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Chemistry: Post your doubts here!

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39)1 and 2
View attachment 61409
40) This is easiest last question, i ever saw xD Sorry to say like this, but just u have to manipulate its structure and match with the given compound in question. All three is having that formula after oxidation. Try once, if u dont get ask me. Bored in making that one in 39 xD :p
39 Okay, got that :)
40 Yeah okay...but I don't get the 3rd molecule ? ANd hey Q 40 is in the examiner report's particularly difficult :p
 
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39 Okay, got that :)
40 Yeah okay...but I don't get the 3rd molecule ? ANd hey Q 40 is in the examiner report's particularly difficult :p
IDK how i found that easiest -_- This was the paper i gave and the only sure answer was this xD
This is your oxidised product of third molecule. CO2 is product as well, i have just shown that main thingy xD
upload_2016-10-31_23-16-23.png
 
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Hey people!
I have done chemistry in A levels. Please let me know if there are any doubts, I would be glad to help
 
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Hi i want to ask why the answer is that for this question
2No2 =N2O4
40% of the moles present at equilibrium is N2O4. If the total pressure of the system is 2 atm, the numerical value of the equilibrium constant, kp is .........the answer is 0.56 why? Thanks
 
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elmansy kp= partial pressure of products^n / partial pressure of reactants^m
partial pressure = total pressure X no.moles of / total number of moles.
pressure = 2atm
partail pressure of N2O4 = 2 X (40/100) = 0.8
partail pressure of NO2 = 2 X (60/100) = 1.2

Kp = 0.8 / (1.2)^2 = 0.56 .

read the book this was a simple question :)
 
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