Chemistry and Physics AS paper 12 MCQS

[Frankuzi]

So

You quoted me guess that means you're answering me.. But My Answer? :O but I didn't ask any questions yet!

Sorry XD~. Hope you don't mind...

 

[syed1995]

So
Sorry XD~. Hope you don't mind...

haha I don't mind at all.. In fact I was about to tag you with a question from chemistry .. And you answered quoting me lol...

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w07_qp_1.pdf Question No. 9 Please..

A detailed explanation on how one solves these questions would be appreciated since I have no idea on how to do these

 

[KWIKIW]

Hi,I found the following qs from may /june /2006 /01 difficult.
Do you know how to do these Qs?
Qs 6, 21, 33
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s06_qp_1.pdf

 

[bogus]

mj 2010 question paper 11 question 40. Help! syed1995

 

[syed1995]

mj 2010 question paper 11 question 40. Help! syed1995

Sorry no clue... All I know is It shows a Distillation process...

Meaning things which are soluble in each other and are boiled at different temperatures can be seperated using this apparatus.. The one with the lower boiling point will get distilled out first.

Now comparing the products reactants..

1.) Bromoethane.. Van der vaal/dipole only.. Lowest MP/BP .. NaBr Ionic .. High MP/BP .. Ethanol Hydrogen Bonding HIGH MP/BP.. so Bromoethane will be first to go out...

2.) CH3CHO .. Van Der Vaal/weak dipole.. so Lowest MP/BP .. Acidified Sodium Dichromate.. (Ionic) Highest MP/BP.. Ethanol Hydrogen Bonding (High MP/BP)

3.) CH2BrCH2Br (Van Der Vaal/Dipole) .. HIGHER MP/BP cuz of Dipole which isn't present in others.. Br2 (Van Der Vaal Only) CH2CH2 (Van Der Vaal Only)

So in 3 Either Bromine or Ethene will get distilled first... cuz they don't have Dipole and only have vander vaal forces... While in 1 And 2 the product will get distilled out.. as it has the weakest bonds and will have the lowest boiling point.. as compared to others...

So B should be the answer.

 

[Frankuzi]

haha I don't mind at all.. In fact I was about to tag you with a question from chemistry .. And you answered quoting me lol...

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_1.pdf Question No. 9 Please..

A detailed explanation on how one solves these questions would be appreciated since I have no idea on how to do these

Whoa~~ This question is sooo hard . But I think I came up with a solution but I am not sure whether it's 100% correct.
First, I attempt to work out the no.of moles of the metallic salt and also sodium sulphite by using the volume and concentration given. Once you worked out the answer, you should get: no.of moles of metallic salt= 5 x 10^-3 moles and no.of moles of sodium sulphite(Sulphite ion)=2.5 x 10^-3 moles. Next you will need to work out the mol ratio of metallic salt to sulphite ion which can be done by simply (5x10^-3)/(2.5x10^-3) = 2/1 so the ratio is 2:1. This means that two moles of metallic salt is required to react with one mole of sulphite ion. Next you will need to look at the sulphite ion half formula and notice that there are 2 electrons on the right side, so you can deduce the half equation for the metallic salt by using the information given (Original Ox.no of metallic salt is +3). My half equation for metallic salt is: 2X3+ + 2e- ---> 2X2+(I put 2X3+ instead of X3+ here because two moles of metallic salt is required to react with one mole of sulphite ion. I put 2e- on the left because when you mix the two half equations together(half equation of sulphite and metallic salt), the 2e- on both side will be cancelled out. The product contain 2+ ions because the total number of charge on left must be equal to the total number of charge on the right. By calculation: ((2x +3) - 2)=+4(charge on left side), (2 x +2)= +4 so the charges on both side are equal). From the half equation of the metallic salt, you can clearly deduced that the new oxidation state of the metal is +2. So answer is B.

So hard~~ Plus the working is quite long. I am afraid I don't have enough time for this during exam...

 

[mariamh]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_12.pdf
Q.6. It's all calculation. We are given the value of V=7000cm^3, Temperature=30 degree C and also the mass of O2 gas. First thing you have to do is to change the unit of volume from cm^3 to m^3 so in this case: 7000cm^3/(100 x 100 x 100) = 7x10^-3 m^3 . Next you will need to find the no.of moles of O2 gas by using the mass given so: no.of moles=0.96/2(16) = 0.03moles. Last step is easy you will only substitute these values to the formula given: Pv=nRT. Since the question is asking for the pressure, so I will make P the subject and the formula will look like this now: P=nRT/v and perform the final calculation using the values: n=0.03moles, R=8.31(Given in data booklet), T=30+273 K, v=7x10^-3 m^3. After you obtain the final answer, divide the final answer by 1000(since the pressure is in Kpa) and round it off to 3s.f. I got 10.8Kpa as my final answer which is C.

Q.7. The key words here are at e.q.m, X moles of R were present and the total moles was (2+x/2). So we will perform the calculation step by step. ("<=> here indicates the equilibrium sign cuz I don't know how to do one )
(a) 1P <=> 2Q + 1R
Initial no.of moles of P=1 (Look at the equation, the initial mole is 1 because of the "1"P)
Initial no.of moles of Q and R=0 moles(No product will be formed at the beginning)
At e.q.m,
No. of moles of R= X(given in the question)
No.of moles of Q = (2 x X) = 2X (I multiplied by 2 because of the "2"Q which means the no.of moles of Q is twice the no.of moles of R)
No.of moles of P= initial moles - moles of the product
= 1 - X
total no.of moles at e.q.m= no. of moles of R + no.of moles of Q + no.. of moles of P
= X + 2X + 1 - X
=2X + 1 moles

(b)2P <=> 2Q + R
initial moles of P=2(because of the "2"P in the equation)
initial moles of Q and R = 0 moles
at e.q.m,
no.of moles of R = X(given in the question)
no. of moles of Q = 2X (multiplied by 2 because of the "2"Q and this shows that the no.of moles of Q is twice the no.of moles of R)
no.of moles of P = 2 - (2 x X) (this time I minus by 2X because of the "2"P there.)
total no. of moles at e.q.m = X + 2X + 2 - 2X
=2 + X moles

(c)2P <=> Q + R
initial moles of P = 2 moles
initial moles of Q and R = 0 moles
at e.q.m,
no.of moles of R = X(given in the question)
no. of moles of Q = X(it's X because Q and R have the same mole ratio 1:1)
no. of moles of P = 2 - (2 xX) (Need to minus by 2X because of the "2"P)
total no. of moles= X + X + 2- 2X
=2

(d)2P <=>Q + 2R
initial moles of P=2 moles
initial moles of Q and R = 0 moles
at e.q.m,
no .of moles of R = X (this time I don't multiply by 2 even tho there is "2"R because the question stated X moles of R will be formed so here we assumed 2R is equivalent to X moles)
no. of moles of Q= X/2 (the mol ratio of Q:R is 1:2 and which means the no.of moles of R is twice the no.of moles of Q so you need to divide the no.of moles of R(X) by 2)
no.of moles of P = 2 - X(I will only minus 2 by X instead of 2X. Notice that P and R have the same mol ratio and since the no.of moles of R is X moles, so you need to minus the initial moles of P by X only)
total no.of moles = X + (X/2) + 2-X
= 2 + (X/2)
And you compared all the results here the obvious answer here would be D~

Thanksssssssssssss you really solved my problems!!

 

[mariamh]

I have decided to answer your question separately so that it won't be that lengthy
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_12.pdf
Q.16. The question is asking what are the reagent and conditions that is needed to change an ELEMENTAL Chlorine(From Cl2) to a compound containing CHLORINE WITH AN OXIDATION STATE OF +5.
(a) will not be an answer because only Cl- ion will react with AgNO3 to gives white PPT but not elemental Chlorine(Cl2)
(b) Same explanation as above. Only Cl- ion will react with H2SO4 to gives white fumes of HCl but not Cl2.
(c)Cold dilute is wrong. This is one of the disproportional reaction of Chlorine. The equation of the reaction will be: Cl2 + 2NaOH ---> NaCl + NaClO + H2O and you will need to find the Ox.no of Cl in both NaCl and NaClO. My results: Ox. no of Cl in NaCl is -1 and in NaClO is +1. So here obviously there is no Chlorine with the oxidation state of +5 so this is why it's wrong.
(d)Another disproportional reaction. Using Hot NaOH, the equation of the reaction will be: 3Cl2 + 6NaOH ---> 5NaCl + NaClO3 + 3H2O. Find the oxidation state of Cl in both NaCl and NaClO3 and my results: Ox.no of Cl in NaCl is -1 and in NaClO3 is +5. So here we have a Chlorine with a +5 oxidation state so this is the correct answer.

Q.24. This question is very hard~ I got it wrong at the first place but finally I can come up with some ideas but I am not sure whether it's 100% correct or not. This question is related to cracking of hydrocarbons. We need to crack oct-1-ene so first we need to count the total number of carbons and hydrogens inside oct-1-ene (C8H16). Then the question is asking which combination of compound will be formed? My working:
When W is produced, (Molecular formula for W=C2H4)
C8H16 - C2H4 = C6H12
now in C6H12, we can form 2 molecules of X (CH3CH=CH2). You can check by simply adding the number of carbons and hydrogens of 2 molecules of X to chek whether it's C6H12 or not.

When X is produced (Molecular formula for X = C3H6)
C8H16 - C3H6 = C5H10
From C5H10, we can form both W and X.(add the carbons and hydrogens of both W and X to check whether it's C5H10 or not)

When Y is produced (Molecular formula for Y=C3H6)
C8H16 - C3H6 = C5H8
From C5H8, you can get:
1/3 X and 1 Z and also 1/2W and 1 Z

when Z is produced (Molecular formula of Z= C4H6)
C8H16 - C4H6 = C4H10
From C4H10, you can get:
1Y and 1/2W and also 1Y and 1/2X

So from all the list here we can conclude that all W,X,Y,Z can be formed during cracking. So this is why the answer is A. (I am not sure about this one :s)

Q.27. This question is very fun~ It's my one of my favorite .
(a) Cannot be Ethane because Ethane is a gas at r.t.p.
(b) Carboxylic acid can fully dissolved in water but will only dissociate partially so no.
(c) Ethanol will mix completely with water because it will form hydrogen bond with the H2O molecules.
(d) Ester cannot mix completely with water even tho there are lone pairs at the oxygen atom. (Try to draw the structure for this ester and you will be able to deduce the position of the lone pairs). This is because Ester does not completely fulfill the requirements to form H-bond. (Recap requirements to form H-bond: 1.) one hydrogen atom covalently bonded to F,O or N. 2.) F,O or N having lone pairs.) so ester only fulfill one of the requirements. This is why it does not mix well with water.

Q.33. The question is asking what is true about sulfuric acid. Hint: Sulfuric acid is a strong acid. Strong acid are acid that dissociate completely when dissolved in water.
1. you compare the [H+] of H2SO4 with the [H+] of HSO4-, H2SO4 will dissociate more than HSO4- since H2SO4 is a strong acid where as HSO4- is a weak ones so therefore the [H+] of H2SO4 would be higher. so 1 is correct.

2. Since H2SO4 dissociate high amount of H+, so higher amount of H+ will cause the equilibrium position of the second equation to be shifted to the left to form more HSO42- so obviously [SO42-] will not increase so it shouldn't be high.

3. When you have a higher concentration of H+, you will get higher [HSO42-] which is obviously higher than the [SO42-]
That should be it and the answer will be D.

Q.37. Which compound will be formed when propene undergo single reaction?
(1) CH2OHCHOHCH3 (this is a mild oxidation reaction where you form diol using reagent: Cold and acidified potassium dichromate. It's a one step reaction because once you add the reagent, you will get the product without using any additional reagent.)

(2)( CH2CH(CH3) ) n
This is additional polymerisation process. Where all the propene monomers link together to form a polymer. This reaction is also a one step reaction.

(3)CH2BrCH2CH2Br
At first, Propene undergo electrophiliic addition reaction with HBr to form CH3CH2Br. Next, this compound will free radical substitution reaction with the presence of UV-light to form CH2BrCH2CH2Br. So this is a two steps reaction.
So final answer is B.

Yey~~ It's done~ But please do double check with your teacher because I am not 100% sure that my explanations are correct. If you have a better explanation would you please kindly reply? Thank you

Thanks alot for your great effort!
Good Luck for ur exams

 

[syed1995]

Whoa~~ This question is sooo hard . But I think I came up with a solution but I am not sure whether it's 100% correct.
First, I attempt to work out the no.of moles of the metallic salt and also sodium sulphite by using the volume and concentration given. Once you worked out the answer, you should get: no.of moles of metallic salt= 5 x 10^-3 moles and no.of moles of sodium sulphite(Sulphite ion)=2.5 x 10^-3 moles. Next you will need to work out the mol ratio of metallic salt to sulphite ion which can be done by simply (5x10^-3)/(2.5x10^-3) = 2/1 so the ratio is 2:1. This means that two moles of metallic salt is required to react with one mole of sulphite ion. Next you will need to look at the sulphite ion half formula and notice that there are 2 electrons on the right side, so you can deduce the half equation for the metallic salt by using the information given (Original Ox.no of metallic salt is +3). My half equation for metallic salt is: 2X3+ + 2e- ---> 2X2+(I put 2X3+ instead of X3+ here because two moles of metallic salt is required to react with one mole of sulphite ion. I put 2e- on the left because when you mix the two half equations together(half equation of sulphite and metallic salt), the 2e- on both side will be cancelled out. The product contain 2+ ions because the total number of charge on left must be equal to the total number of charge on the right. By calculation: ((2x +3) - 2)=+4(charge on left side), (2 x +2)= +4 so the charges on both side are equal). From the half equation of the metallic salt, you can clearly deduced that the new oxidation state of the metal is +2. So answer is B.

So hard~~ Plus the working is quite lenghty. I am afraid I don't have enough time for this during exam...

Wow you're quite good in Chemistry! I couldn't even come that far in the question .... Thanks for the solution.. (Although most of it went over my head? ) Btw Which subjects is it linked from? Only Redox .. or some other as well?

 

[mariamh]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_12.pdf
Q.6. It's all calculation. We are given the value of V=7000cm^3, Temperature=30 degree C and also the mass of O2 gas. First thing you have to do is to change the unit of volume from cm^3 to m^3 so in this case: 7000cm^3/(100 x 100 x 100) = 7x10^-3 m^3 . Next you will need to find the no.of moles of O2 gas by using the mass given so: no.of moles=0.96/2(16) = 0.03moles. Last step is easy you will only substitute these values to the formula given: Pv=nRT. Since the question is asking for the pressure, so I will make P the subject and the formula will look like this now: P=nRT/v and perform the final calculation using the values: n=0.03moles, R=8.31(Given in data booklet), T=30+273 K, v=7x10^-3 m^3. After you obtain the final answer, divide the final answer by 1000(since the pressure is in Kpa) and round it off to 3s.f. I got 10.8Kpa as my final answer which is C.

Q.7. The key words here are at e.q.m, X moles of R were present and the total moles was (2+x/2). So we will perform the calculation step by step. ("<=> here indicates the equilibrium sign cuz I don't know how to do one )
(a) 1P <=> 2Q + 1R
Initial no.of moles of P=1 (Look at the equation, the initial mole is 1 because of the "1"P)
Initial no.of moles of Q and R=0 moles(No product will be formed at the beginning)
At e.q.m,
No. of moles of R= X(given in the question)
No.of moles of Q = (2 x X) = 2X (I multiplied by 2 because of the "2"Q which means the no.of moles of Q is twice the no.of moles of R)
No.of moles of P= initial moles - moles of the product
= 1 - X
total no.of moles at e.q.m= no. of moles of R + no.of moles of Q + no.. of moles of P
= X + 2X + 1 - X
=2X + 1 moles


What is the method you have used to find P moles???

(b)2P <=> 2Q + R
initial moles of P=2(because of the "2"P in the equation)
initial moles of Q and R = 0 moles
at e.q.m,
no.of moles of R = X(given in the question)
no. of moles of Q = 2X (multiplied by 2 because of the "2"Q and this shows that the no.of moles of Q is twice the no.of moles of R)
no.of moles of P = 2 - (2 x X) (this time I minus by 2X because of the "2"P there.)
total no. of moles at e.q.m = X + 2X + 2 - 2X
=2 + X moles

(c)2P <=> Q + R
initial moles of P = 2 moles
initial moles of Q and R = 0 moles
at e.q.m,
no.of moles of R = X(given in the question)
no. of moles of Q = X(it's X because Q and R have the same mole ratio 1:1)
no. of moles of P = 2 - (2 xX) (Need to minus by 2X because of the "2"P)
total no. of moles= X + X + 2- 2X
=2

(d)2P <=>Q + 2R
initial moles of P=2 moles
initial moles of Q and R = 0 moles
at e.q.m,
no .of moles of R = X (this time I don't multiply by 2 even tho there is "2"R because the question stated X moles of R will be formed so here we assumed 2R is equivalent to X moles)
no. of moles of Q= X/2 (the mol ratio of Q:R is 1:2 and which means the no.of moles of R is twice the no.of moles of Q so you need to divide the no.of moles of R(X) by 2)
no.of moles of P = 2 - X(I will only minus 2 by X instead of 2X. Notice that P and R have the same mol ratio and since the no.of moles of R is X moles, so you need to minus the initial moles of P by X only)
total no.of moles = X + (X/2) + 2-X
= 2 + (X/2)
And you compared all the results here the obvious answer here would be D~

 

[mariamh]

Whoa~~ This question is sooo hard . But I think I came up with a solution but I am not sure whether it's 100% correct.
First, I attempt to work out the no.of moles of the metallic salt and also sodium sulphite by using the volume and concentration given. Once you worked out the answer, you should get: no.of moles of metallic salt= 5 x 10^-3 moles and no.of moles of sodium sulphite(Sulphite ion)=2.5 x 10^-3 moles. Next you will need to work out the mol ratio of metallic salt to sulphite ion which can be done by simply (5x10^-3)/(2.5x10^-3) = 2/1 so the ratio is 2:1. This means that two moles of metallic salt is required to react with one mole of sulphite ion. Next you will need to look at the sulphite ion half formula and notice that there are 2 electrons on the right side, so you can deduce the half equation for the metallic salt by using the information given (Original Ox.no of metallic salt is +3). My half equation for metallic salt is: 2X3+ + 2e- ---> 2X2+(I put 2X3+ instead of X3+ here because two moles of metallic salt is required to react with one mole of sulphite ion. I put 2e- on the left because when you mix the two half equations together(half equation of sulphite and metallic salt), the 2e- on both side will be cancelled out. The product contain 2+ ions because the total number of charge on left must be equal to the total number of charge on the right. By calculation: ((2x +3) - 2)=+4(charge on left side), (2 x +2)= +4 so the charges on both side are equal). From the half equation of the metallic salt, you can clearly deduced that the new oxidation state of the metal is +2. So answer is B.

So hard~~ Plus the working is quite long. I am afraid I don't have enough time for this during exam...

My teacher did this this way:

CS^3+Cl2 + Na2Sp3^-2 + H20 -------->2NaCl + CS^+2 SO4^-2
At first CS has 3+ then 2+

 

[Frankuzi]

Wow you're quite good in Chemistry! I couldn't even come that far in the question .... Thanks for the solution.. (Although most of it went over my head? ) Btw Which subjects is it linked from? Only Redox .. or some other as well?

Hmmm~~ Most of it is Redox. Some basic knowledge related to half equations are required as well but I think you should be okay with that right? I know you are good in chemistry too.

 

[Frankuzi]

My teacher did this this way:

CS^3+Cl2 + Na2Sp3^-2 + H20 -------->2NaCl + CS^+2 SO4^-2
At first CS has 3+ then 2+

Thank you for sharing . This working sure save a lot of troubles~

 

[mariamh]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_12.pdf
Q.6. It's all calculation. We are given the value of V=7000cm^3, Temperature=30 degree C and also the mass of O2 gas. First thing you have to do is to change the unit of volume from cm^3 to m^3 so in this case: 7000cm^3/(100 x 100 x 100) = 7x10^-3 m^3 . Next you will need to find the no.of moles of O2 gas by using the mass given so: no.of moles=0.96/2(16) = 0.03moles. Last step is easy you will only substitute these values to the formula given: Pv=nRT. Since the question is asking for the pressure, so I will make P the subject and the formula will look like this now: P=nRT/v and perform the final calculation using the values: n=0.03moles, R=8.31(Given in data booklet), T=30+273 K, v=7x10^-3 m^3. After you obtain the final answer, divide the final answer by 1000(since the pressure is in Kpa) and round it off to 3s.f. I got 10.8Kpa as my final answer which is C.

Q.7. The key words here are at e.q.m, X moles of R were present and the total moles was (2+x/2). So we will perform the calculation step by step. ("<=> here indicates the equilibrium sign cuz I don't know how to do one )
(a) 1P <=> 2Q + 1R
Initial no.of moles of P=1 (Look at the equation, the initial mole is 1 because of the "1"P)
Initial no.of moles of Q and R=0 moles(No product will be formed at the beginning)
At e.q.m,
No. of moles of R= X(given in the question)
No.of moles of Q = (2 x X) = 2X (I multiplied by 2 because of the "2"Q which means the no.of moles of Q is twice the no.of moles of R)
No.of moles of P= initial moles - moles of the product
= 1 - X
total no.of moles at e.q.m= no. of moles of R + no.of moles of Q + no.. of moles of P
= X + 2X + 1 - X
=2X + 1 moles


What method did u use to find the moles of P at equilibrium???

(b)2P <=> 2Q + R
initial moles of P=2(because of the "2"P in the equation)
initial moles of Q and R = 0 moles
at e.q.m,
no.of moles of R = X(given in the question)
no. of moles of Q = 2X (multiplied by 2 because of the "2"Q and this shows that the no.of moles of Q is twice the no.of moles of R)
no.of moles of P = 2 - (2 x X) (this time I minus by 2X because of the "2"P there.)
total no. of moles at e.q.m = X + 2X + 2 - 2X
=2 + X moles

(c)2P <=> Q + R
initial moles of P = 2 moles
initial moles of Q and R = 0 moles
at e.q.m,
no.of moles of R = X(given in the question)
no. of moles of Q = X(it's X because Q and R have the same mole ratio 1:1)
no. of moles of P = 2 - (2 xX) (Need to minus by 2X because of the "2"P)
total no. of moles= X + X + 2- 2X
=2

(d)2P <=>Q + 2R
initial moles of P=2 moles
initial moles of Q and R = 0 moles
at e.q.m,
no .of moles of R = X (this time I don't multiply by 2 even tho there is "2"R because the question stated X moles of R will be formed so here we assumed 2R is equivalent to X moles)
no. of moles of Q= X/2 (the mol ratio of Q:R is 1:2 and which means the no.of moles of R is twice the no.of moles of Q so you need to divide the no.of moles of R(X) by 2)
no.of moles of P = 2 - X(I will only minus 2 by X instead of 2X. Notice that P and R have the same mol ratio and since the no.of moles of R is X moles, so you need to minus the initial moles of P by X only)
total no.of moles = X + (X/2) + 2-X
= 2 + (X/2)
And you compared all the results here the obvious answer here would be D~

 

[mariamh]

Thank you for sharing . This working sure save a lot of troubles~

Hope its rite

 

[syed1995]

Hmmm~~ Most of it is Redox. Some basic knowledge related to half equations are required as well but I think you should be okay with that right? I know you are good in chemistry too.

Guess I need to start revising ASAP then.. I have no clue on how to solve redox questions

My teacher did this this way:

CS^3+Cl2 + Na2Sp3^-2 + H20 -------->2NaCl + CS^+2 SO4^-2
At first CS has 3+ then 2+

Ah Sweet... that does explain quite a bit.. Thanks. btw what happens to H2O?

 

[Frankuzi]

Hi,I found the following qs from may /june /2006 /01 difficult.
Do you know how to do these Qs?
Qs 6, 21, 33
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s06_qp_1.pdf

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

 

[syed1995]

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~

 

[mariamh]

Guess I need to start revising ASAP then.. I have no clue on how to solve redox questions





Ah Sweet... that does explain quite a bit.. Thanks. btw what happens to H2O?

I didnt mention in as we only need metal in the question!!!

 

[mariamh]

October NOvember chemistry 2007 Q 10 Q 26.
October November chemistry 2009 P 12 Q 16, 18, 20,40