text:
Why don't alcohols undergo nucleophilic substitution with ammonia?
Alcohols can undergo nucleophilic substitution with $\ce{PCl5, POCl3, HCl}$.  Halogenoalkanes can undergo nucleophilic substitution with $\ce{NH3}$. The $\ce{C-O}$ (in alcohol) and $\ce{C-Cl}$ (in chloroalkanes) bond energies are comparable. Why don't alcohols undergo nucleophilic substitution with $\ce{NH3}$?

To build on ssavec's answer: Nucleophilic substitution reactions require two species: a nucleophile (a Lewis Base) and a substrate with a leaving group. Nucleophiles are often generically represented as $\ce{Nu}$ and leavings groups as $\ce{LG}$.
All leaving groups are not created equal. The leaving group ability of a leaving group is defined as the relative rates of a particular analogous series of substitutions. For example, the relative leaving group ability might be based on the following reaction, with rates determined for various leaving groups:
$$\ce{EtOH + PhCH(LG)CH3 -> PhCH(OEt)CH3 + H-LG}$$
This is an $\mathrm{S_N1}$ substitution, so the first (and rate determining) step of the mechanism is loss of the leaving group (and is independent of the nucleophile):
$$\ce{PhCH(LG)CH3 -> PhCH+CH3 + LG-}$$
The relative rates of this reaction are influenced by the stability of the $\ce{LG-}$ anion (see the Hammond Postulate, which proposes that the transition state of an endothermic process resembles the products).
The stability of the $\ce{LG-}$ anion can be predicted from the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid of $\ce{LG-}$, which is controlled by the same ability to stabilize a negative charge.
The conjugate acid of $\ce{Cl-}$ is $\ce{HCl}$, which is a strong acid. Using some Hess's Law trickery, the $\mathrm{p}K_\mathrm{a}$ of $\ce{HCl}$ has been predicted to be around $-8$.
The conjugate acid of $\ce{OH-}$ is $\ce{H2O}$, which has a $\mathrm{p}K_\mathrm{a}$ around $+16$.
$\ce{H2O}$ is a much weaker acid than $\ce{HCl}$, which means that the negative charge on  $\ce{OH-}$ is less stabilized than $\ce{Cl-}$. Thus, $\ce{OH-}$ is a poorer leaving group than $\ce{Cl-}$ (by a lot).
Another good discussion of leaving groups (and the $\mathrm{p}K_\mathrm{a}$ data used above) come from this site.
In your example reaction (ammonia + ethanol), the product of the reaction has a better leaving group ($\ce{NH3}$, conjugate base of $\ce{NH4+}$, which has a $\mathrm{p}K_\mathrm{a}$ of $+9.75$) than the $\ce{OH-}$ leaving group in the reactant, so the reaction will also run in reverse, and the equilibrium will strongly favor the reactants.
$$\ce{CH3CH2OH + NH3 <=> CH3CH2}\color{red}{\ce{NH3+}}\ce{+ OH-}\ \ K_\mathrm{a} \ll 1$$
As for your confusion about bond energies...
Bond dissociation energies are defined for homolytic bond cleavage, in which each atom keeps one electron when the bond breaks (see image).

Substitution reactions involve heterolytic bond cleavage, in which one atom gets both electrons:

Nevertheless the question is wrong basicly, because amines are produced from alcoholes and ammonia at multi-thousands of tonnes each year. This is just a question of conditions. Under high enough temperature and pressure (about ammonia synthesis conditions and catalysts) the reaction takes place.

Oh, well, sure. Catalysts change the mechanisms of reactions, so what was once slow/nonproductive starts to become productive. This question appeared to be at the introductory organic level, and I kept my answer there also.

The $\ce{OH-}$ is a very poor leaving group, however, $\ce{H2O}$ is a very good leaving group. Of course, one only needs to acidify to convert one group to the other. For chloride as the nucleophile, this poses no problems; $\ce{HCl}$ is a strong acid and $\ce{Cl-}$ is a weak conjugate base. However, ammonia is a pretty good base, and it converts to the ammonium ion $\ce{NH4+}$ at acidic pH. This greatly reduces its capability as a nucleophile, and the reaction does not proceed.

Sorry I couldn't mark both as the answer. For the benefit of future viewers of this page, this answer is also brilliant.

In substitution, you must think not only about nucleophile, but also about leaving group. That means, alcohols react only with very good nucleophiles, because $\rm{OH^-}$ is so bad leaving group.

Could you tell me why $\ce{Cl}$ in $\ce{PCl5}$, etc. is a better nucleophile than $\ce{NH3}$ then?

@user2246 PCl5first converts OH into OPCl4 and in succesive intramolecular substitution POCl3 acts as very good leaving group.

Simply because ammonia is a great base(due to the presence of 2 lone pairs) and hence the basic character always dominates over the nucleophilic character.

Ammonia isn't a great base 2. Ammonia doesn't have two lone pairs 3. There are a ton of reactions where ammonia preferentially reacts as a nucleophile rather than as a base. In conclusion, literally everything you've said is wrong.


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Can freezing break up sugar polymers?
I recently freezed a huge bunch of apple slices, and thawed them again a few days later. After thawing, I noticed a few things compared to a batch that I had put in the fridge (no freezing)

The apples turned brown far quicker
After pureeing, they tasted sweeter

I know that slow freezingg can destroy cell walls in animals due to crystallization. My guess what has happened here is that the same mechanism also broke up long chain sugar polymers. Thus the sugars are chemically more active, oxidize more easily after contact with air, and taste stronger than in the other batch.
But is my guess plausible? Do polymers actually break under those mechanical stresses? Is this or a similiar effect observed elsewhere?

I think the question has nothing related to organic chemistry and especially to physical chemistry. But the home-experiment tag is appropriate.

physical chemistry - you are probably right. organic - behavior of organic polymers, I think it's relevant. I'll adjust if someone else chimes in.

Generally speaking, the term "organic polymers" is not about polymers observed in living organisms as well as organic chemistry is not a science about processes in living (or natural) organisms. Organic chemistry is a science about hand-made objects and the word "organic" should not confuse you! The sciences about chemical processes in living organisms are enzymology and molecular biology, and NOT organic chemistry.

I thing organic chemistry should stay. I removed physical chemistry and added food chemistry which is appropriate, as well as home experiment.

@AlexeyPopkov - The sciences about the chemical processes in living organisms are biochemistry and molecular biology. I argue that enzymology is a more specialized field lurking in between the two.

@Ben My main point was about organic chemistry since it is very common mistake to think that this is a science about chemical processes in living organisms or in nature. Thanks for clarification.

Freezing breaks the cell structure of an apple, not individual chains of short polysaccharides you can feel the taste. What happens is probably enzymes which the apple contains in cell structures go out of these structures and split long polysaccharides (which have no taste) into short sugars which you feel as sweet.
In other words, the apple digest itself by its own enzyme systems after breaking the cell structures.

sounds plausible but do you have a source?

I cannot recommend you a book where such question is considered. I can only say that I never saw any evidence in scientific literature that freezing can break short polymers. Moreover, the term "mechanical stress" is unsuitable when we speak on molecular level because "mechanical stress" is strictly macroscopic phenomenon.

I don't have a reference handy either, but this is an effect commonly known in kitchens. Potatoes, cabbage, etc. - their starch gets broken down in sugars not only when frozen, but also at normal fridge temperature.

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What are the states of HX when reacting with alcohol and alkenes respectively, and Br2 when reacting with benzene?
I couldn't find this in any textbooks, so I guess I will ask this here. What is the physical state of $\ce{HX}$ when reacting with alcohol (nucleophilic substitution) and alkenes (electrophilic addition) respectively?

For electrophilic substitution of benzene with bromine, what is the state of bromine?  (e.g. aqueous, liquid, solid)

If possible, can you explain why they have the particular state?

By the theory (not practically) what I know, $HX$ and $Br_2$ must be gases (at room temperature) because of lower molecular weights and the covalent bonds. Go through this link http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch4/gases1.html#roomt, you will get idea about which compound or element will be gas. Usually, $HCl$ is gas at room temperature, but while performing experiments in our labs, we use however solutions of $HCl$, so I think both $HX$ and $Br_2$ will be in liquid state, which is most preferred.

The first reaction ($\ce{HX}$+Ethanol) happens in solution.  High concentration of HX mixed with ethanol will produce the product. However, for $\ce{X}$ other than $\ce{Cl}$, more often one put concentrated phosphoric or sulfuric acid with alcohol and a salt of X (for example, $\ce{KBr}$ or $\ce{KI}$).  Concentrated phosphoric/sulfuric acid are easy to obtain and store in high concentration, as opposed to $\ce{HX}$. It is, therefore, a liquid phase reaction.
For ethene + $\ce{HX}$, it can be done in gas-phase, in an aqueous environment, or you can also add $\ce{HX}$ gas to an ethene solution in a solvent such as $\ce{CCl4}$. So, it can be either gas phase or liquid phase.  It is almost always more preferable to do a reaction in solution in a laboratory set up. (So much that we call the experimental part of the chemistry wet chemistry.)  This constraint is, however, lifted in the industrial world, with the availability of expansive specialized reactors.
And for benzene + $\ce{Br2}$, both reactants are liquid and $\ce{Br2}$ is also soluble in benzene, so the reaction is in also the liquid phase.

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How does sodium in ethanol reduce carbonyl compounds?
According to my teacher, Na in ethanol reduces carbonyl compounds (e.g. aldehydes and ketones) to alcohols. However, I can't see where the hydrogen needed for reduction comes from. How does the reduction take place? (or is my teacher wrong?)

Na provides a single electron in one mechanism. and the use of ethanol is as of a proton donor.
The intermediate of Blanc reduction is an aldehyde which is obtained after repeating the above step twice

Sodium is a powerful reducing agent. It wants to donate an electron to things that are electron deficient.
The mechanism of the reduction of aldehydes and ketones by sodium would be similar to that of the Bouveault-Blanc reduction. Ethanol is the source of the protons (at least early in the reaction). The dots in the reactions below signify radicals (unpaired electrons), which are represented to the best mchem and MathJax's ability (which is not great).

Step 1 - one electron transfer from sodium to carbonyl group:

$$\ce{Na. + RCH=O->Na+ + RCH.-O^-}$$

Step 2 - abstraction of a proton from ethanol:

$$\ce{CH3CH2OH + RCH.-O^- <=> CH3CH2O- + RCH.-OH}$$
This step is an equilibrium step, and is driven toward products by the second electron transfer (which consumes the $\ce{RCH.-OH}$.

Step 3 - one electron transfer from a second sodium atom to the organic radical:

$$\ce{Na. +RCH.-OH -> Na+ + RCH:^{-}{}-OH}$$

Step 4 - second proton transfer from ethanol:

$$\ce{CH3CH2OH + RCH:^{-}{}-OH -> CH3CH2O- + RCH2OH}$$
This reaction, while technically an acid-base equilibrium, essentially goes to completion because of the vast disparity in acidity of $\ce{C-H}$ and $\ce{O-H}$.
The net reaction is:
$$\ce{2Na + RCH=O + 2CH3CH2OH -> RCH2OH + 2CH3CH2ONa}$$
$%edit$

Can sodium-ethanol reduce amides,  acyl chlorides, acid anhydrides and esters ?

@anonymous - feel free to ask a new question.

This scheme looks plausible, except that I'd expect $\ce{RCH:^{-}{}-OH}$  to immediately isomerize to $\ce{ RCH2O-}$ which will then be in equilibrium with the ethoxide as pKas are not vastly different anymore.

@BenNorris I've also read in my book that Na+Alcohol is a strong base and is used for dehydrohalogenation. Is is true that the above combination acts as both reducing agent and a strong base? I'm confused since a strong base must accept a proton instead.

Thanks for the explanation sir, but isn't there a major possibility of formation of vicinal diols in these case?.

authors: نبــأ الحسيني


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Detonation and deflagration
I understand that deflagration is slower and smoother than a detonation, but can't a deflagration be used to initiate detonations of other explosive agents?

I don't know enough about combustion dynamics to give a good answer, but the Google search seems to indicate "Yes": http://en.wikipedia.org/wiki/Deflagration_to_detonation_transition

This article may be of interest http://shepherd.caltech.edu/EDL/publications/reprints/jannaf99_paper.pdf


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Why are halogens ortho para directing even though deactivating?
Halogens bonded to benzene ring has three lone pairs. These three electron pairs can cause resonance in benzene ring. But, halogens are also highly electronegative and thus they have strong -I effect. So, they are deactivating groups. But, why are they ortho para directing?

Electronegativity  is a bonded phenomenon; halogens will tend to pull electrons towards themselves when they are involved in a covalent bond. Electronegativity does not mean that they will keep their own lone pairs bound to themselves, they are quite free to delocalize over the molecule to lower it's energy

Let's take as an example a typical halogen, chlorine. Chlorine withdraws electrons through inductive effect and releases electrons through resonance. Through the inductive effect, chlorine destabilises the intermediate carbocation formed during the electrophilic substitution

Image courtesy: NCERT Chemistry Class 11th
Through resonance, a halogen tends to stabilise the carbocation, and the effect is more pronounced at ortho- and para- positions. However, the inductive effect is stronger than resonance here, and causes net electron withdrawal and thus net deactivation. The resonance effect tends to oppose the inductive effect for the attack at ortho- and para- positions and hence makes the deactivation less for ortho- and para- attack.
Reactivity is thus controlled by the stronger inductive effect and orientation is controlled by resonance effect.

"orientation is controlled by resonance effect"- isn't the  para product usually major? That happens because the inductive effect is weaker at para, right(since resonance is the same at ortho and para)? So shouldn't it be the inductive effect  that determines orientation?

+R effect of halogens which increase electron density on o and p position while
overall electron density has reduced on ring because of -I effect.

Firstly, halogen attached to a benzene ring is somewhat deactivating due to it's -I-effect, as a result it tends to withdraw the electrons from the benzene ring, making electrophilic substitution difficult. Next, due to resonance (as it has lone pair of electrons), there is an increase in electron density only at ortho and para positions with respect to the halogen. Hence, electron loving electrophile will be directed towards these positions. So, yes, they act as ortho and para directors.

As. ortho/ para directors are positively charged so halogens, being ortho/ para directors, are positively charged. This positivity is opposed by high electronegativity of the halogens which, make it difficult to orientate negative charge into tha ring. Thus it becomes difficult to add any substituent.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$. I do not understand what you are trying to say, could you be a little more elaborate.

I am also confused by some of the statements. The otho/para directors are not positively charges groups. Cationic functional groups tend to be meta directors (e.g. $\ce{-NR3+}$). Additionally, electronegativity varies greatly over the halogens (from 4.0 for fluorine to 2.4 or 2.5 for iodine). Iodine is approximately the same electronegativity as carbon.

Halogen are unusual in effect on electrophilic aromatic substitution they are deactiviting yet or tho para directing . deactivation is characteristic of electron withdrawal ,where as ortha para orientation is characteristic of electron release . halogen show both characteristics electron withdrawal through , inductive effect and release through resonance effect and the two effect are evenly balanced.

We know that halogen have - I effect which deactivate the benzene ring but halogen are also carried +M effect which is produminished -i effect because +M effect  is greater than -i effect and as we know that all +M-effect operating group are ortho and para directing because in case of +M effect operating group the more electron density are found in ortho and para possition hence further the electrophilic substitution the elecrophile attack in ortho and para position.


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Attempting to balance a redox reaction using half reactions leaves an unbalanced component
I have this redox reaction that I can't balance using half reaction method.
$$\ce{Mg + HNO3 -> Mg(NO3)2 + N2O + H2O}$$
What I get at the end is
$$\ce{2Mg + 6HNO3 -> 2Mg(NO3)2 + 3N2O +3H2O}$$
I can't get the $\ce{N}$ balanced.

It seems likely that you are starting with the wrong half reactions.
Since this reaction happens in water, separate the ionic species into their ions:
$$\ce{Mg(NO3)2 -> Mg^{2+} + 2NO3-}$$
$$\ce{HNO3 -> H+ + NO3-}$$
Now, one half reaction should involve magnesium (and have no nitrogen anywhere). The other half reaction should involve nitrogen only.
$$\ce{Mg -> Mg^{2+}}$$
$$\ce{NO3- -> N2O}$$


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Strange observation!
Every one have observed that when we pour cool water into a transparent glass (or simply glass), some droplets accumulate on the outside part. Did the droplets come pass through the glass? If they didn't, why is there accumulation of water droplets outside the glass?

Please perform a new observation: Pour a ice cold coca cola inside the glass, wait for the outside droplets and lick them to find out, if it is a coca cola passing through the glass. Then edit and improve your question.

I think many people have already tried the experiment @ssavec proposes sometime in their lives. To get some insight, I think you'll find reading this wiki very helpful: http://en.wikipedia.org/wiki/Condensation (pay special attention to the first picture on the right hand side of the page!)

It seems we are making assumptions about the relative humidity in this thought experiment.

This is happening because of moisture present in air.
When you'll pour the cold water in a glass, the outer surface will also show fall in its temperature.
As you know already know that the heat flows from one body to another body until the temperature of both becomes equal.
So in our case the outer surface of the glass will absorb heat from surrounding, thereby causing in condensation of water present in air as moisture.
Another thing, you might notice that when you take out the bottle of cold water from a refrigerator, it have no droplets present on its outer surface because air inside the refrigerator is dry. But after when you place the same bottle out for a few minutes, you will observe the droplets on the outer surface of bottle.
You can also try to perform your experiment again by pouring normal water (at room temperature) in a glass and then placing it in refrigerator and then take it out after 30-40 minutes. Observe that if there are droplets present on the glass just after you take it out.
See this.


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Is the reactivity of Benzene with an -N(R)₃⁺ group greater than its reactivity with an -NH₃⁺ group?
Both are m-directors, and Electron Withdrawing groups, but which one is more reactive? At first sight, one might think clearly $\ce{-N(R)_3^+}$ is more reactive, but when you factor in that $\ce R$ is actually an o,p-director, hence activator, then I think $\ce{-N(R)_3^+}$ is indeed stronger than $\ce{ NH_3^+}$. I can't seem to find any verification on the internet for this.

R being an o-p director doesn't make NR3 a o-p director.

@ManishEarth and David, Isn't $\ce{NR3}$ in $\ce{ph-NR2}$ o-p directing? If I am not wrong, what is deactivating (and probably m-directing) is an $\ce{NR4^+}$ group as in $\ce{Ph-NR3^+}$? In the same way, Aniline has a o-p directing $-\ce{NH2}$ group but $\ce{NH3^+}$ attached to phenyl is deactivating (not sure if m-directing). Correct me if I am wrong.

@SatwikPasani Yeah, I assumed he's talking about $\ce{Ph-NH3^+}$ and $\ce{Ph-NR3^+}$ which are both meta (rather strong, too). David, is this interpretation correct? Or are you asking about $\ce{Ph-NH2}$ vs $\ce{Ph-NR2}$

@ManishEarth correct

$\ce{Ph-NR_3^+}$  is likely weaker than $\ce{Ph-NH_3^+}$ because the mechanism of deactivation is based on the amount of positive charge localized on the Nitrogen:  Alkylgroups will stabilized the positive charge through sigma donation.
That being said, the difference will be small.  Additionally, there are deprotonation side-reactions to consider for anilium ($\ce{Ph-NH_3^+}$) ions, which may compromise any discussion of deactivation.


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Is molarity an intensive property?
$$\mathrm{molarity} = \frac{\text{amount of solute}}{\text{volume of solution}} $$ and amount of substance is based on quantity (larger mass means larger amount), so how come it is an intensive property. Shouldn't it be an extensive property?

Very firstly, no. of moles doesn't depend on weight, it depends on mass. If you will take 12g of Carbon to the moon, it will have same no. of moles on moon as on earth.

Concentration is an intensive property. The value of the property does not change with scale. Let me give you an example:

Let us say you had a homogenous mixture (solution) of sodium carbonate in water prepared from 112 g of sodium carbonate dissolved in 1031 g of water.

The concentration (in mass percent, or mass of solute per mass of solution) is:
$$c=\frac{112\text{ g solute}}{(112+1031)\text{g solution}}=0.09799 =9.799\%\text{ sodium carbonate by mass}$$
The concentration is the ratio of sodium carbonate to the total mass of the solution, which does not change if you are dealing with the entire 1143 g of the solution or if you dispense some of that solution into another vessel.

If you dispense 11.7 g of that solution into a flask for a reaction, what is the concentration of sodium carbonate in that flask?

It is still 9.799% by mass. The ratio of the mass of sodium carbonate present to the total mass present has not changed. The actual mass of sodium carbonate has changed:
$$0.09799\dfrac{\text{g solute}}{\text{g solution}}\times11.7\text{ g solution}=1.15\text{ g solute}$$
The concentration is a property dependent only on the concentration of the solution, not the amount of solution you have. The concentration of a solution with defined composition is independent of the size of the system.
In general, any property that is a ratio of two extensive properties becomes an intensive property, since both extensive properties will scale similarly with increasing or decreasing size of the system.
Some examples include:

Concentration (including molarity) - ratio of amount of solute (mass, volume, or moles) to amount of solution (mass or volume usually)
Density - ratio of mass of a sample to the volume of the sample
Specific heat - ratio of heat transferred to a sample to the amount of the sample (mass or moles usually, but volume also)

Each of these intensive properties is a ratio of an extensive property we care about (amount of solute, mass of sample, heat transferred) divided by the scale of the system (amount of stuff usually). This is like finding the slope of a graph showing the relationship between two extensive properties. The graph is linear and the value of slope does not change based on how much stuff you have - thus the slope (the ratio) is an intensive property.

Consider the following picture:

Break the ice block shown in the picture into two equal halves.Now I hope you would be able to answer the following questions:
1.What are the physical properties of ice block which got halved?
Absolutely mass,volume,etc.(These are all extensive properties.)
2.What are the physical properties of ice block which remained same?
Density,etc.(These are all intensive property.)
If you have the doubt so as to why the density remained same,here is the explanation:
I hope you know basically even if block got halved,mass per unit volume remains the same in either of the pieces.All the way it mean that density remained the same(mass per unit volume).Thus it is an intensive property.
Similarly if you imagine solution instead of ice block,you will find that molarity remains the same even if you divide solution into two equal halves.Thus molarity is a intensive property.

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What does "/2" in the structure of calcium benzoate mean?
Please see the following figure. This is from the high school chemistry book of my son. He wants to know what does that /2 after $\ce{-COOCa}$ mean.

The book is a non English one. And I have changed the relevant caption to English.
Primary searches over Google have not revealed anything usable. And my major not being Chemistry might have to do something with this.

It means half of calcium.
The organic acid is $\rm{COO}^-$, whereas calcium is $\rm{Ca}^{2+}$. For the structure to be fully correct, there are two benzoic acids per one calcium. Other way round this would be $\rm{(C_6H_5COO)_2Ca}$. Full drawing of this will be way to messy, so it is simpler to put half calcium to one acid.


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Which atoms form which elements?
Is there a diagram/flow chart available that shows which atoms fused together to make each of the elements?  I can find examples for some elements, but nothing which shows all the combinations of atoms which make up each of the naturally occurring elements.

All elements can not be formed by fusion of atoms, assuming you are talking about nuclear fusion. Nuclear fusion is restricted to very few elements and so is all forms of nuclear transmutations.

@stochastic13 - there are many possible fusion reactions that we have good cross section data for. Most of them require putting energy into the reaction, but they are fusion reactions none the less. In particular, H, D, or T at sufficient energies often have nuclear reactions to create net-heavier nuclei than those involved in the reaction.

The general consensus is that each element is made up of unique atoms of that element's type. I address nucleosynthesis below, but this is not how elements become elements.
For example, oxygen is made of oxygen atoms. Titanium is made of titanium atoms. Neon is made of neon atoms.
The first good atomic theory, formulated by John Dalton and refined by Amadeo Avogadro and others has a the following principles (some of which are invalidated by modern experiments):

All matter consists of atoms as fundamental building blocks (we now know that subatomic particles are the real fundamental building blocks).
Atoms cannot be created or destroyed (we now have the ability to do this by nuclear fusion and fission).
Each element has its own characteristic type of atom (still valid).
All atoms of the same element element are chemically and physically indistinguishable (the discovery of isotopes invalidated this one).
Chemical reactions occur by the rearrangement of atoms (this one is still valid).

More modern atomic theory takes into account the discoveries of the subatomic particles and the development of quantum mechanics.
Atoms are now defined by their composition of the three principle subatomic particles: protons, neutrons, and electrons. Any atom can be characterized by three numbers derived from this composition:

Atomic Number ($Z$) - equal to the number of protons.
Mass Number ($A$) - equal to the sum of the numbers of protons and neutrons ($A=Z+N$).
Charge ($Q$) - equal to the difference between the number of protons and the number of electrons ($Q=Z-E$).

The atomic number, $Z$, defines to which element the atom belongs. For example, all atoms with $Z=1$ are hydrogen atoms. All atoms with $Z=6$ are carbon atoms. The periodic table is arranged by increasing atomic number. The atomic number controls the number of electrons an atom prefers (see below), and therefor controls the majority of the chemistry of an atom.
The number of neutrons (mass number) defines the isotope. For example, there are three isotopes of hydrogen, with $A=1$, $A=2$, and $A=3$: protium, deuterium, and tritium. These three isotopes have 0, 1, and 2 neutrons respectively, but they each have only one proton, so each of them is a hydrogen atom. The mass differences influence the properties of these isotopes subtly, but not in such a way that is usually noticeable to the average human.
The number of electrons (or more precisely the charge) determines whether the atom is neutral or an ion(http://en.wikipedia.org/wiki/Monatomic_ion). The number of electrons controls the behavior of an atom chemically (for example the number of covalent bonds that can be formed), and the number of electrons of a neutral atom is the same as the atomic number $Z$. For example, a carbon atom has 6 electrons (and you can go further into the quantum mechanics of electron configuration to learn that two of those electrons are less accessible than the other four, which accounts for most of carbon's chemical behavior). Carbon can have a number of electrons not equal to $Z$. If carbon were to have 7 electrons, the charge on carbon would be $Q=Z-E=6-7=-1$, making carbon an anion. Carbon behaves differently as an anion than it does as a neutral atom.
Atoms of some elements can be made by nuclear reactions. For example, in the sun hydrogen is converted into helium by the following pathway (and others). The notation below follows the generic example $^A_Z \ce{X}^Q$, where $X$ is the atomic symbol (and here $n$ means a neutron with $A=1$ and $Z=1$ and $Q=0$ and $e$ means an electron with $A=0$ and $Z=0$ and $Q=\pm1$). A proton is represented as $^1_1 \ce{H}^+$.
$$\ce{^1_1H + ^1_1H -> ^2_2He}$$
$$\ce{^2_2He -> ^2_1H + e+}$$
$$\ce{^2_1H + ^1_1H -> ^3_2He}$$
$$\ce{^3_2He + ^3_2He -> ^4_2He + ^1_1H + ^1_1H}$$
In addition to the nuclear transformations, gamma rays, neutrinos, and heat and being generated and flung around like it was nothing.
The number of possible nuclear reactions is humongous. I would go check out a book on nuclear reactions or nuclear chemistry from a library to learn more.

Thanks, I had a feeling that it wasn't a case of a nice simple table to follow!

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Photochemical ligand substitution of metal carbonyls - role of water cooled lamp-jacket?
Why, in the photochemical substitution of a ligand in a carbonyl complex, is a water-cooled lamp jacket required to separate the solution and the mercury vapour lamp? It's shown in Figure 22.16 of Atkins' Inorganic Chemistry.

For the benefit of those without the book you mention, it would help if you could add a description of the experiment in question. And welcome to Chemistry StackExchange!

The main point is to stop the heat flow from the lamp, while allowing UV through. You want to do photochemistry, not thermochemistry.


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Symmetry and stability of a molecule (Jahn-Teller Distortion)
I've taught in my class that if a molecule posses high degree of symmetry it's stability will be more,in other words symmetry and stability are directly proportional. Is this statement correct? And if this is correct than why Jahn teller distortion stabilizes a molecule when it's symmetry gets low eventually because of the distortion.

The symmetry and stability are not directly related per se; is benzene more stable that naphthalene for example? Benzene is more 'symmetric' than the naphthalene but both are stable molecules; $D_{6h}$ vs $D_{2h}$ point groups.
Jahn  and Teller discovered that in a non-linear molecule in an orbitally degenerate electronic state, say $e_g$ or $t_{2g}$ in an octahedral complex, that there is an interaction between the electronic orbital motion and nuclear vibrational motion, which causes the minimum energy position  to differ from the symmetric structure. This motion lifts the orbital degeneracy and lowers the energy. Thus the lower symmetry state has lower energy.
An example might be the lengthening of a pair of axial bonds in a octahedral complex, $\ce{[CoF6]}^{3-}$ for example, however, the J_T theorem does not give any indication of how big the distortion might be, i.e. it does not predict the geometrical nature of the distortion. There is also a restriction which is that if the degenerate structure has a centre of symmetry so does the distorted structure.
note: The electronic spin of the molecule (singlet, triplet etc.) is not important it is the orbital degeneracy only that matters. The Jahn-Teller effect is a breakdown of the Born-Oppenheimer approximation which is based on separating nuclear and electronic motion.

I don't get that how come elongation of the axial bond, say in $Cu^{+2}(d^9)$ configuration, brings in the stability? How does the bond elongation affecting the interaction of ligand (present on $z$-axis) and the $d_{z^2}$ orbital?

Well generally you can say that symmetry contributed to the stability. But in most cases you will find exception. The Jahn–Teller effect is most often encountered in octahedral complexes of the transition metals. In octahedral complexes, the Jahn–Teller effect is most pronounced when an odd number of electrons occupy the eg orbitals. This situation arises in complexes with the configurations d9, low-spin d7 or high-spin d4 complexes, all of which have doubly degenerate ground states. In such compounds the eg orbitals involved in the degeneracy point directly at the ligands, so distortion can result in a large energetic stabilisation and is not related to symmetry.


text:
Oil :water partition coefficients
Is there a database or a website with a list of oil : water partition coefficients of substances ?
I actually want the oil:water partition coefficients of :
cholesterol, starch , methyl citrate , citric acid , iso- citrate , methyl urea , carotene

There are a wide range of partition coefficient predictors that can estimate logP on the basis of varying structural heuristics - are you potentially interested in these?

@RichardTerrett I really don't know what all this means. I am a biology student and would like to know these values so that I can predict which materials would pass through the plasma membrane.

Two places where you can find predicted values of LogP (the oil:water partitioning coefficient) are chemicalize.org and Wolfram Alpha.  Both provide a predicted value for most of the compounds you are interested in (chemicalize provides a complete set, assuming you mean betacarotene).
Examples
This link provides a direct link to the output of the cholesterol page from Chemicalize.org and shows the logP value on the left hand side (7.11)
To get the logp from Wolfram alpha you must search for logp  and then chose the predicted link that appears when no results are returned.  This link should get you to the value obtained from Wolfram Alpha (7.02).


text:
How do point defects in a crystal relate to Entropy?
I am interested in Non-Stoichiometry and how it relates to crystal structures; In particular I wish to learn about surfaces of metal oxides. I am aware that point defects in such crystals are a result of an entropy driven reactions. So my question is this:
Are there any good introductory resources on this subject? I have basic undergrad chemistry and physics background.

I would strongly recommend this resource:
http://www.tf.uni-kiel.de/matwis/amat/def_en/
It is a good complete on line introduction to defects theory in crystals.
Some key points listed below.
The concentration of point defects $c_{PD}$ at equilibrium can be expressed by an Arrhenius formula:
$$c_{PD}=A \cdot e^{-\frac{G_F}{kT}}=A \cdot e^{-\frac{S_F}{kT}}\cdot  e^{-\frac{H_F}{kT}}$$
where $G_F$ is the free energy of formation, $H_F$ the enthalpy of formation and $S_F$ the entropy of formation. In solids  $H_F$ can be confused with the energy of formation, $S_F$ is associated the variation of the vibration entropy of the crystal due to the introduction of a point defect.
Formation enthalpies, and thus formation energies can be obtained by:
$$ H_F(D) = E_{tot}(D) - E_{tot}(X)+\sum_i n_i \mu_i +q (E_F -E_V) $$

$E_{tot}(D)$  and $E_{tot}(X)$: total energies of the system with and
without the defect D.
$n_i$: number of atom removed from the system,
a negative value denote the addition of one atom.
$\mu_i$: chemical potential of the atom $i$. This should consider the chemical environment at the equilibrium condition (which reservoir is to take into account).
$E_F -E_V$: Fermi energy relative to the valence band maximum.
$q (E_F -E_V)$: energy change due to the exchange of electrons and holes with the carrier reservoirs.

For the equilibrium concentration in the multiple defects case an additional entropy term comes from the standard configuration entropy which can be calculated from the Boltzmann formula
$$S=k_B \cdot \ln P$$
where P is the number of configurations for the same macrostate.

Can you summarize some of the pertinent points in your answer in case the link goes dead?

The answer now is more complete


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How is a reduction in the number of molecules associated with a decrease in Entropy?
How is a reduction in the number of molecules associated with a decrease in Entropy? For example, stoichiometric equations like $\ce{H2 + O -> H2O}$ are enthalpy driven if I am correct. But it also implies that entropy is decreasing because the system is becoming "more orderly." Is this intuition correct?

That is true though the magnitude does depend on what phase are the products. If you somehow use solid reactants to produce a product which is gas, the entropy increases.

Interpreting entropy as "disorder", which comes from thermodynamics applications, is only one way to interpret entropy. Entropy can also be described according to statistical mechanics as the statistical distribution of possible degenerate microstates:
$$S=k_B \ln\Omega$$
where $k_B$ is Boltzman's constant and $\Omega$ is the number of microstates. Thus it is a measure of uncertainty. The more equivalent/degenerate microstates there are, the less likely we know exactly which one of them the system is in, and thus, the higher the entropy. This probabilistic interpretation provides a more intuitive and satisfactory understanding of the relationship between number of molecules and entropy of the system.

A decrease in the number of molecules leads to a decrease in the total number of degenerate microstates, which leads to a decrease in entropy. There are more degenerate ways to distribute N objects in M spaces than there are degenerate ways to distribute (N-1) objects in M spaces.

Let's look at a very simple example, in which we have three molecules $\ce{2H2 + O2}$ in a container with room for four molecules. How many ways can we fill four spaces with these three molecules? There are four spaces in which we could put the $\ce{O2}$ molecule:
$$\begin{pmatrix}
\ce{O2} & ... \\
... & ... \\
\end{pmatrix}
\begin{pmatrix}
... & \ce{O2} \\
... & ... \\
\end{pmatrix}\begin{pmatrix}
... & ... \\
\ce{O2} & ... \\
\end{pmatrix}
\begin{pmatrix}
... & ... \\
... & \ce{O2} \\
\end{pmatrix}
$$
How many ways can we put two $\ce{H2}$ molecules in one of these arrangements? In order to answer this, we need to temporarily distinguish one $\ce{H2}$ from the other $\ce{H2^{\circ}}$ (or to distinguish the one we place first from the one we place second) in order to generate all of the degenerate states.
$$\begin{pmatrix}
\ce{O2} & \ce{H2} \\
\ce{H2^{\circ}} & ... \\
\end{pmatrix}\begin{pmatrix}
\ce{O2} & \ce{H2} \\
... & \ce{H2^{\circ}} \\
\end{pmatrix}\begin{pmatrix}
\ce{O2} & \ce{H2^{\circ}} \\
\ce{H2} & ... \\
\end{pmatrix}\begin{pmatrix}
\ce{O2} & ... \\
\ce{H2} & \ce{H2^{\circ}} \\
\end{pmatrix}\begin{pmatrix}
\ce{O2} & \ce{H2^{\circ}} \\
... & \ce{H2} \\
\end{pmatrix}\begin{pmatrix}
\ce{O2} & ... \\
\ce{H2^{\circ}} & \ce{H2} \\
\end{pmatrix}$$
There are six arrangements of two $\ce{H2}$ molecules for each location of $\ce{O2}$ for a total of $\Omega=24$ arrangements. The entropy $S=k_B\ln(24)=4.39\times10^{-23}) \frac{\text{J}}{\text{K}}$. Note that this happened to be $\Omega=4\times3\times2=12$, or 4 ways of placing the first molecule, 3 ways of placing the second, and two ways of placing the third. We will use this shortcut next time we do this.
Now, let us run the reaction, converting hydrogen and oxygen to water.
$$\ce{2H2 + O2 -> 2H2O}$$
And let us repeat the exercise. How many ways can we fit two $\ce{H2O}$ molecules in four spots? There are 4 ways to place the first one, and 3 ways to place the second for each of those 4 placements of the first. Thus, $\Omega=4\times3=12$ and the entropy is $S=k_B\ln(12)=3.43\times10^{-23}) \frac{\text{J}}{\text{K}}$.

Note that there is a subtle flaw in my explanation. The molecules themselves have an intrinsic entropy hidden in their structures as well. According to this site the standard molar entropies of $\ce{O2},\ \ce{H2},$ and $\ce{H2O}$ are (in$\frac{\text{J}}{\text{K}\cdot\text{mol}}$) 205.1, 130.7, and 188.8, respectively. Thus, we have an entropy change of reaction:

$$\Delta_rS^\circ=\sum S^\circ_{products} - \sum S^\circ_{reactants}$$ $$ \Delta_rS^\circ =\left(2\times188.8 - (205.1+2\times130.7)\right)\frac{\text{J}}{\text{K}\cdot\text{mol}}=-40\frac{\text{J}}{\text{K}\cdot\text{mol}}$$
By comparing standard entropies, we also get a decrease in entropy. Standard entropies are easier to work with because we are frequently working on the molar scale. If we changed our simplified example to be moles of $\ce{H2}, \ \ce{O2},$ and $\ce{H2O}$, in 4 moles of spaces we would have the original number of microstates be:
$$\Omega=(4N_A)(4N_A-1)(4N_A-2)=64N_A^3-48N_A^2+8N_A$$
When you plug in $N_A=6.022\times10^{23}$, all but the most staunch calculators and computer programs are going to give up. Wolfram Alpha can do it because it does so symbolically, and we get
$$\Omega=1.398\times10^{73}$$
$$S=8.314\frac{\text{J}}{\text{K}\cdot\text{mol}}\ln(1.398\times10^{73})=1.400\times10^3\frac{\text{J}}{\text{K}\cdot\text{mol}}$$
(At the molar scale we replace $K_B$ with $R$, the ideal as constant.)
Another apparent weakness of the statistical approach is that it assumes no phase change. As F'x mentioned, there is an entropy change associated with a phase change. Roughly, the number of available spaces increases or decreases.

Gases have much larger molar entropy than liquids and solids, owing to the contribution of translational entropy (gas molecules have much less hindered translational movement than molecules in liquids or solids). Thus, for a reaction that diminishes the total number of gas molecules, you can expect to have $\Delta_r S° < 0$.
This is also true to some extent with other phases, as long as all reactants and products are in the same phase. But, if you have a reaction with some products or reactants in the gas phase, those will determine the sign of the molar reaction entropy.


text:
How do I start studying chemical nomenclature?
Chemical nomenclature seems to be the way chemistry "talks", allied with the mathemathics. But where to start?... and what reasons and theories are behind? Some "names" come from latin... I know there are rules IUPAC etc, but it is so demanding and exaustive to learn all that rules at once, for organic and inorganic chemistry. Maybe there are reasons and rules that help us to name a chemical or compound... Since I am interested in metallurgy, can you give me some advice to start? To associate theory to the formula, ad strutural formula, it would be nice. The implicit "why" in everthing included is always welcome.

Welcome to Chemistry.SE. As you say, there is an awfully large amount of nomenclature, not to mention that additional jargon referring to our methods, techniques, instrumentation, etc. There is a reason that it is normally broken up into smaller chunks and spread out over an entire four year degree in chemistry. Oddly enough, the nomenclature of metals alloys is one of the few types that is rarely taught.

Chemical nomenclature is mixture of historical traits, empirical knowledge and little bit of systematics.
And as you mentioned, it is exhaustive to learn all at once.
To understand the nomenclature, you must understand the systems themselves. Because it's their structure and reactivity which is reflected in the name.
This makes an unfortunate cyclic dependency. To start studying any given chemistry branch, you must know its language. But without knowing the science behind, the language does not make sense. It's like that and you probably cannot do anything about it, just cut this circle and crunch through the nomenclature. It will start to make sense, once you get sufficiently deep. Additionally, the "why" is good indicator. Once you have enough knowledge to understand the answer why it is named so, than you can probably answer yourself.


text:
Favorskii rearrangement
Aren't alpha halo carbonyl compounds extremely fast and reactive via $S_N2$ (the fastest i guess)? If this is so, why isn't the major product in the Favorskii rearrangement an alpha hydroxy carbonyl compound?
And if the alpha carbon(to which the halide is attached) is tertiary, there should be a high chance of an elimination product via $E_2$ (leading to the formation of alpha beta unsaturated carbonyl), right?

This is an old question, but in case anyone comes down this road...

Aren't alpha halo carbonyl compounds extremely fast and reactive via $\ce{S_{N}2}$
  (the fastest i guess)?

I don't think so. In an $\ce{S_{N}2}$ reaction a nucleophile is attacking a positively polarized carbon atom.  The more positive charge on the carbon, the more the nucleophile is attracted.  Look at structure I below, it shows one of the resonance structures for a carbonyl that leads to a polarized carbonyl with positive charge on the carbonyl carbon.  Now to whatever degree the adjacent C-Cl bond is polarized with positive charge on carbon, we wind up with an unfavorable electrostatic situation of adjacent positive charges.  Said differently, the presence of a carbonyl group suppresses polarization of the adjacent C-Cl bond, which will make the C-Cl carbon less prone to nucleophilic attack.

And if the alpha carbon(to which the halide is attached) is tertiary,
  there should be a high chance of an elimination product via E 2  (leading to the
  formation of alpha beta unsaturated carbonyl), right?

No, you want to remove hydrogen D in the diagram below.  Hydrogen D is much less acidic than hyrdrogen A because removal of the latter results in a resonance stabilized carbanion.  The difference in acidity between hydrogens A and D is so large that effectively only hydrogen A will be removed by a nucleophile.
These two arguments effectively eliminate pathways C and D from consideration.  Pathway B is simply an equilibrium that goes nowhere, so it has no real effect on the outcome of the reaction. This leaves pathway A, the Favorskii reaction, as the only viable alternative.  Pathway A, besides being the only game in town, is actually favored by two factors, 1) as noted above, hydrogen A is acidic and it's removal produces a resonance stabilized carbanion, and 2) completion of the Favorskii reaction is strongly favored by entropic factors since the carbanion and the C-Cl carbon are contained in the same molecule and held relatively close together.

I mostly agree with your analysis, but I think there may be some caveats worth noting: (a) while I agree with you about the inductive effect of the carbonyl group on $\ce{C-Cl}$ bond polarity, in the case of a halogen on a primary carbon, $\mathrm{S_N2}$ actually seems plausible to me; (b) I'm inclined to think $\mathrm{E_2}$ is quite possible, though perhaps only favorable under reflux conditions (and obviously dependent on reactivity of the hydrogen $\beta$ to the halogen vs. the $\alpha '$ hydrogen). Of course, I'm referring to different substrates than the one pictured.

I was also thinking about the possibility of aldol condensations (given a less bulky molecule), especially since the carbon bearing the halogen has the more acidic hydrogens, but it's hard to imagine a set of conditions favorable for aldol condensation that wouldn't just give rise to elimination products. If you have any thoughts on that, I'd be very glad to have your input.

@Greg E a) if you agree with the carbonyl inductive effect argument, then you should agree that the $\ce{S_{N}2}$ would be at least slowed down, if not eliminated - right? b) If you agree that H(a) is much more acidic then H(d), then wouldn't a nucleophile strongly prefer to interact with H(a)?  What do you mean by "I'm referring to different substrates than the one pictured"

@Greg E  One thought your comment about removing the alpha hydrogen attached to the C-Cl carbon generated is why not eliminate that alpha hydrogen and chlorine to form a carbene(oid).  It could then insert into H(beta) to produce the "elimination" product and\or insert into the other C(alpha)-CO bond to produce cyclopentane with a ketene attached (methylenecyclopentane with a carbonyl stuck on the exo-methylene - sorry its late for me), which then reacts with the nucleophile to give the Favorskii product.

What I mean by "I'm referring to different substrates than the one pictured" is that my comments were about alternate reaction pathways for $\alpha$-halo ketones in general in the presence of appropriate nucleophiles/bases, not cyclohexanone specifically. As far as $S_N2$ reactions, while I think the carbonyl could plausibly reduce the polarization of the $\ce{C-Cl}$ bond, I would be surprised if that effect were large. Furthermore, the $\delta^+$ of the carbonyl carbon might attract nucleophiles, so even if pathway B in your rxn. scheme leads nowhere, the proximity effect could be relevant.

Also, I believe I've read that the transition states of $S_N2$ reactions can be stabilized by mixing of the developing $\sigma-Nu$ bond with the $\pi^*$ of the conjugated EWG, but I could be mistaken there. With less-substituted ketones, the enolate is also presumably less stable, and the most acidic hydrogen(s) would be on the carbon bearing the halogen.

So far these are all kinetic factors, but substitution immediately leads to a thermodynamically more stable product. For the Favorskii reaction, the first step is an unfavorable proton transfer equilibrium, which must be driven forward by the relative irreversibility of the subsequent steps (I assume the expulsion of the halide with formation of the cyclopropanone is the rate-determining one, aided by it being intramolecular). Ultimately, I'm not sure what conclusions to draw, but I think $S_N2$ needs to at least be considered (for substrates with primary carbons bearing the halogen).

As for the $E_2$ pathway, it's not kinetically favorable, but it yields a very thermodynmically stable product, and is obviously entropically favored. Your idea about the carbenoid is definitely interesting, but without something to stabilize it, it just seems doubtful given that there are lots of alternate pathways. Thank you for your responses, by the way. I hope I'm not a nuisance with these wall-of-text comments.

In extension to ron's answer, there are two reasons, why I think a $\ce{S_{N}2}$ reaction is unlikely to occur.

The LUMO, where the nucleophile will attack, of 2-Chlorocyclohexanone is the antibonding orbital of the carbonyl bond. Also the computed natural charges show a much more polarised carbonyl carbon (DF-BP86/dfe2-SVP).

On the german wikipedia page I found a proposed mechanism, that would make use of the LUMO instead of the acidity of the $\alpha$ hydrogen for the rearrangement. I would suspect, that it heavily depends on the substrate and the base, which one is favoured.

The most stable conformation of this compound is where the chlorine is in equatorial position. The antibonding $\ce{C-Cl}$ bond orbital is sterically shielded by the ring framework. The molecule has to flip the ring to make it accessible. The energy difference is about $1.5~\mathrm{kcal\cdot{}mol^{-1}}$ (DF-BP86/def2-SVP).
I do not have the time to calculate the activation barrier, but I think it is fair to assume an unfavoured equilibrium for the $\ce{S_{N}2}$ state, at room temperature about $1:9$. This would make this reaction not very likely.


text:
Homemade reactor for water cooling
Background:
Well, it's summer here in Brazil, and a great way to keep cool is drinking a so called Tereré, and, of course, we do need cold water for that.
Problem: I was looking for endothermic spontaneous reactions so I can cool the water in a bottle using an inner aluminum bottle in which the endothermic reaction occurs. I already knew that Urea's dissolution is endothermic and spontaneous, but I couldn't find how much endothermic it is, i.e., how many Joules it absorbs per Mol of urea dissolved. Another problem related with urea is that it's dissolution is a slow process (I think, not sure), additionally, the urea I can find is the one use as fertilizer, and this one has lots of impurities that I'll need to take into account when calculating how much I'll need to react.
What I know is that the urea dissolves in the same amount of water in weight, that the reaction time increases as the concentration increases, but have no idea how much energy it absorbs neither the how much time it takes to react, of course it depends on the concentration, granularity, etc.
Question: Considering all of that, are there any other reagent(s) I can buy or synthesize without being a licensed chemist that can be used to cool the water? Or, do you have the information that is missing about urea's dissolution (Joules/Mol absorbed in dissolution, approximated reaction time) so I can figure out if I can use it for this purpose?

Ther is no chemical reaction around! You should learn that dissolution is physics! Last not least, for centuries common salt was used as a coolant, so why would you use the much more expensive urea? Your gueseeing around fertilizer ureas purity is nonsense. All urea is produced by the same plants, impurities might be introduced on distribution and repackaging.

@Georg Please consider the tone of your comments before posting them.  Comment about the posts, not the user that has posted them.  Consider this a warning.

Yeah I am aware of that @Georg, I was just trying not to restrict the answer for dissolution processes, i.e. I could accept a reaction as answer if it was feasible. I really don't know about urea, that's why I made it clear by saying "I'm not sure", what does matter about it is the solution enthalpy that I already managed to find for salts trb456 mentioned on his answer.

@Georg slat does not affect the cooling rate of water but the cooling rate of ice..you can't cool the water with only salt! And in fact reacting hydrated barium hydroxide with ammonium chloride is an endothermic reaction!

Unfortunately, many aqueous endothermic reactions involve some chemicals you may wish to avoid.  Ammonium nitrate is the classic example, but in the US it is becoming more and more unavailable because of its potential use as an explosive.  Another frequently mentioned possibility is reacting hydrated barium hydroxide with ammonium chloride, but $\ce{Ba(OH)2}$ is very caustic and poisonous.
So of what I've read, your best choices might be mixing water with a chloride salt like ammonium or potassium.  I do not know how cold this will get your container. I suspect it will not be as good as ammonium nitrate, which would probably be ideal given the former use of this compound in medicinal cold packs.

There are some Homemade reactor for water cooling  that can do what you want.
Mohammed Bah Abba had the same problem in Niger so he invented the Abba refrigerator: a pot-in-pot refrigerator that uses the evaporation of the water from a porous material to refrigerate the things inside the pot.
The image belowe is not to of the Abba refrigerator but the principle is quite similar and here is well explained.

I think you can make a mod cuia/guampa with a pot-in-pot refrigerator embedded.
To speed up your cooling I should use a pressurized gas, this can cold the thing in two ways:

the gas expansion is an endothermic process
if the flow of gas go above the wet surface you can boost the evaporation process

NOTE: the Abba refrigerator works well in dry and ventilated ambient, I hope you are not in Amazonia!!

This is a great idea, but seems to be very slow process so it does not fit on my needs. I need something to work in a park in some minutes, if it will take too much time I can just go back home and get ice from the refrigerator. Anyway, it would work since I live in the south of Brazil haha

@HericDenis  I have improved the answer is it hot before you cool it?

Beside construction of a solar power driven DIY cooler for beverages (like here), G M's post directing to use the heat of evaporation as a vehicle to cool a liquid may be indeed advantageous.  What about using molar sieves in an outer shell of a barrel -- again, at the place of consumption, no electricity is needed. Yet, CoolKeg claims to be scaleable (7 to 20 L) and both barrel and the active ingredient (zeolithe) is able to be recycled. A video describes how this may be implanted into a production chain, this site around beer provides some to read to understand the function.
Of course, to regenerate the molar sieves, alternative / more ecofriendly sources of heat may be used, too.

I have used very large quantities of urea (fertiliser) mixed with water. No matter how warm the water when mixed thoroughly the water temp will drop to minus 4 degrees.
We had to this as part of the supplementary diet to feed thousands of cattle in drought areas of Australia.
It will chill water extremely effectively for low cost.

I knew about urea, but I didn't know that it could drop the temperature so significantly! -4°C is way below what I expected.

Recently has been developed an ammonium nitrate free cold pack. It consists of a mixing of Water, Urea, and Carbamakool™ this is the secret ingredient you should search for. This is a safe alternative to posioning chemical.
See this site for more information.
Maybe you can try to analyze one of this pack to find the information you need.


text:
What is so special about quasicrystals?
Can whatever is special about quasicrystals be understood at A-Level standard or slightly above it? The material I saw online is either too layman or too advanced for me to follow. What practical uses do they have and why does their structure lead to the properties that allow them to have these practical uses?

For those who don't know: A-level students are 16-18 years old. So advanced high school chemistry equivalent.

It's hard to know exactly what “level” you want address, as the Wikipedia entry seems a decent introduction to me. But in any case, here's what I consider “special” about quasicrystals:
From the fundamental point of view:

Though they are regular (they are formed by symmetry operations that allow them to repeat throughout space), they are not periodic (the symmetry operations are not purely translations). Although there's nothing inherently “wrong” or “impossible” about that from theoretical physics and chemistry, it goes against the classical classification of solids: crystalline or amorphous.
Though hundreds of quasicrystals have been synthesized, there is only naturally occurring quasicrystal phase (icosahedrite)!
All quasicrystals are metal alloys (most of them being binary or ternary intermetallic alloys): given that there are other types of crystalline solids, why aren't there non-metallic quasicrystals?

Regarding applications:

Quasicrystals have found, so far, few practical and industrial applications. In these applications, they are used as other alloys would, and do not have any “magic quasicrystal property”.
One specific feature they have, though, is that quasicrystals tend to be rather strong alloys and that they age well, due to the way they are synthesized


text:
What Does the Equilibrium Constant represent?
If the Equilibrium constant stands for the extent of reaction at equilibrium, whether more of forward or backward reaction is taking place, then why is it not affected by factors like concentration of reactants and product?
For example, if you increase concentration of reactant, equilibrium will shift to the right, which means more of forward reaction is taking place, so shouldn'tthe  equilibrium constant change?

The right explanation lies on very complicated thermodynamic equations, sice I do not know the depth of the answer I'll just say that the $K_{\mathrm{eq.}}$ is only dependent od temperature, because whatever Reactants/Products concentration (activity) you'll have at $T$, the quotioent between those will always reach a fixed value.

Welcome to Chemistry Stack Exchange! This is an excellent question on a topic that confuses many. $K$ defines the allowable values of $Q$ that constitute equilibrium. $Q$ can change, but at fixed temperature, $K$ will not. Because it is so confusing to many, this question has already been asked and answered. See http://chemistry.stackexchange.com/questions/4977/le-chateliers-principle-how-can-the-equilibrium-shift-while-kc-remains-consta and read both answers, which approach the problem in different ways.


text:
Do acids without hydrogen taste sour?
I have heard that acids taste sour because of hydrogen (I don't really know exactly).  According to the definition of acids:

An acid (from the Latin acidus/acēre meaning sour) is a chemical substance whose aqueous solutions are characterized by a sour taste, the ability to turn blue litmus red, and the ability to react with bases and certain metals (like calcium) to form salts

Some acids may not contain hydrogen.  Would they also taste sour?

Could you provide an example of acid, which does not contain hydrogen and you would like to put it in your mouth?

This has to do with molecules interacting with taste receptors, so could it possibly be a question more appropriate for Biology.SE?

Look at the Wikipedia article on sourness for an explanation of that. But you also really need to really read a basic introduction to acid/base chemistry, because the question does not make much sense… Acids are, by definition, substances that increase the concentration of H+ ions in an aqueous solution.

but something without hydrogen can be acidic(low pH and pKa and high pOH and pKb). Several oxides are this way and they don't have hydrogen, just some element(other than hydrogen) and oxygen

@Caters are you referring to cases such as $\ce{P4O10}$ or $\ce{CO2}$? If so, those compounds react with water to create the corresponding hydrogen-containing and hydrogen-releasing acids $\ce{H3PO4}$ and $\ce{H2CO3}$ respectively.


text:
LCAO (Linear Combination of Atomic Orbitals) and Phases
So when combining atomic orbitals to form molecular orbitals, you can either add the wave functions or subtract them. But at the same time, orbitals can exist in opposite phases (say one lobe of the p orbital is '+' and the other '-'). So what happens when you subtract two 1s orbitals of opposite phase, is that the same thing as constructively interference of the wave functions?

What possible phases exist for a 1s orbital?

A single 1s orbital has only one phase, but another 1s orbital can have the opposite phase.

If you substract "s" orbitals of different phase, by which I understand $\psi = $1s$_a - (-$1s$_b)$ then this is the same as $\psi = $1s$_a + $1s$_b$. It will be constructive interference everywhere, but since the probability of finding the particle in all space must be conserved, on places where the original orbitals do not overlap, there will be a small decrease in the probability density.

Otherwise, in general, you can perform any linear combination of the two orbitals, not just a sum or difference. If you want to describe an energy eigenstate, then you have to find the coefficients such that the energy is minimized. For a homonuclear diatomic energy this implies that the square of both coefficients (for say, the 1s$_a$ and 1s$_b$) must be the same, hence you can only sum or substract "equivalent" orbitals.

@perplexity Could you submit your comments as answer, they made most sense to me and it would help to close/solve this question.

@Martin I think that the first part of perplexity's comment is right, but the second part, about change in electron density, depends on too many things a) that there is a bonding interaction between the two atoms, and b) the atom-centered basis functions do not polarize due to other atoms  (to be general).

@EricBrown I hope my answer to the problem is as specific as necessary, while keeping it as general as possible. You'd do me a great favour if you could proof read what I wrote.

The concept of linear combination of atomic orbitals (LCAO) to form molecular orbitals (MO) is probably best understood, while digging a little deeper into quantum chemistry. The method is an approximation that was introduced for ab initio methods like Hartree Fock. I don not want to go into too much detail, but there a some points that need to be considered before understanding what LCAO actually does.
The time independent Schrödinger equation can was postulated as $$\mathbf{H}\Psi=E\Psi,$$
 with the Hamilton operator $\mathbf{H}$, the wave function $\Psi$ and the corresponding energy eigenvalue(s) $E$. I will just note that we are working in the framework of the Born-Oppenheimer approximation and refer to many textbooks for more details.
There is a set of rules, the wave function has to obey.

It is a scalar, that can be real or complex, but the product of itself with its complex conjugated version is always positive and real. $$0\leq \Psi^*\Psi=|\Psi|^2$$
The probability of finding all $N$ electrons in all space $\mathbb{V}$ is one, hence to function is normed. $$N=\int_\mathbb{V} |\Psi(\mathbf{x}_1,\mathbf{x}_2,\dots,\mathbf{x}_N)|^2 \mathrm{d}(\mathbf{x}_1,\mathbf{x}_2,\dots,\mathbf{x}_N)$$
The value of the wave function has to vanish at infinity. $$0 = \lim_{\mathbf{x}\to\infty}|\Psi(\mathbf{x})|$$
The wave function has to be continuous and continuously differentiable, due to the second order differential operator for the kinetic energy $\mathbf{T}_c$ included in $\mathbf{H}$.
The Pauli Principle has to be obeyed. $$\Psi(\mathbf{x}_1,\mathbf{x}_2,\dots,\mathbf{x}_N) = -\Psi(\mathbf{x}_2,\mathbf{x}_1,\dots,\mathbf{x}_N)$$

Also the variational principle should hold for this approximation, stating that the expectation value for the energy of any trial wave function is larger that the energy eigenvalue of the true ground state. One of the most basic methods to approximately solve this problem is Hartree Fock. In it the trial wave function $\Phi$ is set up as a Slater determinant.
$$
\Phi(\mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_N) =
\frac{1}{\sqrt{N!}}
\left|
   \begin{matrix}
      \phi_1(\mathbf{x}_1) & \phi_2(\mathbf{x}_1) & \cdots & \phi_N(\mathbf{x}_1) \\
      \phi_1(\mathbf{x}_2) & \phi_2(\mathbf{x}_2) & \cdots & \phi_N(\mathbf{x}_2) \\
      \vdots               & \vdots               & \ddots & \vdots \\
      \phi_1(\mathbf{x}_N) & \phi_2(\mathbf{x}_N) & \cdots & \phi_N(\mathbf{x}_N)
   \end{matrix} \right|
$$
The expectation value of the energy in the Hartree Fock formalism is set up, using bra-ket notation as $$E =\langle\Phi|\mathbf{H}|\Phi\rangle.$$
Skipping through some major parts of the deviation of HF, well end up at an expression for the energy.
$$E=\sum_i^N \langle\phi_i|\mathbf{H}^c|\phi_i\rangle +\frac12 \sum_i^N\sum_j^N \langle\phi_i|\mathbf{J}_j-\mathbf{K}_j|\phi_i\rangle$$
To find the best one-electron wave functions $\phi_i$ we introduce Lagrange multiplicators $\lambda$ minimising the energy with respect to our chosen conditions. These conditions include that the molecular orbitals are ortho normal. $$\langle\phi_i|\phi_j\rangle =\delta_{ij} =\left\{
\begin{matrix}  0 & , \text{for}~i \neq j\\ 1 & , \text{for}~i = j\\\end{matrix}\right.$$
I will again skip through most of it and just show you the end expression.
$$\sum_j\lambda_{ij}|\phi_j\rangle = \mathbf{F}_i|\phi_i\rangle$$
with the Fock operator set up as
$$\mathbf{F}_i = \mathbf{H}^c +\sum_j (\mathbf{J}_j − \mathbf{K}_j)$$
with $i\in1\cdots{}N$, the total number of electrons.
We can transform these trial wavefunctions $\phi_i$ to canonical orbitals $\phi_i'$ (molecular orbitals) and obtain the pseudo eigenwertproblem $$\varepsilon_i\phi_i'=\mathbf{F}_i\phi_i'.\tag{Fock}$$
This equation is actually only well defined for occupied orbitals and these are the orbitals that give the lowest energy. In practice this formalism can be extended to include virtual (unoccupied) molecular orbitals as well.
Until now we did not use any atomic orbitals at all. This is the next step to find an approximation to actually solve these still pretty complicated systems.
LCAO a superposition method. In this approach we map a finite set of $k$ atomic (spin) orbitals $\chi_a$ onto another finite set of $l$ molecular (spin) orbitals $\phi_i'$. They are related towards each other via the expression
\begin{align}
\phi_i'(\mathbf{x}) &= c_{1,i}\chi_1(\mathbf{x}) + c_{2,i}\chi_2(\mathbf{x}) + \cdots + c_{k,i}\chi_k(\mathbf{x})\\
\phi_i'(\mathbf{x}) &= \sum_a^k c_{a,i}\chi_a(\mathbf{x})\\\tag{LCAO}
\end{align}
From $\text{(Fock)}$ you can see, that there will be $l$ possible equations depending on the chosen set of orbitals, in the form of
$$\sum_a^k\sum_b^k c_{ia}^* c_{ib} \langle\chi_a|\mathbf{F}_i|\chi_b\rangle = \varepsilon_i \sum_a^k\sum_b^k c_{ia}^* c_{ib} \langle\chi_a|\chi_b\rangle$$
or for short there are $l$ equations $$C_{ab}F_{ab}=\varepsilon_i C_{ab}S_{ab}.$$ Since $a,b \in [1,k]$ you can gather $C_{ab}$ in a $k\times k$ matrix $\mathbb{C}$. Since $i\in[1,l]$, there will only be $l$ $\varepsilon_i$ and therefore the matrix of the Fock elements $F_{ab}$ has to be a $l\times l$ matrix $\mathbb{F}$. The whole problem reduces a matrix equation $$\mathbb{F}\mathbb{C}=\mathbb{S}\mathbb{C}\epsilon\!\!\varepsilon,\tag{work}$$
from which it is obvious, that the dimension of the involved matrices have to be the same. Hence $k=l$.
Too long, didn't read
The total number of elements in the finite set of atomic orbitals is equal to the total number of elements in the finite set of molecular orbitals. Any linear combination is possible, but only the orbitals that minimise the energy will be occupied.
The coefficients $c$ from $\text{(LCAO)}$ will be chosen to minimise the energy of the wave function. This will always be constructive interference. This is also independent of the "original" phase of the orbital that is combined with another orbital. In other words, from $\text{(LCAO)}$
$$\phi_i'(\mathbf{x}) = c_{1,i}\chi_1(\mathbf{x}) + c_{2,i}\chi_2(\mathbf{x}) \equiv c_{1,i}\chi_1(\mathbf{x}) - c_{2,i}[-\chi_2(\mathbf{x})].$$

Add or substract the wavefunctions is the first step, then you need to normalize the new wavefunction, so its norm becomes 1.
In your case, you get twice 1s wavefunction before normalization, and result in 1s afterwards. So usually there's no point to combine the same orbitals in LACO.

Since you are talking about s-orbitals and p-orbitals only(which would be a $ \sigma$bond),there is a head-on combination of the orbitals.This means that the orbitals would combine along the "axes"(each p orbital has both of its lobes aligned along the x-axis,or y-axis or z-axis,mathematically).The s-orbitals always combine head-on,since they are spherical.
Now,if two s-orbitals combine,it would be constructive interference if both are of same phase.If they are of opposite phases,there will be destructive interference.What you are talking about when you say "add" or "subtract" means constructive and destructive interference,respectively when the orbitals are of same phase;and destructive and constructive interference,respectively when orbitals are of different phases.

s-orbitals always combine head-on. It is still a model, albeit a very useful one. Perhaps phrase it s-orbitals are combined head-on ?

whats the difference?


text:
Copper oxidation
In an experiment we conducted, we heated 1.6 grams of copper dust, and got 1.7 grams of oxidised copper, (it probably reacted with 0.1 grams of $\ce{O2}$)
and our goal was to find which of the two forms, $\ce{CuO}$ or $\ce{Cu2O}$ we got.
What I did:
\begin{array}{lll}
\hline
 & \ce{Cu} & \ce{O2} \\
\hline
m\ \text{(in g)}& 1.61 & 0.1 \\
M\ \text{(in g/mol)}& 63.5 & 32 \\
n\ \text{(in mol)}& 0.025 & 0.003125 \\
\text{ratio}& 8 & 1 \\
\hline
\end{array}
And the ratio between copper atoms to oxygen atoms is $4:1$.
Thus it seems the form is $\ce{Cu4O}$.
But where is my mistake?

Your calculation looks good. Maybe the issue is not with the calculation, but with the experiment. Did you wait long enough so that all the copper is oxidized?

And then only half of the material reacted? $8Cu + O_2 \rightarrow 2Cu_2O + 4Cu$?

Or only 1/4 of the copper reacted, and formed CuO (which is the expected product for spontaneous oxidation of copper upon heating in air)

Whether Cu2O or CuO is formed depends on temperature. Above about 1100°C you get Cu2O. Another problem is the gas around/within your cruzible, depending on how you heat(gas/electic etc)

When I was at university I used to oxidize metal oxides. In my case was difficult to fully oxidize a metal with only one step. This is what I should do.

First oxidation
weigh it
crush with a mortar what you have oxidized
weigh it
second oxidation
weigh it

If the weight 4. and 6. are the same you can do the calculation if not you have to re-oxidize again.

Good suggestion on how to check for complete oxidation. That's a better answer than mine!


text:
Is H2CO3 (carbonic acid) soluble or insoluble in water?
I am asking this because I know that acids are normally aqueous but according to the solubility rules, all carbonates are insoluble.

Have you looked at general references, such as the Wikipedia page for carbonic acid, before asking here?

Also, all carbonates are not insoluble - the alkali metal carbonates, like [sodium carbonate](sodium carbonate) are soluble.

Please note that Arrhenius' definition of acidity is not the only one.

Yes, it is soluble.
For example, the oceans are constantly getting $\ce{CO2}$ from the atmosphere that gets converted into carbonic acid and thus decreasing pH on a global scale. Or the soft drinks we drink contain carbonic acid in them in dissolved form only due to mixing of $\ce{CO2}$ gas.
Also, not all carbonates are insoluble as Ben Norris has commented.
Hard and fast 'rules' in chemistry tend to be rare.  Consider them guidelines.

The oceans are only "constantly getting CO2" when we've filled the atmosphere with gaseous CO2 to disrupt the equilibrium. Prior to the last century, the oceans were mostly in a nice dynamic equilibrium with atmospheric CO2 (save a few nasty turns here and there)

Yes, it is soluble.
The fizzy part of your soda does not form a layer at the bottom of the bottle. (Personally, I find the empirical solutions best for chemistry)
The carbonate rule only applies to alkali metal carbonates.
Also, as a general rule, anything with the word "acid" is probably soluble in water.  Seeing as acids are sometimes defined as yielding $\ce{H+}$ ions when dissolved.


text:
How can enzyme/substrate reactions that adhere (largely) to quantum theory also require 'Newtonian' consideration of gravity?
I'd just like to ask for a little clarification here due to confusion from interdisciplinary studies.
I'm currently reading the 1976 paper related to the recent 2013 Nobel Prize for Chemistry, by 2 of the laureates Warshel and Levitt — Theoretical Studies of Enzymic Reactions: Dielectric, Electrostatic and Steric Stabilisation of the Carbonium Ion in the Reaction of Lysosyme — and have recently been reading into quantum chemistry, as well as studying biochemistry and have had further lectures in Physics to boot.
I'm wondering if the points made here are comparable to points I've heard in mechanics lectures, about a cannonball not being modelled as a quantum object, since every subatomic particle would be too computationally intensive, so a coarse-grain model is used (i.e. Newton's "classical" laws).
I wouldn't be so quick to draw parallels between mechanics lectures and enzyme biochemistry but it seems to be in the language they are using here (see highlights below).
From what I understand (in mechanics) only 3 of the 4 'fundamental forces' (strong, weak, and EM) only strictly apply at the quantum/microscopic scale, and scientists are as it stands struggling to work gravity into the setup. Is this worth considering in this paper - if an enzyme is considered all of a sudden due to its size as a quantum object does this entail ignoring gravity QM now in favour of 'Newtonian', inertial frame-of-reference type mechanics, or considering it in some other 'Newtonian' manner?
The introductory section is as follows (full paper here):

The conformation of the enzyme-substrate complex can be adequately studied using empirical energy functions based on the “classical” contributions of bond stretching, bond angle bending, bond twisting, non-bonded interactions, etc. The mechanism and energetics of the enzymic reaction can only be studied using a quantum mechanical approach. Previous quantum mechanical calculations on enzymic reactions have been limited in several respects. In the first place, they deal with an over-simplified model system consisting of only small fractions of the atoms involved in the real enzyme-substrate interaction. Second, all available quantum mechanical methods treat the reaction as if in a vacuum and are, therefore, unable to include the dielectric, which has a very important effect on the energy and charge distribution of the system. This limitation does not depend on the actual quantum mechanical scheme, being equally valid for the simplest Hiickel treatment and most extensive ab initio calculation.
Because the treatment of the whole enzyme-substrate complex quantum mechanically is computationally impossible, it is necessary to simplify the problem and use a hybrid classical/quantum mechanical approach. The method described here considers the whole enzyme-substrate complex: the energy and charge distribution of those atoms that are directly involved in the reaction are evaluated quantum mechanically, while the potential energy surfac e of the rest of the system, including the urrounding solvent, is evaluated classically. An important feature of the method is the treatment of the “dielectric” effect due to both the protein atoms and the surrounding water molecules. Our dielectric model is based on a direct calculation of the electrostatic field due to the dipoles induced by polarizing the protein atoms, and the dipoles induced by orienting the surrounding water molecules. With the inclusion of the microscopic dielectric effect we feel that our model for enzymic reactions includes all the important factors that may contribute to the potential surface of the enzymic reaction. Because we use analytical expressions to evaluate the energy and first derivative with respect to the atomic Cartesian co-ordinates, it is possible to allow the system to relax using a convergent energy minimization method. Here we use the method to study the mechanism of the lysozyme reaction, with particular emphasis on factors contributing to the stabilization of the carbonium ion intermediate. We first outline our theoretical scheme, paying attention to the partition of the potential surface into classical and quantum mechanical parts and to details of the microscopic dielectric model. Then we use the method to investigate factors that contribute to the stabilization of the carbonium ion intermediate and deal with the relative importance of each factor. The effect of steric strain is considered for both the ground and the carbonium states and found to be of minor importance. Electrostatic stabilization is treated at some length with special emphasis on the balance between stabilization by induced dipoles (polarizability) and stabilization by charge- charge interactions. The results demonstrate the importance of electrostatic stabilization for this reaction.

Atomistic calculations comfortably neglect 3 out of 4 fundamental interactions regardless of whether they are QM or classical - the strong and weak forces attenuate too rapidly to meaningfully factor into chemical interatomic interactions, whilst gravity is so much weaker than electromagnetism as to be irrelevant.

Nuclei are heavier and more classical than electron. Just like quantum mechanics contains Newtonian mechanics as a classical limit. Although here the Newtonian mechanics does not correspond to gravity, it used other potential (force field). This we aldready saw in the potential energy surface defined by Born-Oppenheimer approximation.

Newtonian gravity is a red herring here as gravity is far too weak to be involved in enzyme chemistry. The Newtonian or classical mechanics involved is a simplification of the molecular structure of most of the enzyme that models is, effectively, as a bunch of atoms connected by springs ( a little like a macroscopic atomic model built from balls with bonds represented by springs). The model uses simple classical approximations of how springs behave (with additions to cope with rotation as well as stretching). This is easier to compute and a good approximation.

Unfortunately there's been something of a nomenclature mismatch here. While classical mechanics refers to Newtonian SUVAT-style things, "classical" molecular dynamics or geometry optimisation refers to, as the paper puts it: "empirical energy functions based on the 'classical' contributions of bond stretching, bond angle bending, bond twisting, non-bonded interactions, etc."
In pure "classical force fields", these functions use parameters produced from analysis of various known geometries, and are collectively optimised using linear algebra optimisation techniques to obtain an approximation of a molecular or ionic system. Simple point-wise electrostatics may also be included.
This paper refers to the use of a combination of quantum mechanical techniques (modelling the complex reaction centre) with classical force field techniques (modelling the rest of the protein) and a dielectric model, modelling the interaction between the two partitions and the effect of the bulk solvent on the protein.
If you'd like to look into this in more detail, I'd suggest looking for mention of "QM/MM" (Quantum Mechanics/Molecular Mechanics) or "ONIOM" techniques, as such models are now known. The use of simulated dielectrics for solvents is now largely covered under "PCM" (polarisable continuum model) and "COSMO" (Conductor-like Screening Model) techniques.
If you're curious about "classical" molecular force-fields, I'd suggest looking up one of the packages that perform such calculations, such as DL POLY or GROMACS.

Thank you, yes I thought it was too simple to be true! Sounds abstract when you're new to it but reading 'parameter sets' as bond parameters helps. I've been looking into the terminology like CFFs, dielectrics etc. as it seems odd not to understand the fundamentals here. I don't even know how I'd begin to get into 'linear algebra optimisation', my maths education only went so far as single variable calculus unfortunately, this seems advanced. DL_POLY looks v familiar from Dalton Trans. figures... V v interesting, thanks

You're basically just trying to find a combination of geometric options that gives you what looks like the minimum energy, but it turns out you can represent that as the approximate solution to a equation multiplying a vector by a matrix. Linear algebra features heavily in computational chemistry (both classical and quantum mechanical), so if you're getting into the field I'd recommend getting familiar with the terminology, even if you don't understand the guts of it.

What they mean by

Because the treatment of the whole enzyme-substrate complex quantum mechanically is computationally impossible, [...]

is basically a consequence of the results from the inner workings of the quantum chemical formulation of the theory. If one goes through the equations (it takes a while), one arrives at an expression that tells you that the required number of calculations scales with $N^4$, where $N$ is the number of basis functions. Imagine the wavefunction being a (hyperdimensional) vector in a coordinate system. Say you had a 3-dimensional vector but only two basis vectors to express it, then your representation of the vector would be very inaccurate, so you add another basis vector to your set of basis vectors and then you can accurately describe the vector by $\vec{v} = a\vec{e}_x + b\vec{e}_y + c\vec{e}_z$.
In enzymes, the huge amount of atoms would lead to gigantic numbers of required calculations which make it largely impossible to calculate completely quantum mechanically. So it's an impracticality, not a theoretical problem.


text:
How did Halogens become known as Halogens? They are not the only elements that form salts!
Having never given it a though before, I recently discovered (in a different context) that the prefix halo- actually means 'salt' or 'sea' and the suffix -gen means 'to form' or 'to generate'. So the Halogens are the elements that 'form salts'. But there are salts that do not have halogens in them, like $\ce{Na2S}$ or $\ce{(NH4)2SO4}$. Was it known at the time that other salts existed, and what other names for this group might have been considered?
On a side note, why aren't Group I metals called halogens? Doesn't it take two to tango (I mean, to form salts?)

This seems like a bit of a rhetorical question, so this isn't a terribly formal or authoritative answer, but anyhow - a lot of chemical nomenclature is like lava flow. It solidified and people just worked around it.
The halogens are so named because they have a rich chemistry of ionic compounds (fluorine through iodine, anyhow). However, both the halogens and the group I metals can form a wide range of things that aren't salts. The noble gases can form compounds with elements of low birth. The rare earth elements aren't particularly rare. Oxygen ('acid-former') is not a necessary component of acids. Technetium ('artificial' + ium) is produced in nature in significant quantities.
In addition, a lot of chemistry defies our human efforts to succintly categorise things, so at some point chemists have tried their best to find pragmatic general descriptors that unite groups of elements or molecules on the basis of the properties or constitution. The divide between organic and inorganic chemistry and the resulting exceptions and edge cases in classification (such as mellitic anhydride) is a good example of this. The map is not the territory.

"nomenclature is like lava flow. It solidified and people just worked around it." - oh, that is a delightful turn of phrase. I'll quote that (crediting you) in my next lecture.

@EnergyNumbers - That was an oblique reference to the lava flow antipattern in software engineering, where it's too costly/complex to overhaul crappy or obsolete code/systems, so people just work around it. So, the phrase is not my invention.

@RichardTerrett Still, you transplanted it over to chemical nomenclature - that's not nothing!

+1 for lava flow. As @hBy2Py points out, you applied it elegantly.

Tc is estimated in uranium about 1mg/10^12 mg of uranium, is that significant?


text:
Yellow-amber debris above water filter: What is it?
I have some weird, yellow-amberish pebbles that have been filtered by my Zero waterfilter. Any idea what they are? Copper maybe?
The black stuff is charcoal from the filter.
The quality of the water in my area is fairly good, with a TDS of 40-60. With the filter, I have a TDS of 0, which is excellent. But I'm still curious as to what those things are.

I would hazard a guess that they are spheres of silica, and are part of the filter that has come loose along with the charcoal.

It looks like ion exchange resin, it is used to specifically capture ions from water and replace them by others (either $\ce{H^+}$ or alkali metal ions.
As such, it is probably also part of the filter.

I think it is the 4th layer of filtering: ion exchange. made of resin. Check the filter image on https://zerowater-eu.com/zero-water-kwaliteit-filter/


text:
How to find the concentration of potassium permanganate in this experiment?
I completed a titration experiment in a lab. The aim was to find the concentration of the titrant $\ce{KMnO4}$ solution by titrating it with a sodium oxalate solution of a known concentration.
I first made a standard solution of sodium oxalate solution, using $\pu{0.826 g}$ of sodium oxalate powder, then poured $\pu{50ml}$ of distilled water along with $15$-$\pu{20ml}$ of $\pu{3 M} \,\,  \ce{H2SO4}$. Then I pipetted out $\pu{25 ml}$ of the solution and poured it in to a volumetric flask. Then I filled the rest of the volumetric flask with distilled water, to form the standard solution. I then did the Mcbride experiment to find $75\%$ of the volume of $\ce{KMnO4}$ need to titrate the solution.
I then did the experiment three times to get an average volume of $\ce{KMnO4}$ to be $\pu{25.48 ml}$. And the volume of sodium oxalate solution is $\pu{25 ml}$.
I should find the concentration of $\ce{KMnO4}$ now.
I know the molar ratio is 2:5. How do I calculate this?
I know I need to find the moles of sodium oxalate then I multiply it by $\frac{2}{5}$ and then divide that value by the volume of $\ce{KMnO4}$ which is $\pu{25.48 ml}$. But how do I calculate the moles of sodium oxalate? What would the concentration be? I know the volume is $\ce{25 ml}$.
This is what I have done, but I am not sure if it is correct:
Moles of sodium oxalate = $\frac{\pu{0.8265 g}}{\pu{133.96 g mol-1}} = \pu{0.006169752 mol}$
$\frac{0.006169752}{\pu{0.025L}}$ = concentration of sodium oxalate is therefore $0.246790086$, $0.0248$ to 4 s.f.
($\pu{0.025 L}$ is the volume of the sodium oxalate)
I then multiplied that by $\frac25$.  $0.0248 \cdot \frac25$ gives me the concentration of $\ce{KMnO4}$ which is $0.09872$.
But I think this is wrong because I have not used the volume of $\ce{KMnO4}$ used, which was the whole point of the experiment.

I think your experiment has one setup mistake.:
You're mixing $\pu{0.8265 g}$ of oxalate with $\pu{50 mL}$ of water and $15$-$\pu{20 mL}$ of sulfuric acid. So your $\pu{25 mL}$ oxalate solution does not contain $\pu{0.8265 g}$  of oxalate (more or less $\frac{25 \cdot 0.8265}{ 65-\pu{70 mL}}$). So you can't know the exact oxalate concentration of this solution.
I recommend you to use a $\pu{100 mL}$ volumetric flask to prepare the oxalate solution. You can then calculate its concentration and the amount of oxalate you have in $\pu{25 mL}$.


text:
Equilibrium between R and S configuration?
We have learnt that the energy evel of both R and S optical isomer of a compoud is same, just as well the entropy of each isomer is also same. If we consider the conversion of R form of a compound to it's S isomer we observe that the change in gibbs free energy is "0". This implies that the two isomers must exist in equilibrium. But we also know that any compound does not change from its R configuration to its S configuration for example inversion of configuration in Sn1 mechanisms lead to only one configuration and not both are obtained after the reaction.
So the question is that why don't any compounds R isomer convert into its S isomer on its own so as to maintain equilibrium ?

Because even though neither the R nor the S isomer is thermodynamically favored (in an achiral environment), the free energy barrier (or activation energy) between those two states is huge. This is the most common case, when R/S interconversion requires breaking and reforming a chemical bond. Then, even though the thermodynamic equilibrium would be to have a racemic mixture, kinetics prevent the solution from reaching this equilibrium state.

But isnt there an energy barrier in all equilibriums ?

Yes (well, almost yes: there are barrierless reactions), but in that particular case the barrier is very high, meaning it will take exponentially large time to pass (thus it does not happen on the experimental time scale)

Great, this is exactly what me and my friend came onto ourselves. We said that maybe the reaction takes too much time and is not observable. Very happy to see that our findings were correct.


text:
Relating a compound's radii and unit cell dimensions
We're asked often to calculate the density of a crystal, and there is a known formula for that related to the unit cell volume.  Now, for simple geometric structures such as the bcc structure (and since it is close packed), I know how to relate the different radii of the constituent compounds with the unit cell's (unique in this case) dimension (the length a), but how do I relate it for such other complex structures?
In general given a certain crystal geometry (cubic , tetragonal , orthorhombic), how does one relate the unit cell dimension with the radii of the constituent atoms?

The ratio of the radius of the cation to the radius of the anion is called radius ratio. The ratio of the radius of cation and the radius of anion i.e, radius ratio ($r_+/r_-$) plays an important role in determining the structures of ionic solids and coordination number of ions. The possible coordination numbers and structural arrangements of anions around  cations for different $r_+/r_-$ values are given below:


text:
Why are the total energies of transition states so commonly corrected for zero-point vibrational energy?
So often I see total energies of transition states corrected for zero-point vibrational energy which always confuses me.  Zero point energy is the lowest energy that a ground state minimum energy configuration can have at 0K in a vacuum.  Transition states are indeed not minima, can never be isolated, have one imaginary mode of vibration, etc.
So why do people constantly factor in zero-point vibrational energy to the total energy for a transition state?  Obviously this is done to have 'comparable' minima and TS energetics (if you are applying ZPE to the minima) but is this even physical?  If zero-point correcting the TS, should the zero-point energy of the imaginary mode be included in the summation or excluded (or should both quantities be provided)?

Transition states are much more real than you think. Read http://chemistry.stackexchange.com/a/16800/186. They can not be isolated, but they can be observed. The 1999 Nobel Prize in Chemistry was awarded for such observations.

@Wildcat I've omitted the word "observed".

So, basically, the answer is, yes, you simply omit the normal mode with an imaginary frequency while calculating ZPE for TS. The mode with imaginary frequency is treated either as a very loose vibration with frequency that tends to zero or even as a translation. And in both cases it won't contribute to zero point vibrational energy.

@Wildcat Right but including ZPVE with a TS goes against its very definition.  I think omitting the imaginary mode is mathematically convenient and it just 'sounds good' but in the end applying ZPVE to a TS is completely unphysical.

Not it is physical, where by "physical" we mean (as always) "in accordance with the laws of some physical theory". Omitting the reaction coordinate mode is not mathematically convenient, it is physically justified in the TST framework as I described above.

Another question is how good is TST theory itself, in particular, how well this specific treatment of reaction coordinate node corresponds to the reality? This can be judged on the basis of comparison theoretically predicted quantities with experimental numbers.

Which quantities? Reaction rates, first of all. And here I can tell you only one thing: for unimolecular reaction rates RRKM theory in which contributions from normal-mode vibrations (and rotations) are included in the model works better than its predecessor, RRK theory.

At the transition state, you have exactly one vibration mode with imaginary frequency (which corresponds locally to the reaction coordinate), and $3N-1$ other “normal” vibration modes. In the transition state theory, you separate out the reaction coordinate (consider it infinitely slow compared to the other movements). Thus, when you consider the energy of the transition state, you have to include the contributions of ground state energy of all the other vibration modes.
The more rigorous way of looking at it is to use statistical physics, and to consider the partition function of the transition state. See here or there for such treatments.

So what you're saying is, we just throw out vibrational modes until we get a 'ground state' molecule? (I say modes because this concept can be applied to higher order saddle points)

@LordStryker you calculate the ZPE by summing over all modes that have an imaginary frequency (i.e. correspond to negative eigenvalues of the Hessian matrix). For a true transition state, this should be all modes but one.

The only thing I'm still unclear about is what consequences are there to throwing away vibrational modes?  Now you'd be comparing the relative energy of a TS with X-1 number of modes to say an intermediate with X number of modes.

@LordStryker that's normal. If you want to know why, read the statistical physics treatment of transition state theory, and you'll see why!

@LordStryker TST has much worse flaws than ignoring the imaginary mode. For saddle points of higher order, TST is not even applicable anymore.// Rotational and translational modes are also projected out, so you end up with $3N-7$ or $3N-6$ (for linear molecules).