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fishD
03-26-2012, 10:03 PM
එළකිරි රසායන විද්‍යාව පන්තිය

EXAM ඉවර වෙලා දෑන් ගොඩාක් නිදහස්

ඉතින් පොඩි වෑඩක් පටන් ගන්න තමා හදන්නෙ !!

ඔක්කොටොම කලින් අද ආපු නන්ගිලා මල්ලීලා හෑමදෙනාටම ජයවේවා කිවුවා !!

හරි වෑඩි කතා ඔනී නෑ

කෙලින්ම වෑඩට බහිමු

ඔයාලා දන්නවා නේ A/L වලට රසායන ව්ද්යාව හරිම වෑදගත්

අපි මේ එකේ ඔයාලට තියෙන රසායන ව්ද්යා ගෑටලු ගෑන කතා කරමු

මම දන්න දන්න විදියට කියලා දෙන්නම් , බෑරි නම් කාගෙන් හරි අහල හරි කියලා දෙන්නම් (අපි එහෙමයි )

THREAD එක BUMPකරලා තියාගමු !!!

දෑන් ඉතින් BUMP එකක්වත් නොදා යනවා නෙමෙයි ඕන් ... ගහනවා ACID !!!

Pink panther
03-26-2012, 10:05 PM
http://i.imm.io/fnVd.jpeg

fishD
03-26-2012, 10:12 PM
RnGu3xO2h74

Balancing Chemical Equations

fishD
03-26-2012, 10:14 PM
bump

yama_palla
03-26-2012, 10:14 PM
gud wrk bro.....keep it up.

fishD
03-26-2012, 10:15 PM
gud wrk bro.....keep it up.

thanks machan
aniwa :)

Mr.Truth
03-26-2012, 10:20 PM
appata.
matanam eda idan pennanna bari wishayak.
ane manda matanam okkoma eka wage. :oo:

Pathum921
03-26-2012, 10:25 PM
WOW maara patta subjct eka bun. Mamanam hena asai :)

fishD
03-26-2012, 10:28 PM
appata.
matanam eda idan pennanna bari wishayak.
ane manda matanam okkoma eka wage. :oo:

:)

fishD
03-26-2012, 10:29 PM
WOW maara patta subjct eka bun. Mamanam hena asai :)

elakiri ah !!!
loweth
:yes::yes::yes:

sameerasadaru
03-26-2012, 10:46 PM
අපෝ මේකෙත් පටන් ගත්තද මේ මගුල :angry:
උබට වෙන විශයක් තිබ්බෙම නැද්ද වෙන :baffled: මදැයි ටිකක් නිදහසේ ඉන්න එලකිරි ආවා :( :rofl:

එල මචන් ගොඩක් අයට වැදගත් වෙයි:yes: දිගටම කර ගෙන යන්න :D

fishD
03-26-2012, 10:56 PM
අපෝ මේකෙත් පටන් ගත්තද මේ මගුල :angry:
උබට වෙන විශයක් තිබ්බෙම නැද්ද වෙන :baffled: මදැයි ටිකක් නිදහසේ ඉන්න එලකිරි ආවා :( :rofl:

එල මචන් ගොඩක් අයට වැදගත් වෙයි:yes: දිගටම කර ගෙන යන්න :D

:D:D:D:D

win7dot
03-27-2012, 05:19 PM
BUMP

fishD
03-28-2012, 10:53 AM
bump

fishD
03-28-2012, 10:57 AM
Using the Oxidation Number Technique

Tip-off – If you are asked to balance an equation and if you are not told whether the reaction is a redox reaction or not, you can use the following procedure.
General Steps
Step 1: Try to balance the atoms in the equation by inspection, that is, by the standard technique for balancing non-redox equations. (Many equations for redox reactions can be easily balanced by inspection.) If you successfully balance the atoms, go to Step 2. If you are unable to balance the atoms, go to Step 3.
Step 2: Check to be sure that the net charge is the same on both sides of the equation. If it is, you can assume that the equation is correctly balanced. If the charge is not balanced, go to Step 3.
Step 3: If you have trouble balancing the atoms and the charge by inspection, determine the oxidation numbers for the atoms in the formula, and use them to decide whether the reaction is a redox reaction. If it is not redox, return to Step 1 and try again. If it is redox, go to Step 4.
Step 4: Determine the net increase in oxidation number for the element that is oxidized and the net decrease in oxidation number for the element that is reduced.
Step 5: Determine a ratio of oxidized to reduced atoms that would yield a net increase in oxidation number equal to the net decrease in oxidation number (a ratio that makes the number of electrons lost equal to the number of electrons gained).
Step 6: Add coefficients to the formulas so as to obtain the correct ratio of the atoms whose oxidation numbers are changing. (These coefficients are usually placed in front of the formulas on the reactant side of the arrow.)
Step 7: Balance the rest of the equation by inspection.

fishD
03-28-2012, 10:59 AM
PART 1
wXM7_54bmNY


PART 2
lF8DnGrhY5Y


PART 3
bk1lsuxQ7IQ

fishD
03-28-2012, 11:03 AM
ANOTHER EXAMPLE

Fe2+(aq) http://www.files.chem.vt.edu/chem-ed/graphics/arrow-right.gif Fe3+(aq) (oxidation)
MnO4-(aq) http://www.files.chem.vt.edu/chem-ed/graphics/arrow-right.gif Mn2+(aq) (reduction)


mewata thamai half reactions kiyanne ,,, eeta passe e half reactions balance kara ganna one
menna mehema


Fe2+(aq) http://www.files.chem.vt.edu/chem-ed/graphics/arrow-right.gif Fe3+(aq) + e-
MnO4-(aq) + 8H+(aq) + 5e- http://www.files.chem.vt.edu/chem-ed/graphics/arrow-right.gif Mn2+(aq) + 4H2O



dan overall reaction eka liya ganna menna mehema




5Fe2+(aq) http://www.files.chem.vt.edu/chem-ed/graphics/arrow-right.gif 5Fe3+(aq) + 5e-
MnO4-(aq) + 8H+(aq) + 5e- http://www.files.chem.vt.edu/chem-ed/graphics/arrow-right.gif Mn2+(aq) + 4H2O
------------------------------------------------------------
5Fe2+(aq) + MnO4-(aq) + 8H+(aq) http://www.files.chem.vt.edu/chem-ed/graphics/arrow-right.gif 5Fe3+(aq) + Mn2+(aq) + 4H2O


:) :) :) :)

mokak hari seen ekak tiyenawa nam PM ekak danna ......... :)

fishD
03-28-2012, 12:34 PM
BUMP

fishD
03-28-2012, 09:10 PM
Compressibility factor


The compressibility factor (Z) is a useful thermodynamic (http://en.citizendium.org/wiki/Thermodynamics) property for modifying the ideal gas law (http://en.citizendium.org/wiki/Ideal_gas_law) to account for behavior of real gases.[1] (http://en.citizendium.org/wiki/Compressibility_factor_%28gases%29#cite_note-McQuarrie-0)[2] (http://en.citizendium.org/wiki/Compressibility_factor_%28gases%29#cite_note-Cengal-1)[3] (http://en.citizendium.org/wiki/Compressibility_factor_%28gases%29#cite_note-Smith-2)[4] (http://en.citizendium.org/wiki/Compressibility_factor_%28gases%29#cite_note-Rao-3)[5] (http://en.citizendium.org/wiki/Compressibility_factor_%28gases%29#cite_note-Mombourquette-4) It is a measure of how much the thermodynamic properties of a real gas deviate from those expected of an ideal gas. It may be thought of as the ratio of the actual volume of a real gas to the volume predicted by the ideal gas at the same temperature and pressure as the actual volume.
For an ideal gas, Z always has a value of 1. For real gases, the value may deviate positively or negatively, depending on the effect of the intermolecular forces (http://en.citizendium.org/wiki/Intermolecular_forces) of the gas. The closer a real gas is to its critical point (http://en.citizendium.org/wiki?title=Critical_point&action=edit&redlink=1) or to its saturation point (http://en.citizendium.org/wiki?title=Saturation_point&action=edit&redlink=1), the larger are the deviations of the gas from ideal behavior.

The ideal gas law is defined as:
http://en.citizendium.org/images/math/1/1/4/1142f61356382badab6f4500c3d131d1.png and the ideal gas law corrected for non-ideality is defined is:
http://en.citizendium.org/images/math/6/d/9/6d95277630cd497b198ae3863960c8fc.png where: http://en.citizendium.org/images/math/4/4/c/44c29edb103a2872f519ad0c9a0fdaaa.png = pressure http://en.citizendium.org/images/math/1/b/0/1b007b2e910d8dbb8ccb4dda34306722.png = molar volume (http://en.citizendium.org/wiki/Molar_volume) of the gas http://en.citizendium.org/images/math/2/1/c/21c2e59531c8710156d34a3c30ac81d5.png = compressibility factor http://en.citizendium.org/images/math/e/1/e/e1e1d3d40573127e9ee0480caf1283d6.png = Universal gas constant (http://en.citizendium.org/wiki/Universal_gas_constant) http://en.citizendium.org/images/math/b/9/e/b9ece18c950afbfa6b0fdbfa4ff731d3.png = temperature and thus:
http://en.citizendium.org/images/math/d/f/0/df0a5ac8b6ac47d01cee7a4e6dbf130e.png

fishD
03-28-2012, 09:15 PM
What is a Mole?

A mole is the amount of pure substance containing the same number of chemical units as there are atoms in exactly 12 grams of carbon-12 (i.e., 6.023 X 1023).



6.023 X 1023 kiyanne constant ekak. e thamai AVOGADRO's number kiyala kiyanne :) :)

fishD
03-28-2012, 09:39 PM
PV=nRT


WScwPIPqZa0

shehan12345
03-28-2012, 11:23 PM
el 3L

fishD
03-28-2012, 11:53 PM
bump

fishD
03-29-2012, 08:17 AM
Aldol Condensation

http://www.organic-chemistry.org/namedreactions/aldol-1.gif
In some cases, the adducts obtained from the Aldol Addition can easily be converted (in situ) to α,β-unsaturated carbonyl compounds, either thermally or under acidic or basic catalysis. The formation of the conjugated system is the driving force for this spontaneous dehydration. Under a variety of protocols, the condensation product can be obtained directly without isolation of the aldol.


AND THE MECHANISM IS



3XiHrsZNZko

fishD
03-29-2012, 08:28 AM
MECHANISM OF THE ALDOL CONDENSATION
Step 1:
http://www.mhhe.com/physsci/chemistry/carey5e/Ch18/aldolmech.gif



http://www.mhhe.com/physsci/chemistry/carey5e/Ch18/aldol-h2omech.gif

fishD
05-26-2012, 10:39 AM
BALANCING REDOX REACTIONS IN BASE SOLUTION

You can balance redox reactions in base solution in almost the same way as if they were in acid solution. In fact, initially we will assume the reaction is taking place in acid solution but eventually switch over to base. We'll go through the sequence of steps one at a time using a sample equation.

12. Check the final equation for mass and charge balance.
Suppose you wanted to balance the following equation:

MnO4- + CN- --> MnO2 + CNO- (in base)

The first step is to split this equation into two half reactions. One half reaction will show the reduction and the other will show the oxidation. It is not important to initially know which half reaction is which.

MnO4- --> MnO2

CN- --> CNO-

We'll deal with one half reaction at a time, starting with the half reaction for permanganate ion.

Check the balance of any atom other than O or H first. In this case the manganese atoms are balanced as written:

MnO4- --> MnO2

Next, balance O's using H2O as if in acid solution by adding H2O's as necessary to the side deficient in O. In this case, 2 H2O's are needed on the right hand side of the half reaction:

MnO4- --> MnO2 + 2 H2O

To balance H's, add as many H+ ions as needed to the side deficient in H, as if in acid solution. In this case, 4 H+ are needed on the left hand side of the half reaction:

4 H+ + MnO4- --> MnO2 + 2 H2O

Now "reality" comes into the picture. We should switch over to base solution by adding the same number of OH- ions as we have H+ ions to both sides of the half reaction. In this case, 4 OH- ions are added to both sides of the half reaction:

4 OH- + 4 H+ + MnO4- --> MnO2 + 2 H2O + 4 OH-

The H+ is now "removed" from the equation by taking the H+ ions and OH- ions found on one side of the half reaction and forming H2O's. In this case, 4 H+ ions and 4 OH- ions make 4 H2O molecules:

4 H2O 4 OH- + 4 H+ + MnO4- --> MnO2 + 2 H2O + 4 OH-

Notice how some water molecules may be cancelled. There are 4 H2O's on the left hand side of the equation, and 2 H2O's on the right hand side of the reaction. The 2 H2O's on the right can cancel 2 of the 4 H2O on the left:

2 4 H2O + MnO4- --> MnO2 + 2 H2O + 4 OH-

Mass balance has now been achieved. Charge balance comes next. Determine the net charge on each side of the half reaction and add electrons to the more positive side so that the charge becomes the same on both sides. In this case, the net charge on the left and side of the half reaction is -1; on the right hand side, the net charge is -4. A charge of -1 is more positive than a charge of -4, therefore, add 3 electrons to the left hand side so that the charge is -4 on both sides:

3 e- + 2 H2O + MnO4- --> MnO2 + 4 OH-

This half reaction, showing the reduction, is now both mass and charge balanced. You can tell it's a reduction because electrons are being gained (they are a "reactant"). We'll put this half reaction aside for now.

Repeat the previous steps to obtain mass and charge balance for the second half reaction. First, check the balance of any atom other than O or H. In this case, both the C and N are balanced as written:

CN- --> CNO-

Next, pretending we are carrying out the reaction in acid solution, balance the O using H2O. In this case, 1 H2O is needed on the left hand side to balance the O:

H2O + CN- --> CNO-

To balance H's, add H+ to the side deficient in H. In this case, add 2 H+ to the right hand side of the half reaction:

H2O + CN- --> CNO- + 2 H+

Now it is time to switch over to base. Recall this is done by adding the same number of OH- as you have H+ to both sides of the half reaction. In this case 2 OH- must be added to both sides:

2 OH- + H2O + CN- --> CNO- + 2 H+ + 2 OH-

Convert the 2 H+ and 2 OH- on the right hand side to H2O's:

2 OH- + H2O + CN- --> CNO- + 2 H+ + 2 OH- 2 H2O

Cancel H2O where possible. In this case, the H2O on the left hand side of the half reaction will cancel one of the H2O's on the right hand side of the half reaction.

2 OH- + H2O + CN- --> 2 H2O + CNO-

The half reaction is now mass balanced. Next, balance the charge as before by adding the electrons to the more positive side so that the charges are the same on both sides. In this case, the net charge on the left hand side of the half reaction is -3 ( 2 hydroxides and 1 cyanide); the net charge on the right hand side is -1 (1 cyanate). A charge of -1 is more positive than a charge of -3, so add 2 electrons to the right hand side of the half reaction:

2 OH- + CN- --> CNO- + H2O + 2 e-

This half reaction showing the oxidation is now both mass and charge balanced. Next, we will retrieve the first half reaction and proceed to follow the steps needed to combine them to for the overall reaction. Before we can add the two half reactions to obtain the overall net ionic reaction, the number of electrons lost must equal the number of electrons gained. In this case, 2 electrons are lost and 3 electrons are gained:

2 H2O + MnO4- + 3 e- --> MnO2 + 4 OH-

2 OH- + CN- --> CNO- + H2O + 2 e-

We need to find the least common multiple (lcm) of 2 and 3. The lcm, in this case, is 6. Multiply through each half reaction by the factor necessary to obtain the least common multiple. Here, the first half reaction must be multiplied by 2 and the second by 3:

2 x ( 2 H2O + MnO4- + 3 e- --> MnO2 + 4 OH- )

3 x ( 2 OH- + CN- --> CNO- + H2O + 2 e- )

The resulting half reactions are:

4 H2O + 2 MnO4- + 6 e- --> 2 MnO2 + 8 OH-

6 OH- + 3 CN- --> 3 CNO- + 3 H2O + 6 e-

Sum the two half reactions together, cancelling any like terms to obtain the balanced, net ionic reaction. In this case, the 6 electrons will cancel, the 3 H2O's on the right hand side will cancel 3 of the 4 H2O's on the left hand side, and the 6 OH-on the left hand side will cancel 6 of the 8 OH- on the right hand side:

1 4 H2O + 2 MnO4- + 6 e- --> 2 MnO2 + 2 8 OH-

6 OH- + 3 CN- --> 3 CNO- + 3 H2O + 6 e-

What remains should be the balanced, net ionic equation. The most important step is to check this final equation for mass and charge balance:

H2O + 2 MnO4- + 3 CN- --> 2 MnO2 + 2 OH- + 3 CNO-

The mass balances: 2 H, 2 Mn, 9 O, 3 C, and 3 N are found on each side of the reaction. For charge balance, the net charge on the left hand side is -5 ( -2 from two permanganate ions and -3 form three cyanide ions), and also -5 on the right hand side ( -2 from two hydroxide ions and -3 form three cyanate ions). Since mass and charge balance has been achieved, the reaction must be correctly balanced.
Summary of Steps


1. Divide the equation into two half reactions.

2. Balance any atom other than O or H first.

3. Balance O using H 2O. Add as many H2O's as necessary to the side deficient in O.

4. As if with acid solution, balance the H's by adding H+ to the side deficient in H.

5. Switch over to base by adding the same number of OH- as there are H+ to both sides of the half reaction.

6. One side of the half reaction will contain H+ and OH-. Convert the n H+ and n OH- to n H2O's.

7. Cancel H2O molecules where possible.

8. Balance the charge by adding electrons to the more positive side to equal the less positive side.

9. Repeat these steps for the second half reaction.

10. If the number of electrons lost does not equal the number of electrons gained, find the least common multiple and multiply each half reaction by the factor necessary to obtain the least common multiple.

11. Add the two half reactions together, cancelling the electrons, and any OH- and/or H2O where possible.

fishD
05-26-2012, 10:41 AM
bump

fishD
05-26-2012, 10:47 AM
Thermodynamics

First law

The first law of thermodynamics (http://en.wikipedia.org/wiki/First_law_of_thermodynamics) may be expressed by several forms of the fundamental thermodynamic relation (http://en.wikipedia.org/wiki/Fundamental_thermodynamic_relation) for a closed system:
Increase in internal energy of a system = heat supplied to the system - work done by the system. U = Q - W For a thermodynamic cycle (http://en.wikipedia.org/wiki/Thermodynamic_cycle), the net heat (http://en.wikipedia.org/wiki/Heat) supplied to the system equals the net work (http://en.wikipedia.org/wiki/Work_%28thermodynamics%29) done by the system. More specifically, the First Law encompasses the following three principles:


The law of conservation of energy (http://en.wikipedia.org/wiki/Conservation_of_energy)

This states that energy can be neither created nor destroyed. However, energy can change forms, and energy can flow from one place to another. The total energy of an isolated system remains the same.


The flow of heat (http://en.wikipedia.org/wiki/Heat) is a form of energy transfer.

(In other words, a quantity of heat that flows from a hot object to a cold object can be expressed as an amount of energy being transferred from the hot object to the cold object.)


Performing work (http://en.wikipedia.org/wiki/Work_%28thermodynamics%29) is a form of energy transfer.

(For example, when a machine lifts a heavy object upwards, some energy is transferred from the machine to the object. The object acquires its energy in the form of gravitational potential energy (http://en.wikipedia.org/wiki/Gravitational_potential_energy) in this example.)



Combining these three principles gives the first law of thermodynamics.






Second law

The second law of thermodynamics (http://en.wikipedia.org/wiki/Second_law_of_thermodynamics) asserts the existence of a quantity called the entropy (http://en.wikipedia.org/wiki/Entropy) of a system and further states that
When two isolated systems (http://en.wikipedia.org/wiki/Isolated_system) in separate but nearby regions of space, each in thermodynamic equilibrium (http://en.wikipedia.org/wiki/Thermodynamic_equilibrium) in itself (but not necessarily in equilibrium with each other at first) are at some time allowed to interact, breaking the isolation that separates the two systems, allowing them to exchange matter or energy, they will eventually reach a mutual thermodynamic equilibrium. The sum of the entropies (http://en.wikipedia.org/wiki/Entropy) of the initial, isolated systems is less than or equal to the entropy of the final combination of exchanging systems. In the process of reaching a new thermodynamic equilibrium, total entropy has increased, or at least has not decreased.
It follows that the entropy of an isolated macroscopic system never decreases. The second law states that spontaneous natural processes increase entropy overall, or in another formulation that heat can spontaneously be conducted or radiated only from a higher-temperature region to a lower-temperature region, but not the other way around.
The second law refers to a wide variety of processes, reversible and irreversible. Its main import is to tell about irreversibility.
The prime example of irreversibility is in the transfer of heat by conduction or radiation. It was known long before the discovery of the notion of entropy that when two bodies of different temperatures are connected with each other by purely thermal connection, conductive or radiative, then heat always flows from the hotter body to the colder one. This fact is part of the basic idea of heat, and is related also to the so-called zeroth law, though the textbooks' statements of the zeroth law are usually reticent about that, because they have been influenced by Carathéodory's basing his axiomatics on the law of conservation of energy and trying to make heat seem a theoretically derivative concept instead of an axiomatically accepted one. Šilahvý (1997) notes that Carathéodory's approach does not work for the description of irreversible processes that involve both heat conduction and conversion of kinetic energy into internal energy by viscosity (which is another prime example of irreversibility), because "the mechanical power and the rate of heating are not expressible as differential forms in the 'external parameters'".[10] (http://en.wikipedia.org/wiki/Laws_of_thermodynamics#cite_note-9)
The second law tells also about kinds of irreversibility other than heat transfer, and the notion of entropy is needed to provide that wider scope of the law.
According to the second law of thermodynamics, in a reversible heat transfer, an element of heat transferred, δQ, is the product of the temperature (T), both of the system and of the sources or destination of the heat, with the increment (dS) of the system's conjugate variable, its entropy (http://en.wikipedia.org/wiki/Entropy) (S)
http://upload.wikimedia.org/wikipedia/en/math/0/6/d/06d4d643bc8b4821d724e0a8e2274bee.png[1] (http://en.wikipedia.org/wiki/Laws_of_thermodynamics#cite_note-Guggenheim_1985-0) The second law defines entropy (http://en.wikipedia.org/wiki/Entropy), which may be viewed not only as a macroscopic variable of classical thermodynamics, but may also be viewed as a measure of deficiency of physical information about the microscopic details of the motion and configuration of the system, given only predictable experimental reproducibility of bulk or macroscopic (http://en.wikipedia.org/wiki/Macroscopic) behavior as specified by macroscopic variables that allow the distinction to be made between heat and work. More exactly, the law asserts that for two given macroscopically specified states of a system, there is a quantity called the difference of entropy between them. The entropy difference tells how much additional microscopic physical information is needed to specify one of the macroscopically specified states, given the macroscopic specification of the other, which is often a conveniently chosen reference state. It is often convenient to presuppose the reference state and not to explicitly state it. A final condition of a natural process always contains microscopically specifiable effects which are not fully and exactly predictable from the macroscopic specification of the initial condition of the process. This is why entropy increases in natural processes. The entropy increase tells how much extra microscopic information is needed to tell the final macroscopically specified state from the initial macroscopically specified state.[/URL]

(http://en.wikipedia.org/wiki/Laws_of_thermodynamics#cite_note-10)
Third law

The third law of thermodynamics (http://en.wikipedia.org/wiki/Third_law_of_thermodynamics) is sometimes stated as follows:
The entropy (http://en.wikipedia.org/wiki/Entropy) of a perfect crystal (http://en.wikipedia.org/wiki/Crystal) at absolute zero (http://en.wikipedia.org/wiki/Absolute_zero) is exactly equal to zero.
At zero temperature the system must be in a state with the minimum thermal energy. This statement holds true if the perfect crystal has only one state with minimum energy (http://en.wikipedia.org/wiki/Microstate_%28statistical_mechanics%29). Entropy is related to the number of possible microstates according to S = kBln(Ω), where S is the entropy of the system, kB Boltzmann's constant, and Ω the number of microstates (e.g. possible configurations of atoms). At absolute zero there is only 1 microstate possible (Ω=1) and ln(1) = 0.
A more general form of the third law that applies to systems such as glasses (http://en.wikipedia.org/wiki/Glass) that may have more than one minimum energy state:
The entropy of a system approaches a constant value as the temperature approaches zero.
The constant value (not necessarily zero) is called the residual entropy (http://en.wikipedia.org/wiki/Residual_entropy) of the system.


[URL="http://en.wikipedia.org/wiki/Laws_of_thermodynamics#cite_note-10"]
(http://en.wikipedia.org/wiki/Laws_of_thermodynamics#cite_note-10)

fishD
05-26-2012, 10:52 AM
bump

KayBX
06-17-2017, 06:08 AM
චරිත දිසානායක සර්ගෙයි අනුෂ්ක ඉදුනිල් සර්ගෙයි නෝට්,ටියුට් ටිකක් හොයාගන්න පුලුවන්ද ?

Don Udugama
06-17-2017, 06:54 AM
bump

spoofywoofy
06-17-2017, 09:35 AM
BUMP

NK123
06-17-2017, 12:44 PM
bp

chamathkara 295
06-22-2017, 08:14 AM
අප්පට සිරි chem is try..හොඳ වැඩක් මල්ලි කරගෙන යන්න අපිත් ගොඩ ගියෙ chemistry හින්ද තමයි.ඕක හරියට ඇද ගන්න පුලුවන් නම්
අනිත් දෙක නිකම් ගොඩ යනව.මම නම් inorganic වලට කවි එහෙම හදල මතක තියාගත්ත.organic ලොවෙත් චිත්‍ර ඇදිල්ලක්නෙ.පාඩම් කරපු හැටි හෙම ඕනෙනම් ඉස්සරහට කතා කරමු.ජය.බම්ප් පිට බම්ප්