Chemistry: Post your doubts here!

[Youssef Tawil]

Q39 the ans is C
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_12.pdf

Draw the 3 displayed formulas , this will help
number 1, carbon number 2 has a methyl group and a hydrogen so it can't be 2,2-methylpentane
number2, carbon number 2 has 2 methyl groups attached and a ch3 at the end
number 3, carbon number 2 has 2 methyl groups attached and a ch3 at the end
so 2 and 3 are correct

 

[papajohn]

Hi guys
How are u all?
i need help in http://papers.xtremepapers.com/CIE/Cambridge International A and 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

For Q8, I recommend u to ask Thought blocker, because he is from Saturn

 

[papajohn]

Draw the 3 displayed formulas , this will help
number 1, carbon number 2 has a methyl group and a hydrogen so it can't be 2,2-methylpentane
number2, carbon number 2 has 2 methyl groups attached and a ch3 at the end
number 3, carbon number 2 has 2 methyl groups attached and a ch3 at the end
so 2 and 3 are correct

Option#1 is 1,1 dimethypentane,right?

 

[Youssef Tawil]

Option#1 is 1,1 dimethypentane,right?

YES

 

[Metanoia]

Can anyone help me with this question?
11 0.144 g of an aluminium compound X react with an excess of water, to produce a gas. This gas
burns completely in O2 to form H2O and 72 cm3 of CO2 only. The volume of CO2 was measured at
room temperature and pressure.
What could be the formula of X?
[C = 12.0, Al = 27.0; 1 mole of any gas occupies 24 dm3 at room temperature and pressure]
A Al 2C3 B Al 3C4 C Al 4C3 D Al 5C3

It can be done by common sense too
see in options you can see only Al and C so what will happen when they react :
View attachment 44371

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

 

[$$AK$$]

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

 

[ALFRED DEI AMOEH]

hello, can anyone help me with multiple choice question 1,8 and 18 of 9701/01/M/J/04

 

[papajohn]

hello, can anyone help me with multiple choice question 1,8 and 18 of 9701/01/M/J/04

Find the number of moles then multiply by the number of atom it contain, to get the moles of each atom.
Q1: A n= 22/44* 3 = 1.5 mol
B- n=8/16 * 5 = 2.5 mol
C- n=20/20 = 1 mol ( it contain the same number of atoms as hydrogen)
D- n= 8/48* 3= 0.5 mol

 

[$$AK$$]

No one

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

 

[080595kat]

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

Thank you ...

 

[$$AK$$]

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)

 

[Metanoia]

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

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.




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)

 

[GCE As and a level]

YouTube Videos on Past Chemistry MCQs
GUYS CHECK THIS ^^

 

[aliciaa]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w08_qp_1.pdf
Help with 39 please! Answer is B, which is I & II.

 

[Youssef Tawil]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s12_qp_11.pdf
Number 9?? Answer B

 

[IGCSE13]

http://papers.xtremepapers.com/CIE/... 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

 

[sweet.sugar]

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)

Thankyou so very much !!!
May God bless you

 

[pikachu]

Could I confirm the date of the chemistry paper?

june 2013

 

[ZaqZainab]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_11.pdf
Number 9?? Answer B

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

 

[Youssef Tawil]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s12_qp_11.pdf
Number 9?? Answer B
Anyone please??