text:
Addition of Br2 to Double Bond and Carboxylic Acid

In the above reaction, I remember reading that Br2 in CCl4 develops a partial polarity and adds on to the double bond, resulting in addition of Br on 2 adjacent carbons.
But, as per the text, the end product is a ester. (As given below)

Why is this the final product, and how to judge the products in cases like this?
Note - If I look at the answer and work backwards I would say that first Br+ attacks, and a carbocation is formed, on which hydroxy group attacks to form the final product. Is this right? And of it is then, why is this happening?

Under basic conditions I would definitely say that this is the product. Under neutral conditions I'm not so sure.

Thanks, @bon, and is the reaction mechanism that I proposed correct?

Yes it is essentially right. The intermediate will be a cyclic bromonium ion.

Just google iodolactonisation. It's basically that but with bromine rather than iodine. Ideally you'd want base involved to deprotonate the carboxylic acid

The first intermediate is indeed Br+ atom attached to the double bond forming a cycle (see step one):

 On the next step, however, oxygen of carboxylic acid attacks the intermediate and forms the product you show on second figure. This happens because carboxylic acid group is in the right position for intramolecular reaction.
First step follows the diagram. In second step the attacking nucleopline is not Br$^-$, but R-COOH or RCOO$^-$ depending on the reaction condition. To obtain high yield of the cyclic product on your second figure you need to slowly add dilute solution of Br$_2$ to a solution of organic reagent. If you add Organic reagent to neat (pure) Br2, then you get a significant amount of the product you predicted on figure 1. Don't do it in real life because a reaction of neat bromine with alkenes is violent and dangerous.

Reaction of bromine with olefins is in no wah dangerous— its a routine reaction. Careful addition and control of temperatire is all that's required

You also need to think about the frontier orbitals or Baldwin's rules and why you'll get the five membered lactone ring rather than a six membered one - bearing in mind that the nucleophilic oxygen on the carboxy group can attack two possible carbon atoms on the bromonium ion. 5 exo tet vs 6 endo tet. Or if you draw a chair confirmation you should be able to see that the anti-bonding sigma orbital on the carbon atom that would make a six membered ring is pointing at a tricky/inaccesible angle compared to the one on the 5 membered ring.


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Is OH the same as O-H?
When 2-hydroxylpropanitrite undergoes hydrolysis by H20, it will become 2-hydroxylpropanoic acid. In the structural formula, I wrote the O-H that are connected to the carbon with double bond as OH. Is it still the same thing or it is incorrect?

$\ce{-OH}$ and $\ce{-O-H}$ are the same.
The single bond between oxygen and hydrogen is implied in the first representation and this representation is the most commonly used because of it's brevity.

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Orbitals SPDF. Why they named like that?
I know there are orbitals named S, P, D and F.

But why they named like that. is there any significance for the letter?
Is there any order in filling the orbitals?

In my old memory they were ; Sharp , Pronounced, Diffuse, and Foggy based on the film images.

From the abstract of Structure of the Line Spectra of the Elements as published in the 1890 Journal of the Chemical Society.

There are three kinds of series — principal, sharp (well-defined), and diffuse (ill- defined). The principal series form the most vivid lines in the spectra, and only occur in the first, periodic group ; next come the diffuse (really double) lines ...

See also On the Structure of the Line Spectra of the Chemical Elements, also from 1890 by Rydberg:

Liveing and Dewar were the first to remark the existence of the series, as well as their different appearances. They are of three kinds : diffuse, sharp, and principal series. The first two are formed by the above-mentioned doublets or triplets ...

Later, "fundamental" was added.
For more information see The Origin of the s, p, d, f Orbital Labels J. Chem. Educ., 2007, 84, 757-758

Can you please add an image of how the different spectra actually look like?

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why does Zn become oxidised in the first place in operation of voltaic cell?
I am so confused that during the operation of a voltaic cell, why does $\ce{Zn}$ gets oxidized? In anode side, there is a zinc metal in $\ce{ZnSO4}$ solution, but why does the $\ce{Zn}$ lost 2 electrons?

I think you need to clarify this question further. Which voltaic cell are you talking about? The reason Zn oxidizes is very coupled to what is reducing on the other side. The reaction is not producing free electrons, those electrons are reducing something else.

A voltaic cell is an electrochemical cell that uses a chemical reaction to produce electrical energy. In anode compartment you have Zn in contact with ZnSO4. In cathode cell you have two other reagents. For example Cu and CuSO4. activity series of the metals suggest that zinc is OK to lose electrons and turn into zinc ion. Copper ions from CuSO4 are strongly willing to turn into reduced copper metal. So zinc gives away electrons to copper ions because Cu$^{2+}$ has higher affinity to electrons than Zn$^{2+}$.
Why Cu$^{2+}$ has higher affinity to electrons than Zn$^{2+}$ is a separate question and it belongs to a different field (quantum mechanics/molecular orbital theory rather than electrochemistry).
In short - there is Zn metal that holds electrons weakly and Cu$^{2+}$ that will hold electrons strongly when it forms Cu metal. Hence, electrons flow to Cu$^{2+}$ from Zn.

Simple one line answer: Because $\ce{Cu^2+}$ is more stable than $\ce{Zn^2+}$.
And by stable I mean it has less energy. Why it has less energy depends on activity of atom, size, and number of electrons. So $\ce{Zn}$ loses its electrons to give $\ce{Cu^2+}$ and then the system loses energy and becomes more stable. That is it. Hope it helps.

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Which hydrogen of 3-methylcyclohex-2-en-1-one is the least acidic?

Which hydrogen of 3-methylcyclohex-2-en-1-one is the least acidic?

According to me it should be 2, as on its removal there won't be any resonance stabilization as would happen in the other cases. In addition to that a negative charge residing on a carbon attached to a double bond would be highly unstable.
But the answer key says that it is 3.
A follow up question: Which is the most acidic hydrogen?
According to me it should be 4 as it leads to a greater number of resonance structures.

For your followup - how many resonance structures can you count for 4? What about 1?

Also, see http://chemistry.stackexchange.com/questions/26610/which-is-more-stable-a-carbanion-on-benzene-ring-or-a-carbanion-on-a-vinylic.

three resonating structures for 4  and two for 1

Did they ask something like "which of sp3 carbons have the least acidic hydrogen?" Or did they ask "which of these four hydrogens is the least acidic?" 2 indeed looks least acidic among all, because it is the only anion that doesn't have appropriate resonance structures.

You are correct. Position 2 is the least acidic.
Position 3 is not the least acidic hydrogen. It might be the most acidic hydrogen. Position 2 is the least acidic hydrogen of the four marked.
What does position 2 have going for it? Why might it not have been marked as the least acidic? The lone pair that would be left in the conjugate base is sitting in a sp² orbital.

All else equal, sp² hybrid orbital containing $\ce{C-H}$ bonds are more acidic than sp³ hybrid orbital containing $\ce{C-H}$ bonds. The $\mathrm{p}K_\mathrm{a}$ of ethane $\ce{CH3CH3}$ is reported to be $\sim 50$, and the $\mathrm{p}K_\mathrm{a}$ of ethene $\ce{CH2=CH2}$ is reported to be $\sim 44$.
However, all else is not equal. Resonance, especially with a carbonyl group, stabilizes conjugate bases more. Acetone $\ce{CH2COCH3}$ has a $\mathrm{p}K_\mathrm{a}$ of $\sim 19$. Position 2 is not stabilized by resonance. The conjugate bases from positions 1, 3, and 4 are all resonance stabilized through the carbonyl group.
Position 1 has two resonance structures.

Position 3 has three resonance structures.

Position 4 has three resonance structures.

It is between positions 3 and 4 for the most acidic. I chose position three because alkyl groups (the rest of the ring) are electron donating by induction, but alkyl group also stabilize pi bonds, so position 4 is a good candidate for most acidic as well.

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Range of distance for Van der Waals force
Is there any range of distance between the nuclei of the atoms (in Angstrom, say between 4 to 12 Angstrom) within which there will be an occurrence of Van der Waals force (attraction) between them?
For example, say within a distance of 4 A between two atoms of same type, there will van der Waals attraction present. Can such a certainty be obtained? If two atoms are very far apart, then there will be not Van der Waals attraction between them, this is known. So my question is how close should the nuclei of two atoms be so that one can say with certainty that there will be Van der Waals attraction?

Van der Waals force is due to interaction of two dipoles. It is never strictly zero, but it diminish very fast with distance.
It is common to look at VdW interaction at Van der Waals radius. The largest radius should be around 2.60 for Fr making the total distance between nuclei more than 4 A. Heavier elements will have larger radius, but heavier elements are not stable. Are we really interested in VdW force between two atoms that will decay in 10^-15 sec?
Also, by "4-12A" do you actually mean distance between nuclei or the surfaces of atoms (VdW radius A + VdW radius B + "4-12 Angstrom")? In the first case iodine has VdW = 2.15A, so nuclei interact on distances more than 4A. This is the distance between atoms of neighboring I2 molecules in the crystal. In the second case you will see some interaction, but it will be quite small.

I meant nuclei of atom .... If it is surface of atom, how does it change

okay so I can understand that each type of atom has their own VdW radius. If two atoms intersect within the radius then there will be force, else no force. Am I correct?


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Is there a structured data source containing information on the World Health Organisation's Stability Testing Policies?
Hoping to automate some product testing guidance software, I've been attempting to find a (hopefully live, up-to-date) Database, Web-service, or other structured data source containing the pharmaceutical Stability Testing policies (zones/countries/conditions) held by the World Health Organization.  But so far, the only format that I can find these policies in is PDF (http://apps.who.int/medicinedocs/documents/s19133en/s19133en.pdf) which is difficult to interpret for my purposes (even more so to automate).
Could anyone confirm the existence or non-existence of a structured data source containing information regarding WHO policy Stability Testing Conditions, Zones and countries within each zone?

I don't know if you have tried, but this sort of question could potentially find an answer on the Biology SE


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Symmetries of the v=0 and v=1 quantum harmonic oscillator wave functions and those of the Cartesian Coordinates (References)
I'm busy with a writing project for my third year Chemistry course and in the project one of the things we need to consider and answer is the symmetries of the v=0 and v=1 quantum harmonic oscillator wave functions and those of the Cartesian Coordinates. This question falls under the suh-heading (in the project) 'Determination of IR-activity of normal modes and the prediction of spectra'.
Now I'm not expecting anyone to give me the answer to this question, but I would greatly appreciate it if you could let me know of websites and/or accessible textbooks that I can use to find this answer. I still have two months to complete the writing project and am essentially done with it, but its just this question that I have been really struggling to answer since I can't seem to find it online or in the textbooks by Atkins and Engel & Reid respectively.
Any help would be appreciated, Thanks in advance.

Check out Molecular Vibrations: The Theory of Infrared and Raman Vibrational Spectra by Wilson, Decius, and Cross. This text is one of the classics and will help get you on your way. Also, it's a Dover series reprint, so even a new copy is quite inexpensive.

In general you need to understand point group tables.
These always have the same form, which is to say that symmetry operations are on the top line, characters (1,-1, 0 etc) in the main body. The most left hand column the Mulliken symmetry labels (also called symmetry species) for each irreproducible representation (one row of characters) .
The right hand two columns, often without header labels contain the operators for different type of transitions. x,y, z are used to dipole transitions because a dipole transforms as x, y or z. (The z direction is usually that of the principal axis). Rx, Ry, Rz are used, for example, spin orbit operators (These are coupling between electron spin and orbital motion, Not rotations).
The last column contains product operators $x^2$, $x^2-y^2$ etc and are generally used to describe Raman transitions since the Raman transition operator depends on a molecules polarisability (a volume) which in projection is proportional to an area hence squared terms.
If you want more complicated operators say $x(x^2-y^3)$ then generally these can be made from simpler operations. When making more complex functions consider the x as a $p_x$ orbital with two lobes of different 'colour'. The subtraction as in $x^2-y^2$ then looks like the $d_{x^2-y^2}$ orbital with opposite lobes coloured in the same way but each pair of a different colour. You can the use the symmetry of this object to determine which irreproducible representation it belongs to. In $C_{2v}$ (below) it is $A_1$.
The C$_{2v}$ point group shows these features. The operators are E, $C_2$ etc,. The Mulliken symbols for the symmetry species, in your case vibrational normal mode symmetries, are $A_1$, $A_2$. The characters are 1, -1 etc.
A totally symmetric normal mode $A_1$ has characters 1,1,1,1, is active to transitions with a dipole on the z direction and is also Raman active, as it has at least one squared term, x$^2$ for example. The $A_2$ symmetry species in not ir active, but is Raman active, due to the xy operator. You can follow the rest.

If you want to work out what the symmetries of a wavefunction are in a given point group, then the easiest way is to sketch the wavefunction and apply the symmetry operations from that point group. Choose a set of axes, z vertical and x horizontal and y out of the page, perpendicular to the other two. Count 1 if the symmetry operation is present and -1 if not.
For example the $v$=0 wavefunction for the harmonic oscillator has a gaussian or bell shape. By applying in turn the symmetry operations in C$_{2v}$ this wavefunction belongs to the $A_1$ symmetry species. The v=1 wavefunction has a node at the origin with an inversion in the values between negative and positive x. In this case C$_2$ is -1, $\sigma _{xy} = 1$ and $\sigma _{yz} = -1$ which is the $B_1$ symmetry species.

Awesome, thank you very much I've got it now.


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How to distinguish between CsCl and NH4Cl by a chemical test?
$\ce{CsCl}$ and $\ce{NH4Cl}$ have similar reactions. For example, they can react with $\ce{AgNO3}$ to produce $\ce{AgCl}$ precipitate.
So my question is:

Can we use a chemical test to distinguish between solid $\ce{CsCl}$ and $\ce{NH4Cl}$?

The $\ce{Cs+}$ ion is very unreactive and I do not know any reactions involving it. I believe the chemical test can somehow convert the $\ce{NH4+}$ ion into $\ce{NH3}$ gas so that we can test it with a moist red litmus paper.

Not a chemical method, but CsCl would give a positive flame test indicative of the presence of caesium. NH4Cl would not

Throw some stuff on a hot pan. $\ce{NH4Cl}$ would evaporate, the other would not.

Correct you can distinguish them by their reaction with a base. You should use a base, such as $\ce{NaOH}$. $\ce{Cs^+}$ wouldn't react with $\ce{NaOH}$, while $\ce{NH4+}$ will undergo reaction $$\ce{NH4+ + OH- -> NH3 + H2O}$$ $\ce{NH3}$ is a gas and can be detected with a moist litmus paper.
If you will, you can actually detect $\ce{Cs+}$ using a procedure from this paper:
"Selective Detection of Cs+ in Water Solutions via One-Step Formation of a New Type of Struvite-Like Phosphate". However, this is an exotic reaction (in practice you would use the flame color test) and you don't have to do it since you only need to distinguish samples.

http://meta.chemistry.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here

@Mithoron fixed it.

Yeah, I improved your post with mhchem later, it looks better. You can also make links and post pictures. See here  or in help.

If you observe the two salt, you can see that the cations are different. So, you will have  to differentiate the cations. There are two test that can differentiate the cations.

Flame test (best differentiating test)

$\ce{Cs+}$ gives blue-Violet color while  $\ce{NH4+}$ don't give flame test. For more information about flame test, see  Jan's answer to this question. It also gives some information about $\ce{NH4+}$ detection.

Specific test (can be cumbersome)

Nessler's reagent test for $\ce{NH4+}$ (from same question). For $\ce{Cs+}$, there is no specific test for its detection and only rely on flame test. But, if it is said to separate $\ce{Cs+}$ from other alkali metal ions, it is separated by  noting the difference of solubility of its platinochlorides, acid tartrates and alum (source) or estimated gravimetrically (source).

Just dissolve some of the stuff in water and apply blue litmus paper or pH paper.  Ammonium chloride is weakly acidic.  Cesium chloride is not.
Similarly magnesium hydroxide will be readily soluble in an ammonium chloride solution which is acidic, but not in a cesium chloride solution.

But this is not really a chemical test.

Why, of course it is.

@IvanNeretin See the edit history.


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Why does water react with H+?
As we all know, if we put $\ce{H^+}$ ions in water then water reacts with them and produces $\ce{H_3O^+}$:
$$\ce{H_2O + H^+ <=> H_3O^+}$$
My question is, why does water react with $\ce{H^+}$ in the first place? It doesn't have an electron, then how can the reaction occur?
And again, it's already stable and adding one more $\ce{H^+}$ will, in my understanding, destabilize it as it will have one more proton in the molecule.

Do you mean "react"?

There's no free H+ (bare proton!), generated via dissotiation, in water or in any solution. Or maybe you're interested in nuclear reactions?

Water has two lone pairs. H$^+$ is a naked nucleus with a positive charge. It is extremely reactive and would bind to any neutral molecule. Water has two lone pairs of electrons and H$^+$ binds one of them to form H$_3$O$^+$.
In absence of water it would bind other things. For example, it would bind methane to form unstable CH$_5^+$ that breaks down to CH$_3^+$ + H$_2$.

so the extra proton forces (in lose terms) the water molecule to make bonds? Does the water molecules get forced to become unstable?

So the lone proton doesn't need an electron to form the bond? It can just get attracted towards the molecule and then stick to it?

Yes. Strictly speaking H+ is a very high energy species. Binding to anything brings its energy down, so it is ready to bind whatever. Now there are molecules that can bind H+ at lower energy cost (like water, or, even more so NH3). For other molecules it is more expensive. But the net energy change for H+ +X  ->HX+ is favoring the products. Once again it is because H+ is so high in energy.

@Adi, If you just look at the water molecule, it seems like it's a "bad deal". The thing is you have to look at the whole system. As much as water is getting more unstable at the $\ce{H3O+}$ molecule, the high energy $\ce{H+}$ is getting stabilized. The final energy in the system is going to be lower than the initial energy (also the reason why this reaction is exothermic, this energy has to go somewhere..).

Just how is $\ce{H3O+}$ unstable? Sure, it would react with certain species, but the same can be said about any molecule or ion.

@IvanNeretin The accuracy of my answer depends on the way the question was asked. Correct explanation would involve Gibbs energy for H$^+$, H2O and H3O$^+$. But that wouldn't really explain anything. I mean, for someone new to chemistry it is just numbers. The matter of fact is H$^+$ reacts  with any neutral molecule (we can discuss HeH$^+$ elsewhere). It even can react with cathions (N2H5$^+$+ H$^+$ -> N2H6$^{2+}$) but there are many neutral molecules and even ions that water wouldn't react with.

@sixtytrees I don't argue with your answer at all. True, naked proton is extremely reactive, to the point of forming $\ce{HeH+}$ (which I myself remember mentioning elsewhere). It is $\ce{H3O+}$ I was talking about. Also, "reactive" is not the same as "unstable".

I agree. H$^+$ is both reactive and stable. Instability refers to ability of a compound to undergo transformation on it's own.

The simple answer is that  H$_3$O$^+$ is more stable that H$^+$+ H$_2$O. The reason for this is that the H$_3$O$^+$ ion is isoelectronic with ammonia and has the same type of geometry, a triangular based pyramid (trigonal pyramid). It is possible to crystallise some salts with H$_3$O$^+$ as a counter ion, but of course these are no longer in water.
Neutral water normally contains a small concentration of ions, 10$^{-7}$ M H$^+$ and OH$^-$ at pH = 7 due to the dissociation/ionisation  2H$_2$O=H$_3$O$^+$+ OH$^-$.  As water is a highly hydrogen bonded liquid there are always H atoms between any two oxygen atoms, and any oxygen has three hydrogens surrounding it, two normal bonds and one Hydrogen bond. Any positive charge now added will form the H$_3$O$^+$ and this has now to be incorporated into the H bonding network. Thus in bulk water H$_3$O$^+$ is probably not a compound in the normal sense of the word because there will only be be a small energy barrier for the charge to move around along the network of hydrogen bonding (all water molecules are the same) as the water molecules jostle due to their thermal energy, i.e. Hydrogen bonds making and breaking. Alternatively you can think of this charge migrating effect  as the charge being spread over more than one water molecule, thus stabilising it. It is possible therefore to imagine complexes of various numbers of waters with a single charge, such as H$_5$O$_2^+$ .

In chemistry course, we are taught at the beginning that a bond is formed between 2 atoms, each of which providing 1 electron to make the bond. But this is more a way to introduce the octet rule and shape of the molecules. In most reactions (as we write them) electrons move by pair.
If a bond is made of 2 electrons, it does not have to be 1+1 (one electron from each atom) but it can be 2+0 (one atom brings 2 electrons, the other one does not).
In the case of $H^+$, it has no electron where it should have 2 when bonded. This empty "slot" is really eager to be filled (octet rule) so it "steals" electrons from water which has 2 big non-bonding pairs of electrons on the oxygen atom. The result is that the positive charge on $H_3 O^+$ is shared by all 3 hydrogen atoms. This dilution of charge is greatly favorable.

The electron configuration of the hydrogens are approximately
$$\ce{\underset{1s}{[\uparrow]}}$$
although some of the charge is pulled off into the oxygen.
The electron configuration of a neutral oxygen atom is
$$[\ce{He}] \underset{2s}{[\uparrow \downarrow]} \underset{2p}{[\uparrow \downarrow \vert \uparrow \vert \uparrow]}$$
In $\ce{H2O}$ each free electron in a hydrogen pairs up with a free electron in an oxygen (approximately.) So how can a $\ce{H3O^+}$ ion form?
In $\ce{H3O^+}$ there is a missing electron so the oxygen could have an electron configuration of
$$[\ce{He}] \ce{\underset{2s}{[\uparrow \downarrow]}} \underset{2p}{[\uparrow \vert \uparrow \vert \uparrow]}$$
which allows for a hydronium ion to form.
Now in an actual hydronium ion the oxygen would have hybridized orbitals of the form
$$[\ce{He}] \underset{(2s)(2p)^3}{[\uparrow \downarrow \vert \uparrow \vert \uparrow \vert \uparrow]}$$
which is a tetrahedral arrangement.
Now this is only an approximation. A more in depth look would go deeper into molecular orbital theory and bond symmetries but that gets annoying for polyatomic molecules.

Also, you are right in saying that hydronium ion is unstable than water. But a system containing hydronium ion is thermodynamically more stable (has lower gibbs energy) system containing a proton and water molecule. The following exposition is based on the subjective knowledge, will be glad to be corrected if I am wrong. This due to the following arguments:
1) number of possible configurations of system increased. In the two particle system, charge was only on proton, which is now shared between several hydrogen atoms bonded to oxygen. but one would argue that number of molecules decreased, but entropy is not only about number of molecules, but it is about number of possible configuration of a system for different parameters
There must be entropy increase due to increased number of ways a charge can distributed, but I am not able to explain how to count this number of ways a charge can be distributed, Please let me know, if anybody understands where and how this entropy increase is coming from.


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Free energy change in collagen hydrolysis
In now-deleted comments on this answer, a user with a self-proclaimed chemical engineering background claimed that the reactions that take place when slow-cooking meat are endothermic and require far more energy than actually heating the meat. My intuition says this can't be - the meat would rapidly cool itself down if removed from the heat source, the sous vide circulator would have to work hard the whole time not just during initial heating, the bonds in the collagen are presumably not terribly strong, and, well, heating mostly-water things takes a lot of energy! But I don't want to assume too much.
I believe the primary interesting reaction here is collagen conversion, hydrolizing it into gelatin, and the collagen is something like a few percent of the meat by mass. (When cooking something like this, the rapid reactions have already taken place: the meat looks cooked, and it gets slowly more tender over time as the collagen breaks down.)
So, how much energy does that actually require?

tldr: reaction is exothermic because the heat of reaction is -2.5 Kcal/mol.
(1) Peptides do hydrolize in presence of enzymes. It means that Gibbs energy of hydrolysis is negative. However, Gibbs energy consists of $\Delta$ H and $\Delta$ S*T. $\Delta$ S of hydrolysis is favorable (more disorder), but $\Delta$H can be either lower or higher.  $\Delta$ H is what we are discussing here (the enthalpy of the reaction).
(2) Some endothermic reactions are really very endothermic (like dissolving NH4NO3 in water). Other reactions are mildly endothermic.  Hydrolysis of meat doesn't have a large $\Delta$ H change. "HEATS OF HYDROLYSIS OF AMIDE AND PEPTIDE BONDS" paper has exact value for the reaction: -2.5 Kcal/mol of peptide.  A negative value indicates the reactants have greater enthalpy, or that it is an exothermic reaction. So the reaction is slightly exothermic.

Awesome! If the amino acids are 110 grams/mol and you're breaking all the bonds, that's 23 cal/gram of collagen, and if the meat is 5% collagen, that's only 1.1 cal/gram. So in addition to it being exothermic, it's also on a similar scale to heating 1°C, pretty small compared to the 50-100°C change to go from refrigerator to cooking temperature.

2.5 Kcal/mol is roughly the energy of a hydrogen bond. So, not much, you are correct. Meat cooks faster in boiling water mostly due to Arrhenius law.


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Evaporation Rate of Water
In AP Chem we did a lab investigating the relationship between vapor pressure and water temperature. The relationship is linear and as the vapor pressure approaches atmospheric pressure, it begins to boil. Obviously warm water will evaporate faster than cold water, so it made me wonder how do you calculate the evaporation rate of water? Is there a relation between the evaporation rate and temperature just like there is a relation of temperature to vapor pressure? Of course the surface area needs to be taken into account, but I'm more focused on the relationship between temperature and the rate of evaporation.

Things get complicated. E.g. the rate might be zero if the gaseous phase is saturated with water. Then lowering the temperature will let condense water, not evaporate.

It gets complicated. A drop of water (30 mg) will evaporate completely in several hours under standard conditions. You need to consider these factors:
(1) Temperature
(2) Relative humidity
(3) Wind
(4) Surface area
(5) Thermal flux to water (container walls -> water heat flux)
By all means you can do it, but it is a question for physics forums. The main idea would be to write to equations:
(a) Heat flux +walls_to_water + air_to_water -heat_of_evaporation = 0
(b) Diffusion balance. Water pressure at its surface, vs water pressure at the top of the flask is governed by diffusion laws.
(c) Material balance: amount_of_water(t)=amount_of_water(t$_o$)-evaporated_water.
Those are differential equations, so it would be hard to work with them without a solid understanding of calculus.


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Why does dissolving a salt in water get a high pH?
I dissolved some sodium carbonate in water and I measured the $\mathrm{pH}$. It turned out to be 11. I don't really understand why dissolving it in water increased the $\mathrm{pH}$. I mean, I know that $\mathrm{pH}$ of water was 7, so it must have something to do with the $\ce{H+}$ ions. We just covered acid and bases today so please bear with me.
Here are my thoughts. My salt had some $\ce{OH-}$ salt and this combined with the $\ce{H+}$ to produce $\ce{H2O}$ thereby decreasing the amount of $\ce{H+}$ ion. Or maybe one of my salt ions combine with the dissolved $\ce{H+}$, again thereby decreasing the $\ce{H+}$ concentration and increasing the $\mathrm{pH}$, so I'd have something like this could have something like:
$$\ce{Na+ + H+ -> NaH2+}$$
or maybe
$$\ce{CO3^2- + H+ -> HCO3-}$$
Which one do you think happened?

Last one is OK.

Could you tell me why? Is it just more likely?

This salt doesn't have any OH- and protonation of Na+ is nonsense in any solution.

It's a little more complicated than that.. When you dissolved $\ce{Na2CO3}$ in water you had a solution of ions: $\ce{Na+}$, $\ce{CO3^2-}$, $\ce{H3O+}$ and $\ce{OH-}$ (the later are present in water even when there is no salt, since it dissociates too, however they are in equal quantity in pure water so the pH is 7).
Now when you mix all those ions, the equilibrium isn't only the equilibrium of water dissociation ($\ce{H2O + H2O <=> H3O+ + OH-}$). Now you have a whole series of equilibria occurring in the solution simultaneously:
1) Water dissociation:
$\ce{H2O + H2O <=> H3O+ + OH-}$
2) Salt dissociation:
$\ce{Na2CO3 <=> 2Na+ + CO3^2-}$
3) Base equilibrium*:
$\ce{NaOH <=> Na+ + OH-}$
4) Acid equilibrium*:
$\ce{CO3^2- + 2 H3O+ <=> H2CO3 + 2 H2O}$
The salt dissociation doesn't affect the pH directly. We have two equilibria though that do unbalance the proportion of $\ce{H3O+}$ and $\ce{OH-}$ ions, which are the "base equilibrium" and "acid equilibrium". We don't think about them at first sight because we didn't add any $\ce{NaOH}$ or $\ce{H2CO3}$ to the solution, but everything needed to form them is right there.
$\ce{NaOH}$ is a strong base, however $\ce{H2CO3}$ is a weak acid. So it's likely (and you witnessed it in experience) that more $\ce{H3O+}$ will be consumed in comparison with the $\ce{OH-}$, making the solution more alkaline.

Um, could you clarify on what exactly an equilibrium is?

I'm thinking of a simplified way of explaining it. A chemical reaction can have a direct path (of product formation) and a reverse path (where the products react to form the reactants again). Say we have $\ce{A+B <-> C}$. Just as A and B can react and form C, C also can react and form A and B again. But there is a tendency of each of them to happen and one of the variables that usually changes that tendency is the concentration of A, B and C. [continue...]

Say you have 1M of A, 1M of B and 0M of C. The rate of $\ce{A +B -> C}$ will be much higher than the rate of $\ce{C -> A + B}$. But as time passes the concentration of C will increase, while the concentration of A and B will decrease. This will decrese the rate of $\ce{A+B -> C}$ and increase the rate of $\ce{C -> A + B}$, until the rates are the same. When this happens you will have reached the static equilibrium state of the reaction.
In reactions like $\ce{NaOH <-> Na+ + OH-}$ equilibrium tend to the right ($\ce{Na+}$ and $\ce{OH-}$ are likely to be present in higher quantities than NaOH).

But why is a reverse reaction possible? What causes this to happen?

This video makes a good analogy on reverse reaction and equilibrium: https://www.youtube.com/watch?v=dUMmoPdwBy4

Does "disassociating" mean the same as dissolving?

No, those terms do not mean the same, but they're related. If something (the solute) "dissolves", it forms a solution (in the solvent). "Dissociation" on the other hand simply means "breaking up". However, an ionic compound, like salt (solute), will dissociate (break up) upon dissolving in water (solvent), resulting in salty water (solution).

@IanC I have feeling you're confusing OP even more.

@Mithoron Well, I tried to simplify it as I could.. But it's hard to explain the reason behind salts acidity/basicity without mentioning chemical reactions/dissociation equilibrium. I was hoping the youtube video would clarify things a little for him

You have some detailed answers, but the principles are quite simple to grasp.
A 'strong' acid or base is one in which there is complete, i.e. 100% dissociation of the molecule when dissolved in water. Examples are $\ce{HCl}$ which dissociates into $\ce{H+}$ and $\ce{Cl-}$, and NaOH into $\ce{Na+}$  and $\ce{OH-}$. If 1 mole of each is added to pure water the solution remains neutral ($\mathrm{pH = 7}$) because there are equal numbers of $\ce{H+}$ and $\ce{OH-}$.
A 'weak' acid or 'weak' base is one which does not dissociate completely, so, for example, some fraction of $\ce{H2CO3}$ molecules remain in solution and the rest dissociate producing $\ce{H+}$ and $\ce{CO3^2-}$ ions. Because some protons remain bound in the $\ce{H2CO3}$ when 1 mole of strong base is added to 1 mole of the acid, there is an excess of $\ce{OH-}$ and the solution is alkaline. A similar argument applies to a weak base and strong acid.

TL;DR: Because there is an equilibrium between anion of $\ce{CO3^2-}$ and $\ce{HCO3^-}$ that depletes $\ce{H+}$ and that causes more water to dissociate and form free $\ce{OH-}$.

You should start by reading about hydrolysis and buffers (chemistry). Briefly, you have these equilibria:
\begin{align}
\ce{M+ + OH- &<=> MOH}& &\text{(limiting cases: $\ce{M+}$ = $\ce{Na+}$ or $\ce{NH4+}$)} \\
\ce{H+ + A- &<=> HA}& &\text{(limiting cases: $\ce{A-}$ = $\ce{I-}$ or $\ce{CH3COO-}$)} \\
\ce{H+ + OH- &<=> H2O} &
\end{align}
Here $[\ce{Y}]$ means concentration of $\ce{Y}$. Note that $\frac{[\ce{MOH}]}{[\ce{M+}][\ce{OH-}]} = K_\mathrm{b}$, $\frac{[\ce{HA}]}{[\ce{H+}][\ce{A-}]} = K_\mathrm{a}$ and $[\ce{H+}][\ce{OH-}] = 10^{-14}$.
Depending on the $K_\mathrm{a}$ and $K_\mathrm{b}$ you have several situations. Acids and bases are called strong if they favor dissociation on ions and weak if they favor association of ions.
If both your acid and base is strong, then they both dissociate and the released $\ce{H+}$ and $\ce{OH-}$ are recombining to form water. If acid is strong and a base is weak then acid dissociates fully and yield a lot of $\ce{H+}$, but base doesn't dissociate fully and yield less $\ce{OH-}$. As a result of recombination of $\ce{H+}$ and $\ce{OH-}$, you get excess of $\ce{H+}$ and acidic condition. If the acid is weak and the base is strong you have more free $\ce{OH-}$ and basic reaction much by the same logic. If both acid and base are weak $\mathrm{pH}$ depends on relative strengths on acid and base.
In your case acid is weak ($\ce{CO2}$ forms slightly acidic solution in water) and base is strong ($\ce{NaOH}$ is a strong base). Thus, you have a basic solution.
This story is overly simplified because you have an acid with two stages of dissociation:
\begin{align}
\ce{H2CO3 &<=> H+ + HCO3-} \\
\ce{HCO3- &<=> H+ + CO3^2-}
\end{align}
This adds one equation to the system but similar logic applies.


text:
How do technicians treat spilt mercury and bromine?
As we know, mercury and bromine are in liquid state at room conditions and they are hazardous. The technicians should have ways to treat them if they are spilt.

How can we safely treat spilt $\ce{Hg}$ and $\ce{Br}$?

I have thought of using dry ice to cool the mercury down into a solid, which can then be easily removed. But I do not think that this is really practical. Maybe some chemical reactions can help to somehow convert the them into a solid in order to remove them.

Do you mean spilt instead of split?

Split mercury could be dangerous! I'd read an msds on the materials involved before I was using them and work out a plan of action from there.

Any reputable chemical supplier will normally sell a mercury spill kit, and anyone working with mercury these days should have one. (Either a commercially prepared one, or an equivalent one that's assembled from individual supplies.) Generally speaking, they contain some sort of absorbent/adsorbent to pick up mercury, and a chemical which will react with the mercury and keep it from volatilizing. Sulfur is a common choice (recommended by the EPA), as mercury sulfide is pretty darn inert, as well as being very insoluble in water. (Also, "flowers of sulfur" does double duty as an absorbent.)
After the cleanup, the absorbent is collected (by sweeping - vacuuming can spread mercury vapors), packaged securely, and sent to a hazardous waste disposal company.

As elemental bromine is less widely used than elemental mercury, designated spill kits for bromine are less widely available, but the procedure is generally the same: spread a chemical to react with the bromine and render it more inert, then spread an absorbent to contain the liquid. Elemental bromine is rather reactive (an oxidizer), so a number of common reducing agents can be used. A 5% solution of sodium thiosulfate is a common choice (1) (2).
Again, if you're working with elemental bromine, your lab should have a spill response kit already assembled, with sufficient pre-prepared sodium thiosufate and absorbent to handle the amount of bromine you normally deal with.

The previous was for general information purposes only. If you are looking to work with elemental mercury or bromine, consult with your institution's chemical safety officer before purchasing for more information and restrictions which may apply in your area. Your institution likely also has chemical safety and spill response training, which will inform you of current best practices in handling mercury and bromine spills.
As always with accident handling, the most important thing is not to put yourself in greater danger by attempting to fix it. If the size or extent of the spill is such that you're unsure you can handle the spill safely, the best course of action is to evacuate the premises and let the hazardous materials team do the cleanup. (They'll likely use the same techniques of reacting and absorbing the material, but they'll be wearing more extensive personal protective equipment while doing it.)

Mercury has been removed from practically any modern laboratory since ten years, but bromine is absolutely elementary for organic cemistry, every second year student does bromination reactions! But yes, thiosulfate solution. A large, freshly prepared beaker of thiosulfate solution. Never seen a "spill kit".

First, spilled mercury and bromine are emergencies. If this is happening to you - inform colleges and call specialists.
Second, Mg is magnesium, Hg is mercury.
Bromine is very volatile and irritating. It can kill or injure people very fast. If small amount was spilled - leave the open windows, leave the room, close doors, wait. If spilled one someone (this is a real emergency) you should apply 95-100% ethanol to the injured part of the body. This method helps the best.
Hg is chronically toxic (acute toxicity of metallic mercury is low). You can absorb it on iron iron chloride (III) powder to absorb Hg or apply sulfur to the area of the spill.
Mg is relatively safe and can be collected with with a floor mop.

@The worst , what about the 'split' ?

iron dust || mercury does not adhere iron. Copper wire should be used.

Iron does not dissolve in mercury, so this procedure is useless. And mixing iron dust with sulfur is dangerous, it can result in a fire!

Also, in case of skin contact with bromine, use a lot of water to wash it, not ethanol! Ethanol may cause the bromine to penetrate deeper into the skin causing more severe burns!

https://www.researchgate.net/post/How_can_I_remove_the_Hg_from_the_floor Yes, yo ucan use FeCl3 to absorb mercury.

@vapid you probably never spilled BR2 on yourself. Water will do nothing. Br2 dissolves in skin and doesn't partition in water. Use ethanol to treat Br2.

Well I didn't spill bromine on myself, probably because I always familiarize myself with msds's. There is a general rule that you never use organic solvents to treat chemical burns. In case of bromine I would use solution of sodium bicarbonate. Some msds's suggest also soap water in that case. Yes, ethanol may seem to work if you have bromine absorbed only on your epidermis, because the alcohol will extract it. But for deeper burns it will only worsen things.

@vapid A large beaker of thiosulfate solution MUST be present whenever you work with bromine. Immediately stick your finger in there for a few seconds, wash off with plenty of water. Using ethanol is of course dangerous nonsense, but only water doesn't help either.

@Karl I agree, thiosulfate is a good idea but water is better than nothing.


text:
What is this "bulbous"/"blood sugar" test tube used for?
I was at an antique and bric-a-brac store, and I found several pieces of glassware. They were pretty cheap, so I purchased some of the more interesting ones. Among them were a few somewhat strange "bulbous test tubes". I purchased 2 of them, and on sale for $2.50 each.(the blue line means half off).
Here they are:

After looking up "kimax 47121" I found an amazon listing for some, labeled as a "blood sugar tube". I could not find anything else besides this. Are they particularly useful, for anything besides stuff relating blood sugar? How are they used? Can I heat them without fear of it breaking? Would they be worth trying to sell if I cannot use them?
Answers to any of these questions would be great.


text:
Why is there no salt bridge in electrolytic cell?
I was just wondering why is there no salt bridge in electrolytic cell? It is said that electrodes are in a common electrolyte solution without salt bridge.
However, when I searched the image of the difference between electrolytic cell and galvanic cell, the salt bridge are in both of the cells.
The reference website http://mcat-review.org/electrochemistry.php

Electrolytic and galvanic cells have separation of electrolytes for different reasons. In galvanic cell you wouldn't have current if you mix electrolytes from both cells. In electrolytic cell you can dip both anode and cathode in the same solution. Then yes, you don't have a salt bridge because you use the same electrolyte.
Salt bridge passes ions at a certain rate that is not too high in order to prevent electrolyte mixing. In electrolysis you can allow mixing so you don't need a bridge.
In electrolytic cell you have two transports. Lets say you have Zn/ZnSO4 and Cu/CuSO4
(1) Electrons are flowing from more active (Zn) to more passive metal (Cu). This is essential for creating the electric electric current in the cell. In solution without a salt bridge you have constant concentrations of anion (SO4$^2-$). However, concentration of Cu2$^+$ is decreasing (Cu deposits on metal) and concentration of Zn${2+}$ increases (Zn dissolves). This creates charge imbalance that will quickly compensate difference in potentials of Zn/Zn${2+}$ and Cu/Cu${2+}$.
(2) You cave a salt bridge. It has to pass ions at a right speed. Not to slow and not too fast. If ions can travel too fast (no wall) then diffusion will mix electrolytes in two cells, copper will precipitate on both sides and there will be no current. If ions travel to slow (solid wall) then the charge difference on both sides of the electrolyte will prevent further current.
historically, it was just a piece of fabric between two compartments. Diffusion through it is slow, but drift along the electric potential is fast.


text:
Why are substituted biphenyls optically active?
Why are some substituted biphenyl compounds optically active and what kind of substituted biphenyls are optically active . I read this in a book but no satisfactory reasons were given and no examples were given thanks in advance .

Related: http://chemistry.stackexchange.com/questions/5489/racemization-of-biphenyl-compounds/5499#5499

In general, compounds without improper axes of rotations are chiral. More concretely (and less correctly), compounds with less symmetry are more likely to be chiral.
Substituted biphenyl compounds generally have little symmetry. Why is this? Steric hindrance causes the two phenyl rings to be tilted relative to each other and serves as a barrier to rotation. This introduces dimensionality to the compound and breaks symmetry in the plane of the ring. If, in addition, there are substituents located asymmetrically on the rings, then the two rings can be differentiated from one another and thus lead to chirality. A similar situation is observed for spiro compounds.
BINAP is an example:

Remark. Most textbooks mention chiral centers as a sufficient (but not necessary) condition for chirality, and fail to emphasize other possibilities. Substituted biphenyl compounds, for example, may be chiral even without chiral centers.

"If, in addition, there are substituents located asymmetrically on the rings, then the two rings can be differentiated from one another and thus lead to chirality. " - doesn't the symmetry breaking due to strain suffice to enforce chirality? Are substituents necessary for that?

@HarryHolmes, the activation energy required for rotation in an unsubstituted biphenyl is low enough to be overcome quite readily, so the two stereoisomers (more specifically, atropisomers) can interconvert. Substituents induce chirality by increasing steric hindrance and hence the activation energy for rotation. I think the substituents do not have to be asymmetrically distributed on the rings, but this is a five-year-old answer so I’m not certain what exactly I meant.


text:
Is cobalt (II) chloride in silica gel dangerous when heated?
Today I decided to reheat 3 x 40 gram silica gel canisters with the blue/pink silica gel for the first time, and after I decided to order some more. While shopping, I noticed people say how it is a carcinogen, and that my family and I could now be at risk for cancer. Is this true?
It was in the toaster oven of the kitchen for 5 minutes, 2x each since by the time the canisters cooled off they became pink again before I put them where I needed to.
So am I looking at a risk, or is what I have done too minimal to worry about?

The colouring used to indicate whether or not the beads are dry does indeed have some associated toxicity.
With the amount you have, however, I don't think it should be a problem.  Labs commonly keep large quantities of the stuff in the oven, and even then the only precaution we take is heatproof gloves to protect from the heat

Can you please explain your procedure? Did you reheat it in sealed canisters? If yes, then your silica gel didn't get drier(water just stayed there), but you are safe. If canisters are open then were you using a separate oven (the one that is not used for food)? Is the oven electric or gas?

Related: Blue silica gel more harmful than the orange ones?

The canisters were inside of the Hydrosorbent 40 Gram Canisters. I put them into the small toaster oven, where we sometimes warm food up, and it is electric.

I have to assume things to answer this question. It would be great if you could include a picture of the actual stuff you have so that I may assess whether my assumption was correct or not.
The only aggregate state of desiccant silica gel, whether it be blue gel, orange gel or a colourless one, that I have seen to date is best described as a very coarse granulate. From these, unless you vividly shake the canisters repeatedly, there should be no dust evolution. Stuff like that is routinely stored and handled in chemical laboratories without anybody ever giving a damn because of dust — much unlike the very fine silica gel used for chromatography which can very much, as sixtytrees points out, cause silicosis.
The carcinogenic stuff inside blue gel is a cobalt compound. The salts of cobalt are only carcinogenic if you indigest them, and they do not permeate well through skin. They are even less prone to produce dust or carcinogenic volatiles even when heated to a hundred or so degrees. What they can do is stay on the skin after you touched the gel and then be transferred onwards, for example in your mouth on the crunchy apple you decided to eat — so wash your hands thoroughly if you accidentally handled the actual gel and did not use gloves.
Orange gel is slightly different, it uses organic moisture indicators. These have a much more significant vapour pressure than inorganic salts do. On the upside, these are also less carcinogenic. So the overall effect is probably similar.
Aside from that, you are at a very low risk. Sitting by a campfire is very likely more carcinogenic than putting blue gel desiccant into the oven for a few minutes.

I used them to absorb moisture of a product that in itself was full of moisture and it's intended job was to remove moisture from the guitar, they are called Daddaro humidipaks. I put them in an airtight container along with the humidipaks. After reading your answer I am going to throw the the silica gels away. But if I touch the humidipaks now and then touch my mouth, will I be exposing myself to cancer?

Main danger of silica is that it can cause silicosis. https://en.wikipedia.org/wiki/Silicosis This is because silica is not soluble in water, cannot be broken down by enzymes and silica dust can get into air very easily. Silica isn't carcinogenic, but silicosis is a bad disease too.

The silica was inside of the canisters in their silica form. Can dust be created from the toaster oven heating it up, and is the amount I listed something that should make me worried? Edit: also I heard that the blue color meant toxic/carcinogen, which Is why I worried.

From an above comment from the OP that clarifies the problem:
The canisters were inside of the Hydrosorbent 40 Gram Canisters. I put them into the small toaster oven, where we sometimes warm food up, and it is electric. – tacoman50
Good lab safety requires that you never mix chemicals and food in any manner. Thus you should never use an oven for both chemicals of any sort and food.
The whole point is to draw an absolute line rather than trying to determine that something is "not too dangerous." Who knows what the silica gel may have adsorbed besides water.

The thing is whether these were sold for lab use or for non-lab use.


text:
I have a sodium silicate sealant in my car and need to find a way to soften it
Poured Blue devil sealant and found that it clogged my engine. Can someone who knows about chemistry and cars maybe give me any ideas on how to solve this issue?

I do not know much about engines, but from a chemical point of view, the clog that formed in your engine is most likely a hydrated silica. This compound is very insoluble and unreactive - there are only few things that can dissolve silica, e.g., hydrofluoric acid or a hot concentrated solution of sodium/potassium hydroxide, but I wouldn't pour any of these to an engine. Therefore, it seems that your only option is to dismantle the engine and try to remove the clog mechanically.

Radiator sealants can cause engine damage by clogging the cooling passages in the engine block and cylinder head(s). If you try to use some technique such as forcing water through the block  it's likely some passages will still be left clogged, causing later over-heating and warping of the block. or heads.

Added sodium hydroxide, aka drain cleaner. Worked well by dissolving the sodium silicate. Thanks though!


text:
Why are Cu2+ ions blue?
I figured that it is probably having to do with the wavelengths of visible light that are absorbed and transmitted by the ions, but I'm not entirely sure.

Your assumption about the absorption of light at different wavelengths is on the right track


text:
How to Protect one -OH group in 1,4-benzenedimethanol using TBDMS
Actually i want to protect one -OH group using using 1.1Eq of TBDMSCl and 1.5 Eq of Imidazole. Is is possible to protect one OH group using these condition. Even if i get mixture of Products ( one protected OH and Both OH protected compounds), separation by column would be easy and what would be approximate yield of both products.
any suggestions.?

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. This appears to be a homework question, please share your thoughts and attempts towards the solution. If you have a procedure you are working on, please provide a citation to it. Possibly the context of it might already explain the selectivity of such a reaction.

The product is symmetrical so it doesn't really matter which alcohol gets protected, you'll end up with the same product.
There are many examples of doing this kind of mono protection of simple substrates. Quite often a fairly large excess of the diol is added to try and ensure no bis protection takes place. In your case the starting material should be easily separable from product by column chromatography.
A potentially better method is to formally deprotonate the diol with 1 eq of a base such as sodium hydride or butyl lithium then trapping with a solution of the silyl chloride. This often works well in cases where bis protection may occur, as energetically its unfavourable for both oxygens on the molecule where there's molecules that are neutral

Actually I tried Mitsunobu reaction of 1,4-benenedimehtanol (1eq) with Imidazole (1eq) using Ph3P and DIAD 2 eq each. I was expecting to  get mixture of Products  ( one Imidazole group replace OH group of diol or both OH are replaced by imidazole). when i performed the experiment I only able to get product in which both OH group are replaced by Imidazole. what would be the reason for that, Due to this i thought better to protect one OH group to successfully carried out Mitsunobu reaction.

If all of the hydroxyl groups have reacted then you must have added too much reagent. Maybe try looking on Reaxys for your specific reaction but I've did the silyl protection on many similar sunstrates and never had an issue as long as I was careful with weighing


text:
Reaction of H2o2 on Ozone in jar washing Tank in packaged drinking Water Company
What is reaction of H2o2 when added in Jar Washing Tank  where ozone is also being dosed at 0.5ppm in a Packaged Drinking Water Industry ?

When would I receive answer to my Question

What are you actually asking here? Are you looking for the reaction between hydrogen peroxide and ozone?

Both ozone and H2O2 are reactive oxygen species. They don't necessarely react with each other, but oxidize organics and act as bacteriocides. Majority of bacteria are killed by H2O2 (which is cheaper and safer). In case some spots were not cleaned ozone is added. Producing enough ozone to clean it all would be difficult. Guaranteeing that H2O2 reached every spot is also hard. this is why they use two chemicals.
0.5ppm (1/2 parts per million) is a relatively low dose, but working with ozone at higher concentrations is dangerous. Ozone is more toxic than chlorine.
So, this is much like mixing dishwashing liquid with laundry detergent. There is no reaction between them.
Note that in a lab you can have O3+H2O2 -> H2O3 + H2O, but you have to run it at low temperature, the yield is low and the product decomposes fast.


text:
What does K3Fe(CN)6 become?
What does the salt $\ce{K3Fe(CN)6}$ change to when it is dissolved in water? I mean, I think its going to ionize, but in what way? Is there a way to know this? I tried to search the web but wasn't able to find the answer.

In water it should dissociate into $\ce{K+}$ and $\ce{[Fe(CN)6]^3-}$. This is because the $\ce{Fe}$ forms an octahedral complex with $\ce{Fe}$ at the centre and six $\ce{CN-}$ ions distributed around this each at 90 degrees to one another. This ion is sufficiently stable that it will not dissociate in water. The bonding is due to $\mathrm d$ orbitals on the $\ce{Fe}$. All undergraduate inorganic text books will give details of this. The Wikipedia page on ferricyanide gives pictures of the complex's geometry.


text:
Is there a reaction to get hydrogen from hydrocarbons?
Combustion of hydrocarbons reveals hydrogen, but not sure at what efficiency. Is there a reaction to get hydrogen from hydrocarbons?

Combustion of hydrocarbons generally does not produce hydrogen because hydrogen itself will combust.

You could considering cracking. It is an industrial process of getting hydrogen gas. However, the main purpose of cracking is usually to get some light alkene molecules such as ethene from higher petroleum fractions

Hydrocarbons are in fact the main source of hydrogen gas. According to Wikipedia, 95% of all hydrogen is produced by the steam reforming reaction:
$$\ce{CH4(g) + H2O(g) -> 3H2(g) + CO(g)}$$
This is a very important industrial process which has been studied in depth. It is done at very high temperatures (~1000 °C) and under pressure (~20 atm), and can be "up to 75%
efficient" (though efficiency here may not be defined as solely reaction yield).
As an aside, the fact that so much hydrogen is obtained from methane (itself generally sourced from natural gas, a fossil fuel) is one reason why hydrogen fuel cells currently have limited environmental benefit, though there is a huge amount of research into more efficient techniques for obtaining hydrogen from water splitting (especially using light) which could be carbon-neutral.

Downside: Release of a great deal of carbon Monoxide.

@Shadur Not really. Any plant producing hydrogen is either going to use the CO in some synthesis reaction, or use it to perform the water gas shift reaction to produce more hydrogen.

is there any chance to get C and CO2 instead of CO?

@shadur the reaction mix is cooled, then passed through another reactor where the shift reaction CO + H2O -> CO2 + H2 occurs. After excess H2O has been condensed, the mixture goes to a pressure swing absorbtion (PSA) unit,  where everything except H2 is absorbed onto molecular sieves. Once the sieves are fully charged the pressure is released and all the byproducts are flushed out, and sent to the furnace for burning. They therefore are not wasted but supply part of the heat needed to make the steam reforming reaction go. The PSA normally has 10 or more vessels to enable continuous operation.


text:
Safe silver electroplating solutions
Are there any practical, sane and safe solutions that can be used for silver elctroplating/electrolysis that are purchasable?
Objective is to create silver crystals, but without having evil cyanide floating around the house... copper I can handle with copper sulphate.  Silver has me stuck...
Any ideas?

Silver nitrate?

I've never tried to grow big crystals or anything, but we commonly use 30 mM silver nitrate in 1 M $\ce{NH4OH}$ for electrodepositing silver onto metal (usually gold) electrodes.
If you don't care what kind of surface finish you end up with, it will work, but it makes a fairly rough film that can be somewhat easily removed with abrasion.
The tricky part is trying to get very smooth and adherent films.
It seems that there are still not any good replacements for electroplating from cyanide-containing solutions, for decorative or engineering purposes where the deposit will be subject to wear or has to look nice.

As both responders above mention, silver nitrate should work well, especially fo small crystals. See https://www.youtube.com/watch?v=fnyBldC4Ra4

There isn't a safe option is what I was afraid of... but I wasn't enough of a chemist to be sure :(


text:
How to prevent Ag2O from decomposing to Ag on Aluminum
I have silver oxide in an ink formulation. When I apply it (in small quantities) over aluminum and heat the resulting (to evaporate the solvent), the monovalent silver oxide seems to decompose to Ag. The temperature I am evaporating the solvent is around 30-40˚C. The decomposition of Ag2O to Ag occurs at a temperature over 200˚C. However, I am aware that aluminum is a good conductor of heat, so even though the temperature was only 30-40˚C, Ag2O could have deposed because the temperature on the aluminum surface could have been around (or higher than) 200˚C.
My question is, is there a way to solve this problem. How can I still use aluminum and not worry about decomposing Ag2O to Ag when evaporating the solvent? Do I have to further reduce the temperature and is this the only solution? I tried to research this concept but couldn't find anything.
Thank you so much for your help.

First,  as you mention, $\ce{Ag2O}$ starts to decomposes from ≥200 °C. Measure the temperature of the aluminum substrate with an IR thermometer or attached thermocouple to be sure you stay well below that. You can substantially speed drying of the ink by lowering the air pressure, e.g. using an aspirator pump, perhaps eliminating need to heat the substrate.
You may also need to anodize the aluminum prior to printing  to prevent chemical reaction with the $\ce{Ag2O}$ in the ink. Aluminum is much higher in the electromotive series and would rapidly replace silver in the oxide.

Thank you so much for your answer. I was researching a bit and found out to switch aluminum for copper (for several other issues). Aside from the anodizing, is there anything to keep in mind with copper. Do I just worry about reducing the air pressure or can I do something else with copper. Thank you so much!

Copper is less reactive than aluminum, but is still more reactive than silver, so there is some chance it will displace it in the oxide... it depends on how soluble the silver oxide is, and the cleanliness of the copper. The hotter the surface and the cleaner the copper, the faster will be the displacement. Anodized aluminum might work just as well as copper, particularly if you use mild vacuum to speed evaporation.


text:
Why are acid catalysed reactions of carbonyl compounds reversible?
I am currently studying aldehydes and ketones and their reactions. I've noticed that acid catalysed reactions are represented with reversible arrows, while base catalysed reactions are shown with regular arrows. Why is this so? Is this because the addition of the nucleophilic base stabilises the electrophilic carbon? Or is there some other reason?
Update: Sorry, "all" would not be appropriate here. In my text-book, it is mentioned only once. Nucleophilic substitution by weak nucleophiles is carried out in a weakly acidic medium, while a basic medium is used for strong nucleophiles. The corresponding mechanism use the convention I've spoken about. I can't see the use of reversible arrows elsewhere. I wanted to know why is the nucleophilic addition reversible (as per my book) in a weakly acidic medium.
My text book (not a standard one): Nootan ISC Chemistry XII

The pictures of the reactions (I'm not able to rotate them for some reason)

Can you give us examples of the sort of reactions you have been studying?

@bon Sorry for the late reply. I wrongly used the word "all". I've seen it only once.

@ambigram_maker- any chance of you taking a picture/drawing out the actual reactions in your book... it might be easier for you to get a useful answer if the question is specific.       (By weak and strong nucleophiles I'm imagining you're talking about something like hydroxide vs something like a Grignard? )

Given the mechanisms in your post, I don't think it's appropriate to use the reversible arrows, and I don't believe the mechanism is quite accurate. In the acid-catalyzed case, the first step should really be protonation of $\ce{O}$ on the carbonyl group; in addition, negatively charged intermediates should not be present in acidic media.

This is a picture from wiki (Simple aldol reaction). Both Acid and base catalyzed reactions are reversible up to aldol intermediate. Dehydration is the only irreversible step. So this statement " I've noticed that all acid catalysed reactions are represented with reversible arrows, while base catalysed reactions are shown with regular arrows. " is incorrect.
Single arrow means that the step is irreversible. Reversible arrow means that reaction can go both forward and backward. Usually reversible steps are happening before the irreversible step and a happening faster than irreversible step.
This is easy to demonstrate with a pool table that nave a small obstacle near the hole. Reversible reactions are balls that move over the the surface of the table. They can get from any point to any other point of the table. Sometimes they get on that small obstacle near the hole and roll back to the table (they were not able to overcome activation barrier). Irreversible reactions are represented by balls that were able to overcome the barrier and now ended up in the hole. Now they have no chance to go back to the table on their own.
What makes reaction irreversible vs reversible. From thermodynamics perspective irreversible reactions are reactions with a large change in $\Delta$G. From organic chemists perspective reversible reactions are typically those that include protonation/deprotonation steps. Irreversible reactions are those where a strong bond like C-C or O-P forms.
Reversibility is a subject to conditions. 2H -> H2 (recombination of radicals) is effectively irreversible at room temperature, but is reversible at 3000K.
Why do we need to account for reversible reactions? (1) To understand the mechanism. (2) To know how to increase concentration of the reagents right before the first irreversible step to accelerate the overall reaction.
tl;dr: in reactions you are asking about formation of C-C bond is effectively irreversible step.

Im not sure this really answers why acid catalysed mechanisms have reversible steps whilst base catalysed ones have unidirectional steps in some cases...

$\ce{C-C}$ bonds can form in both acid and base catalysed mechanisms (aldol reaction for example). This doesn't actually answer the question.

@NotNicolaou because the assumption is incorrect, that's why.


text:
Are these metamers, if not why?
Please tell me if these are metamers ie Metamerism: In Chemistry the type of isomerism in which chemical compounds have identical proportions of the same elements and the same molecular weight, but have radicals differing in type or position, with resulting differences in chemical properties.

The term metamer isn't terminology that is widely used today, and indeed doesn't appear in the IUPAC Gold Book (the International Union of Pure and Applied Chemistry publishes guidelines for standardised nomenclature in chemistry and is widely adopted).
The description you're looking for is structural isomerism, with the compounds you show being examples of positional isomers (isomers which differ by the position of the functional groups, for instance in this case the relative positions of the methyl and methoxy moieties).

Thank you,but can you still answer it while referring to metamerism.

Well, yes, they are metamers...you're just unlikely to hear them referred to as such.


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Polarimetry test for cholesterol?
If I have a pure substance which may be cholesterol can I test that by polarimetry? In other words the idea would be to dissolve the sample in dichloromethane or ethanol, then test the rotation using a polarimeter.
My concern is that I read that cholesterol will not actually dissolve, only form a colloid, so that may invalidate the possibility of a test.

A polarimeter alone isn't a very definite test to identify a compound. Think of this from the point of view of information theory. Let's assume that the value can be +/- 360 degrees. That is 720 buckets to put billions of  billions of compounds. It would like trying to identify a person from the entire earth population by their height.

Taken from the website of a chemical supplier, the following solubility data can be easily obtained. Historically benzene and chloroform are commonplace for rotation measurements, however it can be any solvent in which the sample is soluble (or mixtures of solvents). You might be best to find a literature value for cholesterol (scifinder/reaxys) and then measure in the same solvent so at least you have a reference point to compare to.

One gram dissolves in 2.8 ml of ether, in 4.5 ml of chloroform, and in 1.5 ml of pyridine. The product is soluble in acetone, dioxane, ethyl acetate, benzene, petroleum ether, oils, fats, and in aqueous solutions of bile salts. Solutions should be protected from light.
Source Sigma Aldrich

Generally speaking, optical rotation isn't a method commonly used to determine whether or not you have a given compound. Methods like NMR, IR, MS, or even just TLC comparison are far more suited to this.
Optical rotation is useful if you already know you have the right thing, and want to know what your optical purity is like. Without having already confirmed that the sample is 1. clean and 2. the right thing, you'll struggle to make sense of the data since the polarimeter will simply give you a number.
Depending on where you bought your cholesterol from, and various other things, this number may be close to the literature value (maybe some confidence in it being the right thing), however even then you cant be sure its cholesterol as many other things will have similar values.

I know the substance is pure to begin with because it is crystalline

Just because its a crystal doesn't mean its cholesterol, it just means its a crystal.         You might try taking the melting point too, at least then you have two pieces of data (the standard amount for known compounds)

I said I know it is pure because it is crystal. I don't know what it is composed of.

Again, just because its a crystal, doesn't mean its pure. Many things co-crystallise etc.

The proton NMR should be unambiguous and published.


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How can I determine if a chemical will dissolve aluminum oxide, on paper?
I am trying to clean some corroded aluminum. In my quest to find out how, I've stumbled across a lot of bewilderment and misinformation, and a lot of different feasible techniques as well.
My question is: How can I determine, on my own, whether something will dissolve aluminum oxide (room temperature) or not? Like, just by looking at chemical formulas.
I'd also need to be able to check to make sure a chemical that dissolves aluminum oxide doesn't also dissolve the good aluminum.
My goal is to be able to figure out not only if a given chemical will dissolve it, but also to come up with an idea of which chemicals that I don't know about may also dissolve it.
As a bonus it'd be nice to determine the temperature at which a reaction would occur if it's a bit higher than room temperature so I can determine on my own if I can apply a reasonable amount of heat to make something work.
I know very little about chemistry except a basic understanding that things react and can be described by equations, and that reactions absorb or release some amount of quantifiable energy.
Aluminum oxide is Al2O3.
Sorry I do not know how to tag this.
Also I'm not sure if "dissolve" is the right word. To be clear I use it in a way that means if I apply the chemical to corroded aluminum, it'll make the corrosion go away into the air, or turn back into aluminum, or at least make it easy to scrub or wipe off.

As a good rule of thumb for alumina, nothing dissolves it at room temperature. Even HF requires elevated temperature.

I'm afraid you're quite confused - Al2O3 is nothing like rust and it actually protects aluminium as it's less reactive then metal.

Normally you would learn about properties of chemicals from general chemistry textbook. Al2O3 is pretty stable and is hard to dissolve in anything at room temperature.
It also depends a lot on what is the physical state of Al2O3. Fresh fine Al2O3 is much more reactive than aged (or coming in big chunks) Al2O3.
Searching in general chemistry literature is the best way to go, but Al2O3 is really robust.


text:
Can you electrolyse a single H2O molecule?
I don't know why I'm asking or why you'd want to but I'm just curious.
Every resource I read about electrolysis says the reaction is $\ce{2H_2O (l) \rightarrow 2H_2 (g) +O_2 (g)}$.
Can you electrolyse a single molecule?

In chemical equations you have stoichiometric coefficients not numbers of molecules in reaction mixture.

In a traditional experimental setup this would never happen, as there are always several water molecules. But lets just think about it for the sake of it.
If you wanted to design an experiment where you only have one water molecule, you would have to get two electrodes extremely close together. Lets say you find a way to levitate the molecule between the two electrodes. You might be able to then apply a voltage and tear the molecule apart, forming one H$_2$(g) molecule at the cathode. The oxygen radical would then most likely immediately react with the anode to which it is attracted by the Lorentz force and form an oxide.
So the result of electrolysing several single molecules of water might result in gaseous hydrogen and no oxygen, but a buildup of oxide at the anode.
However, having two electrodes so close together would probably create a series of problems. Not only to hold the single water molecule in place (which could be achieved by optical levitation) but because quantum mechanical effects such as tunneling currents would occur. Probably there are several more complications that would by all means make it nearly impossible to actually observe this. The knowledge gain derived from such an experiment would likely not justify the effort and expenses.
EDIT
I just revisited this question and thought about it again:
it's not absolutely necessary that the two H atoms would form an H$_2$ molecule. Depending on the way the cathodes are formed, the electric field might not pull the charged particles together, but to a collision with something else in the vicinity; the cathode or any ambient gas.
Furthermore, my notion above that "H$_2$(g)" would form is a bit nonsensical, as a single molecule cannot really be attributed a state of matter (i.e., being gaseous in this case).

I wonder if something of the sort hasn't been performed using an AFM tip and a molecule of water adsorbed on a surface. Chemical reactions have been triggered like that, previously.

@NicolauSakerNeto I  do also wonder that. Rather than an AFM I could actually imagine something like that happening with an STM (tunneling current rather than force). Atoms most certainly have been rearanged with STMs, see the famous IBM logo image. Not sure wheter molecules have been "torn apart" or electrolysed in that way though. Maybe somebody knows a reference.


text:
Lysol and bleach
I mixed Lysol with bleach and observed that the container became very hot. What reaction occurs between Lysol and bleach, and could it be dangerous or explosive?

I would dispose of it in the cleanest way possible. Keep any fumes away from you and go to the doctor if you start feeling very sick. This may be of some use http://skeptics.stackexchange.com/questions/295/what-happens-when-you-mix-lysol-and-bleach

This question should stay open. It is not unclear what you're asking. The OP wants to know if $$\ce{Lysol + bleach->bad}$$ We can answer this pretty definitively, and the link to the Skeptics.SE post will help.

I'm voting to close this question because we have no way to know the particulars of "the" container that the OP mixed and whether or not it will really cause an explosion or injury.  As such, as currently worded this is an off-topic medical question.

Rewording it to be about the general results of mixing lysol and bleach, not about any specific container of lysol and bleach, would make it (IMO) on topic.

You have to add the components of you lysol if you want help

I guess it is due to the hydrogen peroxide content in Lysol and of hypochlorite in bleach.
If I'm right, what is going on is a reaction like:
$$\ce{OCl^- + H2O2 -> O2 + H2O + Cl^-}$$
Be careful: oxygen enhances fire.
Anyway, being hypochlorite an oxidizing agent, some other reactions could be happening: it would still be safe not to mix them.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I have updated your post with chemistry markup. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details.

For the benefit of future readers it would be worth including a clear outline of the key ingredients in Lysol and bleach. We can work it out but it would be better to state it clearly for the record.


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Purification/Extraction of Sodium Permanganate
I read this question and answer, and it seemed like a rather nice way to make sodium permanganate.  Using the method with sodium hydroxide, I produced a solution. Once filtered and decanted, it was dark green.
At first this seemed somewhat strange, as I would expect to see a red color for sodium permanganate. Though, after a bit of research it seems this is actually sodium manganate. The color matches, and considering it only exists in very basic conditions, and since there is a lot of sodium hydroxide in the solution, this would make sense. To test this further, I added some sodium bicarbonate to, oddly enough, act as an acid. The sample of solution quickly turned the pink color I had expected.
So, I currently have some amount of sodium manganate, sodium hypochlorite, sodium hydroxide, and sodium chloride in this solution.
At first I hoped I might be able to extract this using organic solvents, but unfortunately wikipedia claims potassium permanganate decomposes in organic solvents and alcohol. I assume the sodium permanganate or manganate would do the same.
Upon boiling down the solution I would end up with a mess of crystals of various shades of blackish and whitish.
So, my real question here is, if there is a way or ways to remove the other substances (sodium hypochlorite, sodium hydroxide, sodium chloride) and attain more or less pure sodium permanganate crystals.

The easiest way to get some permanganate is to go with potassium salt. Potassium permanganate is fairly soluble in warm water, but only moderately soluble in cold water. No so luck with the sodium salt. Afafik, the sodium salt in clean state may be acquired by electrolysis of sodium hydroxide solution with manganese anode.

Yes, permeakra is correct. Sodium permanganate has solubility of 900 g/L so it will always come last from the solution. And there is no simple way to separate it from other salts. Also, it's hygroscopic. BTW, use less concentrated sodium hydroxide for oxidation, so you will get directly sodium permanganate and not manganate.

Solid potassium permanganate isn't heat stable, so I wouldn't try to boil off all the water.

KMnO4 is likely to react with organic solvents rather than react with them...

You are correct. First step produces $\ce{K2MnO4}$, then you need to oxidize it.
My first guess was that adding $\ce{H2SO4}$ will cause disproportionation of $\ce{K2MnO4}$ into $\ce{KMnO4}$ and $\ce{MnO2 + K2SO4}$. Filtering removes $\ce{MnO2}$. Crystallization will give you pure product. Or you can use a mix of
$\ce{KMnO4 + K2SO4}$ for many purposes.
Industry uses more efficient method: electrolysis of $\ce{K2MnO4}$.

Wouldn't the addition of sulfuric acid produce manganese heptoxide?

It wouldn't. Mn$^{5+}$ (as in K**$_3$MnO4) and Mn$^{6+}$ (as in K$_2$**MnO4) are only stable under basic conditions. In acid they disproportionate to yield MnO2 and KMnO4. If you add too much H2SO4 you can get Mn2O7 - a very reactive and dangerous Mn$^{7+}$ oxide that inflames EtOH on contact.


text:
Why is allyl chloride more reactive towards substitution than alkyl chloride?
I can understand that the reactivity of alkyl halide is more than vinyl and aryl halides because the halogen atom in alkyl halide is connected to the sp3 carbon.
Whereas in vinyl or aryl the halogen atom is connected to the sp2 carbon.
But how allyl halide is more reactive than alkyl halide?

Reactive towards what? It's all relative unless you have a specific reaction in mind

Reactive towards substitution.

Substitution with .....

It depends on the reagents. Because the answer would be different if the mechanism was Sn1 or Sn2. For Sn1 the answer would be the stability of the allylic carbocation INTERMEDIATE. However for Sn2 there is allylic stabilisation of the TRANSITION STATE. But either way it's essentially conjugation

But how allyl halide is more reactive than alkyl halide?

Reactivity towards $\ce{S_N1}$
The key to reactivity towards $\ce{S_N1}$ is the stability of the formed carbocation. Allyl system stabilises the carbocation through overlap with the vacant p orbital (@gsurfer999 has shown the resonance structures in his answer below). However, note that any allyl halide wouldn't be better at $\ce{S_N1}$ than any alkyl chloride.
In fact, the tertiary alkyl chloride's rate of $\ce{S_N1}$ is faster than that of allylic chloride which is secondary at one end. (due to $\ce{+I}$ effect of 3 methyl groups as well as hyperconjugation)
Visual aid for stabilisation through hyperconugation (from Clayden):

For such comparison, here's a useful table again from Clayden:
Reactivity towards $\ce{S_N2}$:
In $\ce{S_N2}$ the most important factor is the substrate. The reaction proceeds through a single transition state with trigonal bipyramidal geometry and $\ce{\sim sp^2}$ hybridisation of the electrophilic carbon. The p-orbital makes two partial bonds, one with the nucleophile and another with the leaving group. Thus, it's electron deficient.  In such a case, the allyl system provides the additional electron density through conjugation.
However, again note that primary alkyl chloride is better at $\ce{S_N2}$ than allylic system due to least steric hindrance!

But why is carbonyl group more effective ?

The allyl cation has delocalisation which decreases the energy of the system. This makes it easier to remove a halide.

I didn't get how energy of the system is related to removal of halide...

@Shuvam Shah The stability of the allyl cation means the chloride can leave more readily. A less stable cation would react rapidly with the chloride to re-form the reactant.


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What is the position of hydrogen in the periodic table?
Hydrogen is not in the first group as it was before and it is now placed above the periodic table? So why is it still categorized as a metal in some books?

Which books will categorize a reactive gas as a metal? Theoretically there does seem to be metallic hydrogen, but so far experimental efforts to produce a sample have either failed or returned questionable results.

There is a lot that could be said about this question, but a simple explanation goes like this:

Some periodic tables categorize hydrogen as a metal (or at least an alkali metal) because it has one electron in its outermost orbital - just like Lithium, Sodium, Potassium, etc., which are obviously metals.
Other authors of chemistry textbooks realize that hydrogen seems to love the gas-phase so much over the solid phase that it just doesn't make sense to call it a metal, irrespective of what atomic orbital pattern may tell us.

Hence the differences between authors of texts and where they put Hydrogen on their periodic table.


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Why is the first member of group IVA and the second member of group VIA show stronger tendencies to catenate compared to the other group members?
catenation is a process by which a molecule or an atom forms a straight or repeating chain with other atoms.in carbon it has 4 electrons in the outer most energy level hence can form chains repeatedly

The ability to catenate appears to be based on several factors:

The bond energy of the element to itself - stronger bonds promote catenation
Atom size - smaller atoms promote catenation due to lower steric demand and better orbital overlap (especially p-p)
Degree to which orbital hybridization is important in the structure of that elements compounds (for example, the bond angle in $\ce{NH3}$ is $\sim 108^\circ$ while that of $\ce{AsH3}$ is more like $90^\circ$
Electronegativity of that element - atoms with electronegativity in the middle of the scale appear to catenate best.

Here is a plot of bond energy and electronegativity for the first 36 elements. Note that there appears to be a sweet spot of electronegtaivity between ~2 and ~3 that produces high bond energies. In the box I drew are all of the elements that are known to have covalent compounds containing more than 3 catenated atoms of that element. In case you were wondering, I used bond energies for single bonds only so comparison was better.

Note that hydrogen is in this box. Hydrogen cannot catenate since it has only one valence electron, but if it had electrons in the second shell, it probably would catenate.

Hydrogen can catenate, to a degree.  The cyclic trihydrogen cation is an important constituent of interstellar gas clouds (https://en.wikipedia.org/wiki/Trihydrogen_cation).


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How could uranium be extracted from ore to be melt?
I would like to go hunting for uranium ore, and because it has a relatively low melting point (1132.2 °C), I thought it would be cool to melt it just like any other metals. According to this video extracting yellowcake (mainly U3O8, UO2 and UO3) is relatively easy, however, how could I remove the oxigen from the oxides? Or is it easier to obtain the uranium directly from the ore in another way?
EDIT: because of all of the comments I've received I'm going to explain better the question:

I'm well aware of the dangers of handling uranium, a weak radioactive materials, the fact that it's dangerous as an heavy metal and that the ore is more radioactive than the pure metal because there are products of the decaying chain that are more active tha the uranium itself.
I know all the regulations and laws in my country, so that I know how much I can keep it (about 0.1 curie) and what I can and what I cannot do with my uranium source. I also know where I can get it withouth getting in trouble.
what I need to know is what chemical peocess I could use to extract pure uranium metal (238U + 235U) that I can melt in whatever shape I please. And no, I don't want to make a nuke because that would be impossible

Assuming that you do know what you are doing, you should go for reference chemical synthesis books. The readily available sources point that the metal may be produced by electrolisys of molten salt mixture or by reaction of halides with alkali or alkali-earth metals. Neither process looks like something I would consider safe and worth the effort, and definitely neither should be employed without proper description of experimental procedure, that should be obtained in specialized literature.

There is another process for extraction of Uranium from the uranium ore in the form of UO2(NO3)2 by dissolving the Uranium ore in nitric acid and extract of uranium  using 20% solution of Tributyl phosphate/dodecane. TBP form strong complex with uranium nitrate and thus you can easily remove uranium from other impurities. Obtained UO2(NO3)2 can easily be converted to (NH4)2U2O7 by reacting with ammonia and finally reduction with hydrogen gave you UO2.
You can't obtained pure Uranium because it is susceptible to oxidation easily.

From theoretical perspective:
For a long time elemental uranium wasn't known. Scientists tried to heat the ore with carbon, but the result was uranium carbide, not uranium. Eventually it was obtained by first into UO$_2$(NO$_3$)$_2$ and then to UO2 and then reduction with hydrogen gas.
From practical perspective (I would assume you want to make a nuke):
Uranium ore is a strategic resource and is well guarded. Mines always belong to someone and guards are dead serious there. If they catch you - you are in trouble. Uranium ore is dangerous. Not even because of decay of U (it lives billions of years), but because of the product of it's decay. You will get a dose of radiation if you are not well equipped.
Even if you succeed, you will blow yourself up when trying to reduce it in the flow of hydrogen at 500C. This will be a simple chemical explosion, you will be cut by shattered glass and cleaning of the place from radioactive contamination is expensive. In US they will take away your house because loan contract says that you promise not to do it at home. Seriously.
Finally, if you magically get pure uranium it is quite useless. You now need to extract $^{235}$U, or, better, turn some into Pu$^{239}$. There is zero chance to succeed in it. It took US 3 years, an army o0f engineers and 1 billion dollars (back then it was a lot more than now) to get the first nuke.

I don't want to make a nuke, I'm not inrerested in enriching uranium and I know that it's extremely difficult. I would just like to get some 238U and melt it something I can bring around like a coin. Also uranium is not that dangerous: it has a low radioactivity and mainly of alpha particles, not dangerous until eaten or inhaled, it is only tpxic as an heavy metal. Lastly I know avandoned uranium mine

Low grade uranium ore is all over the place.

@MauroF. Uranium cannot be carried around like a coin. It oxidizes on air, but unlike most metals does not form a protective oxide film, so the "rust" (in this case it will be yellow uranium oxide) will eat through your coin. Uranium samples have to be kept either under oil or in an inert atmosphere.

@MauroF. While having lead-enforced pants is not a strict requirement for work with uranium, it certainly helps. Wearing such a coin with yourself all the time is not a wise idea. Radiation damage accumulates, so given enough time it will cause trouble. Luckily, most uranium radioactivity is rather soft, especially if we consider U-238. Uranilacetate was fairly available in the past as  analytical reagent, so chances are you can hit it, given enough time.

@MauroF. Probably the safest route would be to make a ceramic/glass piece using uranium oxide and then enclose it into some glass shell.

I'm well conscious of the dangers of radiation and the regulamentation in my country, and I've never said I would keep it around with myself. It would be just a collection item. Could we please just stick to the question?


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Is it possible to curve fit GC, Mass-spectrometry and NMR peaks?
I am trying to measure the amount of compounds by the peak area and I want to increase accuracy to 0.1% by curve fitting. Can I do it and what functions should I use?

Yes, you can curve fit all three of them. For NMR use lorentzian shape.For other two use Gaussian peak shape.
But remember: measurements in chemistry are typically 10% accurate. Increasing accuracy beyond that is difficult.
Mass-spectrometry is only quantitative if you study isotopic molecules. You cannot compare alcohol with an acid and amine using MS. Amine will dominate in positive mode and acid will dominate in negative mode.
GC is the most reliable but you have to find the right dose. Not to small, so that your peak is visible, and not to high to avoid asymmetry. Tecknically, you can use asymmetric fitting curve, but those fits are less stable (less reliable, even though the R is better).

For MS and for GC, fitting Gaussian functions to the data is possible but is completely unrelated to quantifying the amount of the substance.

You should not assume any given peak shape as in practice this may not be what it should be from theory.
You have a set of data points so once you have determined the baseline a numerical integration should work well. There are many of these in the literature. When fitting use the chi squared statistic as a measure of fit, the Pearson R is just useless and should never be used in fitting. (This is because a poor fit has, say, an R = .99 and a good fit .999 so its not anywhere near discriminating enough). You should also look at the shape of the residuals, autocorrelation function, runs tests etc to make sure you have a good fit. A one parameter good/bad test is often not good enough.
If you have overlapping peaks then you will have to assume some shape simply to use non-linear least squares to estimate one peak from another. This will lead to some more uncertainty but depending on what you are trying to determine this may not matter.
In gc data the peak shape is not necessarily gaussian, often it is lopsided so a numerical method is better here.
It may not be necessary to fit your data to try to get such high precision, the data simply may not allow it due to instrumental factors. But what is important is results not data, i.e. its what the data tells you that is important; this may not need such high precision as you seek. I'm absolutely not being patronising, but I would go as far as to say that if you need 0.1% accuracy you should perhaps think of another way of trying to learn what you want to know.


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Why is BuLi a better nucleophile than base?
In Carey, Sundberg Advanced Organic Chemistry it is written:

The relative slowness of the abstraction of protons from carbon acids by organolithium
  reagents is probably also due to the compact character of the carbon-lithium
  clusters. Since the electrons associated with the carbanion are tightly associated with
  the cluster of lithium cations, some activation energy is required to break the bond
  before the carbanion can act as a base. This kinetic sluggishness of organometallic
  compounds as bases permits important reactions in which the organometallic species
  acts as a nucleophile in preference to functioning as a strong base.

Won't the same reasons apply for nucleophilic displacement? Compactness of the aggregates should affect nucleophilicity, too. So why will BuLi add to a carbonyl instead of deprotonating the alpha carbon atom?

tBuLi and LDA? Both organolithiums that act solely as bases (or at least never as nucleophiles)

They act so because of they are bulky. I am more interestsed in BuLi and similar organolithiums.

'Why are organolithiums better as nucleophiles than bases'    Maybe update your title and question to be more specific. The statement isn't really a general one other than n-Buli

Edited. Hope that it is more clear now.

This is a trick question, which explains why the first sentence from the book takes so many precautions. It is very difficult to provide a general explanation because there are many factors influencing the result.

Solvent: the solvent is extremely important in such reactions. It
plays an important role in the order of oligomerization of the
organolithium reagent; monomers, dimers, tetramers or polymers. Of
course, the reactivity is really different.
Co-solvent: a co-solvent like HMPT or TMEDA can drastically change
the course of the reaction by changing the order of oligomerization.
Temperature: it is well known that n-butyllithium is a very strong
base at low temperatures (<-60°C) but will become a good nucleophile
at higher temperatures. It can even open a THF ring.
Thermodynamic or kinetic control: organolithium compounds are
usually used under kinetic control, but...
HSAB: how the oligomerization affects the hardness/softness of the
base and its nucleophilicity?
3D shape of the organolithium compound: it greatly influences the
shape and order of oligomers.
...

So now you feel completely depressed, I can try to answer your question. I think that a key factor is actually that in a cluster (dimer, tetramer and so on), charges are more diluted which enhances the nucleophilic behavior. Also, the geometry of the cluster and of its 3D-complex with the substrate could explain why a reaction is favored over the other.

Great answer! :D I concluded that oligomeriation softens the organolithium.