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P1 MCQ's preparation thread for chemistry ONLY!!!!

hino

hino

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bamteck it was a really good effort on your part,if it is possible for you,can u post solutions to some more papers ,it ll be really helpful
Thankyou
 
hino

hino

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81
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bamteck it was a really good effort on your part,if it is possible for you,can u post solutions to some more papers ,it ll be really helpful
Thankyou
 
H

hela

Messages
355
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hela said:
lease help with the following questions:​
Q23 Q36​
Q13 Q22​
Click to expand...
 
N

Nightmare***

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34
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autumnsakura said:
Q.26: You have to consider that there's an intermediate so there's two humps. The front hump is always higher than the second hump. I'm not sure why but it's just a basic rule, I think. XD
Q.27: Hydrolysis means substituting Br with OH group. So if all the Br are substituted, they would give you the same diol, wouldn't they? It's a nucleophilic substition reaction. And only the second organic compound can form H bond due to presence of OH group.
Q.33: CAtalyst does not increase KE of molecules. They lower the activation e so that molecules with lower KE can react. Catalyst also increases the rate of reaction whether it's backward or forward. It doesn't affect enthalpy change, same amount of energy released or used by reaction whether catalyst is there or not. It just speeds up the reaction.
Q.35: For the first: CaO + SO2 = CaSO3 thats correct
second: SO2+O2 = 2SO3 due to excess air then CaO + SO3 = CaSO4
third: CaO + CO = CaCO3 (lazy to balance) this is not a likely reaction as CO is neutral (but CO2 is acidic!) My lecturer helped with this question :)
Click to expand...

the front hump is higher bcoz the first part of the reaction is always slower (the rate determining step) and hence its activation energy is higher thn that of the second step(faster one)
 
h4rriet

h4rriet

Messages
325
Reaction score
215
Points
53
hela said:
Please help with the following questions:​
Q23 Q36​
Q13 Q22​
Click to expand...

23. The mole fraction of ethene is 1/2. If the products can be only methane, ethene and propene, there must be 2 ethene molecules, 1 methane molecule and 1 propene molecule, so that the mole fraction of ethene = 1/2. What you can do now is draw 2 ethene molecules, 1 methane molecule and 1 propene molecule and stick these molecules together (i.e. make them all one big molecule).
36. Y is a non-metal oxide. The non-metal X can be either S or N, therefore Y can be SO2 or NO/NO2. Since oxidation of Y occurs in the atmosphere, you can eliminate SO2 because for further oxidation of it to happen, you need a catalyst. Y can now only be NO (1 mole of NO reacts with 1/2 mole of O to yield NO2). X, Y & Z are all known now. The oxidation number of N is NO = +2 and in NO2 = +4. NO has an unpaired electron (in NO, O has a full shell & N is missing 1 electron). The molecule is non-polar, because we know that N and O have extremely high electronegativites.

13. For this question, you need to know that Group II metal carbonates decompose to give CO2, and that the temperature of decomposition increases as you go down the group. The answer can't be A because if CaCO3 is heated, all you'd have left will be Ca and there won't be any effervescence with HCl. B is wrong too because it doesn't contain two Group II metal carbonates; that leaves us with C & D. D: CaCO3 and MgCO3 will both decompose when heated by a Bunsen flame because they're right under each other in the Periodic table. C: Ba is way down under Ca in the Periodic table. That means BaCO3 will need a temperature much higher than the flame of a Bunsen burner can provide for it to decompose.
22. Ethyl propanoate will give ethanol and sodium propanoate upon hydrolysis with NaOH. All you have to do is calculate the mass of ethanol and divide that with (mass of ethanol + mass of sodium propanoate) and multiply that times 100 to get its percentage by mass.

10. The enthalpy of neutralisation is the heat released when 1 mole of water is formed. In the second equation, 2 moles of water are formed, so -114 x 2 = -288 kJmol-1.
20. = 22 above.
 
H

hela

Messages
355
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95
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38
t
h4rriet said:
23. The mole fraction of ethene is 1/2. If the products can be only methane, ethene and propene, there must be 2 ethene molecules, 1 methane molecule and 1 propene molecule, so that the mole fraction of ethene = 1/2. What you can do now is draw 2 ethene molecules, 1 methane molecule and 1 propene molecule and stick these molecules together (i.e. make them all one big molecule).
36. Y is a non-metal oxide. The non-metal X can be either S or N, therefore Y can be SO2 or NO/NO2. Since oxidation of Y occurs in the atmosphere, you can eliminate SO2 because for further oxidation of it to happen, you need a catalyst. Y can now only be NO (1 mole of NO reacts with 1/2 mole of O to yield NO2). X, Y & Z are all known now. The oxidation number of N is NO = +2 and in NO2 = +4. NO has an unpaired electron (in NO, O has a full shell & N is missing 1 electron). The molecule is non-polar, because we know that N and O have extremely high electronegativites.

13. For this question, you need to know that Group II metal carbonates decompose to give CO2, and that the temperature of decomposition increases as you go down the group. The answer can't be A because if CaCO3 is heated, all you'd have left will be Ca and there won't be any effervescence with HCl. B is wrong too because it doesn't contain two Group II metal carbonates; that leaves us with C & D. D: CaCO3 and MgCO3 will both decompose when heated by a Bunsen flame because they're right under each other in the Periodic table. C: Ba is way down under Ca in the Periodic table. That means BaCO3 will need a temperature much higher than the flame of a Bunsen burner can provide for it to decompose.
22. Ethyl propanoate will give ethanol and sodium propanoate upon hydrolysis with NaOH. All you have to do is calculate the mass of ethanol and divide that with (mass of ethanol + mass of sodium propanoate) and multiply that times 100 to get its percentage by mass.

10. The enthalpy of neutralisation is the heat released when 1 mole of water is formed. In the second equation, 2 moles of water are formed, so -114 x 2 = -288 kJmol-1.
20. = 22 above.
Click to expand...
thanks very helpful
 
mannat0033

mannat0033

Messages
21
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5
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3
one mole of Cl reacts with H2 are twice the moles of Cl reacting with KBr ? can someone explain this ?
 
O

Omar99

Messages
41
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7
Points
8
can some one please help me with few questions,
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w07_qp_1.pdf
question no. 9, 32
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w09_qp_11.pdf
question no. 19 and 36
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s10_qp_11.pdf
question no. 11, and 21
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w10_qp_11.pdf
question no. 4, 34, 35 and 36
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w10_qp_12.pdf
question no. 7, 8, 26 and 39
please help me with these big doubts.......
 
h4rriet

h4rriet

Messages
325
Reaction score
215
Points
53
Omar99 said:
can some one please help me with few questions,
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_1.pdf
question no. 9, 32
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_11.pdf
question no. 19 and 36
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_11.pdf
question no. 11, and 21
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_qp_11.pdf
question no. 4, 34, 35 and 36
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_qp_12.pdf
question no. 7, 8, 26 and 39
please help me with these big doubts.......
Click to expand...

9. The ox. number of S in SO32- is +4 and is SO42- is +6. +6-+4=+2.
32. A. The ox. state of C in CO is 2+, and in CO2 it's 4+. Since CO readily turns into CO2, CO2 is more stable than CO.
B. An exothermic reaction yields stable products. If the enthalpy of formation of a compound is very negative, that compound is very stable.
C. A reaction with a high Kc value has a high yield of products. They already told us in the question that CO reacts readily with O2, therefore it gives a high yield of CO2.

19. They said there's no by-product, so if CO(NH2)2 is the only product, the elements in ammonium cyanate must balance the elements in this compound. We already know that ammonium is NH4+; if you add a CNO- ion to it, the reaction will be balanced. You can't add the 2- ion to it because NH4+ has a 1+ charge, not 2+; the charges wouldn't cancel.
36. The ox. state of S in H2SO4 is +6, and in SO2 is +4; it is therefore oxidised, not reduced. And bromide ions are oxidised to bromine, not reduced.

11. By acidifying the pool water, you'll be adding excess H+ ions which can react with the OCl- to form more HOCl.
21.

4.
34. This question tests whether or not you know that Group II hydroxides get more soluble as you go down the group. Barium hydroxide dissolves; calcium hydroxide forms a suspension.
35. 1 and 2 are known already to be disproportionation reactions. You have to balance the 3rd equation, using HNO2, and you'll see that it too is a disproportionation reaction.
36. Element X is sulphur. Molecule Y is SO2; if you draw its structure, you'll see it has a lone pair. It is known that SO2 needs a catalyst to turn into SO3. SO2 is colourless; it's known.

7. Use p1v1+p2v2=p3v3. (12 x 5)+(6 x 10)=15 x p3.
8. You'll have to draw a Hess cycle:
2ymvecz.jpg

26. Aldehydes get oxidised by Tollens' and Fehling's reagents.
39. Find the molecular masses of butanone, butanoic acid and 2-methylpropanoic acid. Then calculate the % yield uses the masses given.
 
H

hela

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355
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38
hela said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s09_qp_1.pdf Q9 A Q20 C

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_12.pdf Q7 D

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_12.pdf Q9 D Q25 C

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w12_qp_13.pdf Q22 B Q25 A
Click to expand...
can someone please explain few Q

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s09_qp_1.pdf Q9 A Q20 C

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_12.pdf Q7 D

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_12.pdf Q9 D Q25 C

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w12_qp_13.pdf Q22 B Q25
 
h4rriet

h4rriet

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325
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215
Points
53
bogus said:
nov 08 mcq 23 help plz h4rriet ... anyone?
Click to expand...

The cloudiness occurs because of the AgX. The rate of formation of cloudiness depends on how fast X- ions are released into the solution, so that Ag+ ions get attracted to them. How fast X- ions are released depends on the R-X bond energy.
 
h4rriet

h4rriet

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325
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215
Points
53
hela said:
can someone please explain few Q

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s09_qp_1.pdf Q9 A Q20 C

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_12.pdf Q7 D

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_12.pdf Q9 D Q25 C

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w12_qp_13.pdf Q22 B Q25
Click to expand...

9. Ag is under Cu in the reactivity series, so electrons are given to Cu in preference to Ag at the cathode.
20. At each C=C double bond, a cis-trans isomer exists. That makes it 6. No idea where the other 2 come from.
7. P <-> Q + 2R
Initial moles 2 0 0
Change in moles -x +x +2x
Final moles 2-x x 2x
If 2x=x, then x=0.5. The new final moles: 2-x, 0.5x & x. x+0.5x+2-x=(x/2)+2.
9. P is wrong; a strong acid cannot have a pH of 6. You can tell that Y dissociates more than X because both of them have the same concentration and yet Y has a pH farther away from 7 than X. That is, the difference between neutral and substance X's pH is 7-6=1. The difference between neutral and substance Y's pH=9-7=2. Therefore, Y dissociates more than X.
25. Don't know, unfortunately
 
P

polniks

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6
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can anyone please explain these questions:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s03_qp_1.pdf Q 2,16,18

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s04_qp_1.pdfQ 12,32

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s05_qp_1.pdf Q 10,13,18

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s06_qp_1.pdf Q10

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s07_qp_1.pdf Q 34

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_1.pdf Q 22,25,27,29,30
 
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