Chemistry and Physics AS paper 12 MCQS

[KWIKIW]

Hello~ I think I can help you out with these questions.

Q.6. Which gas least resembles ideal gas? To answer this question I think you need to recall the kinetic theory of gases and one of them will be "No intermolecular forces between gas molecules."
(a) Ammonia(try to draw the structure) is capable of forming hydrogen bond. Requirement to form H-bond: 1.) F,O or N atom contains lone pairs 2.)F,O or N is covalently bonded to H atom. And since ammonia fulfills all the requirements here, Ammonia is capable of forming one of the strongest intermolecular forces which is H-bond.

(b)Helium is a noble gas which is unreactive(Does not form any bonds with other gas molecules) so this makes it behave more like ideal gas.

(c)H2 gas are held together by weak V.D.W forces only.

(d)Trichloromethane contains permanently induced dipole-dipole between the C-Cl due to the presence of the electronegative atom Cl. However, this intermolecular force is still weaker than H-bond.
So the most attractive force can be found in Ammonia so the answer will be A.

Q.21. Chiral centres: When a carbon atom is covalently bonded to 4 different groups. So any carbon bonded to double bonds should be ignored. And here, we must consider the benzene ring as one group. Do not attempt to find chiral centres from the benzene ring. And the final answer that I obtained is 1 chiral centre which is B.

Q.33. (1) This is correct because when the activation energy is too high, more energy need to be supplied in order for the reaction to occur so high activation energy may be one of the reason why hydrazine does not react well with oxygen.

(2) Hydrazine only involved N=N and not N triple bonds. So this is not a right choice.

(3) Being a liquid does not determines how good or bad a reaction is. Example: Alcohol is a liquid and you can react alcohol with oxygen(Process is known as combustion). So 3 is not a wise choice either.
You have only one correct answer here so it's D.

Hope it helps

Your explanations were very helpful.
Thanks

I have a small doubt regarding intermolecular forces and about 31 qs.
*Generally between which types of molecules are van der waal's forces made, and between which type of molecules are permanently induced dipole-dipole made?I am a bit confused.

*And in 31 qs I think that a triple bond is present between N atoms as in my diagram.
In the examiner's report it said,

Question 33 explored the fact that hydrazine does not burn spontaneously in oxygen. 28% of candidates correctly appreciated that this is a mark of high activation energy and gave the key D, but 40% gave B as the key, incorporating the high strength of the N≡N triple bond as being a factor. This would be relevant to a calculation of the overall change in enthalpy, but not to the rate of the reaction.

But I didn't understand the last statement.
Do you know what it means?

 

[iKhaled]

can someone please explain to me question 7 and question 10 of this paper ??

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s12_qp_12.pdf

 

[Frankuzi]

Your explanations were very helpful.
Thanks

I have a small doubt regarding intermolecular forces and about 31 qs.
*Generally between which types of molecules are van der waal's forces made, and between which type of molecules are permanently induced dipole-dipole made?I am a bit confused.

*And in 31 qs I think that a triple bond is present between N atoms as in my diagram.
In the examiner's report it said,

Question 33 explored the fact that hydrazine does not burn spontaneously in oxygen. 28% of candidates correctly appreciated that this is a mark of high activation energy and gave the key D, but 40% gave B as the key, incorporating the high strength of the N≡N triple bond as being a factor. This would be relevant to a calculation of the overall change in enthalpy, but not to the rate of the reaction.

But I didn't understand the last statement.
Do you know what it means?

Intermolecular forces: (These are from my notes. XD)
Van der waals forces: also known as the 'Temporarily dipole-dipole forces'. It is the temporary electrostatic attraction between atoms or molecules due to temporarily dipoles. Dipoles are only formed temporarily because the electrons inside an atom or molecule are free to move around. So any molecules or compounds with electrons are capable of forming Van der Waals forces.

Permanent induced dipole-dipole: It is the electrostatic attraction between polar molecules. Polar molecules are molecules that contains an electronegative atom. Example of polar molecules are HCl, CHCl and e.t.c. The electronegative atom in this case is Cl but of course it does not necessarily has to be Cl. It can be other atom as well such as F,O,N,Br and e.t.c.(You can do some internet search on this). So this permanent induced dipole-dipole can be formed when you have polar molecules.

Q.31 ??

Q.33. You sure you got an N triple bond in your Hydrazine compound? My structure will look like this: H2N=NH2(which total molecular formula is N2H4). Actually you have many ways in answering this question. My method was looking at the N bonds present. The question is asking for N triple bonds but N2H4 does not contains any N triple bonds but only double bonds instead.

Hmmm... I think the examiner report is trying to tell us that the rate of reaction will not be affected by the presence of N triple bonds there. It's true that you can only work out the enthalpy change by using bond energies but not the rate of reaction. We have another method of measuring the rate of reaction but clearly not using this N triple bonds.

I am not sure about this actually~ But this is my understanding so far. If you have a better explanation please kindly reply me. Thank you~

 

[Frankuzi]

Frankuzi

Wow you're awesome at explaining chemistry! Thanks a lot for all your effort in helping others... I know It takes a lot of time to answer and I really appreciate what you're doing

<3 ya~

Thank you and all the best for your exam~

 

[hope4thebest]

Help

 

[syed1995]

HelpView attachment 27828

B?

 

[ahmed abdulla]

any one help ?
phy(1) +chem (2)
syed1995

 

[syed1995]

any one help ?
phy(1) +chem (2)
syed1995

No Clue about physics ..

Chem 5 C .. the greatest rise is between 5 and 6 meaning it's a 5th Group Element .. so XCl3

28 B I think .. cuz there 3 OH groups which are (Primary and Secondary) .. And deutrium will substitute the H in OH to form OD.

3 D.. Well 10cm^3 of Meth will react with 30cm^3 of O2 according to the equation .. leaving 30cm^3 of O2 in excess.

The moles of CO2 = 1 meaning 10cm^3 of CO2 formed. same for SO2 10cm^3 and 2 moles of H2O which equal 20cm^3

now add them.. 10+10+20 = 40cm^3

There is 30cm^3 of excess O2 also .. so answer is 70cm^3

 

[ahmed abdulla]

No Clue about physics ..

Chem 5 C .. the greatest rise is between 5 and 6 meaning it's a 5th Group Element .. so XCl3

28 B I think .. cuz there 3 OH groups which are (Primary and Secondary) .. And deutrium will substitute the H in OH to form OD.

3 D.. Well 10cm^3 of Meth will react with 30cm^3 of O2 according to the equation .. leaving 30cm^3 of O2 in excess.

The moles of CO2 = 1 meaning 10cm^3 of CO2 formed. same for SO2 10cm^3 and 2 moles of H2O which equal 20cm^3

now add them.. 10+10+20 = 40cm^3

There is 30cm^3 of excess O2 also .. so answer is 70cm^3

i now its in the fifth qp .. but how did u now the eqn ?
ur mole question is Wrong ..

 

[ahmed abdulla]

:?

 

[syed1995]

i now its in the fifth qp .. but how did u now the eqn ?
ur mole question is Wrong ..

WELL CAN YA TELL ME THE ANSWER instead of saying it's wrong?

 

[ahmed abdulla]

WELL CAN YA TELL ME THE ANSWER instead of saying it's wrong?

C ...50 cm^3

 

[Frankuzi]

any one help ?
phy(1) +chem (2)
syed1995

Hi~ I think I can give a try on phy question.
Q.25. I think the answer is B because in transverse wave, the particles are oscillating perpendicularly to the direction of energy transfer so we assume that the movement of the particle is up and down. Therefore at the max. displacement Q, the movement is already at its maximum and cannot go up anymore but will try to go down instead. You only have one choice for Q moving downward so the answer is B.
Tell me the answer please~~

 

[Frankuzi]

HelpView attachment 27828

Student X said that all the 17 carbons on the ring lies on the same plane. Hmmmm~~, I don't think so because there are carbons covalently bonded to 4 groups which makes it tetrahedral(angle:109.5 degree) so they shouldn't be lying on the same plane except when they are planar(angle: roughly 120 degree) but in this case, they are not.

Student Y suggested that cis-trans isomers could be formed at the C=C double bond. yet again, I don't think so. The C=C double bond are all bonded to different groups. (3 different big groups and one Hydrogen) In order to have cis-trans isomers you need to have 2 groups which are the same.
So both students are wrong~ B will be the answer?

 

[fishcook]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w04_qp_1.pdf
Q12. why is the answer B? and not D

 

[syed1995]

C ...50 cm^3

Ah didn't read the part about cool to room temperature then there won't be any 20cm^3 of H2O gas formed.. so vol. will be 50 instead of 70cm^3 ...

And Valency can either be -3 or +5 in group 5.. PCl3 and PCl5 example.

 

[ahmed abdulla]

Ah didn't read the part about cool to room temperature then there won't be any 20cm^3 of H2O gas formed.. so vol. will be 50 instead of 70cm^3 ...

And Valency can either be -3 or +5 in group 5.. PCl3 and PCl5 example.

what about this bro >?

 

[KWIKIW]

Intermolecular forces: (These are from my notes. XD)
Van der waals forces: also known as the 'Temporarily dipole-dipole forces'. It is the temporary electrostatic attraction between atoms or molecules due to temporarily dipoles. Dipoles are only formed temporarily because the electrons inside an atom or molecule are free to move around. So any molecules or compounds with electrons are capable of forming Van der Waals forces.

Permanent induced dipole-dipole: It is the electrostatic attraction between polar molecules. Polar molecules are molecules that contains an electronegative atom. Example of polar molecules are HCl, CHCl and e.t.c. The electronegative atom in this case is Cl but of course it does not necessarily has to be Cl. It can be other atom as well such as F,O,N,Br and e.t.c.(You can do some internet search on this). So this permanent induced dipole-dipole can be formed when you have polar molecules.

Q.31 ??

Q.33. You sure you got an N triple bond in your Hydrazine compound? My structure will look like this: H2N=NH2(which total molecular formula is N2H4). Actually you have many ways in answering this question. My method was looking at the N bonds present. The question is asking for N triple bonds but N2H4 does not contains any N triple bonds but only double bonds instead.

Hmmm... I think the examiner report is trying to tell us that the rate of reaction will not be affected by the presence of N triple bonds there. It's true that you can only work out the enthalpy change by using bond energies but not the rate of reaction. We have another method of measuring the rate of reaction but clearly not using this N triple bonds.

I am not sure about this actually~ But this is my understanding so far. If you have a better explanation please kindly reply me. Thank you~

Thanks.
If I have a better explanation I'll definitely reply to you.

 

[hope4thebest]

Student X said that all the 17 carbons on the ring lies on the same plane. Hmmmm~~, I don't think so because there are carbons covalently bonded to 4 groups which makes it tetrahedral(angle:109.5 degree) so they shouldn't be lying on the same plane except when they are planar(angle: roughly 120 degree) but in this case, they are not.

Student Y suggested that cis-trans isomers could be formed at the C=C double bond. yet again, I don't think so. The C=C double bond are all bonded to different groups. (3 different big groups and one Hydrogen) In order to have cis-trans isomers you need to have 2 groups which are the same.
So both students are wrong~ B will be the answer?

Thanks

 

[hope4thebest]

B?

Correct its B