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Chemistry and Physics AS paper 12 MCQS

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53
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48
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28

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.
 
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415
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53
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.

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
 
Messages
53
Reaction score
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Points
28


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!
 
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86
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28
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!

shukran! good luck to u too if ur appearing for any exam:)
 
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3
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11
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!


thanks soo much this helped
 
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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).

Thanks :)
 
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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.
 
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17
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9
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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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Please Solve paper 11 22, 34
Paper 12 question 5, 8, 14
Please please, i am struggling terribly :cry::(

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
 
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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.

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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