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Author Topic: CHEMISTRY PAPER 1 HERE  (Read 51114 times)

Offline Bani

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Re: CHEMISTRY PAPER 1 HERE
« Reply #240 on: June 11, 2009, 05:34:47 pm »
please someone answer my previous question and this one too: Q 24 M/J 2007  ???
for Q24, how do we know that chlorine would displace potassium from potassium bromide and iodine won't
Please help...
Thanks in advance
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Offline girl_92

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Re: CHEMISTRY PAPER 1 HERE
« Reply #241 on: June 11, 2009, 05:36:41 pm »
Quote from: Bani on June 11, 2009, 05:34:47 pm
please someone answer my previous question and this one too: Q 24 M/J 2007  ???
for Q24, how do we know that chlorine would displace potassium from potassium bromide and iodine won't
Please help...
Thanks in advance
reactivity decreases down the group in grp7
a more reactive halogen displaces a less reactive one 4rm its compound
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Re: CHEMISTRY PAPER 1 HERE
« Reply #242 on: June 11, 2009, 05:37:06 pm »
Quote from: Bani on June 11, 2009, 05:34:47 pm
please someone answer my previous question and this one too: Q 24 M/J 2007  ???
for Q24, how do we know that chlorine would displace potassium from potassium bromide and iodine won't
Please help...
Thanks in advance
Chlorine is more reactive than bromine, so it will displace bromine from its compounds. Iodine is less reactive so won't displace bromine.
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Offline sweetsh

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Re: CHEMISTRY PAPER 1 HERE
« Reply #243 on: June 11, 2009, 05:39:37 pm »
According to the halogens..
F
Cl
Br
I
At
So the KBr can react with Cl making a displacement rxn but the Kbr cant react with I because I is less reactive than Br..
So the answer is B
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Offline AS girl

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Re: CHEMISTRY PAPER 1 HERE
« Reply #244 on: June 11, 2009, 05:43:45 pm »
f is the most reactive halogen and so can displace all the others...as u go down reactivity decreases

F
Cl
Br
I
At

so Cl would be able to displace all the halogens which are below them but not above and so on!
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Offline Bani

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Re: CHEMISTRY PAPER 1 HERE
« Reply #245 on: June 11, 2009, 05:47:16 pm »
thanks a lot everyone!
someone please try answering my previous question (M/J 2007, Q16-----the one about redox reaction) Why is the answer C?
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Offline sweetest angel

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Re: CHEMISTRY PAPER 1 HERE
« Reply #246 on: June 11, 2009, 05:51:43 pm »
Quote from: Bani on June 11, 2009, 05:47:16 pm
thanks a lot everyone!
someone please try answering my previous question (M/J 2007, Q16-----the one about redox reaction) Why is the answer C?
since in equation 3 no reactants have lost or gained electrons, oxygen or hydrogen!
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Offline Bani

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Re: CHEMISTRY PAPER 1 HERE
« Reply #247 on: June 11, 2009, 06:00:38 pm »
Quote from: sweetest angel on June 11, 2009, 05:51:43 pm
Quote from: Bani on June 11, 2009, 05:47:16 pm
thanks a lot everyone!
someone please try answering my previous question (M/J 2007, Q16-----the one about redox reaction) Why is the answer C?
since in equation 3 no reactants have lost or gained electrons, oxygen or hydrogen!

Ya,but why is equation 1 and 2 REDOX?
1 and 2 are only oxidation, not reduction...
please help...
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Offline sweetest angel

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Re: CHEMISTRY PAPER 1 HERE
« Reply #248 on: June 11, 2009, 06:06:16 pm »
Quote from: Bani on June 11, 2009, 06:00:38 pm
Quote from: sweetest angel on June 11, 2009, 05:51:43 pm
Quote from: Bani on June 11, 2009, 05:47:16 pm
thanks a lot everyone!
someone please try answering my previous question (M/J 2007, Q16-----the one about redox reaction) Why is the answer C?
since in equation 3 no reactants have lost or gained electrons, oxygen or hydrogen!

Ya,but why is equation 1 and 2 REDOX?
1 and 2 are only oxidation, not reduction...
please help...
i dnt quite get it but i guess since 1 and 2 r reversible reactions they r oxidised in forward and reduced in backward rit? :-\
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Offline Ghost Of Highbury

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Re: CHEMISTRY PAPER 1 HERE
« Reply #249 on: June 11, 2009, 06:15:43 pm »
S has many oxidation states including -2,0,2,4,6...
in this particular reaction the oxidation state of S is "+2"
a decrease in oxidation state - reduction
an increase in oxidation state - oxidation

1.) S + O2 -> SO2
     here at the left handside, as sulphur exists itself (not compound), the oxidation state is 0
          at the right hand sidde it forms a compound and gains an oxidation state of +2 = this is oxidation
          whereas O has an oxidation state of "0" in the left hand side and decreases to -4 in the right hand side
               this is reduction

therefore its redox
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nid404

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Re: CHEMISTRY PAPER 1 HERE
« Reply #250 on: June 11, 2009, 06:16:31 pm »
Quote from: Bani on June 11, 2009, 06:00:38 pm
Quote from: sweetest angel on June 11, 2009, 05:51:43 pm
Quote from: Bani on June 11, 2009, 05:47:16 pm
thanks a lot everyone!
someone please try answering my previous question (M/J 2007, Q16-----the one about redox reaction) Why is the answer C?
since in equation 3 no reactants have lost or gained electrons, oxygen or hydrogen!

Ya,but why is equation 1 and 2 REDOX?
1 and 2 are only oxidation, not reduction...
please help...

it's a little difficult to explain
but i'll try...in the first one Sulphur gains oxygen atoms(oxidation) but oxygen gains electrons(reduction).
in the second one sulphur gains oxygen but oxygen gains electrons
whereas in third one although oxygen is gained, there is no transfer of electrons and hydrogen does not lose oxygen but combines with SO4
It's very difficult to explain.

* Oxygen's oxidation state reduces in the covalent bond it forms with sulphur
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Offline girl_92

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Re: CHEMISTRY PAPER 1 HERE
« Reply #251 on: June 11, 2009, 06:18:48 pm »
its not a que which can be explaind the best thing u can do is to go through ur notes n eat ur teachers head if u get her n u have time
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Offline Ghost Of Highbury

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Re: CHEMISTRY PAPER 1 HERE
« Reply #252 on: June 11, 2009, 06:21:53 pm »
Quote from: eddie_adi619 on June 11, 2009, 06:15:43 pm
S has many oxidation states including -2,0,2,4,6...
in this particular reaction the oxidation state of S is "+2"
a decrease in oxidation state - reduction
an increase in oxidation state - oxidation

1.) S + O2 -> SO2
     here at the left handside, as sulphur exists itself (not compound), the oxidation state is 0
          at the right hand sidde it forms a compound and gains an oxidation state of +2 = this is oxidation
          whereas O has an oxidation state of "0" in the left hand side and decreases to -4 in the right hand side
               this is reduction

therefore its redox
for the second equation
i can notice the decrease in oxidation state of oxygen
but not sulphur
the oxidation state of sulphur doesn't change
---
maybe this too is called a redox...
sry
but i think the first one's explanation is correct
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Offline girl_92

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Re: CHEMISTRY PAPER 1 HERE
« Reply #253 on: June 11, 2009, 06:22:42 pm »
bye ppl g2g bst of luck 4 2morrow
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nid404

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Re: CHEMISTRY PAPER 1 HERE
« Reply #254 on: June 11, 2009, 06:30:06 pm »
Quote from: eddie_adi619 on June 11, 2009, 06:21:53 pm
Quote from: eddie_adi619 on June 11, 2009, 06:15:43 pm
S has many oxidation states including -2,0,2,4,6...
in this particular reaction the oxidation state of S is "+2"
a decrease in oxidation state - reduction
an increase in oxidation state - oxidation

1.) S + O2 -> SO2
     here at the left handside, as sulphur exists itself (not compound), the oxidation state is 0
          at the right hand sidde it forms a compound and gains an oxidation state of +2 = this is oxidation
          whereas O has an oxidation state of "0" in the left hand side and decreases to -4 in the right hand side
               this is reduction

therefore its redox
for the second equation
i can notice the decrease in oxidation state of oxygen
but not sulphur
the oxidation state of sulphur doesn't change
---
maybe this too is called a redox...
sry
but i think the first one's explanation is correct

sulphur gains oxygen atoms..so that is oxidation whereas oxygen's oxidation state falls which is reduction
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