy, 93% is emitted as thermal, and 7% is emitted as light. since we have to calculate the efficiency as the percentage of ELECTRICAL ENERGY converted to LIGHT ENERGY,
it will be:
(7/100)x100=7%
Correct me anyone if i am wrong.

 

[ahmed abdulla]

no problem
For Q1
Sn1 is two step reaction
sn2 is one step reaction
even in 3rd option u have cl attached to primary carbon so it is Sn2 reaction
since option 2 is incorrect so we are left with 1 being correct
intermediate means there is a two step reaction
for Q2
well for option b
u do know that aldehyde is oxidized and a silver mirror is formed right
silver mirror is Ag atoms
silver has +1 in ion
the change of +1 to 0 is reduction right?
so there is redox

Thanks <
well, whats the linkage between the graph and intermediate ,,,? how did u now that the graph code for an intermediate rn

 

[snowbrood]

Thanks <
well, whats the linkage between the graph and intermediate ,,,? how did u now that the graph code for an intermediate rn

u see that two maximas exo and endothermic reaction one reaction cant be exothermic and endothermic at the same time is it possible? ZARA sochiyay

 

[IGCSEstudent2012]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s08_qp_1.pdf
ques 6 prettyplease!
snowbrood

 

[KWIKIW]

Intermolecular forces: (These are from my notes. XD)
Van der waals forces: also known as the 'Temporarily dipole-dipole forces'. It is the temporary electrostatic attraction between atoms or molecules due to temporarily dipoles. Dipoles are only formed temporarily because the electrons inside an atom or molecule are free to move around. So any molecules or compounds with electrons are capable of forming Van der Waals forces.

Permanent induced dipole-dipole: It is the electrostatic attraction between polar molecules. Polar molecules are molecules that contains an electronegative atom. Example of polar molecules are HCl, CHCl and e.t.c. The electronegative atom in this case is Cl but of course it does not necessarily has to be Cl. It can be other atom as well such as F,O,N,Br and e.t.c.(You can do some internet search on this). So this permanent induced dipole-dipole can be formed when you have polar molecules.

Q.31 ??

Q.33. You sure you got an N triple bond in your Hydrazine compound? My structure will look like this: H2N=NH2(which total molecular formula is N2H4). Actually you have many ways in answering this question. My method was looking at the N bonds present. The question is asking for N triple bonds but N2H4 does not contains any N triple bonds but only double bonds instead.

Hmmm... I think the examiner report is trying to tell us that the rate of reaction will not be affected by the presence of N triple bonds there. It's true that you can only work out the enthalpy change by using bond energies but not the rate of reaction. We have another method of measuring the rate of reaction but clearly not using this N triple bonds.

I am not sure about this actually~ But this is my understanding so far. If you have a better explanation please kindly reply me. Thank you~

I asked one of my friends about the 33 Qs and my friend also said the same thing that you said about N2H4 having only N=N.
So you're method for 33 was correct. Thanks.

Do you know how to do Qs 33 in 01/M/J/2003?
In Qs 33 why is the third statement, "the enthalpy change of combustion of diamond is greater than that of graphite" correct?
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s03_qp_1.pdf

And do you know how to do Qs 17 in 01/O/N/2005?
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w05_qp_1.pdf

 

[HubbaBubba]

Can anybody explain this to me? It's q1 in May June 2004..

1 Which of these samples of gas contains the same number of atoms as 1g of hydrogen (Mr: H2 = 2)?
A) 22g of carbon dioxide (Mr: CO2, 44)
B) 8g of methane (Mr: CH4, 16)
C) 20g of neon (Mr: Ne, 20)
D) 8g of ozone (Mr: O3, 48)

Aren't A and B the same answer?