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Chemistry P4| A2 only

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can sum1 plzzz help in these questionss....i dont get how electrode potentials increases or decreases when u increase conc of of the reactants..these are the papers.q3 of oct/2010/41 and q5c of oct/2010/43..
The general idea is like this:
X(n) = X(n-m) + m e(-) [numbers in the round brackets represent charge] There is an equilibrium position in the middle.
According to Le Chatelier's Principle, increasing the conc. of one side would favour the reaction and shift the euqi. position towards the other side.
In p41 Q3... Can you tell me which part is concerned with concentration? Sorry. :p
In p43 Q5(c), Cl2 + 2e(-) = 2Cl(-)
Increasing [Cl(-)] favours the backward reaction so comparatively, Cl2 is less willing to be reduced. The E(red) for Cl2 decreases.
The O2 is not affected by this since redox of oxygen does not involve any chlorine, so its E(red) experiences no change.
 
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First, we need to get the Ka value from the pKa ..
pKa=-log(Ka)
So, Ka = 10 ^(-4.76)

The acid equation is : CH3COOH <---> (H+) + CH3COO-
So, Ka = ([H+]*[CH3COO-]) / [CH3COOH]
Re-arranging the equation: [H+] = (Ka*[CH3COOH]) / [CH3COO-]


But, we don't know what [CH3COO-] is,
So, we hav to find it with the formula C = {n(H+) - n(OH-)} / V
So, C = (0.1*0.01-0.01*0.25) / 0.02 = 0.075

Then, we can calculate for the H+ with the above reaction. And then pH. My teacher just used some pH = 4.76 + log(0.05/0.075), but I don't understand why.

:ROFLMAO:Well i did the same way but i dint understand this part :(
But, we don't know what [CH3COO-] is,
So, we hav to find it with the formula C = {n(H+) - n(OH-)} / V
So, C = (0.1*0.01-0.01*0.25) / 0.02 = 0.075

explanation bro :p
 
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(b)(iii)
Thickness can be one. The thickness represents how hard it is for the drug to go through the hydrogel. Also the more thick the more time it takes
Using different chemicals or materials to make the hydrogel can affect the rate of reaction with HCl and so the rate of release.
Making holes on the coat serves to increase the contact area with HCl. It is another way to increase the rate of reaction.
 
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(c) Use only one kind of monomer and "poly" it to get a homopolymer. I recommend that you use CO2HCHRNH2. It's simpler to draw.
BE AWARE: you cannot choose ethan-1,2-diol, OHCH2CH2OH, since it does not possess the functional group CO2H to initiate esterification.
The "poly" version of this molecule is then ...-NH-R-CH-CONH-R-CH-CONH-R-CH-CO-...

Use two different (you can use all the three... if you dare try ;)) monomers to construct a heteropolymer.
BE AWARE: ethan-1,2-diol and the amino acid cannot be polymerised. -OH and -NH2 cannot react.
You can choose 2-hydroxy-butan-1,4-dioic acid and ethan-1,2-diol.
Then the polymer looks like this:
...-O-CH2-CH2-O-CO-CH(OH)-CH2-CO-O-CH2-CH2-O-...
 
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can ny1 tell me what are the important aspects or topics should i cover for the Bio-chemistry?
nd plz dont say me the syllabus just some of the most frequently asked sections!! pls
 
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can any one give me a full explanation that how br2 is produced instead of o2 in electrlysis of aq MgBr2..............
plz so confused............
 
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For alcohol + PCl5 and acid + PCl5, do we have to heat ? Why some marking scheme say needed & some say don't need ?
 
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For alcohol + PCl5 and acid + PCl5, do we have to heat ? Why some marking scheme say needed & some say don't need ?
No You do not need. In Chemistry Coursebook Pcl5+alcohol rxn is done in room tempr and for carboxylic acid + pcl5 it has explicitly been mentioned that heat in not required.However, for reaction involving carboxylic acid and Pcl3 heat is required.
 
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For alcohol + PCl5 and acid + PCl5, do we have to heat ? Why some marking scheme say needed & some say don't need ?
no u dont have to heat it with PCl5 or SOCl2, but if u're using PCl3 u need to heat it
 
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can any one give me a full explanation that how br2 is produced instead of o2 in electrlysis of aq MgBr2..............
plz so confused............
i have the same question!!
why is Br2 produced at the anode when aqueous MgBr2 is used for electrolysis. It doesnt say 'concentrated'...
 
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i have the same question!!
why is Br2 produced at the anode when aqueous MgBr2 is used for electrolysis. It doesnt say 'concentrated'...
what i conclude is o2 must be produced because in other two with the same theory o2 is produced........
 
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can any one tell me conditions for protien hydrlysis,,,,,,,,,,
is it correct....boil with dil acid H+/H2O?????????????????/
 
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what i conclude is o2 must be produced because in other two with the same theory o2 is produced........
With aqueous solution containing halogens > halogens are always released at anode unless it is very dilute in which case you will obtain Oxygen. standard electrode potential value does not support this because there is something known as 'over voltage efect(which i think is out of syllabus scope)' so its better to remember that halogens are discharged at anode in most case(exception: solution is highy dilute)
 
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