Chemistry: Post your doubts here!

[Ahmed Aqdam]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s04_qp_1.pdf

Q3

Mr of NaN3=65
Moles of NaN3=0.77

Moles of N2=0.77*3/2=1.15
Volume=1.15*24=27.7 dm^3

 

[Rockstar RK]

just for the record if i was being rude who was the one who started it:

Hey bro, will u be available here till 10:00pm IST(Indian Standard Time)..... PLEASE BRO!!!!

 

[Suchal Riaz]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s04_qp_1.pdf

Q3

 

[Suchal Riaz]

Hey bro, will u be available here till 10:00pm IST(Indian Standard Time)..... PLEASE BRO!!!!

I will try but i think this community has more harmony and peace without me.

 

[crazytaylorfanXD]

Can someone help me with a few questions from :
JUN 08 QS31
NOV 08 QS 2 , 31
JUN 11 QS 24 , 31

 

[MYLORD]

Mr of NaN3=65
Moles of NaN3=0.77

Moles of N2=0.77*3/2=1.15
Volume=1.15*24=27.7 dm^3

Can u help me with the same year q1

 

[Rockstar RK]

Q25) When heated with chlorine, the hydrocarbon 2,2-dimethylbutane undergoes free radical
substitution.
In a propagation step the free radical X• is formed.
2,2-dimethylbutane+ Cl● → X ●+ HCl

How many different forms of X• are possible?
A 1 B 2 C 3 D 4

 

[Rockstar RK]

Q25) When heated with chlorine, the hydrocarbon 2,2-dimethylbutane undergoes free radical
substitution.
In a propagation step the free radical X• is formed.
2,2-dimethylbutane+ Cl● → X ●+ HCl

How many different forms of X• are possible?
A 1 B 2 C 3 D 4

Suchal Riaz pls solve dis 1

 

[crazytaylorfanXD]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s08_qp_1.pdf
qs 31


http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w08_qp_1.pdf
qs 2 , 31

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s11_qp_11.pdf
qs 24 , 31

 

[Haya Ahmed]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s04_qp_1.pdf

Q18 and Q14

 

[ZaqZainab]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s04_qp_1.pdf

Q18 and Q14

2NaCl+2H2O----> Cl2+H2+ 2NaOH
Mr of NaCl=58.5
Mr of NaOH=40
Mr of Cl2=71
Mr of H2=2
Moles of NaCl=1
so moles of Cl2=0.5
H2 is 0.5
and
NaOH is 1

mass of Cl2=35.5
mass of H2=1
by now we know the answer is A
mass of NaOH is 40

 

[Youssef Tawil]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s13_qp_13.pdf
Number 18??? Answer is A
anyone please?

 

[ZaqZainab]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s04_qp_1.pdf

Q18 and Q14

q14 A because it is the most electro positive and so it going to draw more negative charges than the other 3 choices will.

 

[Metanoia]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_1.pdf
qs 31


http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w08_qp_1.pdf
qs 2 , 31

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s11_qp_11.pdf
qs 24 , 31

Hi, please include the answers for your respective questions to allow us to reply easier.

s08qp1

Q31. Balance the equations below to see which 1 mol of substances require 2.5 mol of O2
C2H5OH + O2 --> CO2 + H2O
C2H2 + O2 --> CO2 + H2O
CH3CHO + O2 --> CO2 + H2O

w08qp1

Q2


Q31.
Statement 1: Justify the strength properties
Statement 2: Justify the lightweight properties

s11qp11

I have uploaded videos on the s11 paper, clarify again if the explanations don't make sense.

 

[ZaqZainab]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s13_qp_13.pdf
Number 18??? Answer is A
anyone please?

ammonium cyanate has ammonia and cynate ions
we know ammonium is NH4+ we are left with CON-
CO(NH2)2

 

[Snowysangel]

can u help me solve this
View attachment 45138 ans is c

Ally PV=nRT where n= 1/18

 

[Ahmed Aqdam]

Can u help me with the same year q1

Moles of H2=0.5
Atoms in one mole=2*0.5*6.02*10^23 (No need to multiply 6.02*10^23 i every step as it is same, just find the ratio)
So here atoms=1

A: 0.5*3=1.5
B: 0.5*5=2.5
C: 1*1=1
D: 1/6*3=0.5

So C has same no of atoms.

 

[NIM]

guyzz need help!!!
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s07_qp_1.pdf
1, 18, 34
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w06_qp_1.pdf
3, 5 (why cant it be D), 11, 14, 21, 37
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s06_qp_1.pdf
1, 3, 10

 

[Metanoia]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s04_qp_1.pdf

Q18 and Q14

s04qp1

Q14. Al3+ is most polarising. Highest charge density will be most polarising (big charge/small radius)

Q18. Use mole calcualtions to solve the amount of products, basing on 58.5 kg of NaCl

2NaCl + 2H2O --> H2 + Cl2 + 2NaOH

 

[NIM]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s04_qp_1.pdf
9, 37 (i didnt get the 2nd and 3rd statement as only one reaction can take place)