Chemistry: Post your doubts here!

[AbbbbY]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s09_qp_1.pdf Q9 A Q20 C

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_12.pdf Q7 D

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_12.pdf Q9 D Q25 C

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w12_qp_13.pdf Q22 B Q25 A

You need explanations for these or are you answering them for someone?

 

[hela]

You need explanations for these or are you answering them for someone?

YES I NEED EXPLANATION PLEASE

 

[hela]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s09_qp_1.pdf Q9 A Q20 C

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_12.pdf Q7 D

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_12.pdf Q9 D Q25 C

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w12_qp_13.pdf Q22 B Q25 A

can anyone explain thanks

 

[rawaaaan]

hello
plz help this this question looks easy but i cant solve
here u go ...Which mass of gas would occupy a volume of 3dm3 at 25°C and 1 atmosphere pressure? [1mol of gas occupies 24dm3 at 25°C and 1 atmosphere pressure.]
a)3.2g O2 gas
b)5.6g N2 gas
c)8.0g SO2 gas
d)8.0g SO2 gas

 

[omarjaved619]

W11_qp11 Q27

A reaction between chlorine and propane in ultraviolet light produces two isomeric
monochloropropanes, C3H7Cl, as products. Which information about this reaction is correct?

Help?

 

[AbbbbY]

hello
plz help this this question looks easy but i cant solve
here u go ...Which mass of gas would occupy a volume of 3dm3 at 25°C and 1 atmosphere pressure? [1mol of gas occupies 24dm3 at 25°C and 1 atmosphere pressure.]
a)3.2g O2 gas
b)5.6g N2 gas
c)8.0g SO2 gas
d)8.0g SO2 gas

Simple!
1 mol : 24 dm3
x mol : 3 dm3
x= 0.125 mol
Next, you simply see which gas has 0.125 moles.
So the answer is C/D [8g of SO2] whichever isn't the incorrectly written 8g of SO2.

 

[AbbbbY]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s09_qp_1.pdf Q9 A Q20 C

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_12.pdf Q7 D

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_12.pdf Q9 D Q25 C

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w12_qp_13.pdf Q22 B Q25 A

ON09
Q9 - Sorry I suck at Electo. Maybe someone else can answer this for you and invariably for me.

MJ12 P12
Q7
P -> <- Q + R
Start: 2 mol : 0 mol : 0 mol
Eq: 2-x : x/2 : x mol

Total moles, 2+x/2
Moles of P and R = (2-x)+x = 2
Moles of Q: 2+(x/2)-2 = x/2

Having done such questions enough times, I can simply deduce the ratio is 2: 1 : 2

However, if you can't, use any value of X under 2 and over 0.

1.5 will work too, but to simplify things, lets take x=1
1 : 1/2 : 1
= 2 : 1 : 2
So 2P -> <- Q + 2R

MJ10 P12
Q9

pH 6 is a very weak acid, so Student P HAS to be wrong
pH 9 is a weak base, but relative to a neutral of 7, it's stronger than the acid.

Change of 1 under neutral pH for the Acid, but change of 2 over the neutral pH for the base. Since concentrations are the same, the base will dissociate more than the acid, thus Student Q is corrent.

Since student Q only is correct, it's D



Sorry, I can't answer the Organic chem questions. I've only done Physical and Inorganic. Maybe someone else here can help you with that

 

[hela]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_12.pdf Q9 D Q25 C
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w12_qp_13.pdf Q22 B Q25 A
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_11.pdf Q14 B Q27 D Q36 A
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_11.pdf Q9 B Q13 C Q18 C Q22A Q38 A

 

[rawaaaan]

hey help in this please
in leaded petrol there i an additive composed of lead,carbn and hydrogen only.this compound contains 29.7% and 6.19% hydrogen by mass.
what is the value of x in the empiricl formula pbc8hx
a)5
b)6
c)16
d) 20

 

[rawaaaan]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w09_qp_11.pdf
question 11,19,20,21,23,28,2,30,36,35,31 i know its too much :/ but plz help

 

[bogus]

mj 2010 question paper 11 question 40. Help!

 

[rawaaaan]

Simple!
1 mol : 24 dm3
x mol : 3 dm3
x= 0.125 mol
Next, you simply see which gas has 0.125 moles.
So the answer is C/D [8g of SO2] whichever isn't the incorrectly written 8g of SO2.

thnx a lot

 

[AbbbbY]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_11.pdf
question 11,19,20,21,23,28,2,30,36,35,31 i know its too much :/ but plz help

2- D!
Two methods. I'll tell you the quicker one.

CxHy -> xCO2 + y/2H2O
35.2/44 = 0.8 mol
14.4/18 = 0.8 mol
0.2 mol -> 0.8 mol + 0.8 mol
= 1 -> 4 : 4

Next, you write up the combustion equations and balance them to see which one falls under the molar ratio.
simply by looking at the ratio I can tell that it has to be CnH2n so it's A or D.

Next, since xCO2 = 4 mol, so C4 has to be the answer. You're left with C4H8.

If this is too hard or confusing, you can write the combustion equations for each and cross check. Will waste a lot of time though.

11 A!

A-> Cu+ is changing to Cu2+ in Reaction two which in turn reacts in 3 to form Cu+ again in the final step, so it's technically unused. Hence it is acting as a catalyst.

B-> Cu+ ions are oxidised so act as reducing agents

C-> Changes w.r.t light are due to Silver atom concentration. It can become dark and light so reaction is reversible, hence A is the reaction absorbing light. Besides, Reaction 2 has no Ag so it's out automatically

D-> Ag+ is reduced to Ag


19, 20, 21, 23, 28, 30, 31
All Organic. I only know Physical and Inorganic. Someone else will answer these for you


35 C
Easy method:

Simple neutralization reaction so Sulfuric acid can never form! 1 is out so Option C is left.

Time breaking method:
CaO + (NH4)2SO4 -> CaSO4 + NH3 + H2O


36: D

Just by reading, eliminate 3 because Only I- is reduced to I2 in such reaction.

NaBr + H2SO4 -> HBr + Na2SO4

2 and 3 are incorrect.

Left with D
[Please check marking scheme if I'm correct].

 

[AbbbbY]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_12.pdf Q9 D Q25 C
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w12_qp_13.pdf Q22 B Q25 A
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_11.pdf Q14 B Q27 D Q36 A
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_11.pdf Q9 B Q13 C Q18 C Q22A Q38 A

Sorry just saw your questions.

ON11 P11
Q14 B

CaCO3 -> CaO + CO2

Mass of Limestone: 1000+200 = 1200m ton
Mol = 1200/100 = 12 mol
1 mol = 44m ton
12 mol = 44*12 mton = 528m tonnes.

I must've rounded off something in the Ar but the closest answer gets it hence B. Do it properly and you'll get it to 527. [Didn't have data booklet]



Q36 A

Nitrogen is the element (Not metallic so not lead)

Oxide = NO
Oxidation = NO2
Verification: NO + 0.5O2 -> NO2

1- Yes. NH3 is formed.
2- N2 exists, so yes.
3- N is the third most polar so defo yes. (NOF)

Hence A
The rest I'll answer later. Too tired.

 

[AbbbbY]

Also, guys, if you have OCRs in your PDF readers and it allows you to copy, please copy the question and put it here. So much quicker to solve that we! More questions can be solved that way and can solve even when short on time.

 

[Omar99]

Re: All Chemistry help here!! Stuck somewhere? Ask here!

tonnes of CHEMISTRY WORKSHHETS
http://www.chemactive.com/a_level_chemi ... tions.html

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s11_qp_12.pdf
question no. 11 and 32

 

[AbbbbY]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s11_qp_12.pdf
question no. 11 and 32

11- Oo that's a tough one!

X is an Al C compound so Al(OH)3 and CH4 will be formed

X -> Al(OH)3 + CH4

CH4 + O2 -> 2H2O + CO2

CH4 : CO2
CO2 moles = 1.7 mol


Fk it man. Can't be this long a solutoin.
Eventually you'd get the answer by making each reaction balacing it's equation and seeing if the ratio of X:CH4 is that of the eqation but they can't take you that long in an MCQ I think it's a trap.

It has to be something simple in that case. My arrows point to Al4C3 because C has 4 outer electrons and Al has 3. Can't be this long.

What's the answer? Maybe someone else can correct me here if I'm wrong.


32:

1- Ideal behavior says it has negligible volume. This negates ideal behavior, hence true. Option C is eliminated.

2- My logic tells me this could be possible but 6000kPa is a measly 60atm and essentially the change is 45atm's. Would be too less to liquify a gas (given that NH3 exists as gas at 200atm). I know it's not an accurate measure. Tough call to rule this one out, but If I had to choose, I'd say it wasn't liquifying.

3- That's a ridiculous statement.

We're left with B and D. I'd pick D because of the pressure-vs-liquify logic I posted up there. But, that's just me. Can't open the marking schemes somehow. What's the answeR?

 

[Student12]

Some Doubts..
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s06_qp_1.pdfhttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s06_ms_1.pdf
Questions 1 , 3 , 10

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s07_qp_1.pdfhttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s07_ms_1.pdf Questions 8 , 9 , 26

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_1.pdfhttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_ms_1.pdf Questions 1, 9 , 10

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_1.pdfhttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_ms_1.pdf Questions 1,2,6,15,31

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w08_qp_1.pdfhttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w08_ms_1.pdfQuestions 1,2,4,30

I know these are aloot in one go.. but i need help! =)

 

[AbbbbY]

Some Doubts..
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s06_qp_1.pdfhttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s06_ms_1.pdf
Questions 1 , 3 , 10

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s07_qp_1.pdfhttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s07_ms_1.pdf Questions 8 , 9 , 26

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_1.pdfhttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_ms_1.pdf Questions 1, 9 , 10

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_1.pdfhttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_ms_1.pdf Questions 1,2,6,15,31

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w08_qp_1.pdfhttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w08_ms_1.pdfQuestions 1,2,4,30

I know these are aloot in one go.. but i need help! =)

MJ 06
1
N2O4 : 2NaOH
0.02 mol : 0.04 mol

conc = mol / vol
0.5 = 0.04 / vol
vol = 0.08 dm3
Hence, D

3-

Nitrogen atom: 1s2 2s2 2p3
Ga3+ means it's losing 3 electrons, so Nitrogen becomes 1s2 2s2 2p6
Hence D

10:

Partial Pressure of Products: (2/3)(2/3)*1
Partial Pressure of Reactants: (1/3)*1

Divide them, you get 1.3 (since I had to do it on windows calc)
so that's 4/3 atm


MJ07

8-

By defination, Standard Enthalpy change of Formation is the energy change associated with the formation of ONE MOLE of a substance from it's constituent elements in standard states.

This reaction is the formation of 2 moles of Fe2O3 so enthalpy change / 2 = enthalpy of formation

Hence B

9-

For simplicity's sake over here, I'll write :
X2 = A
Y2 = B
X2Y = C

Reac 1: [C][C]/[A][A] = 2
Reac 2: [sqrtB][A]/[C]

Resolve them and you should get 1/2 or 1/sqrt2 don't have a paper please check and if you can't figure it out let me know


ON07P1

Q1

1 mol has 6.02x10^23 atoms
Mass of Ag = 0.216g
Mass of Ag per cm3 = 0.216/150 = 0.00144
Moles of Ag per cm3 = 0.00144/108 = 0.0000133
Atoms of Ag per cm3 = 6.02x10^23 * 0.0000133 = 8.023 x 10^13
so, A


The rest, I'll answer in the evening. Gotta rush.

 

[ahmed abdulla]

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