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if anybody else could help too,it ll be really kindbamteck 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
lease help with the following questions:Q23 Q36Q13 Q22
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
Please help with the following questions:Q23 Q36Q13 Q22
thanks very helpful23. 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.
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.......
can someone please explain few Qhttp://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
nov 08 mcq 23 help plz h4rriet ... anyone?
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
anything that is more electronegative will go on cathodr and for 20 read the examiner report ... it explains it wellll
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