Chemistry: Post your doubts here!

[Xaptor16]

http://cieoandalevelnotes.blogspot.in/2015/05/chemistry-level-paper-5-p5-solved-past.html Follow the link in that website ... it has solved chemistry P5 ... might help you out.

yes i've visited that site thankyou but i still need a bit of help with the specific question i mentioned

 

[Copy Cat]

(iV) part?

 

[Wolf fangs]

Those whom have given AS before already. Are you guys gonna read AS material again for A 2 session?

 

[Konstantino Nikolas]

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the balancing of the equation in c(ii) pls with the working

 

[Konstantino Nikolas]

Those whom have given AS before already. Are you guys gonna read AS material again for A 2 session?

You kinda have ... especially Organic chem, group 2,17 and periodic table, and well, the basics about bond angle, shape, orbitals and stuff ... they ask everything
There was this paper 4 ... w13 i guess ... it had soo much of AS in it

 

[qwertypoiu]

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the ratio is 9:6:1 but how does that come about?

There are two types of Cl: Cl35 and Cl37

The relative abundance of Cl is 3:1. So Cl35 occurs three times as much as Cl37.

Now when there are two Cl atoms in a molecule, the combinations (permutations??) possible are as follows:

1) Cl35 Cl35
2) Cl35 Cl37 (same as Cl37 Cl35)
3) Cl37 Cl37

The chances of no. 1 is 3*3 = 9
The chances of no. 2 is 3*1 + 1*3 = 6
The chances of no. 3 is 1*1 = 1

Hope that makes sense

 

[Wolf fangs]

You kinda have ... especially Organic chem, group 2,17 and periodic table, and well, the basics about bond angle, shape, orbitals and stuff ... they ask everything
There was this paper 4 ... w13 i guess ... it had soo much of AS in it

Thank you so much.
I ll check the paper right away.
And Can you tell we have to revise for Other subjects? Like physics mathematics and biology?

 

[qwertypoiu]

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part (iii) pls

(ii)
Concentration of X in hexane = 0.4g/20cm3 = 0.02
Concentration of X in water = 0.1g / 100cm3 = 0.001

Kpc = 0.02/0.001 = 20

(iii)
Kpc is now fixed.
Kpc between hexane and water = 20

So step 1:
Adding 10cm3 of hexane to [100cm3 water with 0.5g X in it].
If x is the mass of X removed:
Concentration of X in hexane = x/10
Concentration of X in water = (0.5-x)/100

Kpc = x/10 ÷ (0.5-x)/100 = 10x / (0.5 - x)
20 = 10x/(0.5-x)
10 - 20x = 10x
x = 1/3 = 0.333g

So in the first step, 0.333g of X was extracted by hexane, leaving 0.1667g in water.

Now step 2:
Adding 10cm3 of hexane to [100cm3 water with 0.1667g X in it].
If x is the mass of X removed:
Concentration of X in hexane = x/10
Concentration of X in water = (0.1667-x)/100

Kpc = x/10 ÷ (0.1667-x)/100 = 10x / (0.1667 - x)
20 = 10x/(0.1667-x)
3.333 - 20x = 10x
x = 1/9 = 0.1111g

So this time the hexane removed 0.111g of the organic substance X from water for us.

In total, the amount extracted = 0.3333g + 0.11111g = 0.4444g

What is the conclusion??

If, someday, you have some organic substance stuck inside water and you wanna extract it using hexane, trying to remove it all in one go will extract LESS, compared to extracting step by step (despite using the same amount of hexane overall)

In this case, using 20cm3 of hexane extracted 0.4g, but extracting it 10cm3 then 10cm3 made you extract more (0.4444g)
In fact, the more 'splits' you do, the more you will extract (eg. 5cm3, 5cm3, 5cm3, 5cm3)

Of course, there is a limit to how many splits you can do as it wastes time so in an industry process they look for the optimum amount.

 

[Konstantino Nikolas]

Thank you so much.
I ll check the paper right away.
And Can you tell we have to revise for Other subjects? Like physics mathematics and biology?

In math, even if there's anything from AS, it will be something you need for solving A2 problems. For example, trigonometry graphs and solving basics are needed from AS but I don't think you need to revise AS for that. In one p3 question, we had to apply sequence formula ... but even that formula was given in formula sheet. So just remember that sometimes, you might have to think in the AS way.
For physics, you might just want to go through formulae and definitions ... and also calculations of uncertainties, taking measurements, units, suffixes ...
Sorry, I don't know nothing about Bio bruh.

 

[qwertypoiu]

View attachment 59299

how is 81Br+ one species and 79Br+ another??

I am not sure what is supposed to be surprising. Is it the plus charge?
The plus charge is there because when Br2 enters the ionization chamber, electrons are knocked off from it to form Br2 +. (this is two Br atoms with 1 plus charge)

Br2 + is not particularly stable. Some of it will fall apart to give a bromine atom and a Br+ ion. This is called fragmentation.
The Br+ ion can be attracted by a negatively charged plate, but the Br atom cannot be attracted. It would either be ionized first (in the chamber) and THEN be attracted, or it will simply not be attracted and so it would not be detected.

There are two isotopes of bromine: Br79, and Br81. This is the reason for the two species you mention.

 

[qwertypoiu]

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View attachment 59304

Firstly, they told us that benzene is planar and cyclohexane CANNOT be planar.
Secondly, they told us butane can be planar but methylpropane cannot. This means that in a linear chain, if the carbon atoms are joined to maximum two other carbon atoms, they can be made to be planar (as in the case of butane). But the moment one carbon atom is joined to three carbons, it'll have to be tetrahedral. (as in the case of methylpropane)

A is made of benzene rings only. Obviously it would be planar.

B contains cyclohexane. Obviously it CANNOT be planar.

C looks like it has tetrahedral structure, but the carbon atoms CAN be arranged in a planar form, since each of the carbons on the side is only attached to two other carbon atoms of the methyl groups. So yes, it can be planar.

D is linear and again obeys the rule about having less than two carbons attached to each carbon, so it can be linear.

E again is a benzene ring with two carbons attached on the sides; those two carbon atoms are not bonded to three Cs so yeah looks all great and can be planar.

Note that I made up that rule about 2C's based on the information given in the question. It may not be strictly correct but I got the right answer from this line of thinking so I decided to share it here. If anyone has a better answer they should tell us

 

[qwertypoiu]

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View attachment 59301

the table 2nd and 3rd row ?? why is a=1 in the 2nd??

Step 1 is slowest: So reaction rate depends on concentration of H2O2 and I-, both one mole, so a=1, b=1, c=0 (no dependence on H+)

Step 2 is slowest: reaction rate is dependent on IO- and H+. Now we have the H+ in the rate equation, so we can set c=1, but where is IO-?? How do we take that to account? Why isn't it even in the rate reaction? Because it is an INTERMEDIARY!.
You must always account for intermediaries in slow steps by tracing it back and seeing what produced it.
So our reaction depends on IO-, and IO- is generated by step1, so IO- depends on H2O2 and I-.
So we depend on three things: H2O2, I-, and H+. Each is 1 mole. So a=1, b=1, c=1.

Step 3 is slowest. Same method as above.
Dependent on HOI, H+ and I-.
HOI was made by IO- and H+, so lets replace the sentence above:
IO-, H+, H+, I-

Replace IO- by what produced it: (H2O2 and I-):

H2O2, I-, H+, H+, I-

So you can see there are two moles of H+ being dependent upon, and two moles of I-, so b=2, c=2, and a =1.

 

[a_wiserME!!]

Step 1 is slowest: So reaction rate depends on concentration of H2O2 and I-, both one mole, so a=1, b=1, c=0 (no dependence on H+)

Step 2 is slowest: reaction rate is dependent on IO- and H+. Now we have the H+ in the rate equation, so we can set c=1, but where is IO-?? How do we take that to account? Why isn't it even in the rate reaction? Because it is an INTERMEDIARY!.
You must always account for intermediaries in slow steps by tracing it back and seeing what produced it.
So our reaction depends on IO-, and IO- is generated by step1, so IO- depends on H2O2 and I-.
So we depend on three things: H2O2, I-, and H+. Each is 1 mole. So a=1, b=1, c=1.

Step 3 is slowest. Same method as above.
Dependent on HOI, H+ and I-.
HOI was made by IO- and H+, so lets replace the sentence above:
IO-, H+, H+, I-

Replace IO- by what produced it: (H2O2 and I-):

H2O2, I-, H+, H+, I-

So you can see there are two moles of H+ being dependent upon, and two moles of I-, so b=2, c=2, and a =1.

thank you so much for a really clear explanation!

 

[Salonee B]

Hello! I have doubts in question 1 of this paper

http://studyguide.pk/Past Papers/CIE/International A And AS Level/9701 - Chemistry/9701_w04_qp_4.pdf

A) Why is the concentration of H2SO4 halved in q1 a (ii)??
Mark scheme: (pH = 0.70) ⇒ [H+
] = 10-0.7 = 0.20 (mol dm-3) [1]
∴ [H2
SO4] = (0.10 mol dm-3)


B) I dont understand how to solve q1 a (iii)
Mark scheme : (end point is at 34.0 cm3
(± 0.5 cm3),
so amount of H+ used = 0.2 x 25/1000 = 0.0050 mol ecf from (ii) [1]
moles of guanidine = moles of H+ = 0.0050 mol
[guanidine] = 0.005 x 1000/34.0 = 0.147 (mol dm-3) [1]


please if anybody could explain I would be very very thankful to you!!

 

[The Sarcastic Retard]

Hello! I have doubts in question 1 of this paper

http://studyguide.pk/Past Papers/CIE/International A And AS Level/9701 - Chemistry/9701_w04_qp_4.pdf

A) Why is the concentration of H2SO4 halved in q1 a (ii)??
Mark scheme: (pH = 0.70) ⇒ [H+
] = 10-0.7 = 0.20 (mol dm-3) [1]
∴ [H2
SO4] = (0.10 mol dm-3)


B) I dont understand how to solve q1 a (iii)
Mark scheme : (end point is at 34.0 cm3
(± 0.5 cm3),
so amount of H+ used = 0.2 x 25/1000 = 0.0050 mol ecf from (ii) [1]
moles of guanidine = moles of H+ = 0.0050 mol
[guanidine] = 0.005 x 1000/34.0 = 0.147 (mol dm-3) [1]


please if anybody could explain I would be very very thankful to you!!

1a)ii)
pH = -log[H+]
0.7 = -log[H+]
-0.7 = log[H+]
10^-0.7 = [H+]
[H+] = 0.2 moldm^-3
Ratio of H2SO4 and H+ is 1 : 2
so 1 mol of H2SO4 ----> 2 mol of H+
so, how many moles of H2SO4 will be produce by 0.2 mol of H+ ? ---> crossmultiply and it will be 0.2/2 = 0.1 mol hence its halved.

iii) amount of H+ used = 0.2 * 25 * 10^-3 = 0.005 mol (When a 25.0 cm3 sample of dilute sulphuric acid was titrated against a solution of guanidine, the following titration curve was obtained.)
so now moles of B = moles of H+ used as ratio is 1 : 1 = 0.005 mol
So concentration of B = 0.005 / (34 * 10^-3) = 0.147 moldm^-3 (From graph u can read that end point is at 34cm^3)

 

[Salonee B]

1a)ii)
pH = -log[H+]
0.7 = -log[H+]
-0.7 = log[H+]
10^-0.7 = [H+]
[H+] = 0.2 moldm^-3
Ratio of H2SO4 and H+ is 1 : 2
so 1 mol of H2SO4 ----> 2 mol of H+
so, how many moles of H2SO4 will be produce by 0.2 mol of H+ ? ---> crossmultiply and it will be 0.2/2 = 0.1 mol hence its halved.

iii) amount of H+ used = 0.2 * 25 * 10^-3 = 0.005 mol (When a 25.0 cm3 sample of dilute sulphuric acid was titrated against a solution of guanidine, the following titration curve was obtained.)
so now moles of B = moles of H+ used as ratio is 1 : 1 = 0.005 mol
So concentration of B = 0.005 / (34 * 10^-3) = 0.147 moldm^-3 (From graph u can read that end point is at 34cm^3)

thank you so much for that clear explanation!! really helped

 

[The Sarcastic Retard]

thank you so much for that clear explanation!! really helped

^_^

 

[Eugene99]

 

[Catherine_1]

Does anyone have some good resources for NMR and infrared spectroscopy, please?

 

[Saad the Paki]

Basically we have to do this for all the options and check if the mass matches the one given in the question