text:
What will be the major product of of the Wurtz reaction of 3,6-dibromocyclohexa-1,4-diene?

According to me, the answer should be (A) because that's a Wurtz reaction, but the answer given in the book is (B).

The answer is (B) because (B) is benzene, which is the prototypical aromatic hydrocarbon. The aromaticity of (B) increases its thermodynamic stability relative to (A), which is called Dewar benzene (mistakenly attributed to James Dewar), which is notoriously difficult to synthesis and isolate. It reverts back to benzene with a half-life of two days.
Generally, if a pathway exists for a reaction to produce an aromatic product, then the reaction likely will do so. In this particular case, the diradical intermediate that might be produced as part of the Wurtz reaction's mechanism, is the triplet excited state of benzene.

In addition. Even if (A) is immediate product, it quickly converts to (B). It is known.

From Wikipedia:

Since the reaction involves free radical species, a side reaction occurs to produce an alkene. This side-reaction becomes more significant when the alkyl halides are bulky at the halogen-attached carbon. This is because the activation energy required for the SN2 reaction in the second step becomes significantly high, so the alternate elimination mechanism is favoured.

I can't see the Wurtz reaction happening to make (A) in this case with the two double bonds, there isn't enough flexibility to get the radical to attack across the ring. Also, (B) is making an aromatic ring, which is normally the most stable product when given the option.
So I'll go with (B) for this.

The answer is (B). (B) is an aromatic Kelukes benzene where as (A) is dewar benzene. Even if (A) is formed it quickly converts into Kelukes benzene ie (B).

Actually the conversion to benzene is quite slow and takes several days.


text:
What will be the product when 3,4-dibromocyclohexene is reacted with sodium followed by heating with selenium?
I'm tutoring few Class 12 students for their chemistry exams. I'm stuck with few problems. I don't know, if the key is wrong or I am wrong.

According to me product (A) must be the cyclic diene, as given in option (A). Now, I don't know, what happens when you heat cyclic diene with selenium. But I do know that,  $\ce{SeO2}$  oxidizes allylic positions of alkenes into alcohols. But, none of the options given are fitting into that argument.
The answer given in the book is option (B).
How can that be explained?

Some pre-aromatic hydrocarbons can be easily oxidized by sulfur or similar rather weak oxidants into aromatic hydrocarbons. It is different type of reactions and it is not general

Assuming your Product A is correct then product B should be benzene. Selenium act as a mild oxidizing agent at temperature 220 °C to 330 °C. In the above reaction selenium oxidizes cyclohexadiene to form benzene.
Below is a reaction in which 1-methyl-1,3-cyclohexadiene is converted into toluene.

you can also go to this link for more info

Since the treatment of your compound with sodium will result the formation of diene as both bromine elements have same affinity to remove otherwise as per wurtz reaction sodium generates alkyl compound . The diene here forms benzene as the se acts as the oxidizing agent As per the options available the answer is benzene

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text:
What will be the Ozonolysis product?
Problem:

My Answer:
According to me, both the left side and the right side double bonds undergo cleavage and hence the product must be, two molecules: Keto aldehyde and Ester aldehyde. But since, such product is not given in options, let us assume that only one bond gets oxidized. But which one? Which bond is the most reactive? I don't know.
Answer given in the book:

(B)

How come?

There are two types of control, that governs reaction path:

kinetic control (path with least activation energy) and
thermodynamic control (path with least energy of products).

Given temperature favors kinetic control.
Ozonolysis starts as 2+3 electrocyclic addition (with later regroup of the product), similar to Diels-Alder reaction . Please, read about this reaction in good organic chemistry textbook, especially about what controls geometry of products, it will help to understand what I'm going to say next.
Ozone has low energy of LUMO, so alkene part here provides HOMO, that is still lower then Ozone's LUMO. Oxygen from $OMe$ group donates some electronic density to $C-C$ double bound, so this bond becomes more active, then opposite one.

I didn't actually understand your last paragraph. To understand what you said, I've read about MO theory. Most of the books I referred to were giving examples of diatomic molecules like $Be_2$, $O_2^+$ etc. none of them discussed polyatomic molecules. Anyway, even now I still don't understand your last paragraph. Could you kindly suggest some good book or online material which explains (with examples) the kind of reasoning that you are giving in your last paragraph

Basically, electrocyclic addition starts with HOMO of one molecule interacting with LUMO of another molecule. Usually LUMO is pi* orbital of alkene (which is lowered by substitution C-atoms in alkene with N or O atoms) and HOMO is higher pi-orbital of diene. This is why Diels-Alder reaction works well with methoxy-butadien+acetilenedicarboxylic acid. Role-reversion is possible, and first stage of ozonolisys mechanism is an example. As I remember, J. March in his 4-book textbook  and later 1-book reedition covered the basics pretty well.

Any reaction in general will try to attain maximum possible stability.
The reactivity of a molecule or a group is inversely proportional to its stability.
The general order of reactivity of carbonyl compounds is as follows:
$$\small\ce{Aldehyde > Ketone > Acid halide > Anhydride > Ester > Carboxylic Acid > Acid Amide}$$
(Reference)
So if the left bond breaks it results in a ketone which is more reactive(hence less stable) as compared to the product formed by the cleavage of the right bond which is an ester.
Hence the bond on the right would break to form an ester.

My answer may seem simple(because it only relies on logic) but @permeakra 's answer actually states the reason why it happens. In my answer some one can argue that how will the reacting molecule know if the formed product is more stable or less, it's like looking into the future.

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text:
Why are carbonyl compunds more reactive towards catalytic hydrogenation than alkynes and alkenes?
Alkynes are more reactive towards catalytic hydrogenation than that of alkenes, because they are more unsaturated. This is the correct reasoning, right?
But why are carbonyl compounds more reactive towards catalytic hydrogenation than alkynes? The $\ce{C=O}$ in carbonyl group is polar, but how does that affect reactivity?

Example:

Answer: Option (D), $c>a>b$

I doubt that it is carbonil group that is hydrogenated. It is more likely, that $C-Cl$ bond will be cleaved here. $COCl$ group is known to be very ready to react.

That is not true in general: the stronger nature of the C=O bonds (larger bond energy) means that they are not as easily hydrogenated as C=C bonds. However, your compound (c) is not a regular carbonyl group, but an acyl chloride, which is a quite reactive species and thus very easily hydrogenated.
For a bit of reading, see here the section on “Catalytic Hydrogenation”:

As a rule, the carbonyl group does not add hydrogen as readily as do the carbon-carbon double and triple bonds. Thus, it is fairly easy to reduce an alkene or alkyne function without affecting any carbonyl functions in the same molecule. By using a platinum catalyst and increased temperature and pressure, it is possible to reduce aldehydes and ketones to alcohols, but carboxylic acids, esters and amides are comparatively unreactive. The exceptional reactivity of acyl halides, on the other hand, facilitates their reduction under mild conditions, by using a poisoned palladium catalyst similar to that used for the partial reduction of alkynes to alkenes. This reduction stops at the aldehyde stage, providing us with a useful two-step procedure for converting carboxylic acids to aldehydes, as reaction #1 below demonstrates. Equivalent reductions of anhydrides have not been reported, but we might speculate that they would be reduced more easily than esters. The only other reduction of a carboxylic acid derivative that is widely used is that of nitriles to 1º-amines.


text:
What would be the mechanism of the given reaction?
Problem:

$\ce{A}$ can be?

concentrated sulphuric acid
alcoholic potassium hydroxide
triethylamine
potassium tert-butoxide

My comments:
Option (A): Conc. $\ce{H2SO4}$, is an acid, it gives $\ce{H+}$ ions which must act like electrophile and add to the double bond, which must produce a carbocation and then, $\ce{HSO4^-}$ ion must act like a nucleophile and form a bond with the carbocation. Hence, the reaction must be addition of $\ce{H+}$ and $\ce{HSO4^-}$
Options (B),(C),(D): are bases, with basicity order: (D) > (B) > (C). But what would they do to an alkene?
Answer given in the book is A. How is this so?

Under (A) you have heat added. What is to stop a second molecule reacting with the first under electrophilic addition to remove the H2SO4 as a good leaving group and give the product shown in the question?

Bottom line: $\ce{HSO4-}$ is a really bad nucleophile, and it doesn't enter in the picture.

(I realize after scanning that I'm missing the "+ H+" alongside the final product.)

Question for extra credit: after the second step, why is the third step (carbocation migration) favorable rather than directly reforming the double bond on the right-hand side ring?

Answer to extra credit: Because of the hydride migration, it forms a tertiary carbocation which is very much stable than secondary carbocation

In you second last step. You've removed a proton. I strongly advice my students not to just protons getting away. Instead, I ask them to show a base picking up a proton. Who took that proton? Last step is like an Elimination 1 reaction. Which usually needs a strong base, but I don't see a strong base here.

@claws the last step has nothing to do with an elimination. The carbocation is the conjugate acid of the alkene, so under acidic conditions it is formed in small amounts (equilibrium with alkene). The proton is not “picked up” by the base: this is an acidic solution, protons come and go all the time.


text:
Removing Epoxy Glue from a ceramic piezostack
I am going to stick a strain gauge to the side of an expanding piezocrystal using a basic 2 part epoxy glue. I am eventually also going to be sticking other things to it. Does anyone know a safe way to remove the glue without damaging the piezostack? The manual says specifically not to use acetone.


text:
Ionic Radius of Iodine in High Temperature Form
The ionic radius of iodine in the high temperature form can be determined once the unit cell is known, but how can I calculate the radius of $\ce{I}$ given that the cubic lattice parameter is 6.10 Å.

Considering by cubic lattice you mean a Simple Cubic structure, you could do the following:

Reference image1
Reference image2

(could not put it here because of copyright)
In a cubic lattice the atoms touch each other along the edges(see image1 and image2), hence the edge length/parameter will be equal to twice the radius (see image2).
$$\therefore\ \ce{Edge\ Length(a)=2\times radius(r)}$$
$$\therefore\ a \ = \ 2\times r$$
This method can be adopted for any lattice, you only need to find the axis along which the atoms touch each other(image1 bottom row).

If they exhibited an fcc close packing structure, you would notice that the atoms touch each other along a face-diagonal.
$$
\begin{align}
\therefore\ce{Length\ of\ face\ diagonal&\ =\ radius\ of\ top\ left\ atom +\\
                                        &\quad\ diameter\ of\ atom\ at\ face-center + \\
                                        &\quad\ radius\ of\ bottom\ left\ atom}\\
\therefore\ce{Length\ of\ face\ diagonal&\ =\ r\ + ( 2\times r ) + r}\\
\therefore\ce{\sqrt{2}\times a&\ =\ 4\times r}\\
\therefore\ce{a&\ = 2\sqrt{2}\times r}
\end{align}
$$

Similarly for bcc structure the atoms touch each other along the body-diagonal
$$\therefore\ce{\sqrt{3}\times a\ =\ 4\times r}$$

So, the answer must be in terms of the diagonal unit of the cube.

@ordinarychemistrystudent I thought you stated it followed a cubic lattice, in which case it would be the first section, that is $a\ =\ 2\times r$ where $a$ is edge length


text:
"Tie-breaker" rules for IUPAC nomenclature of organic compounds
I have three questions:

To find the parent chain of a saturated hydrocarbon, we have to find the longest chain and then if we have a choice, we choose the one with the more substituents. But what if we still had a choice? What would we do next?
Similarly, for numbering the chain, the last step is to number the chain such that the substituent which comes alphabetically first gets the lower number. But what if we again had a choice? I'm not sure here, but I think we can't possibly have another choice. Can anyone confirm this?
If an alkyl substituent is attached to the parent chain by a double or triple bond, how would we name it?

Examples: Here are some examples which demonstrate the above 'tie-breakers'.
For question 1:

The (green + blue) and (green + magenta) chains are the longest and both have two substituents.

The (green + blue) and (green + yellow) chains contain the maximum double bonds and are also the longest. They both have one substituent.
For question 2: As I said, I'm not sure if there can possibly be a 'tie-breaker' here.
For question 3:

The longest chain is the green one. I'm not sure how to name the substituent attached by the double bond.

could you atleast name any conflicting molecules for each question, it will also make it a lot easier for us to answer.

In question 3 do you mean something like this

@AdityaSriram Okay. And for your example: yes, I do mean something like that.

For Question 1
Example 1

You should select the (green + blue) chain because, no matter which among the blue and yellow you pick, the numbering would not start from the green chain rather the yellow or blue.
Now if you pick the blue chain the substituent ($\ce{methyl}$) would get number 4, whereas if you picked the green chain the substituent($\ce{isopropyl\ or\ methylethyl}$) would get the number 5, so in accordance to the First Point of Difference Rule you should select the green + blue chain.
Also Worth noticing is the fact that no matter which chain you pick, the other chain which becomes the substituent would always get the number 7
So locants (4,7) are better than (5,7).
$$\begin{matrix}
\text{Better name:}& \text{7-(2-isopropylhexyl)-4-methylheptadecane}&\\
\text{Rather than:}& \text{5-isopropyl-7-(3-methylhexyl)heptadecane}&
\end{matrix}$$

Example 2

In this case according to the above mentioned rule, you should check the position of the double bond, now since there is a double bond already in the green chain at the first position it would get number 1 and the numbering would start from there.
If you choose the green + blue chain the other double bond would get number 5 and for the green + yellow chain the other double bond would get number 6. So you should go for the green + blue chain in this case.
So locants (1,5) are better than (1,6) with reference to double bonds.
$$\begin{matrix}
\text{Better name:}& \text{(5}\textit{Z}\text{)-5-ethenylhepta-1,5-diene}\\
\text{Rather than:}& \text{(3}\textit{Z}\text{)-3-ethylidenehepta-1,6-diene}
\end{matrix}$$

For Question 2
If the chain has to be numbered and two substituents have the same locant the n you see the alphabetical order of the names of the groups/substituents and if they are also the same it only means that they are the same substituent and hence are written together with there locants separated by a (comma),.
For example in you would name the below compound as
$\text{2,4-dichloropentane}$

For Question 3
Example 1

The substituent in this example would be named $\ce{ethylidene}$.
$$
\begin{matrix}
\text{Final name:}& \text{5-ethylidenenona-1,8-diene}
\end{matrix}
$$
Similar substituents can be named as
$$
\begin{array}{|c|c|c|c|}
\hline
\ce{Substituent& Saturated\ form& Name\ of& Name\ of \\
&& saturated\ form& substituent \\
\hline
R=CH2& R-CH3& Methyl& Methylidene\\
\hline
R=CH-CH3& R-CH2-CH3& Ethyl& Ethylidene\\
\hline
R-CH=CH2& R-CH2-CH3& Ethyl& Ethenyl\\
\hline
R=CH-CH2-CH3& R-CH2-CH2-CH3& Propyl& Propylidene\\}
\hline
\begin{align}
\ce{R=C&-CH3\\
|& \\
C&H3}
\end{align}&
\begin{array}{rl}
\ce{R-C&-CH3\\
|\ &\\
C&H3}
\end{array}&
\ce{Isopropyl& Isopropylidene}\\
\hline
\end{array}$$

Could you tell me your sources?

Question 1 , Example 1 and 2 only rely on the First point of difference rule,  for question 2, it was trivial.

Dont you think in example 2, numbering should start from the yellow double bond rather than the green one? This would give its substiuent(the blue chain) a lower number and still have the same number of carbon atoms in the parent chain when numbering srarts from green chain.

Note that the preferred prefix is ‘propan-2-yl’ rather than ‘isopropyl’.

Also, magenta ≠ yellow.

For Example 2, the name “5-ethenylhepta-1,5-diene” (not “3-ethylidenehepta-1,6-diene”) is correct. However, the given explanation based on a “First Point of Difference Rule” is misleading since the locant set “1,3,6” would be lower than “1,5,5”.

I think in case of #1, you should keep the chain which contains the overall less-complex substituents (which would be green and magenta). Stating it into a rule I guess it would mean comparing the longest substituent of both alternatives.
For question #2, you should chose the numbering that gives an overall lower numbering result, even if you had a choice based on the alphabetically first substituent.
I have no idea for question #3. My instinct tells me that naming that =C-CH3 as a substituent feels wrong and I should choose the shorter chain even if its against the rules. But there's nothing scientific about that.

Just to make things clearer, here are the IUPAC rules to be applied. Each rule must be evaluated in order until there is only one possibility for the main chain.
(For hydrocarbons, there is no principal group so we do not evaluate rule #1.)

Maximum number of substituents corresponding to the principal group.
Maximum length.
Maximum number of double and triple bonds considered together.
Maximum number of double bonds.
Lowest locants for the principal groups (that is, for the suffix).
Lowest locants for multiple bonds.
Lowest locants for double bonds.
Maximum number of substituents cited as prefixes.
Lowest locants for all substituents in the principal chain cites as prefixes.
The substituent first cited in alphabetical order.
Lowest locants for the substituent cited first as prefix in alphabetical order

Edited to reflect changes in 2013 rules (thanks to Loong)

Could you cite a source? Also, what if there are still ties after evaluating all 11 rules?


text:
How do you create primary amines from alcohols?
By a primary amine I mean a compound where the hydroxyl group of the corresponding alcohol is replaced with an amine group.

For  aliphatic compounds you can use alkylation
$$\ce{R-OH + NH3 -> R-NH2 + H2O}$$

For aromatic compounds like phenol
$$
\begin{matrix}
\ce{C6H6O &\quad +& Zn& \ ->& C_6 H_6\\
Phenol&  & (dust\ form)& \ & Benzene&}
\end{matrix}
$$
Zinc dust reduces the OH group in phenol

$$\begin{matrix}
\ce{C6H6& + &HNO3& ->[H2SO4]& C6H5NO2 + &H_2 O&\\
Benzene& & (high\  concentration)&& Nitrobenzene&  water&
}
\end{matrix}$$
Nitration

$$\begin{matrix}
\ce{C6H5NO2& +& Sn& +& \qquad HCl_{(aq)} ->& C6H5NH2&\\
Nitrobenzene& & Tin& &Hydrochloric & Aniline&\\
&&&&acid}
\end{matrix}$$
$ $  Tin and hydrochloric acid act as reducing agents

I think you are oversimplifying and skipping a few steps and issues by just saying "alkylation". For a good answer I'd like to see the details and complications. Sure it is possible, but it isn't easy and that's what usually matters in chemistry.

(-1) That "alkylation" doesn't work.

In the conversion of primary alcohols to primary amines
$$\ce{R-CH2OH -> R-CH2NH2}$$
direct alkylation of ammonia normally is the last thing you want to do in the lab.
Under conditions where $\ce{OH}$ is a good leaving group, i.e. in acidic medium, the nucleophilicity of ammonia is reduced due to protonation. Moreover, there's little chance to prevent further alkylation of the initial product.
More promising strategies are:

Gabriel synthesis: Hydroxy-halogen-exchange, followed by reaction of the primary halide with phthalimide and subsequent hydrazinolysis
Mitsunobu reaction ($\ce{Ph3P, DEAD}$) of the primary alcohol using hydrazoic acid $\ce{HN3}$ as a nucleophile, followed by reduction of the alkyl azide
An (almost-one-pot) combination of the Appel and the Staudinger reaction

Heat the primary alcohol ($\ce{R-CH2OH}$) with $\ce{Ph3P}$  and bromotrichloromethane $\ce{CBrCl3}$ in benzene, then cool down to RT (room temperature) to yield $\ce{R-CH2Hal}$
Add $\ce{NaN3}$,tetrabutylammonium bromide and N,N-dimethylformamide and reflux again to obtain the azide $\ce{R-CH2N3}$, then cool down to RT, pour on water, extract with benzene and dry the organic layer
Add triethylphosphite $\ce{(EtO)3P}$ to the organic layer, stir at RT to yield the iminophosphorane $\ce{R-CH2N=P(OEt)3}$
Hydrolyze the iminophosphorane by refluxing with hydrochloric acid, then neutralize to release the amine

Mitsunobu was my first though. Appel is nice, too. Interestingly, I'd never heard of the Gabriel Synthesis; nice.

In place of using hydrazoic acid, DPPA might be a slightly less dangerous alternative (albeit more expensive and less atom economic).

In addition to the method provided by t3st you could also try the following for
$$\begin{matrix}
\ce{Ph-OH& +& Zn& ->& C6H6 \\
Phenol& & Zinc& & Benzene&}
\end{matrix}$$
(Just like the first step of t3st 's answer.

Then you could do Electrophilic halogenation of benzene using $\small\ce{AlCl3\ or\ FeBr3}$ as catalyst.
$$\begin{matrix}
\ce{C6H6& ->[\ce{Cl2/AlCl3}]& Ph-Cl\\
Benzene& & Aryl\ halide&}
\end{matrix}$$

Finally you could react aryl-halide with soda-amide($\small\ce{NaNH2}$) to substitute $\ce{Cl}$ with $\ce{NH2}$[citation needed]
$$\begin{matrix}
\ce{Ph-Cl& ->[\ce{NaNH2}]& Ph-NH2&\\
Aryl halide& & Aniline&}
\end{matrix}$$

Low yields all the way with this route

For PhOH to PhNH2 Pd catalysis offers a better route than Zn reduction etc.
Formation of the triflate by reaction with triflic anhydride then reaction under Pd catalysis with a source of N (I've used benzophenone imine or acetamide but there are others) then finally aqueous hydrolysis gets to the aniline in fairly respectable yield.


text:
Why is there not a strict correspondence between electronegativity and ionic character?
Why is the Electronegativity difference for atoms in bonding uncertain while determining what the compound will be?
According to the IB(International Baccalaureate) they say that the Electronegativity difference required for Ionic bonds is 1.7 though compounds like Hydrogen Fluoride are covalent even though they have a Electronegativity difference of 1.9. If the IB scale of above 1.7 being ionic and below 1.7 being covalent then why are there many exceptions to this rule? Also, why does Electronegativity not change, since having more electrons in atom would probably cause repulsion of new electrons?
If you search for what the Electronegative difference to predict bonding and were to look at Yahoo.answers you will see 3 different answers to how to predict these reactions at 1.7,1.8 or 1.9 electronegative difference Yahoo question. Another exception is Calcium Carbide which forms a ionic bond at 1.5 electronegative difference.

There are several methodics of assgning Electronegativity, each giving slightly different results. There is no strict border between ionic and covalent bound, and it usually have some weigh of covalent bonding too.
Electronegativity of atom in molecule changes in response to its neighborhood
Hydrogen without electrons will be naked proton. Because of its size it produces so strong electrostatic field, that it ruptures electronic shell of other atoms. So, hydrogen never forms positive ions, it is always attached to some other atom(s) covalently.

Lots of materials show a bonding character different from what would be expected from electronegativity difference alone. Here are a few:
• Hydrogen fluoride has an electronegativity difference greater than 1.7, yet it is generally considered polar covalent.
• Borides and hydrides of alkali and most alkaline earth metals are found to be predominantly ionic when charge separation is measured. The case of magnesium diboride ($\ce{MgB2}$), with an elecronegativity difference less than $1.0$, is discussed here.
• Alkyl fluorides and alkyl lithium have nearly the same electronegativity difference in the bond between carbon and the respective other element, but the alkyl lithium compounds show much more ionic character as evidenced by their reactivity.
The crux of the matter is that ionic character is determined not by electronegativity difference but by molecular orbital structure. Electronegativity difference can, of course, have a major influence on this structure but is not the sole determinant. Magnesium diboride is more ionic than hydrogen fluoride bcause in the former, the boron atoms associate to produce molecular orbitals that accept electrons from the magnesium (or any other strongly electropositive element) more readily than individual boron atoms. Largely covalent alkyl fluorides have good carbon-fluorine covalent overlap, whereas more ionic alkyl lithiums have poor overlap between the carbon and the bulky, diffuse valence orbitals on lithium.
The balance between ionic and covalent bonding is thus not a simple matter of looking up table valued, despite what many an introductory textbook might say. A good grasp of molecular orbital theory (which itself can get complicated), and preferably some experimental verification, should be brought to bear.


text:
What is the nature of chemical energy?
Is chemical energy basically stored energy?
I'm not sure how to define "chemical energy" while I know many physical formulas for energy e.g. potential, movement and other.
Is chemical energy an energy that is "conserved" and if so, why is chemical energy not just potential energy?
Additional questions:

If chemical energy is an "own" form of energy, why does it not have a specific equation?
Did we have chemical energy before the big bang or when did it start?
Can you have chemical energy at or close enough to absolute zero?
How much of the total energy in the universe is chemical energy?
What's the point of many different definitions of energy?

Chemical energy is really nothing but electric potential energy.
The crucial difference is that chemical energy involves energy that can be extracted by breaking/making bonds.
For example, we say that the reaction $\ce{2H2 + O2->H2O}$ is accompanied by a release of energy. This energy is the difference of potential energies of the bonds in the reactants and products. Which potential energies? The bond potential energies. Bonds are, at the most basic level, formed due to attraction and repulsion of electrons and nuclei (coupled with some quantum mechanical effects). This carries some potential energy. The change in this potential energy achieved by breaking the bond is the "bond energy", and forms part of your "chemical energy". Some other parts are lattice energy, hydration energy etc, which are (in the end), due to electromagnetic forces/electromagnetic potential.
So yeah, chemical energy is just a "subset" of potential energy,

If chemical energy is an "own" form of energy, why does it not have a specific equation?

Since chemical energy is an amalgamation of various types of electric potential energy in atoms, it cannot be given a formula (though you can calculate it for some compounds if you know some standard values).

Is chemical energy an energy that is "conserved" and if so, why is chemical energy not just potential energy?

Chemical energy isn't really conserved anywhere unless there is no change to the chemical composition of the system. However, in a chemical reaction, most of the time any changes in energy are compensated for by absorption/release of heat. But it is just potential energy, as mentioned.

Did we have chemical energy before the big bang or when did it start?

Meaningless question, there was no "before" the big bang. Also, energy really isn't something that you should classify into compartmentalized types.
But, since atoms formed a few minutes after the Big Bang, there wasn't anything to facilitate the existence of any chemical energy -- so the first occurrence of chemical energy was then.

Can you have chemical energy at or close enough to absolute zero?

Yep. Bonds exist at absolute zero. Note that we can never reach absolute zero, but bonds exist both close to and hypothetically "at" absolute zero.

How much of the total energy in the universe is chemical energy?

I can't answer that, and I doubt it can be answered to any reasonable approximation. I think the percentage would be rather low -- our universe is mainly stars (besides being empty space). The average star doesn't have many chemical bonds--it contains hydrogen/helium plasma, and plasma can't exactly form bonds.

What's the point of many different definitions of energy?

In certain situations, you know that there is no method to facilitate conversion of one form of energy into another. In such a case, one can separately conserve that form of energy. So the forms of energy help in analyzing/solving a situation, but not much more.

“at absolute zero, there is no thermal motion of atoms” is not true. See zero-point energy

“chemical energy is conserved in chemical reactions”: energy is conserved, but it can be converted (and usually is) to/from heat. Exothermic reactions release heat, endothermic reactions consume heat.

@F'x: re:zero point: Hmm, didn't think of that. re:conserved: Oops, I think I temporarily considered the energy released in a reaction as "a release of chemical energy"(that's how I've read it) and proceeded accordingly. My bad :s Thanks for pointing that out! :)

@F'x "At absolute zero, there is no thermal motion of atoms" Actually, this statement is accurate. The motion of the classical approximation of the zero-point energy is non-thermal, deriving from heat-/temperature-independent quantum effects.

I heard that the concept of chemical energy being "stored" in the bonds was wrong, and that it was the forming of bonds that released energy in an exergonic reaction (I think a Veritassium video, but might have misunderstood). The view that the energy released is the change in potential across all the bonds formed (and broken) across all atoms involved, makes more sense.

Does the oxidation of glucose release energy because C-C and C-H bonds are broken, becasue O-O bonds are broken, or becasue C-O and H-O bonds are formed? Neither, and all?


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Dissolving Organic Tissues
Watching some movie/TV this question came to mind, purely theoretical of course, if one wanted to dissolve a human corpse, like getting rid of a body after killing someone, is it better to use an acid or a base? In the movies (as well the popular perception for some reason) they always use acid. The movie I was watching they had a bathtub full of acid in which they throw in the body. But I also remember my chem teacher telling me (a long long time ago) that if she had a choice of an acid or a base in her eye, she would take acid any day because a base will dissolve organic tissue faster. So assuming that the acid and base have the same level of strength (say a pH of 1 and 13 respectively) which is better for getting rid of a corpse?
Thanks!

A note: The people who watch movies may not know what a "base" is, which is why they always use acids (a term known to everyone).

The movie was The American version of La femme Nikita, Point of no return, correct? Harvey Keitel, the "cleaner" comes in to get rid of the bodies. he uses acid.

@Tyberius , since this question was asked before the one in your link, shouldn't the linked question be considered the possible duplicate?

@airhuff I did not look at the dates carefully enough so that's my bad. But both of the questions came up as recently active and I think one of these should become the primary question and the other a duplicate to avoid multiple questions on the same topic.

@Tyberius , I completely agree.

@airhuff regarding closing an older question as a dupe of the newer question, you may be interested in https://meta.stackexchange.com/q/147643/312770

@Tyberius see above comment ^

I would say that acids are better.
Why? Bases and acids dissolve flesh pretty easily (I'm thinking of $\ce{NaOH}$ and $\ce{HCl}$ here, weaker acids/bases -- not so much). Most of our body is made up of proteins, which are overall pretty much neutral (generally). Both acids and bases can dissolve protein.
On the other hand, there is one major component of our body which is easier to dissolve with acid. Bones. The major component of bone is $\ce{CaCO3}$. The reaction $\ce{2HCl +CaCO3 -> H2O + CO2 ^ + CaCl2}$ is much more efficient/faster than $\ce{2NaOH + CaCO3 -> Na2CO3 + Ca(OH)2}$. Why is this? Well, for one, in the acid reaction, a gas is being released. This speeds up the reaction, by Le Châtelier's principle.
Apparently, NaOH is not able to dissolve bone completely, you are left with a fragile shell. Either way, it will be much slower than $\ce{HCl}$ while dissolving bone.
Here's an interesting article (no claims on validity) of why lime shouldn't be used to dissolve a corpse.

While (as far as I can tell) acids may be the better way to dissolve a body, I feel that filmmakers use acids because the term is known to most of the general public (unlike "bases", which isn't)

And if you pick nitric or sulfuric acid to dissolve bone, then you get $\ce{Ca(NO3)2}$ or $\ce{CaSO4}$, both of which are even more soluble than $\ce{CaCl2}$.

... Except those acids are harder for the average cleaner to get their hands on. Muriatic acid is available at any home improvement store, for a variety of uses, primarily pool chemistry.

Chop the body into smaller pieces, using a sturdy butcher axe.
Use a pressurized stainless steel autoclave (preferable with stirrer) or comparable professional kitchen equipment, suspend the pieces in concentrated $\ce{NaOH}$ and apply heat!
Cool down, release pressure and check. Saponification of body fats releases fatty acids that might form protective films and thus prevent further rapid decomposition of the "biological material". If necessary, stir and repeat until the meat is converted to a sludge.
Dispose the sludge, wash the bones, crush them to smaller pieces (= larger surface) and dissolve them in conc. $\ce{H2SO4}$.
Bring something to read - just in case you have to spend some 30+ years in a locked room - a book on Adolph Luetgert would be an obvious choice :D

Hot concentrated aqueous lye will dissolve tissue, aided by the process making soap.  Bone is about 50 wt-% hydroxylapatite, calcium phosphate, $\ce{Ca10(PO4)6(OH)2}$.  That is a problem.  Calcine, then make
http://en.wikipedia.org/wiki/Bone_china
http://en.wikipedia.org/wiki/MythBusters_(2013_season)
Mythbusters (Episode 208) demonstrated that acid as such is not effective for dissolving bulk flesh. "Piranha solution" (concentrated hydrogen peroxide in concentrated sulfuric acid) eats organics, oh yes indeed, and can explode while doing it; while making it, too.

There's an interesting documentary on how to dissolve organic tissue on YouTube: How to Commit the Perfect Murder, a BBC Horizon Documentary
They demonstrate that a washing powder is the best for dissolving organic tissue provided you can heat it up a bit.


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Can bromide ion ever act like a base?
I was reading MECHANISM OF α-HALOGENATION OF ALDEHYDES AND KETONES:

As you can see in the third step. Bromide ion is picking up a base from protonated acetone. Doesn't that step mean that protonated acetone is stronger acid than Hydrobromic acid? Is that even possible?

Come on, you can figure out this one on your own! How do you compare acidities? Here's a helpful hint.
Armed with that information, I have to add something. I believe the mechanism as drawn above is not universal: it sure depends on the conditions and solvent used. Some other proposed mechanisms involve the enol, rather than the protonated ketone, which might be more plausible in mild conditions. In water, anyway, HBr is fully dissociated and it shouldn't be “reformed” like that. The proton will return to be solvated in the solution.

I did looked up for pKa values and compared. According to the pKa values that I've seen, HBr was 1000 times stronger acid than protonated acetone. That's why the question. The pKa values given in your chart are different from what I've in my hand.

@claws okay, that's weird, but evaluation of pKa values outside 0–14 is tricky business, so why not. In that case, it would have helped to include that information in your question, so we know what you have tried and not. In any case, I do agree that Br– picking up a stray H+ looks weird. I wouldn't write it that way.


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Fourier transform for spectroscopy spectra?
I was thinking back to my chemistry major days and remembering the bunch of cool spectra we produced by various means (HNMR, IR, UV/Vis) and I remember all the cool 'self-interference' like couplets, triplets doublets, etc.
My question is basically motivated by the complexity of spectra, and the reductionist instinct to 'break it down' into simpler components. The FFT or DFT is known to separate out some periodic components of signals, as are SVD or tensor decompositions known to isolate independent components.
I am completely not an expert in any of this but I am interested to hear from any coal-face chemists if they regularly use some 'spectra decomposition' method to aid in interpretation or analysis. I remember being taught how to read various peaks in spectra and analyze them, but I can't help thinking, for some very complex molecules, surely it gets pretty hard. And what if, we don't actually know the structure beforehand?
So given that Fourier transform is used in NMR, my question is : is there ever a need to take a spectrum and do a computer analysis of it that isn't just about matching peaks to a database of known signals? What other nifty things can be done with a spectrum once it is obtained?

I'm not sure what exactly you are asking, NMR spectra e.g. are already fourier-transformed (unless you have a really antique spectrometer).

Well I did not know that. I will tack on a more specific question to the end.

As Mad Scientist said, many spectroscopy methods are already based on Fourier transform of a time-dependent observable. (This is not true of the basic spectrophotometry measurements, where you directly measure the transmittance as a function of frequency.) NMR is a method based on Fourier transform, as is FTIR.
Regarding NMR signal processing, it's a very wide field, which includes interesting research ranging from academic interest to software development. So, it's hard to answer your question other than saying “yes, of course, further processing of NMR signals is sometime possible”. One thing you have to realize is: while you have probably learnt proton NMR during your studies, this is only one NMR technique, and the simplest of them all. There are plenty of different NMR techniques, depending on the type of excitation and sequence of RF pulses, many of them being multidimensional. Each method can use a variety of signal processing techniques.

Cool. It sounds like something interesting to look into more. Thanks for pointing the way.


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Does citric acid react with carbonated water?
I've noticed that when adding lemon juice to carbonated water, large 'soapy' bubbles appear on the surface. Since I'm adding another acid to a solution of carbonic acid, I would maybe expect a faster release of $\ce{CO2}$, but the bubbles I'm seeing do not resemble the usual fizzy drink bubbles.
What's going on?

Lemon juice is more than just citric acid. Have you tried the same experiment with pure citric acid (should be available in the baking section of a supermarket)?

The chemistry you're describing the the one you expect: CO2 is formed, and released from the liquid. However, your question is more about kinetics than about the final product. We can assume that slow release of gas leads to larger bubbles, because nucleation rate for the bubble formation is lower. Now, in fizzy drink, there is a large quantity of carbon dioxide accommodated by the liquid due to pressurization. In your case, it will depend on the concentration of your lemon juice (natural lemon juice has pH ~ 2), and more importantly on that of the carbonated water: I just tried with a few grams of sodium bicarbonate dissolved in a glass of water in my kitchen, and the kinetics and bubble size really depends on initial carbonate concentration. (I haven't been able to make pictures or video, because I had to handle my kids who were very fast drawn to this impromptu experiment.)
Also, I second Richard's comment: there are many additional chemicals in lemon juice, some of which surely act as surfactants… which changes bubble formation and kinetics.


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Why do Organolithium or Grignard reagents act as nucleophiles and not as bases with aldehydes and ketones
I've read entire Chapter 14: Organometallic Compounds of  Francis Carey's "Organic Chemistry" but I still didn't get an answer to my question.
Quote from the book:

Because of their basicity organolithium compounds and Grignard reagents cannot be prepared or used in the presence of any material that bears a $\ce{-OH}$ hydroxyl group. Nor are these reagents compatible with $\ce{-NH}$ or $\ce{-SH}$ groups, which can also convert an organolithium or organomagnesium compound to a hydrocarbon by proton transfer.

Fine. No arguments with that.

Terminal alkynes also give alkanes with Organolithium and Organomagnesium compounds.

Must be true if he says so. Now coming to the point,

Ammonia and Amines have $\mathrm{p}K_\mathrm{a}$ around 36.
Terminal alkynes have $\mathrm{p}K_\mathrm{a}$ of 25.
Aldehyde $\mathrm{p}K_\mathrm{a} = 17$, Ketone $\mathrm{p}K_\mathrm{a} = 19$ and an Ester $\mathrm{p}K_\mathrm{a} = 25$.
Water & Alcohols have $\mathrm{p}K_\mathrm{a}$ of 15–18.

If organolithium compounds and Grignard reagents act like bases with alcohols, terminal alkynes, and amines/ammonia, why do they act like nucleophlies and give addition reaction with carbonyl compounds? Why don't they act as bases and give alkanes instead? What am I missing here?

One more comment, why don't you simply draw an equation contains an organolithium reagent and an aldehyde to product the deprotonated aldehyde as you proposed? You would like to use pKa table to figure out the easiness of deprotonation at any position.

In fact, both reagents you noted here have quite complex structure and are not nucleophiles at all: both are electrophiles, because metallic atom here has too little neighbors to draw electrons from.
Let me explain, using Grignard reagent $\ce{EtMgCl}$ . In reality it has complex structure with $\ce{Mg}$ atom coordinated to alkyl fragment and two diethyl ether molecules. When reacts, it coordinates first with most nucleophilic atom. In case of alcohol it is oxygen of $\ce{OH}$ fragment, and of course it  reacts with closest easily migrating atom: hydrogen, producing alkane. In case of ketone it coordinates with $\ce{CO}$ group, and alkyl group attacks closest electrophile : carbon from $\ce{CO}$ group, whose electrophility was just boosted.
If right compound, effectively bounding to metal atom, is added and no other way of reactions is readily available, both reagents can work as bases. For example, organolithium compounds can replace hydrogen atoms in alkynes, alkenes and arenes as you noted earlier.

Could kindly expand/explain your answer little bit. It sounds little complex. I didn't understand it clearly. If its not too much to ask, figures would be great!

Technically, you are comparing apples and oranges. Grignards and organolithium compounds are good nucleophiles (with the caveat permeakra highlighted) towards carbonyl groups. But they cannot attack alcohols, water, amines or alkynes nucleophilicly. There is simply no attack mechanism possible that would result in a more stable addition product. In fact, alcohols amines and water are pretty much entirely unimpressed by nucleophilic attacks. All of these can react as bases, though, so when presented with one of the two compounds mentioned, they will gladly give up their proton. (Remember that butane has a $\mathrm{p}K_\mathrm{a}$ value of $\approx 40$, so n-butyllithium is a very strong base.)
Carbonyls simply react much faster under nucleophilic attack than under deprotonation.

But isn't it true that acid-base reactions are much much faster than nucleophilic attack?

@ShoubhikRajMaiti Not necessarily. In non-protic solvents, there is no rapid proton shuffling mechanism as in water. This, acid-base reactions are much slower and comparable to regular reactions.

Well, in fact, the alpha-proton of aldehyde and other carbonyl/carboxylic acids are acidic, across a large range. The alpha-proton has a chance to be deprotonated by organolithium and magnesium reagents and forms a so called "enolate", which undergoes "aldol condensation" reaction with residual aldehyde.
Some knowledge derived from practice but not covered in text book is, if alkyl lithium reagent is added to an enolizable aldehyde, the nucleophilic additon and enolization-aldol condensation of aldehyde would take place together, and the product will be a mixture of alcohol and aldol condensation in various ratio depending on substrates. Some aldehydes, e.g. phenylacetaldehyde, provides aldol condensation product in major due to their high enolizabization property.
Mg/Na/Li alkoxides and amides also can deprotonate the alpha-proton of carbonyl as well as attacking certain certain carbonyls, like carboxylic ester/amide and nitrile. The carbonyl of aldehydes and ketone are not attacked.

Butyllithium is a powerful nucleophile much faster than it is a strong base.  Add two equivalents of dry N,N,N',N'-tetramethylethylenediamine to coordinate the lithium.  Now you have an extraordinary base (it will deprotonate THF at room temp) but a very poor nucleophile.  One then presumes that nucleophilic attack upon a carbonyl may be a four-center intermediate in which C-O-Li coordination sets up and directs attack of the alkyl anion.
BuLi plus Li tert-butoxide is a strong enough base to initiate anionic polymerizations.  The alkyllithium cage polymer is disaggregated.


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How can I find whether there is oxygen present inside a well of diameter 1 m at 50 m's deep?
How can one find whether or not there is enough oxygen for breathing present inside a (dry) well of diameter 1 m at a depth of 50 m?
Many people in Nepal are killed while cleaning wells because of a lack of oxygen.

If I understand you correctly, you want to ask if there is free oxygen for breathing present deep in a dry well, right?

If this is for an actual application, please use whatever answer you get as a starting point for a consultation with an expert on such matters.

is this a theoretical problem or practical one, the data seems as if it is for theoretical purpose.

Dangle a candle into the well and observe whether it goes out at the bottom.

The ideal solution would be a single gas personal detector, such as this one (it's only the first example I found, I don't know this brand, but there are many equivalent apparatuses from other vendors).
While digital personal detectors are expensive ($200+ range), there are simpler systems which also provide some basic safety (better than nothing). The Davy lamp is perhaps the best known example of these, and it was a huge breakthrough in mine safety when it was popularized after 1815.

(Source: http://lindal-in-furness.co.uk/MinersLamps/MinersLamp.jpg)

The Davy lamp isn't an oxygen detector (or wasn't designed to be). It is designed to prevent the lamp flame from igniting flammable gasses that are often present in mines. If the lamp goes out it is a sign there isn't enough oxygen, but you would probably already be unconscious at that point so use as a detector of low oxygen is a little futile.


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How do you Synthesise Glycosides of flavonoids and stilbenoids?
I would like to know how you go about synthesising glycosides of flavonoids and stilbenoids from their respective sugars and flavonoids/stilbenoids. I read http://www.netsci-journal.com/97v1/97005/005p20.htm but I found it too specific (as it only applied to aliphatic compounds) and required an enzymatic catalyst that is impossible to predict. I would like a general formula that if it does contain an enzymatic catalyst the enzyme is either constant or at least usually predictable.

Could you clarify what you mean by a 'constant or usually predictable enzyme'?

An enzyme that doesn't change (i.e. is constant) or is predictable by some mathematical model.

My educated guess would be that such an enzyme does not exist. Enzymes are usually very specific to their substrates. As for predictability, it is possible to model an enzyme to a desired substrate using computational methods (i.e. Molecular Modelling). However you might be better off looking into classical organic synthesis methods, retrosynthesis specifically.

In terms of process, sugars tend to be exceedingly insoluble in organic media.  Simple per-acetylation of glucose is not a high yield reaction. If the glucose is ball-milled overnight (lapidary tumbler no more than 1/3 full including grinding balls), the micronized glucose reacts in very good yield.  There are other tricks to helping things move along one you have the synthesis itself.

I suggest to forget about the enzymes.
Pepare for a probably lengthy synthesis - you will need patience and a clever protection/deprotection scheme.
The following is definitely not applicable to all starting materials but might give some directions.

If the flavonoid contains a catechol moiety, just like quercetin (1), the vicinal phenolic $\ce{OH}$ groups can be protected as an acetal by heating the starting material with $\alpha,\alpha$-dichlordiphenylmethane in diphenylether (DOI). The protecting groups is easily removed by hydrogenation over $\ce{Pd/C}$ at 1 atm.
Note that the PMB (4-methoxybenzyl) group is another protecting group for (phenolic) $\ce{OH}$ that gets off by hydrogenation.
As far as selectivities for the protection of the other $\ce{OH}$ groups in 1 are concerned, have a look at the 1962 paper by Leonard Jurd, who described that the methylation of quercetine pentaacetate in dry acetone predominantly furnishes 7-O-methylquercetin. There are quite some more recent articles citing this!


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How would you go about synthesising amino sugars from sugars?
How would you go about synthesising amino sugars from sugars? e.g. How would you synthesise the compound below from D-glucose?

Would adding ammonia ($\ce{NH3}$) to it work?

First, the anwer is “Would glucose and ammonia work?” is: no, definitely not. The $\ce{OH}$ group is a poor leaving group, and alcohol + ammonia definitely won't yield an amine. Moreover, in your case, you have many hydroxyl groups and you only want one transformed into an amine, which is tricky.
I don't know how the synthesis of the above compound can be achieved, but I know one way to help you start a literature search: your expected compound is called an amino sugar. In this particular case, it's the 6-amino-D-glucose.


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Is it possible to estimate a substance's water solubility from data like the substance's partition coefficient, dissociation constants, etc.?
If it is possible, how would you estimate a compound's water solubility from data like its partition coefficient, $ \log P $, its dissociation constant, $pK_a $, its molar mass, etc.?

I think it can be done, but first you have to have model, so you can gather your data via http://www.vcclab.org/lab/alogps/ -> lots of compounds -> lots of descriptors (as you mentioned) -> output logS (solubility) -> heuristic(?) model (maybe ANN would be useful?) -> then input data -> output (predicted solubility)

Moreover similar things have been done at http://onschallenge.wikispaces.com/Predictive+Solubility

It could certainly be estimated from those properties, the structure, etc.  A lot of companies have developed proprietary algorithms for things like that:  solubility, diffusion coefficient, polarity.... (For example of a company, see "ChemAxon")
The water solubility for many chemicals is already known, though.  Search for a chemical ChemSpider, and look under the "Properties" tag.
They've even got an API, in case you're interested in using water solubility in an app.


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How would you synthesise tetrahydrohyperforin and octahydrohyperforin from hyperforin?
The chemical structure of the compounds in question are:

Figure 1: Hyperforin

Figure 2: Tetrahydrohyperforin

Figure 3: Octahydrohyperforin
Would heating hyperforin in the presence of water manage to saturate the required bonds and thus allow one to synthesise these compounds?

That route won't work.
You need a hard base reduction that does not attack the enone, or you need to protect the enone to make Tetrahydrohyperforin.
Catalytic hydrogenation that is selective against carbonyl reduction is necessary to make Octahydrohyperforin.


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How do the molecules know the future? How do they proceed to form "Most Stable" compound?
I'm tutoring few students for Chemistry. During the course, I many times use the phrase "Out of these possibilities this one is stable. So it is formed". I get questions like, "How do the molecules know?"
Honestly I don't know. How do the molecules know the future? How do they proceed to form "Most Stable" compound?
Cross Post: https://physics.stackexchange.com/questions/51894/how-do-the-molecules-of-reacting-compounds-proceed-to-form-most-stable-molecul

Chemical reactions occur because the thermal energy in a substance makes molecules bump into each other, and each time they bump into each other there is some chance for a reaction to take place.  This is an oversimplified model, but it suffices to get some intuition for what's going on.
The molecules of course do not "know" that some configuration is more stable than others.  What happens is that the molecules react randomly, again and again; they take a sort of random walk through the available configurations.  Every now and then, a molecule will purely randomly find its way into the stable state.  Once it's there, it's "stuck" - it has a much smaller chance of getting out of the stable state.  That's more or less what it means for a state to be stable.
Over time, more and more molecules will find their way into the stable state and get stuck, so the result looks as if the molecules "knew" which state to go to and just went straight there.  But actually they only got there through the process of randomly bouncing around the configuration space, plus the stable state acting as a trap that molecules can enter but not (as easily) leave.

This is a good answer, but one thing to add.  The random action of molecules works so well because there are usually a huge number of them interacting, and thus the Law of Large Numbers from probability theory works extremely well.  For example, one mole of any molecule involves over 600 sextillion molecules (that's 600 with 21 additional zeros after it).  Even if trillions of molecules fail to react, this is insignificant compared to sextillions of molecules. So yes, randomness is a great reason why molecules don't need to "know the future".

Clearly, molecules are not in any sense volitional agents. They're simply subject to the basic and immutable laws of thermodynamics, the second of which states that any isolated system always spontaneously develops toward a state of equilibrium in which entropy (which can be informally described as "disorder" or "randomness," meaning thermal energy distributed in such a way that it cannot be used to perform useful work) is maximized. Increased stability is really another term for the minimization of free energy, i.e., energy available to perform useful work. All irreversible spontaneous chemical reactions are ones in which the free energy change is negative, either because the chemical reagents involved experience an increase in entropy, or because the thermal energy released by the reaction sufficiently increases the entropy of the surroundings. The basic, simplified equation for describing this situation is:
$$\ce{\Delta G\ =\ \Delta H\ -\ T\Delta S}$$
Where $\small\ce{\Delta G}$ is the change in free energy, $\small\ce{\Delta H}$ is enthalpy (change in heat energy), $\small\ce{T}$ is absolute temperature, and $\small\ce{\Delta S}$ is the change in entropy. If $\small\ce{\Delta G}$ is negative for a particular reaction at a given temperature, the reaction will proceed spontaneously. In essence, the entire universe is evolving toward a state in which entropy is maximized and free energy is minimized, and only chemical reactions which contribute to that phenomenon can spontaneously take place.

If I understood you correctly, what  you are saying is that reaction proceeds in that direction, which reduces free energy. This is Second Law of thermodynamics. But "Law" is an action/process. Its not the cause. right? I feel my question is still unanswered, How does the reactants know that going in a particular direction reduces $\Delta G$?

Well, it's a law because it's been empirically demonstrated and mathematically determined to be the fundamental, underlying force at work in driving reactions in a particular direction. The reactants cannot "know" anything, they don't have any sort of intentional drive, they're simply at the mercy of probability and thermodynamic imperatives. There's really no deeper meaning to be discovered here, at least as far as chemistry is concerned. For that, you can look to metaphysics. The laws of thermodynamics are axiomatic, they're first principles, so there's no deeper meaning to search for here.

To expand: when you have molecules constantly in motion, colliding with great frequency and in very large numbers, then it becomes statistically inevitable that they will form bonds (and break others). However, only certain combinations yield products that are stable (i.e., lower in free energy), and these are the combinations that are observed to exist with at least some degree of persistence. Still, there's no way to answer the question without recourse to principles of quantum mechanics, probability, thermodynamics, and reaction kinetics. This is simply the nature of our universe.


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Law of Multiple Proportions: What is the significance of small whole numbers?
I'm new to Chemistry and in my textbook, it describes the Law of Definite Proportions (aka Proust's Law) and then goes on to describe the Law of Multiple Proportions.
The example they give is carbon monoxide and carbon dioxide, where the mass ratio of oxygen to carbon in carbon dioxide is $2.67:1$ and the mass ratio of oxygen to carbon in carbon monoxide is $1.33:1$.
I understand the above example, and I understand that when you compare them in the following way:
$2.67~\mathrm{g}$ (proportion of oxygen to $1~\mathrm{g}$ carbon in carbon dioxide)$/ 1.33~\mathrm{g}$ (proportion of oxygen to $1~\mathrm{g}$ carbon in carbon monoxide) $= 2.00$, a small whole number.
$$\frac{\text{Mass oxygen to 1 g carbon in carbon dioxide}}{\text{Mass oxygen to 1 g carbon in carbon monoxide}} = \frac{2.67~\mathrm{g}}{1.33~\mathrm{g}} = 2.00$$
My textbook doesn't describe what a small whole number is, or what it means … just that, in this example, $2.00$ is a small whole number.
What is the significance of the $2.00$?
I understand that carbon dioxide has double the oxygen. Is this where the $2.00$ plays a role?

There's no universally accepted hard cut-off, as far as I'm aware, which determines the range of numbers that would constitute "small" mass ratios. Generally, all textbook examples select compounds that yield single-digit mass ratios.
The significance of the ratios being small (as well as their constancy and the limited number of different mass ratios for binary compounds actually found in nature) is that they allow for reliable conclusions to be made about the stoichiometric proportions in which elements combine based on crude empirical data alone, as well as leading to conjectures about the number of possible configurations, bond arrangements, and oxidation states yielding said combinations.
The law is valid for a large variety of binary compounds simply by virtue of the fact that there is typically a limited number of permutations in which two given elements can be combined to yield a stable compound. For example, C and O have only two, N and O have only four, metals have a limited number of oxidation states, therefore giving rise to relatively few possibly formulas for binary ionic compounds, etc. The law breaks down for cases in which two elements can combine to form long chains or complex molecules (e.g., large hydrocarbons and [mostly hypothetical] silicon analogues), when the mass ratios can sometimes become quite large.

If SQL can treat numbers in the range –32,767 to 32,767 as "small integers", the mass ratios (like the one obtained here) are indeed small whole numbers.

The law of multiple proportions is largely obsolete these days because we all believe that atoms exist.  Prior to about 1900 this and similar laws were used to show that, for example, the ratio of oxygen in $\ce{CO}$ to oxygen in $\ce{CO2}$ is 2:1.
Of course, that's trivially obvious when we write the formulas, but before we believed in atoms the "small whole number" ratios were used as an example of an experimental situation (the 2:1 ratio) could be explained by atoms (i.e. quantizing atomic masses.)
Today, the law causes more confusion than it cures.  Atoms are here to stay.

When teaching beginners, usually non-stoichiometric chemical compounds are ignored. The reality is more complex than what they teach undergraduates and high school students. One important point about chemistry is that it is the study of atoms (mostly of the "naturally occurring" 92 elements) and their bonding and reactions.
The ideas of indivisible units (atoms) and shapes of molecules (geometry) are foundational in the science. An added confusion is that atoms (and molecules) are too small to count (or even to weigh, except with very expensive equipment in very special conditions). So we invented a number (similar to the number "dozen", except much much much bigger) and use Avogadro's Number as the count of the defined quantity of a "mole". We can do almost all (but not quite all) chemistry by considering one atom or molecule or one mole of atoms or molecules as being what is represented in a chemical equation A + B → C + D.
When we're talking about doing lab experiments, our only means to measure quantities is not to count atoms but to weigh them - large numbers of them. So, most chemistry is done in the lab using moles, not indivisible units (atoms or molecules). This works out pretty well.
Usually, the average isotopic composition is accurate enough so that 1.0000 moles of carbon weighs the same here or in Beijing. (In reality, the exact mass of 1 mole of C depends on its source: the ratio of 12C to 13C varies between C found in limestone, in plants, in seawater and in the atmosphere (slightly). but the kind of accuracy/ precision we generally work at in the lab allows us to ignore these fine points). Average atomic weight is good enough. Since this is so, for stoichiometric compounds the ratio of carbon (by mass) in C2H6 (ethane) to C7H5N3O6 (trinitrotoluene, T.N.T.) is 2:7 (or 7:2) a ratio of "small integers".
The small whole number thing follows directly from the atomic weights being (nearly) constant and the stoichiometric nature of the compounds being considered. Dealing with more complicated situations which don't follow that "law" just usually isn't done in beginning chemistry.
In science, almost everything you learn is wrong. Wrong in the sense that it is only approximately correct. So, your knowledge will be like an onion, with a layer under a layer under a layer.
For this law, all it is really saying (using atom's mass rather than their count) is that the counts will be in ratios of small integers. Given a set of atomic weights, you can't give me an example of their mass ratios violating the law. Because for element Z, all chemical equations and formulae, whether expressed in atoms of Z or in moles of Z will be in multiples of either it's count (which is pretty useless in the laboratory or in the real world) or the atomic weight ( we can typically weigh amounts to 4 decimal digits precision, and 6 isn't uncommon and 8 is possible). So that means mass (weight) is very useful for determining the counts (hence the formula).
Looking at the periodic table, you'll notice that (in a good one) the atomic weights do NOT all have the same precision. The reason that C is 12.011 but He is 4.0026 and Li 6.94 is not because we can't determine the weight (mass) accurately, but because it varies depending on the source of the element.
The mass of each atom depends on four things: its atomic number (which tells you the number of electrons and protons present in the (electrically neutral) atom), the number of neutrons, the binding energy (remember E = mc² or rearranging m = E/c² where m is binding mass and E is binding energy of the quark-gluon cloud which makes up the nucleus), and any kinetic or potential energy due to its speed or location in a field.
So, no, you can't calculate an element's average atomic weight by simply multiplying the number of neutrons by a factor and the number of protons by a factor and adding them up (although it's not a terrible approximation for nonrelativistic atoms).
There are several examples of non-stoichiometric materials. Metamaterials, which are (usually) built up of layers of atoms, so the atom count isn't a useful measure of the material. Polymers (where you almost always have a range of molecular sizes, so that the typical formula of AxBy where x and y can be any rational number since a polymer is a distribution of molecules of different sizes, it isn't quite what a purist would say is a chemical compound, but most of us would accept it as one especially for large x + y without many (or any) small mw polymer molecules present.
And finally there are many many examples of substitution compounds, consider electronics. The "brains" of a computer is made of billions of transistors made by adding very small amounts (doping) of one element to another. Silicon chips, for instance, would be worthless without doping, but the amount of dopant is almost never a small integer SiMx where x almost never an integer, (and M is generally a group 13 or 15 element.)

Incredibly, this answer is so long and yet manages to completely evade the topic being asked.


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What commonly availible salts are strong dessicants?
I'm looking into building an absorption chiller as a decidedly non-professional DIY project. For safety reasons and because temps. > 0°C are sufficient, I won't be using ammonia but some saline solution as absorber and water as cooling agent.
The obvious choice for the salt would be LiBr, as this is used in comercial installations. However, I'm not sure I can procure it (chemicals suppliers don't sell to private persons in germany) and it's not cheap (~ 30€/100g).
What other salts are strongly hydroscopic and commonly availible?
Alternativly, where can I find a concise list with the physical properties of many common (=household) salts?

Have you looked into silica gel or zeolites? Both are non-toxic, and thus likely obtainable by private persons. Zeolites are used in self-refrigerating beer kegs.

I'm looking to create a liquid aquaous solution of the salt, so silica gels or zeoliths won't work for me. I know of absorption chillers using solid absorbers, but I'm looking to build something mechanically simpler.


text:
Does an "excited atom" lose energy gradually?
When an electron absorbs energy and is in a higher energy orbit (I guess the atom would be in an unstable state), when the electron releases this energy, would all of this energy be released at the same time, or could it release energy one "level" at a time?
So for example if a Hydrogen atom absorbed a lot of energy and its electron went to some level n = 5, could it slowly make its way back down to level 1?

Yes, it certainly can. In fact, it can do either, traverse gradually across excitation states down to its ground state, or jump from its current state to its ground state, or to any lower state for that matter. So essentially, as long as the target state is equal or lower in energy than the current one the electron can jump to it.
Those jumps have different probabilities though. Determining which process is more likely to occur, and what the probabilities are is a matter of quantum mechanic calculations. Have a look here for example: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/fermi.html

In presence of an environment (a solvent), it is also possible that the excitation energy is dissipated as heat (vibrations) to the solvent… this is called non-radiative relaxation.

To add to @Tanith Rosenbaum's answer, the energy can be lost to other energy levels by emitting a photon to that other level, say n = 4 to 3. The total energy is conserved, thus the energy of atom in n=3 + photon is the same as that of the atom in n = 4. Emission (fluorescence) is a spontaneous process, and the lifetime of an excited atom can vary widely, typically from nanoseconds to seconds. The reason for this depends on symmetry and angular momentum of different energy levels (generally called selection rules).
To loose energy in any other way, say by collision with another atom, then that atom has to carry away the difference in energy just as the photon did. Note that a collision can also add energy by the same reasoning, as can absorbing a photon so the atom will now contain even more energy.  In a more detailed examination, properties such as symmetry and angular momentum have to be considered.


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Why is there a max radius for an orbit of an electron around a nucleus?
Why is there a max radius for an orbit of an electron around a nucleus ?
I had a course in electromagnetism but I do not get this.

Exact orbits literally can't be known even if they can be really called orbits.

Perhaps you can clarify your question a bit, but if you are referring to the size of the radial distribution of electron density, this increases with increasing principal quantum number, up to the ionisation limit, at which point the spacing between energy levels trends towards 0 and the interaction energy trends towards zero. After this, the electron is for all intents and purposes unbound. This is illustrated for the hydrogen atom in the Lyman series (image borrowed from Wikipedia):

The energy of the largest transition $n=\infty \rightarrow n=1$ is 13.6 eV, or half a Hartree (this is the energy of a photon with a wavelength of 912 Ångström). This is the binding energy of the hydrogen atom. Even the Lyman gamma line - representing a $n=4 \rightarrow n=1$ transition - is quite close to the ionisation limit. As n increases, from a strictly electrostatic point of view the orbital becomes far less stable, and this is reflected in the general negative trend of ionisation energies across the periodic table, however the complete picture is a bit more complex than that because electrons can interact with electrons as well as the nucleus.
The key point is that as n increases, the energy required lose an electron entirely becomes smaller and smaller, and this fixes a practical limit on where we can say the electron is actually bound. This has a caveat though - cold gasses can form unusually long-lived Rydberg states with ridiculous principal quantum numbers with correspondingly enormous orbital radii (up to micrometers!*) and semiclassical electron orbits. The reasons for the stability of these states is outside of my knowledge.
**To paraphrase, Alisa Bokulich points out in 'Re-examining the Quantum-classical Relation' that Rydberg atoms are of comparable size to grains of (fine) sand, which is pretty neat*


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Measuring distance in a conductive liquid
What is needed to  measure distance between two electrodes imersed in seawater ?
I have  tried measuring the voltage across the path (a suposed resistance) with a constant current source (about 10uA) passing the water to GND of circuit. (the resistance is suposed proportional to distance of electrodes) The probes are 100Hz reversing switched to avoid polarization. And I am using AG/ AGCl pellets as electrodoes. But until now I could not have stable results.
What may be wrong ?  Could someone help ?
Thanks

The AgAgCl sinterized disks was choose because it is very used as reference electrode for seawater potential measurements (cathodic protection) with very good results and low maintenance, but I really dont know if it is the the best for my application. The suface area is about 1 cm2 then we are expecting about 20 Ohms per cm of distance in seawater. With a 10uA as (AC square wave) current source  between electrodes voltage will range from about few milivolts up to 1 volt at 50m.  As you see the potential is bellow of water electrolisys. What you mean "with a lot of corrosion protection" for Stainless Steel plates ?
I understand that salinity, temperature etc could vary in the media but these factors are not expected  to change so quickly and I also intend to use a known distance sensors for reference as well averaging the measures.
I intend use this sistem in a dynamic environment where the reletive speed between sensors or the flow may be about 10 cm / sec.
What you think about the method ? There are a better one ?

Can you provide some more information about the instability of your results, such as a plot of resistance over known distance, and/or resistance over time?

(Initial, preliminary answer, here.)
There are several possible reasons for your unstable results:
Possibility 1:  Wrong electrodes
Ag|AgCl pellets seem like an odd choice of electrodes for this application.  What are the reasons for using them as electrodes?  What is their surface area?  Their complex shape means the electrical current maybe isn't uniform between the two electrodes.  Also, the fact that you're drawing current through an electrode with an AgCl coating may mean the coating is building-up/breaking-down (which could mess up your measurements) ...  In your current application, it seems like you'd want to use two, big, parallel stainless steel plates (with a lot of corrosion protection!) as your electrodes.
Possibility 2:  Changing salinity
Depending on the local conditions, the conductivity of the seawater may be changing.  For example, if you are by a river delta, the tides may be washing fresh water between your plates.  Or, if the electrodes are at the water surface in really sunny location, day-time evaporation may increase salinity ...  If either of those are the case, install a second set of plates with a known, fixed electrode spacing to measure salinity, and use that to modify the results from the original set of electrodes.
Possibility 3:  Convection Currents
Is the seawater between the two plates moving?  If so, how fast?  Does the speed change over time?  All of these factors could contribute to unstable electrochemical readings, too ... If so, you may add mesh or baffles to create more-consistent flow conditions.

(Follow-up answer, here)
As far as I can tell, you're method sounds alright from an electrochemical perspective.  Cathodic protection is fine for minimizing corrosion.  Having "known distance sensors" will avoid problems with changing salinity, etc (as you mentioned).


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Palladium hydride forming water and heat at room temperature
I just read that saturated palladium hydride, $\ce{PdH}$0.7, when exposed to oxygen will generate heat and water on it's surface, $\ce{4 H}$(Pd) $\ce{+O2}$(g) $\ce{\to 2 H2O}$(g, l).
What is going on?

I'm not an expert, but typically there are a series of surface reactions where the reactive species are all adsorbed on the surface. So atomic hydrogen wouldn't "leave" the palladium before it reacts, it just diffuses from an internal crystal site to a surface site.
Likewise, the oxygen would adsorb onto the Pd surface, I'll assume dissociatively. So now there is atomic O and H diffusing around the surface sites of the Pd. These can react and form adsorbed OH, which can further react and form adsorbed H2O, which can then desorb and give you your water.
Your enthalpy calculation is off, the equation for water already includes the fact that an oxygen molecule must break apart to form it.
If you include the hydrogen gas reaction with Pd in your picture, all you're doing is turning hydrogen gas and oxygen gas into water, using Pd as a catalyst. So the total enthalpy change is just -285.83kJ/mol. You don't have enough information to determine how much heat is released (or absorbed possibly) in the hydrogen absorption reaction and how much is released in the water production reaction, but the total of the two will be -285.83kJ/mol.

I'm guessing that when the $\ce{H}$ atoms leave the palladium are they still monatomic for an instant, but without the $\Delta H^0_f = 436 kJ/mol$ extra energy that free  $\ce{H}$ atoms would normally have, as bypassed by the $\ce Pd$ somehow. Most of the $\ce{H}$ atoms simply fall in step with one another, reforming $\ce{H2}$ without a peep. But some latch onto nearby $\ce{O2}$ molecules, disassociating them and forming $\ce{H2O}$.
The heat generated is the difference between heat required to dissociate the $\ce{O2}$ and the heat generated by the formation of $\ce{H2O}$.
$\ce{O2 \to 2 O}$   $\Delta H^o_f = 247.5 kJ/mol$
$\ce{ H2 + 1/2 O2 \to  H2O}$ $\Delta H^o_f = -285.83 kJ/mol$
$∴ \ce{2H_{(Pd)} + 1/2O2_{(g)} -> H2O_{(g, l)}} \quad \Delta H^o_{\mathrm{f}} = -38.33 \mathrm{kJ/mol}$
Being as this is only$\ 5.68 kJ/mol$ short of the heat required to vaporise $\ce{H2O}$, it explains why some of the product would be gas and some liquid.
This is a complete guess, am I hilariously wrong here?


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How might one reduce the concentration of bubbles in their ice, from ice cube trays?
So I have been freezing ice a lot recently, and I will come out and say it, as it turns out it is, albeit unlikely, it is sometimes possible for ice to "explode." I was not so aware of this before, but from now on I consume my ice beverages with deliberate care.
The first action I have taken is to make sure no bubbles are visibly present before freezing.
I was thinking about how one might reduce these frozen bubbles when making ice with ice trays and a freezer (the bubbles are sometimes a few millimeters across). I was thinking perhaps the bubbles are more likely to form if there are more dissolved minerals and other compounds commonly found in tap water. I think it is time to review osmosis to see if anything will shed light on this idea. My first instinct was to distill water and freeze it to see if there would be less bubbles, but since there will always be dissolved gases even in a freezer it does not follow that this will greatly affect anything.
Is it the case that at lower temperatures water will not be able to hold as many dissolved gases? If that is the case, then it might be worthwhile to lower the temperature of ice in the fridge or to a temperature just above freezing and allow the solution to stabilize, and then put it in the freezer.
Your insight is appreciated, thank you.

IIRC, the solubility of virtually all gases in water decreases with temperature. This is the reason why heating water will result in bubble nucleation, and this can be exploited to produce very clear ice by boiling water, perhaps several times, before putting it into the trays.

To minimize the air bubbles in ice, either 1) hold the water under a vacuum or 2) 'degas' the water with an ultrasonic cleaner before freezing.
1) Holding the water under a vacuum will pull the dissolved gases out of solution.  The will diffuse to the air-water surface, and then be vacuumed away.
2) An ultrasonic cleaner will vibrate the liquid at ultrasonic frequencies.  Vibrating water-gas mixtures at the right frequency will drive out dissolved gases in much the same way that an earthquake will drive ground water out of water-saturated soil (ie liquifaction).  Many such ultrasonic cleaners even have a 'degas' setting for just such a purpose.
Note also:
* Water will freeze as soon as it dips below 0 degrees C (assuming it's pure water, it's not being agitated, and it's at standard atmospheric pressure).  So, cooling the ice at a lower temperature freezer will make the ice form quicker.  Thermodynamically then, it's creating a greater temperature difference between the air-water interface and the center of the water, so the heat flow out of the ice will be greater, and water will reach 0 degrees C sooner.
* Gases do not immediately dissolve into water.  Assuming you had water with absolutely no dissolved gas it takes some time for the gas atoms/molecules at the air-water interface to dissolve into the center of the water.  -- So I guess this means, assuming you degas the water, forming the ice in a lower temperature freezer will help, too.


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Do metallic bonds contain London dispersion forces?
On our chemistry exam, a question asks "Which force is present between Fe particles? a) Van der Waals b) Metallic"
At first it seemed obvious that it was metallic; however, upon closer reading, it seemed like a trick question. First of all, there's no such thing as a metallic force. In addition, I'm not sure if Van der Waals forces exist for metallic bonds. If they do exist, then the answer was definitely "Van der Waals".
So my question is, would Van der Waals/London dispersion forces exist between particles of iron?

If by "Fe particles" you mean distinct particles (as opposed to the atoms inside a continuous Fe crystal structure), then the answer would be VdW/London forces.
From wikipedia:
This is the only attractive intermolecular force present between neutral atoms

Which, in a way, can be applied to particles or particle surfaces as well, as long as they're not charged.
Whereas, if you are talking about a continuous iron crystal structure, and by particles you mean atoms within that structure, then you have a metal-type bond between them caused by delocalization of electrons into the conductive band. See wikipedia for further details.


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How to compare viscosity of alcohol to water?
Are there any simple at-home experiments to test the relative viscosity of alcohol to water?
If they are the same or practically the same, are there any household liquids that are less viscous than water? If so, how can I demonstrate this to first-graders?

There aren't many common liquids much less viscous than water.

A common experiment would be to have two concentric cylinders. The outer one can rotate almost freely (with some slight and constant retardation/breaking applied), while the inner one is driven by a motor, like so. Between them is a layer of the liquid that you wish to determine the viscosity of. The speed of the outer cylinder is a measure for the viscosity.
However, since you want to explain this to first-graders, you might wanna start with something more obvious than the difference in viscosity between ethanol and water. How about the difference in viscosity between water and honey? As a demonstration experiment, you could use a funnel in which you pour an equal amount of either liquid, and then measure the time it takes until it has passed through the funnel. This should be more tangible and easier to understand. You could also prepare beakers with water, honey and maybe peanut butter, and have your students run a spatula through all of them. That way they can experience the difference in force that is necessary to move the spatula first hand.


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Software for chemical reaction simulation in ICE
I want to study what happens in a internal combustion engine in which fossil fuels are mixed along with other chemical additives.
Example:
Acetone drastically reduces surface tension.
Most fuel molecules are sluggish when it comes to their natural frequency. But acetone has inherent molecular vibration that "stirs up" the fuel molecules, to break the surface tension.
This results in a more complete vaporization with the various other factors remaining the same. More complete vaporization means less wasted fuel, hence the increased gas mileage from the increased thermal efficiency.
Along with this I also want to study the emission of $\ce{CO2,\  O2\,\ H2O}$ and more.
Is it possible to carry out the above phenomena/process in a virtual simulator ?

""Acetone drastically reduces surface tension.
Most fuel molecules are sluggish when it comes to their natural frequency. But acetone has inherent molecular vibration that "stirs up" the fuel molecules, to break the surface tension. "" 1st acetone does not reduce surface tension "drastically". And the "theory" You give with those natural frequencies is plain nonsense.

I am not sure about the phenomena you are describing but simulators are just tools with built-in models and thermodynamic data together.
What you need is, for example, to find some kinetic (combustion kinetic) that it has in its equation a parameter regarding the acetone concentration.
Try to find at the literature some of these kinetic equations, if you are student you can access with your university credentials into sciencedirect.com papers.
If not, check arxiv.org or google ;)
Once you have these, then you can use simulators to reproduce your "special" kinetic.
Another possibility is to describe better your system behaviour (including graphs) and try to fit them suing a correlation or a polynomial fit. Or ask the mathematicians, they could recognise models just seeing the patterns and give you an equation type that you can use.
Check this paper as an example;
http://www.combustion.org.uk/ECM_2007/ecm2007_papers/1-20.pdf


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Chemical diagram for cummingtonite?
Any suggestions on diagramming the chemical structure for cummingtonite?
$$\ce{(Mg,Fe)_7 Si_8 O_{22} (OH)_2}$$
As I understand it, the $\ce{(Mg,Fe)_7}$ says the molecule has seven atoms that could be either Magnesium or Iron, with probably more Magnesium than Iron, meaning the structure isn't precisely defined, i.e. a mineral not a specific molecule.  Is there a suitable way to diagram this that looks nice but not excessively complex?
In fact, I'm rather ignorant about chemistry, but started screen printing subtlety goofy neck ties.  I'm envisioning doing essentially this but vertically and adding a chemical structure diagram.  I'll do the final pattern in latex if possible.
p.s.  If I were making ties professionally, maybe I'd diagram an assortment of different cummingtonite molecules as a background for the formula.  I'm not doing that, way beyond my screen printing ability.  Just curious though, how does one know the probability distribution for the Magnesium and Iron?  Just varies by sample perhaps?

Crystalline structures of inorganic materials aren't usually represented by diagrams, but rather by 3D renderings. First, you need to find a crystallographic structure file, which are usually in CIF format. A Google search leads to this one for cummingtonite.
Then, you need some molecular visualization software. Here, I use CrystalMaker, which can create nice photorealistic 3D images, and has a simple tool to record animations. The result is here:

($\ce{Fe}$ is pink, $\ce{Mg}$ is yellow, $\ce{Si}$ is blue; both $\ce{Mg}$ and $\ce{Si}$ are at the center of the tetrahedra, and oxygen atoms correspond to all the tetrahedra edges)

Alright thanks!  I'll play around with these tools.  So there aren't any standard diagrams that represent the same information more efficiently?

@JeffBurdges what do you call more efficiently? If you're interested, open the CIF file in a text editor and see how it goes: list of symmetry operators and atomic positions of the asymmetric cell


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How to find the formula of this alkane?
It says that the mass of Carbon and Hydrogen are 8:45. We are talking about an alkane here.
General formula is CnH2n+2. What is the particular formula?

I presume that you mean the ratio by mass of hydrogen to carbon is 8:45. Depending on how many significant figures you select for your carbon and hydrogen masses, as well as the tolerable degree of error in your calculation, you may have several reasonable answers. If we take the mass of one mole of carbon to be 12g, and one mole of hydrogen to be 1g, then the following equality should hold:
$$
12n/(2n+2) = 45/8
$$
This follows naturally from the general formula for alkanes. Solving the equation, we find $n = 15$, yielding the chemical formula C15H32.


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How do you Benzylate (add a benzene group to) cyclic compounds?
How do you add a benzene group to cyclic compounds like piperidine, piperazine and phenyl derivatives?
Piperidine

Piperazine

You could try halogenation on it (to create partial positive on some carbon ($\large\ce{\delta}$ $^+$)) and then add $\ce{Phenyl\ Magnesium\ Halide\  (Ph-Mg-X)}$(Grignard's reagent) to subtitute halide group on ring nucleophilically with Benzene, although that would be difficult/not feasible to carry out in phenyl derivatives.

I'd just like to point out that benzylpiperazine is a schedule 9 (banned) substance in Australia and whilst good faith is assumed (nothing wrong with academic interest), people might be loath to address such a specific reaction.

Your good faith is right I have no intention in synthesising such a compound. This is such a simple curiousity

Do you mean a compound like this, or this or this ot this, out of piperidine.

(1) & (3), no, that's a phenylpiperidine, (2), compounds like, yes, but not (2) because that's 4-benzylpiperidine -- a stimulant and likely addictive drug, (4), yes.

Well, (4) is pretty easy to make, you take Benzyl-halide along with a Lewis Acid(like $\ce{AlCl3}$ or $\ce{FeCl3}$) and since Nitogen is a nucleophile it will easily attack on piperidine on the generated carbocation. By the way (4) is also a benzyl derivative not phenyl derivative

Brenton, the term "benzene group" is a bit misleading; it seems that you'd like to add a benzyl group, i.e. $\ce{C6H5-CH2}$.
Typically, these reaction are carried out using freshly distilled benzyl halides. In the case of the diamines, monoalkylation is usually achieved using the bis-hydrochloride of the amine and a stochiometric amount of base, thus keeping one nitrogen atom protected.
Note however, that the Australian regulations, kindly mentioned by Richard Terrett, are likewise true in the EC: 1-benzylpiperazine, an infamous psychoactive party drug, is a controlled substance!


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Is it possible to crystallize a solid ingot of metal from an (acidic) aquaous solution?
To my knowledge, metalic ions only precipitate from a solution as salts & and a process that could possibly produce monocristalline metal would be very interesting indutrially - I've never heard of one, so I don't think it is possible.
But: is it not conceivable to dissolve a metal in an acid (or maybe base), and then (later) neutralize the acid, without resulting ions that form a salt with the metal?
What am I missing?
As an aside, I arrived at this question while thinking if there could be a biological, and thus wet-chemical, pathway from ores to pure metals (in metallic form). The last step in a such a process would be the precipitation/crystallization as a metal.

The easiest way to produce big crystals of metal from water solution is reactions like this
$\ce{Fe_{(s)} + CuSO_4_{(aq)}\to Cu_{(s)} + FeSO_4_{(aq)}}$
performed at very low concentration of the salt. The lesser know method is usage of reducer, like hydrazine
$\ce{CuSO_4_{(aq)} + N_2H_4_{(aq)} \to N_2_{(g)} + H_2_{(g)} + H_2SO_4_{(aq)} + Cu_{(s)}}$
This kind of reactions usually produces fine powder (when I tried it, the resulting powder oxidezed with air in less then half of hour), but it is possible to produce visible crystals this way, working in gel with reagent transfer by diffusion.
For industrially valuable path from ores to metals it not absolutely necessary to get metal directly as result of the process. For example, iron oxalate when heated slightly decomposes, releasing metallic iron
$\ce{FeC_2O_4_{(s)} \to Fe_{(s}) + 2 CO_2_{(g)}}$
The process gives very fine powder, like in previous example.

thx - just to get this straight: getting pure metal is not a very big challenge, but crystal growth in metals works such that the crystals are very small/it's very hard to grow large crystals and the binding that happens between monocrystallines in a macroscopic lump of iron does no arise  in a situation as described here?

@martin Well, the first path easily gives crystals 1-5 mm in size. Crystals with size 1-10 cm may be obtained, if molted metal is cooled slowly. Sure, they are tied together in metal slab, but they exists. Actually, many processes are focused on making this crystals smaller.

This reaction isn't too far away from biology. Hydrazine can be biosynthesized in small amounts https://www.researchgate.net/publication/226213418_Improved_nitrogen_removal_by_application_of_new_nitrogen-cycle_bacteria. So maybe somewhere a microbe is making it's own reduced (metallic) copper nanoparticals for whatever reason.

I think you've just described the general concept of Electrowinning?
There are a large number of ways to recover metals from solutions or molten salts.
Producing monocrystalline metals is a different matter - crystal grains in metals tend to be microscopic, but there are techniques that permit growing large crystals of metals such as the Czochralski process or zone melting. Monocrystalline metals sometimes have attractive mechanical properties not achievable in polycrystalline metals - for example, Callister Materials Science and Engineering 6e p. 225 has a nice photo of a single-crystal jet turbine blade that is highly resistant to creep failure with respect to conventional blades.