Chem Paper 1!

[Adahshan]

nvm, i got it

 

[Adahshan]

http://www.xtremepapers.me/CIE/Cambridg ... 8_qp_1.pdf

3
8
21
22
23
29
39

Thanks guys, your the best!

 

[xIshtar]

http://www.xtremepapers.me/CIE/Cambridge%20IGCSE/0620%20-%20Chemistry/0620_w08_qp_1.pdf

3
8
21
22
23
29
39

Thanks guys, your the best!

3. I think 3 is B, because lead nitrate forms a precipitate, the same as silver nitrate.
8. D is an ion, as it has more protons than electrons, meaning that it has lost them in a reaction, as an ion is an atom which carries an electrical charge.
21. Litmus paper turns red, and magnesium fizzes. This immediately creates the assumption that it is an acid. With the addition of barium nitrate, a white ppt is formed. This implies that there are sulphate ions present.
22. The answer is D. This is because in the first reaciton, ammonia is produced, which turns red litmus blue. When sulphuric acid is added to a neutral ammonium chloride, there is no reaction, thus no colour change.
23. Definitely A. As I know that magnesium definitely reacts, as does the oxide and carbonate, and thus A is the only acceptable answer.
29. Definitely B. This is because divalent means that there are two electrons in it's outer shell. This means that it is definitely a group 2 metal, and thus forms an ionic compound.
39. B. This is because alkenes take part in addition reactions, as their double bonds are broken.


If you have any further questions, I'd love to help...


Don't forget the thank button is always there guys

 

[xIshtar]

In addition to my explanation for question 3.. Lead (ii) Nitrate, only reacts with halides. Thus a precipitate of lead iodide is formed.

 

[Adahshan]

Bro, Thanks and i Thanked you <3

but i didn't understand number 3

for 29, How do you know that in 29, oxide will react? I'm confused why won't magnesium chloride react?

 

[xIshtar]

Bro, Thanks and i Thanked you <3

but i didn't understand number 3

for 29, How do you know that in 29, oxide will react? I'm confused why won't magnesium chloride react?

Magnesium chloride is a salt, and thus neutral, so in the same type of reaction as that of number 23, it will not react..

As for number 3.. Basically, they can get rid of the iodine ions without affecting the barium ions, by forming an iodide precipitate. And this is done by adding lead (ii) nitrate. If you read through the anion tests in your book, you will find that the observations shown are due to the formation of precipitates. So, as lead ioidide is formed, it is yellow.

If you still don't understand, kindly elaborate on which part, I mean, you may get why sulphuric acid cannot be used instead, etc..

Well, taking if you don't understand it that way, sulphuric acid cannot be used, as a barium salt would be formed, and you wish to keep the barium ions, so the lead nitrate method is the only acceptable way.


And thanks to you too



Oops. Forgot to explain why the oxide will react. Basically, metal oxides are basic, with the exception of Zinc and Aluminium. So, an acid will react with a base, forming a salt and water.

 

[Adahshan]

Ahh, I'm confused as fk for number 3.

Now if we add lead (II) nitrate, it forms Barium nitrate?

 

[xIshtar]

Ahh, I'm confused as fk for number 3.

Now if we add lead (II) nitrate, it forms Barium nitrate?

No, if we add lead (ii) nitrate, it forms a precipitate of lead (ii) iodide. Removing the Iodine ions, and isolating the Barium ions.

 

[Adahshan]

I get it like this:

We want to remove yellow color so basically it's B or D. and we want to remove I and leave Ba ions. So when we add sulphuric acid, It will react giving Barium sulphate and iodide so still a yellow color will be there, so only chance we have is B and that is to add lead (II) nitrate forming lead(II) iodide and barium nitrate. Correct?

 

[Adahshan]

http://www.xtremepapers.me/CIE/Cambridg ... _qp_11.pdf

16
26
please

 

[xIshtar]

I get it like this:

We want to remove yellow color so basically it's B or D. and we want to remove I and leave Ba ions. So when we add sulphuric acid, It will react giving Barium sulphate and iodide so still a yellow color will be there, so only chance we have is B and that is to add lead (II) nitrate forming lead(II) iodide and barium nitrate. Correct?

Yep Exactly.

http://www.xtremepapers.me/CIE/Cambridge%20IGCSE/0620%20-%20Chemistry/0620_s10_qp_11.pdf

16
26
please

16. The rate of reaction in P is slower than Q, but the same volume is produced, so no extra reactants are added.
So, the answer is D.

26. The answer is D. This is because, the more reactive halide displaces the other in the salt. The higher up in group VII, the more reactive, so chlorine would displace the iodine, causing a yellowy brown colour, iodine to form.

 

[Adahshan]

for 16, How bro?

There are 4 choices, could you explain why each is not the right answer? thnx.

 

[xIshtar]

Right, compare each of the choices.. P is slower than Q, as I stated before. So analyse the effect of each option...

If a catalyst was used in P, it would be faster than Q, so it is wrong..
If a higher temperature was used in P, the reaction would again be faster than Q, so it is wrong..
If bigger chips were used in Q, the rate would be slower, so again, it is wrong..
Thus that only leaves D to be acceptable, which comparing the two reactions, it is correct

 

[MohammedNoor]

http://www.xtremepapers.me/CIE/Cambridg ... 8_qp_1.pdf




plz can anyone make me understand the question number 3............plz

 

[xIshtar]

http://www.xtremepapers.me/CIE/Cambridge%20IGCSE/0620%20-%20Chemistry/0620_w08_qp_1.pdf




plz can anyone make me understand the question number 3............plz

My full explanation, you could have found it if you just read back a page..

As for number 3.. Basically, they can get rid of the iodine ions without affecting the barium ions, by forming an iodide precipitate. And this is done by adding lead (ii) nitrate. If you read through the anion tests in your book, you will find that the observations shown are due to the formation of precipitates. So, as lead ioidide is formed, it is yellow.

If you still don't understand, kindly elaborate on which part, I mean, you may get why sulphuric acid cannot be used instead, etc..

Well, taking if you don't understand it that way, sulphuric acid cannot be used, as a barium salt would be formed, and you wish to keep the barium ions, so the lead nitrate method is the only acceptable way.

 

[MohammedNoor]

http://www.xtremepapers.me/CIE/Cambridge%20IGCSE/0620%20-%20Chemistry/0620_w08_qp_1.pdf




plz can anyone make me understand the question number 3............plz

My full explanation, you could have found it if you just read back a page..

As for number 3.. Basically, they can get rid of the iodine ions without affecting the barium ions, by forming an iodide precipitate. And this is done by adding lead (ii) nitrate. If you read through the anion tests in your book, you will find that the observations shown are due to the formation of precipitates. So, as lead ioidide is formed, it is yellow.

If you still don't understand, kindly elaborate on which part, I mean, you may get why sulphuric acid cannot be used instead, etc..



thnx alot!!!!!!!!!!!!!!!!!!!!!!!

Well, taking if you don't understand it that way, sulphuric acid cannot be used, as a barium salt would be formed, and you wish to keep the barium ions, so the lead nitrate method is the only acceptable way.

 

[MohammedNoor]

http://www.xtremepapers.me/CIE/Cambridge%20IGCSE/0620%20-%20Chemistry/0620_w08_qp_1.pdf




plz can anyone make me understand the question number 3............plz

My full explanation, you could have found it if you just read back a page..

As for number 3.. Basically, they can get rid of the iodine ions without affecting the barium ions, by forming an iodide precipitate. And this is done by adding lead (ii) nitrate. If you read through the anion tests in your book, you will find that the observations shown are due to the formation of precipitates. So, as lead ioidide is formed, it is yellow.

If you still don't understand, kindly elaborate on which part, I mean, you may get why sulphuric acid cannot be used instead, etc..

Well, taking if you don't understand it that way, sulphuric acid cannot be used, as a barium salt would be formed, and you wish to keep the barium ions, so the lead nitrate method is the only acceptable way.

thnx alot

 

[MohammedNoor]

http://www.xtremepapers.me/CIE/Cambridg ... 7_qp_1.pdf


in q 3 answer is A but it should be B as i think can anyone help me out



plz

Noor

 

[SmartNour]

pleasee guys itss rely urgentt
this question suckss a lottttttttttttttttttttt
May June 2007 q16 chemistry paper 1 plzz any1 help
hw can we find the answer
i knw the answer is C
but how??? please any1 explain!
http://www.xtremepapers.me/CIE/index.ph ... 7_qp_1.pdf

 

[ajmanutd]

pleasee guys itss rely urgentt
this question suckss a lottttttttttttttttttttt
May June 2007 q16 chemistry paper 1 plzz any1 help
hw can we find the answer
i knw the answer is C
but how??? please any1 explain!
http://www.xtremepapers.me/CIE/index.ph ... 7_qp_1.pdf

this is because 3 is not a redox reaction as for a reaction to be redox, oxidation states must change.sincce all H's are at +1; all S's are at +6, and all O's are at -2 on both sides of the arrow the reaction is not a redox reaction..