Chemistry: Post your doubts here!

[XPFMember]

..
hope it helps =)

Aoa wr wb!

JazakAllah khairen...It did help a lot..

thnx again ..

 

[namraahmad]

http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_w10_qp_43.pdf Que 10 a (iii) please someone explain???

 

[rehan751]

] hey for the pic im seeing a photo of a frog in an icecube im not seeing anything else?

Oh I don't know its opening fine here but I have attached it with this post also. Try now.

 

[melly713]

Oh I don't know its opening fine here but I have attached it with this post also. Try now.

View attachment 8295

oh ok i see now thanx !

 

[namename]

Can anyone please tell me why the mark scheme for w09 p4 4c) for alkaline aqueous iodine (I2 +OH-) and mark scheme for w10 p42 5b) iv) are contradicting??
Does a phenol have reaction w/ I2+OH- ??

NObody can answer my question?

 

[namename]

Hey please help question b)
I'm quite sure Ecell = Ereduction - Eoxidation
then it becomes -0.83-1.23 = -2.86V and -0.83 -1.36 = -2.19V

However the ms says +2.86V and +2.19V
why is it so?

does Ecell becomes reversed for electrolysis?

Help

 

[saudha]

hi
how can i look for a chiral if there is a double bond ? im only seeing 3 groups?

dont take the double bond C as the chiral centre...... take the second carbon as the chiral centre...the for different groups are.... Br,H,CH=CH2,and CH3

 

[saudha]

The density of ice is 1.00gcm–3
.
What is the volume of steam produced when 1.00 cm3
of ice is heated to 323 °C (596 K) at a
pressure of one atmosphere (101kPa)?
[1mol of a gas occupies 24.0dm3
at 25°C (298K) and one atmosphere.]
A 0.267dm3
B 1.33dm3
C 2.67dm3
D 48.0dm3
please help me somebodyy.....
i cant get the ans.......

 

[leosco1995]

bt its da sme i all...if u c da increase in mols..it increases by 0.25..so y cnt it b A??

They said only 1 mole of the metals was burnt. A is wrong because there would be problems with the vertical scale. The first reading would HAVE to be 0.5, but the other 2 would have to be increased by 0.25.

 

[Nibz]

The density of ice is 1.00gcm–3
.
What is the volume of steam produced when 1.00 cm3
of ice is heated to 323 °C (596 K) at a
pressure of one atmosphere (101kPa)?
[1mol of a gas occupies 24.0dm3
at 25°C (298K) and one atmosphere.]
A 0.267dm3
B 1.33dm3
C 2.67dm3
D 48.0dm3
please help me somebodyy.....
i cant get the ans.......

Mass of 1 cm3 of steam (H20) = 1 g
Mol of H20 = 1g / 18 = 0.0555556
1 mol of H20 occupies 24dm^3 at 298K
0.0556 would occupy = 24 x 0.0556 = 1.33 dm^3
Volume of steam at 298K = 1.33 dm^3
Volume of steam at 1 K = 1.33/298 = 0.00446 dm^3
Volume of steam at 596 K = 0.00446 x 596 = 2.668 dm^3 ~ 2.67 dm^3
Answer C.

 

[saudha]

Mass of 1 cm3 of steam (H20) = 1 g
Mol of H20 = 1g / 18 = 0.0555556
1 mol of H20 occupies 24dm^3 at 298K
0.0556 would occupy = 24 x 0.0556 = 1.33 dm^3
Volume of steam at 298K = 1.33 dm^3
Volume of steam at 1 K = 1.33/298 = 0.00446 dm^3
Volume of steam at 596 K = 0.00446 x 596 = 2.668 dm^3 ~ 2.67 dm^3
Answer C.

thxx a lot Nibz

 

[leosco1995]

The density of ice is 1.00gcm–3
.
What is the volume of steam produced when 1.00 cm3
of ice is heated to 323 °C (596 K) at a
pressure of one atmosphere (101kPa)?
[1mol of a gas occupies 24.0dm3
at 25°C (298K) and one atmosphere.]
A 0.267dm3
B 1.33dm3
C 2.67dm3
D 48.0dm3
please help me somebodyy.....
i cant get the ans.......

PV = nRT
V = nRT/PV

Moles of water used = Mass/Mr = 1/18 = 0.056...

V (in cm^3) = (0.056 * 8.314 * 596) / (101 x 10^3)
V = 2.72 dm^3

So C is right. It's off by a little but the only option close to it is C. Also, remember that you have to use SI units in the formula.

 

[Student12]

http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_s06_qp_1.pdf
Q2 ,10 ,11, 16

http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_s07_qp_1.pdf
8,9

http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_w06_qp_1.pdf
1,4,11,14
Can Someone please EXPLAIN these questions ?

 

[melly713]

oh okay thanx

 

[roadtrip9o9]

http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_s10_qp_12.pdf 3,12,16,27

 

[Nibz]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s06_qp_1.pdf
Q2 ,10 ,11, 16

Q2.
3 peaks.
1 peak - Cl 35 Cl 35
2 peak - Cl 35 Cl 37
3 peak - Cl37 Cl37

Q 10. N2O4 -> 2NO2
Initila pressures : 100% -> 0%
Pressure at 60 : 50% -> 2 (50%)
Mol ratio(pressure/total pressure) = 50/150 -> 100/150
= 1/3 -> 2/3

Kp = (2/3)^2 / (1/3) = 4/3 atm Answer.C


Q.11

Option B is out of the question, because the reaction involving Chloride ions is not reversible.
Option A is correct because it will move the equilibrium position to the right, hence highest conc. of HOCl.

Q.16
Thermal decomposition of Calcium Nitrate:
2Ca(NO3)2 -> 2Cao + 4NO2 + O2
Solid residue in the product side is CaO only. The rest are gases.
Ca(NO3)2 and CaO are in 1:1
Mol of Ca(NO3)2 = 4.10/164.1 = 0.02498
Mol of CaO = 0.02498
Mol = mass/mr
o.02498 = Mass/56.1
= 1.40 g Answer C

 

[Nibz]

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

Q.8 Enthalpy of formation is always for one mole a substance. Since Enthalpy change for 2 moles of Fe2O3 is given. One mol of it would give you the enthalpy change of formation of Fe2O3.
= -1648/2 = -824 Answer B.

9. Involves simple calculations. Try it yourself.

 

[smartangel]

please explain working for Q2.. i dont get it and Q17 http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w08_qp_1.pdf

 

[Nibz]

please explain working for Q2.. i dont get it http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w08_qp_1.pdf

CS2 and O2 are in 1:3, Correct?
10 cm^3 of CS2 has been completely burned. So this would 'eat up' 10x3 = 30 cm'3 Oxygen. Remaining oxygen would be 50-30 = 20 cm'3
Also, 10 cm^3 CS2, when burned, would produce 10cm'3 of CO2. Because both are in 1:1
And 10cm^3 CS2 , when burned, would produce 10x2 = 20cm'3 of SO2. Because the ratio is 1:2

Now, CS2 + 3O2 -> CO2 + 2 SO2
Initally: 10 + 50 -> 0 + 0
Later: 0 + 20 -> 10 + 20
Volume of gas remaining = 20 (O2) + 20(SO2) + 10(CO2) = 50 cm'3

NaOH is an alkali, when added, it would eat up all the 'acidic gases'. Acidic gases present are CO2 and SO2.
So when SO2 and CO2 are eaten up, the only remaining gas would be O2, and it's volume is 20cm^3.

So 50 cm^3 and 20 cm^3 is your answer.

 

[smartangel]

CS2 and O2 are in 1:3, Correct?
10 cm^3 of CS2 has been completely burned. So this would 'eat up' 10x3 = 30 cm'3 Oxygen. Remaining oxygen would be 50-30 = 20 cm'3
Also, 10 cm^3 CS2, when burned, would produce 10cm'3 of CO2. Because both are in 1:1
And 10cm^3 CS2 , when burned, would produce 10x2 = 20cm'3 of SO2. Because the ratio is 1:2

Now, CS2 + 3O2 -> CO2 + 2 SO2
Initally: 10 + 50 -> 0 + 0
Later: 0 + 20 -> 10 + 20
Volume of gas remaining = 20 (O2) + 20(SO2) + 10(CO2) = 50 cm'3

NaOH is an alkali, when added, it would eat up all the 'acidic gases'. Acidic gases present are CO2 and SO2.
So when SO2 and CO2 are eaten up, the only remaining gas would be O2, and it's volume is 20cm^3.

So 50 cm^3 and 20 cm^3 is your answer.

thank you so much! i got it now! please explain Q17 also.