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Chemistry and Physics AS paper 12 MCQS

1

1597.2217

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cute97 said:
HELP PLS
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w06_qp_1.pdf
Q11 and Q14
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w07_qp_1.pdf
Q9
Thanks is advance:D
Click to expand...

Couldn't do Q11.

Q14:

2 X(NO3)2 → 2 XO + 4 NO2 + O2

The gases formed are NO2 and O2. 3.29g are formed. So moles = 3.29 (2x (14+16+16) + 16) = 0.03046 Moles [2 Moles of nitrate gives 4 moles of NO2 and 1 mole of O2. 1 Mole of Nitrate would give 2 mole of No2 and half mole of O2.]

We know that n=m/Mr
0.03046 = 5.00/Mr
Mr = 164.13

Now we open the periodic table and find the Mr of the given nitrates. To make this easier, we can subtract the Mr of (NO3)2 from 164.13. 164.13-124 = 40.1
Calcium is 40.1 so answer is (B).


W07-
Q8:

Can't do this.
 
ahmed abdulla

ahmed abdulla

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1597.2217 said:
Couldn't do Q11.

Q14:

2 X(NO3)2 → 2 XO + 4 NO2 + O2

The gases formed are NO2 and O2. 3.29g are formed. So moles = 3.29 (2x (14+16+16) + 16) = 0.03046 Moles [2 Moles of nitrate gives 4 moles of NO2 and 1 mole of O2. 1 Mole of Nitrate would give 2 mole of No2 and half mole of O2.]

We know that n=m/Mr
0.03046 = 5.00/Mr
Mr = 164.13

Now we open the periodic table and find the Mr of the given nitrates. To make this easier, we can subtract the Mr of (NO3)2 from 164.13. 164.13-124 = 40.1
Calcium is 40.1 so answer is (B).


W07-
Q8:

Can't do this.
Click to expand...

A student calculated the standard enthalpy change of formation of ethane, C2H6, using a method
based on standard enthalpy changes of combustion.
He used correct values for the standard enthalpy change of combustion of ethane
(–1560 kJ mol–1) and hydrogen (–286 kJ mol–1) but he used an incorrect value for the standard
enthalpy change of combustion of carbon. He then performed his calculation correctly. His final
answer was –158 kJ mol–1.
What did he use for the standard enthalpy change of combustion of carbon?
A –1432 kJ mol–1
B –860 kJ mol–1
C –430 kJ mol–1
D –272 kJ mol–1
 
1

1597.2217

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IGCSEstudent2012 said:
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_s09_qp_1.pdf
questions...11//13//14/26//31//33
iknow all these ques are simple but pls anyone ..help me:/
1597.2217
Aries_95
Click to expand...


Q11
Q and S are same.
However R is greater than P. As depth increases, the pressure increases. R exerts a greater force on the ball.

Q13
Couldn't do it.

Q14
F=kv^2
P=Fv [Power = Force x velocity.]
P/v = F [Now take this and replace F in the first equation]
P/v = kv^2 [Now make k the subject.]
P/v^3 = k

Answer is (C).

Q26
Speed = Wavelength x Frequency
Wavelength = 2/3 x 2.1 [Because there are 3/2 waves in 2.1 meters.]
Speed = 1.4 x 80
= 112 m/s which is (B).

Q31
Energy = V x I x t
Charge = I x t
We have energy and V. Energy/V = I x t which is charge.
So 72000/12 = 6000 C.

Q33
The direction of battery connected is opposite. So instead of adding the total voltage, we substrate greater by smaller. The voltage is 2V. We know that in parallel circuit the current divided but the voltage remains same, so I=V/R. 2/9 = o.22 which is close to (B) as 0.2. Answer is (B) of 0.2
There are other methods too, but I suggest you to only stick to one method.


Good luck!
 
IGCSEstudent2012

IGCSEstudent2012

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1597.2217 said:
Q11
Q and S are same.
However R is greater than P. As depth increases, the pressure increases. R exerts a greater force on the ball.

Q13
Couldn't do it.

Q14
F=kv^2
P=Fv [Power = Force x velocity.]
P/v = F [Now take this and replace F in the first equation]
P/v = kv^2 [Now make k the subject.]
P/v^3 = k

Answer is (C).

Q26
Speed = Wavelength x Frequency
Wavelength = 2/3 x 2.1 [Because there are 3/2 waves in 2.1 meters.]
Speed = 1.4 x 80
= 112 m/s which is (B).

Q31
Energy = V x I x t
Charge = I x t
We have energy and V. Energy/V = I x t which is charge.
So 72000/12 = 6000 C.

Q33
The direction of battery connected is opposite. So instead of adding the total voltage, we substrate greater by smaller. The voltage is 2V. We know that in parallel circuit the current divided but the voltage remains same, so I=V/R. 2/9 = o.22 which is close to (B) as 0.2. Answer is (B) of 0.2
There are other methods too, but I suggest you to only stick to one method.


Good luck!
Click to expand...

shukran! good luck to u too if ur appearing for any exam:)
 
C

carilove

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1597.2217 said:
Q1:

The volume of Oxygen used is Initial-final = 70-20 = 50 Cm^3

This means 10cm^3 of X will react with 50cm^3 of O2 to give 30cm^3 of CO2 and H2O
With this we can make a equation giving the moles:

X + 5 O2 ---> 3 CO2 + y H2O [We can find y by 5x2 O2 - 3x2 O2 giving us 4 O which are in the H2O. So y=4)
X + 5 O2 ---> 3 CO2 + 4 H2O [Now calculate the number of C and H]

This gives us 3 C and 8 H. So answer is C3H8.


Q2:

From equation 1 we see that 2 moles of Sodium Azide gives us 3 moles of Nitrogen gas. So 1 mole gives 1.5 moles of Nitrogen gas.
From equation 2 we see that 10 moles of Sodium give 1 mole of Nitrogen. However, 2 Mole of Sodium Azide gives 2 Mole of Sodium. So 1 Mole of Sodium Azide will give 1 mole of Sodium. You divide the moles in second equation by 10 to get this:

Na + 0.2KNO3→ 0.1K2O + 0.5Na2O + 0.1N2

So your answer is 1.5 moles from first equation plus 0.1 moles from second equation to give 1.6 moles of Nitrogen gas, which is (B).

Hope you get it!
Click to expand...


thanks soo much this helped
 
Ahmedraza73

Ahmedraza73

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M

mariamh

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1597.2217 said:
I couldn't do Q16. Sorry.

As for Q28:
d= 0.08
L= 0.03

dSinx=nL
Sinx = nL/d
Sinx = 2x0.03/0.08
x= 46.6

However, we do not have to find the angle between incident ray and second order diffraction. We have to find between first order diffraction and second order diffraction. So, 46.6-22 gives 26.6 degree which is (B).
Click to expand...

Thanks :)
 
IGCSEstudent2012

IGCSEstudent2012

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Ahmedraza73 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_11.pdf
help with Q:10,27

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_12.pdf
help with Q:15,22,29


http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_12.pdf
help with Q:12,21,26,34


http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_1.pdf
help me with Q:13

Please anyone can help me?:(
Click to expand...


for s09 ques 13..
see the moment produced by the weight will be
900x0.2=180Nm
the torque by F will be Fx1.2=180
hence F will be 150N
 
Aries_95

Aries_95

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Ahmedraza73 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_11.pdf
help with Q:10,27

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_12.pdf
help with Q:15,22,29


http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_12.pdf
help with Q:12,21,26,34


http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_1.pdf
help me with Q:13

Please anyone can help me?:(
Click to expand...

For May/June 2010:
Q.10:
Using the law of conservation of momentum:
Before collision= after collision.
m(60)+m(-40)=(m+m)v
20m=2mv
v=10 cm/s^-1
Hence the answer is A.

Nov 10:
Q.29:
F=qV
mg=q(V/d)
q=mgd/V
Hence the answer is A.
 
L

LeoMessi

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Please Solve paper 11 22, 34
Paper 12 question 5, 8, 14
Please please, i am struggling terribly :cry::(
 

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M

mehranshaikh

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LeoMessi said:
Please Solve paper 11 22, 34
Paper 12 question 5, 8, 14
Please please, i am struggling terribly :cry::(
Click to expand...

Paper 12.

Q5:

x= 1.28-0.83 = 0.44
Error in x = 0.02 + 0.01 = 0.03
Uncertainty= 0.03/0.44 * 100 = 6.88
Answer D.

Q14:

Length of Ramp= 1.5/sin30 = 3m
Work Done against Friction = 150*3 = 450 J
Work done against Gravity = 200*1.5 = 300

total= 450+300 = 750 J
 
cute97

cute97

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1597.2217 said:
Couldn't do Q11.

Q14:

2 X(NO3)2 → 2 XO + 4 NO2 + O2

The gases formed are NO2 and O2. 3.29g are formed. So moles = 3.29 (2x (14+16+16) + 16) = 0.03046 Moles [2 Moles of nitrate gives 4 moles of NO2 and 1 mole of O2. 1 Mole of Nitrate would give 2 mole of No2 and half mole of O2.]

We know that n=m/Mr
0.03046 = 5.00/Mr
Mr = 164.13

Now we open the periodic table and find the Mr of the given nitrates. To make this easier, we can subtract the Mr of (NO3)2 from 164.13. 164.13-124 = 40.1
Calcium is 40.1 so answer is (B).


W07-
Q8:

Can't do this.
Click to expand...

Thank you :D and thnkss for your tym but i didn't understand ""moles = 3.29 (2x (14+16+16) + 16) = 0.03046 Moles"" what is this Mr isn't it total Mr of the two gases No2 and O2 which is supposed to be ( 14+16+16)+(16+16)=78 ??? pls answer and thnkss again
 
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