text:
Relation between thermodynamic reversible process and reversible reaction
I know it seems to be a weird question. But for long I have been thinking whether there is any relation between thermodynamic reversible process and reversible reaction. Do they have any connection and if so how?

To complement other answers, I provided here definitions extracted from official IUPAC sources:
reversible process
A definition comes via the concept of entropy (Ref. 1):

Quantity the change in which is equal to the heat brought to the system in a reversible process at constant temperature divided by that temperature. Entropy is zero for an ideally ordered crystal at $\pu{0 K}$. In statistical thermodynamics
$$S=k \ln W$$
where $k$ is the Boltzmann constant and $W$ the number of possible arrangements of the system.

The second sentence is a formulation of the third law, while the last is a definition of entropy viewed from the context of statistical thermodynamics. The first sentence describes the thermodynamic entropy. Coupled with the second law expressed as an inequality relation between heat and entropy,
$$\Delta S \ge \frac{q}{T} \tag{const. T}$$
it provides a functional definition of a reversible process (it is a process that maximizes $q$).
reversible reaction
The definition provided by IUPAC is related to microscopic reversibility:

In a reversible reaction, the mechanism in one direction is exactly the reverse of the mechanism in the other direction. This does not apply to reactions that begin with a photochemical excitation.

Such a definition implies thermodynamic reversibility.
References

IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). Online version (2019-) created by S. J. Chalk. ISBN 0-9678550-9-8. https://doi.org/10.1351/goldbook.

Reversible reaction
According to Merriam Webster, a reversible reaction is:

a reaction that takes place in either direction according to conditions (as the formation of hydriodic acid by union of hydrogen and iodine or its decomposition into these elements)

For this type of reaction, you would use a double-harpoon to write the chemical equation:
$$\ce{H2(g) + I2(g) <=> 2HI(g)}$$
and would be able to write an equilibrium constant expression
$$K = \frac{[\ce{HI}]^2}{[\ce{H2}][\ce{I2}]}$$
with [] denoting activity or fugacity. If you start with pure reactants, the reaction would go forward (decrease in reactants, increase in products). If you start with pure products, the reaction would go backwards (increase in reactants, decrease in products). Once the reaction has reached equilibrium, concentrations would no longer change.
In any case, reactions in both directions would occur at the particular level. When the reaction is at equilibrium, you just won't know at the macroscopic level and would have to disturb the equilibrium to see macroscopic changes again.
Reversible process
Again according to Merriam Webster, a reversible process is:

an ideal process or series of changes of a system which is in complete equilibrium at each stage such that when the process is reversed each of the changes both internal and external is reversed but with the amount of transferred energy unaltered

Applied to chemical reactions, a reversible process is one where the Gibbs energy is constant throughout (or where the entropy of the universe is not increased by it). We would not expect any macroscopic changes (the process is at equilibrium), which why this is an ideal process (not real). The closest real process is one where the system is near equilibrium, and the increase in entropy of the universe is minimal.
What does IUPAC say?
The IUPAC Gold book makes reference to a paper (DOI: https://doi.org/10.1351/pac199466051077) offering a glossary of terms used in physical organic chemistry. In their definition of chemical equilibrium, it says:

While the definition of reversible process matches the one I cite above, and they use the same notation for the equilibrium reaction. However, the connection between reversible process and chemical equilibrium surprises me. When a reaction approaches equilibrium, the Gibbs energy does change, and it will not return to the initial state unless work is done on the system. Maybe the confusion regarding reversible process vs. reaction stems in part from this.

I agree with both the MW and the IUPAC definitions, but perhaps you can help me checking if I am getting this right.
If a reactive system involving a reaction that is reversible in the chemical sense (i.e. it "goes both ways" around its equilibrium (i.e. the  activities ( ~ concentrations) of its components can match the value of the equilibrium constant $\mathrm{K(T)}$ ), then as a whole there are changes within the system that satisfy $\Delta{S_{system}} = 0$ (i.e. reversible in a thermodynamic sense) because the reversible reactions within have molar changes of entropy + $\Delta S$ (for the forward reaction, typically non-zero) and $- \Delta S$ (for the backward reactions, with a minus sign in front), with forward and backward reactions at the same rate (dynamic equilibrium).
In a previous reply, I attempted the following wording, which perhaps is more detailed but does not make explicit the distinction between entropy of reaction and entropy of the system:
With $\Delta G = \Delta H - T\Delta S$ for a reversible reaction, where $\Delta G$ is the change in Gibbs energy, $\Delta H$ the enthalpy of reaction and $\Delta S$ the change in entropy (and $T$ and $P$ constant), I think what is going on then for a system where a dynamic equilibrium is established between backward and forward reaction, which could qualify as a reversible process, once the equilibrium is reached. $T$ sets the value of the equilibrium constant and hence $\Delta G$ for these conditions. $\Delta H$ and $\Delta S$ typically are considered to be fairly constant (you could see how $\Delta S$ may in fact be dependent on temperature though, just through different vibration energies of different bonds).
In the special case where $\Delta G = 0$ (when equilibrium constant $K = 1$), with $\Delta G = \Delta H - T\Delta S$, one has to assume that the $\Delta S$ is fully reversible, which I think assumes faultless conversion of $T\Delta S$ to $\Delta H$ and vice-versa, between reacting species.


text:
Who Was the First Chemist?
I know chemistry has been around for a long time, but I was curious who the first chemist in recorded history. I am aware that Antoine Lavoisier was one of the first modern chemistry and is considered to be the "Father of Modern Chemistry", but was he is the first chemist?

I'll write it up later, but Robert Boyle is often considered the first modern chemist.

That depends how you define "chemist."
Today we think of chemists as working at the molecular level. However Lavoisier didn't have any concrete evidence of an atom's existence. He worked with Sulfates (SO4-2), metal oxides, and other molecular groups in an empirical an algebraic fashion.
Throughout much of the 1800s there was enormous debate about the very existence of atoms. Much of the problem arose from confusion over molecular weights and stoichiometry. Physicists were the most stubborn opponents of atoms, which we still can't fully explain even today. Eventually Advodagro's number and accepted stoichiometry combined to form irrefutable evidence of molecules. Many scientists were completely convinced that atoms existed long before they were proven though. Dalton was the first to introduce stoichiometry so he would be a good candidate as well.

The first chemist? The first person who learned that if you mix crushed plant root with the runoff from fermented vegetation (alcohol) you got something you could use to make your robe a bit more colourful? If he experimented with various plants and liquid ratios he would meet the modern definition of "chemist". I recall doing similar things in grade 9 Chemistry class.
That individual is long lost in the mists of time. The main problem with history is we only know what people bothered to write down, or left behind in a durable form. In the days before patents or research for it's own sake all this would have been closely guarded secrets recorded (if at all) in shorthand on small pieces of whatever passed as paper in those days.
The first chemistry research paper was possibly found in the Great Library of Alexandria, which unfortunately burned down on more than one occasion.

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how to measure dissolved oxygen?
How do you measure dissolved oxygen reliably in a culture of cells? This page has info on one sensor: http://water.me.vccs.edu/concepts/domeasure.html
how would you know that this sensor is working - meaning that only dissolved oxygen triggers the electrical change in the sensor, and not other atoms? and how would you calibrate the sensor to translate changes in electric signal to the change in number of dissolved oxygen atoms?
also, how would one get pure dissolved oxygen in varying amounts as 'positive control' for the sensor?

To start, the figure and explanation in the link are almost comically misleading. Potassium chloride solution does not "attract" oxygen. The oxygen flowing through the wire is pretty silly and

Where the probe joins the wire, oxygen mingles with the electricity.
  Oxygen is not very ionized, meaning that it does not have a negative
  charge as electricity does, so the oxygen dilutes the current at the
  electrode beyond the probe.

is just plain ridiculous. This one is better.
The way this type of probe works is by amperometric reduction of $\ce{O2}$, the first example being the Clark electrode. Essentially, an electrolyte is separated from the measurement solution by a porous film of poly(tetrafluoroethylene) (Teflon). These pores are small enough that only small substances like oxygen can diffuse through. On the electrolyte side, there are two electrodes in one of two possible configurations:

In the original Clark-type electrode, the working electrode is platinum, which is chemically-inert but catalyzes the reduction of oxygen. The counter electrode is a AgCl-coated Ag electrode that acts as a reference for the measurement. (Its potential is supposed to remain constant) To reduce the oxygen, an external potential must be applied across the two electrodes (normally using a potentiostat).
The galvanic cell–type electrode has one electrode made of silver (where the oxygen reduction occurs) and one made of an easily oxidized metal like lead. It works the same way as the Clark electrode, but instead of applying an external potential, the current required to reduce the oxygen is supplied by the oxidation of the lead electrode.

Any oxygen that reaches the working electrode is reduced and produces a current that is measured to produce the analytical signal. The amount of current produced depends on how much oxygen diffuses through the membrane, which depends on the concentration of dissolved oxygen in the test solution.

Now, for your questions:

how would you know that this sensor is working - meaning that only
  dissolved oxygen triggers the electrical change in the sensor, and not
  other atoms?

This one is a bit tricky. You're right that oxygen isn't the only thing that could produce a signal—amperometry isn't a very selective technique on its own. You can control the redox reactions that occur at the working electrode, to a limited degree, by changing the potential you apply, i.e. different substances require different potentials to be reduced and won't be reduced if an insufficient potential is applied. In the case of this specific type of electrode, selectivity comes mainly from the PTFE membrane. Since it limits diffusion to only a few very small molecules, that greatly increases the selectivity of the electrode, though depending on the solution you're trying to measure, there still may be interferences.

and how would you calibrate the sensor to translate changes in
  electric signal to the change in number of dissolved oxygen atoms?

In principle, you could calculate the number of reduction reactions happening at the surface from Faraday's laws, but to turn this into a measure of the exact concentration in solution requires a lot of thought of how the transport of oxygen to the electrode occurs, from diffusion from the electrolyte to the electrode surface and transport from the test solution into the electrolyte. In practice, the assumption that the measured current is directly proportional to the concentration in solution is made (and is valid under many conditions), and a calibration is made using a standard solution.

also, how would one get pure dissolved oxygen in varying amounts as 'positive control' for the sensor?

Calibration solutions seem to be commonly made using tables of oxygen saturation at different temperatures. It's also possible to measure dissolved oxygen through other means, such as titration. Solutions can be made up, titrated, then the value used to calibrate an electrode.

Now, for the measurement specifically in cell cultures, a membrane electrode may not be the best bet, at least not without sample pre-treatment. The PTFE membrane is quite susceptible to adsorption of proteins and the like, which will probably affect the electrode's response. It may be necessary to somehow remove these contaminants (without affecting the dissolved oxygen concentration). Other options that might be viable are colourimetric measurements or titration, though there are problems with these, as well.

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What do chemical constituent measurement units 'g%g' and 'µg g⁻¹' mean?
I'm trying to understand the following data being presented in this paper measuring the mineral content of the Cannabis sativa plant (leaves and inflorescences):

I need to interpret these data readings in terms of simple percentages.
What would the first two examples - sodium and potassium - translate to in simple percentage / ppm etc.?

I am not sure if Mathematics would be a better site for this question, (the theme/context is chemistry though the question itself specifically on mathematical units of data measurement), suggestions welcome about that!

I'm not sure I've seen the g%g, but micrograms/gram is interpreted as micrograms of the element per gram of the sample, so in parts per million.  I would lean toward g%g being a percentage, so parts per hundred.

$\pu{g\%g}$ literally means grams per cent grams or grams of mineral per $\pu{100g}$ of sample. So, in population 1 there is $\pu{2.33g}$ of $\ce{K}$ per $\pu{100g}$ of leaves. Not bad when you consider cooked spinach only has about$\pu{0.05\%}$ $\ce{K}$! Eat your greens, kids. About $\pu{2\%}$ $\ce{K}$ in plants is average across a broad spectrum of plants.
$\pu{ug g-1}$ is micrograms of mineral per gram of sample. Population 1 has 23 micrograms of $\ce{Na}$ per gram of leaves.
You should be able to convert these to percents or ppm, as needed, and check in the text as to whether this refers to dry weight of plant or not. I'm guessing with this multivariate array of mineral analyses, the origin of particular cannabis crops could be determined.


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Confusion with Law of Equivalence
Law of equivalence states that

One equivalent of an element combines with one equivalent of the other.

Equivalent of $\ce{Na}$ is $22.99$ and that of $\ce{O}$ is $8$. By the law of equivalence, $1$ equivalent of $\ce{Na}$ combines with $1$ equivalent of $\ce{O}$. But in $\ce{Na2O2}$, $$ 23 \times 2 ~\text{parts by weight of sodium combines with } 16 \times 2~\text{parts by weight of oxygen} \\\implies 23~\text{parts by weight of sodium combines with } 16~\text{parts by weight of oxygen}$$ which means one equivalent of sodium combines with two equivalents of oxygen which is contradiction to law of equivalence. So why does this compound not follow the law of equivalence?

The idea of a chemical equivalent relies on a compound's properties with respect to the usual reactions that occur with a few select species that tend to behave in very predictable ways.  In this case, the number of grams of sodium that would react with 8 grams of elemental oxygen, 23g Na, are one equivalent of sodium for oxidation-reduction purposes.
$\ce{Na_2O_2}$ is sodium peroxide.  It contains oxygen (normally a very well behaved element, chemically speaking) behaving in an unusual way, bonding not only with sodium but also with another oxygen atom.  The oxygen can be thought of as part of a peroxide ion, $\ce{O2^2-}$.
If you know about oxidation and reduction, then you'll remember that oxygen usually has an oxidation number of -2, but when it is in a peroxide it has an oxidation number of -1.  Because this particular compound involves the -1 state, it takes twice as much oxygen to react with a given amount of sodium (which in both compounds has an oxidation state of +1.
Finally, and perhaps more important in answering the root source of your trouble here, a scientific law is a thing that describes the behavior of the universe in specific conditions.  Outside of those conditions the law does not apply.  Even inside those conditions laws are often incorrect to some degree (though still useful approximations.)  It's not unexpected that we would find some exceptions to this law.

Sir,what is the equivalent of sodium in sodium peroxide? My book says sodium doesn't exhibit variable equivalent weight so eq.wt of $\ce{Na}$ is $23$ . But to me, it is $11.5$ . Which is right?

@user36790 One equivalent of sodium would still be 23 grams.  In this case it is oxygen that is behaving oddly.  One equivalent of oxygen in a peroxide would be 16 grams instead of 8 grams.


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How can I estimate reaction rates in water with computational chemistry?
I want to estimate rates of substitution reaction of esters with nucleophiles in water.
How can I estimate them with a computational chemistry software (preferably GAMESS(US))?
Or is it practically impossible?
I know that this is possible in gas phase, but I don't know how to do it in solution.

Finding solvent  effect in computational chemistry is still a crucial part and subject to deep research effort. Now there are two types of models that are being used in computational chemistry.
One is explicit solvation model: Here you will use explicit solvent molecule with your reactant or products and then using first principles you will find the equilibrium configuration but the problem is, it is very expensive and the configuration space is so much high that it is practically impossible. For that you have to use molecular dynamics or monte carlo techniques. In our group we use 1000 to 5000 explicit water molecule but we can't do it for large reaction.
Another method is implicit solvation model: Here electrostatic potential is applied to account the solvent effect but this procedure is so expensive that now that potential is replaced by the solvent induced reaction field energy.
In our group we are using a hybrid approach that involves communication of several software at different level of theory. In my quantum chemistry class i used spartan and qchem and they have some implicit solvation but i didn't use GAMESS.
From GAMESS manual i found that they have solvation module. They are using QM/MM approach. That's actually a very good approach. Please see page 9 of this file:  http://www.sdsc.edu/~kimb/gamesstut.pdf
In this following file at page 3 & 4 you will also find the list of keywords that are needed to start a solvation model run
http://www.msg.ameslab.gov/gamess/GAMESS_Manual/input.pdf


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Why is active mass of a pure solid or liquid always taken as unity?

Active mass is defined as the molar concentration ie. number of Gram-moles per litre.

My book then wrote

Active mass of pure solid/liquid is always 1 .

The book reasoned that

Molar concentration is directly proportional to density. Since density of solid or liquid always remains constant, the active mass is taken as 1 .

Really? Doesn't the density of the solid reactant decrease when the forward reaction is about to reach the equilibrium??
In a word, I am greatly confused. What is the reason for taking the active mass 1 ? Please help.

Does the activity of a solid or liquid change over the course of a reaction?
The density of a solid or liquid reactant doesn't change over the course of a reaction. The mass and volume do as it is consumed, but the ratio of the two is constant. If the reaction causes a temperature change then there are small changes in density, but that would also alter the equilibrium constant and the concentrations of the species involved.
The equilibrium constant is based on activities, which you can think of as the ratio of the concentration (molar density) at standard state to the concentration in the reaction. Since solid/liquid density is constant at a given temperature, activity is too.  It takes quite a large change of temperature or pressure to make a substantial change in the density of a substance, assuming no phase changes occur.
What about gases, why don't they show the same phenomenon and have fixed activities? (Asked in comments.)
For solids or liquids, as mass decreases the volume also decreases by a proportional amount. For example, $\mathrm{100mL}$ $\ce{H2O}$ $\mathrm{= 100g}$ and half as much, $\mathrm{50mL}$ $\ce{H2O}$ $\mathrm{= 50g}$. Both have molar densities of $\mathrm{55.5 mol/L}$.) Gases keep the volume of their container, so if you remove some via a reaction, the mass of the gas decreases, but its volume remains the same, and its concentration decreases as a result.
Why is active mass given a value of 1 then?  Is this simply a convention?  Shouldn't its value depend on the compound? (Asked in comments.)
No, it's not a convention. Activities, active mass or any other activities, are really ratios. The ratio here is a comparison between the activity under experimental conditions and the activity at standard state under standard conditions.  For gases that is 1 bar of pressure at 25°C, for solutions that is 1 molar at 25°C and 1 bar, and for solids and liquids that is the molar density of the substance in its most stable phase and allotrope at 25°C and 1 bar.  The activities of real substances require some additional work and there are particular compounds that do not behave quite as nicely as expected, but the basic idea of activity being the ratio between experimental state and standard state is sound.
With that in mind, consider the activity of solids and liquids.  Their standard state is their density at standard conditions.  Their experimental state is their density at experimental conditions.  It takes significant pressure and temperature changes to cause substantial changes in the density of a solid or liquid, so their experimental state is approximately equal to their standard state, and the ratio of these two states is therefore quite close to 1.

can you please provide some relevant text/reference to read more on the same topic.

@Md_Asher I'm sorry, I don't have a good current reference, I just remember this from my schooling and later use as a chemistry teacher.  As is often the case, the Wikipedia article on the topic of Thermodynamic Activity is well written.  A good college level Physical Chemistry text should include a discussion of activity.

The activity of a liquid can change over the course of a reaction, for instance, for a liquid-liquid equilibrium, where both liquids are miscible.

See also https://chemistry.stackexchange.com/a/59331/104336

Density is an intensive property that is:

An intensive property is a bulk property, meaning that it is a physical property of a system that does not depend on the system size or the amount of material in the system. Examples of intensive properties include temperature, refractive index, density, and hardness of an object.

Probably that answers your question, for more detail see this

This one as well.  It's quoted fine, but please cite the source.

Does the activity of a solid or liquid change over the course of a reaction?
The density of a solid or liquid reactant doesn't change over the course of a reaction. The mass and volume do as it is consumed, but the ratio of the two is constant. If the reaction causes a temperature change then there are small changes in density, but that would also alter the equilibrium constant and the concentrations of the species involved.
The equilibrium constant is based on activities, which you can think of as the ratio of the concentration (molar density) at standard state to the concentration in the reaction. Since solid/liquid density is constant at a given temperature, activity is too. It takes quite a large change of temperature or pressure to make a substantial change in the density of a substance, assuming no phase changes occur.
What about gases, why don't they show the same phenomenon and have fixed activities? (Asked in comments.)
For solids or liquids, as mass decreases the volume also decreases by a proportional amount. For example, 100mL H2O =100g and half as much, 50mL H2O =50g. Both have molar densities of 55.5mol/L.) Gases keep the volume of their container, so if you remove some via a reaction, the mass of the gas decreases, but its volume remains the same, and its concentration decreases as a result.
Why is active mass given a value of 1 then? Is this simply a convention? Shouldn't its value depend on the compound? (Asked in comments.)
No, it's not a convention. Activities, active mass or any other activities, are really ratios. The ratio here is a comparison between the activity under experimental conditions and the activity at standard state under standard conditions. For gases that is 1 bar of pressure at 25°C, for solutions that is 1 molar at 25°C and 1 bar, and for solids and liquids that is the molar density of the substance in its most stable phase and allotrope at 25°C and 1 bar. The activities of real substances require some additional work and there are particular compounds that do not behave quite as nicely as expected, but the basic idea of activity being the ratio between experimental state and standard state is sound.
With that in mind, consider the activity of solids and liquids. Their standard state is their density at standard conditions. Their experimental state is their density at experimental conditions. It takes significant pressure and temperature changes to cause substantial changes in the density of a solid or liquid, so their experimental state is approximately equal to their standard state, and the ratio of these two states is therefore quite close to 1.

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text:
Why fossil fuel is too dangerous?
This seems stupid question, but from many years I am studying that burning/combustion fossil fuel cause pollution because it releases $\ce{CO2}$.
I have recently read about $\ce{CFC}$ here, which is more dangerous than $\ce{CO2}$ and seems it is not release by burning of fossil fuel then why are scientist more concerned about fossil fuel instead of $\ce{CFC}$?
I want proper explanation Why fossil fuel is too dangerous?

Fossil fuels are burned at the rate of billion of barrels a day. The production of the more hazardous CFC's has been banned. See http://en.wikipedia.org/wiki/Chlorofluorocarbon

Why fossil fuel is too dangerous?

Combustion of fossil fuel does not only release $\ce{CO2}$ but many other harmful substances(pollutants) like oxides of sulphur ($\ce{SO_{x}}$), oxides of nitrogen($\ce{NO_{x}}$), Carbon Monoxide($\ce{CO}$), etc...
oxides of sulphur ($\ce{SO_{x}}$):
When sulphur containing fossil fuel is burnt, then oxides of sulphur is produced. Sulphur Dioxide($\ce{SO2}$) is generally produced.
Low concentration of $\ce{SO2}$ causes diseases like asthma, swelling and irritation of respiratory tract.
High concentration $\ce{SO2}$ leads to stiffness of flower buds, which eventually fall off!
Oxides of Nitrogen($\ce{NO_{x}}$):
When in automobile engine, fossil fuel is burnt at high temperature significant amount of nitric oxide($\ce{NO}$) and nitrogen dioxide($\ce{NO2}$) is released.
$\ce{NO2}$ is lung irritant. it is highly toxic to living tissue.
$\ce{NO2}$ decrease the rate of photosynthesis, cause respiratory diseases in children, etc.
Carbon monoxide($\ce{CO}$):
$\ce{CO}$ is produced due to incomplete combustion of carbon. It is major constituent in smoke of vehicles.
$\ce{CO}$ binds with haemoglobin and form carboxyhemoglobin, which is 300 times more stable than usual oxygen-haemoglobin! If concentration of carboxyhemoglobin reaches 3 to 4% then oxygen carrying capacity of blood reduces.
Few more!
$\ce{SO2}$ with smoke and fog is responsible for classical smog(London smog)
$\ce{NO2}$ and $\ce{NO}$ is responsible foe photochemical smog(Los-Angeles smog)
Update:
I was reading more about environmental chemistry and i discovered that $\ce{NO_{x}}$ is responsible for 40% depletion  of concentration of Ozone layer!

why are scientist more concerned about fossil fuel instead of CFC?

As commented by LDC3 The production of the more hazardous CFC's has been banned.
There are 95 derivatives of chloro and bromo fluorocarbon which are listed as ODS(Ozone Depleting Substances).
On 16th September, 1987 93 countries sign Montreal Protocol and accept not to use ODS.

Reference: GSEB textbook pdf

Don't forget that even though other gases may be better greenhouse gases, their concentrations are so low compared to that of water and CO2, they're not significant contributors to the greenhouse effect.

@MichaelD.M.Dryden Yes are not in significant amount. But I think OP wanted to know about other harmful substances.

Could you cite some of the sources for your numbers that you use?

@jonsca I m afraid, this is from my textbook. But I will try to find references.

@Freddy Including the title and author and year of publication is fine.  Ideally we should use the ACS format for citations, but it's not absolutely necessary

@jonsca Sorry for delay. But i have added pdf link of book. :)


text:
Can ketones next to a nitrogen hetero atom tautomerise?
I worry that the electronegative N will stabilise keto form so much that equilibrium will be so far towards keto it would be unreasonable to call the enolisiation process a factor

If I understand your question correctly, I think you're asking if an amide can enolize like a ketone.  The answer to that question is yes.

The enol content of most organic compounds is very low and special methods must be used to measure the enol content.  Perhaps the question becomes, "is there more or less enol content in an amide than in an ester, for example"?
We can draw resonance structures for an amide (or ester) showing that the nitrogen stabilizes the amide.  But above we also drew resonance structures showing how the nitrogen can also stabilize the enol.  We can draw similar structures for both an ester and its enol as well.  Although in the ester case we wind up putting a positive charge on oxygen in some of the resonance structures, so they might not count for as much as they do in the amide case - but both the starting ester and its enol would both be affected similarly.  So as a first approximation I might guess that the enol content in an ester and an amide would be similar.

Indeed the enol content in malonamide is similar to the enol content in diethyl malonate (see page 34 here)

With a nitrogen adjacent to a carbonyl moiety, the functional group is an amide, and the tautomer involving a $\ce{C=N}$ bond is called an imidic acid. The amide is usually vastly more stable and typically predominates by many orders of magnitude. This can be rationalized by several considerations. First, the two tautomers differ by the exchange of a $\ce{C=O}$ bond with a $\ce{C=N}$ bond and an $\ce{N-H}$ with an $\ce{O-H}$, but the $\ce{C=O}$ bond is particularly strong, making the amide enthalpically more stable. Second, if you consider the resonance structures (i.e., mesomers) available to the two respective tautomers, the amide has a contributing structure in which a negative formal charge is placed on the more electronegative oxygen atom, whereas the corresponding structure in the imidic acid has the charge placed on a less electronegative nitrogen atom. Granted, resonance is not a real physical process, but examination of these structures gives some idea of the distribution of electron density in the frontier orbitals, which too points to favorable increase in stability for the amide. You can apply similar logic to the case of tautomers involving a $\ce{C=C}$ bond, where the ultimate conclusion should be the same (and the difference in stability between tautomers even greater).

Greg, maybe I'm misunderstanding the question, but isn't the OP asking about enolization away from amide nitrogen?  In that case which tautomer has a C=N  double bond?

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text:
Binding forces in crystalline solids
I am referring to rocks in this instance. Individual crystals grow together in an igneous rock as it solidifies. I know that the $\ce{SiO4}$ tetrahedron is the basic building unit of silicate minerals, and metal cations attach to ensure electroneutrality.
What I am wondering, however, is what force keeps the crystals together (as in bound to one another) in the solid? For comparison, sediment grains are held in place by a silica or calcite cement in fragmental rocks, but what stops individual crystals from breaking free in crystalline solids?

So, more 'Binding forces in polycrystalline solids' then?  The energy to introduce a grain boundary into a single crystal depends on the relative orientation of the two resulting grains, but there is some atomic rearrangement (strains or reconstruction), and you will end up with a number of bonds across the grain boundary.  The boundaries are usually weaker than the bulk crystal, which is why they preferentially fracture along the grain boundary.

Converting to answer: While the energy to introduce a grain boundary into a single crystal depends on the relative orientation of the two resulting grains, there will be some atomic rearrangement (strains or reconstruction), and you will end up with a number of bonds across the grain boundary. The boundaries are usually weaker than the bulk crystal, which is why they preferentially fracture along the grain boundary.


text:
Yttrium -- rare earth or transition metal?
The element yttrium is called a rare earth element, yet periodic tables label it as a transition metal.
Why is that?
Thanks

Do you perhaps mean to ask why Y is counted amongst the rare earths even though it isn't an f-element?

Uh, why not both?

Comparing "transition elements" to "rare earth elements" (REE) is a bit like comparing apples and oranges. It all boils down to the different ways in which the elements can be classified.
Let's take a look at the periodic table:

(modified from this)
The rare earth elements are marked with a red outline.
Why are all rare earth elements are technically transition elements?
Strictly speaking, transition elements are the elements in groups 3 to 12 of the periodic table. The rare earth elements, in their widest sense, are Sc, Y and the lanthanides, La to Lu. You can see that if you group the lanthanides as one element, they fit nicely below Y in group 3 of the periodic table. The lanthanides are occasionally called the "inner transition elements".
Why is yttrium a rare earth element?
There are chemical and historical to the inclusion of yttrium in the rare earth elements.
Historically, yttrium was among the first elements to be isolated from the REE-bearing minerals of the Ytterby mine in Sweden. What was though as first to be "yttria" was later shown to contain also the rest of the heavy REE. It just took a longer time to purify them.
Chemically, yttrium commonly occurs with the other REE in the same mineral deposits around the world. You probably know the the chemical properties of the lanthanide series vary continuously from La to Lu. Yttrium also follows that pattern, and in terms of chemical properties it sits very nicely somewhere between Dy and Ho (that's why it's associated in mineral deposits with the heavy REE).
Yttrium is commonly used in the same high-tech applications as the other REE: it is commonly alloyed with other REE for speciality materials. It is also commonly used in combination with the other REE in lasers and various detectors in electron microscopy and analytical chemistry.
A different look
If you take a look at this figure, taken from here: https://earthscience.stackexchange.com/a/2951/725

You can see that most transition elements, as they appear in nature, usually have a cationic charge between 60 and 80 pm. The REE (including Y) are somewhat larger than that, resulting in slightly different properties.
You may also be interested in reading the answers to the following questions, also here on Stack Exchange:
What are rare earths and why do they cluster near alkaline magmatism?
Why are the rare earths erbium (Er), terbium (Tb), yttrium (Y) and ytterbium (Yb) named like that?


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How to efficiently reduce of ilmenite to pure titanium?
I am looking for a process in which reduces ilmenite to pure titanium with the highest yield possible. We are to use a process of our choice or one that we have created in conjunction with FactSage to calculate the yield of titanium.
the materials we have available are:

Pure Metals

$\ce{Al, Ca, Mg, Na, Si}$

Pure Non-metals

$\ce{Ar, C, Cl2, H2, N2, O2, S }$

Molecules

$\ce{CH4, CO, CO2, HCl, H2O, H2S, NH4}$

There are a few process that come to mind such as reduction using Magnesium, $\ce{H2SO4}$, high temperature $\ce{CO2}$, but any help on which would be the most efficient combined with the highest yield would be great.

@Micheal it would be better, if you would use the mhchem package to format chemistry, rather than HTML.

Thermite Reduction is easiest.

Ilmenite mainly consists of $\ce{TiO2}$ and iron oxides. The idealized formula of ilmenite is approximately $\ce{FeTiO3}$.
Note that the aluminothermic process would yield ferrotitanium alloys, and the reduction with carbon would yield titanium carbide.
The titanium ore can be converted to titanium tetrachloride according to
$$\ce{TiO2 + 2Cl2 + 2C -> TiCl4 + 2CO}$$
or
$$\ce{2FeTiO3 + 7Cl2 + 6C -> 2TiCl4 + 2FeCl3 + 6CO}$$
(If a large content of iron oxides causes problem, it can be partially removed first by partial reduction with carbon.)
Titanium tetrachloride (boiling point: 136.5 °C) can be purified by fractional distillation.
Finally, titanium tetrachloride can be reduced to titanium sponge with magnesium (Kroll process):
$$\ce{TiCl4 + 2Mg -> Ti + 2MgCl2}$$
The remaining $\ce{Mg}$ and $\ce{MgCl2}$ can be removed with hydrochloric acid.


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How to find the mass spectrum of Cl3+?
The setting of this problem is on the moon of Jupiter: Io.

Given that chlorine exists in two isotopes: $\ce{{}^{35}_{17}Cl}~(33.3\%)$ and $\ce{{}^{37}_{17}Cl}~(66.7\%)$, what is the mass spectrum of $\ce{Cl3+}$?

What I know now is that there are 3 peaks in the mass spectrum of $\ce{Cl2+}$ because there are 3 possible combinations ($\ce{{}^{35}_{17}Cl}$ + $\ce{{}^{35}_{17}Cl}$ / $\ce{{}^{35}_{17}Cl}$ + $\ce{{}^{37}_{17}Cl}$ / $\ce{{}^{37}_{17}Cl}$ + $\ce{{}^{37}_{17}Cl}$). Therefore, there are 4 possible combinations and 4 peaks for $\ce{Cl3+}$.
However, I do not know how to make a mass spectrum graph with only this much information.

Consider Pascal's Triangle.

Okay, let's tackle this. First of all you worked out correctly that there are four different combinations, resulting in four different peaks. You have also given the probability for finding a $\ce{{}^{35}_{17}Cl}$ is $a=33.3\%\approx\frac13$ and the probability for finding a $\ce{{}^{37}_{17}Cl}$ is $b=66.7\%\approx\frac23$. Since you have three positions to fill, this will result in a binomial formula of third order, $(a+b)^3$. You can solve this easily with Pascal's triangle, $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$. You spectrum should therefore look like this table:
\begin{array}{lrr}\hline
   & \text{m/z} &\text{abundance}/\%\\\hline
\ce{[{}^{35}_{17}Cl_3]^{+}}                & 105   &  3.7  \\
\ce{[{}^{35}_{17}Cl_2{}^{37}_{17}Cl]^{+}}  & 107   & 22.2  \\
\ce{[{}^{35}_{17}Cl{}^{37}_{17}Cl_2]^{+}}  & 109   & 44.4  \\
\ce{[{}^{37}_{17}Cl_3]^{+}}                & 111   & 29.7  \\\hline
\end{array}
The following snowball diagram should further demonstrate this:

I don't know mass spectro stuff but I guess in this line - "since you have three positions to fill, this will result in a binomial formula of third order" you probably meant "since you have four positions to fill..." (as there are four rows in your table)

@GaurangTandon No. Three atoms, three positions, three rows in the diagram, four different isotope combinations.

Aren't there eight possible combinations of isotopes? (as I see from the bottom row of your graph, and as is also easily observed from the fact there are three atoms in $\ce{Cl3+}$ each having two choices)

@GaurangTandon there are four different masses you can measure, because positions are irrelevant for that. There are six possible combinations, as there are symmetry equivalent ones.

"Six possible combinations" of what...?

@GaurangTandon Last row count all and substract the summery equivalent ones, you have 6 different ones left; four different sum formulas.


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What happens when adding swimming pool salt to a freshwater pool?
Why does adding swimming pool salt to a freshwater swimming pool cause a white sediment to build up and stick to the floor of the pool?

Because the salt did not dissolve?

First of, what do you understand by "swimming pool salt"? There are millions of different "salts".

Salt used in swimming pools is exactly like table salt.  The only difference is, the salt is ground more finely to motivate faster, more complete dissolution.  Clearly some salt did not dissolve fast enough.  Pools that use chlorinators (a device built into the pump's pipeline) generally require salt to produce chlorine.


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Compressible Factor
I 've a soft question on the terminology of compressibility factor. It is termed compressible factor because, its value determines how any reals gas is compressible with respect to an ideal gas according as its value-less than or greater than 1. If $Z>1$ then $\frac{pV}{nRT}>1$ and so $p>\frac{nRT}{V}$ which means it is more compressible than ideal gas. Because for ideal gas it is $p = \frac{nRT}{V}$. Simillar for negative deviation. But the question arises why we have not taken $V$ on LHS in these equations and inequations. I mean why does scientists doesn't named this as something like "volumetric factor" (actually can't find any specific word, but it is not important, hope readers can understand my question.) I mean something like Volume is greater than of ideal gases.
And another one is we always have graphs of $Z \rightarrow p$ to understand the deviation in real gases from ideal gases. (I'm posting a graph for instance.)

Why we don't take $T$ or $V$ on the X-axis to understand this concept.

The compressibility factor measures the ratio of actual volume of the gas (under the given conditions of temperature and pressure) to the volume the gas must have occupied if it were an ideal gas (under the same temperature and pressure). This ratio tells us how much compressible is a gas (more compressible if the volume is lesser) by a given pressure applied on it. At this point, we tend to take the temperature as constant and assume the volume is a function of pressure alone. Then, suppose changing the pressure and observe how the volume changes for the gas. This is what the graph in your question shows. It illustrates the variation in compressibilities of different real gases with pressure. The term as well as the graph makes sense in that they show the willingness of a given real gas to get compressed upon subjecting them to various pressures.

Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site.

I look at it in a different way; not about volume but rather the relationship between molar volume and 'ideal' molar volume, $\frac{V}{n}$  and $\frac{RT}{P}$.
From $PV = ZnRT$ write
$$\frac{V}{n} = Z\frac{RT}{P}$$
The reason the compressibility factor Z is called 'compressibility' factor is that it relates these two quantities like a 'spring' constant. It relates how many molecules you can actually 'squeeze' or rather  'compress' into a volume at some particular state of temperature and pressure.      For Z > 1 the number of molecules, n is reduced for the same volume, temperature and pressure than when Z = 1.

authors: Богдан Войтенко


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Why does delocalization of π electrons bring stability?
My book mentions that

Delocalization of $\pi$ electrons makes a compound stable as the process of delocalization involves loss in internal energy. This is the synopsis of resonance.

But how does shifting of $\pi$ electrons let the body lose internal energy? What is the cause?
While studying resonance, I observed that one structure possesses charge while in other no charge. But in the final structure, there is, according to the book, partial charge present.
Eg.
In phenol, three resonating structures contain '+' charge on $\ce{O}$ . Thus, in the final structure, there is a slightly positive charge on oxygen and also, small negative charge on the main body.
Now my question is, why there is partial negative charge on the body? And why does oxygen carry partial positive charge? What is the cause?

@edkinsella  The solution to the "particle in a box" problem shows that energy levels are a function of 1/L^2, where L is the length of the box.  So even with just 1 electron in a molecular orbital, where electron repulsion doesn't come into play, the longer we make the box (e.g. the longer the delocalization path) the more stable (lower in energy) the electron.

In simple terms electrons repel each other, so distributing them over a geater number of bonds will reduce the repulsive force between them so resulting in a lower energy state. In the case of phenol the oxygen p orbital lone pair at right angles to the ring can overlap with the delocalised p orbital above and below the ring enabling a degree of delocalisation into the ring system so leaving a slight + charge on the oxygen. This in turn will reduce the electron density in the OH bond so weakening it and accounting for the weak acidity of phenol.

@user36790 You were probably thinking of a,b-unsaturated carbonyl that is commonly drawn as the oxygen carrying the partial charge. Now think of a,b-saturated carbonyl. It has the same amount of electrons but more space for the electrons to fly in. The oxygen does not draw more electrons than it can handle, the partial charge at the oxygen will not increase to infinity when you add double bonds to the other end.

Q1.How does shifting of π electrons let the body lose internal energy? What is the cause?
Delocalization of electrons constitute a current in the molecule which leads to an increase in randomness/ entropy in the molecule that leads to stability.
Also, according to molecular orbital diagram, the pi-electrons finally achieve an overall lower energy level on interaction with other pi-electrons

Q2. Why there is partial negative charge on the body? And why does oxygen carry partial positive charge? What is the cause?
The first thing you need to understand is that the resonance structures (canonical structures or contributing structures) are hypothetical and individually do not represent any real molecule. They contribute to the actual structure in proportion to their stability
In phenol, three resonating structures contain '+' charge on O . Thus, in the final structure, there is a slightly positive charge on oxygen and also, partial negative charge on the benzene ring

"They contribute to the actual structure in proportion to their stability" that is absolutely wrong. You say before that, that they are purely hypothetical and do not represent the molecule. They are not stable in any sense of the word. And please do not call them resonating structures. The concept is called resonance. The structures do not resonate with each other. It is a mathematical trick.

@Martin-マーチン link Head to page 20 of the pdf.

It's very likely that ncert is getting this wrong; they are one of the worst offenders when it comes to outdated, deprecated, sloppy, and incomplete information. Also, it should be fairly obvious, that in any sense of the words, hypothetical and stable are mutually exclusive.


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How do I approach drawing all of the possible constitutional isomers?
Does anyone have any tips on how to approach these types of questions?
A specific example would be:

Draw all constitutional isomers of $\ce{C2H4O}$

I can get the easier ones like vinyl alcohol, but it's hard to get ethylene oxide and it takes a long time for me to get acetaldehdye. And how would I know if I got all of the possible constitutional isomers?

I approach questions like this by considering all possibilities of arrangements of the non-hydrogen atoms. Obviously, there's C-C-O, C-O-C, and then the cyclic ethylene oxide.
Then I make sure I've matched bond orders to get the hydrogen count.
So I get vinyl alcohol, acetaldehyde, and ethylene oxide. H2C-O-CH2 would only work for a biradical, which is probably not expected.
The systematic fashion then is to arrange the "heavy" atoms, first in all linear forms, then all cyclic forms, then all two-ring forms, etc. Of course with large formulas, the number of isomers goes up exponentially.


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Resonance structures of lysine and cause of its basic nature
Lysine is one of the most important amino acids. IUPAC name it as 2,6 - diaminohexanoic acid . But what are its resonance forms?? And can resonance tell why it is basic??? Please help.

The group that constitutes the side chain is nothing but a primary amine. That explains the high pKa of ~ 10.5 (compare propylamine (pka ~10.9).  No need to invoke resonance.  Are you perhaps thinking of arginine?

+1. Sir, first of all, thanks for the answer. I did want to know the resonating structures. So, can you please??

There is just one Lewis structure.  The charge is confined to the side-chain nitrogen.

Sir,can you tell why there is dot-dot bond between $O$ atoms? Doesn't this mean resonance??

That's the carboxylate functionality, which explains the smallest pKa of ~ 2.2

The alpha amine is also basic, though not as much.


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Do each and every resonant structures contribute to a resonance hybrid?

There can be many resonating forms for a compound. It is to be noted that they are not in any equilibrium to each other. Each and every structure will contribute to the ultimate hybrid, according to their stability. The most important point is that resonating hybrid is not a mixture of the resonating structures, but rather an intermediate of them.

This is what my book writes. But if it is intermediate, it must contain at least some properties of each and every structure. Now, one structure has no charge, while other structure has one positive charge and one negative charge. In the final structure, it still contains both the charges. So, how can it be an intermediate as it got no contribution from the first?

What molecule are you referring to?

Brief answer about charge - if the molecule has no net formal charge, then it will always have no net formal charge. Every resonance structure must always have no net formal charge. However, individual atoms may have partial formal charges, and the net partial charge on an atom will be the weighted average over the formal and partial charges on that atom in each resonance contributor.

@Marko: I asked the quo thinking about phenol. Phenol has 5 resonating str. 3 of them contain $+$ & $-$ charges;the other being neutral. In the hybrid also, charges remain present. So, if the hybrid is intermediate, what does it take from the two neutral strs.?

You can understand the relationship between the resonating forms for a compound and the ultimate hybrid by comparing it with the relationship between the different projections of a vector on the three coordinate axes and the vector itself. None of the projection can lonely represent the vector. But all of them can define the vector precisely in the space. Each projection contributes in defining the modulus of the vector according to its own value.

That's a great analogy. Take an example: phenol. It has 5 resonating structures. 3 of them are charged; the other 2 being neutral. Now, like the original vector, the hybrid must be contributed by the 5 strs. But the hybrid is also charged. So, how can it be an intermediate, since it has no contribution from the 2 neutral strs??

And neutral str. is more stable than dipolar str. So, why don't the more stable neutral structures contribute to the hybrid of phenol??

First of all the the neutral stru. is not more stable than dipolar str. (phenol is a slightly polar molecule). Secondly, you have three charged resonating structures counter two uncharged. So, it's normal that the charged str. weighed more in the hybrid of phenol. You should also remember, that I've given you an analogy to explain the mesmerism. So, it has its own limits. It cannot explain everything.

Sir, firstly, your analogy was great & has no flaw. Actually both the neutral ones contribute as the hybrid contains only partial charge,not whole charge. Now, regarding neutral being more stable, my  book...

...cites an example: there are two resonanting strs of 1,3 di pentene . First,neutral and second the polar. Then the book says, the neutral one is more stable.

Regarding 1,3 di pentene,  the neutral resonating str is,of course, more stable because this molecule is non-polar. This is not the case for phenol.


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Why is bipolar resonating structure less stable than monopolar structure?
In my book, it is written:

Resonance energy of a system involving monopolar resonance structures is greater than that involving bipolar resonance structures.

The book cites example:

Carboxylate ion which contains negative monopolar charge is in fact more stable than the corresponding carboxylic acid which contains both negative and positive charge.

But what's the reason?? What is going on?? Please help me with a math-free explanation.

This sounds weird to me, and googling for "monopolar resonance structures" gives only this page (besides antenna construction).

@user36790 Could you cite the source of the quote, too, please.

In addition to Max's good answer, it is a complete logical fallacy to go into "this resonance contributor is better than that resonance contributor". Neither contributor is a real thing. However, you can talk about "which contributor more closely resembles the true structure of the hybrid".

In the diagram below, the 'bipolar resonance structure' refers to the resonance structure with both positive and negative partial charges (left), while the monopolar resonance structure refers to the anionic form in the right. I don't think this terminology is appropriate, and I have never encountered it before. I will refer to the structures as 'protonated' (left) and 'deprotonated' (right).

The diagram above illustrates the scenario. While it is true that the resonance stabilization is greater for the deprotonated form, the protonated form is still overall more stable. The book is inaccurate in that regard.
The "resonance stabilization" refers to electrons being donated into empty orbitals within the molecule. In this case, there is a lone pair in the hydroxide oxygen atom being donated into the antibonding pi orbital of the C = O. A way to rationalize the effect is that, when the proton is attached to the oxygen, you have both the H and the O pulling on the electrons. When the H is removed the electrons will more readily flow into the C = O pi*. Since both O atoms are equivalent, you end up forming an orbital that looks like an extended pi bond.
The stabilization due to the resonance of the deprotonated form is of greater magnitude than the stabilization due to resonance of the protonated form. However, the proton-oxygen bond is more stabilizing than the resonance, so the protonated form is overall more stable.
Diagram soure: http://chemwiki.ucdavis.edu/Organic_Chemistry/Carboxylic_Acids/Properties_of_Carboxylic_Acids/Physical_Properties_of_Carboxylic_Acids


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Stereo Isomerism - syn/anti
I was studying the types of stereo-isomerism E-Z, Cis-Trans and then I come across syn-anti stereo-isomerism this one is giving me a bit of fight in understanding, especially where lone pairs are taken into consideration.How do classify the molecules with lone pairs in consideration as syn and anti Please help me out here!

Use of the terms "syn" and "anti" to describe the geometry about double bonds is no longer encouraged, but you'll still come across it, particularly in older articles.  Syn and anti are identical to Z(usammen) and E(ntgegen) and were often used to describe the geometry about carbon-nitrogen double bonds. In such cases, the lone pair of electrons is given the lowest priority and the sequence rule applied as usual.  For example, in the following N-methyl imines, the imine on the left has the highest priority group attached to the carbon in the double bond (ethyl) on the same side as the highest priority group attached to the nitrogen in the double bond  (the methyl has higher priority than the lone pair). Therefore the molecule is the syn isomer (but would now be called the Z isomer).

By the same reasoning, the molecule on the right is the anti of E isomer.  Again, the key point is that the lone pair has a lower priority than any atom.

do we have to cut the double bond perpendicularly and then see the priority groups?

@HydrousCaperilla Yes, cut the double bond perpendicular to the $\ce{C-N}$ axis, then figure out which group has the highest priority at the carbon, then do the same at the nitrogen and see if those two highest priority groups are on the same or opposite sides of the double bond.

There is another use of the terms anti and syn that is prevalent among steroid chemists. While the relationships between rings A and B/ B and C are described as trans in tricycles 1 and 2, the relationship between rings A and C in the former is anti while in the latter it is syn.


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In the wet medium of an ocean, how does a hard shell form?
In general, if you were to try make something hard in a liquid medium (especially water) it is quite difficult to make the material solid.
Things such as mollusks though, have no problem generating hard shells in a wet medium.
What exactly are the physics behind this formation?

You know cement hardens under water. And your body is a pretty wet place but your bones are still solid. So molluscs aren't particularly unique.

Should the question be rephrased to what is the physics behind making hard substances under water?

I think this would be a relevant video from Minute Earth.

Have you noticed lime deposits? hard stuff sticking to the bath , the kettle etc? They come out of the water. It is called precipitation of dissolved elements, and they form a lattice given a seed crystal and the proper temperatur to do so. This is a chemistry question really. Cells in sea organisms utilize the chemical possibilities of building lattices out of available solvents or food elements.

Carbon dioxide from the atmosphere (or from decaying matter in the ocean) reacts with water to form carbonic acid:
$$ \ce{ CO2 + H2O -> H_2CO_3 }$$
and this reacts with calcium ions to form calcium carbonate:
$$ \ce{H2CO3 + Ca^{2+} -> CaCO3 + 2H+ }$$
The solubility of calcium carbonate is about $13~\mathrm{mg \over L}$, so if the concentration of calcium carbonate is greater than this the excess will precipitate out as solid calcium carbonate.
Shell-forming organisms actively absorb calcium from the water around them, so they are able to increase the concentration of calcium carbonate to above $13~\mathrm{mg\over L}$ and precipitate the excess to form their shells. They can get the carbon dioxide from the water around them or from their own metabolism. The actual details of shell formation is far more complex than this as it's controlled by processes with the cells of the organism rather than just being uncontrolled precipitation. I'm not sure how well the details are understood even today.
Incidentally, this is why shelly fauna are not keen on the acidification of the oceans that results from increased atmospheric carbon dioxide. The solubility of calcium carbonate is strongly $\mathrm{pH}$-dependent and rises sharply as the water gets more acid. Given that the shell is usually intended to stop other animals eating you, having your shell dissolve is generally not good for life expectancy.


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how protons(H+) react with matter and cause the solution to be acid?
For example if my hands were put in a HCl solution how does the protons in the solution which in high amounts will interfere/react with my hand(skin->proteins..) and cause it to burn...or how protons react with a plastic bag and cause it to melt.

Well HCl won't desolve all plastics. In case of your hand there are complex proteins involved containing a variety of elements e.g, carbon, nitrogen, oxygen..., when your hand comes in contact of a proton, the proton is a electrophile and seeks negative charge centers, where ever on a protein it finds the negative charge center e.g, on oxygen atom in a hydroxyl(-OH) group in some amino acid in some protein in your hand, it will attach itself to the oxygen and eventually causing it(oxygen) to loose off from the protein as water, because the ion H20^+ isn't stable and the newly formed Oxygen Hydrogen bond isn't easy to break, so the part breaks off as water. millions of such reactions or even trillions happen eventually corroding off the hand or al large part of hand exposed to protons in a mobile medium.
An acid burn isn't actually a burn but corrosion of proteins, and it doesn't either melt things away rather disintegrates them.

it will attach itself to the oxygen and eventually causing it(oxygen) to loose off from the protein as water Do you have a reference to support this claim? I'm no expert, but I'm not sure wether the proteins actually decompose when brought into contact with acid. I would have found it more likely that the change in pH and the accompanying protonation of hydroxyl and amino groups leads to irreversible changes in the conformation of the proteins and thus inactivates them.

The example I gave below is Alanine(An amino acid) reacting with HCl.   _________CH3CH(NH2)COOH + HCl -> H2O + CH3CH(NH2)COCl
As it's a neutralisation reaction (products must be water and a salt).

this reaction is a dehydration reaction that is the reacting molecule looses water molecule

change in pH and the accompanying protonation of hydroxyl and amino groups leads to irreversible changes in the conformation of the proteins and thus inactivates them -  the proteins get decomposed and a part of the protein is detached off as water molecule(in above given reaction of Alanine) which is seen eventually as the protein structure disappearing off. the protein part that still sticks to hand is eventually inactivated as its structure is no longer that of the reacting protein

also when the acid has corroded a part of skin whcih is a protective covering the sensitive proteins inside turn into zwitterions helping the reaction to occur easily.

In the above example I took Alanine found on the inside of our body, a better example would have been Arginine in our skin, but that molecule is a bit large to be shown in the comments.

if you find an answer useful mark it as answered, it helps future users and prevents duplication of questions


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Does radioactivity affect chemical reactions?
Do compounds of radioactive elements show a different behaviour? What happens when the radioactive part of the given compound decays? e.g. Radium chloride; what happens to it when radium decays?

As pure speculation I assume the compound falls apart. The chemical bonding is related to the number of electrons around an atom, and number of electrons is related to number of protons. Let's say an atom goes through alpha decay and loses 2 protons, this changes the charge balance and might disrupt the chemical bonding. If I had sources I'd try to make an answer.

@tschoppi Radioactive decay can be used to intentionally trigger reactions.

During the nuclear reaction, the atomic number Z (and therefore chemical identity) changes (transmutation).
In your example, $\ce{RaCl_2 -> \alpha^{2+} + RnCl_2^{2-} -> He + RnCl_2}$.
Radon being noble gas would react further, e.g. $\ce{RnCl_2 -> Rn + Cl_2}$
If the decaying atom would be a transition metal with ligands, it's properties  should not change so drastically so that the ligands will stay attached.

poor answer, the recoil energy when a decay occurs can break bonds cause other changes to occur.

I'd like to add to the other answers. Although not directly an answer to your question, I do feel it's relevant. Radioactive decay can cause damage to crystalline solids.
Take the example of apatite $\left(\ce{Ca5[PO4]3[F,Cl,OH]}\right)$, which can host a bit of uranium in the crystal structure. When this uranium decays, the resulting particle damages the host crystal. The crystal can then be polished and etched to reveal these so called "fission tracks":

(source: Stanford Fission Track Thermochronology Laboratory)
Heating of a damaged crystal will cause the damage to "heal" and the tracks to disappear. Thus, radioactive decay can indirectly induce chemical reactions. This routinely occurs in the deep Earth. It also has some very useful applications in geochronology, which you can read about in the Wikipedia article about fission tracks.

I think we need to avoid concetrating too much on the atom at which the decay occured, it is true that many radioactive decays will cause bonds to break as a result of the recoil energy. But alpha, beta and gamma radiations coming from the radioactive atom can deposit energy in the solution, liquid, gas or solid where the radioactivity is present and thus cause chemical reactions to occur.
If you irradate water then the first products are the solvated electrons and H2O cations, these can then react to form a range of reactive species such as HO and H radicals. These can then react further to give things like hydrogen gas and hydrogen peroxide. All of these reactive species can cause chemical reactions to occur.
It is interesting that a solution of a radioactive chemical will undergo autoradiolysis which can change the chemical form of the radionuclide. For example if I was to start with a concentrated plutonium solution in the +6 oxidation state in an aqueous solution. Then if I was to seal it up then the radiation could cause the formation of other oxidation states of Pu, so I would soon end up with a mixture of Pu(III), Pu(IV), Pu(V) and Pu(VI). It is possible with additves to try to control the oxidation state of the plutonium (or other metal), so in some systems the chemistry can be tamed.

In addition to radioactive decay, which would, in the case of radium, give radon and an alpha particle (helium nucleus) that will strip 2 electrons from chloride ions or anything else nearby, the radioisotope will have a different mass from the stable isotope. This in turn may affect its reaction rate . The so-called isotope effect is largest in the case of hydrogen (tritium being the radioactive and three times heavier isotope, and it reacts slower in reactions involving the C-H bond.

What about the fundamentally different chemical reactivities of the decay products? I fail to see how this question has anything to do with isotope effects.

One part of the question was do radioactive compounds show a bit different behavior. Isotope effects are part (but not all) of the answer.

I am not asking about reactions they show, but their properties How would example of NaCl (sodium-24, half life 15 hours) follow.

Until the moment a radioactive isotope decays, it behaves like a stable isotope. You do not seem to be asking about how the radioisotope differs up until the moment of decay, and yet after it decays, it is now a different element. Looks like some of the other forum members have covered it pretty well by now.


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Why isn't water an ionic compound?
If two alkali metal atoms join with an oxygen atom, an ionic bond forms. Since hydrogen has the same number of valence electrons as alkali metals, why can't water be ionic?
This is what I'm thinking:
$$\ce{(H+)2O^2-}$$

In other words.. Is water ionic or molecular/covalent?

It seems water can be even "superionic" under high pressure and temperature: http://chemistry.stackexchange.com/questions/26863/what-is-the-meaning-of-superionic

http://chemistry.stackexchange.com/questions/17064/why-is-kf-the-most-ionic-compound

First of all, the difference between ionic and covalent bonds is not sharp. As electronegativity differences increase, you move away from covalent and towards ionic bonds. There are "in between" states like polar covalent, where one side of the bond is stronger but not fully ionic. And this I think is the main reason: hydrogen has fairly high Pauling electronegativity (2.20), rather close to oxygen (3.44), which seems polar covalent overall (and why we get hydrogen bonding with water). In contrast, the alkali metals all have electronegativity less than 1.00, a much bigger difference versus oxygen and thus a more ionic bond.

What do you mean that one size of a polar bond is stronger?

Side, not size. One atom is more electronegative and has the electron more often.

This video from Khan Academy explains this gradual move from ionic to covalent bonding due to differences in electronegativity: https://www.youtube.com/watch?v=126N4hox9YA

No compound is purely ionic. Water is about 33% ionic.
Linus Pauling discusses the ionic character of water in his famous "The Nature of the Chemical Bond" [1] at pages 100-101.
Pauling says as a first approximation water should be considered as having contributions from four resonance structures:
\begin{array}{lc}
&37\% \qquad & \ce{H-O-H} \\
&24\% \qquad & \ce{(H-O)^-~H+} \\
&24\% \qquad & \ce{H+~(O-H)-} \\
&15\% \qquad & \ce{H+~O^{2-}~H+}
\end{array}
However, he goes on to explain that this is based upon considering each $\ce{O-H}$ bond independently, and the fact that the two $\ce{O-H}$ bonds are not independent reduces the contribution of $\ce{H+~O^{2-}~H+}$ somewhat and increases the other three contributions.
In any case, the purely convalent form $\ce{H-O-H}$ should be considered as contributing only somewhere between 37% and 44% according to Pauling.
References

Pauling, L. The Nature of the Chemical Bond and the Structure of Molecules and Crystals: An Introduction to Modern Structural Chemistry, 3rd ed.; Cornell University Press, 1960. ISBN 978-0-8014-0333-0.


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Why is osmium the densest known element?
Why is osmium so dense despite there being heavier elements after it in the periodic table?

Is there any agreement as to whether Iridium is denser or Osmium? They're very close.

The density of an element is related to how many atoms can be placed in a given volume and the weight of the nuclei.  Therefore, the smaller the atomic radius of an atom and the higher the atomic number of the nucleus, the greater the density of the element. The very small atomic radius of osmium results in a small metal-metal separation.  This small atomic separation along with osmium's relatively high atomic number gives rise to osmium's high density.
The small atomic radius can be attributed to the following factors:

f orbitals are very diffuse and therefore result in poor screening of the electrons further out.  In the case of osmium ($\ce{[Xe] 4f^{14} 5d^6 6s^2}$) the poor screening by its 4f orbitals leads to a contraction in the n=5 and n=6 orbitals.
Due to osmium's high atomic number, relativistic effects come into play.  These relativistic effects have been discussed previously on SE Chem for gold and mercury.  A general discussion as well as the mathematics describing the relativistic effect can be found in these earlier posts.  Basically, for heavy nuclei the s electrons must circulate at relativistic speeds in order to maintain a stable orbit (not fall into the nucleus).  Therefore the electron mass increases and the s orbital radius decreases (the p orbital radius also decreases, but to a lesser degree).

The orbital contraction caused by these two effects leads to a small atomic radius for osmium.  Therefore, short metal-metal bonds result.  This is reflected in the small unit cell volume for osmium (27.96 cubic angstroms).  For comparison the unit cell volume for lead is 121.3 cubic angstroms! As a result, more osmium atoms can pack into a given volume compared to other elements
Osmium's relatively high atomic number along with its small atomic radius, as explained above, gives rise to osmium's high density.

That, and the fact that it has a close-packed structure.

Do the relativistic effects reduce the OVERALL size if the atom or just the inner s orbitals (and slightly p)? so basically, is this effect more significant it increasing the mass of the atom or decreasing it's size?

@Michael  Lead has a face centered cubic crystalline structure while osmium crystallizes in a hexagonal close packed structure.  In both structures the lead and osmium atoms occupy around 74% of the volume, yet the densities of lead and osmium are very different (11.34 and 22.61 g/cm^3 respectively).  So I don't see how the crystal structure, per se, plays into the answer.  I fall back to unit cell size, which reflects osmium's small radius, and atomic weight. If I'm missing something, please elaborate.

@user3764899  The relativistic effects reduce the size of all orbitals that have significant density at the nucleus.  Therefore, all s and, to a lesser degree, p orbitals are contracted.  Osmium's 6s orbital (its outermost orbital) is consequently contracted, so overall osmium's size (atomic radius) is reduced.  Look here and notice how osmium's radius sticks out like a sore thumb.  Its radius is by far the lowest for its atomic number.

@ron true. I wasn't saying that it's the only reason - but it definitely helps. A small size does not help if the structure is not close-packed.

Adding a discussion of packing fraction for various crystal structures would make this answer an A+. https://en.wikipedia.org/wiki/Atomic_packing_factor

The argument about cell volumes is fallacious. Looking at molar volumes gives a more reasonable perspective. Pb = 207.2/11.34 = 18.27 cm^3/mole. Os = 190.23/22.59 = 8.421 cm^3/mole. So Os has a molar volume which is 46.09% of that for Pb.


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What type of compound/element might be in these glowing tiles
I saw these tiles and these as well while surfing the net, they obiviously glow when their electrons are exited to a higher state and then de-exite over a long period to give off visible light.
My question is which chemical might get exited by visible light and them emit back visible light for a long time.

Glow in the dark materials are materials that exhibit phosphorescence. The most common materials used for this effect are zinc sulfide and strontium aluminate, doped with various elements that affect the colour of luminescence.
Phosphorescence is similar to fluorescence, where an electron is excited to a higher energy state, but instead of dropping back immediately by emitting a photon, an intersystem crossing event occurs where the electron's spin flips resulting in a triplet state (S=0 to S=1 in the figure). Because the electron has the same spin as the other electron it left behind in the ground state, the transition to the ground state is forbidden and thus occurs much more slowly. In practice, a given material usually exhibits both fluorescence and phosphorescence. Materials that are very phosphorescent have a high probability of intersystem crossing so that when light is shined on them, a large population builds up in triplet states that will release light slowly. What makes a given material have a high intersystem crossing probability is pretty complicated, but it tends to be higher in molecules with large atoms—even things like fused silica glasses are phosphorescent, but very weakly.

Source


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How does a catalytic poison reduce a catalyst's activity?
How does a catalytic poison reduce a catalyst's activity, such as palladium used in the hydrogenation of alkenes or any other such catalyst?

Basically the catalytic poison reduces the activity by slowing or inhibiting the reaction by occupying the surface of the catalyst and therefore the surface area available for the reactants reduces. You can get a general idea from the following text.

Substances which themselves are not catalysts, but when mixed in small quantities with the catalysts increase their efficiency are called as promoters or activators.
Catalyst poisoning refers to the partial or total deactivation of a catalyst caused by exposure to a range of chemical compounds. Poisoning may be desirable when it results in improved selectivity (e.g.Lindlar's catalyst) but may be undesirable when the catalyst is rendered ineffective (e.g. in catalytic converters). Poisoning refers specifically to chemical deactivation, rather than other mechanism of catalyst degradation such as thermal decomposition or physical damage.
Poisoning involves compounds which bonds chemically to the active surface sites of a catalyst. This may have two effects: the total number of catalytic sites or the fraction of the total surface area that has the capability of promoting reaction always decreases, and the average distance that a reactant molecule must diffuse through the pore structure before undergoing reaction may increase. Poisoned sites can no longer accelerate the reaction with which the catalyst was supposed to catalyze. Large scale production of substances such as ammonia in the Haber–Bosch process include steps to remove potential poisons from the product stream.

For more info, read the Wikipedia article on Catalytic poisoning.


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What type of Organic Substance is used in LCD
I know that the liquid crystal displays contain some class of organic compound, which is somehow used to polarise light to make contrast variations(In calculator displays).
My Question is What type of Organic Compound is used in this case and How does it polarise light only when stimulated by external agents such as current in LCD's

Liquid crystals are usually rigid rod-like molecules that tend to show some degree of intermolecular order even when notionally liquids (there is a good summary on Wikipedia). This happens because the rigid polarisable parts of the molecules can have relatively strong directional interactions even in the liquid phase.
Some of the structures they form can rotate the plane of polarised light You can get some intuition about why knowing that polarising lenses consist of highly ordered molecules whose interaction with light is highly direction dependent. In normal liquids the molecular orientations are effectively random so, even in each molecule interacts with light depending on its orientation, this effect washes out due to the randomness of the overall structure. Liquid crystals form structures where there is the possibility or large-scale order allowing macro-effects on polarisation (just like the rigid structure in a polaroid film).
The other thing that makes liquid crystals useful is that the details of the ordering in the liquid can be affected by external electric fields, especially in thin films of the compounds. In principle this is possible when any molecule is polar (that is it has a net electrical dipole). Of course, making an effective display requires a lot of fine tuning of the electrical, mechanical and chemical structure, but the basic details are simply that the molecules are affected by electrical fields and this allows some control of the macro-structure which, in turn, alters the way the liquid crystal interacts with light. the effect is not "turned on" by the field, just altered and that is enough to make a viable display technology.
Liquid crystals have a long history. The first compounds known to show liquid crystal behaviour is cholesterol benzoate, the first molecule shown below, which was dicoveredn in 1888. The other structures show some actual molecules from the early days of LCD displays to modern developments. For more examples of the history there is a great summary from Merck (one of the major manufacturers) in this slideshow.


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How do I find the oxidation number of a coordination complex
I actually know how to calculate the oxidation number for coordination complexes but some complexes like the below are giving some reistance
$\ce{[Cu(NH3)_2][Fe(H2O)_3]}$ (diamminecopper(?) triaquoferrate(?))
I just wrote this example now and don't know wether it exists or not I am rather concerned about how do I find their oxidation numbers

Think of it like this. Ammonia and Water are both neutral molecules, therefore they won't contribute to the charge of the two coordination spheres. There are also no subscripts, which indicates that iron and copper both have the same magnitude of charge. Now copper here could be +1 and iron could be in -1, or copper in +2 and iron in -2. (Of course, this isn't possible since this molecule doesn't exist.) I cannot conclusively say which though, without any further data, such as maybe their spin only angular momenta.
For other complex compounds, i suggest you practice, and eventually you'll get the hang of what oxidation states are common for different metals. Also use the concepts you use for normal covalent compounds. Trust me, it's very easy!


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Lewis Acid Catalysts
I am studying for the International Chemistry Olympiad and some of questions involve synthesis of complicated organic compounds and i have no idea how to do them.
The setting of the questions is usually like this: they give you a rough idea of the end product however some of the groups are replaced with R. They replace all the other intermediates with A, B, C ... They also give the reactants and catalysts for each step.
Most of the time the catalysts are lewis acids. I wondering if you were able to provide mechanisms and general resulting reactions of these catalyses so i could predict what the product will be.
Also, is $\ce{NaBH4}$ a lewis acid?

Welcome to chemistry.se! I am afraid, I do not quite understand what you are asking. I might be better, if you could provide us with a specific question, an example of what you are struggling with. In its current form, this question might be too broad and is likely to be closed. You also might want to consider to ask two separate questions for the two problems you face.

Ask yourself: can NaBH4 accept anymore electrons? Remember the octet rule.

Can you provide a specific example of a question, so we know the exact context?

Hi, thanks for reading my question and sorry for being really confusing. I cant write the question down as it is really long and i cant draw the structures here. Here is the link to the question- file:///C:/Users/pc/Downloads/volume3-icho41-45.pdf

It is the 7th question from the 2011 paper on page 1272.

@the The "link" you posted is a pathname on your computer. It will not help us and we cannot help you.

@bon The edit you suggested was not really necessary, I approved it, but generally I would consider this to be too minor.

Sorry, i am not the best when it comes to computers. However it doesn't really matter, all i want to know is general mechanisms for reactions involving lewis acid catalysts.

Firstly, $\ce{NaBH4}$ is not a Lewis acid because it cannot accept any more electrons. Aqueous $\ce{NaBH4}$ is a commonly used reducing agent often for reduction of carbonyls to their respective alcohols.
The mechanism of Lewis acid catalysis depends on the reaction but often Lewis acids are used to generate cationic electrophiles such as in the Friedel-Crafts reaction.

Thank you for answering my  question. However is there any other reaction mechanisms involving lewis acid catalysts?

@the any reaction where you want a cationic electrophile could quite easily involve a lewis acid. are you thinking of any specific example?


text:
Practical safety of storing potassium
Potassium ($\ce{K}$) seems to be the transitional alkali metal in terms of safety. Lighter than $\ce{K}$ and we can store under oil with no worries. Heavier than $\ce{K}$ and only ampules will work. $\ce{K}$ is such that even under oil it can absorb enough oxygen to form superoxides that are contact explosives. But this is the theory: does anyone have any experience with extended proper storage of potassium?
I ask for two reasons. First, there are lots of online videos showing $\ce{K}$ under oil that the scientist then proceeds to cut, smash, and otherwise manhandle with no fear of detonation. Second, my $\ce{K}$ arrived tarnished, which I have stored under oil, and then vacuumed sealed in a bag, which has shown zero change in visual oxidation. So while I get that leaving an open jar of $\ce{K}$ under oil is problematic, I'm less clear on well stored $\ce{K}$.
So I'd appreciate any comments from practitioners that use $\ce{K}$ regularly. What situations are genuine dangers? Thanks!

@venture: Please undelete your answer. There is nothing wrong with it.

You should have sufficient reputation to cast an undelete vote. If I remember correctly, it needs 3, mine is already on it.

I actually don't have enough reputation--need 4k.

I have never worked with potassium personally, but have experience with other alkali metals.   You are spot on about potassium superoxide.  I have direct knowledge of an explosion that occurred when potassium metal was stored for an extended period of time in mineral oil.  I would therefore be extremely cautious if you intend to store it in this manner.  Factors such as length of storage time, permeability of your container seals and storage temperature can all affect the amount of superoxide that forms.  Based on the evidence I have seen, it is not safe to assume that there will be no transport of oxygen through a mineral oil barrier.  The only way I would work with this metal is in an inert glovebox, period.  That is my own preference, of course, and I always tend to err on the side of caution.

Right, all of this is what I know. And yet, every demonstration I can document seems to only be storing under oil, whereas I am doing that and using inert gas and then sealing the bottle in a vacuum bag (poor man's glove box). So it either suggests that poorly stored $\ce{K}$ is a bit more resistant to manhandling that we might think, or that lots of what we see on YouTube is rather freshly prepared. The source I bought my $\ce{K}$ from said they had stored for an extended period with no ill effects, which suggests 1). But what you say reads true.

I would suspect that the risk increases with the amount of superoxide that forms on the metal surface and at some point reaches a critical level that, if ignited, imparts sufficient energy to start the metallic substrate to ignite. Cutting could be a trigger, and I've read accounts of self ignition. I have personally worked with $K$ but have never experienced autoignition or explosions.

The following excerpt regarding the management of alkali metals is from:
Management of time-sensitive chemicals (I): Misconceptions leading to incidents by Jim Bailey, David Blair, Lydia Boada-Clista, Dan Marsick, David Quigley, Fred Simmons and Helena Whyte
"Slow reactions of alkali metals and their alloys with oxygen to form oxides and superoxides have been well documented*. Even when these metals are stored under mineral oil, oxygen can dissolve in the mineral oil and react. Potassium, under these circumstances, forms yellow or orange coating that can explode or catch fire upon cutting. NaK, a eutectic mixture of potassium and sodium, can undergo a similar reaction. Lithium stored under dry nitrogen can react with the nitrogen to form a nitride. Formation of the nitride is autocatalytic and can eventually lead to autoignition of the lithium. Fires of alkali metals are extremely dangerous and difficult to extinguish."

Management of time-sensitive chemicals (II): Their identification, chemistry and management,Jim Bailey,David Blair,Lydia Boada-Clista,Dan Marsick,David Quigley,Fred Simmons,Helena Whyte: Chemical Health and Safety, Volume 11, Issue 6, November–December 2004, Pages 17–24

I've come to the realization that if you plan to keep chemicals around you need to know all the risks and provide effective management including a plan for periodic inspection.

I have used potassium metal in glass containers stored in mineral oil for 30 years.  You are correct that there is residual oxygen in mineral oil that will "rust (oxidize)" the metal, but I have never seen the yellow/orange superoxide layer that some have described.  If you need it unreacted, then it needs to be stored so as to not come in contact with oxygen. If you are using it to remove residual oxygen from a glove box, then you probably don't want to even open it until you get ready to use it and then open it in the glove box.  If you are using it to demonstrate periodic reactivity, then storing it in mineral oil in a sealed glass container has worked well for me over hundreds of demos for three decades with no issues, even with potassium that has been stored this way for three or more years.


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Enantiomerism and Diasteriomerism
I am really confused between them. I know that an enantiomer is an isomer having optical activity, non super-imposability of image on original configuration, same molecular formula(Correct me if I am wrong). The thing that is causing me trouble is what diffrentiates this from diasteriomerism. It would be appreciated if you provide an example to clarify like a fischer projection

Your definition of enantiomer is correct.
This simple further classification might be helpful.  Let's take 2 molecules with the same molecular formula.  These molecules are either constitutional (or structural) isomers or stereoisomers.  If they have different structures (or connectivity patterns)  and are separated by an energy barrier, then they are constitutional isomers.  If they have the same structure or connectivity pattern, then they are stereoisomers.
Among stereoisomers there are 3 classes that we can place a pair of molecules into.  Take any two stereoisomers and compare them, they are either

the same (2 identical molecules)
mirror images of one another (enantiomers)

If they are not the same or enantiomers, then they are

different, and we call them diastereomers.

One additional thing to add to @ron's answer is that diastereomers have different chemical properties .. they can be distinguished using conventional spectroscopic techniques, or even by measuring thermodynamic parameters (at least in principle .. sometimes the differences are very small and hard to resolve).  They can also have different chemical reactivities.  Apart from their different optical activities, enantiomers have identical chemical properties ... that is to say, you have to do an experiment that exploits their chiral natures to tell them apart.
Another point is that diasteromers always have at least two chiral centers ... a molecule with a single chiral center will have two enantiomers.  A molecule having two chiral centers will have two enantiomers and two disasteromers.  There is a nomenclature convention that can be used to index the various choices: each stereocenter is labeled "R" or "S" depending on whether the components are arranged clockwise (R) or counter-clockwise (S) around it, according to a conventional order of preference.  In the case of a molecule with two stereo centers, the (R,R) and (S,S) stereoisomers are mirror images, and therefore are enantiomers, and the (R,S) and (S,R) stereoisomers are not mirror images, and therefore are usually diastereomers. As pointed out by @ron in the comments, if the molecule is highly symmetrical, the (S,R) and (R,S) forms may actually be indistinguishable, which would be a "meso" compound.

Good point. When I said diastereomers were different, that means in every way imaginable, physical properties, spectrally, reactivity, etc.  BTW, your sentence, "A molecule having two chiral centers will have two enantiomers and two disasteromers" (and some of what follows after that) is not quite correct.  It is possible that the (R,S) and (S,R) forms are identical - this gives rise to a "meso" compound.

Good point about mess compounds .. I had neglected that special case .. I have edited my answer accordingly.

@dtmoore, +1, Your answer helped me in understanding ron's answer well


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Can we make a rusting battery
Since in rusting (oxidation) of Iron, transfer of 4 electrons takes place is it possible to use this reaction under catalytic conditions to create a simple electric cell, even if it just gave a millivolt of potential.
$\ce {Fe + H2O -> FeOH3 + H2}$
Considering that we have complete control over the surroundings, like volume expansion, disposing hydrogen and other such factors.

Yeah in theory is possible, but the efficiency would be se low that it will be practicaly useless :x

@Babounet, I know the efficiency is less but I require it to perform a very small task, that's why. and if we catalyse the reaction it can give good results.

Sure! A nickel-iron battery uses the oxidation of iron at the negative plate to produce about half the voltage. Though invented by Jungner, it's often called an Edison battery. See http://en.wikipedia.org/wiki/Nickel%E2%80%93iron_battery.
There are very low power semiconductor devices which could be powered by an iron battery. You could make three or four small cells from iron nails and a more electronegative element, e.g. lead, to power a digital watch; even without optimization to remove layers of hydrogen and rust, it should operate for months.
For some other low power ideas, see https://www.enocean.com/en/energy-harvesting/.


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What will eventually happen to uncovered vegetable (seed) oil exposed to the air?
If I leave some olive oil uncovered and fully exposed to indoors air for enough time (though away from UV exposure or any other environmental factors), what will happen to the oil?
I read here that some of its constituents ('lighter' molecules) will start to evaporate but that heavier ones will be left behind, which will 'not quite' evaporate.
If that is true (it may not be), would the non-evaporative part of the oil eventually harden and assume a cake-like consistency, almost like wax or hardened animal lard stored at refridgerator temperatures?

Some of the lighter constituents could evaporate. There are a lot of minor components in olive oil, and they are largely responsible for the fragrance and flavor. As these evaporate, the fragrance and flavor would change. It is unlikely that they would all evaporate though, as many of these components likely have high boiling points/low vapor pressures.
The oil itself would not evaporate, but it would not solidify either. At most you might notice a thickening of the oil.
Over long periods of time the oil may react with oxygen to form various oxidation products. This process is so slow, however, that it is definitely not a concern for olive oil you might leave out in your kitchen. The oil will start tasting and smelling stale long before significant oxidation occurs.

Just to be clear, the oil will taste and smell stale (then rancid) due to oxidation.  I realize that you were referring to a more thorough version of the process.  :)

The funny thing is, I was wondering if it may actually be useful for something (as a material) - rather than worrying about it - and upon googling 'epoxidized' I discover: 'Epoxidized soybean oil, better known by its acronym, ESBO, is a plasticizer used in polyvinyl chloride (PVC) plastics' - so there you go...

epoxidized olive oil || that's fantasy. Oxidation does happen, but via radical route, leading to hydroperoxides and products of their decay.

@permeakra, you are correct. I think I just got distracted by the prevalence ESBO and didn't think through the actual mechanisms of oxidation applicable here.

The key to the answer is what kind of oil do you have, there are differences. In general, oils with big amount of double bonds MAY oxidize by oxygen from air. The process is radical, UV-induced, and very slow. Some other oils, with no or too little double bonds do not oxidize. and remain mostly same.
The process of oxidation (drying) initially produces hydroperoxides, later able to serve as a source of more free radicals, leading eventually to inter-molecular linkages. I found contradicting  notions for what these linkages consist of (it is either oxygen bridges or direct links similar to polymers), but the end process if well known and used in varnishes. However, varnishes has extra components: they are usually diluted by some solvent to reduce viscosity and always contain a metal catalyst to accelerate drying of the oil, that also is usually heated for prolonged period of time, again, to increase the speed of drying.
The considerations above ignore obvious process of evaporation of light impurities and possibility of bio-degradation (I'm not aware of bio-degradation processes for pure fatty oils, but there are bacteria that live in almost undiluted toluene and there are bacteria that thrive on petrol, so I'm open to this idea).

Vegetable oils do lose some lighter constituents through evaporation, particularly substances responsible for aroma, but the primary degradation pathways, making it rancid, are through oxidation, hydrolization and microorganism metabolism; see http://www.ivannikolov.com/how-cooking-oil-goes-rancid/ and http://en.wikipedia.org/wiki/Rancidification.
Some tree-nut oil, in particular, quickly develops a noticeable off-taste. BHT, BHA and other additives (often applied to packaging rather than to the food itself) help decrease oxidative rancidity.
That said, vegetable oil used for lubrication in old machine parts certainly does thicken to the consistency of wax, though I do not know how much of that is due to evaporation vs. degradation.