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Chemistry: Post your doubts here!

080595kat

080595kat

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Metanoia said:
The "crossing of charges method" does give us the correct answer in this case, so if all else fails, I guess we can use it. However, lets see if we can make us of the other information given.

Understand that whatever moles of C we have at the beginning will eventually be the same number of moles of C in CO2 at the end.

Let x and y be the ratio of Al and C respectively.
AlxCy --->???---> CO2

Working backwards from CO2,
moles of CO2 = 72/24 000 = 0.003 moles = moles of C in AlxCy

Going through the options
A Al 2C3, moles of Al2C3 = 0.144/90 = 0.0016, moles of C = 0.0016 x 3 = 0.0048 (incorrect)
B Al 3C4 , moles of Al3C4 = 0.144/129 = 0.0016, moles of C = 0.0016 x 4 = 0.0045 (incorrect)
C Al 4C3 , moles of Al4C3 = 0.144/144 = 0.001, moles of C = 0.001 x 3 = 0.003 (BINGO!)
D Al 5C3
Click to expand...
Thank you ...:)
 
Metanoia

Metanoia

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$$AK$$ said:
Hi guys
How are u all?
i need help in http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w13_qp_12.pdf
Q8)
Q9)
Q13)
Q16)
Q17)
Q20)
THANK YOU GUYS
Even if u can help me only in few of them , I have NO PROBLEM :D
Click to expand...

w13qp12

Do try to include the answers next to your selected questions, it would make it easier for us to reply. :)

Q8. To find the "heaviest atmosphere", we can use the weighted average of the Mr.
Using option D as example.
Weighted Mr of D
= 0.825 x Mr of H2 + 0.152 x Mr of He + 0.023 x Mr of CH4
=0.825 x 2+ 0.152 x 4 + 0.023 x 16
= 2.63

Work out the other 3 options, the highest one is the most dense.

Q9.

Picture 10.png


Q13.
White precipitate means it forms an insoluble sulfate, which would be barium sulfate.

Q16.
Reaction 1: O.S of sulfur +6 (in H2SO4) --> +6 (in K2SO4), no change
Reaction 2: O.S of sulfur +6 --> +4 (in SO2) , change of 2 units
Reaction 3: O.S of sulfur +6 (in H2SO4) --> -2 (in H2S), change of 8 units

Q17. Ammonium chloride solution is slightly acidic as the NH4+ can donate protons (acidic), so it reacts with the alkaline magnesium hydroxide.

Q20.
1st reaction: oxidation (loss of H)
2nd reaction: nucleophilic addition (the lone pair on OH attracts the slightly positive carbon on the aldehyde)
3rd reaction : oxidation (loss of H)
 
Last edited:
sweet.sugar

sweet.sugar

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Metanoia said:
Q1.

CxHy + O2 --> CO2 + H2O

from info, 10 cm3 of CxHy reacts with 50 cm3 of O2 to produce 30 cm3 of CO2

1CxHy + 5O2 --> 3CO2 + ?H2O

Comparing both sides of equation,

balancing carbon : x =3
balancing oxygen : 10 = 6 + ? , therefore ? = 4

1CxHy + 5O2 --> 3CO2 + 4H2O

Finally, balancing H: y = 8.

Q2.
From 1st reaction , 1 mole of NaN3 produces 1 mole of Na and 1.5 mole of N2.

The 1 mole of Na produced in 1st reaction will be used in the 2nd reaction, producing 0.1 mol of N2.

So total N2 produced = 1.5 (from 1st reaction) + 0.1 (from 2nd reaction) = 1.6 moles

Q13.

Be and Al has what we call a diagonal relationship.

Be2+ has a smaller charge than Al3+, but Be2+ has a smaller radius than Al3+. This causes their charge density (charge/radius) to be similar which in turns allows them to have similar chemical properties.

This also applies to Li and Mg and other pairings..

Google "diagonal relationship" for more details.
http://chemistry.tutorvista.com/inorganic-chemistry/s-block-elements.html

Q14. Unlikely that Ca's IE is higher than Mg, as the electrons of Ca are in a further shell.
Sum of Mg 1st 2 IE = 736 + 1450
Sum of Ca 1st 2 IE = 590 + 1150

Q38.

C2H4 + H2 --> C2H6
C2H4 --> polyethene (polymerisation)
C2H4 + cold KMno4 --> HOCHCHOH (forming diol)
Click to expand...
Thankyou so very much !!!
May God bless you :)
 
ZaqZainab

ZaqZainab

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Youssef Tawil said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_11.pdf
Number 9?? Answer B
Click to expand...
A: initial: 2 moles of P
equilibrium: x moles of R, therefore there must be 2x moles of Q (as the ratio of Q to R is 2:1). P will be 2-x moles (we subtract moles of R from moles of P since they are the same in equilibrium)
Add them up: x + 2x + 2 - x = 2x + 2 so A is incorrect

B: initial: 2 moles of P
equilibrium: x moles of R, therefore there must be 2x moles of Q (as the ratio of Q to R is still 2:1). P will be 2-2x moles (we subtract Q from P this time because P is the same as Q in equilibrium in this equation)
Add them up: x + 2x + 2 - 2x = x + 2 so B is the correct answer
 
Jelleh Belleh

Jelleh Belleh

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$$AK$$

$$AK$$

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Metanoia said:
Do try to include the answers next to your selected questions, it would make it easier for us to reply. :)

Q8. To find the "heaviest atmosphere", we can use the weighted average of the Mr.
Using option D as example.
Weighted Mr of D
= 0.825 x Mr of H2 + 0.152 x Mr of He + 0.023 x Mr of CH4
=0.825 x 2+ 0.152 x 4 + 0.023 x 16
= 2.63

Work out the other 3 options, the highest one is the most dense.

Q9.

View attachment 44395


Q13.
White precipitate means it forms an insoluble sulfate, which would be barium sulfate.

Q16.
Reaction 1: O.S of sulfur +6 (in H2SO4) --> +6 (in K2SO4), no change
Reaction 2: O.S of sulfur +6 --> +4 (in SO2) , change of 2 units
Reaction 3: O.S of sulfur +6 (in H2SO4) --> -2 (in H2S), change of 8 units

Q17. Ammonium chloride solution is slightly acidic as the NH4+ can donate protons (acidic), so it reacts with the alkaline magnesium hydroxide.

Q20.
1st reaction: oxidation (loss of H)
2nd reaction: nucleophilic addition (the lone pair on OH attracts the slightly positive carbon on the aldehyde)
3rd reaction : oxidation (loss of H)
Click to expand...
Thnx man
It was very helpful (y)(y)
Sure,next time i will post the answer :D
 
Metanoia

Metanoia

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IGCSE13 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_12.pdf Q3 ans D ,Q26 ansA , Q31 ansA , Q35 AnsB , Q36 ansD Q37 ansD please help if you can
Click to expand...
w11qp12

Q3.
Cl- : 1s2 2s2 2p6 3s2 3p6
F- : 1s2 2s2 2p6
K+ : 1s2 2s2 2p6 3s2 3p6
Na+ : 1s2 2s2 2p6

We need to focus on F- and Na+ as the electrons removed are from a shell nearer to the nucleus.
Between F- and Na+, Na+ has more protons, so will hold on to its electrons more tightly.

Q26.
C2H5OH --> CH3CHO (note that aldehyde is collected, not acid)

mass of C2H5OH converted = 2.30 x 0.7 = 1.61 g
moles of C2H5OH converted = 1.61/46 = 0.035 mol
moles of CCH3CHO formed = 0.035 mol
mass of CH3CHO formed = 0.035 x 44 = 1.54 g

Q31.
Picture 3.png

Q36.
X is actually N2.
X--> Y--> Z
N2 --> NO --> NO2

Q37. C4H10O , note that the C4H10 is following the general formula CnH2n+2, so it means that the carbon are all single bonded. The oxygen atom would be part of an alcohol group and not ketone/aldehyde.

Since not ketone/aldehyde, we will not get observation 1.
If tertiary alcohol, we will get observation 2.
If primary or secondary alcohol, we will get observation 3.
 
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