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

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6. An ideal gas has negligible intermolecular forces, so the gas with the strongest intermolecular forces will least resemble an ideal gas.
21.
O4snYKo.jpg

33. They're asking why it doesn't burn spontaneously, not why it has a high activation energy. It has a high activation energy because of the high bond energy.
 
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mj 2010 question paper 11 question 40. Help!h4rriet

The apparatus shown is used for reflux. Reaction 1 is nucelophilic substition and reaction 2 is oxidation of an alcohol; both these need reflux. The 3rd is the bromine water tes; it occurs at room temperature.
 
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2. 1 mole of H2 removes one C=C bond. At the beginning, there are 9 C=C bonds in the molecule; at the end, there are 4 (2 in one side chain, 1 in each of the other 2 side chains). 9-4=5.
16. Cl2 has to react with hot, concentrated NaOH to give ClO3- ions. The ox. no. of Cl in ClO3- is +5.
18. NH4+ compounds will liberate NH3 whenever reacted with any alkali/base.

12. Most rapidly=small activation energy. Good yield=exothermic, because exothermic reactions always yield stable products.
32. The energy needed by graphite to turn into diamond is very little, and yet the reaction doesn't occur easily. Why? Because of the high activation energy required.

10. If the temperature is increased, it'll favour the forward reaction because the forward reaction is endothermic. Therefore more molecules of gas will be made. The pressure will increase. An increase in pressure favours the side of the reaction with the lower number of molecules.
13. Al is the first metal in Group 3; Be is the first metal in Group 2. There's a rule that says that metals at the top of Groups will be similar chemically to metals at the top of adjacent Groups. In other words, it isn't something you have to deduce, it's something you have to memorise.
18. 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.

10. If it's 50% dissociated, that means 50% of 1 mole of N2O4 exists at equilibrium. That's 0.5 moles. And 50% of 2 moles of NO2 exist; that's 1 mole. The partial pressure of N2O4 at equilibrium = 0.5/(0.5+1) and of NO2 = 1/1.5. Kp=(pNO2)x(pNO2)/(pN2O4).
 
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6. An ideal gas has negligible intermolecular forces, so the gas with the strongest intermolecular forces will least resemble an ideal gas.
21.
O4snYKo.jpg

33. They're asking why it doesn't burn spontaneously, not why it has a high activation energy. It has a high activation energy because of the high bond energy.


Thanks.


(21)Why can't the carbon labelled as A in my diagram be a chiral carbon ?

(6)Does trichloromethane have van der waal's forces or permanent dipole-permanent dipole forces?
 

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Thanks.


(21)Why can't the carbon labelled as A in my diagram be a chiral carbon ?

(6)Does trichloromethane have van der waal's forces or permanent dipole-permanent dipole forces?

You labelled a bond, not the Carbon atom. The Carbon atom you meant can't be chiral because is had a double bond with O. Trichloromethane has only van der Waals and perhaps very weak permanent dipole forces, but the strongest intermolecular force is the hydrogen bonding.
 
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I need help with J12/P12 question no. 7 :( need full explanation, would be very grateful :cry:
a>

The area of a v/t graph is the displacement; the 1st half of the wave is positive displacement, the 2nd half is negative displacement. Since the positive and negative displacements are the same, the overall displacement is zero. At 1/2T, the displacement is maximum; in B, C & D, the displacements are all 0 at 1/2T. A is the correct answer.
 
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The area of a v/t graph is the displacement; the 1st half of the wave is positive displacement, the 2nd half is negative displacement. Since the positive and negative displacements are the same, the overall displacement is zero. At 1/2T, the displacement is maximum; in B, C & D, the displacements are all 0 at 1/2T. A is the correct answer.

I'm sorry if I'm confusing you or anything but the answer doesnt seem to match the question. This is actually the Qcapture-20130528-190747.png

and thanks for trying to help tho :)
 
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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.


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. can u please solve for me
 
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I'm sorry if I'm confusing you or anything but the answer doesnt seem to match the question. This is actually the QView attachment 27765

and thanks for trying to help tho :)

I opened the Physics paper, lol, sorry. Here's the answer to your question:
P <-> Q + 2R
Initial moles 2 0 0 (respectively)
Change in moles -x +x +2x (respectively)
Final moles 2-x x 2x (respectively)
Add the final moles, and you'll get the number of moles they've given in the question.
 
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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. can u please solve for me

The ox. no. of N in NO2 is +4. In HNO3, it's +5 and in HNO2, it's +3. It undergoes disproportionation.
 
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w08 : 1 : 13 and 14 please help i think it's kind of a base knowledge question yet i don't know it help

no 13 : which oxide will produce the most acidic solution with water ?
A-co B-CO2 C-SiO2 D- P2O5

n0 14 : which salt is produced when amonia is added to aqueus sulpher dioxide until jest alkaline ?
A- NH4SO3
B- NH4 SO4
C - (NH4)2SO3
D- (NH4)2SO4

answer to 14 is c why sulphite not sulphate ?!
 
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You labelled a bond, not the Carbon atom. The Carbon atom you meant can't be chiral because is had a double bond with O. Trichloromethane has only van der Waals and perhaps very weak permanent dipole forces, but the strongest intermolecular force is the hydrogen bonding.


Thanks .
(21)
Sorry I wasn't very clear with my diagram.
But the carbon atom that I actually referred to is the carbon atom in the benzene group as shown in my new diagram.
I thought that this carbon is a chiral centre because there's a H atom attached to it, 2 parts of the benzene ring and another group.
But actually this carbon isn't a chiral carbon.

I am confused why this carbon isn't a chiral carbon?
 

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Thanks .
(21)
Sorry I wasn't very clear with my diagram.
But the carbon atom that I actually referred to is the carbon atom in the benzene group as shown in my new diagram.
I thought that this carbon is a chiral centre because there's a H atom attached to it, 2 parts of the benzene ring and another group.
But actually this carbon isn't a chiral carbon.

I am confused why this carbon isn't a chiral carbon?
A benzene molecule is like this:
stereochem_str5.GIF
. The Carbon atom you marked has a double-bond, therefore has only 3 groups attached to it & can't be chiral. It helps to remember that benzene isn't chiral, no matter which Carbon you refer to (they all have double-bonds anyway).
 
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