Chemistry: Post your doubts here!

[anastasia grey113]

http://justpastpapers.com/cie-9701-...question-paper-and-mark-scheme/9701_m17_qp_12 in Q39 why would propan-2-0l give yellow ppt with iodine?

it gets oxidised to propanone by NaOH first
this propanone formed gives a positive test

 

[Metallic9896]

wow !!!!! THANKYOU SO MUCH!!

Haha no problem at all, happy to be of help.

 

[darks]

http://justpastpapers.com/cie-9701-...question-paper-and-mark-scheme/9701_m17_qp_12 in Q39 why would propan-2-0l give yellow ppt with iodine?

CH3CHOH/CH3CO branch (Or should i say group?) will always give a yellow ppt.

 

[Zaki ali asghar]

Sorry to post it here but can someone explain why is option A right? for Q7 papers.gceguide.com/A%20Levels/Physics%20(9702)/9702_m16_qp_12.pdf

 

[Metallic9896]

http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_s16_qp_11.pdf
Q4 I dont understand what Np Pu means??
Q6 ??
Q17 ??
Q30??

http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_s16_qp_13.pdf
Q9 ???
Q21 ??
Q33 Statement 2 ; No particles have zero energy at either temperature. ?? What does this mean in context?

Q17: You have to mentally or physically make a note of the equation of hydrochloric acid and calcium carbonate.

2HCl + CaCO3 = CaCl2 + CO2 + H2O

In this question, 36/1000 x 0.5 = 0.018 moles of HCl reacted with CaCO3.

As per equation, 2 moles of HCl reacted with 1 mole of CaCO3. How many moles of CaCO3 react with 0.018 moles of HCl then? That turns out to be, 0.018/2 = 0.009 moles of CaCO3. How much is the mass of 0.009 moles of CaCO3? Moles = mass/Mr. So, 0.009 = mass/Mr of CaCo3 = mass/100. Mass = 100 x 0.009 = 0.9 grams. This rock contains 0.9 g of calcium carbonate, and itself the entire thing weights 2 grams. So % of CaCO3 = 0.9/2 x 100 = 45%. Answer should be B.

 

[Metallic9896]

http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_s16_qp_11.pdf
Q4 I dont understand what Np Pu means??
Q6 ??
Q17 ??
Q30??

http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_s16_qp_13.pdf
Q9 ???
Q21 ??
Q33 Statement 2 ; No particles have zero energy at either temperature. ?? What does this mean in context?

Q30: At first it seems like, "Okay all C=C double bonds means cis trans isomer!" But no. Start by the extreme left C=C. The left C in that C=C has two CH3 groups attached to it, so it doesn't qualify for cis-trans isomerism. The only other C=C is in the middle so that gives us 2 cis-trans isomers. Options C and D are ruled out. Only question now is how many chiral centres. As far as X is concerned, when Br2 is added the C=C will turn into Br-C-C-Br. Draw it out on the diagram yourself. You will find that, from the extreme left, on the main chain, the third C will become a chiral centre, the sixth C will become a chiral centre, and the seventh C will become a chiral centre. Basically, out of the four C involved in the first C=C and the second C=C, the left most WON'T be a chiral centre, while all the other 3 C will be. Answer will be A.

 

[Metallic9896]

http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_s16_qp_11.pdf
Q4 I dont understand what Np Pu means??
Q6 ??
Q17 ??
Q30??

http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_s16_qp_13.pdf
Q9 ???
Q21 ??
Q33 Statement 2 ; No particles have zero energy at either temperature. ?? What does this mean in context?

Now to variant 13.

Q9:

Kc = [products]/[reactants] (each raised to appropriate power from equation)

For this equation:

Kc = [NH3]^2 / [N2] x [H2]^3

They gave you equation, Kc, and conc. of reactants, so:

2 = [NH3]^2 / [2] x [2]^3

2 = [NH3]^2 / 2 x 8

2 x 2 x 8 = [NH3]^2

32 = [NH3]^2

sqrt 32 = [NH3]

Answer should be C.

 

[Metallic9896]

http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_s16_qp_11.pdf
Q4 I dont understand what Np Pu means??
Q6 ??
Q17 ??
Q30??

http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_s16_qp_13.pdf
Q9 ???
Q21 ??
Q33 Statement 2 ; No particles have zero energy at either temperature. ?? What does this mean in context?

Q21:

These are the 3 isomers I drew up. You should be able to do this. Draw it out on the question paper if you have to. The original molecule was CH2=CHCH=CH2, and you had 2 HBr molecules you could add. So they could be added in this fashion (same order as drawn above) (From left most C to right most C):

H then Br, then H then Br
OR H then Br, then Br then H
OR Br then H, then H then Br

(there is no Br then H, then Br then H cause that's the same as the first one, just a mirror image, hence why total 3 isomers).

In these 3 the first 2 contain chiral centres as I've underlined.

 

[Metallic9896]

http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_s16_qp_11.pdf
Q4 I dont understand what Np Pu means??
Q6 ??
Q17 ??
Q30??

http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_s16_qp_13.pdf
Q9 ???
Q21 ??
Q33 Statement 2 ; No particles have zero energy at either temperature. ?? What does this mean in context?

Q33: As anastasiagrey113 said, although going by the graph it looks like there is a o point at the origin. Still, that's what "appears" to be the case like when you see a curve that almost becomes and appears to become horizontal but never actually is horizontal if you zoom it up. Similarly, I believe, you can zoom in as much as you want to the least energy possessing particle but still discover that it has SOME energy in it. The only point when a molecule has no energy whatsoever is absolute O K, for which as Wikipedia says gorgeously, "Absolute zero cannot be achieved". So here you even have some temperature probably much higher than absolute 0, so of course there'll be molecules that have a very low energy, but energy nonetheless. Nothing will have 0 energy.

 

[amina1300]

Q30: At first it seems like, "Okay all C=C double bonds means cis trans isomer!" But no. Start by the extreme left C=C. The left C in that C=C has two CH3 groups attached to it, so it doesn't qualify for cis-trans isomerism. The only other C=C is in the middle so that gives us 2 cis-trans isomers. Options C and D are ruled out. Only question now is how many chiral centres. As far as X is concerned, when Br2 is added the C=C will turn into Br-C-C-Br. Draw it out on the diagram yourself. You will find that, from the extreme left, on the main chain, the third C will become a chiral centre, the sixth C will become a chiral centre, and the seventh C will become a chiral centre. Basically, out of the four C involved in the first C=C and the second C=C, the left most WON'T be a chiral centre, while all the other 3 C will be. Answer will be A.

Thankyou so muchh again!! I have a Q though can the Br be attached to the one end of broken C=C with 2 methyl groups attached (leftmost side) and the other Br at other end of the broken C=C at the same time? [ One as markovnikov's rule product and one not?]

 

[Metallic9896]

Thankyou so muchh again!! I have a Q though can the Br be attached to the one end of broken C=C with 2 methyl groups attached (leftmost side) and the other Br at other end of the broken C=C at the same time? [ One as markovnikov's rule product and one not?]

Wow, nice question. Well that is precisely what happens when you just add Br2 and not HBr. In Br2 since the only atom we have is Br, both Br are attached to the same C=C carbon atoms. However, when we have an HBr, the default programming is that the HBr will split as H+ and Br-, and because C=C is electron-rich and negative, Br- will automatically be repulsed and H+ will attach to one C from C=C, leaving the other and ONLY C open for Br. Even if you have two HBr, the MOMENT the HBr splits, the Br- from both HBr is going to be repelled by the C=C, and when it is able to come back, it's ONLY when one C is occupied by an H, leaving no possibility for 2 Br to attach to the same C=C pair. I hope this made sense. Basically, in HBr addition, Br will only exist as Br-, and Br- ain't got no chance with C=C unless HC already forms. And you're welcome!

 

[Metallic9896]

Thankyou so muchh again!! I have a Q though can the Br be attached to the one end of broken C=C with 2 methyl groups attached (leftmost side) and the other Br at other end of the broken C=C at the same time? [ One as markovnikov's rule product and one not?]

With the Br2 alone, there isn't much of a Markavnikov's rule going. They are all just going to saturate the C=C. And you need not worry about the mechanism if it isn't clear cause it's p1 tomorrow so probably nothing like that will show up.

 

[DeadbeatCIE]

Anyone?

 

[Thelastmoment]

Anyone?
View attachment 62475

KNO3 = 101.1g contains one N
CO(NH2)2 = 60g and 2N

So for it to have same mass of N as in KNO3 the mass of CO(NH2)2 must halve so you have 30g

Hope this helped

 

[amina1300]

Please post the hardest MCQs and Answer (A,B,C,D )from years 2014 and 2015 pleaseeeeeeeee!

 

[SohaibButt]

H always has a +1 state
so here it will be +1 too

n also recall the Cl reaction with NaOH
it makes NaClO right?
Cl has +1 state n Na has +1 too

HClO is a compound just like it
Cl reacts with water to make HClO in which Cl has +1 oxidation state n H also has +1
this can be found in the bleaching properties of Cl section
bleaching is in our syllabus

Exactly!

 

[SohaibButt]

Anyone?

 

[amina1300]

It was tough paper *sigh*

Can you tell the answers too plz.

25 B?
The one with Si02 and CO2 --> A?
Q32 D?

 

[Brisingr]

7 What is the volume of steam produced when 1.00 g of ice is heated to 323°C at a pressure of
101 kPa?
A 0.27 dm3 B 1.3 dm3 C 2.7 dm3 D 48 dm3
anyone?how to do this qs i use pv=nrt but still ans not matched.ans is c

 

[amina1300]

7 What is the volume of steam produced when 1.00 g of ice is heated to 323°C at a pressure of
101 kPa?
A 0.27 dm3 B 1.3 dm3 C 2.7 dm3 D 48 dm3
anyone?how to do this qs i use pv=nrt but still ans not matched.ans is c

Mass of 1 cm3 of steam (H20) = 1 g
Mol of H20 = 1g / 18 = 0.0555556
1 mol of H20 occupies 24dm^3 at 298K
0.0556 would occupy = 24 x 0.0556 = 1.33 dm^3
Volume of steam at 298K = 1.33 dm^3
Volume of steam at 1 K = 1.33/298 = 0.00446 dm^3
Volume of steam at 596 K = 0.00446 x 596 = 2.668 dm^3 ~ 2.67 dm^3