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Chemistry P41 How was it?

How hard was it?

  • Easy

    Votes: 1 2.1%
  • medium

    Votes: 27 57.4%
  • Hard!!!

    Votes: 19 40.4%

  • Total voters
    47
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Yup, the gt would be around the same range that it has been around the last few years. If not higher, I don't know.
This paper was easier compared to last years- no NMR, mass spectroscopy. But still some parts were VERY tricky and carried a lot of marks,
i found the whole paper was tricky to be honest!!
 
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it was great, probably the easiest paper in years...no enthalpy, no moles...and no NMR man I do not like that topic
Expect the GT to be 60+
 
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Well, many people I've seen got 25. Still trying to figure out what reasoning did they approach.
My Kc was 2.5x10^9. Did anyone get that value also? Or did anyone know somebody got that value?
 
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Well, many people I've seen got 25. Still trying to figure out what reasoning did they approach.
My Kc was 2.5x10^9. Did anyone get that value also? Or did anyone know somebody got that value?
What was your fe^2* value of con centration, was it same for that of fe^3+ given in the question. And how did u get the answer for this part??
What about the rate question what did get for the order of reaction of hcl.
 
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What was your fe^2* value of con centration, was it same for that of fe^3+ given in the question. And how did u get the answer for this part??
What about the rate question what did get for the order of reaction of hcl.
ws ur answer k=25?
becuase that is what i got
 
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What was your fe^2* value of con centration, was it same for that of fe^3+ given in the question. And how did u get the answer for this part??
What about the rate question what did get for the order of reaction of hcl.

No, it's not the same as fe3+,

[Fe3+]=[I-]. And [Fe2+]= 2[I2]

I used mole ratio between reactants. And mole ratio between products.
 
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No, it's not the same as fe3+,

[Fe3+]=[I-]. And [Fe2+]= 2[I2]

I used mole ratio between reactants. And mole ratio between products.
bro it is the same becuse yo dont see the overall equaton see the individual half equation hence 1:1
 
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