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Chemistry and Physics AS paper 12 MCQS
- Thread starter MariamHASAN
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Q.26 Will not react with Na means that there is no alcohol or Carboxylic acid group present in the compound. So if you try to draw the structure of every molecular formula, you will get (A) Aldehyde group (B) Alcohol and Aldehyde group (C) Two Carboxylic acid group and (D) Carboxylic acid group. So obviously the answer will be A.
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Q.28 First thing that you have to do is to break every double bond into their respective compounds because here involved the use of hot and concentrated potassium permanganate(ALL THE DOUBLE BONDS IN THE COMPOUND MUST BREAK). Once you break the double bonds in the compounds, you will get 3 ketones group and these ketones group will react with the 2,4 DNPH. Since there are 3 ketones group present in the compound, therefore 3 molecules of 2,4 DNPH is required to react with one molecule of X. So I guess the answer will be C.
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Q.20 is too lengthy~~ plus my working is quite complicated. Hopefully somebody can help us with this question.
Thanks i got it
yes!! but are all these 3 ketones?
I was getting 6 isomers but the answer is 8
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s09_qp_1.pdf
questiion 40 and 20??
questiion 40 and 20??
question 27 and 38
- http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_s10_qp_12.pdf
- question 12, 25 ,26
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Very sure of it. Because no hydrogen is bonded to the C=O carbon and we have 4 C=O bonds when we break all the double bond and since it's a strong oxidation reaction, one of them should include a C.A group. So overall we have 1 C.A group and 3 Ketones group and C.A will not react with 2,4 DNPH. (I am not really good in making things clear so please forgive me if I've made any mistakes
)
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For Q.40, We will have to look at the hydrolysis of esters using base(NaOH). Try to separate or split the esters into Sodium salt(Not C.A because we are using NaOH) and also alcohol. (1) is actually the alcohol product produced by base hydrolysis of ester. (2) is the sodium salt produced by base hydrolysis of ester. (3) is totally out because aqueous NaOH cannot be used to hydrolyse the C=C double bond so hence the answer will be B.
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http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s08_qp_1.pdf
Q.27 unreactive towards mild oxidation reagent simply means that the compound cannot be oxidised. So since the question is asking for the product after dehydration, this shows that there must be an alcohol group present in the compound. Primary alcohol and secondary alcohol can be oxidised but tertiary alcohol will not have any desired effect so we must look for the compound that contains tertiary alcohol. Try to draw the structures for every compounds before they undergo dehydration(Simply add H20 to the alkene). So you will get D as the final answer.
Q.38 monosodium glutamate contains an extra carbon compared to the initial compound. So we will need to replace the Cl with a nitrile group(CN) to the compound which can be done by using ethanolic KCN or NaCN. Since monosodium glutamate has a sodium salt group, this means that we have to do base hydrolysis and we usually use NaOH. so (1) is correct.
(2) is incorrect because during the reaction NaCl will be formed and the resulting ions(The carbocation and the HCO2-) will react together to form a new compound which is obviously not monosodium glutamate since Na is a leaving group.
(3) Heat under reflux with NaOH will only replace the Cl with OH (Nucleophilic substitution ofHalogenoalkane).
Answer is D and good luck~
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NEED HELP IN PHYSIC MCQ'S
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you can post your doubts anytime. I believed all of us here are willing to help.
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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
). So 3 will be the best answer which is C.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.
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Q
Q.36. For the 3 springs in parallel, you will need to find the spring constant first for each spring. To do so you will need to split the weight of the load(W) into 3 because there are 3 springs. So each spring will carry the weight of W/3. Use the formula F=KX (Where F=Weight of the load, K=Spring constant and X=extension of the spring) to determine the spring constant of each spring. For each spring, K=W/3X.
Now we will look at the second scenario where the middle spring and weight are removed and the weight of 2W is hung. We now have 2 springs in parallel so we are asked to find the new extension of each spring. Same thing, split the weight of the load into 2 because we have 2 springs so each spring will now carry the weight W. So apply the formula F=KX to determine the new extension. We will use the spring constant K=W/3K because we are using the same spring here so perform the calculation Xnew = 3X and therefore the answer is D.
I am not sure of my answer because I haven't looked at the marking scheme. Please reply if you do have any questions. Thank you~~
Q.35 R1 requires 2v, R2 requires 1v and R3 requires 2v and in series the current should be the same at all point. Since R1 and R3 have the same potential difference, I would say that Resistance of R1 = Resistance of R3 based on the formula R=V/I where V is the same and I is constant. so my answer would be C.
Q.36. For the 3 springs in parallel, you will need to find the spring constant first for each spring. To do so you will need to split the weight of the load(W) into 3 because there are 3 springs. So each spring will carry the weight of W/3. Use the formula F=KX (Where F=Weight of the load, K=Spring constant and X=extension of the spring) to determine the spring constant of each spring. For each spring, K=W/3X.
Now we will look at the second scenario where the middle spring and weight are removed and the weight of 2W is hung. We now have 2 springs in parallel so we are asked to find the new extension of each spring. Same thing, split the weight of the load into 2 because we have 2 springs so each spring will now carry the weight W. So apply the formula F=KX to determine the new extension. We will use the spring constant K=W/3K because we are using the same spring here so perform the calculation Xnew = 3X and therefore the answer is D.
I am not sure of my answer because I haven't looked at the marking scheme. Please reply if you do have any questions. Thank you~~
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i should say thank youQ
Q.35 R1 requires 2v, R2 requires 1v and R3 requires 2v and in series the current should be the same at all point. Since R1 and R3 have the same potential difference, I would say that Resistance of R1 = Resistance of R3 based on the formula R=V/I where V is the same and I is constant. so my answer would be C.
Q.36. For the 3 springs in parallel, you will need to find the spring constant first for each spring. To do so you will need to split the weight of the load(W) into 3 because there are 3 springs. So each spring will carry the weight of W/3. Use the formula F=KX (Where F=Weight of the load, K=Spring constant and X=extension of the spring) to determine the spring constant of each spring. For each spring, K=W/3X.
Now we will look at the second scenario where the middle spring and weight are removed and the weight of 2W is hung. We now have 2 springs in parallel so we are asked to find the new extension of each spring. Same thing, split the weight of the load into 2 because we have 2 springs so each spring will now carry the weight W. So apply the formula F=KX to determine the new extension. We will use the spring constant K=W/3K because we are using the same spring here so perform the calculation Xnew = 3X and therefore the answer is D.
I am not sure of my answer because I haven't looked at the marking scheme. Please reply if you do have any questions. Thank you~~
both of ur anwers are rightmy doubt is how did u now that r1 and r3 have a same voltage of 2v ?
2nd question.. after finding K value in first case ... did u substitute it in the new scenario .... and shoudnt the equation be K=3W/x?
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You will need to look at the voltage output. At first the voltage is 5v and after it passes through R1 the voltage output dropped from 5v to 3v and this means that 2v voltage is used in the R1 resistor and this is why the voltage of R1 is 2v. Same goes for R2, after passing through R2, Voltage output dropped from 3v to 2v which means 1v is used in R2. Lastly for R3, voltage dropped from 2v to 0v and 2v is used in R3. You get the idea right?i should say thank you
my doubt is how did u now that r1 and r3 have a same voltage of 2v ?
2nd question.. after finding K value in first case ... did u substitute it in the new scenario .... and shoudnt the equation be K=3W/x?
Yes substitution is required. The weight that each spring hold is W/3 and the extension of each spring is x so i got (W/3)=Kx. Rearrange will give you K=W/3x. Hope this can clear your doubts.