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Chemistry : P1 and P2 doubts only (Only AS allowed)

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X --- > Y + 2Z ( i dont knw jow to show reversible sign)
Initiall: moles 2 0 0 ( x disociated by 20% ... 0.4 of x converted to product)
2-0.4 0.4 0.8 ( I got Y and Z by mole ratio.. if 0.4 reacts then it will produce 0.4Y and 0.8Z)

P (x) = 1.6/2.8 x Tp
P (y ) =0.4/2.8 x Tp Tp = Total Pressure
P(z) = 0.8/2.8 x Tp

Just plug the values in the equation

Kp = P(y ) x (P(z))^2/P(x) (P(z))^2 because coefficients become power

AH SWEET>. that clears it up ...
 
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Test for Alkene (the Br2 water test) .. the condition would be in darkness yeah? and what'd be the product? di-bromoalkane? or bromo-alcohol?

what about Br2 in CCl4 .. it will be at room temperature and only di-bromoalkane as product yea?
 
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Test for Alkene (the Br2 water test) .. the condition would be in darkness yeah? and what'd be the product? di-bromoalkane? or bromo-alcohol?

what about Br2 in CCl4 .. it will be at room temperature and only di-bromoalkane as product yea?
first the condition is rtp not darkness formation formation willl be bromo alcohol but u might get away wiith using just bromo alkane
 
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guys any thing u found difficult any question ?????? anything new???????
 
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guys can anyone write here the full equation of reaction od H2SO4 with hydrogen bromidee and iodide

H2SO4 + NaBr --> NaHSO4 + HBr

2HBr + H2SO4 --> Br2 + SO2 + 2H2O

8HI + H2SO4 --> 4I2 + H2S + 4H2O
 
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Haloalkane + aq NaOH is a reaction to make alcohol.. but is there a reaction between Haloalkanes and Water to make alcohol when heated?
 
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T
Haloalkane + aq NaOH is a reaction to make alcohol.. but is there a reaction between Haloalkanes and Water to make alcohol when heated?
That OH in water is not much effective,the OH is more effective in NaOH
 
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Here u go

PRODUCTS^moles/REACTANTS^moles

[CH3CO2C2H5][H2]/[CH3COOH][C2H5OH]

Kc = 4
Initial
Reactants=moles Ethanol=Ethanoic Acid = 0.5 moles
Products=moles of Water=Ester = 0.1 moles

Final would be
Reactants - x = ethanol-x , acid-x = (0.5-x) for both
Products +x = water+x,ester +x = 0.1+x for both

Now substitute that in the equation above...

(0.1+x)^2/(o.5-x)^2 = 4

take square root on both sides..

o.1+x/0.5-x = 2
0.1+x=1-2x
3x=0.9
x=0.3

Now calculate the no of moles.. by substituting x ..

Reactants = 0.5-x = 0.2 each
Products = 0.1+x = 0.4 each

Answer correct?
 
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