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

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

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

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

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

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

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

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

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

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

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


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How does dimethyl sulfoxide transport other molecules through skin?
Wikipedia says:

Use of DMSO in medicine dates from around 1963, when an Oregon Health & Science University Medical School team, headed by Stanley Jacob, discovered it could penetrate the skin and other membranes without damaging them and could carry other compounds into a biological system.

How does this work? Is there a term for this property? Can DMSO transport molecules through other barriers? What are the limitations on which compounds DMSO can transport? Which other compounds share this property with DMSO?

DMSO interacts with the outer membrane of skin, the stratum corneum.  The stratum corneum is a typical lipid bilayer and DMSO can solvate the exposed hydrophilic lipid head.  This results in a weakening of the lateral forces between the individual phospholipid molecules.  The weakened lateral forces result in a widening of the channel between the phospholipid molecules and the widened channel allows small molecules to pass through the lipid layer.  After the DMSO is removed the lipid bilayer restores itself back to normal.

Although it is agreed that small molecules can pass through the DMSO-affected bilayer, there is some evidence that larger molecules such as viral agents might also pass through.  There is also some controversy over the effect of DMSO on the blood-brain barrier.  Here is a link to a nice review of DMSO's pharmacological effects, including transport of larger molecules and transport across the blood-brain barrier.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


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How to determine rotational states of a molecule?
How can you determine the possible rotational states of a molecule? Atkins Physical Chemistry book says that for linear molecules there are two possible rotational states, but what is the general method for finding the amount of rotational states for other relatively more complex molecules, say for example $\ce{H2O}$ and $\ce{CH4}$?

Molecules have only 3 (or 2 if linear) rotational modes.  A molecule with n atoms has 3n-6 vibrational states, 3n-5 if it is linear.  The "6" or "5" that is subtracted represents the 3 translational modes that all molecules can move in (x, y, and z directions) plus the 3 rotational modes (rotation about the x, y and z axes).  The 3 rotational modes reduce to 2 if the molecule is linear since you can't observe rotation about its molecular axis (the axis that contains the $C_{\infty}$ symmetry element).

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


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Which has a greater entropy?
Which of the following has greater entropy?
a.  $\ce{CH3COOH~ (aq)}$
or
b.  $\ce{CH3COO^{-}~ (aq) + H+~(aq)}$
They are both in the same state, and have the same number of particles.. so how do I figure this out? My instinct says a, because the products will be more "random", but I'm not completely sure.. Any ideas?

Your welcome, glad to help.

In the gas phase I would suspect that the entropies would be equal as both systems have the same number of translational, rotational and vibrational modes (18).
However you specifically ask about the liquid phase system, since you show that both "a" and "b" are solvated in aqueous media. In this case, the ionized molecule (b) has charges on both ions and this will cause increased ordering of the solvent molecules surrounding the ions, more so than neutral molecule "a".  This increased ordering of the system in case "b" will reduce the entropy of the system.  Consequently system "a" will have greater entropy.

I don't think your statement for the gas phase is correct. Even though both systems might have the same number of degrees of freedom the entropy is not the same for translation, rotation and vibration. I plan to provide an answer to your question "Can Entropy Increase During Bond Formation?" when I get the time and this answer will also deal with this topic.

@Philipp  Great, I look forward to reading it!

entropy != disorder.  I also hope you are not seriously suggesting orderly arrangement has any meaning in a normal aqueous solution.  The solvent separated ion pair will have more solvent molecules involved in the interaction than the neutral molecule leading to a greater number of available microstates, leading to a higher entropy.

@Lighthart  Yes, I am suggesting that the orderly arrangement has entropic significance.  Are you saying that system B is less ordered than A and consequently B has higher entropy?  If so, perhaps you could elaborate your comment into an answer.

I am not saying anything about order.  Order is not related to entropy in any way.  Entropy is the number of available microstates-- the number of possible ways to arrange items in the system.  When there are more parts, there are necessarily more ways to arrange the parts, and hence larger entropy.  Thus, entropy favors dissociation.

@Lighthart  Due to the electrostatic charges on the 2 ions when the acetic acid is ionized, aren't there fewer microstates available to the solvent molecules then there are for the solvent molecules when the acetic acid is unionized?

I would say no.  The covalent bond severely restricts the number of available cofigurations as compared to the dissociated, solvent-solvated ion pair.  Furthermore, in aqueous solution, the cation and anion may 'switch partners' with other solvated ion pairs, further increasing the number of configurations.  Similarly the solvent molecules may exchange amongst themselves.  A slightly different question is proton exchange between hydronium ions and neutral waters, but that process is fast on the timescale of defining microstates.  Unfortunately, I can't seem to find a tabulated value.

This page: https://www.chem.tamu.edu/class/majors/tutorialnotefiles/entropy.htm  suggests solvation of ammonium nitrate is entropically favored (ie entropy increases).  It is a poor model at best.

The problem posed by the OP compares acetic acid already in solution vs. ionized acetic acid in solution, so there is no entropic contribution due to addition of acetic acid to the water to consider.  This link in the section "Entropy in Solution Formation" notes that "there is a loss of entropy associated with the water molecules organizing their 'solvent cages' around the ions". I was guessing that

the entropic decrease due to this fairly long-range ordering outweighs the entropic increase due to ionization, but it is just an educated guess on my part.

Similar viewpoint, just opposite.  I also no longer have a lab to run the experiment, which given the two positions presented, I think is necessary to resolve.


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How to simulate chemical reactions in GAMESS?
I am rather newb in GAMESS field but finally, learned to make input files for most molecular simulations and could even use TD-DFT for excited states (special thanks to Geoff Hutchison for his help).
Now I want to start simulating simple chemical reactions. I looked at the GAMESS manual and think I need to make transition states and find least energy pathways but have no idea how to make input file for that.
Can anyone give me a simple input file example for a simple reaction?

As in geometry optimization, you are searching for stationary point on the potential energy surface (PES). Not for local minimum, but for saddle point, therefore in GAMESS, you specify RUNTYP=SADPOINT. You would also need the correct (non-guess) Hessian matrix, which you can calculate separately with RUNTYP=HESSIAN. In transition state (TS), you should have zero gradients (stationary point) and one imaginary eigenvalue of Hessian (the reaction coordinate in TS).
For working example, see e.g. SN2 transition state. In principle you need very good guess at TS geometry, which you obtain by scanning along some chosen coordinate. The point highest in energy should be reasonably similar to the TS, and you optimize to saddle point starting from there. Already at the guess geometry you must have one imaginary Hessian eigenvalue (resembling the reaction you are looking for, not methyl rotation somewhere else).
But beware, finding the transition states is one of the most difficult topics in computational chemistry and you are not guaranteed to find one, no matter how hard you try. One of the biggest problem is solvation, as you often wish to find reaction barrier in solvent (where majority of chemistry happens), but you can calculate only in vacuum. Just think of how carefully the workbench chemist finds the correct solvent for given reaction, in computer you have only very crude approximations to it (and most often just vacuum).
That said, it is great fun and rewarding to find the transition states. For best results, use the simplest theory available, as it is better to have smooth and well behaving PES, even though it is slightly wrong, so it is no harm to start with HF or BLYP (or B3LYP). To get some ideas, start with simple reactions, i.e. where no charged species are involved, so electrocyclic reactions are probably best choice, e.g. Diels-Alder.

(+1) I am not an expert in GAMESS, but what you say makes sense in general. As for the procedure, I agree with what you say. I would recommend to start with BP86, because it is a pure functional, quite robust and pretty fast. Best also use a basis set of double zeta quality, like cc-pVDZ or def2-SVP. Better avoid the quite old Pople basis sets, like 6-31G*. Also use a fine integration grid and tight convergence criteria. It takes longer, but usually the optimizations don't run in a undesired direction. A Diels-Alder reaction is certainly a good training exercise.

Thank you !   This was a great answer and I learned a lot with that. As I realized from that SN2 Transition State example, we should already know ( or at least guess ) the result and then make our energy path on that.  Is there any way to do it in reverse? I mean doing the simulation to make some crude guess about what "may" happen in a new reaction ?

@Aug: in general, you cannot ask computer: "I have those two molecules, how will they react?" If you assume that new bond will be formed between them, than the computer should try to propose MxN reaction, where M, N is number of atoms in each molecule. But what if it is electrocyclic reaction? Or elimination? So it is up to you, chemist with brain, to propose something and ask computer: "Does this reaction look reasonable?" and get answer "Yes/No/Depends"

People are starting to put the results of their calculations on figshare.com, eg http://chemapps.stolaf.edu/jmol/jmol.php?source=http://files.figshare.com/1860026/ethen_01oIRC.cml To see the reaction, (i) right click somewhere next to the molecule and a menu will pop-up; (ii0 go to animation -> animation mode and press loop; (iii) right click somewhere next to the molecule to get the pop-up menu and press play.  There are many other examples.


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In the following reaction, will the ratio between substances be 3:1?
In the reaction
$$\ce{3 C2H2(g) <=> C6H6(g)}$$
will the ratio of $\ce{[C2H2(g)]}$ to $\ce{[C6H6(g)]}$ at equilibrium be 3:1?
How is the mole ratio related to the equilibrium (or lack thereof) of the reaction? How do I know if 3:1 is indeed the ratio?
Thanks.

No. The location of the equilibrium is dependent on many factors, most prominently temperature and pressure.
A chemical equilibrium appears to have a static composition, which is of course not true. The point in question is where the rates of the reaction equal in both directions. The equilibrium can be described by the equilibrium constant $K$, which is defined as the product of partial pressure $p_{\ce{B}}$, concentration $c_{\ce{B}}$, [...], chemical activity $a_{\ce{B}}$, in general $x_{\ce{B}}$ of the reactants and products, with respect to their stoichiometric number $\nu_{\ce{B}}$.
$$ K_x = \prod_{\ce{B}}(x_{\ce{B}})^{\nu_{\ce{B}}}$$
The standard equilibrium constant $K^\circ$ is given through the standard reaction Gibbs energy $\Delta_r G^\circ$ and the thermodynamic temperature $T$, $\mathcal{R}$ is the gas constant.
$$K^\circ = \exp\left\{\frac{-\Delta_r G^\circ}{\mathcal{R}\cdot T}\right\}$$
Since all equilibrium constants are proportional to each other it follows for a constant temperature, $T = \mathrm{const.}$, that the location is dependent on the Gibbs energy:
$$ \prod_{\ce{B}}(x_{\ce{B}})^{\nu_{\ce{B}}} \propto \exp\left\{-\Delta_r G^\circ\right\}$$

In your case $$\ce{3 C2H2(g) <=> C6H6(g)}$$ you would probably choose a pressure based approach $$K_p = \frac{p(\ce{C6H6})}{p^3(\ce{C2H2})} \propto \exp\left\{-\Delta_r G^\circ\right\}$$
And you can see the temperature and pressure dependency in the original definition:
\begin{align}
G(p,T) &= H - TS\\
H(p,T) &= U + pV\\
\end{align}
So you will only know the product ratio by very accurate calculations or empirical measurement.
For the system in question, I believe it has been done and the results may be found in W. J. Taylor, et. al., J. Res. Natl. Bur. Stand., Vol. 37, No. 2, p. 95.


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Is this reaction at equilibrium?
The following equilibrium occurs:
$$\ce{2 NOCl(g) <=> 2 NO(g) + Cl2(g)}$$
A gaseous mixture of $\ce{NOCl}$, $\ce{NO}$ and $\ce{Cl2}$ is put in a container. After a few minutes it is found that two moles of $\ce{NOCl}$ react for every three moles of products which react. Is the mixture at equilibrium? Why? In other words, how does the mole ratio relate to whether or not the reaction is at equilibrium? Thanks.

The information you provided ("two moles of NOCl react for every three moles of products which react") confirms that the reaction occurs as depicted, but is insufficient to establish that the system is at equilibrium. The important missing information is not the relative proportions of these three species in the mixture, but whether those proportions are changing (system is not at equilibrium) or staying the same (system is at equilibrium).

When a chemical system is at equilibrium the rate of the forward reaction equals the rate of the reverse reaction. The equilibrium expression is a ratio of the conentrations of the products raised to their stoichiometric coefficients over the concentrations of the reactants raised to their stoichiometric coefficients. So, by knowing the equilibrium constant for the reaction and the concentrations of the species you can tell in what direction the reaction is going to proceed.


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Which chemicals could be used to dissolve Polybutylene Terephthalate (PBT), but not dissolve brass?
This is for a recycling project, I am looking to recover brass pins from electronic connectors made of glass reinforced Polybutylene Terephthalate (PBT).
The objective is to find a chemical which would be able to dissolve Polybutylene Terephthalate (or PBT, a common type of plastic) so that i can separate the brass pieces inside, without dissolving them too. I think that if it dissolves PBT the glass reinforcement will become dust so it won't be an issue.
So far my search has been unsuccessful, either it was acids which were too strong and would also dissolved] the brass, or common solvent which had no effect on the PBT (tested Acetone). Even 30% Caustic soda doesn't attack the PBT.
I found a method with high temperature methanol under pressure into a closed stainless steel shape, but it's way too complex for me to make. Similarly chemicals which are too cancerogenous are not worth it (at least given that i am not a specialist capable of handling the most dangerous ones).
If anyone has an answer or some ideas i'd love to hear it!
Thanks a lot!

PBT is a thermoplastic and its melting point is only 223 °C. Depending on its composition, brass doesn't melt until 900 °C or so. High enough heating may change the structural properties of the brass, but perhaps there is a temperature to which you can soften the plastic enough to detach the pins without damaging them.
Unfortunately, brass is easily attacked by many chemicals and PBT is resistant to many solvents, but something like hot DMSO or N-methyl-2-pyrrolidone may swell the polymer enough to separate the parts.

Thanks! I'll try these 3 possibilities and tell you how it worked.


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Elasticity of latex
Latex is a natural resource and the elasticity can be increased by heating it with sulfur. My question is why they are have different elasticity than that of latex? I mean like rubber gloves have a good elasticity while rubber bands not really elastic. Is it cause by the amount of sulfur that we heat?

The 'elasticity' of a material is property which describes how it changes shape in response to a tensile stress. Specifically, when a material is placed under a tensile stress, it experiences some sort of elongation (strain). If removing the stress results in a return to the original size, the material is described as 'elastic' over this range of stress (note: under sufficiently high stress, all materials will eventually 'yield' to permanent deformation and ultimately 'break').
In 'engineering materials terms' the ratio of tensile stress (in MPa) to strain (in mm) within its elastic region is called the 'modulus of elasticity' of the material (aka 'elastic modulus' or 'Young's modulus').
In the case of latex rubber, it has an elastic modulus of about 0.1 GPa.
Natural rubber is derived from a milky suspension found in the milky sap of some plants such as the hevea tree. The sap which contains a polymer of isoprene called 2-methyl-1,3-butadiene, as well as a small percentage of proteins and acetone-soluble materials such as resins, fatty acids, and sterols.
Most of the isoprene consists of the long-chained stereoisomer, cis-1,4-polyisoprene.

The 'cis-' indicates that substituent groups are on the same side of the C=C double bond, as shown above. The high cis- content (~99%) of natural rubber latex is what gives its elastic properties. Natural rubber in this form is a thermoplastic, that is, it is soft at higher temperatures but becomes hard and brittle when cooled.
Vulcanization involves heating the natural rubber with sulfur at $140-180~\mathrm{^\circ C}$ which encourages the formation of cross-links between the long-chained polymers, resulting in a 'thermosetting' elastopolymer. Other additives are also sometimes used, such as N-phenyl-P-naphthylamine antioxidants to reduce deterioration by atmospheric oxygen or ozone, or 'accelerators' such  2-mercaptohenzothiazole, used to speed up the reaction or allow it to occur at lower temperatures. Reinforcing agents such as carbon or silica can also be introduced to added the 'stiffness' and tensile strength as well as resistance to abrasion. Softeners such as fatty acids, wood resins, pine oils or coal tar byproducts are also sometimes added. Finally, colouring agents and fillers (barium sulfide, calcium carbonate) are also sometimes used for aesthetic or to reduce the cost, but sometimes at the expense of reduced strength.
So by combining small amounts of other chemicals to latex, chemists are able to adjust the properties of the rubber in subtle ways to achieve the most desirable mechanical characteristics.


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Hypothetical: What happens to water as pressure increases to infinity?
I've asked a similar question here but the answer given shows the behaviour of water under general conditions.
I'd like to know what the behaviour of water is like as pressures increase towards infinity without being able to escape it's confinement.. i.e. a ball of water at the core of a galactic mass.. maybe this question is more for theoretical physics since we can't really measure or experiment?

Nuclear fusion?

Ron's answer gives us a good idea of what might happen in terms of "normal" chemistry, but if you really mean "increases to infinity", some very strange stuff happens. Suffice it to say, it doesn't stay water after a certain point. The intense temperatures created by the compression will cause the water to break apart, eventually no longer even having oxygen atoms due to nuclear reactions. Because we're talking about an externally applied pressure, the Chandrasekhar limit doesn't apply, so there is a point at which electrons and protons combine (when the electron degeneracy pressure is overcome) and a mass of neutrons remains. Neutrons themselves also have a degeneracy pressure (though we don't have good models to predict the exact pressure that has to be overcome). From here, we don't know what happens with as much certainty, but the formation of quark matter has been predicted.
Eventually, we reach a singularity. We can think of this as all the matter we had before being compressed into an infinitesimal volume with infinite density and our applied pressure ceases to mean anything. If we started with enough water, this would behave much like any other black hole, though micro black holes are hypothesized to have some special properties.

Great answer, Michael. I see how the lattice of water rearranges itself up to a point and wonder if that process continues through fusion to ever-denser lattice or would you think water will never separate under these conditions and remain H2O until some of the H2O in the middle begins to shmush into a clump of dense neutrons? In your last sentence 'If we started with enough water..', how much water might be enough to kick-off a singularity? Thx

When you actually reach singularity, there is no pressure anymore (such physical quantities are undefined there). But approaching singularity we have +∞ limit for pressure, and when gravitation became strong enough an external (applied) pressure becomes irrelevant, right.

@irth It wouldn't take much to actually get some kind of singularity; basically anything over the Planck mass (~22 µg). Very small black holes are thought to evaporate very quickly from Hawking radiation, though, which is why I distinguished them. As for when it stops being water, certainly once the nuclear electrostatic repulsion is overcome, fusion will occur, which should happen well before we're converting things to neutrons. (in white dwarfs, it takes the gravitational force of ~1.4 stellar masses to overcome this)

@MichaelD.M.Dryden, thanks for extrapolating. Sounds like singularities could be so common that they could be at the centre of any mass, be that atom or galaxy. wild to ponder. just to clear up this question tho, will water fuse with other water before hydrogen and oxygen split in this quesiton's conditions?

The only reason we can think of micro black holes in this thought experiment is because we're talking about a magical external pressure. There's no reason to expect a singularity would occur at the centre of just any mass in particular. As for whether fusion would occur before water breaks into individual atoms, I don't know. Pressure presumably has some effect on water's thermal decomposition, but I don't have enough information to speculate. Whether the heat produced by the compression is removed or not probably makes a big difference.

@MichaelD.M.Dryden water becomes superionic much earlier http://chemistry.stackexchange.com/questions/26863/what-is-the-meaning-of-superionic

The phase diagram for water (reported in your link and reproduced below) is a good starting point.

The diagram shows us that at pressures around 1 terapascal (about 10 million atmospheres) ice is a solid, at least up to 400 C. It has been predicted (reference_1, reference_2) that at higher pressures, somewhere between 1.5 and 6 tera pascals, solid ice will undergo an insulator to metal transition and display properties typically associated with metals (band structure, electrical conduction, etc.). That's around 15-60 million atmospheres.

Would you happen to know whether all materials eventually undergo metallization at sufficiently large pressures? It seems to be a universal feature, and it makes sense (core electron orbitals are forced to interact and overlap due to proximity, creating lots of overlapping electron bands), but I've never seen that mentioned explicitly.

@NicolauSakerNeto  Interesting question. I don't know the answer but I poked around and found this, Under sufficiently high pressures, just over half of the nonmetals, starting with phosphorus at 1.7 GPa,[92] have been observed to form metallic allotropes. Maybe at even higher pressures the other non-metals would also show metallic behavior.  I see that they predict helium would be metallic on massive planets.  If helium can be metallic, then wouldn't you expect everything to be metallic at high enough pressure, like you suggest?

Any idea what kind of temperatures would exist between 1.5 and 6 tera pascals of pure water. Trying to understand if H2O would still possess movement.

@irth  The above chart suggests that all temperatures would be accessible at 1 TPa, so my guess would be the same for somewhat higher pressures if adequate temperature control  is available.

Reading up on the research, at those high pressures, actual temperature stops playing a role in increased pressure is found through increasing the speeds of molecules, this makes the molecules more dense and increases pressure.
Another interesting topic on the subject is triple points. I understand most elements and molecules have various triple points and wonder if the chart above is accurate in showing only one for water at 0 celcius.

Am I the only one who noticed that ice turns in liquid at about 500kPa? Or do you guys not find it strange?

@TomášZato I don't follow. It turns into ice a bit below 1 kPa

@ron I meant 500MPa. Here's the enlarged picture: Water around 500MPA seems to be liquid even at -10°C (263K)

For example here ICE VII - phase of water on Super-Earthes - how behaves itself water under big pressure.

Ice VII is a cubic crystalline form of ice. It can be formed from
  liquid water above 3 GPa by lowering its temperature to room
  temperature, or by decompressing (D2O) ice VI below 95 K.
  ©Wikipedia

This could be equivalent to the compression generated by a black hole, hopelessly water as any other fluid is compressed to a high density solid state


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How to make permanent ink that appears after some time?
I'd like to ask,  how to make permanent ink, that will be invisible when used, but after like 30 minutes it will appear (preferably dark color)? Thanks.

Depending on what you want to do (what do you want to do?) you might be looking for gall ink.  Briefly, the chemistry works like this: the recipe consists of soluble polyphenols (typically gallic acid) and a Fe(II) salt.  The preparation must be stored in an air-tight container and even then decomposes over time.  As soon as it is applied to paper and exposed to air, the Fe(II) is oxidized to Fe(III), which forms insoluble deeply coloured complexes with your polyphenol.  The permanent colour takes about a day to develop.  That's why often a non-permanent dye is added so people can see what they are writing.

You are looking for a Vat dye.  Speed of color appearance may vary, so further research according to your requirement is needed.

I do not think there is a precise answer to your question, but I would use some form of pigment/dye in a precursor state which is able to react (slowly) with air, either oxygen or water vapour, to produce the coloured form of the pigment/dye. Given the type of reaction, especially if it is a form of oxidation, this could be irreversible, therefore making the ink permanent.

Thanks, but I'm a total beginner, could anyone please make a bit more detailed "tutorial" for me? (Yes, I tried googling, but I didn't find anything)

I'm sorry but I do not think that's possible. It's not that easy. It'd require at least some research to find candidates, experiments to assess their reactivity and suitability, etc.

Well, It is possible, because friend of mine uses this kind of ink, that appears after about 30 minutes and has dark brown color, but he doesn't want to tell me the recipe...

Please re-read my answer. I never said it's not possible. I just said it's not that straightforward going from a “request” to an actual formulation. Unless something already exists, as noted above.

Sure, sorry.. Also I have just found this experiment, where you mix methylene blue with something so it loses it's color and after exposure to oxygen, it regains it. Could that be used for this purpose?

Methylene blue is usually used as a redox indicator as it turns colourless when exposed to a reducing agent. So you would need something that is turned into a reducing agent by oxygen (which I am not sure about).

This (http://www.nuffieldfoundation.org/practical-chemistry/blue-bottle-experiment) is the experiment... Do you think, if I put the solution in a felt marker and then wrote something with it, it would appear after few minutes?


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Nomenclature for different types of ice
How many types of ice exist and what is their nomenclature? I know that the ice on Earth is $I_h$, but why so?

You can have a look at the phase diagram of water provided in this answer. There you can see which types of ice exist under different conditions.

The first thing to note about ice water is:
All ice is not crystalline and not all crystalline water is ice
There appear to be fifteen crystalline phases of ice, named 'Ice 1' to 'Ice 15' using Roman numerals $I$ to $XV$. There also appears to be three 'polyamorphic' states:.
$LDA$ : Low Density Amorphous
$HDA$ : High Density Amorphous
$VHDA$ : Very High Density Amorphous
All crystalline ice is tetrahedrically coordinated (where the oxygen atoms are in fixed positions relative to each other but the hydrogen atoms may or may not be disordered but obeying the 'ice rules'), and the polymorphs differ primarily in H-bond angles and next-nearest neighbour separations.
Ice I encompasses all ice found on naturally on earth (in a stable form).
This is usually split into two phases called $I_h$ and $I_c$.
Hexagonal ice (ice $I_h$) is 'ordinary ice', the form of all natural snow and ice on Earth. This is evidenced in the six-fold symmetry in ice crystals grown from water vapor (that is, snow flakes).
Cubic ice (ice $I_c$) is a solid form of water that can form in high clouds. Ice Ic is metastable, and irreversibly converts to ice $I_h$. Both forms contain similar layers of hexagonal rings, but differs in the way these layers are connected.
For more information on their structure and properties of ice, check out 'the Many Phases of Ice'.
Martin Chaplin maintains a fairly thorough and well referenced online resource on water, which includes a comprehensive section on the ice phases.
Finally, Google Books has the definitive 'the Physics of Ice' by Petrenko & Whitworth available online.

I see there few forms of ice for each amorphous state, so there are even more than 20 forms of ice, right?

That depends on how you want to distinguish one form of ice from another (crystal structure, symmetry or density) and on how 'stable' (or 'metastable') the form in question is. The resources described at the end of my answer give more details.

There are different types of ice, depending on its cristalline structure.
My guess would be that Ice $I_h$ and $I_c$ are named directly after their structure, which is respectively hexagonal and cubic.
The numbering, which makes Ice I, Ice II, Ice III and so on till Ice XV, seems to be arbitrary, probably ordered in sequence as they have been discovered. This also tells you about the number of different types of ice known: 16 (or 19 if we considering amorphous ice in its three forms).

Isn't it 19 then?

Yes, sorry. Miscalculation. :)


text:
Any home made tricks to fill up a baloon that floats?
These days there are a number of fancy balloons available, some of them are very beautiful. As always, they loose their gas overtime and don't rise anymore.
Some of those new type balloons appear to have been made of plastic and are quite easily re-usable but then you can't be going to the balloon guy every time you want a refill.
Are there any home made techniques that can let you fill those up (so they float again) and which generate a gas which is safe to handle at home without any protective gear?
P.S: This question is not motivated by any urge to save extra money, it's just that I find so many of them lying at home useless when they can be filled up again. Looking for something that's simple to make if possible.

There are some options, not all easily available or safe: Helium is the most used and it most certainly makes them float, being safe at the safe time. However, it's not an home-made trick. Methane is more easily available (in many cities, methane is piped into homes for domestic heating and cooking purposes) but it's very unsafe for such an use, as it could explode. Hot air is also used (think of hot-air balloons) but I wouldn't know how to make it last for a long time as the air would cool down to room temperature very quickly. Lastly, there's hydrogen which could also be produced with some reagents available at home, but it's as dangerous as methane.

I styled with a line over the two solutions of methane and hydrogen as a “disclaimer”.

Thanks @entropid. That's a bummer but thank you for your safety concerns :)

The only think I've found is how to fill them with carbon dioxide which doesn't definitely float. You can still produce hydrogen from simple chemicals from home or use methane but you'd need to be very careful. :(


text:
Do chemicals have an expiry date?
A bonus question would be, if so, why? Is it because it decomposes and hence the expiry date is calculated out?

Whether manufacturers put one on is a separate question to whether all chemicals will decay in some way when stored. Some will expire quickly, others slowly and other not at all. And some will last indefinitely if stored appropriately but only for a short time if not. There is no general answer.

They do indeed. It could be for different reasons:

A chemical might undergo chemical changes due to different storage conditions: it could react with other chemicals present in the air or in the environment. This could change its nature (reaction to create other compounds), properties and/or purity.
A chemical might undergo physical changes as well: the temperature and pressure vary, as well as other conditions like humidity, light, electromagnetic fields. This could change it as stated above.
Certain chemicals are simply unstable and decay or degradate over time to something else, without the need of externals input / changes. This could affect the chemical as stated above.
Certain chemical suffers transpiration losses.
Human factors, like contaminations, spillings, loose caps, etc.

However, it might be noticed that these reasons could be divided to produce two different dates: shelf life vs expiration date.

And even if chemicals don't have an expiry date the manufacturer still prints one on the box.  That's because the date serves as the batch number.

@AbelFriedman so the expiry date its more for stock taking than anything else?

Not really, it depends on the nature of the chemical.

You'd think if it's sodium chloride or suchlike (it does have an expiry date, do check!), it's more to be able to recall a bad batch.

Even alcohol has expiry data... Anyway, it is a bad practice to keep old chemicals around ad infinitum. I happened to work in a lab with plenty of ancient chemicals around. Even with the stable one after a while the labels peels off, the lid get stuck, etc..

I found an old bottle of magnesium chloride in one of the places I worked. The problem was that it had absorbed so much water from the air, it was all in solution when I found it.

Nice one. Cite some sources regarding your answers, which will be more helpful @entropid


text:
Electron shell, how do they know it?
I started to learn Chemistry with something like Hydrogen having 1 electron in the first shell, Oxygen having 2 electron in the first shell and 6 electron in the second shell... But I don't really understand how scientists know it? How do they know there are how many shells and how many electron per shell? Is that something visible? Or just an imagination theory or they using a method to measure?

Electrons are not actually separated in shells. That is a simplified model to explain the atom's structure when people first start approaching Chemistry or Physics.
In reality, electrons' movements around the nucleus are governed by complex equations (some of them not even resolvable so approximations are needed) to form what is called an electron cloud. This cloud is confined in proximity of the nucleus by the electromagnetic force. Every electron is defined by its wave equation, which tells us the probability that the electron appears to be at a particular location when its position is measured. These equations allows us to define regions in space where the electron spend the most part of its time around the nucleus, and they can be visualised as atomic orbitals, as shown below. Orbitals can have one or more ring or node structures, and they differ from each other in size, shape and orientation.

Each orbital (which is defined by a different equation) has different associated parameters, called quantum numbers, and a particular energy. These properties allow us to divide them into levels and sublevels, to form a structure that resembles multiple shells one inside each other. But as you can see from what I explained the structure is in fact far more complex.
This knowledge has been gained over time both from the mathematical and physical analysis of different equations (like Schröndiger equation) and by experimental data collected by the means of different techniques. Starting from Thomson and Rutherford the model of the atomic structure has seen many changes (Bohr, Moseley, Van der Broek, de Broglie, etc.) and it has come closer to a realistic representation of the actual atom. Take a look at their experiments to understand how it has been studied. However, they often involve particle physics and are not really straightforward to understand.

"This knowledge has been gaining over time both from the mathematical and physical analysis of different equations (like Schröndiger equation) and by experimental data collected by the means of different techniques.". Can you give me some example of these techniques ? I really want to focus on which techniques and how they knowledge it.

As I said above, you should look at their experiments. Just Google any of the aforementioned scientists plus ‘experiment’ (e.g. Rutherford's experiment, https://www.google.co.uk/search?q=rutherford+experiment&rlz=1CDGOYI_enGB608GB608&oq=rutherford+ex&aqs=chrome.0.0j69i57j0l2&sourceid=chrome-mobile&espv=1&ie=UTF-8&hl=en-GB) and you'll find more on these techniques. They're mostly single and specifically designed experiments rather than particular instruments used (however, nowadays some of them could be used to have a better understanding of the theory).

One way to think of this is to think of the total number of bonds that atoms can make. You can then use experimentation to look at the structure of the compounds, and that will tell you a lot about how the electrons are configured around the nucleus. For instance, Carbon compounds tend to take on tetrahedral shape if the bonds are saturated (single covalent), this means that carbon bonds most favourably by forming 4 bonds. If the moieties that are bound are all the same such as hydrogen, the angle of the bonds should be the same, and using x-ray crystallography for instance shows us that this is the case. We understand chemical bonding quite well, and we can deduce that each hydrogen, donates its single valence electron to a single valence electron from the carbon to form a covalent bond, the total number of valence electrons for carbon must therefore be 4. This is because the 2p shell is higher energy than the 2s shell and only the 2p electrons are free to bond. (This isn't really the case, the bonds are hybrid sp bonds but I won't get in to that right now)


text:
How can I find the stoichiometry of a reaction from the volumes of the reactants and products given?

I have volume $V$ of gas $\ce{A}$ and an equal volume $V$ of gas $\ce{B}$ at room temperature and pressure. I mix, ignite, and find I have a gas (mixture of $\ce{A}$ and product $\ce{C}$) at room pressure, $325~ \mathrm{^\circ C}$, and total volume $3V$. What was the stoichiometry of the reaction?
  \begin{align}
 \text{a)}&&\ce{2A + B &-> C}\\
 \text{b)}&&\ce{A + 2B &-> 2C}\\
 \text{c)}&&\ce{2A + B &-> 2C}\\
 \text{d)}&&\ce{A – 2B &-> C}\\
\end{align}

The answer is $\text{b)}$ but I can't seem to get the logic.
$V$ vol of $\ce{A}$ plus $V$ vol of $\ce{B}$ reacts to give $2V$ vol of $\ce{C}$. How did we find stoichiometry from this? What about some unreacted $\ce{A}$ ?

In a problem like this I think that the best way to answer the question is to set up the equations and try all the answers. However, in this case we can infer from the text that $B$ need to be the limiting reacting, as all of it reacts but some of $A$ is left unreacted. The only answers providing this are answer b and answer d. However, the latter contains a minus sign which doesn't makes sense chemically, so let's focus on the former.
We can compute the moles involved in the reaction using the ideal gas equation $PV = nRT$. Let's assume the volume is $1 \mathrm{L}$, we can calculate the amount of moles for both $A$ and $B$ (as they have the same volume and are present in the same conditions):
$$n = \dfrac{PV}{RT} = 40.88 \mathrm{mmol}$$
If the stoichiometry is the one proposed in answer b ($\ce{A + 2B -> 2C}$)
then the whole quantitity of $B$ will react with half the quantity of $A$ to give the same amount of moles of compound $C$, which will be $n = 40.88 \mathrm{mmol}$. Half of $A$ will be left unreacted, which is $n = 20.44 \mathrm{mmol}$. Using the aforementioned equation we can now compute their volumes:
$$V = \dfrac{nRT}{P}$$
Which is $V_a = 1 \mathrm{L}$ and $V_c = 2 \mathrm{L}$, which in total is $V_{tot} = 3 \mathrm{L}$, exactly three times more than the original volume.

If you're referring to the different mole value, it's just a typo. It's $n = 40.88$ in the product too.

How did you know $ n=40.88$ in the product too ? *

Also there we started with V vol of A and V vol of B, so how can $final volume = 3V $ $>$ $2V$ ?

It's basic stoichiometry: the stoichiometric coefficients can also be considered as the number of moles reacting. In addition, the volume changes also because the reaction conditions (the temperature) has increased.

If we calculate $V(b)$ = $V$(a)$ = 1 L , with each   n=$40.88$ And $V(c)$=$2$L$. So unless whole of A reacts with B, how can we get 2L of C ?

Please re-read my answer. The passages are explained. 40.88 mol of $C$ are produced, which converted using the last equation in the given conditions gives 2L.

I agree. But how can 0.5L of A , if only half of A reacted, and 1 L of B give 2L of product C ?

Because you don't consider the volumes within reactions. The volumes depends on multiple factors like pressure and temperature. You should consider only the amounts of reagents, using the moles, for stoichiometric computations. Then from those derive the volume or mass of your compound. Using the volumes along with the stoichiometric coefficients is wrong and will lead to non-sensical results.

Ok. I think I got it.


text:
Cathode region in corrosion
The anode region is on the iron where the following reaction occurs:
$$\ce{Fe(s) -> Fe^2+(aq) + 2e-}$$
and at the cathode region, the following reaction occurs
$$\ce{O2(g) + 4H+(aq) + 4e- -> 2H2O(l)}$$
However, the textbook does not explicitly state what the cathode region is. Can someone provide an answer?

I would assume that the cathode region is a source of $\ce{O2(g)}$, which is probably the air in contact with the iron. Therefore the reacting should occur at the interface iron/air.

There's a pretty good diagram of the process here. Where the reduction reaction actually occurs basically depends on where the oxygen can reach the metal.


text:
Is there a material that reduces copper melting point and completely dissolves?
Is there any material that you mix/alloy with copper that

will result in something that behaves similar to plastic (that is at room temperature it is solid, at maybe 500K or similar it can be easily formed/3D printed)
When heated even more (700K?) it will make the the included copper melt, and dissolves itself, leaving only solid copper.

Is this even theoretically possible? The main goal is to significantly reduce the temperatures involved, 700K is a ballpark number here. Also it doesn't need to be very thick if this makes things easier.

When heated even more (700K?) it will make the the included copper melt, and dissolves itself, leaving only solid copper. You may be able to make an alloy with a melting point below 500K, but you cannot remove the other metals simply by applying heat. Carbon compounds are burn away above 700K, which is why the plastic binders work.

@LDC3: The problem with those though seems to be that what is left isn't solid copper, is it?

That's what I implied, alloys are not going to give you the result that you specified. The plastic binder does, since the carbon compounds are burned away.

The usual metal 3D-printing processes usually involve metal powder and a plastic binder.  After printing, the green part is sintered.  See this.

I know that very well, but I wasn't asking how it is currently being done, I was merely exploring (theoretical) alternative methods.

You are asking for some material/process that can (a) make some metal powder hold shape and (b) upon heating, goes away.  A plastic binder that burns off while the green piece is sintering together fits the requirement rather nicely.  I can't think of anything that dissolves in the metallic phase to reduce processing temperature that subsequently evaporates because generally speaking, only metals dissolve in other metals.  There isn't much available with boiling point below the melting point of copper, except, of course, mercury.

Maybe the question isn't clear then, but the point is to do this at significantly lower temperatures than copper melting point. Ideally at around 700K.

There has been lots of good work of late in using nanoparticles to reduce the melting point in the direct write process. More classical metallurgy (using eutectics and whatnot) can not accomplish what you want.


text:
Procedures in experimental chemistry
An experiment was carried out involving $\alpha$-bromoheptaldehyde dimethyl acetal. The experimental details are given below.

A solution of enol acetate  (15.6 g, 17.7 mL) in chloroform (20 mL) was cooled
  to 0°C. A solution of bromine (16 g, 5.1 mL) in chloroform (5 mL) was added
  slowly with stirring with the rate of addition being controlled to keep the
  reaction temperature below 10°C. After the completion of addition, anhydrous
  methanol (60 mL) was added to the mixture and it was stirred for 48 hours.
  The mixture was diluted with water (200 mL) and ether (50 mL). The organic
  layer was separated, washed with water (100 mL) and with 5% aqueous
  sodium carbonate (100 mL). The organic layer was dried using sodium sulfate and
  filtered. The solvents were removed by distillation under reduced pressure.
  The residual oil was distilled under reduced pressure in the presence of a
  small amount of sodium carbonate. The fraction boiling at about 118°C at 17
  mmHg was collected to give bromo acetal as a colourless oil (19.5 g).
(a) Do you think that this reaction should be done using a drying tube, under
  nitrogen or open to the air? Explain your answer.
(b) Without doing TLC or any other form of spectroscopy or chromatography,
  how would you be able to tell if sufficient Bromine had been added?
(c) Will the organic layer be the upper or the lower layer? How can you
  quickly check this experimentally?
(d) Sodium carbonate is added just before the final distillation. Why?
(e) The procedure calls for removal of the organic solvents by “distillation
  under reduced pressure”. Which one piece of laboratory apparatus would you
  use for this?
(f) Suggest two serious and different hazards associated with this experiment.

My attempt
(a) It can be done in drying tube, as the reactant have to be prevented from coming into contact with water vapor from the air as the water vapor could potentially react with the reactant.
(b) By observing the colour of the reactant mixture. Since bromine is brown in colour, Once sufficient bromine has been added, there will then be excess bromine if bromine was continued to be added, hence the reactant will turn brown once enough bromine is added.
(c) The organic layer will be at the bottom. Since it is know that the density of the organic layer is more that that of the ether.The density of the organic layer can be measured and compared to that of the ether.
(d) To remove any excess acid that might be present in the mixture.
(e) Rotatory evaporator
(f) Bromine is toxic if inhaled, while methanol is a carcinogen
I am unsure of my answers though. Especially part (c) as I can't decide whether the aqueous or organic layer will be at the top. Could anyone explain this.

I believe that when you say "a solution of enol acetate" you are refering to a solution of the corresponding enol-acetal of hetanal? You should use full names of the compounds used, if not, it might be confusing.
My opinions

(a) Do you think that this reaction should be done using a drying tube, under nitrogen or open to the air? Explain your answer.
  It can be done in drying tube, as the reactant have to be prevented from coming into contact with water vapor from the air as the water vapor could potentially react with the reactant.

I don't see a reason you would need any inert conditions for this experiment. I think you will be fine if you perform it open to the air.

(b) Without doing TLC or any other form of spectroscopy or chromatography, how would you be able to tell if sufficient Bromine had been added?
  (b) By observing the colour of the reactant mixture. Since bromine is brown in colour, Once sufficient bromine has been added, there will then be excess bromine if bromine was continued to be added, hence the reactant will turn brown once enough bromine is added.

Indeed. The end point of the reaction is reached when the calculated amount is absorbed and the bromine is no longer decolorized.

(c) Will the organic layer be the upper or the lower layer? How can you quickly check this experimentally?
  The organic layer will be at the bottom. Since it is how that the density of the organic layer is more that that of the ether.The density of the organic layer can be measured and compared to that of the ether.

This is wrong. The organic layer IS the one which is will be a mixture formed by chloroform, unreacted enol acetate and ether. The other phase will be the one composed mainly by water and methanol.
Methanol+water solution will probably be less dense than chloroform+ether+enolacetate solution, and therefore, organic layer will be the upper one. I cannot tell for sure, since you have mixtures in high proportions of compounds with quite different densities, but and easy way to tell is:
- Separate both phases using an extraction funnel. Put some tap water on the separated phase you suspect is the aqueous phase. It dissolves? That is the aqueous phase. It does not dissolve and forms a two-phased system? It was the organic phase, extract it again.
You can double check adding water to the other phase you extracted, and see that the opposite effect should be observed.

(d) Sodium carbonate is added just before the final distillation. Why?
  To remove any excess acid that might be present in the mixture.

Yeah. The acetal must be free of acid; otherwise decomposition takes place during distillation.

(e) The procedure calls for removal of the organic solvents by “distillation under reduced pressure”. Which one piece of laboratory apparatus would you use for this?
  Rotatory evaporator

Correct. Rotovap is your guy.

(f) Suggest two serious and different hazards associated with this experiment.
  Bromine is toxic if inhaled, Methanol is a carcinogen

Bromine will definitely be your top priority concern. Even more if you have to prepare the solution from pure bromine. Also, it has a very strong unpleasant smell, which gives it the name it has (from Greek: βρῶμος, brómos, meaning "strong-smelling" or "stench")
It is corrosive very toxic, volatile (fumes a lot, MUST be handled on a fumehood).
Chloroform is also toxic; Heptanal is volatile and flammable.
Methanol, even though you should not drink it wouldn't be of an important concern.
To sum up, some of the reactants are of unpleasant odor or have lachrymatory effects it is best to use a good hood.
And also a reference: Organic Syntheses, Coll. Vol. 3, p.127 (1955); Vol. 29, p.14 (1949).

I have removed the link http://orgsyn.cambridgesoft.com/orgsyn/prep.asp?prep=cv3p0127 as it is dead and web archive doesn't have a copy of it archived.


text:
Does inversion of configuration necessitate reversed optical rotation?
Say, optically pure 2-bromobutane undergoes $\mathrm{S_N2}$ reaction with an alkali to form 2-butanol. The configuration about chiral carbon will now be inverted, but is it necessary that the sign of optical rotation will be opposite to that of substrate? i.e. if optical rotation of optically pure 2-bromobutane is $+x^\circ$, will optical rotation of the product be $- y^\circ$?

if optical rotation of optically pure 2-bromobutane is some (+X)
  degrees, will optical rotation of the product be some (-Y) degrees?

No, there is no correlation between the configuration at a chiral carbon and the sign of optical rotation.
For example (from the Aldrich catalogue), the following chiral bromo compound, (R)-(+)-2-Bromo-3-methylbutyric acid,

has an optical rotation of [α]22/D +21°, c = 37 (c is the sample concentration in grams/100 ml solution) in benzene.  The inverted hydroxyl analogue, (S)-(+)-2-Hydroxy-3-methylbutyric acid,

also rotates light in the (+) direction.  Its optical rotation is [α]20/D +19°, c = 1 in chloroform.

Optical activity determination is only by experiment(polarimetric) , it's not necessary that a dextro form should give laevo form after inversion of configuration, product may have dextro or leavo rotation.

At least the first part is not completely true; optical rotation can be calculated. http://gaussian.com/polar/


text:
Good book for foundational chemistry
I need a good book that explains the basics of chemistry in a lucid and importantly, systematic manner. Most books on this subject are not comprehensive and assume that the reader is well acquainted with certain topics. Also, most chemistry books have references to later chapters. And later chapters, references to earlier ones. This makes it very confusing.
I would like a book that assumes (effectively) no prior knowledge and builds up the basics needed to understand chemistry (structural theory, acid-base theory, electrode potentials) to a level adequate to understand undergraduate level organic and inorganic chemistry.
The reason I need this book is because frequently, while studying inorganic and to a certain extent organic chemistry, I am unable to understand the rationale behind a certain reaction. And most textbooks do not take the effort of explaining it. So, chemistry, for me,  just reduces to memorizing a set of reactions. I think solid basics should remedy that.

This is one of those things that is going to hinge on personal preferences more than anything.  In addition to being opinion-based, it's also quite broad.  In these situations I always recommend that people go to a local university library and flip through a few textbooks to find one that they can learn from.  If you don't have access to a university library, your local library can usually request the books to be sent interlibrary loan.

For the concepts underlying General Chemistry find that Huheey's book gives very lucid explanations.  As far as Organic Chemistry goes, vol. A of Carey-Sundberg's opus might be what you are asking for.  What level of text are we looking at?

The level of text of would be typically undergraduate or even a bit lower.


text:
7,7-dichlorobicyclo[4.1.0]heptane NMR solving
i made a synthesis of this material in the lab and now i got the H-NMR specta , and by litreture i know that i got the right product.
im having trouble to match the hydrogens to the peaks,i tried to look at the molecule in it chair form, i know that the 2 hydrogens(A) attached to the cl cl group are in the highest ppm than the other 4 (B), but on the last 4 i got 2 different multiplets, which is a match to which hydrogens? i added the  spectrum

I would start by looking at the cis rather than trans isomer as the starting point for interpreting the NMR spectrum. It is much more likely to be the product you got from this reaction. Also, be sure to identify all the chemically distinct hydrogens in your structures.


text:
What's the difference between a nucleophile and a base?
Obviously it depends on the context whether you would call a particular species a nucleophile or a base but are the two terms largely synonymous or is there a difference?

Basicity is simply a special kind of nucleophilicity, namely towards protons. So, both terms are quite related but in organic chemistry the term nucleophilicity is usually used with respect to carbon. The reason why basicity and nucleophilicity correlate rather well but not completely comes from the differences between the electrophilic properties of carbon atoms and protons (using the HSAB principle one might describe the difference like this: ...

... protons are hard electrophiles and carbon centers are rather soft electrophiles, so hard nucleophiles are better bases than soft nucleophiles and soft nucleophiles react better with carbon).

The two are related, in that most nucleophiles are (Lewis) bases and vice versa. Some good nucleophiles are also strong bases, e.g. $\ce{HO-}$. However, a species can be a good nucleophile and a weak base, e.g. $\ce{I-}$; or a species can be a weak nucleophile and a strong base, e.g. $\ce{t-BuO-}$. How can we separate this behavior?

Nucleophilicity is a kinetic phenomenon.

Nucleophilicity is most often defined based on the relative rate of the reactions of nucleophiles with a standard substrate in a standard solvent.
For example, a standard reaction might look like:
$$\ce{CH3I ->[Nu-][H2O] CH3Nu}$$
The nucleophilicity will be related to the relative rate constant of reaction with the nucleophile (relative to the rate constant of the reaction with water $\equiv 1$).

Basicity is a thermodynamic phenomenon.

Basicity is based on the position of equilibrium:
$$\ce{B + HSol <=> BH+ + Sol-}$$

There is a difference indeed: basicity is a particular kind of nucleophilicity. A nucleophile is a chemical species that donates an electron pair to an electrophile. A nucleophile can also be called a base when this donation occurs towards a particular electrophile, which is an hydrogen ion (a proton).

So is it always the case that a strong base is a good nucleophile? e.g the hydroxide ion is strongly basic and it's a good nucleophile?

No, not really. Read the other answer for a more specific explanation of the phenomena.

$\ce{H2O}$ has a $\mathrm{p}K_\mathrm{a}$ of approximately 15, giving us $\ce{OH-}$ a reasonably strong base but also good nucleophile which you'll find competing between elimination and substitution reactions. Now $\ce{H2}$ has a pka of 30 which makes its conjugate base ($\ce{H-}$) an even stronger base. $\ce{CH4}$ which has a $\mathrm{p}K_\mathrm{a}$ of 50, and its conjugate base ($\ce{CH3-}$) which is an extremely strong base, in fact it's strong enough to deprotonate any acid below it.
Think of a strong base being equal to or stronger than $\ce{OH-}$. Furthermore, the higher the $\mathrm{p}K_\mathrm{a}$ of the acid the greater the strength of the conjugate base.

When the size of a negatively charged molecule is small then it behaves as a nucleophile, but when size of a negatively charged molecule is big then it behaves as a base.  This is because of crowding around the central atom of other reactant species.

What do you mean by crowding? Does it have anything to do with charge density?

The hydride ion is pretty small but it is a strong base. This seems like a major oversimplification.

It's the other way round actually...


text:
How would you find the mass in grams, if you only know the number of particles of a portion of the formula?
If you were given a question like below:

What is the mass in grams of a sample of $\ce{Fe2(SO4)3}$ that
  contains $3.59 \times 10^{23}$ sulfate ions, $\ce{SO4^{2−}}$ ? The molar mass
  of $\ce{Fe2(SO4)3}$ is $399.91 \, \text{g}/\text{mol}$.

How would you go about solving a question like that?
I'm confused about how to use only the quantity of sulfate ions to find the mass in grams of the whole sample, given it's grams per mole.

You have to guess, how much full formulas correspond to this number of formula portions. Obviously.

The formula $\ce{Fe2(SO4)3}$ tells us that 3 molecules of $\ce{SO4^{2-}}$ exist for every $\ce{Fe2(SO4)3}$ molecule.
$\frac{\ce{1 Fe2(SO4)3 molecule}}{\ce{3 SO4^{2-} molecule}} * 3.59*10^{23} \ce{SO4^{2-} molecule}$ can be used to find the number of molecules of ferric sulfate.
You can then use $\frac{1 \ce{mole}}{6.022*10^{23} \ce{molecule}}$ to find the moles of ferric sulfate.
You can then use the given information to solve for grams of ferric sulfate.
Write a comment if there's any ambiguity.

Basically the premise behind my approach is abusing conversion factors to get from atoms to grams. If you write everything out first it might be easier to see how the units cancel.


text:
Half cell method of voltage calculation in an electrochemical cell

A cell consists of a copper electrode $\ce{Cu}$, in contact with $1.800\cdot10^{-2}~\mathrm{M}$ cupric ion and a platinum electrode, $\ce{Pt}$, in contact with $1.400\cdot10^{-1}~\mathrm{M}$ ferrous ion and $3.700\cdot10^{-2}~\mathrm{M}$ ferric ion in $1~\mathrm{M}$ $\ce{HCl}$.
  What is the potential of cell?

I would have an idea of what to do with this question if it had a concentration for copper in the cupric half reaction; I would just use the nernst equation, rearranged to solve for $E$:
$$E = E^\circ - \frac{\mathrm{R}T}{n\mathrm{F}}\log Q$$
Then plug in the concentration values for the cupric half-reaction into $Q$ to get the $E$ value for this half reaction. I'm assuming that the half reaction taking place here is:
\begin{align}
\ce{Cu^{2+} + 2e^- &-> Cu^0} &E^\circ &= +0.337~\mathrm{V}
\end{align}
and the other half reaction taking place would be:
\begin{align}
\ce{Fe^{3+} + e^{-} &-> Fe^{2+}} &E^\circ &= +0.771~\mathrm{V}
\end{align}
Then, I would plug the concentration values, along with their standard potentials, into the rearranged nernst equation to get the $E$ values for each half, then subtract the larger value from the smaller. I already did this, assuming that the concentrations of $\ce{Cu^0}$ and $\ce{Cu^{2+}}$ were the same, giving a logarithm of $0$ and making $E = E^\circ$. The other equation remained the same. I got an answer of $0.399~\mathrm{V}$, which is wrong.
Also, I'm not sure how the concentration for $\ce{HCl}$ factors into this question, because the half reaction for $\ce{Fe}$ doesn't have any hydrogens in it.
Can anyone give me a hand with this one?

For the reaction quotient Q, solids are omitted. That would mean that the copper electrode concentration would be taken out of the equation.

Think about what state $\ce{Cu^0}$ is in. $\ce{Cu^2+}$ is dissolved in the solution, but is this true of the unionized copper? Try to write out the half-reactions with the physical states included. What activity (effective concentration) do solids have in an equilibrium expression?
Also, though there are some reactions that could happen involving the acid, and I don't know whether they actually contribute to the cell potential in this case, the way the question is written, it looks like the two half-cells you have are the significant ones.

Don't know why but you're just making some value errors, happens to everybody once in a while:
$$\ce{2Fe^{3+} + Cu -> 2Fe^{2+} +Cu^{2+}}$$
$$\mathcal{E}\approx(0.771-0.337)-\underbrace{\frac{0.059}{2}}_{\large\frac{\mathcal {RT}}{\mathcal F}\ln{10}\approx0.059,\nu=2}\cdot\log_{10}\underbrace{\overbrace{\frac{0.018\cdot0.14^2}{1\cdot0.037^2}}^{\ce{[Cu^2+][Fe^2+]^2}}}_{\ce{[Cu][Fe^3+]^2}}$$
So:
$$\mathcal{E}\approx0.451\cdot$$


text:
Find H3O+ of weak acid salt solution which compose of solute that can gain and give protons
Find $[\ce{H3O+}]$ of 0.100 M $\ce{NaHCO3}$ ($K_{\ce{H2CO3},1}=4.45×10^{-7}$, $K_{\ce{H2CO3},2}=4.7×10^{-11}$)
My work is: $$\ce{HCO3- + H2O<=>H2CO3 + OH-}$$
Initial: $[\ce{HCO3-}]$ = 0.100 M, $[\ce{H2CO3}]$ = 0, $[\ce{OH-}]$ = 0
Change: $[\ce{HCO3-}]$ = -X, $[\ce{H2CO3}]$ = +X, $[\ce{OH-}]$ = +X
Equilibrium: $[\ce{HCO3-}]$ = 0.100 - X, $[\ce{H2CO3}]$ = X, $[\ce{OH-}]$= X
$K_{b2}=\frac{Kw}{K_{a1}} = 2.25×10^{-8}$
$K_{b2} < 10^{-3}: 0.100 - x = x$
$K_{b2} = \frac{x^2}{0.100}$
$X = [\ce{OH-}] = 4.47×10^{-5}$ so $[\ce{H+}] = \frac{10^{-14}}{[\ce{OH-}]}$, $[\ce{H+}] = 2.11×10^{-10}$
And the second equation is
      $$\ce{HCO3- + H2O<=>CO2 + H3O+}$$
For this equation $K_{a2}=4.7×10^{-11}$
So when I compared $K$ value between both of these equation, I will get $K_{b2} >> K_{a2}$ (more than $10^3$ times). So I think increase $[\ce{H3O+}]$ which occur or $[\ce{CO2}]$ can be ignored. So my answer $[\ce{H3O+}]$ is equal to $2.11×10^{-10}$ M.
But the solution in my book is:
From the formula:
$[\ce{H3O+}]= \sqrt{\dfrac{K_1 K_2 × [\ce{NaHCO3}] + k_1 k_w}{[\ce{NaHCO3}] +K_1}}$
$[\ce{H3O+}]=4.6×10^{-9}$ (solution answer)
Do I approximately second equation false or $K$ value is false or another thing?

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The two equilibrium equations you need to worry about are:
$$\ce{H2CO3 + H2O <=> HCO3- + OH-}\ \ \ \ K_{a1} $$
$$\ce{HCO3- + H2O <=> CO3^{2-} + H3O+}\ \ \ \ K_{a2}$$


text:
How should the hydrated proton be represented in chemical equations?
I learned the equation
$$\ce{H2O + H+ -> H3O+}$$
And I know $\ce{H+}$ and $\ce{H3O+}$ really mean the same thing. But I am confused as to when I should use $\ce{H+}$ and when I should use $\ce{H3O+}$? Is any one of them a more accurate representation?

See http://chemistry.stackexchange.com/questions/487/stability-of-h%E2%81%BA-ion

In addition to entropid's answer, let's remember why we invoke the hydronium ion $\ce{H3O+}$ in the first place.
We use $\ce{H3O+}$ as a shorthand for $\ce{H+(aq)}$, which looks more like protonated water clusters of the generic formula $\ce{H+.(H2O)_{$n$}}\equiv \ce{H_{$2n+1$}O_{$n$}+}$.
Almost ten years ago, a very interesting paper appeared in Science that examined the structure of these clusters ($n=2,3,4,5,6,7,8$). The linked article has been made freely available by Science, but you need to register for a free account at sciencemag.org.
One of the key points of the article was the determination of the inner structure of these clusters. For $n=1$, you have $\ce{H3O+}$, the hydronium ion, which we are reasonably familiar with. When $n=2$, you have $\ce{H5O2+}$, which has a different structure: the proton is evenly shared between two water molecules: $\ce{[H2O\bond{...}H\bond{...}OH2]+}$. Higher order clusters have these two structures, termed the "Eigen" and "Zundel" ions, respectively, at their cores.
The study used a combination of theory (MP2/aug-cc-pDVZ) and experiment (photodissociation vibrational spectroscopy). They found the for $n=2,6,7,8$, the clusters have a Zundel core, while the Eigen core exists for $n=3,4,5$. Whoa! And it gets more complicated as you increase the number of water molecules in the clusters (might be behind a pay wall) - though it turns out the Eigen ion is slightly more common in larger clusters.
In aqueous solutions, we can have an extensive dynamic network of clusters always in flux. In which case, we have a mix of solvated Eigen $\ce{H3O+}$ ions representing protons closely associated with water molecules and Zundel $\ce{H5O2+}$ ions representing protons in transit.
So, we use $\ce{H3O+}$ to keep things simple, and also to satisfy the needs of Brønsted-Lowry acid-base theory, in which every acid-base reaction needs a proton-donor and a proton-acceptor. Thus, the "dissociation" of acids in water is not really a dissociation, but an ionization reaction - an acid-base reaction in which water is the base and hydronium is the conjugate acid.
$$\ce{HA + H2O <=> A- + H3O+}$$
Of course, it is only appropriate to use $\ce{H3O+}$ in aqueous solution. In other solvents, the "proton" has a different structure. In general, the proton is attached to a basic site on the solvent molecule. For example, in methanol:
$$\ce{HCl + CH3OH <=> Cl- + CH3OH2+}$$

Just putting it out there: Some newer research using IR spectroscopy and X-ray crystal structures of $\ce{H(CHB11I11).8H2O}$ indicates that the most important value of $n$ is 6. See JACS 2010, 132, 1484 and JACS 2009, 131, 17540; there's also a nice review of the topic here: Acc. Chem. Res. 2013, 46, 2567. Of course it invalidates absolutely nothing of what you have written.

There is no real difference between the two notations. As you know, $\ce{H+}$ cannot exist on its own in solution as it gets hydrated to $\ce{H3O+}$, but in most cases it can be used when writing equations or working with reactions. To my knowledge, $\ce{H3O+}$ is slightly preferred when dealing with acid/base reactions,  e.g.
$$\ce{H2O + HCl -> H3O+ + Cl-}$$
while $\ce{H+}$ is used more – for simplicity – with redox reactions, e.g.
$$\ce{Zn + 2H+ -> Zn^{2+} + H2}$$


text:
What is the effect of conjugation of a carbonyl group in IR spectroscopy?
Doing an exchange term so I've missed some classes that the others have attended, and unfortunately we're getting contradictionary explanations, so I thought I'd try and ask you.
Would be extremely thankful if someone could explain the effect of conjugation of a carbonyl group on the resonance frequency in IR spectroscopy.
Our 'assignment' is basically to figure out if it's supposed to be lower or higher than an unconjugated carbonyl.
Thanks in advance!

As a short answer: conjugation (with a $\ce{C=C}$ bond) moves the IR peak of a carbonyl group to lower wave numbers.

It would shift the IR peak to lower wavenumbers. You can rationalise this by understanding the bonding in the carbonyl group; in essence, by conjugating the carbonyl, the $\ce{C=O}$ bond is weakened, which shifts the IR peak down. You can think of it in terms of, if there is little or no conjugation, the electrons are firmly based in the $\ce{C=O}$ bond. As soon as we introduce some resonance structure, or conjugation, now those electrons can delocalise over 3-4 atoms or even more. This weakens the $\ce{C=O}$ bond, moving the IR peak down to lower wavenumbers.


text:
Sunscreen or sunblock
When I went to the supermarket, I noticed that some of the product say sunscreen and some of them say sunblock.
My question split into two part.
1) What are the different chemical substances use in both item? (In general)
2)What is more effective to prevert sunburn? (In general)

Giving advice on preventing sunburns may fall under health advice, which is not permitted on this site. For advice, you should ask a medical professional.
The following information is from The Melanoma Foundation.
What is the difference between sunscreen and sunblock?

Sunscreens can be classified into two major types: chemical and physical. Chemical sunscreens contain special ingredients that act as filters and reduce ultraviolet radiation penetration to the skin. These sunscreens often are colorless and maintain a thin visible film on the skin. These sunscreens usually contain UVB absorbing chemicals and more recently contain UVA absorbers as well.
Physical Sunscreens, most often referred to as sunblocks, are products containing ingredients such a titanium dioxide ($\ce{TiO2}$) and zinc oxide ($\ce{ZnO}$) which physically block ultraviolet radiation (UVR). Sunblocks provide broad protection against both UVB and UVA light. They can be cosmetically unacceptable to many people, because they are often messy, visible and do not easily wash off. However, some new zinc oxide products are available in brightly colored preparations which are popular with young people. The amount of sun protection these sunblocks provide, while potentially high, cannot be quantified in the same manner as sunscreen SPFs. Physical sunscreen is recommended for individuals who have unusual sensitivity to UVR. Most recently on the sun protection scene is sun-protective clothing designed to block UVA and UVB radiation. The effective SPF is greater that 30.

When should you use a sunscreen?

Sunscreens should be used daily if you are going to be in the sun for more than 20 minutes. Most people will receive this amount of sun exposure while performing routine activities. They can be applied under makeup. There are many cosmetic products available today that contain sunscreens for daily use because sun protection is the principal means of preventing premature aging and skin cancer. Sunscreen used on a regular basis actually allows some repair of damaged skin. Because the sun's reflective powers are great - 17 percent on sand and 80 percent on snow - don't reserve the use of these products for only sunny summer days. Even on a cloudy day 80 percent of the sun's ultraviolet rays pass through the clouds. Skiers beware, ultraviolet radiation increases 4 percent for every 1,000-foot increase in altitude.

To add to this last part, sunblock manufacturers usually include instructions with their products on the proper way to use them.

One thing to add, the U.S. Food and Drug Administration (FDA) recently changed the labeling requirements for sunscreens.  As part of it the word "sunblock" will no longer be allowed.  You'll still find some sunscreen's using the term "sunblock" and this is usually because they were manufactured before the labeling changes went into effect.
Going forward everything will be labeled "sunscreen" whether it is a mineral or chemical sunscreen.  So to tell in the future you'll need to look at the "Active Ingredients" in a sunscreen (the FDA requires sunscreens to list all their ingredients).  If you see only zinc and/or titanium there then it's a mineral sunscreen.  Any other ingredient listed under "Active Ingredients" is a chemical UV filter as currently 17 ingredients (2 minerals and 15 chemicals) are approved by the FDA as sunscreen filters.
Something you definitely want to look for on a sunscreen's label is the term "Broad Spectrum".  According to the new FDA regulations only sunscreen's that protect against both UVA and UVB rays can be labeled as such.
If you're interested, here's a bit more about the new sunscreen labeling laws from the Skin Cancer Foundation: "New Sunscreen Labeling Rules Issued by FDA".
One other interesting thing is that some UV filtering ingredients protect against the full UVA and UVB spectrum while others protect against only parts.  This is why some sunscreens use multiple active ingredients. If you're interested, we have bit more about that here "UVA, UVB and UVC Rays: What They Are and How Sunscreen Protects You".


text:
Some abnormalities in the Periodic Table
I observe some elements like $\ce{Cr}$ and $\ce{Cu}$. Some of them have strange oxidation numbers that don't really fit the purpose of making the element as stable as possible. Just Chromium, for example, it has one valence electron in 4s, the oxidation tells that it is gonna lose 2 electrons. Which obviously breaks the stability of half filled 3p orbitals.
If this can be explained by the fact that $\ce{Cr}$ has weak electronegativity, then what about $\ce{Cu}$? It is among the most electronegative transition metals, but still might lose two electrons despite having fully filled 4d orbitals.

While neither an answer nor a duplicate, the weirdness of chromium and copper have been previously addressed on this site: http://chemistry.stackexchange.com/questions/151/why-do-elements-in-columns-6-and-11-assume-abnormal-electron-configurations

Electronegativity is not a good concept to rely on for explaining ground-state electron distributions, due to the simple fact that electronegativity itself is not precisely and unambiguously defined, especially for $d$- and $f$-block elements. A good answer is quite a bit more involved, and even then only qualitative. See this previous question and the questions linked within for some necessary background. I'd be happy if someone can round it all up.

How do you know they won't reach stability? It's all to be blamed on the more complicated chemistry of d orbitals; the rules we learn about p block and s block elements don't hold there.


text:
Is is possible to do mass spectrometry without coupled liquid chromatography?
It's possible to identify a solution with several compounds* without the coupled liquid chromatography (LC) before to isolate the compounds?
On the university where I study, the LC is broken, so all we can do is inject the solution directly on the mass spectrometer. I wonder if I will be able to identify the metabolites? I don't need quantitative analysis, I just want to know what's in there.
(*) To be precise: A bacterial culture, that has gone under organic solid phase extraction, freeze dried and resuspended in methanol.

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The practical answer really depends on two things: do you know exactly what you're looking for and can your instrument do tandem MS?
If you know exactly which metabolites you're looking for and they're present in great enough abundance, it may be possible to resolve them from the the background, depending on the resolving power of the instrument and what else is in there (if you have an ICR instrument, it may not be a problem at all, but if it's just a single quadrupole, you may be out of luck).
If you know exactly which metabolites you're looking for (or know that they share some common structural feature that can be knocked off by CID or the like) and your instrument can do tandem MS, you're in a much better position. Tandem MS is often used to avoid needing to do a slow LC step as it's possible to identify two compounds that aren't resolved on a precursor ion scan by looking at their fragment ions: A messy primary ion scan can be dealt with if you know that your analyte has a [M-18] transition or whatever, by CID. Also, a given class of compounds often shares certain CID reactions, which can be picked up with a neutral loss scan.
If you have tandem, but aren't sure exactly what you want to find, it's a bit more work, but it just means you need to look at product ion scans of peaks that are ambiguous and try to work out what they are.

It all depends on the selectivity of your SPE for the class of compounds you're looking for.  If you have some confidence that the SPE has let only a few components of the bacterial lysate pass into the eluate, then you might see few enough m/z signals on the MS that qualitative analysis would be possible.  If your SPE is relatively unselective, though, you will probably have a dense array of peaks that won't tell you much.
It (probably?) can't hurt to try running the sample(s), though.  If you see relatively few peaks, then you could rerun the samples, spiked with known amounts of known compounds, to see whether:

A new peak shows up (the spiked chemical is absent in the sample);
An existing peak gets bigger (modest indication that the chemical is present); or,
No change is observed (the spiked chemical is absent in the sample or the spike was too small)

If you're looking for more than a few chemicals, this promises to be a pretty drawn-out process, with lots of samples pushed through the MS. Might still be faster than waiting for the LC, though!


text:
Why is it that every snowflake is unique?
It's pretty hard to believe in around 4.5 billion years of this planet's existence, no snowflake pattern was repeated, and that's not including potential earth-like planets.
Is there a proof to the structure or is the probability even more statistically insignificant than I thought, or what?

I am not sure on the statistics of the snowflakes but have heard that the (artificial) snow flakes are very hard to make and do not always have the beautiful symmetric structure the well know pictures have. Could not find a reference for that fast enough, sorry.

Remember that the expressions for permutations and combinations involve factorials, which for the number of water molecules in a snow flake becomes mind-boggling large really fast.

Combinatorics, one of the scarier branches of mathematics...

It turns out that 2 identical snowflakes have been observed, but...
Two Identical Snowflakes
Although when we think "snowflake" we usually picture an object with 6-fold radial symmetry, snowflakes actually come in many different shapes (reference 1).  Many sites report that in 1988 an NCAR researcher found 2 identical snowflakes of the hollow columnar variety (reference 2 - see "Uniqueness" section, reference 3).  This pattern is much simpler than the 6-fold radial design, so the odds of finding 2 identical flakes are vastly improved.
But
If you wanted to confine the question to snowflakes of the 6-fold radial variety, then an argument can be made that 2 identical snowflakes will not be found.

Due to the vapor pressure difference between ice and water, vapor is pushed from water towards the ice (hence, the ice grows) in an ice-water droplet falling through the atmosphere.  As it falls through different temperature zones, the rate of ice crystal growth will change leading to different dendritic patterns along the radial arms.  Further, due to perturbations caused by temperature and wind gradients across a single crystal, each radial arm will have a slightly different pattern.
The mathematical argument against identical (6-fold radial) snowflakes is, as Jon Custer suggested in his comment, based on factorial probabilities.  The argument goes something like this.  If you have 10 different pictures to put in an album you could arrange them in 10! (10 factorial) or 3,628,800 different ways.  If you look at a snowflake under magnification, there are easily over a hundred different features that are identifiable.  In this case there are over 100! or greater than $10^{158}$ possible arrangements of these identifiable features.  Estimates are that there are around $10^{80}$ atoms in the universe.  Each visible feature in a snowflake contains a large number of water molecules and an even larger number of atoms.  Hence, the number of possible arrangements is far, far greater than the number of available atoms, making it extremely improbable that a duplicate of a complex snowflake would be created.
If you're still not convinced, we can add another level of probabilistic protection.  Remember that there are isotopes of hydrogen and oxygen.  The probability of finding two 6-fold radial snowflakes with the same dendritic pattern and the same isotopic substitution pattern is even closer to zero.

Statistically, the probability of finding two 6-fold radial snowflakes is very close to zero, although not actually zero. We could say that one will almost surely never find identical snowflakes, but there is still a possibility of it happening. Zero-probability events


text:
Do water hybrid cars really exist?
I have heard a few claims about water-fueled cars. Being a student of Engineering I totally understand that water-fueled cars (using electrolysis to generate Hydrogen) are nothing more than a hoax, unless  you are giving electricity from an outer source.
However, I have heard quite knowledgeable people that water hybrid cars are quite a reality and you can save up to 60% on your fuel costs using some kind of kits. I have difficulty agreeing with this claim as well. I think inclusion of another step (hydrolysis) in the energy cycle of car power system will only make it less efficient. Reason being you can achieve at most as much energy from making a bond (burning Hydrogen) as much you spent on breaking it (Electrolysis).
Can someone please help me understand if I am missing some point? Or I am correct?

The idea of splitting water using electrolysis and pumping it into an engine's air intake has been around for a long time. The idea is that it's supposed to increase the efficiency of burning gasoline. However, even assuming that this is true (there are many reasons why it may not be—modern gas engines are in the range of 95–98% efficient at combustion), electrolysis is very inefficient, so I'm quite dubious of any potential gains.
Hunting around on the internet, it's obvious that a lot of people seem to think this works (and that people who say it doesn't are oil company shills or some such), but I wasn't able to find any compelling evidence to support it. How this kind of system interacts with a modern computer controlled engine may perhaps even make efficiency worse. There doesn't seem to be much academic research into it, though a Google Scholar search brings up a lot of kooky patents. This kind of thing is pretty difficult to test well—tests need to be performed under very well controlled conditions to get accurate results and is thus always subject to confirmation bias and the like.
I did find a single report done by Environment Canada on one of these products for diesel engines and they found that adding the hydrogen generator device didn't make any statistically significant change to combustion efficiency, exhaust emission rates, or fuel consumption. They calculated that the device was about 60% efficient at making hydrogen, at best, and used about 8 mL of diesel fuel every 100 km, so it doesn't really make efficiency worse to any appreciable degree, but doesn't help either. (Source: ERMD Report # 2004-032 If you search, it's on scribd, but I couldn't find any permanent sources)
In short, a water-fuelled car on its own or as a hybrid with gasoline or diesel don't seem to be possible. Water is a very stable compound and it takes a great deal of energy to break it into hydrogen and oxygen gas. At most, you can get the energy you put in back, and realistically, a lot will be lost along the way. This makes a solely water fuelled car a perpetual motion machine of the first kind as it violates the law of conservation of energy—one cannot burn hydrogen and oxygen to produce energy to make the same amount of hydrogen and oxygen consumed and still have energy left over to move a vehicle. A hybrid vehicle faces a similar problem. Because the part of the system that involves water consumes energy rather than contributes it to motion, the gas or diesel part has to produce more energy to make up for the difference. Given that modern gas and diesel engines burn almost all the fuel put into them, there doesn't seem to be any mechanism by which adding a small quantity of hydrogen and oxygen could produce significantly more energy.

Thanks for your answer. Modern IC engines are 25-30% efficient. And yes, my research on the topic concluded that it is not possible to have a hybrid car with water. Definition of hybrid vehicle tells us why. Water is not being used as a source of power. So we can not call above mechanism as hybrid. Furthermore, Inclusion of electrolysis and burning of Hydrogen loop is only going to make efficiency worse, so it is not feasible either.

In short, there is no possibility to make an engine which can use water as a fuel (neither as only source nor as hybrid). Please read this and include your findings in your answer and I will accept it as correct answer. So that it can help someone in the future. :)

By 95–98% efficient, I'm not talking about thermal efficiency, I mean that 95–98% of fuel added to the cylinder is actually combusted.

A small amount of hydrogen injection greatly increases the octane rating of the fuel. The idea may well beat carrying around a hydrogen tank.

@joshua I have nothing on which to dispute that claim, but even if true: higher octane is not necessarily any better for the car. I mean perhaps adding hydrogen will let you use a lesser grade of gasoline on a car with a high compression engine that needs (or at least benefits from) premium gasoline, but a car not designed to take advantage of higher octane fuel may not really benefit at all.

@KevinCathcart: Hint: lower octane gasoline is usually more energy dense. I could start filling up with 85 instead of 87.

If you actually had a source for 85 octane fuel. In practice 85 is only available in high altitude regions where non-fuel-injected naturally aspirated vehicles can operate with lower than normal octane rating, due to the sparser air causing less compression heating. I mean if you buy your fuel by the tanker truck-load, then great, you can presumably get whatever octane rating you want, but most people don't do that, and only can buy that the local gas stations actually offer.

For any engine that does not have a sophisticated ECU then the addition of hydrogen into the air intake does produce significant savings on fuel and reductions in pollution.
The modern ECU will offset this effect but older engines will perform well.
The greater risk is the hydrogen that remains in the engine, but this is present at 0.001% parts per air and since I believe (though please correct me if I am wrong) hydrogen is explosive at a min of 18% even this should not be an issue


text:
Hydrolysis under basic conditions
Which route actually occurs? Is it really true that the carboxylic acid never forms under basic conditions?
Why not just form the -COOH and deprotonate it afterward?? Or is there a higher activation every barrier to the second route?
I understand that carboxylic acids don't stick around long in basic solutions but that doesn't necessarily mean they don't exist in basic solutions at all.

Wouldnt the mechanism depend upon the concentration of strong base taken?

I find the top mechanism unlikely for two reasons.
1) In the top mechanism, the proton removed in the first step should be way less acidic than a typical hydroxyl since the intermediate is already negatively charged. Going to a di-anion should be very far uphill, and therefore, slow. If that di-anion was ever formed, the next step would be exceedingly fast
2) In the bottom mechanism, the first step should be fast. It's intramolecular and entropically favored. The second step is instantaneous.
I can't imagine any text book includes the top mechanism. Check your book for saponification of esters.

Can I ask what book you are using?

Also if I'm not mistaken I've seen the first mechanism in Organic Chemistry by Clayden.

March's Advanced Organic Chemistry shows eight different mechanisms for ester hydrolysis under acidic and basic condtions. Some of these I would consider very unlikely, but the top mechanism that you've drawn is not included.

Neither Loudon or Maitland Jones show the top mechanism, only the bottom mechanism. Are your notes a direct copy of something your prof provided to you (on the board or otherwise)?

Let us continue this discussion in chat.

Another argument favoring @jerepierre 's position is: Why should the mechanism for basic hydrolysis of an ester be different than the mechanism for transesterification under basic conditions?  In this latter case, there is no proton to remove from the tetrahedral intermediate, so the dioxa anion cannot form.  The 2 reactions are very similar ($\ce{OH^{-} vs. OR^{-}}$), why would you invoke different mechanisms for 2 reactions so seemingly similar?

@ron would making the first mechanism described in my post "concerted" make a difference? Or does everything just boil back down to the same issue: that proton we're taking off just doesn't want to be taken off? Thinking about it, it might just make a difference. I'll open a new question.

I think it's the latter too.  Isn't there a way to ask your prof what argument he used to get the textbook folks to accept his mechanism as correct where he views your question as positive rather than confrontational?

The reaction proceeds by a mechanism named $B_{AC_2}$ (Base-Acyl Bond-2 molecularity).
The rate is given by: (note that it is order 2 and molecularity 2) $$r=k\ce{[ester][OH- ]}$$
The $\bf rds$ is the attack of hydroxide ion. Since the last step is an acid-base reaction, it is unidirectional and reverse esterification is not possible under general conditions (I think you understand what I mean by this). Thus relative rates are also given which are well explained by this mechanism (kinda proof, considering $\bf rds$):

Acid base reactions are always unidirectional?

yes where weaker conjugate acids and weaker conjugate bases are formed.

Determining the rate law often can distinguish between mechanisms, but in this case, both proposed mechanisms could follow that rate law.

@jerepierre oh jerrie perry, This is the actual mechanism (surely), I just gave the related info to prove that this is not wrong.It's not the other way where I'm trying to prove this mechanism with the help of rate law

@ADG I agree with you that the addition/elimination followed by acid-base reaction is the mechanism that operates here. However, none of the information you have provided helps distinguish between the two mechanisms.  The relative rates would hold for either mechanism.

@jerepierre now you want me to prove an already proved and named mechanism again? maybe you are not going to accept it until i provide some source?

@ADG It wasn't my question, so it's not up to me to accept your answer. This answer repeats the mechanism that is given in the question and gives some rates that show that mechanism is reasonable. However, a full answer also has to say why it is more reasonable than the alternative. That is what the questioner wants to know. As I read it, this answer argues, "That's just the way it is" without considering the alternative mechanism. Keep in mind that I agree with you that the second mechanism is more reasonable.


text:
Does product induced poisoning qualify as a rate limiting step?
I have a question about whether a particular situation qualifies as a rate limiting step. I have a catalyst that under normal conditions gets poisoned when products accumulate on the catalyst surface. I interpret that as the product desorption as being a rate limiting step, but is that true?
I also thought that that a rate limiting step means that the rate of the reaction equals the rate of that step, but in reality what I see is that the initial rate in the case where the product sticks and when you can desorb the product is the same, but just that you get an abrupt halt in reactivity once the catalyst is full of product. Conversely, by tweaking the conditions where i can cause the product to desorb it just will keep that initial rate up for 10,000 times longer.
Does that still mean that product desorption is a RLS or is there a better way to describe this poisoning event?
Thanks!