Chemistry Applications and Physical Chemistry:Please help

[Illuminati]

Please help me with apps and physical chem.Kc ka Kb etc all are very confusing.Any better way to memorise them and their applications?
Please suggest me any good resource for chem app revision.
Can any one explain different structures of protein as required by Cie?
waiting for your reply...

 

[Hanna92]

Re: Chemistry Applications and Physical Chemistrylease hel

I know what you mean about the applications!!....The first one is simple for me....i dont even have to study it because its AS biology syllabus....im guessing you havent taken AS biology then!!

well Kc and all those are also fairly easy...i mean theirs always tougher things in chemistry!!...but here are some links for chemistry ....they are basically e-books....and i dont know exactly which ones are the most useful and which ones work the best...but hey at least im sharing them...

Links:
http://books.google.com.au/books?id=q1U ... &q&f=false

http://books.google.com.au/books?id=tgv ... hs&f=false

http://books.google.com.au/books?id=q1U ... &q&f=false

http://books.google.com.au/books?id=4vt ... &q&f=false

http://books.google.com.au/books?id=25q ... &q&f=false

http://books.google.com.au/books?id=ZCk ... ry&f=false

http://books.google.com.au/books?id=WZI ... ry&f=false

http://books.google.com.au/books?id=lUv ... ry&f=false

http://www.chemguide.co.uk/index.html#top

www.s-cool.co.uk

WELL....thats all that i know....the last two are the best ones....i recommend them...

For protein structure:
There are four basic types.....you dont need to know the fourth one that well but it is still required....the four protein structures are:
1. Primary structure
2. Secondary structure
3. Tertiary structure
4. Quaternary structure

Primary Structure: It is the simple chain and length of amino acids.....consisting of hydrogen bonds and peptide bonds.... along with the N- terminal ( the end with the nitrogen group) and the C terminal ( the end with the carboxylic acid group)

Secondary structure: It is either in two forms the alpha helix or the beta plated sheet....both of these are formed due to hydrogen bonds.....the alpha helix is like the DNA shape....and the beta plated sheet like /\/\/\/\ <- Like that

Tertiary Structure: this forms a more rounded protein shape with many bends and bonds.....which inculde....hydrogen bonds, van der waals bonds, ionic interactions and the sulphide bonds which lock the structure in place....

NOTE: all these structures are for one polypeptide chain...meaning one protein chain ONLY!!!

Quaternary structure: this is when TWO or more protein chains interact to form a complex structure......this structure involves the same type of bonds as the tertiary structure.....
you basically need to know the quaternary structure vaguely for referring to the haemoglobin molecule required in the syllabus.....thats all!!

Well i hoped that helped!!.....

 

[mista.lova.lova]

Re: Chemistry Applications and Physical Chemistrylease hel

great!! its really some bit of info...

 

[princesszahra]

Re: Chemistry Applications and Physical Chemistrylease hel

do application booklet in detail
its more than enough!

 

[Jazib]

Re: Chemistry Applications and Physical Chemistrylease hel

quatimary structure (or watever it is) is not included in the syllabus!

 

[Jazib]

Re: Chemistry Applications and Physical Chemistrylease hel

Please help me with apps and physical chem.Kc ka Kb etc all are very confusing.Any better way to memorise them and their applications?
Please suggest me any good resource for chem app revision.
Can any one explain different structures of protein as required by Cie?
waiting for your reply...

if u want to get an A only, then u can follow the syllabus n LEARN(with understanding ofcourse) everything written in that! will get u an A for sure, or maybe an A* if ur lucky!

 

[mawak]

Re: Chemistry Applications and Physical Chemistrylease hel

Equilibria, Acids And Bases
1. At equilibrium, there is no net change in the concentrations of reactants and products, due to equal
and opposite tendencies for change in both directions.
2. A homogeneous system is one in which all the species present are in the same phase.
3. For any reversible reaction at equilibrium, e.g. aA + bB cC + dD, the concentrations of the
components are related by:
a b
c d
c [A]
K = [C] [D]
4. The value of Kc gives an indication of the position of equilibrium:
a. Kc < 1 – the equilibrium position will be towards the left (more reactants).
b. Kc = 1 – the equilibrium position will be in the middle (equal reactants and products).
c. Kc > 1 – the equilibrium position will be towards the right (more products).
5. For a reaction completely in the gaseous phase, e.g. aA(g) + bB(g) cC(g) + dD(g), the
equilibrium constant Kp can be used, whereby p(X) denotes the partial pressure of X:
a b
c d
p p(A) p(B)
K = p(C) p(D)
6. The partial pressure of a gas is the pressure that a gas would exert if it were the only gas in the
system. The sum of partial pressures gives the total pressure:
a. The number of moles of a species A is denoted by n(A).
b. The mole fraction of a species A, x(A), is defined as
n tot
x(A) = n(A) .
c. The partial pressure of a species A, p(A), is defined as p(A) = x(A)´Ptot .
questionbase.50megs.com A2-Level Revision Notes
7. The value of Kc/Kp is constant for a specific reaction at a specific temperature. The temperature is
the only condition that will cause a change in the value, and this relationship is exponential
(pressure and concentration changes will change the position of equilibrium, but not the
equilibrium constant):
a. For an exothermic reaction, raising the temperature will result in fewer products and more
reactants, hence Kc/Kp is smaller.
b. For an endothermic reaction, raising the temperature will result in more products and
fewer reactants, hence Kc/Kp is bigger.
c. A catalyst does not affect the position of equilibrium or the value of Kc/Kp – it only
affects the reaction kinetics.
8. An acid is defined as a species that releases H+(aq) ions, and has a pH lower than 7. A base will
neutralise an acid to form a salt and water only (normally metal oxides or hydroxides). An alkali is
a soluble base.
9. The Brønsted-Lowry definition of acids and bases:
a. An acid is a proton donor.
b. A base is a proton acceptor.
10. The Lewis definition of acids and bases:
a. An acid is an electron acceptor.
b. A base is an electron donor.
11. An H+ ion by itself is not able to exist (it is not stable), so it will combine with water molecules to
form hydronium ions, H3O+(aq).
12. Any reaction between an acid and a base forms an acid-base equilibrium, whereby the acid donates
a proton to form its conjugate base, and the base accepts the proton to form its conjugate acid. A
conjugate (acid/base) pair differ only by one proton.
13. pH is defined as: pH = –log[H3O+(aq)]. pH values are usually given to 2 decimal places.
14. For a strong acid, always assume complete dissociation into ions. The value of [H3O+] will
therefore be given by the concentration of acid, multiplied by the number of moles of protons
released by one mole of acid.
15. The ionic product of water is defined as Kw = [H3O+(aq)][OH–(aq)]:
a. At 298K, Kw = 1.0 ´ 10–14 mol2dm–6.
b. The conditions are neutral when [H3O+] = [OH–], but the pH changes with temperature as
the value of Kw changes (the pH is only exactly 7 at 298K).
c. For a strong alkali, the concentration of OH– ions will be proportional to the concentration
of alkali, hence the concentration of H3O+ ions can be calculated using the ionic product
of water, and consequently the pH can be calculated.
16. For a mixture of an acid and an alkali, calculate the number of moles of H3O+ from the acid, and
OH– from the alkali, then calculate the excess after neutralisation. This can consequently be used
to calculate the pH of the solution.
17. A weak acid or base only partially dissociates (ionises) in aqueous solution.
18. For the weak acid equilibrium, HA(aq) + H2O(l) H3O+(aq) + A–(aq):
a. The acid dissociation constant is defined as [HA(aq)]
[H O (aq)] [A (aq)]
K 3
a
+ -
= .
b. pKa = –log Ka
c. The smaller the value of Ka and the higher the value of pKa, the weaker the acid.
d. You can assume that [H3O+] = [A–], and that [HA] at equilibrium is approximately the
same as that before dissociation, resulting in the expression [H3O+]2 = [HA]tot Ka.
19. A buffer solution is one that is able to oppose changes in pH when small amounts of acid or base
are added, or if it is diluted with water:
a. An acidic buffer is a mixture of a weak acid and its salt (e.g. ethanoic acid and sodium
ethanoate), maintaining a pH below 7:
i. Ethanoic acid – CH3COOH + H2O H3O+ + CH3COO– (equilibrium to the left).
ii. Sodium ethanoate – CH3COONa Na+ + CH3COO– (equilibrium to the right).
iii. When the ethanoate ions from the salt are added to the weak acid, the position of
equilibrium of the acid is ‘pushed’ even further to the left – resulting in an
excess of both the acid and its conjugate base.
iv. Added H+ ions will react with the ethanoate ions, but as they are in excess, the
equilibrium will only shift slightly, and the pH is maintained.
v. Added base will react with the ethanoic acid, but as the acid is also in excess, the
equilibrium will only shift slightly, and the pH is maintained.
questionbase.50megs.com A2-Level Revision Notes
b. A basic buffer is a mixture of a weak base and its salt (e.g. ammonia and ammonium
chloride), maintaining a pH above 7:
i. The equilibrium can be written as NH4
+(aq) NH3(aq) + H+(aq).
ii. From this,
[NH ]
[H ] [NH ]
K
4
3
a +
+
= .
iii. If [NH3] = [NH4
+] then pH = pKa for the equilibrium.
c. The pH of a buffer solution is dependant upon the value of Ka, and the ratio of the
concentrations of the conjugate acid/base pair.
20. The equivalence point for an acid-base titration corresponds to the mixing together of
stoichiometrically equivalent amounts of acid and base. At equivalence, the pH will be 7.0 for a
strong acid and a strong base, less than 7.0 for a strong acid and a weak base, and greater than 7.0
for a weak acid and a strong base. A graph of pH against volume added during a titration results in
a pH curve:
21. Weak, diprotic, acids and bases will produce two end points in a titration, as the ions form
separately and sequentially (for a strong diprotic acid such as H2SO4, it is fully ionised, so there is
only one end point). Different indicators can be used to distinguish between the end points, which
occur in a 1:2 ratio. For sodium carbonate, the equilibria are:
a. Na2CO3 + H+ NaHCO3 + Na+ (1st end point)
b. NaHCO3 + H+ Na+ + CO2 + H2O (2nd end point)
22. An indicator is a weak acid, whose conjugate base is a different colour:
a. HIn + H2O H3O+ + In–
b. Adding acid moves the equilibrium to the left to form colour A. Adding alkali pushes the
equilibrium to the right to form colour B.
c. At the end point, [In–] = [HIn], so pH = pKa, and the pH at which the indicator changes
colour is determined by its Ka value.
d. Phenolphthalein is colourless in acids (HIn), and purple in bases (In–), and has a pH range
of 8.2 to 10.0. It is used for weak acid–strong base titrations.
e. Methyl Orange is pink in acids (HIn), and yellow in bases (In–), and has a pH range of 3.2
to 4.4. It is used for strong acid–weak base titrations.

 

[mista.lova.lova]

Re: Chemistry Applications and Physical Chemistrylease hel

Please help me with apps and physical chem.Kc ka Kb etc all are very confusing.Any better way to memorise them and their applications?
Please suggest me any good resource for chem app revision.
Can any one explain different structures of protein as required by Cie?
waiting for your reply...

if u want to get an A only, then u can follow the syllabus n LEARN(with understanding ofcourse) everything written in that! will get u an A for sure, or maybe an A* if ur lucky!

in bhai ka phir bhee nai aana...!!
saara saal to skool mai shakal he nai dikhai...

 

[Illuminati]

Re: Chemistry Applications and Physical Chemistrylease hel

@lova.lova aka shaheer tarik bss fsd...I dont let schooling interfere with my education
@mawak and hanna.THANKS ALOT FOR THE HELP
@hanna.Is this enough for A levels?
@zahra.Dont have time to revise Booklet as it is very lengthy and complicated
@Lova.Lova..._|_

 

[Illuminati]

Re: Chemistry Applications and Physical Chemistrylease hel

And please keep checking this topic as i may ask for more help.

 

[mista.lova.lova]

Re: Chemistry Applications and Physical Chemistrylease hel

hahaha begairat!!
tariq mai Q ata hai not K !!
and is sign ko pechay lay lo..

 

[Hanna92]

Re: Chemistry Applications and Physical Chemistrylease hel

@ Illuminati: Dont mention it .....you're welcome....and yes that is enough for A levels!!!....good luck

 

[Illuminati]

Re: Chemistry Applications and Physical Chemistrylease hel

hmm..Thanks again
@lova.lova.watch your a*s before you speak.This froum is not for stupid catfight

 

[mista.lova.lova]

Re: Chemistry Applications and Physical Chemistrylease hel

shut up aouno!!

 

[Illuminati]

Re: Chemistry Applications and Physical Chemistrylease hel

jugtology?:O:S..get a life