Chemistry and Physics AS paper 12 MCQS

salvatore

salvatore

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IGCSEstudent2012 said:
salvatore
Click to expand...
(16) You know that the weight always acts at the centre.. so lets assume the distance XY to be 1m.
The horizontal component of force F is Fsin30.

Sum of clockwise moments = sum of anti-clockwise moments
0.5 x 10 = 1 x Fsin30
F = 10N

(17) The tension (4N) is the same on both sides; XY and XZ.
First, we have to find the angle YXP
Sin theta = 40/50
theta = 53

Next, find the horizontal component of the tension:
= 4cos53
= 2.4N

The same horizontal component will be experienced on XZ

Hence, 2.4 X 2

=> 4.8 N
 
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1597.2217

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@IGCSEstudent2012
Sorry my bad. I tried but I couldn't solve them and sent them in Email to someone who will hopefully reply with the answers soon.

stareaj said:
Somebody please helppppppppp these are my problems don't have any link pls
Physic As oct 2010 / 12 mcq 22 and 34
May 2011/12 mcq 15,16,24
May 2011/ mcq 14,15
Oct2011/13 mcq 17,25
Oct 2011/12 mcq 26
May 2012/12 mcq 5
Click to expand...


12/ON10-
Q22-

Low K means high extension (F=Kx).
So, first we need to have high sensitivity at low mass (means lower K spring to be extended first). This isn't possible in figure (B) and figure (D) because of the way they are arranged. In figure (C) the lower K spring is outside the box, which means it will get extended later, and the higher K will be extended first. This means low sensitivity at small masses. This is totally opposite of what we need. In figure (A), the low K spring would stretch first giving a sensitive result, and the higher K later. Hence (A) is the answer.

Q34-

R=ρL/A
L and A is constant for all three wires. So, the ratio of R is ratio of ρ of the wires. Which is 1:2:3.
We know that V=IR. With higher R, the V will be higher.
The line cannot be smooth, so rule (A) out. The line in (D) is ridiculous so rule that out too. Fig (C) shows that the voltage increases after every joint, which is not true. The voltage lost in the wire will increase in the wire at every joint, so output voltage will decrease after every joint. Answer is (B).

12/MJ11-

Q15-

For a couple, the two forces must be of same magnitude and their direction should be opposite at 180 degree. (A) and (D) get cancelled. (C) however, has the angles right but there isn't any perpendicular distance between the forces hence it is stationary. Answer is (B).

Q16-

(As replied to the person above, I will respond as soon as I get the answer in Email)

Q24-

A- Using more springs in layers mean each spring will get same force and will compress more.
B- More springs per area means force is divided and the compression is less. [This is hence the answer because we need more compression.]
C- [F=Kx.] Lower K will give higher x.
D-[Y=Fx/AL]. Lower Y will give higher x.

13/MJ11-

Q15-

I've sent this question to my teacher to answer, which I will post here later.

Q16-

Easy.
Ek=0.5 x m x v^2
We have Ek as 1500 and v as 10. We can hence find mass (30). With the mass and new speed, we can find the Ek.
Ek = 0.5 x 30 x 40^2
Ek = 24,000 (C).

13/ON11-

Q17-

Ek before collision = 0.5 x m x v^2 + 0.5 x m x 0 [0.5 x m x v^2]
Ek after collision = 0.5 x 2m x v^2 (assuming they have the same speed).
Now subtract both and you will get answer as (C).

Q25-

No idea, sorry.

12/ON11-

Q26-

The phase difference at 18s is 1/2 of a cycle. We can use this and ratio:
0.5 --- 18
0.125 --- x
x = 18 x 0.125 / 0.5

Answer is (B) 4.5s.

12/MJ12-

Q5-

Volume of cylinder = Pie x r^2 x l
So we need to find radius and length.
Answer is (B).


Good luck. Remember me in your prayers.
 
IGCSEstudent2012

IGCSEstudent2012

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salvatore said:
(16) You know that the weight always acts at the centre.. so lets assume the distance XY to be 1m.
The horizontal component of force F is Fsin30.

Sum of clockwise moments = sum of anti-clockwise moments
0.5 x 10 = 1 x Fsin30
F = 10N

(17) The tension (4N) is the same on both sides; XY and XZ.
First, we have to find the angle YXP
Sin theta = 40/50
theta = 53

Next, find the horizontal component of the tension:
= 4cos53
= 2.4N

The same horizontal component will be experienced on XZ

Hence, 2.4 X 2

=> 4.8 N
Click to expand...

thanks a lot!
 
K

kingos3111

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2.76 g of ethanol were mixed with an excess of aqueous acidified potassium dichromate(VI). The
reaction mixture was then boiled under reflux for one hour. The organic product was then
collected by distillation.
The yield of product was 75.0 %.
What mass of product was collected?
 
cute97

cute97

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1597.2217 said:
Because 2 Mole of Salt gives 4 Mole of NO2 and 1 mole of O2. We need to calculate for 1 mole of salt, so all the moles will get divided by two. Hence we took the Mr for 2 moles of NO2 and 0.5 mole of O2. Hope you get it :)
Click to expand...

yhh thank you:D
 
IGCSEstudent2012

IGCSEstudent2012

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cute97 said:
PLEASE HELP ME :cry:
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s12_qp_11.pdf
Q21, Q22, Q23
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s12_qp_12.pdf
Q16, Q20, Q28
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w12_qp_11.pdf
Q23, Q17
THANKS IN ADVANCE :)
Click to expand...


MJ 2012 P11
Q21 See Na react with OH of both acid n alcohol
But Naoh reacts with acid only
When these react with one OH it get a charge of -1
...initially Na reacts with 3 Oh to -3.... That OH vould be of anyone of alcohol or acid.... Then Naoh reacts with one OH to give -1....
Now we know that Naoh only reacts with acid so there is one acid group..... N other r two acohols groups...so ans is c

Q22 We get ethanol n sodium ethanoate on hydrolysis
Get the Mr of both
Add them
This is ur total product mass
Now take % for each

Q23 See the product is an ester
And it is forming a ring
That means k it is joining with itself
To do that it must contain both alcohol n acid
So B is the one....
 
IGCSEstudent2012

IGCSEstudent2012

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Propanone has the molecular formula C3H6O.
The enthalpy change of combustion of hydrogen is –286kJmol–1
.
The enthalpy change of combustion of carbon is –394kJmol–1
.
The enthalpy change of combustion of propanone is –1786kJmol–1
Using this information, what is the enthalpy change of formation of propanone?
I SUCK IN INTHALPIES!!!:confused:
ANYONE PLEASEEEEEE HELP!
 
K

kingos3111

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On collision, airbags in cars inflate rapidly due to the production of nitrogen.
The nitrogen is formed according to the following equations.
2NaN3 → 2Na + 3N2
10Na + 2KNO3 → K2O + 5Na2O + N2
How many moles of nitrogen gas are produced from 1 mol of sodium azide, NaN3?
 
cute97

cute97

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IGCSEstudent2012 said:
MJ 2012 P11
Q21 See Na react with OH of both acid n alcohol
But Naoh reacts with acid only
When these react with one OH it get a charge of -1
...initially Na reacts with 3 Oh to -3.... That OH vould be of anyone of alcohol or acid.... Then Naoh reacts with one OH to give -1....
Now we know that Naoh only reacts with acid so there is one acid group..... N other r two acohols groups...so ans is c

Q22 We get ethanol n sodium ethanoate on hydrolysis
Get the Mr of both
Add them
This is ur total product mass
Now take % for each

Q23 See the product is an ester
And it is forming a ring
That means k it is joining with itself
To do that it must contain both alcohol n acid
So B is the one....
Click to expand...

Thank you :D and good luck for tom;)
 
IGCSEstudent2012

IGCSEstudent2012

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kingos3111 said:
On collision, airbags in cars inflate rapidly due to the production of nitrogen.
The nitrogen is formed according to the following equations.
2NaN3 → 2Na + 3N2
10Na + 2KNO3 → K2O + 5Na2O + N2
How many moles of nitrogen gas are produced from 1 mol of sodium azide, NaN3?
Click to expand...


find moles of sodium azide=m/mr
then look at the mole ratio....its 2:3
hence find moles by ratio method.
 
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