Chemistry: Post your doubts here!

[sweet.sugar]

Am i

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w05_qp_1.pdf
Q1-c
i got D cz pb(c2h5)4 +o2 -->pbo + co2
2*4=8 c and 5*4=20h on react side so if we balance i got 8 oxy frm co2 and 10 oxygens from h2o n 1 oxy from pbo total 27 :/
Q4-c (explain it)
Q6-a (h2=hr-h1 so -395+297=-98 so y is it a ?)
Q12-d (i wrote the eq n for mg n s it is 1 mol of o2 n for al it is 3 moles (3o2) so y is the curve for s the highest ie ans shud be c ) :/
Q20-b (plz write the isomers )
Plz help

invisible :/

 

[ZaqZainab]

Am i

invisible :/

Q1
Pb(C2H5)4 +13.5 O2= 8 CO2+10 H2O+PbO
this is the formula what about H2O??
Q4 C as Hydrogen atom is attached to Oxygen atom
Q6
Because of the moles
the enthalpy change is -297 then 1 mole of SO2 is formed but we need 2 moles to be formed and so enthalphy change wil be (-297*2)
and so for the SO3 it will be (-395*2)
now do you your calculations
Q12 with Mg it is 1/2 mole O2 Mg+1/2O2=MgO
with Al it is 3/4 O2 Al+3/2 O2= 1/2 Al2O3
with Sulpur it is 1 mole of O2 S+O2=SO2
Q20
CHCl=CHCl
CH2=CCl2
And a cis trans
for CHCl=CHCl
like in one of them H will be on the top while in another at the bottom for the second carbon

 

[sweet.sugar]

Q1
Pb(C2H5)4 +O2= 8 CO2+10 H2O+PnO3
this is the formula what about H2O??
Q4 C as Hydrogen atom is attached to Oxygen atom
Q6
Because of the moles
the enthalpy change is -297 then 1 mole of SO2 is formed but we need 2 moles to be formed and so enthalphy change wil be (-297*2)
and so for the SO3 it will be (-395*2)
now do you your calculations
Q12 with Mg it is 1/2 mole O2 Mg+1/2O2=MgO
with Al it is 3/4 O2 Al+3/2 O2= 1/2 Al2O3
with Sulpur it is 1 mole of O2 S+O2=SO2
Q20
CHCl=CHCl
CH2=CCl2
And a cis trans
for CHCl=CHCl
like in one of them H will be on the top while in another at the bottom for the second carbon

Yeah thanks tht helped
Bt a small doubt for q-12 u said for sulphur only 1 mol of o2 is used ryt n thts wt i said too bt in the graph it says sulphur used the most moles o o2 as the curve for S is the highest

 

[ZaqZainab]

Yeah thanks tht helped
Bt a small doubt for q-12 u said for sulphur only 1 mol of o2 is used ryt n thts wt i said too bt in the graph it says sulphur used the most moles o o2 as the curve for S is the highest

we have 1/2 3/4 and 1
which one is the most? 1 right?

 

[sweet.sugar]

Q1
Pb(C2H5)4 +O2= 8 CO2+10 H2O+PnO3
this is the formula what about H2O??
Q4 C as Hydrogen atom is attached to Oxygen atom
Q6
Because of the moles
the enthalpy change is -297 then 1 mole of SO2 is formed but we need 2 moles to be formed and so enthalphy change wil be (-297*2)
and so for the SO3 it will be (-395*2)
now do you your calculations
Q12 with Mg it is 1/2 mole O2 Mg+1/2O2=MgO
with Al it is 3/4 O2 Al+3/2 O2= 1/2 Al2O3
with Sulpur it is 1 mole of O2 S+O2=SO2
Q20
CHCl=CHCl
CH2=CCl2
And a cis trans
for CHCl=CHCl
like in one of them H will be on the top while in another at the bottom for the second carbon

Besides im impressed by the explantion of q1 cz there is no suh thing pno3 n i managed to get the ans by myself we divides 27 by 2 n ans is 13.5 cz o2 molecule has a 2 so tht would be something lile 13.5 * 2 = 27
Thx anyway

 

[ZaqZainab]

Besides im impressed by the explantion of q1 cz there is no suh thing pno3 n i managed to get the ans by myself we divides 27 by 2 n ans is 13.5 cz o2 molecule has a 2 so tht would be something lile 13.5 * 2 = 27
Thx anyway

There such thing as called PbO3 it was typo
and i wanted you to get it yourself that's why i stated just the equation

 

[sweet.sugar]

There such thing as called PbO3 it was typo
and i wanted you to get it yourself that's why i stated just the equation

Haha np bt they said lead(ii)oxide which is PbO

 

[ZaqZainab]

Haha np bt they said lead(ii)oxide which is PbO

My apologies on that one though

 

[Youssef Tawil]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s06_qp_1.pdf
Number 10??? answer is 4/3 how????

 

[sweet.sugar]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s06_qp_1.pdf
Number 10??? answer is 4/3 how????

Q10
Initial moles of N2O4 as 1 and that of NO2 as zero.

50% of N2O4 dissociates means that 1 moles of NO2 is formed. So total number of moles = 1.5

PArtial pressure of N2O4 = 0.5/1.5 = 1/3
Partial pressure of NO2= 1/1.5 = 2/3

So Kp = (2/3)^2 divided by (1/3). So 4/3 is the answer.

 

[Sharmin Zinia]

Oct/Nov 2009 P11 Q2. Help needed asap! I know its lame but my head is not working right now :/ please help!

 

[kingo44]

C6H12HOW MANY STEREIOSOMERS ND STRUCTURAL SAY

 

[kingo44]

I

1 aldehydes and ketones actually contain the same no of C and H so its defiantly a yes
2 take a carboxylic acid you decide?
for example ethanoic acid C2H4O2
or just CnH2nO2
C:H= n/2n=1/2
now lets take an ester
methyl ethanote C3H6O2
or just CnH2nO2
C:H= n/2n=1/2

3 alkenes CnH2n (n/2n)
ketones CH3COCH3
C3H6O
CnH2nO
C:H= n/2n

IDIDINT UNDERSTAND POLEASE HELO

 

[ZaqZainab]

I

IDIDINT UNDERSTAND POLEASE HELO

Take examples and then make ratio using the actual general formula may get you confusing

 

[Hadi Murtaza]

Oct/Nov 2009 P11 Q2. Help needed asap! I know its lame but my head is not working right now :/ please help!

Q.2
Unbalanced equation:
R + O₂ -------> CO₂ + H₂O

moles of hydroCarbon R = 0.2 mol
moles of CO₂ = (35.2/44) = 0.8 mol
moles of H₂O = (14.4/18) = 0.8 mol

total atoms of O = (0.8 × 2) + 0.8 [ 2 from CO₂ n 1 from H₂O ]
total atoms of O = 2.4 mol
total moles of O₂ = 1.2 mol

Balanced equation:
0.2R + 1.2O₂ -------> 0.8CO₂ + 0.8H₂O

V need 2 find atoms in 1 moles of R, so divide equation by 0.2
R + 6O₂ -------> 4CO₂ + 4H₂O

Now v can c dat, burning 1 mole of hydroCarbon R gives 4 moles of CO₂, hence 4 moles of C, n 4 moles of H₂O hence 8 moles of H.
So molecular formula = C₄H₈
Answer: D

 

[ZaqZainab]

GCE As and a level
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w12_qp_11.pdf

DOUBTS:

Q1 ----> Calculate the oxidation number before and after
A) Cl2 undergoes oxidation from 0 to 1
B) again Cl2 undergoes oxidation from 0 to 5
C) Cr from 6+ to 3+ (reduction) and Fe from 2+ to 3+
D)Mn from 6 to 3 (only reduction)
B has the biggest oxidation change
Q3 ----> There is not explanation for this buddy i explained this before remember using the same theory here

Q6 ----> use PV=(mass/Mr)RT
103*1000*5.3*10^-3/(8.31*(273+60))=(mass/17)
mass=3.4

Q9 ----> ANSWER A
All the other are wrong
all particles of the particular element don't always have the same mass maybe it is an isotope then they have diffrent
yet again an isotope of one element can have a mass of other element
They can be!

Q11 --> Remember what bond energy is,it is the energy released when a covalent bond of a gaseous molecule is broken to give gaseous atoms and not molecules or compounds or molecules. So in A only is this happening in the rest of the options ∆H not only contains the bond breaking energy but also bond formation so the resulting∆H for B,C and D is lesser than the value of X----Y bond energy !

Q12 --> ANSWER B
water is bent and CO2 is linear
B is correct as both are bent(Sulfur has six valence electrons, plus 14 for two chlorine for a total of 20 electrons. When you construct the Lewis structure you will see two bonds and two lone pairs on sulfur. This will give tetrahedral electron pair geometry and bent molecular geometry. Plus it is bent at a little less than 109.5 degrees. )

Q13 -->
The Sodium (Na) atom is much larger by comparison to the Chlorine (Cl) atom.
The periodic trend of atomic radii gets smaller as you move accross a row. Since Sodium is at the start of the row and Chlorine at the end, Sodium is larger.
This has to due to with the largest shell holding electrons, both Na and Cl's largest shell is the 2S orbital, but since Chlorine has many more protons in its nucleus, those electrons are held more closely than those of Sodium.
On the other hand, when an atom becomes an ion, it either gains or looses an electron (protons dont change). When Sodium becomes an ion, it looses 1 electron to become Na+ and because of this, it looses the electron in its outer shell which is making it so big (ion is smaller than the atom).
When Chlorine becomes an ion, it gains an electron which goes into completing its Octet and the ion is larger than the atom.
When all is said and done, the Chloride ion (Cl-) is much larger than the Sodium ion (Na+).

Q17 --> n = PV/RT
n = (1 x 0.3)/(0.0821 x 298)
n = 0.0123 moles of O2
moles of O^2- ions = 0.0246
moles of metal = 0.0246
molar mass of metal = 1.15/0.0246 = 46.75g.mol
this molar mass is more closer to the molar mass of Ca. So this metal should be Ca and the oxide will be CaO

Q21 --> For a compound to have cis trans

here the second carbon already has CH2CH3 which is C2H5 so you cant have it again if you do then no Cis and trans
Q24 --> ANSWER C
How to get
HO2CCOCH2CO2H using ho acidified KMnO4
If you have the course book check out page 226
go back wards

Q26 --> ANSWER C
F usually doesn't have a free radical
and Cl does do C

Q30 --> ANSWER B
C is correct ans, as C6H12O is an alcohol, so it ll oxidise to carboxylic acid and on reaction with ethanol, ester produce
C6H12O will form when propanoic acid react with entanol

Q31--> ANSWER B
can you mention wha is wrong it is quite a straight forward question

Q38 --> ANSWER B
Regent Y is PCl3
and 1 and 2 are correct

Q40 --> ANSWER D
1- because it has double bond C=C
2- it is for alcohols for Primary and secondary but here we have carboxylic acid
3- it is not an aldehyde it is Carboxylic acid group

 

[GCE As and a level]

GCE As and a level
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w12_qp_11.pdf

DOUBTS:

Q1 ----> Calculate the oxidation number before and after
A) Cl2 undergoes oxidation from 0 to 1
B) again Cl2 undergoes oxidation from 0 to 5
C) Cr from 6+ to 3+ (reduction) and Fe from 2+ to 3+
D)Mn from 6 to 3 (only reduction)
B has the biggest oxidation change
Q3 ----> There is not explanation for this buddy i explained this before remember using the same theory here

Q6 ----> use PV=(mass/Mr)RT
103*1000*5.3*10^-3/(8.31*(273+60))=(mass/17)
mass=3.4

Q9 ----> ANSWER A
All the other are wrong
all particles of the particular element don't always have the same mass maybe it is an isotope then they have diffrent
yet again an isotope of one element can have a mass of other element
They can be!

Q11 --> Remember what bond energy is,it is the energy released when a covalent bond of a gaseous molecule is broken to give gaseous atoms and not molecules or compounds or molecules. So in A only is this happening in the rest of the options ∆H not only contains the bond breaking energy but also bond formation so the resulting∆H for B,C and D is lesser than the value of X----Y bond energy !

Q12 --> ANSWER B
water is bent and CO2 is linear
B is correct as both are bent(Sulfur has six valence electrons, plus 14 for two chlorine for a total of 20 electrons. When you construct the Lewis structure you will see two bonds and two lone pairs on sulfur. This will give tetrahedral electron pair geometry and bent molecular geometry. Plus it is bent at a little less than 109.5 degrees. )

Q13 -->
The Sodium (Na) atom is much larger by comparison to the Chlorine (Cl) atom.
The periodic trend of atomic radii gets smaller as you move accross a row. Since Sodium is at the start of the row and Chlorine at the end, Sodium is larger.
This has to due to with the largest shell holding electrons, both Na and Cl's largest shell is the 2S orbital, but since Chlorine has many more protons in its nucleus, those electrons are held more closely than those of Sodium.
On the other hand, when an atom becomes an ion, it either gains or looses an electron (protons dont change). When Sodium becomes an ion, it looses 1 electron to become Na+ and because of this, it looses the electron in its outer shell which is making it so big (ion is smaller than the atom).
When Chlorine becomes an ion, it gains an electron which goes into completing its Octet and the ion is larger than the atom.
When all is said and done, the Chloride ion (Cl-) is much larger than the Sodium ion (Na+).

Q17 --> n = PV/RT
n = (1 x 0.3)/(0.0821 x 298)
n = 0.0123 moles of O2
moles of O^2- ions = 0.0246
moles of metal = 0.0246
molar mass of metal = 1.15/0.0246 = 46.75g.mol
this molar mass is more closer to the molar mass of Ca. So this metal should be Ca and the oxide will be CaO

Q21 --> For a compound to have cis trans
View attachment 44768
here the second carbon already has CH2CH3 which is C2H5 so you cant have it again if you do then no Cis and trans
Q24 --> ANSWER C
How to get
HO2CCOCH2CO2H using ho acidified KMnO4
If you have the course book check out page 226
go back wards

Q26 --> ANSWER C
F usually doesn't have a free radical
and Cl does do C

Q30 --> ANSWER B
C is correct ans, as C6H12O is an alcohol, so it ll oxidise to carboxylic acid and on reaction with ethanol, ester produce
C6H12O will form when propanoic acid react with entanol

Q31--> ANSWER B
can you mention wha is wrong it is quite a straight forward question

Q38 --> ANSWER B
Regent Y is PCl3
and 1 and 2 are correct

Q40 --> ANSWER D
1- because it has double bond C=C
2- it is for alcohols for Primary and secondary but here we have carboxylic acid
3- it is not an aldehyde it is Carboxylic acid group

THNX ALOT

 

[kruti]

Hi guys,
Can someone please look into the following questions
Paper - June 2011 QP12
Questions - 6,11,15,17,24,29,35,40.

 

[GCE As and a level]

Hi guys,
Can someone please look into the following questions
Paper - June 2011 QP12
Questions - 6,11,15,17,24,29,35,40.

right now i am solving the paper

 

[kruti]

right now i am solving the paper

Alright please let me know how well you do at those questions.