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Chemistry and Physics AS paper 12 MCQS
- Thread starter MariamHASAN
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Q.30 Is about how we can observe interference pattern. According to my understanding, we can only see interference clearly on the screen which in this case is line XY. We cannot see any interference along line RS. Imagine using a monochromatic light(Laser), you can only observe the interference pattern on the screen but not in the distance between the screen and the slits.(RS)
Please do correct me if I've made any mistakes TQ
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Qn 12-it says the ice rink is frictionless so there shouldnt be any resistive forces and therefore momentum remains constant (no change in momentum). after it collides with the wall, it rebounds in the opposite direction so Velocity is negative. In addition to that since the collision is inelastic, energy(kinetic) is dissipated or lost and therefore Velocity decreases resulting is a smaller momentum. Negative velocity does not mean there is less velocity, just a CHANGE IN DIRECTION.
Qn 20
-It says that the mass moves from P to Q and it losses Potential Energy so according to our knowledge of gravity, the direction should be --> (moving closer towards = lost in PE and vice versa). The magnitude of acceleration can be found with 2 equations. E(lost)=MGH which is E=MGX and F=MA. Modify the equation F=MA into A=F/M. add displacement to both sides so you get AX=FX/M. sub E=FX and you get AX=E/M and move X over to get A=E/MX. This equation is correct as if you sub MGX=E you will get A=G which is viable.
Qn 24
-As you know strain energy is area under graph of force against extension therefore the energy that the rubber band will have will be area X+Y when it is stretched to e. So in order for the rubber to have an extension e, an equivalent work has to be done so minimum energy required will be the area under the graph. Do not get confused with the graph of Stress against Strain that is different
Qn 27
-Fact. Memorise the wavelenghts of the wave spectra RMIVUXG . Lengths(1 x 10 to the power of) 1,-1,-5,-6,-7,-9,-10 p.s these are only an approximate value as there are a range for each em wave.
Qn 29
-In order for stationary waves to be formed, there must be two coherent waves of same frequency and wavelength moving in opposite directions that meet and interfere causing a resultant wave. So P is possible as the wave is reflected at the end of the tube. Q is a little bit special. It is an effect of the boundary conditions imposed at the end of the tube(the fact that pressure at the end of a tube must be equal to atmospheric pressure at the other end forming a boundary and therefore enabling the wave to reflect)
haha these are my answers and may not necessarily be 100% correct
Thanks a lot!
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12. Horizontal, frictionless surface - constant p. Therefore B & C can be ruled out. Rebounds along its original path - the final p has to be negative.
20. We know that GPE increases when an object's height increases, on the Earth's surface. The Earth's gravitational field lines are downwards. We can conclude that PE decreases as an object moves in the direction of the field lines of the field it is in. The mass's PE decreases for P to Q, therefore the direction of the field lines of this particular field is to the right. Now, we know that acceleration due to gravity on Earth = g. We use g to calculate GPE. GPE=mgh, therefore g=GPE/(mh). That's the same as E/mx.
24. A is wrong; the energy which 'heats the band' is the same as the energy required to stretch it to e. That energy is not area X but area X+Y.
B is correct. C is wrong; area X+Y is the EPE stored in the band when it is stretched to E. D is wrong; the net WD on the band is X. WD in stretching=Y+X & energy lost as heat=Y, so net WD=Y+X-Y=X.
26. Delta l is compression & l is the original length. If compression increases, the new original length decreases. That's inverse proportionality. We can put it mathematically in this way: strain = e/L. e=extension, so 1/e=compression. Now, strain =1/e/L=1/eL. You can see from this new equation that e & L are inversely proportional.
27. Lambda decreases in the order RMILUXG. Tip: you don't have to memorise all the wavelengths, only the visible light wavelengths. All you need to know is that they're always to the order 10^-7. Therefore we can rule out C & D. U is right under L, so U can be 10^-8 instead of 10^-10. So the answer's A.
28. The centre of the fringe patterns = constructive interference, i.e. the waves are exactly in phase, so the resultant amplitude is 2A. When one slit is covered, only one wave reaches the centre of the screen. Therefore the amplitude then becomes 1A.
29. A stationary wave in a closed end tube always has an antinode at the open end, so the fundamental mode will be 1/4th of a lambda. A stationary wave in an open end tube too always has an antinode, this time at both ends, so the fundamental mode will be 1/2 a lambda. For tube P: If 1 lambda = 20cm, 1/4 lambda = x. x = 5. That's a multiple of 35, so stationary waves can be formed in P. For tube Q: If 1 lambda = 20cm, 1/2 lambda = x. x = 10. That's a multiple of 50, so stationary waves can be formed in Q.
30. The key to answering this question is reading the question properly. They say that the line RS is halfway between PQ. That means waves from P & Q reaching any point on RS will be exactly out of phase, i.e. the phase difference will be 1/2 lambda. There will always be destructive interference on the line RS.
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yeah thanks alot.
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no problem
Glad to be of help .. this is like the only topic in Physics that I somewhat understand http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_12.pdf
Q. 12. (A) Acid will dissociate into H+ when dissolved in the pool water. The H+ ions will then react with OCl- to form more HOCl.
(B) Adding solution of chloride ions will not have any effect because Cl- ions does not play any role in the reaction between OCl- and H2O to produce HOCl.
(C) Adding OH- will cause the equilibrium position to be shifted to the left(Based on Le Chatelier's principle that a reaction will try to minimise a change in chemical reaction by producing an effect that opposes it) so more reactants will be formed(OCl- and H2O).
(D)Bubble air through the water will not have any effect on the concentration of HOCl. All the reactants and products are in aqueous and liquid so obviously we are not dealing with any gaseous compounds or elements.
Suggested answer will be A.
Q.25. Free radical substitution involved the remove of one H from the hydrocarbon compound and form HCl and also the hydrocarbon radical. So if you try to draw the structures you will get these: CH2.CH2C(CH3)3 , CH3CH.C(CH3)3, CH3CH2C.(CH3)3 (Hopefully you can understand my structural formula
Q. 26. Unaffected by changes in concentration of OH- means that the rate of reaction will not be altered due to changes in the concentration of OH-. You also know that there are 2 nucleophilic substitution reaction for halogenoalkane (Sn1 and Sn2). Sn1 is called Sn"1" because only one factor can alter the rate of reaction which is the concentration of the halide ions present. The reason why Sn2 is labelled "2" because there are two factors that can alter the rate of reaction which is the concentration of halide ions present and also the concentration of OH-. So here we are looking for halogenoalkane which will undergo Sn1 reaction. (Note: Tertiary halogenoalkane undergo Sn1 RXN, Primary halogenoalkane undergo Sn2 RXN and secondary halogenoalkane will undergo either Sn1 or Sn2 RXN) Here we do not consider secondary halogenoalkane even though it undergoes Sn1 because there is also a probability that Sn2 may occur so we do not want that. Attempt to draw the compounds and the answer will be D.
Thanks a lot! I got it u are really good at explaining these. If u could solve the rest of the problems i have mentioned it would be a great help
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https://www.xtremepapers.com/community/attachments/capture-9-png.27610/no problem
Can you answer it please?
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Don't mention it. It's my pleasure.
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m/j/2012 p12 q.26: this question requires you to draw different isomers that contains pi bonds.(Double bonds) so I will show you my structural formula. CH2=CHCH2CH2CH3, CH2=CHCH(CH3)CH3, CH2=C(CH3)CH2CH3, don't forget the cis-trans isomers too ~ H(CH3)C=CH(C2H5), H(CH3)C=CH(C2H5) and lastly (CH3)(CH3)C=CH(CH3) (This is not a cis-trans because both CH3 are on the same side). so total isomers are 6 and the answer will be D.
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http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_s11_qp_11.pdf question 2
To answer this question you will need to apply your oxidation number knowledge. firstly, you find the change in oxidation state of N when it changes from NO3- to NO2- (Ox.no of N in NO3- =5 and Ox.no of N in NO2- =3 so change in Ox.no is 5-3=2). Do the same thing for the other as well. From NO2- to NO(Ox.no of N in NO2-=3 and Ox.no of N in NO=2 so change in Ox.no= 3-2=1). From NO to N2O(Ox.no of N in NO=2 and Ox.no of N in N2O=1 so change in Ox.no=2-1=1). Lastly, From N2O to N2 (Ox.no of N in N2O=1 and Ox.no of N in N2=0 and change in Ox.no will be 1). phew.. Quite lengthy ehh
so the unique one here is change in Ox.no of 2 so the answer will be A. I hope my explanation can clear your doubts and good luck
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mcq 23 nov 08 ... chem?
I get itm/j/2012 p12 q.26: this question requires you to draw different isomers that contains pi bonds.(Double bonds) so I will show you my structural formula. CH2=CHCH2CH2CH3, CH2=CHCH(CH3)CH3, CH2=C(CH3)CH2CH3, don't forget the cis-trans isomers too ~ H(CH3)C=CH(C2H5), H(CH3)C=CH(C2H5) and lastly (CH3)(CH3)C=CH(CH3) (This is not a cis-trans because both CH3 are on the same side). so total isomers are 6 and the answer will be D.
Thanks !http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_s11_qp_11.pdfquestion 2
To answer this question you will need to apply your oxidation number knowledge. firstly, you find the change in oxidation state of N when it changes from NO3- to NO2- (Ox.no of N in NO3- =5 and Ox.no of N in NO2- =3 so change in Ox.no is 5-3=2). Do the same thing for the other as well. From NO2- to NO(Ox.no of N in NO2-=3 and Ox.no of N in NO=2 so change in Ox.no= 3-2=1). From NO to N2O(Ox.no of N in NO=2 and Ox.no of N in N2O=1 so change in Ox.no=2-1=1). Lastly, From N2O to N2 (Ox.no of N in N2O=1 and Ox.no of N in N2=0 and change in Ox.no will be 1). phew.. Quite lengthy ehh
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23 D.
It's a fact. That the reason why the compounds HX or any other halide decomposes easily down the group is because of the low Bond Energy between them. That's why putting a Rod in HI decomposes it.. as it requires less Activation energy to break the bond.
This is the reason why you never see HF or flourides decompose because their Bond energy is very strong so they don't decompose.
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http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s12_qp_12.pdf
Q.6. It's all calculation. We are given the value of V=7000cm^3, Temperature=30 degree C and also the mass of O2 gas. First thing you have to do is to change the unit of volume from cm^3 to m^3 so in this case: 7000cm^3/(100 x 100 x 100) = 7x10^-3 m^3 . Next you will need to find the no.of moles of O2 gas by using the mass given so: no.of moles=0.96/2(16) = 0.03moles. Last step is easy
you will only substitute these values to the formula given: Pv=nRT. Since the question is asking for the pressure, so I will make P the subject and the formula will look like this now: P=nRT/v and perform the final calculation using the values: n=0.03moles, R=8.31(Given in data booklet), T=30+273 K, v=7x10^-3 m^3. After you obtain the final answer, divide the final answer by 1000(since the pressure is in Kpa) and round it off to 3s.f. I got 10.8Kpa as my final answer which is C. Q.7. The key words here are at e.q.m, X moles of R were present and the total moles was (2+x/2). So we will perform the calculation step by step. ("<=> here indicates the equilibrium sign cuz I don't know how to do one
)(a) 1P <=> 2Q + 1R
Initial no.of moles of P=1 (Look at the equation, the initial mole is 1 because of the "1"P)
Initial no.of moles of Q and R=0 moles(No product will be formed at the beginning)
At e.q.m,
No. of moles of R= X(given in the question)
No.of moles of Q = (2 x X) = 2X (I multiplied by 2 because of the "2"Q which means the no.of moles of Q is twice the no.of moles of R)
No.of moles of P= initial moles - moles of the product
= 1 - X
total no.of moles at e.q.m= no. of moles of R + no.of moles of Q + no.. of moles of P
= X + 2X + 1 - X
=2X + 1 moles
(b)2P <=> 2Q + R
initial moles of P=2(because of the "2"P in the equation)
initial moles of Q and R = 0 moles
at e.q.m,
no.of moles of R = X(given in the question)
no. of moles of Q = 2X (multiplied by 2 because of the "2"Q and this shows that the no.of moles of Q is twice the no.of moles of R)
no.of moles of P = 2 - (2 x X) (this time I minus by 2X because of the "2"P there.)
total no. of moles at e.q.m = X + 2X + 2 - 2X
=2 + X moles
(c)2P <=> Q + R
initial moles of P = 2 moles
initial moles of Q and R = 0 moles
at e.q.m,
no.of moles of R = X(given in the question)
no. of moles of Q = X(it's X because Q and R have the same mole ratio 1:1)
no. of moles of P = 2 - (2 xX) (Need to minus by 2X because of the "2"P)
total no. of moles= X + X + 2- 2X
=2
(d)2P <=>Q + 2R
initial moles of P=2 moles
initial moles of Q and R = 0 moles
at e.q.m,
no .of moles of R = X (this time I don't multiply by 2 even tho there is "2"R because the question stated X moles of R will be formed so here we assumed 2R is equivalent to X moles)
no. of moles of Q= X/2 (the mol ratio of Q:R is 1:2 and which means the no.of moles of R is twice the no.of moles of Q so you need to divide the no.of moles of R(X) by 2)
no.of moles of P = 2 - X(I will only minus 2 by X instead of 2X. Notice that P and R have the same mol ratio and since the no.of moles of R is X moles, so you need to minus the initial moles of P by X only)
total no.of moles = X + (X/2) + 2-X
= 2 + (X/2)
And you compared all the results here the obvious answer here would be D~
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I have decided to answer your question separately so that it won't be that lengthy
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s12_qp_12.pdf
Q.16. The question is asking what are the reagent and conditions that is needed to change an ELEMENTAL Chlorine(From Cl2) to a compound containing CHLORINE WITH AN OXIDATION STATE OF +5.
(a) will not be an answer because only Cl- ion will react with AgNO3 to gives white PPT but not elemental Chlorine(Cl2)
(b) Same explanation as above. Only Cl- ion will react with H2SO4 to gives white fumes of HCl but not Cl2.
(c)Cold dilute is wrong. This is one of the disproportional reaction of Chlorine. The equation of the reaction will be: Cl2 + 2NaOH ---> NaCl + NaClO + H2O and you will need to find the Ox.no of Cl in both NaCl and NaClO. My results: Ox. no of Cl in NaCl is -1 and in NaClO is +1. So here obviously there is no Chlorine with the oxidation state of +5 so this is why it's wrong.
(d)Another disproportional reaction. Using Hot NaOH, the equation of the reaction will be: 3Cl2 + 6NaOH ---> 5NaCl + NaClO3 + 3H2O. Find the oxidation state of Cl in both NaCl and NaClO3 and my results: Ox.no of Cl in NaCl is -1 and in NaClO3 is +5. So here we have a Chlorine with a +5 oxidation state so this is the correct answer.
Q.24. This question is very hard~
I got it wrong at the first place but finally I can come up with some ideas but I am not sure whether it's 100% correct or not. This question is related to cracking of hydrocarbons. We need to crack oct-1-ene so first we need to count the total number of carbons and hydrogens inside oct-1-ene (C8H16). Then the question is asking which combination of compound will be formed? My working:When W is produced, (Molecular formula for W=C2H4)
C8H16 - C2H4 = C6H12
now in C6H12, we can form 2 molecules of X (CH3CH=CH2). You can check by simply adding the number of carbons and hydrogens of 2 molecules of X to chek whether it's C6H12 or not.
When X is produced (Molecular formula for X = C3H6)
C8H16 - C3H6 = C5H10
From C5H10, we can form both W and X.(add the carbons and hydrogens of both W and X to check whether it's C5H10 or not)
When Y is produced (Molecular formula for Y=C3H6)
C8H16 - C3H6 = C5H8
From C5H8, you can get:
1/3 X and 1 Z and also 1/2W and 1 Z
when Z is produced (Molecular formula of Z= C4H6)
C8H16 - C4H6 = C4H10
From C4H10, you can get:
1Y and 1/2W and also 1Y and 1/2X
So from all the list here we can conclude that all W,X,Y,Z can be formed during cracking. So this is why the answer is A. (I am not sure about this one :s)
Q.27. This question is very fun~ It's my one of my favorite
.(a) Cannot be Ethane because Ethane is a gas at r.t.p.
(b) Carboxylic acid can fully dissolved in water but will only dissociate partially so no.
(c) Ethanol will mix completely with water because it will form hydrogen bond with the H2O molecules.
(d) Ester cannot mix completely with water even tho there are lone pairs at the oxygen atom. (Try to draw the structure for this ester and you will be able to deduce the position of the lone pairs). This is because Ester does not completely fulfill the requirements to form H-bond. (Recap requirements to form H-bond: 1.) one hydrogen atom covalently bonded to F,O or N. 2.) F,O or N having lone pairs.) so ester only fulfill one of the requirements. This is why it does not mix well with water.
Q.33. The question is asking what is true about sulfuric acid. Hint: Sulfuric acid is a strong acid. Strong acid are acid that dissociate completely when dissolved in water.
1. you compare the [H+] of H2SO4 with the [H+] of HSO4-, H2SO4 will dissociate more than HSO4- since H2SO4 is a strong acid where as HSO4- is a weak ones so therefore the [H+] of H2SO4 would be higher. so 1 is correct.
2. Since H2SO4 dissociate high amount of H+, so higher amount of H+ will cause the equilibrium position of the second equation to be shifted to the left to form more HSO42- so obviously [SO42-] will not increase so it shouldn't be high.
3. When you have a higher concentration of H+, you will get higher [HSO42-] which is obviously higher than the [SO42-]
That should be it and the answer will be D.
Q.37. Which compound will be formed when propene undergo single reaction?
(1) CH2OHCHOHCH3 (this is a mild oxidation reaction where you form diol using reagent: Cold and acidified potassium dichromate. It's a one step reaction because once you add the reagent, you will get the product without using any additional reagent.)
(2)( CH2CH(CH3) ) n
This is additional polymerisation process. Where all the propene monomers link together to form a polymer. This reaction is also a one step reaction.
(3)CH2BrCH2CH2Br
At first, Propene undergo electrophiliic addition reaction with HBr to form CH3CH2Br. Next, this compound will free radical substitution reaction with the presence of UV-light to form CH2BrCH2CH2Br. So this is a two steps reaction.
So final answer is B.
Yey~~ It's done~ But please do double check with your teacher because I am not 100% sure that my explanations are correct. If you have a better explanation would you please kindly reply? Thank you
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I have decided to answer your question separately so that it won't be that lengthy
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_12.pdf
Q.16. The question is asking what are the reagent and conditions that is needed to change an ELEMENTAL Chlorine(From Cl2) to a compound containing CHLORINE WITH AN OXIDATION STATE OF +5.
(a) will not be an answer because only Cl- ion will react with AgNO3 to gives white PPT but not elemental Chlorine(Cl2)
(b) Same explanation as above. Only Cl- ion will react with H2SO4 to gives white fumes of HCl but not Cl2.
(c)Cold dilute is wrong. This is one of the disproportional reaction of Chlorine. The equation of the reaction will be: Cl2 + 2NaOH ---> NaCl + NaClO + H2O and you will need to find the Ox.no of Cl in both NaCl and NaClO. My results: Ox. no of Cl in NaCl is -1 and in NaClO is +1. So here obviously there is no Chlorine with the oxidation state of +5 so this is why it's wrong.
(d)Another disproportional reaction. Using Hot NaOH, the equation of the reaction will be: 3Cl2 + 6NaOH ---> 5NaCl + NaClO3 + 3H2O. Find the oxidation state of Cl in both NaCl and NaClO3 and my results: Ox.no of Cl in NaCl is -1 and in NaClO3 is +5. So here we have a Chlorine with a +5 oxidation state so this is the correct answer.
Q.24. This question is very hard~
When W is produced, (Molecular formula for W=C2H4)
C8H16 - C2H4 = C6H12
now in C6H12, we can form 2 molecules of X (CH3CH=CH2). You can check by simply adding the number of carbons and hydrogens of 2 molecules of X to chek whether it's C6H12 or not.
When X is produced (Molecular formula for X = C3H6)
C8H16 - C3H6 = C5H10
From C5H10, we can form both W and X.(add the carbons and hydrogens of both W and X to check whether it's C5H10 or not)
When Y is produced (Molecular formula for Y=C3H6)
C8H16 - C3H6 = C5H8
From C5H8, you can get:
1/3 X and 1 Z and also 1/2W and 1 Z
when Z is produced (Molecular formula of Z= C4H6)
C8H16 - C4H6 = C4H10
From C4H10, you can get:
1Y and 1/2W and also 1Y and 1/2X
So from all the list here we can conclude that all W,X,Y,Z can be formed during cracking. So this is why the answer is A. (I am not sure about this one :s)
Q.27. This question is very fun~ It's my one of my favorite
(a) Cannot be Ethane because Ethane is a gas at r.t.p.
(b) Carboxylic acid can fully dissolved in water but will only dissociate partially so no.
(c) Ethanol will mix completely with water because it will form hydrogen bond with the H2O molecules.
(d) Ester cannot mix completely with water even tho there are lone pairs at the oxygen atom. (Try to draw the structure for this ester and you will be able to deduce the position of the lone pairs). This is because Ester does not completely fulfill the requirements to form H-bond. (Recap requirements to form H-bond: 1.) one hydrogen atom covalently bonded to F,O or N. 2.) F,O or N having lone pairs.) so ester only fulfill one of the requirements. This is why it does not mix well with water.
Q.33. The question is asking what is true about sulfuric acid. Hint: Sulfuric acid is a strong acid. Strong acid are acid that dissociate completely when dissolved in water.
1. you compare the [H+] of H2SO4 with the [H+] of HSO4-, H2SO4 will dissociate more than HSO4- since H2SO4 is a strong acid where as HSO4- is a weak ones so therefore the [H+] of H2SO4 would be higher. so 1 is correct.
2. Since H2SO4 dissociate high amount of H+, so higher amount of H+ will cause the equilibrium position of the second equation to be shifted to the left to form more HSO42- so obviously [SO42-] will not increase so it shouldn't be high.
3. When you have a higher concentration of H+, you will get higher [HSO42-] which is obviously higher than the [SO42-]
That should be it and the answer will be D.
Q.37. Which compound will be formed when propene undergo single reaction?
(1) CH2OHCHOHCH3 (this is a mild oxidation reaction where you form diol using reagent: Cold and acidified potassium dichromate. It's a one step reaction because once you add the reagent, you will get the product without using any additional reagent.)
(2)( CH2CH(CH3) ) n
This is additional polymerisation process. Where all the propene monomers link together to form a polymer. This reaction is also a one step reaction.
(3)CH2BrCH2CH2Br
At first, Propene undergo electrophiliic addition reaction with HBr to form CH3CH2Br. Next, this compound will free radical substitution reaction with the presence of UV-light to form CH2BrCH2CH2Br. So this is a two steps reaction.
So final answer is B.
Yey~~ It's done~ But please do double check with your teacher because I am not 100% sure that my explanations are correct. If you have a better explanation would you please kindly reply? Thank you
You quoted me guess that means you're answering me.. But My Answer? :O but I didn't ask any questions yet!