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Chemistry: Post your doubts here!

B

Browny

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http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w02_qp_1.pdf

Can anybody please explain questions 5,13,23,36 and 39?

The problems are as follows:

In 5 why can't A be correct as it is symmetrical.
In 13 why can't A(Mg) be the answer as when it dissolves the pH is around 6.5 which suggests hydrolysis.
In 23 why can't D(180) be the answer as the chain is like -C-C-C-C-C- which shows the C-C-C bond to be 180.
In 36 why is 2(NH3 behaves as a base) is correct.
In 39, actually I don't get a single point!

Thanks in advance and please try these questions out if possible as they are confusing and they might even help you out!:)
 
I

IGCSE13

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Metanoia said:
Q10. Graphite is made up of sheets of hexagonal rings, the angle among the carbon atoms are 120 degrees.

Q18.
A. (Wrong) Br2 is non-polar molecule (ID-ID intermolecular attractions), should be soluble in non-polar solvents.
B. (Correct) Br2 molecule mass is 80 x 2 = 160, which is heavier than air (Mr approximately 30).
C. (Wrong) Br2 vapourise easily (low boiling point) due to ID-ID intermolecular attraction.
D. (Wrong) Gaseous bromine is reddish brown.

Q20. Questions like these are quite time consuming if many atoms are involved. Might be more sensible to spend time on other questions first during the actual exam.

View attachment 44361

Q25. First 3 reactions require heating. The 4th one is reaction of 2,4-DNPH with carbonyl, which can be carried out at room temperature.

38.

I seemed to have 2 and 3 as the answer. The answer provided was A?
Click to expand...
Sorry you are right the answer to 38 is C but in Q20 there are 3 carbon atoms not 4
 
S

Snowysangel

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Browny said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w02_qp_1.pdf

Can anybody please explain questions 5,13,23,36 and 39?

The problems are as follows:

In 5 why can't A be correct as it is symmetrical.
In 13 why can't A(Mg) be the answer as when it dissolves the pH is around 6.5 which suggests hydrolysis.
In 23 why can't D(180) be the answer as the chain is like -C-C-C-C-C- which shows the C-C-C bond to be 180.
In 36 why is 2(NH3 behaves as a base) is correct.
In 39, actually I don't get a single point!

Thanks in advance and please try these questions out if possible as they are confusing and they might even help you out!:)
Click to expand...
Is 39 C?
 
Metanoia

Metanoia

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Browny said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w02_qp_1.pdf

Can anybody please explain questions 5,13,23,36 and 39?

The problems are as follows:

In 5 why can't A be correct as it is symmetrical.
In 13 why can't A(Mg) be the answer as when it dissolves the pH is around 6.5 which suggests hydrolysis.
In 23 why can't D(180) be the answer as the chain is like -C-C-C-C-C- which shows the C-C-C bond to be 180.
In 36 why is 2(NH3 behaves as a base) is correct.
In 39, actually I don't get a single point!

Thanks in advance and please try these questions out if possible as they are confusing and they might even help you out!:)
Click to expand...

Q5. The molecule is not symmetrical, you have to view it in its 3D shape, it is tetrahedral.

Q6. MgCl2 dissolves, SiCl4 hydrolyses.

Q23. Each carbon has 4 single bonds, they are tetrahedral shape around each carbon.

Q36. NH3 accepts a H+ to form NH4+, it is a proton acceptor (base)

Q39.
H2SO4 (acid) donates a proton to C2H5OH.
 
080595kat

080595kat

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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
 
♣♠ Magnanimous ♣♠

♣♠ Magnanimous ♣♠

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080595kat said:
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
Click to expand...
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 :
Untitled.png
 
Metanoia

Metanoia

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Hi all,

in view of the lack of solutions for past year MCQ papers, I've attempted to create a youtube playlist to explain the answers.

It is a long and laborious process, it is not possible for me to cover a large number of practice papers before your exams on 10 June, I view it as more of a long term project for future batches.

I'm sure that there will be mistakes in my videos as they are done in 1 long sitting, so feel free to point out any mistakes or clarify concepts.

If the videos prove to be useful, I will slowly add on to them in the future.

Here is the link to the June 2013 P11.
http://www.youtube.com/playlist?list=PL4Jmce8VJnNXQDPOYEr7j99fO9nIy55eh

I'll create start a thread elsewhere so the video links can be neatly maintained over time.
 
$$AK$$

$$AK$$

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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
Metanoia said:
Hi all,

in view of the lack of solutions for past year MCQ papers, I've attempted to create a youtube playlist to explain the answers.

It is a long and laborious process, it is not possible for me to cover a large number of practice papers before your exams on 10 June, I view it as more of a long term project for future batches.

I'm sure that there will be mistakes in my videos as they are done in 1 long sitting, so feel free to point out any mistakes or clarify concepts.

If the videos prove to be useful, I will slowly add on to them in the future.

Here is the link to the June 2013 P11.
http://www.youtube.com/playlist?list=PL4Jmce8VJnNXQDPOYEr7j99fO9nIy55eh

I'll create start a thread elsewhere so the video links can be neatly maintained over time.
Click to expand...
 
Youssef Tawil

Youssef Tawil

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papajohn

papajohn

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Youssef Tawil said:
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
Click to expand...
Option#1 is 1,1 dimethypentane,right?
 
Metanoia

Metanoia

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080595kat said:
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
Click to expand...

♣♠ Magnanimous ♣♠ said:
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
Click to expand...

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
 
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